Chemistry and Chemical Reactivity [11 ed.] 9780357851401, 9780357851418

Citation preview

Periodic Table of the Elements Hydrogen 1

H

1

2

3

4

5

6

7

MAIN GROUP METALS

1.008 1A (1)

2A (2)

Lithium 3

Beryllium 4

Li

TRANSITION METALS

Uranium 92

U

METALLOIDS

Be

6.94 9.0122 Sodium Magnesium 12 11

Na

Mg

22.990

24.305

Potassium 19

Calcium 20

39.098

40.078

K

Ca

Rubidium Strontium 38 37

Rb

Sr

3B (3)

4B (4)

5B (5)

6B (6)

7B (7)

Scandium Titanium Vanadium Chromium Manganese 25 22 23 24 21

Sc

44.956 Yttrium 39

Y

Ti

47.867

V

50.942

Cr

51.996

Mn

54.938

Zr

Nb

Hf

Ta

Mo

Tc

W

Re

87.62 Barium 56

88.906 91.224 92.906 Lanthanum Hafnium Tantalum 57 72 73

132.91 Francium 87

137.33 Radium 88

138.91 178.49 180.95 183.84 186.21 Actinium Rutherfordium Dubnium Seaborgium Bohrium 105 107 104 106 89

(223)

(226)

Fr

Ba Ra

La

Ac

(227)

Note: Atomic weights are IUPAC values. For elements for which IUPAC recommends ranges of atomic weights, conventional values are shown. Numbers in parentheses are mass numbers of the most stable isotope of an element.

Atomic weight

8B (8)

(9)

(10)

Iron 26

Cobalt 27

Nickel 28

55.845

58.933

58.693

Fe

Co

Ni

Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium 45 42 43 40 44 46 41

85.468 Cesium 55

Cs

Symbol

238.03

NONMETALS

Atomic number

Rf

(267)

Lanthanides

Db

(268) Cerium 58

Ce

140.12 Actinides

95.95 (98) Tungsten Rhenium 75 74

Sg

(269)

Bh

(270)

Ru

101.07 Osmium 76

Os

Rh

102.91 Iridium 77

Ir

Pd

106.42 Platinum 78

Pt

190.23 192.22 195.08 Hassium Meitnerium Darmstadtium 109 110 108

Hs

(277)

Mt

(276)

Ds

(281)

Praseodymium Neodymium Promethium Samarium Europium 59 60 61 63 62

Pr

140.91

Nd

144.24

Pm

(145)

Sm

150.36

Eu

151.96

Thorium Protactinium Uranium Neptunium Plutonium Americium 92 94 91 90 93 95

Th

232.04

Pa

231.04

U

238.03

Np

(237)

Pu

(244)

Am

(243)

For the latest information see: https://iupac.org/what-we-do/periodic-table-of-elements/ and https://www.nist.gov/pml/periodic-table-elements

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8A (18) Helium 2 4A (14)

5A (15)

6A (16)

7A (17)

4.0026

He

hydrogen atoms

Boron 5

Carbon 6

Nitrogen 7

Oxygen 8

Fluorine 9

Neon 10

oxygen atoms

10.81 Aluminum 13

12.011 Silicon 14

14.007 Phosphorus 15

15.999 Sulfur 16

18.998 Chlorine 17

20.180 Argon 18

Al

C

Si

N P

O S

F

Cl

Ne Ar

1B (11)

2B (12)

26.982

28.085

30.974

32.06

35.45

39.95

Copper 29

Zinc 30

Gallium 31

Germanium 32

Arsenic 33

Selenium 34

Bromine 35

Krypton 36

63.546

65.38

69.723

72.630

74.922

78.971

79.904

83.798

Silver 47

Cadmium 48

Indium 49

Tin 50

Iodine 53

Xenon 54

107.87 Gold 79

112.41 Mercury 80

114.82 Thallium 81

Ag Au

Zn

Cd

Hg

carbon atoms

3A (13)

B

Cu

Standard Colors for Atoms in Molecular Models

Ga In Tl

Ge Sn

118.71 Lead 82

Pb

As

Se

Antimony Tellurium 51 52

Sb

121.76 Bismuth 83

Bi

Te

Br I

127.60 126.90 Polonium Astatine 84 85

Po

At

nitrogen atoms

chlorine atoms

Kr

Xe

131.29 Radon 86

Rn

200.59 204.38 207.2 208.98 (209) (210) (222) 196.97 Roentgenium Copernicium Nihonium Flerovium Moscovium Livermorium Tennessine Oganesson 114 111 112 113 115 116 117 118

Rg

(282)

Cn

(285)

Nh

(286)

Fl

(289)

Gadolinium Terbium Dysprosium Holmium 66 67 65 64

Gd

157.25 Curium 96

Cm

(247)

Tb

158.93

Dy

162.50

Ho

164.93

Mc

Lv

(290)

(293)

Erbium 68

Thulium 69

167.26

168.93

Er

Tm

Ts

(294)

Og

(294)

Ytterbium Lutetium 71 70

Yb

173.05

Lu

174.97

Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium 97 100 98 99 101 102 103

Bk

(247)

Cf

(251)

Es

(252)

Fm

(257)

Md

(258)

No

(259)

Lr

(262)

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Chemistry

11th Edition

& Chemical Reactivity

Kotz Treichel Townsend Treichel

Australia • Brazil • Canada • Mexico • Singapore • United Kingdom • United States

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Chemistry and Chemical Reactivity, ­Eleventh Edition

© 2023, 2019, 2015 Cengage Learning, Inc. ALL RIGHTS RESERVED.

John C. Kotz, Paul M. Treichel, John R. Townsend, and David A. Treichel

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Brief Contents

Part One The Basic Tools of Chemistry

Part Five The Chemistry of the Elements

1 Basic Concepts of Chemistry  2 1R Let’s Review: The Tools of Quantitative Chemistry 30 2 Atoms, Molecules, and Ions  62 3 Chemical Reactions  138 4 Stoichiometry: Quantitative Information about Chemical Reactions  190 5 Principles of Chemical Reactivity: Energy and Chemical Reactions  254

20 21 22 23 24 25

Part Two Atoms and Molecules

A Using Logarithms and Solving Quadratic Equations A-2 B Some Important Physical Concepts  A-6 C Abbreviations and Useful Conversion Factors  A-9 D Physical Constants  A-13 E A Brief Guide to Naming Organic Compounds A-15 F Values for the Ionization Energies and Electron Attachment Enthalpies of the Elements  A-18 G Vapor Pressure of Water at Various Temperatures A-19 H Ionization Constants for Aqueous Weak Acids at 25 °C A-20 I Ionization Constants for Aqueous Weak Bases at 25 °C A-22 J Solubility Product Constants for Some Inorganic Compounds at 25 °C  A-23 K Formation Constants for Some Complex Ions in Aqueous Solution at 25 °C  A-24 L Selected Thermodynamic Values  A-25 M Standard Reduction Potentials in Aqueous Solution at 25 °C  A-32 N Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions A-36

6 The Structure of Atoms  304 7 The Structure of Atoms and Periodic Trends  344 8 Bonding and Molecular Structure  386 9 Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals  458

Part Three States of Matter 10 11 12 13

Gases and Their Properties  498 Intermolecular Forces and Liquids  540 The Solid State  580 Solutions and Their Behavior  626

Part Four The Control of Chemical Reactions 14 Chemical Kinetics: The Rates of Chemical Reactions 672 15 Principles of Chemical Reactivity: Equilibria  736 16 Principles of Chemical Reactivity: The Chemistry of Acids and Bases  776 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria  830 18 Principles of Chemical Reactivity: Entropy and Free Energy 886 19 Principles of Chemical Reactivity: Electron Transfer Reactions 932

Nuclear Chemistry  992 The Chemistry of the Main Group Elements  1038 The Chemistry of the Transition Elements  1104 Carbon: Not Just Another Element  1150 Biochemistry 1206 Environmental Chemistry—Earth’s Environment, Energy, and Sustainability  1244

List of Appendices

Index of Names  I-1 Index and Glossary  I-4

iii

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Contents Preface  xii

Part One The Basic Tools of Chemistry

1 1.1 1.2 1.3 1.4 1.5 1.6 1.7

Basic Concepts of Chemistry  2

Chemistry and Its Methods  3 Sustainability and Green Chemistry  7 Classifying Matter  8 Elements  12 Compounds  13 Properties and Changes  15 Energy: Some Basic Principles  20 Applying Chemical Principles 1.1: CO2 in the Oceans  21

2.3

A Closer Look: Isotopic Abundances and Atomic Weights  70 Key Experiments: The Nature of the Atom and Its Components 72 The Periodic Table  74 A Closer Look: Mendeleev and the Periodic Table  78

2.4 2.5 2.6

A Closer Look: Marie Curie (1867–1934)  82 Molecules: Formulas, Models, and Names  84 Ions  89 Ionic Compounds: Formulas, Names, and Properties  93 A Closer Look: Hydrated Ionic Compounds  98

2.7 Atoms, Molecules, and the Mole  99 A Closer Look: Amedeo Avogadro and His Number  100 A Closer Look: The Mole, a Counting Unit  103

2.8 Chemical Analysis: Determining Compound

Review: The Tools of 1R Let’s Quantitative Chemistry 30 1R.1 Units of Measurement  31

2.9

 

A Closer Look: The SI Base Units  34 A Closer Look: Energy and Food  37

1R.2 Making Measurements: Precision, Accuracy, 1R.3 1R.4 1R.5 1R.6

Experimental Error, and Standard Deviation  37 Mathematics of Chemistry  41 Problem Solving by Dimensional Analysis  47 Graphs and Graphing  48 Problem Solving and Chemical Arithmetic  49 Applying Chemical Principles 1R.1: Out of Gas!  51 Applying Chemical Principles 1R.2: Ties in Swimming and Significant Figures 52

2

Atoms, Molecules, and Ions 62  

2.1 Atomic Structure, Atomic Number, and Atomic Mass 63

2.2 Atomic Weight  67

Formulas  106 Instrumental Analysis: Determining Compound Formulas  114 Applying Chemical Principles 2.1: Using Isotopes: Ötzi, the Iceman of the Alps  117 Applying Chemical Principles 2.2: Arsenic, Medicine, and the Formula of Compound 606  118 Applying Chemical Principles 2.3: Argon—An Amazing Discovery  118

3

Chemical Reactions  138

3.1 Introduction to Chemical Equations  139 3.2 3.3 3.4 3.5 3.6

A Closer Look: Antoine Laurent Lavoisier (1743–1794) 141 Balancing Chemical Equations  142 Introduction to Chemical Equilibrium  145 Aqueous Solutions  147 Precipitation Reactions  151 Acids and Bases  156 A Closer Look: Sulfuric Acid  162

iv Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

3.7 Acid–Base Reactions  163 3.8 Oxidation–Reduction Reactions  167 3.9

A Closer Look: Are Oxidation Numbers Real? 171 Classifying Reactions in Aqueous Solution  173 A Closer Look: Alternative Organizations of Reaction Types  174 Applying Chemical Principles 3.1: Superconductors 177 Applying Chemical Principles 3.2: Sequestering Carbon Dioxide  177 Applying Chemical Principles 3.3: Black Smokers and Volcanoes  178

4 4.1

Stoichiometry: Quantitative Information about Chemical Reactions  190

Mass Relationships in Chemical Reactions: Stoichiometry 191

5

5.1 Energy: Some Basic Principles  255 5.2 5.3 5.4

5.5 5.6 5.7 5.8

4.5

Is Present in Limited Supply  195 Percent Yield  200 Chemical Equations and Chemical Analysis  202 A Closer Look: Nuclear Magnetic Resonance (NMR) Spectroscopy  208 Measuring Concentrations of Compounds in Solution  209 A Closer Look: Serial Dilutions  215

4.6 pH, a Concentration Scale for Acids and

Bases  215 4.7 Stoichiometry of Reactions in Aqueous Solution—Fundamentals  217 4.8  Stoichiometry of Reactions in Aqueous Solution—Titrations  220 4.9 Spectrophotometry  227 Applying Chemical Principles 4.1: Atom Economy  232 Applying Chemical Principles 4.2: Bleach 232 Applying Chemical Principles 4.3: How Much Salt is There in Seawater? 233 Applying Chemical Principles 4.4: The Martian  234

A Closer Look: What Is Heat? 257 Specific Heat Capacity: Heating and Cooling  258 Energy and Changes of State  262 The First Law of Thermodynamics  266 A Closer Look: P–V Work  268

4.2 Reactions in Which One Reactant 4.3 4.4

Principles of Chemical Reactivity: Energy and Chemical Reactions  254

A Closer Look: Enthalpy, Internal Energy, and Non-Expansion Work  270 Enthalpy Changes for Chemical Reactions  271 Calorimetry  274 Enthalpy Calculations  278 A Closer Look: Hess’s Law and Equation 5.7 284 Product- or Reactant-Favored Reactions and Thermodynamics  285 Applying Chemical Principles 5.1: Gunpowder 286 Applying Chemical Principles 5.2: The Fuel Controversy—Alcohol and Gasoline 287

Part Two Atoms and Molecules

6

The Structure of Atoms  304

6.1 Electromagnetic Radiation  305 6.2 Quantization: Planck, Einstein, Energy, and 6.3 6.4 6.5 6.6 6.7

Photons  308 Atomic Line Spectra and Niels Bohr  312

A Closer Look: Niels Bohr (1885–1962)  314 Wave–Particle Duality: Prelude to Quantum Mechanics  319 The Modern View of Electronic Structure: Wave or Quantum Mechanics  321 The Shapes of Atomic Orbitals  325 A Closer Look: More about H Atom Orbital Shapes and Wavefunctions  329 One More Electron Property: Electron Spin  330 Applying Chemical Principles 6.1: Sunburn, Sunscreens, and Ultraviolet Radiation 330

Contents v

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Applying Chemical Principles 6.2: What Makes the Colors in Fireworks? 331 Applying Chemical Principles 6.3: Chemistry of the Sun  332

7

8.9

A Closer Look: Measuring Molecular Polarity and Debye Units  426

The Structure of Atoms and Periodic Trends  344

7.1 The Pauli Exclusion Principle  345 7.2 Atomic Subshell Energies and Electron

8.10

7.3

Assignments  347 Electron Configurations of Atoms  350

8.11

7.4

A Closer Look: Orbital Energies, Z*, and Electron Configurations 359 Electron Configurations of Ions  360 A Closer Look: Questions about Transition Element Electron Configurations  361

7.5 7.6

A Closer Look: Paramagnetism and Ferromagnetism 364 Atomic Properties and Periodic Trends  364 A Closer Look: Photoelectron Spectroscopy  371 Periodic Trends and Chemical Properties  374 Applying Chemical Principles 7.1: The Not-So-Rare Earths  375 Applying Chemical Principles 7.2: Metals in Biochemistry  376

8

Bonding and Molecular Structure  386

8.1 Lewis Electron Dot Symbols and Chemical Bond 8.2 8.3 8.4 8.5

Formation 388 Electronegativity and Bond Polarity  390 Lewis Structures of Molecules and Polyatomic Ions  393 Common Patterns of Bonding in Lewis Structures  399 Resonance  404 A Closer Look: Resonance 405

8.6 Exceptions to the Octet Rule  406 8.7

8.8 vi

A Closer Look: Structure and Bonding for Hypervalent Molecules  409 Formal Charges in Covalent Molecules and Ions  410 A Closer Look: Comparing Oxidation Number and Formal Charge  411 Molecular Shapes  416

A Closer Look: A Scientific Controversy—Resonance, Formal Charges, and the ­Question of Double Bonds in Sulfate and Phosphate Ions  417 Molecular Polarity  424

A Closer Look: Visualizing Charge Distributions and Molecular Polarity—Electrostatic Potential Surfaces and Partial Charge  428 Bond Properties: Order, Length, and Dissociation Enthalpy  432 DNA  437 Applying Chemical Principles 8.1: Ibuprofen, A Study in Green Chemistry  441 Applying Chemical Principles 8.2: van Arkel Triangles and Bonding  442 Applying Chemical Principles 8.3: Linus Pauling and the Origin of the Concept of Electronegativity  443

9

Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals  458

9.1 Valence Bond Theory  459 9.2 Molecular Orbital Theory  473 9.3

A Closer Look: Molecular Orbitals for Molecules Formed from p-Block Elements  481 Theories of Chemical Bonding: A Summary  483 A Closer Look: Three-Center Bonds in HF22, B2H6, and SF6 484 Applying Chemical Principles 9.1: Probing Molecules with Photoelectron Spectroscopy 485 Applying Chemical Principles 9.2: Green Chemistry, Safe Dyes, and Molecular Orbitals 486

Part Three States of Matter

10 Gases and Their Properties 

498

10.1 Modeling a State of Matter: Gases and Gas Pressure 499 A Closer Look: Measuring Gas Pressure  500

10.2 Gas Laws: The Experimental Basis  502

Contents

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10.3 10.4 10.5 10.6 10.7 10.8

A Closer Look Studies on Gases—Robert Boyle and Jacques Charles  508 The Ideal Gas Law  508 Gas Laws and Chemical Reactions  513 Gas Mixtures and Partial Pressures  514 The Kinetic-Molecular Theory of Gases  517 Diffusion and Effusion  521 A Closer Look: Surface Science and the Need for Ultrahigh Vacuum Systems  523 Nonideal Behavior of Gases  524 Applying Chemical Principles 10.1: The Atmosphere and Altitude Sickness  526 Applying Chemical Principles 10.2: The Chemistry of Airbags  527

Forces and 11 Intermolecular Liquids  540 11.1 States of Matter and Intermolecular Forces  541 11.2 Interactions between Ions and Molecules with a Permanent Dipole  543

11.3

11.4 11.5

11.6

A Closer Look: Hydrated Salts: A Result of Ion– Dipole Bonding  545 Interactions between Molecules with a Permanent Dipole  546 A Closer Look: Hydrogen Bonding in Biochemistry 551 Intermolecular Forces Involving Nonpolar Molecules  552 A Summary of van der Waals Intermolecular Forces  555 A Closer Look: Geckos Can Climb Up der Waals 556 Properties of Liquids  557 Applying Chemical Principles 11.1: Chromatography 567 Applying Chemical Principles 11.2: A Pet Food Catastrophe  568

12

The Solid State  580

12.1 Crystal Lattices and Unit Cells  581 12.2

A Closer Look: Packing Oranges, Marbles, and Atoms 588 Structures and Formulas of Ionic Solids  589 A Closer Look: Using X-Rays to Determine Crystal Structures 593

12.3 Bonding in Ionic Compounds: Lattice Energy  594

12.4 Bonding in Metals and Semiconductors  596 12.5 Other Types of Solid Materials  602 12.6

A Closer Look: Glass 605 Phase Changes  609 Applying Chemical Principles 12.1: Lithium and Electric Vehicles  613 Applying Chemical Principles 12.2: Nanotubes and Graphene: Network Solids 614 Applying Chemical Principles 12.3: Tin Disease  615

13 Solutions and Their Behavior 

626

13.1 Units of Concentration  627 13.2 The Solution Process  630 A Closer Look: Supersaturated Solutions  631

13.3 Factors Affecting Solubility: Pressure and 13.4

Temperature  637 Colligative Properties  640

A Closer Look: Growing Crystals  641 A Closer Look: Hardening of Trees  646

13.5

A Closer Look: Reverse Osmosis for Pure Water 649 A Closer Look:  Osmosis and Medicine  651 Colloids  655 Applying Chemical Principles 13.1: Distillation 659 Applying Chemical Principles 13.2:  Henry’s Law and Exploding Lakes  660 Applying Chemical Principles 13.3:  Narcosis and the Bends  661

Part Four The Control of Chemical Reactions Kinetics: The Rates 14 Chemical of ­Chemical Reactions  672 14.1 14.2 14.3 14.4 14.5

Rates of Chemical Reactions  673 Reaction Conditions and Rate  678 Effect of Concentration on Reaction Rate  680 Concentration–Time Relationships: Integrated Rate Laws  686 A Microscopic View of Reaction Rates  694

Contents vii

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14.6 14.7

A Closer Look: Rate Laws, Rate Constants, and Reaction Stoichiometry  695 A Closer Look: More about Molecular Orientation and Reaction Coordinate Diagrams  697 Catalysts  702 A Closer Look: Thinking about Kinetics, Catalysis, and Bond Energies  702 Reaction Mechanisms  706 A Closer Look: Organic Bimolecular Substitution Reactions 709 Applying Chemical Principles 14.1: Enzymes–Nature’s Catalysts  716 Applying Chemical Principles 14.2: Kinetics and Mechanisms: A 70-Year-Old Mystery Solved  717

of Chemical Reactivity: 15 Principles Equilibria  736 15.1 Chemical Equilibrium: A Review  737 15.2 The Equilibrium Constant and Reaction Quotient  738

16.7 Calculations with Equilibrium Constants  796 16.8 Polyprotic Acids and Bases  805 16.9 Molecular Structure, Bonding, and Acid–Base Behavior  807

A Closer Look: Acid Strengths and Molecular Structure 811 16.10 The Lewis Concept of Acids and Bases  812 Applying Chemical Principles 16.1: Would You Like Some Belladonna Juice in Your Drink? 816 Applying Chemical Principles 16.2: The Leveling Effect, Nonaqueous Solvents, and Superacids  817

of Chemical Reactivity: 17 Principles Other Aspects of Aqueous Equilibria  830

17.1 Buffers  831 17.2 Acid–Base Titrations  843 17.3 Solubility of Salts  853 A Closer Look: Minerals and Gems—The Importance of Solubility  859

A Closer Look: Activities and Units of K 740 A Closer Look: Equilibrium Constant Expressions for Gases—Kc and Kp 742 15.3 Determining an Equilibrium Constant  746 15.4 Using Equilibrium Constants in Calculations  748 15.5 More about Balanced Equations and Equilibrium Constants  754 15.6 Disturbing a Chemical Equilibrium  757 Applying Chemical Principles 15.1: Applying Equilibrium Concepts— The Haber–Bosch Ammonia Process  762 Applying Chemical Principles 15.2: Trivalent Carbon  763

of Chemical Reactivity: 16 Principles The Chemistry of Acids and

17.4 17.5

of Chemical Reactivity: 18 Principles Entropy and Free Energy  886 18.1 Spontaneity and Dispersal of Energy: 18.2

Entropy  887 Entropy: A Microscopic Understanding  889

18.3 18.4

A Closer Look: Reversible and Irreversible Processes 890 Entropy Measurement and Values  894 Entropy Changes and Spontaneity  898

Bases  776

16.1 The Brønsted-Lowry Concept of Acids and

Bases  777 16.2 Water and the pH Scale  780 16.3 Equilibrium Constants for Acids and Bases  783 16.4 Acid–Base Properties of Salts  789 16.5 Predicting the Direction of Acid–Base Reactions  791 16.6 Types of Acid–Base Reactions  794

viii

A Closer Look: Solubility Calculations  860 Precipitation Reactions  863 Equilibria Involving Complex Ions  867 Applying Chemical Principles 17.1:  Everything that Glitters . . .  871 Applying Chemical Principles 17.2: Take a Deep Breath  872

18.5 18.6 18.7

A Closer Look: Entropy and Spontaneity? 899 Gibbs Free Energy  902 Calculating and Using Standard Free Energies, DrG°  906 The Interplay of Kinetics and Thermodynamics  914

Contents

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Applying Chemical Principles 18.1: Thermodynamics and Living Things  916 Applying Chemical Principles 18.2: Are Diamonds Forever? 917

19 19.1 19.2 19.3 19.4 19.5 19.6 19.7

19.8 19.9

Principles of Chemical Reactivity: Electron Transfer Reactions  932

Oxidation–Reduction Reactions  933 Voltaic Cells  940 Commercial Voltaic Cells  946 Standard Electrochemical Potentials  951 A Closer Look: EMF, Cell Potential, and Voltage 952 Electrochemical Cells Under Nonstandard Conditions  960 Electrochemistry and Thermodynamics  964 Electrolysis: Chemical Change Using Electrical Energy  968 A Closer Look: Electrochemistry and Michael Faraday 970 Counting Electrons  973 Corrosion: Redox Reactions in the Environment  975 Applying Chemical Principles 19.1:  Electric Batteries versus Gasoline  978 Applying Chemical Principles 19.2:  Sacrifice! 978

Part Five The Chemistry of the Elements

20 Nuclear Chemistry 

992

20.1 Natural Radioactivity  994 20.2 Nuclear Reactions and Radioactive Decay  995 20.3 20.4 20.5 20.6 20.7 20.8 20.9

A Closer Look: Radioactive Decay Series  997 Stability of Atomic Nuclei  1000 Origin of the Elements: Nucleosynthesis  1006 Rates of Nuclear Decay  1008 Artificial Nuclear Reactions  1014 Nuclear Fission and Nuclear Fusion  1018 A Closer Look: Lise Meitner (1878–1968)  1020 Radiation Health and Safety  1021 Applications of Nuclear Chemistry  1023 Applying Chemical Principles 20.1: A Primordial Nuclear Reactor  1027

Applying Chemical Principles 20.2:  Technetium-99m and Medical Imaging  1028 Applying Chemical Principles 20.3:  The Age of Meteorites  1029

Chemistry of the Main 21 The Group Elements 1038 21.1 Abundance of the Elements  1039 21.2 The Periodic Table: A Guide to the Elements  1040

21.3 Hydrogen  1046 21.4 21.5

A Closer Look: Hydrogen in Transportation 1049 The Alkali Metals, Group 1A (1)  1049 A Closer Look: The Reactivity of the Alkali Metals 1054 The Alkaline Earth Elements, Group 2A (2)  1054 A Closer Look: Alkaline Earth Metals and Biology 1057

A Closer Look: Cement—The Second Most Used Substance 1058 21.6 Boron, Aluminum, and the Group 3A (13) Elements  1059 A Closer Look: Complexity in Boron Chemistry 1064 21.7 Silicon and the Group 4A (14) Elements  1065 21.8 Nitrogen, Phosphorus, and the Group 5A (15) Elements  1070 A Closer Look: Alchemists Making Phosphorus 1072 A Closer Look: Ammonium Nitrate—A Mixed Blessing 1075 21.9 Oxygen, Sulfur, and the Group 6A (16) Elements  1079 21.10 The Halogens, Group 7A (17)  1082 A Closer Look: Iodine and Your Thyroid Gland 1084

21.11

A Closer Look: The Many Uses of FluorineContaining Compounds  1085 The Noble Gases, Group 8A (18)  1087 A Closer Look: The Noble Gases—Not So Inert 1088 Applying Chemical Principles 21.1:  Lead in the Environment  1089 Applying Chemical Principles 21.2:  Hydrogen Storage  1090

Contents ix

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Chemistry of the 22 The Transition Elements  1104 22.1 Overview of the Transition Elements  1105 22.2 Periodic Properties of the Transition 22.3 22.4

Elements  1107 Metallurgy  1111 Coordination Compounds  1114

A Closer Look: Hemoglobin: A Molecule with a Tetradentate Ligand  1118 22.5 Structures of Coordination Compounds  1122 22.6 Bonding in Coordination Compounds  1128 22.7 Colors of Coordination Compounds  1133 Applying Chemical Principles 22.1:  Blue!  1137 Applying Chemical Principles 22.2:  Cisplatin: Accidental Discovery of a Chemotherapy Agent  1138 Applying Chemical Principles 22.3: The Rare Earth Elements  1139

23

Carbon: Not Just Another Element  1150

23.1 Why Carbon?  1151

23.3 23.4

A Closer Look: Writing Formulas and Drawing Structures 1153 Hydrocarbons  1155 A Closer Look: Flexible Molecules  1161 Alcohols, Ethers, and Amines  1170 Compounds with a Carbonyl Group  1176

23.5

A Closer Look: Omega-3 Fatty Acids  1180 Polymers  1184

23.2

A Closer Look: Green Chemistry: Recycling PET 1190 Applying Chemical Principles 23.1:  An Awakening with l-DOPA 1192 Applying Chemical Principles 23.2: Green Adhesives  1193 Applying Chemical Principles 23.3: Bisphenol A (BPA)  1193

24

Biochemistry  1206

24.1 Proteins 1208 24.2 Carbohydrates  1216

x

24.3 Nucleic Acids  1219 24.4 24.5

A Closer Look: Genetic Engineering with CRISPR-Cas9 1222 Lipids and Cell Membranes  1225 A Closer Look: mRNA Vaccines  1228 Metabolism  1230 Applying Chemical Principles 24.1: Polymerase Chain Reaction  1236

Chemistry—Earth’s 25 Environmental ­Environment, Energy, and Sustainability  1244

25.1 The Atmosphere  1246 A Closer Look: The Earth’s Atmosphere  1247 A Closer Look: Methane Hydrates  1256

25.2 The Aqua Sphere (Water)  1256 25.3

A Closer Look: Perfluoroalkyl Substances (PFAS) 1263 Energy  1263 A Closer Look: Fracking 1266

25.4 Fossil Fuels  1267 A Closer Look: Petroleum Chemistry  1271

25.5 Environmental Impact of Fossil Fuels  1271 25.6 Alternative Sources of Energy  1277 25.7 Green Chemistry and Sustainability  1281 Applying Chemical Principles 25.1:  Chlorination of Water Supplies  1282 Applying Chemical Principles 25.2: Hard Water 1283

List of Appendices  A-1 A

Using Logarithms and Solving Quadratic Equations  A-2

B C D E F

Some Important Physical Concepts 

G

Vapor Pressure of Water at Various Temperatures  A-19

H

Ionization Constants for Aqueous Weak Acids at 25 °C  A-20

A-6

Abbreviations and Useful Conversion Factors  Physical Constants 

A-9

A-13

A Brief Guide to Naming Organic Compounds 

A-15

Values for the Ionization Energies and Electron Attachment Enthalpies of the Elements  A-18

Contents

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I

Ionization Constants for Aqueous Weak Bases at 25 °C  A-22

J

Solubility Product Constants for Some Inorganic Compounds at 25 °C  A-23

K

Formation Constants for Some Complex Ions in Aqueous Solution at 25 °C  A-24

L

Selected Thermodynamic Values 

A-25

M

Standard Reduction Potentials in Aqueous Solution at 25 °C  A-32

N

Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions   A-36

Index of Names  I-1 Index and Glossary  I-4

Contents xi

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Preface

David Kotz

The first edition of this book problems as a scientist. We have was conceived over 40 years ago. tried to provide the tools to help Since that time, there have been you become a chemically and ten editions, and over one milscientifically literate citizen. lion students worldwide have used the book to begin their Audience for study of chemistry. Although the Chemistry & details of the book and its orgaChemical Reactivity nization have changed over the years, our fundamental goal has This textbook is designed for remained the same: to provide a students interested in furbroad overview of the principles ther study in science, whether of chemistry, the reactivity of the that science is chemistry, biolchemical elements and their comogy, medicine, e­ ngineering, pounds, and the applications of geology, physics, or related chemistry. To reach this goal, we subjects. Our assumption is have tried to show the close relathat students in a course using tionship between the observathis book have had some preptions of chemical and physical aration in ­algebra and general changes made by chemists in the science. Although u ­ ndeniably laboratory and in nature and the helpful, a ­previous ­exposure to way these changes are viewed at chemistry is neither assumed the atomic and molecular levels. Fireworks. See Chapter 6 for the chemistry of nor required. We have also tried to convey the ­fireworks. sense that chemistry not only has a lively history but is also interesting and dynamic, with Philosophy and Approach of important new developments occurring every year. Further- Chemistry & Chemical Reactivity more, we wanted to provide some insight into the chemical aspects of the world around us. We have had several major but not independent objectives The authors of this text have collectively taught chem- since the first edition of the book. Our first goal has been istry for over 100 years, and we have engaged in years of to write a book that students will find useful and interestfundamental research. Like countless other scientists, our ing and that presents chemistry and chemical principles in goals in our research and in writing this textbook have a format and organization typical of college and university been to satisfy our curiosity about areas of chemistry, to courses today. Second, we want to convey the utility and document what we found, and to convey that to students importance of chemistry by introducing the properties of and other scientists. Our results, and many others, may the elements, their compounds, and their reactions. be used, perhaps only years later, to make a better mateThe American Chemical Society has been urging edurial or better pharmaceutical. Every person eventually cators to put chemistry back into introductory chemistry benefits from the work of the worldwide community of courses. We agree wholeheartedly. Therefore, we have scientists. tried to describe the elements, their compounds, and In recent years, when people around the world have their reactions as early and as often as possible by: experienced various epidemics and increasing evidence of climate change has been published, science has come under • Bringing material on the properties of elements and compounds into the Examples and Study Questions. attack. Some distrust the scientific community and dismiss the results of carefully done research. Therefore, a key objec- • Using numerous photographs of the elements and comtive of this book, and of a course in general chemistry, is to mon compounds, of chemical reactions, and of comdescribe basic chemical “facts”—chemical processes and mon laboratory operations and industrial processes. principles; how chemists came to understand those prin• Each chapter incorporates Applying Chemical Principles and new ideas; how they can be applied in industry, ciples questions that delve into the applications of medicine, and the environment; and how to think about chemistry.

xii Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

General Organization

units are introduced in Chapter 1, and thermochemistry is introduced in Chapter 5.

Through its many editions, Chemistry & Chemical ­Reactivity has had two broad themes: Chemical ­Reactivity and ­B onding and Molecular Structure. The chapters on ­P rinciples of Reactivity introduce the factors that lead chemical ­r eactions to be successful in converting ­reactants to products: common types of reactions, the energy involved in reactions, and the factors that affect the speed of a reaction. One reason for the enormous advances in chemistry and molecular biology in the last several decades has been an understanding of molecular structure. The sections of the book on Principles of ­Bonding and Molecular Structure lay the groundwork for understanding these developments. Particular attention is paid to an understanding of the structural aspects of such biologically important molecules as hemoglobin and DNA.

Sections of the Book — Organization and Purpose Part One:  The Basic Tools of Chemistry The basic ideas and methods of chemistry are i­ ntroduced in Part One. Chapter 1 defines important terms, and the accompanying Chapter 1R reviews measurement units and some fundamental mathematical methods used throughout the text. Chapter 2 introduces atoms, molecules, and ions as well as the most important ­organizational device in chemistry, the periodic table. In Chapter 3, we begin to discuss the principles of chemical reactivity. Writing chemical equations is covered here, and there is a short introduction to the concept of ­chemical equilibrium. In addition, some major types of chemical reactions in aqueous solution are introduced. Then, in Chapter 4, we describe the numerical methods used by chemists to extract quantitative information from chemical reactions. Chapter 5 is an introduction to the energy changes involved in chemical processes.

As we look at the introductory chemistry texts currently available and talk with colleagues at other universities, it is evident there is a generally accepted order of topics in the course. With minor variations, we followed that order. That is not to say that the chapters in our book cannot be used in some other order. We have w ­ ritten this book to be as flexible as possible. An example is the flexibility in covering the behavior of gases (­Chapter 10). It has been placed with the chapters on liquids, solids, and solutions (Chapters 11–13) because it logically fits with those topics. However, it can also be read and ­understood after covering only the first four chapters of the book. Similarly, the chapters on atomic and molecular structure (Chapters 6–9) can be used in an atoms-first approach before the chapters on stoichiometry and common reactions (Chapters 3 and 4). To facilitate this, there is an introduction to energy and its units in C ­ hapter 1. Also, the chapters on chemical equilibria (Chapters 15–17) can be covered before those on solutions and ­kinetics ­(Chapters 13 and 14). Although organic ­chemistry (Chapter 23) is one of the final chapters in the textbook, the topics of this chapter can also be presented to ­students following the chapters on structure and bonding (Chapters 9 and 10). The order of topics in the text was also devised to introduce as early as possible the background required for the laboratory experiments that are usually ­performed in introductory chemistry courses. For this reason, ­chapters on chemical and physical properties, ­common reaction types, and stoichiometry begin the book. In addition, because an understanding of energy is very important for the study of chemistry, energy and its

© Charles D. Winters/Cengage

Flexibility of Chapter Organization

Elemental sulfur.

Part Two:  Atoms and Molecules The current theories that explain the arrangement of electrons in atoms and monatomic ions are presented in Chapters 6 and 7. This discussion is tied closely to the arrangement of elements in the periodic table and to their periodic properties. In Chapter 8, we discuss the details of chemical bonding and the properties of these bonds. In addition, we show how to derive the threedimensional structure and charge distribution of simple molecules. Finally, Chapter 9 considers the major theories of chemical bonding in more detail.

Preface xiii

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Part Four: The Control of Chemical Reactions

© Charles D. Winters/Cengage

This section is wholly concerned with the Principles of Reactivity. Chapter 14 examines the rates of chemical processes and the factors controlling these rates. Next, Chapters 15–17 describe chemical equilibrium. After an introduction to equilibrium in Chapter 15, we highlight reactions involving acids and bases in water (Chapters 16 and 17) and reactions leading to slightly soluble salts (Chapter 17). To tie together the discussion of chemical equilibria and thermodynamics, we explore entropy and free energy in C ­ hapter 18. As a final topic in this section, we describe in ­C hapter 19 chemical reactions that involve the transfer of e­ lectrons and the use of these reactions in electrochemical cells.

Liquid oxygen is attracted to a strong magnet. See ­Chapter 9 for an explanation of the magnetic properties of oxygen.

Part Three:  States of Matter

© Charles D. Winters/Cengage

The behavior of the three states of matter—gases, liquids, and solids—is described in Chapters 10–12. The discussion of liquids and solids is tied to gases through the description of intermolecular forces in Chapter 11, with particular attention given to liquid and solid water. In Chapter 13, we describe the properties of solutions— intimate mixtures of gases, liquids, and solids.

The explosive reaction of hydrogen and oxygen.

John C. Kotz

Part Five: The Chemistry of the Elements

Hot air balloon takes off. See Chapter 5 for an introduction to transfers of energy as heat and work. See Chapter 10 for a discussion of the gas laws.

xiv

Although the chemistry of many elements and compounds is described throughout the book, Part Five considers this topic in a more systematic way. ­Chapter 20 is an overview of nuclear chemistry. C ­ hapter  21 is devoted to the chemistry of the main group elements, and Chapter 22 is a discussion of the transition elements and their compounds. Chapter 23 is a brief introduction to organic chemistry with an emphasis on molecular structure, basic reaction types, and polymers, and Chapter 24 is an introduction to biochemistry. Finally, Chapter 25 brings together many of the

Preface

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

What’s New to this Edition?

concepts in earlier chapters into a discussion of “Environmental Chemistry—Earth’s Environment, Energy, and Sustainability.”

The entire book was thoroughly reviewed. Many parts are rewritten, and new Study Questions were added. In addition, there are several new features that occur in each chapter.

Think–Pair–Share Questions

values per mole and per gram. Which provides more energy

For the purposes of this analysis, octane (C8H18) is used as a substitute for the complex mixture of hydrocarbons in gasoline. Data needed for this question (in addition to the data in Appendix L) are: ∆fH° [C8H18()] = −250.1 kJ/mol Density of ethanol = 0.785 g/mL Density of octane = 0.699 g/mL 1. Calculate ∆rH° for the combustion of ethanol and octane producing carbon dioxide gas and liquid water, and compare the

Elements: yellow phosphorus in water (left) and shiny ­potassium under oil (right).

Photos: © Charles D. Winters/Cengage

Questions

Many instructors have with moving per mole? Which provides more experimented energy per gram? beyond lecturing in liter general One 2. Comparesimply the energy produced per of the chemistry. two fuels. Which produces more energy for a given volume approach that has worked especially well(something has been to take useful to know when filling your gas tank)? class time for students to work together on ­questions that 3. What mass of CO2, a greenhouse gas, is produced per liter of help learn course ­topics. This is the purpose of the fuelthem (assuming complete combustion)? Think–Pair–Share ­q uestions. Studentsbasis. work indepen4. Now compare the fuels on an energy-equivalent What volumeon of the ethanol would havefirst, to bethen burnedform to getsmall the same dently questions groups to energy as 1.00 L of octane? When you burn enough ethanol discuss their answers, and finally, present their results to to have the same energy as a liter of octane, which fuel protheduces class. These more CO2? questions do not generally require many ­calculations. Instead, they focus on helping ­students to think more deeply about the ­concepts of the chapter. They are placed after the A ­ pplying Chemical ­P rinciples questions and before the Chapter Goals Revisited.

Think–Pair–Share 1. You are tasked with determining the specific heat capacity (in J/g ? K) for an unknown metal. The following equipment and supplies are available: a 25.0-g piece of the metal, a 200-mL insulated container, a graduated cylinder, water, ice, and a thermometer that can measure temperature changes accurately to ±0.02 °C. (a) Outline the steps for an experimental procedure to determine the specific heat capacity of the metal using the available equipment and supplies. (b) Identify possible sources of error in your experimental procedure that might cause an inaccurate result. (Assume the graduated cylinder and thermometer are both accurate and precise, and that human error is not a source of error.) (c) Suggest some ideas for how to correct for some of these error sources. 2. For introductory laboratories, resealable plastic bags are a convenient way to conduct experiments involving gas-forming reactions. Energy as heat or work can transfer in or out of the plastic bag, but reactants and products remain trapped. In an experiment, nitric acid reacts with a small amount of copper in a sealed plastic bag according to the following balanced chemical equation: Cu(s) + 4 HNO3(aq) n Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O() As the reaction proceeds, the contents of the bag become warm, and the bag inflates with a brown gas (NO2). (a) Define the system and the surroundings for this experiment. (b) Does energy as heat (q) flow into the system or out of the system? What is the sign of q? (c) Is work (w) done in this experiment? If so, is work done by the system on the surroundings, or by the surroundings on the system? What is the sign of w? 3. The following table shows the enthalpy of combustion for some fuels in units of kJ/mol, kJ/L, and kJ/g. The enthalpy of combustion per volume assumes a temperature of 25 °C and atmospheric pressure (1 atm). Note that although gasoline is a mixture of many hydrocarbons, it is often represented as octane.

288

Fuel

∆H° (kJ/mol)

∆H° (kJ/L)

∆H° (kJ/g)

Hydrogen, H2(g)

−285.8

−11.7

−141.8

Methane, CH4(g)

−890.2

−36.4

−55.5

Ethane, C2H6(g)

−1560.6

−63.8

−51.9

Ethanol, C2H6O()

−1376.5

−23,590

−29.9

Octane, C8H18()

−5490

−33,814

−48.1

(a) When determining which fuel provides the most energy in a car engine, is it best to compare the enthalpy change in a combustion reaction by moles, volume, or mass? Explain your reasoning? Based on your decision, which fuel provides the most energy (at the given temperature and pressure). (b) Gasoline sold in the United States is often a blend of 10%, 15%, or even up to 85% ethanol by volume. Identify at least one advantage and one disadvantage of using gasoline blended with ethanol versus ethanol-free gasoline? (c) Why are the enthalpies of combustion per liter of hydrogen, methane, and ethane much lower than those of ethanol and octane? (d) The enthalpy of combustion of hydrogen per gram is nearly three times that of a hydrocarbon. And unlike a hydrocarbon, which produces the greenhouse gas CO 2 upon combustion, the combustion of hydrogen produces only water. Unfortunately, there are multiple issues in replacing gasoline with hydrogen as a fuel. What are some of the problems that must be overcome if hydrogen is to compete with, or possibly replace, gasoline as a fuel in vehicles?

Chapter 5 / Principles of Chemical Reactivity: Energy and Chemical Reactions

Preface xv

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Dr. Jessica N. Isaac

Dr. Jessica N. Isaac

Jessica N. Isaac is a Clinical Assistant Professor at at the chemical interactions between medications Binghamton University’s School of Pharmacy and and the chemicals in the human body; creating and refining drug development; and Pharmaceutical Sciences, where her work is “a comstudents that they have amolecules place infor chemistry and STEM Chemistry in Your Career bination of administration, teaching,courses. research, clin- combining, mixing, or altering chemicals to create All of us have students who have gone on to wonderful ical pharmacy, and mentorship.” As an instructor, a medication tailored to the needs of a patient.  careers, some in traditional chemistry careers, Dr. Isaac describes how her identity as a an one of Dr. Isaac’s goalsbut is tooften help “student pharmaRedesigned Strategy Maps in some other field where acists background in chemistry develop the skills necessary to safely and African American/Jamaican American woman has For many students, a visual Strategy and Mapprofessionally. can be a useful is useful. Each chapter features a short biography of a shaped her both personally “Being accurately prepare medications.” tool in problem (as onand page 194). These have been person who studied chemistry and perhaps Black woman a first-generation student in “Thenow studyworks, of chemistry is integral to the study a solving pharmacy many obstacles; of pharmacy,” explains Dr. Isaac, including “stoiredesigned to be more school fully presented incorporated into thehowever, Soluas a chemist, but more generally in a profession where experiences also helped me toare develop certain dimensional analysis, acid–base prop- of these tion section the Example problems. There 44 Strategy they use their background chiometry, in chemistry. This feature . .including social intelligence, perseverance/ erties, and physiochemical properties.” The study strengths. Example problems in the book. highlights people from diverse backgrounds to show all Maps accompanying of pharmacy has three core components: looking resilience, creativity, gratitude, and critical thinking.”

Exam pl e 4 .1

Mass Relations in Chemical Reactions

Strategy Map Problem Calculate mass of O2 required for combustion of 25.0 g of glucose. Data/Information Formulas for reactants and products and the mass of one reactant (glucose)

Problem Glucose, C6H12O6, reacts with oxygen to give CO2 and H2O. What mass of oxygen (in grams) is required to completely react with 25.0 g of glucose? What masses of carbon dioxide and water (in grams) are formed?

What Do You Know? You are given the mass of one of the reactants (glucose) and are asked to determine the masses of the other substances in the reaction. You know formulas for the reactants and products and need to calculate their molar masses. Strategy Write the balanced chemical equation for this reaction. Then, follow the scheme outlined in Problem Solving Tip 4.1 and in the Strategy Map for this example.

Solution Step 1

Write the balanced equation. C6H12O6(s) + 6 O2(g) n 6 CO2(g) + 6 H2O(ℓ)

Step 2

Convert the given mass (grams) to amount (moles). Use the molar mass of glucose to convert its mass (25.0 g) to the amount of glucose. 25.0 g glucose 

Step 3

1 mol glucose  0.1387 mol glucose 180.2 g glucose

Use the stoichiometric factor to convert the amount (moles) of glucose to the amount (moles) of O2. Use the coefficients of the balanced equation to obtain the stoichiometric factor of 6 mol O2 per 1 mol glucose, and then convert the amount of glucose to the amount of O2. 0.1387 mol glucose 

Step 4

6 mol O2  0.8324 mol O2 1 mol glucose

Convert the amount (moles) of the requested substance to its mass (grams). Use the molar mass of O2 to convert from the amount of O2 to the mass of O2. 0.8324 mol O2 

32.00 g O2  26.6 g O2 1 mol O2

194

Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions

xvi

Preface

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Some of the Other Changes for This Edition

•

Recent changes to the definitions of the SI Base Units are explained in a new Closer Look box (page 34).

•

The atomic weights of the elements on the periodic tables and other tables have been updated based on values from the International Union of Pure and Applied Chemistry (IUPAC) and the National ­Institute of Standards and Technology (NIST).

environmental chemistry can serve as an overarching conclusion to the book.

•

A new Closer Look box about radioactive decay series (page 997) was added to Chapter 20.

•

The section on the origin of the elements (in Chapter 20, “Nuclear Chemistry”) was expanded and revised.

•

In Chapter 24, “Biochemistry,” new material has been added on mRNA vaccines, electron carriers in biochemical oxidation–reduction reactions, and the metabolism of glucose in respiration.

•

Chapter 25, “Environmental Chemistry—Earth’s Environment, Energy, and Sustainability” has been updated and reorganized.

•

All Examples have been reviewed, some have been revised to make the steps of the strategy clearer, and nine new Examples were written (Example R.3 “Precision and Standard Deviation;” Example 2.5 “Binary Molecular Compounds;” Example 2.8 “Naming Ionic Compounds;” Example 3.5 “Separating a Mixture by Selective Precipitation;” Example 3.9 “Recognizing Oxidation–Reduction Reactions;” Example 4.3 “Calculating Percent Yield;” Example 4.8 “Relating Amount and Molarity;” Example 4.12 “Acid–Base Titration;” and Example 6.6 “Orbitals and Quantum Numbers”).

•

The issue of ranges being recommended by IUPAC for the atomic weights of some elements is discussed in a new Closer Look box (page 70).

•

References to element groups in the periodic table are now given in both the traditional system used in the United States (Group 5A, for example) and the 1–18 system recommended by IUPAC.

•

Biographies of some important scientists and their discoveries have been prepared (Marie Curie [page 82]; Antoine and Marie-Anne Lavoisier [page 141]; Niels Bohr [page 314]; James Watson, Francis Crick, and Rosalind Franklin [page 439]; and Lise Meitner [page 1020]).

•

A new subsection “Naming Common Acids” (page 157) was added.

•

The classification scheme for acid–base and gas-­ forming acid–base reactions (Sections 3.6–3.7) has been revised as well as the overall classification scheme of reactions in aqueous solution (Section 3.9).

•

There are now over 2650 Study Questions in the book. Of these, over 360 are either new or revised in this edition.

•

•

All appendices have been updated to ensure they contain the latest information.

A Closer Look box (page 208) on nuclear magnetic resonance spectroscopy has been added.

•

•

A greater distinction between heat capacity and ­specific heat capacity was made in Chapter 5.

•

A new Closer Look box “Enthalpy, Internal Energy and Non-Expansion Work” (page 270) was added.

Appendix N, “Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions,” has been accuracy checked by the book authors and the author of the Student Solutions Manual, Professor Charles Atwood.

•

•

A new Problem-Solving Tip (page 355) was included about the different methods used for writing the electron configuration of an atom.

An Index of Names has been added so readers can find the contributions of generations of chemists.

Features of the Book

•

Chapter 8, “Bonding and Molecular Structure,” was restructured.

•

The story of the unraveling of the structure of DNA was expanded in Chapter 8.

•

Two Closer Look boxes were added in Chapter 12, one on using X-rays to determine crystal structure and the other about glass.

•

The positions of the chapters about nuclear chemistry (Chapter 20 in the current edition) and environmental chemistry (Chapter 25) were swapped so that information about nuclear chemistry can inform the content of later chapters and so that the chapter on

Some years ago, a former student of one of the authors, now an accountant, shared his perspective on his experience in general chemistry. He said that, while chemistry was one of his hardest subjects, it was also the most useful course he had taken because it taught him how to solve problems in addition to having learned to appreciate a bit of chemistry. We were certainly pleased because we have always thought that an important goal in general chemistry is not only to teach students chemistry but also to help them learn critical thinking and problem-solving skills. Many of the features of the book are meant to support those goals.

Preface xvii

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Problem-Solving Approach: Organization and Strategy Maps

A Closer Look Essays and Problem-Solving Tips

Worked-out Examples are an essential part of each ­chapter. To better help students follow the logic of a solution, all Examples are organized around the following outline:

As in the tenth edition, there are boxed essays titled A Closer Look that take a more in-depth look at relevant chemistry. While retaining and updating many from previous editions, we wrote several new ones and ­h eavily revised some others including: “The SI Base Units” (Chapter 1R), “Isotopic Abundances and Atomic Weights” (Chapter 2), “Marie Curie (1867–1934)” (Chapter 2), “Amedeo Avogadro and His Number” (Chapter 2), “Nuclear Magnetic Resonance (NMR) Spectroscopy” (Chapter 4), “Enthalpy, Internal Energy, and Non-­E xpansion Work” (Chapter 5), “Niels Bohr (1885–1962)” (Chapter 6), “Lise Meitner (1878–1968)” ­(Chapter 20), “Hydrogen in Transportation” (Chapter 21), “mRNA Vaccines” (Chapter 24), “Methane Hydrates” (Chapter 25), and “Perfluoroalkyl Substances (PFAS)” (Chapter 25). From our teaching experience, we have learned some “tricks of the trade” and try to pass on some of those in Problem-Solving Tips.

Problem: A statement of the problem. What Do You Know?: The information given is outlined. Strategy: The information available is combined with the objective, and we begin to devise a pathway to a solution. Solution: We work through the steps, both logical and mathematical, to the answer. Think About Your Answer: We ask if the answer is reasonable or what it means. Check Your Understanding: This is a similar problem for the student to try. A solution to the problem is in Appendix N.

Chapter Goals Revisited The learning goals for each chapter section are listed at the beginning of the section. The goals are revisited on the last pages of the chapter, and specific end-of-chapter Study Questions are listed that can help students determine if they have met those goals.

End-of-Chapter Study Questions There are between 48 and 178 Study Questions for each chapter, and answers to the odd-numbered questions are given in Appendix N. Questions are grouped as follows: Practicing Skills: These questions are grouped by the topic covered by the questions. General Questions: There is no indication regarding the pertinent section of the chapter. They generally cover several chapter sections. In the Laboratory: These are problems that may be encountered in a laboratory experiment on the chapter material. Summary and Conceptual Questions: These questions use concepts from the current chapter as well as preceding chapters. Finally, some questions are marked with a small red triangle (▲). These are more challenging than other questions.

xviii

Applying Chemical Principles At the end of each chapter are longer ­questions that use the principles learned in the chapter to study examples of forensic chemistry, environmental chemistry, ­medicinal chemistry, or other areas. Examples are “Atom Economy” (Chapter 4), “What Makes the ­C olors in Fireworks” (Chapter 6), “A Pet Food ­C atastrophe” (Chapter 11), “Lithium and Electric ­Vehicles” (Chapter 12), “The Age of Meteorites” ­(Chapter 20), and “Blue!” (Chapter 22).

Online Learning Created by teaching chemists,  OWLv2  is a powerful online learning solution for chemistry with a unique Mastery Learning approach. It enables students to practice at their own pace, receive meaningful feedback, and access a variety of learning resources to help them master chemistry and achieve better grades. The textbook’s Study Questions are available in the OWLv2 online learning system. OWLv2 now has over 1800 of the roughly 2650 Study Questions in the book. The OWLv2 course and MindTap eReader both contain nearly 300 videos on specific topics narrated by the authors to help students visualize concepts and master difficult problems by watching them be solved on screen.

Preface

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Acknowledgments

Preparing this new edition of Chemistry & Chemical ­Reactivity took about two years of continuous effort. As was true for our work on the first ten editions, we have had the support and encouragement of our colleagues at Cengage and our families, friends, faculty colleagues, and students.

Cengage The ten previous editions of this book have been published by Cengage and its predecessor companies, and once again we had an excellent production team in place for this, the eleventh edition. Maureen McLaughlin and Helene Alfaro led the team with Mona Zeftel overseeing many aspects of book design. Various people helped with content organization: James Nash, Breanna Holmes, and Kelly Aull. They were invaluable. The first half of the book in this edition was thoroughly reviewed and edited by Margy Kuntz. This is the eleventh edition of a book that has been used successfully in its previous editions by over a million students. Nonetheless, Margy found ways to better organize and clarify sections in these chapters. The Chemistry in Your Career boxes are a new feature of the book, and we want to acknowledge the many people who told us their stories. We hope these will help the many students who take a chemistry course see how it can be important in their careers. Rebecca Heider at Cengage was masterful in putting their stories into small, readable vignettes. Chemistry & Chemistry Reactivity has been supported by OWL for many editions. The relationship of the book and OWL has continued to be very well managed by ­Theresa Dearborn.

Art, Design, and Photography Many of the color photographs in this book have been beautifully created by Charles D. Winters over many years and ten editions.

The book still profits from the design and illustration skills of Patrick Harman. Pat designed the first edition of our Interactive General Chemistry CD-ROM (published in the 1990s). For the fifth through the tenth editions of the book, Pat revised many of the figures in the book to bring a fresh perspective to ways to communicate chemistry. All these illustrations remain in use in this edition.

Other Collaborators We have been fortunate to have had several colleagues play valuable roles in this project over its many editions. One who has been especially important to this edition is Professor Charles (Butch) Atwood. He has been very helpful in ensuring the accuracy of the Study Question answers in the book and producing the Student Solutions Manual.

Eleventh Edition Reviewers We encourage users, both faculty and students, to contact us about book content and with suggestions for improvement. There have been many instances of this over the years and they have improved the book. In particular, we would like to thank Roger Barth (West Chester University of Pennsylvania) for many useful comments that assisted us as we planned changes for this edition. The following reviewed the book for this edition:

• • •

John M. Farrar, Northern Kentucky University

• • • • •

Jessica A. Parr, University of Southern California

Bernard Majdi, South Georgia State College Danica A. Nowosielski, Hudson Valley Community College Dr. Jeff Seyler, University of Southern Indiana Jeffrey Stephens, North Iowa Area Community College Tarek Trad, Sam Houston State University Saul R. Trevino, Houston Christian University

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About the Authors

John (Jack) Kotz graduated from Washington and Lee University in 1959 and earned a Ph.D. in c­ hemistry at Cornell University in 1963. He was a National ­Institutes of Health postdoctoral fellow at the ­University of ­M anchester in England and at Indiana University. He was an Assistant Professor of Chemistry at Kansas State University before moving to the State University of New York at Oneonta in 1970. He taught general chemistry and inorganic chemistry, and in 1986 was appointed a State University of New York Distinguished Teaching Professor of Chemistry. He retired from active teaching in 2005. He is the author or coauthor of sixteen chemistry textbooks, among them two in advanced inorganic chemistry, two introductory general chemistry books in numerous editions, and various manuals and study guides. The general chemistry book has been published as an interactive CD-ROM, as an interactive ebook, and has been translated into ­ ublished research papers five languages. He has also p in organometallic chemistry, and among his awards are the SUNY Award for Research and Scholarship and the Catalyst Award in ­Education from the Chemical Manufacturers Association. He was a ­Fulbright Senior Lecturer in Portugal and a mentor for the U.S. National Chemistry O ­ lympiad team. He has served on the boards of trustees for the ­C ollege at Oneonta Foundation, the Kiawah Island Nature C ­ onservancy, and Camp Dudley. He is also an avid ­photographer, primarily of wildlife (www.greensward.smugmug.com). His email address is [email protected].

Emeritus Professor of ­Chemistry. During his faculty career he taught courses in general ­chemistry, ­inorganic chemistry, ­organometallic c­ hemistry, and s­ cientific ethics. Professor Treichel’s research in ­organometallic and metal-cluster ­chemistry and in mass spectrometry, aided by 75 graduate and undergraduate students, has led to more than 170 papers in scientific journals. He may be contacted by email at [email protected].

Paul M. Treichel received his B.S. degree from the ­ niversity of Wisconsin in 1958 and a Ph.D. from H U ­ arvard University in 1962. After a year of postdoctoral study in London, he assumed a faculty position at the University of Wisconsin–Madison. He served as department chair from 1986 through 1995 and was awarded a Helfaer Professorship in 1996. He has held visiting faculty positions in South Africa (1975) and in Japan (1995). Retiring after 44 years as a faculty member in 2007, he is currently

David A. Treichel, Professor of Chemistry at Nebraska Wesleyan University, received a B.A. degree from ­Carleton College. He earned a M.S. and a Ph.D. in analytical chemistry at Northwestern University. After postdoctoral research at the University of Texas in Austin, he joined the faculty at Nebraska Wesleyan University. His research interests are in the fields of electrochemistry and surface laser spectroscopy. He may be contacted by email at [email protected].

John R. Townsend completed his B.A. in C ­ hemistry as well as the Approved Program for Teacher ­Certification in Chemistry at the University of Delaware. After a career teaching high school science and mathem­atics, he earned his M.S. and Ph.D. in biophysical c­ hemistry at Cornell University, where he also received the DuPont Teaching Award for his work as a teaching assistant. After teaching at Bloomsburg University of P ­ ennsylvania, he joined the faculty at West Chester University of ­Pennsylvania where he coordinated the chemistry e­ ducation program for prospective high school teachers and the general ­c hemistry lecture programs, taught undergraduate courses in general chemistry and biochemistry, and was the university supervisor for 78 prospective high school chemistry teachers during their student teaching semester. In 2021, he was the recipient of the Award for Excellence in Undergraduate Teaching in Chemical Science from the Philadelphia (Pennsylvania) S­ ection of the ­American Chemical Society. Retiring in 2021, he is an Emeritus Professor of Chemistry at West Chester ­University. He may be contacted by email at [email protected].

xx Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

About the Cover Kotz • Treichel • Townsend • Treichel

Chemistry

11th Edition

& Chemical Reactivity

Chemistry & Chemical Reactivity 11th Edition

Ralph Lee Hopkins

Kotz Treichel Townsend Treichel

SE/Kotz • Treichel • Townsend • Treichel, Chemistry & Chemical Reactivity, 11th Edition ISBN 9780357851401 Printer: Binding: Casebound Trim: 8.5” x 10.875” CMYK

©2024

Designer: Chris Doughman

Cover photo by Ralph Lee Hopkins, Fagradalsfjall volcano, Iceland, 2021.

Fagradalsfjall volcano, on the Reykjanes Peninsula not far from Reykjavik in Iceland, started erupting on March 19, 2021, and continued to erupt for six months. The volcano erupted again August 3, 2022, but went quiet after only 10 days. For many observers, the most spectacular part of a volcanic e­ ruption is the lava or molten rock that comes from the Earth’s mantle during an ­e ruption. For others, particularly scientists, volcanic activity provides ­opportunities to explore the chemical makeup of the Earth’s mantle and to study the effects of volcanic activity on the environment. Volcanoes are part of the story of the history and chemistry of the Earth. Volcanic events have always occurred and continue to occur around the globe as the result of tectonic plate activity. One massive volcanic event occurred in 1883 on the island of Krakatoa in Indonesia. Thousands of people in the volcano’s vicinity perished quickly, but the eruption also had global consequences. For example, temperatures across the northern h ­ emisphere dropped by an average of 0.4 °C in the year following the eruption. Environmental scientists study the effects of volcanic activity. ­Scientists know that volcanic activity injects large amounts of water vapor, carbon dioxide, and sulfur dioxide into the atmosphere. Other gases released include hydrogen chloride and stinky “sewer gas” or hydrogen sulfide. There is a significant cooling effect in the atmosphere, partly from the conversion of sulfur dioxide to sulfuric acid and sulfate aerosols. These reflect sunlight back into space and cool the Earth. You might think that the ­carbon ­dioxide from volcanoes would increase global warming, a topic of current concern. However, scientists have shown that volcanoes emit less than 1% of the massive amount of carbon dioxide put into the atmosphere by human activities today. The elements and compounds that arise from volcanic activity are ­fundamentally important. You will encounter these substances and many others in this general introduction to the field of chemistry. Personal statement from Ralph Lee Hopkins, the photographer: It was a dream come true for a geologist-photographer to trek to an active volcano in ­Iceland. Words can’t describe the sights, sounds, and smells of new earth being created. I spent two weeks and made five treks in between bad weather and poison gas warnings. I was very lucky to witness flowing lava up close and a sinuous river of lava spilling from the crater at sundown.

xxi

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Dedication

We wish to dedicate this edition of Chemistry & Chemical Reactivity to our colleagues who have contributed to our knowledge of chemistry and teaching and to our many students, some of whom became good friends and who helped us understand better how to communicate our science. We also acknowledge and thank Professor Paul Treichel who helped shape this book with his work on many of the previous editions. His expertise, good humor, and friendship over the years are appreciated. And, finally, we thank our families who supported the years of work needed to produce this book and for their support throughout our careers.

xxii Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

1 Basic Concepts of Chemistry

Dnsphotography/iStock/Getty Images

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C hapt e r O ut li n e 1.1 Chemistry and Its Methods 1.2 Sustainability and Green Chemistry 1.3 Classifying Matter 1.4 Elements 1.5 Compounds 1.6 Properties and Changes 1.7 Energy: Some Basic Principles

Chemistry is the scientific study of the composition, structure, and properties of matter and the changes in both composition and energy that matter undergoes during reactions. Although chemistry is endlessly fascinating—at least to ­chemists— why should you study chemistry? Each person probably has a different answer, but many students take a chemistry course because those who are professional scientists, who teach and are involved daily in scientific work, realize how important chemistry is in any curriculum leading to a career in a science-related discipline. You will come to appreciate that chemistry is central to understanding disciplines as diverse as biology, geology, materials science, medicine, physics, and some branches of engineering. This is why chemistry is sometimes referred to as the central science. In addition, chemistry plays a major role in national economies, and chemistry and chemicals affect our daily lives in a wide variety of ways. A course in chemistry can also help you see how a scientist thinks about the world and solves problems. The knowledge and skills developed in chemistry courses will benefit you in many career paths and help you become a better-informed citizen in a world that is becoming technologically more complex—and more interesting.

Matter  Anything that occupies space and has mass — all substances and mixtures in the universe are composed of matter.

1.1 Chemistry and Its Methods Goal for Section 1.1 • Recognize the difference between a hypothesis and a theory, and understand how laws are established.

This book provides a foundation for learning chemistry, a discipline that has ­d eveloped over many centuries through the work of people around the planet. H ­ owever, chemistry is about far more than historical knowledge. New ◀ Methane Bubbles Trapped in Ice.  Bodies of water are often filled with and surrounded by

vegetation. Over time, the vegetation will decay, slowly being digested by bacteria that release methane, a greenhouse gas, as a product of the digestion. Some of the methane bubbles rise to the surface, and in the winter the bubbles can be trapped in ice. The white patches in the photo are trapped methane bubbles in a lake in Alberta, Canada.

3

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discoveries occur frequently, and many recent discoveries are highlighted in this book. As you read, please do not overlook the special features that explore some of these d ­ iscoveries, in particular “A Closer Look” boxes and “Applying Chemical Principles” sections. Are you interested in medicine or medical advances? Do not miss the story on the development of mRNA vaccines (“A Closer Look: mRNA Vaccines,” page 1228) that have been valuable in the fight against the COVID-19 viruses or the story on how gene editing holds the promise for correcting genetic mutations that lead to diseases (“A Closer Look: Genetic Engineering with CRISPR-Cas9,” page 1222). Chemists, physicists, and material scientists work together to develop electrical devices using atomically thin films of pure carbon (“Applying Chemical Principles 12.2: Nanotubes and Graphene: Network Solids,” page 614), create high-­ temperature superconductors that may one day replace inefficient power transmission lines (“Applying Chemical Principles 3.1: S­ uperconductors,” page 177), and find ways to create naturally occurring, but rare materials, such as diamonds in laboratories (“Applying Chemical Principles 18.2: Are Diamonds Forever?,” page 917). Perhaps most importantly, scientists across multiple disciplines are s­ tudying ways to slow climate change. Increasing levels of carbon dioxide, methane, and other greenhouse gases are changing the conditions on earth, both on land and in the oceans (“Applying Chemical Principles 1.1: CO 2 in the Oceans,” page 21). It is accepted by the scientific community that significant changes to the ­e nvironment will continue to occur if greenhouse gas emissions are not reduced. Some hope comes from sequestering greenhouse gases (“Applying ­ ioxide,” page 177) and reducChemical Principles 3.2: Sequestering Carbon D ing our reliance on fossil fuels (“Applying Chemical Principles 5.2: The Fuel Controversy—Alcohol and Gasoline,” page 287). The environment is so important that an entire chapter (Chapter 25) is devoted to the subject. As you use this book in your study of chemistry and chemical principles, be sure to understand that it is just the beginning. It provides an introduction to the most important topics of chemistry, but we hope that it will also help you appreciate those topics and their interconnections as well as their uses and importance in your lives.

Cinnabar

Mercury droplets

Figure 1.1  Cinnabar and mercury.  Heating cinnabar

(mercury(II) sulfide) in air changes it into orange mercury(II) oxide, which, upon further heating, decomposes into the elements mercury and oxygen gas.

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© Charles D. Winters/Cengage

Chemistry and Change Chemistry is about change. It was once only about changing one natural substance into another—wood and oil burn, grape juice turns into wine, and cinnabar (Figure 1.1), a red mineral, ultimately changes into shiny quicksilver (mercury) when heated. The emphasis was largely on finding a recipe to complete a desired change with little understanding of the underlying structure of the materials or explanations for why particular changes occurred. Chemistry is still about change, but now chemists focus on the change of one pure substance, whether natural or synthetic, into another and on understanding that change (Figure 1.2). Chemists now picture an exciting world of submicroscopic atoms and molecules interacting with each other, and they have developed ways to predict whether or not a particular reaction may occur.

Methods of Science Neil deGrasse Tyson, noted physicist, author, and TV personality once said “Science is a method of inquiry. Science is a way of expressing doubt and knowing when it’s time to embrace what’s discovered and move on to something else to doubt.” While there is no one scientific method by which all scientists conduct their studies, there are certain common practices. You almost always start the process by asking questions. These can be questions of your own choosing or ones that someone else poses. Having posed a reasonable question, the next step is often to look at the experimental work done in the field so that you have some notion of the possible answers. Based on this work, you may form a hypothesis, a tentative explanation or prediction of experimental observations.

Chapter 1 / Basic Concepts of Chemistry

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Martyn F. Chillmaid/Science Source © Charles D. Winters/Cengage

Sodium solid, Na

Sodium chloride solid, NaCl

Figure 1.2  Forming a chemical compound.  Combining sodium

Chlorine gas, Cl 2

metal (Na) and yellow chlorine gas (Cl2) gives sodium chloride.

After formulating a hypothesis, systematic investigations are conducted, which may include formal experiments designed to give results that will confirm or invalidate the hypothesis. Systematic investigations require the collection of information or data, which may be either quantitative or qualitative. Quantitative information is numerical data, such as the mass of a substance (Figure 1.3) or the temperature at which it melts. Qualitative information, in contrast, consists of nonnumerical observations, such as the color of a substance or its physical appearance. The data from your investigations must be analyzed and interpreted in order to derive meaning. Based on the analysis of your investigations, and perhaps studies from other researchers, you may have evidence supporting your hypothesis. However, it is also possible, and quite common, that you will need to revise your hypothesis and continue to test it with more experiments, or that the investigation will end up raising more questions for you to answer. After you have checked to ensure that your results are truly reproducible, a pattern of behavior or results might begin to emerge. At this point, you may be able to summarize your observations in the form of a law, a concise verbal or mathematical statement of a relation that is always the same under any condition.

Figure 1.3  Qualitative and quantitative observations.  ​

Weighing a compound on a laboratory balance.

Quantitative: mass is 28.331 grams



© Charles D. Winters/Cengage

Qualitative: blue, granular solid

1.1  Chemistry and Its Methods

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© Charles D. Winters/Cengage

Figure 1.4  The metallic element sodium reacts with water.

Much of the work in science is based on laws because they help predict what may occur under a new set of circumstances. For example, chemists know from ­experience that if the element sodium comes in contact with water, a violent r­ eaction occurs and new substances are formed (Figure 1.4). They also know the mass of the substances produced in the reaction is the same as the mass of the sodium and water used in the reaction. That is, mass is always conserved in c­ hemical ­reactions, a statement of the law of conservation of matter. Once enough reproducible studies are conducted and experimental results generalized as a law or general rule, it may be possible to conceive a theory to explain the observations. A theory is a well-tested, unifying principle that explains a body of facts and the laws based on them. Theories can suggest new hypotheses that can be tested experimentally. Sometimes nonscientists use the word theory to imply that someone has made a guess and that an idea is not yet substantiated. To scientists, however, a theory is based on carefully determined and reproducible evidence that is being continuously tested. Theories are the cornerstone of our understanding of the natural world at any given time. Remember, though, that theories are inventions of the human mind. Theories can and do change as new facts are uncovered.

Goals of Science Scientists, including chemists, have several goals. Two of these are prediction and control. Scientists do experiments and look for generalities because they want to predict what may occur under other circumstances. They also want to learn how to control the outcome of a chemical reaction or process. Understanding and explaining are two other important goals. For example, certain elements such as sodium react vigorously with water. But why is this true? To explain and understand this, you need a background in chemical concepts.

Dilemmas and Integrity in Science You may think research in science is straightforward: Do experiments, collect information, and draw a conclusion. But, research is seldom that easy. Frustrations and disappointments are common, and results can be inconclusive. Experiments always have some level of uncertainty, and sometimes the data collected are contradictory. For example, suppose you do an experiment expecting to find a direct relation between two experimental quantities. You collect six data sets. When plotted on a graph, four of the sets lie on a straight line, but two others lie far away from the line. Should you ignore the last two sets of data? Or should you do more experiments when you know that others could publish their results first and thus get the credit? Or should you consider that the two points not on the line might indicate that your original hypothesis is wrong and abandon a favorite idea you have worked on for many months? Scientists have a responsibility to remain objective in these situations, but sometimes it is hard to do. It is important to remember that a scientist is subject to the same moral pressures and dilemmas as any other person. To help ensure integrity in science, some simple principles have emerged over time that guide scientific practice:

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Experimental results should be reproducible. Furthermore, these results should be reported in the scientific literature in enough detail to be used or reproduced by others.

•

Research reports should be reviewed before publication by experts in the field to ensure that the experiments were conducted properly and that the conclusions are logical. (Scientists call this peer review.)

•

Conclusions should be reasonable and unbiased.

•

Credit should be given where it is due.

Chapter 1 / Basic Concepts of Chemistry

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Chemistry in Your Career

Darius Z. Brown

Darius Z. Brown adequate industry experience in various fields as a c­ hemist, has allowed me to focus on the relationships needed to teach high school students,” says Brown, who believes that his own disadvantaged background helps him connect with students who face similar challenges. “I believe that once you can see the world in terms of atoms and chemical reactions, your perspective . . . changes, and you become more conscious and aware of the little things in life, which ultimately helps . . . with problem-solving and working together.”

Darius Z. Brown (he/him/his) began his journey in the world of chemistry by obtaining a B.S. (University of Buffalo) and M.S. (University of ­ ­Illinois), with a focus on materials chemistry. He first put his degrees to use in a variety of industrial and research settings, ­including with a food manufacturer, university, and a paint producer. Missing something in his industry work, Brown returned to school to become a high school chemistry teacher. “Having a solid educational background in chemistry, along with

1.2 Sustainability and Green Chemistry Goal for Section 1.2 • Understand the principles of green chemistry.

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“It is better to prevent waste than to treat or clean up waste after it is formed.”

•

“Synthetic methods should be designed to maximize the incorporation of all materials used in the final product.”

•

Synthetic methods “should be designed to use and generate substances that possess little or no toxicity to human health or the environment.”

•

“Chemical products should be designed to [function effectively] while still reducing toxicity.”



© Charles D. Winters/Cengage

The world’s population is over 8.0 billion people, with about 99 million added per year. Each new person needs shelter, food, and medical care, and each uses increasingly scarce resources like fresh water and energy. And each produces by-products in the act of living and working that can affect our environment. With such a large population, these individual effects can have large consequences for our planet. The focus of scientists, planners, and politicians is increasingly turning to the concept of sustainable development. James Cusumano, a chemist and former president of a chemical company, said that “On one hand, society, governments, and industry seek economic growth to create greater value, new jobs, and a more enjoyable and fulfilling lifestyle. Yet, on the other, regulators, environmentalists, and citizens of the globe demand that we do so with sustainable development—meeting today’s global economic and environmental needs while preserving the options of future generations to meet theirs. How do nations resolve these potentially conflicting goals?” This conflict is even more evident now than it was in 1995 when Dr. Cusumano made this statement in the Journal of Chemical Education. Much of the increase in life expectancy and quality of life, at least in the developed world, is derived from advances in science. But it comes at a cost to the environment, with increases in polluting gases such as nitrogen oxides and sulfur oxides in the atmosphere, acid rain falling in many parts of the world, and waste pharmaceuticals entering the water supply. Among many others, chemists are seeking answers to these problems, and one response has been to practice green chemistry. The concept of green chemistry began to take root more than 30 years ago and now leads to new chemical methods and lower pollutant levels. Paul Anastas and John Warner stated 12 principles of green chemistry in their book Green Chemistry: Theory and Practice (Oxford, 1998) that have become hallmarks for chemists attempting to devise processes and products that are more environmentally sustainable. Among these are

GREEN

C H E M I S T RY

1.2  Sustainability and Green Chemistry

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7

•

“Energy requirements should be recognized for their environmental and economic impacts and should be minimized. Synthetic methods should be conducted at ambient temperature and pressure.”

•

Raw materials “should be renewable whenever technically and economically practical.”

•

“Chemical products should be designed so that at the end of their function, they do not persist in the environment or break down into dangerous products.”

•

“Substances used in a chemical process should be chosen to minimize the potential for chemical accidents, including releases, explosions, and fires.”

You will be reminded about these principles at various points in Chemistry & Chemical Reactivity as they are applied to modern applications in chemistry. Stating these fundamental ideas is good, but the real challenge is to put them into practice.

1.3 Classifying Matter Goals for Section 1.3 • Understand the basic ideas of kinetic-molecular theory. • Recognize the importance of representing matter at the macroscopic, submicroscopic, and symbolic levels.

• Recognize the different states of matter (solids, liquids, and gases) and know their characteristics.

• Recognize the difference between pure substances and mixtures as well as the difference between homogeneous and heterogeneous mixtures.

This section is an introduction to how chemists think about science in general and about matter in particular. Terms such as atom, element, molecule, and compound may appear to describe similar things, but each term has a unique definition. It is important that you know these definitions as they will be used throughout the book.

States of Matter and Kinetic-Molecular Theory

Solid

Bromine solid and liquid

Bromine gas and liquid Liquid

Figure 1.5  States of matter— solid, liquid, and gas.  Elemental

bromine exists in all three states near room temperature.

8

© Charles D. Winters/Cengage

Gas

One key property of matter is its state—that is, whether a substance is a solid, liquid, or gas (Figure 1.5). A solid has a rigid shape and fixed volume that changes little as temperature and pressure change. Like solids, liquids have a fixed volume, but a liquid is fluid—it takes on the shape of its container and has no definite shape of its own. Gases are fluid as well, but the volume of a gas is determined by the size of its container. The volume of a gas varies with changes in temperature and pressure. At low enough temperatures, virtually all matter is in the solid state. As the temperature is raised, solids usually melt to form liquids. Eventually, if the temperature is high enough, liquids evaporate to form gases. Volume changes typically accompany changes in state. For a given mass of material, there is usually a small increase in volume upon melting—water being a significant exception—and then a large increase in volume occurs upon evaporation. The kinetic-molecular theory of matter helps you interpret the properties of solids, liquids, and gases. According to this theory, all matter consists of extremely tiny particles (atoms, molecules, or ions) in constant motion. Solids: In solids, particles are packed closely together, usually in a regular pattern. The particles vibrate back and forth about their average positions, but seldom do particles in a solid squeeze past their immediate neighbors to come into contact with a new set of particles.

Chapter 1 / Basic Concepts of Chemistry

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Liquids: The particles in liquids are arranged randomly rather than in the regular patterns found in solids. Liquids and gases are fluid because the particles are not confined to specific locations and can move past one another. Gases: Under normal conditions, the particles in a gas are far apart. Gas molecules move extremely rapidly and are not constrained by their neighbors. The ­molecules of a gas fly about, colliding with one another and with the container walls. This ­random motion allows gas molecules to fill their container, so the volume of the gas sample is the volume of the container. There are net forces of attraction between particles in all states—they are ­g enerally small in gases and large in liquids and solids. These forces have a ­significant role in determining the properties of matter. An important aspect of the kinetic-­molecular theory is that the higher the temperature, the faster the particles move. The energy of motion of the particles (their kinetic energy, Section 1.7) acts to ­overcome the forces of attraction between particles. A solid melts to form a liquid when the ­temperature of the solid is raised to the point at which the particles vibrate fast enough and far enough to push one another out of the way and move out of their regularly spaced positions. As the temperature increases even more, the particles move faster still until finally they escape the clutches of their neighbors and enter the gaseous state.

Matter at the Macroscopic and Particulate Levels The characteristic properties of gases, liquids, and solids can be observed by the unaided human senses. They are determined using samples of matter large enough to be seen, measured, and handled. You can determine, for example, the color of a substance, whether it dissolves in water, whether it conducts electricity, and if it reacts with oxygen. Observations such as these generally take place in the ­macroscopic world of chemistry (Figure 1.6). This is the world of experiments and observations. Now imagine taking a macroscopic sample of material and dividing it again and again, past the point that the sample can be seen by the naked eye, and past the point where it can be seen using an optical microscope. Eventually, you reach the level of individual particles that make up all matter, a level that chemists refer to as the submicroscopic or particulate world of atoms and molecules (Figure 1.6).

Figure 1.6  Levels of matter.  Chemical and physical

A beaker of boiling water can be modeled at the particulate level as rapidly moving H2O molecules.

C PI

PA

© Charles D. Winters/Cengage



The process is symbolized by a chemical equation.

L E V E L S O F M A T T E R

R

T ATE UL IC

Observe

M ACR O

S

C

O

processes are observed at the macroscopic level. To understand or illustrate these processes, scientists often imagine what has occurred at the particulate atomic and molecular levels and write symbols to represent these observations.

Imagine

S Y C M B O L I

H2O (liquid) 88n H2O (gas)

Represent 1.3  Classifying Matter

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9

Chemists are interested in the structure of matter at this particulate level. Atoms, molecules, and ions cannot be seen in the same way that one views the macroscopic world, but they are no less real. Chemists imagine what atoms must look like and how they might fit together to form molecules. They create models to represent atoms and molecules (Figure 1.6)—where tiny spheres are used to represent atoms—and then use these models to think about chemistry and to explain their observations about the macroscopic world. For example, the macroscopic properties of gases are explained using the kinetic-molecular theory that assumes the existence of submicroscopic particles in motion. Chemists carry out experiments at the macroscopic level, but they think about chemistry at the particulate level. They then record their observations as symbols, formulas (such as H2O for water or NH3 for ammonia molecules), and drawings that represent the elements and compounds involved. As you study chemistry, try to make the connections in your own mind among the symbolic, particulate, and macroscopic worlds of chemistry.

Pure Substances A chemist looks at a glass of drinking water and sees a liquid. This liquid could be the pure chemical compound water. However, it is also possible the liquid is a mixture of water and dissolved substances. Specifically, matter can be classified as either a pure substance or a mixture (Figure 1.7). A pure substance has a set of unique properties by which it can be recognized. Pure water, for example, is colorless and odorless. If you want to identify a s­ ubstance conclusively as water, however, you would have to examine its properties more c­ arefully and compare them against the known properties of pure water. For example, pure water melts at 0 °C and boils at 100 °C under normal atmospheric pressure. If you can show that the substance melts and boils at these temperatures, you can be certain it is water. No other known substance melts and boils at precisely those temperatures. A second feature of a pure substance is that it cannot be separated into two or more different species by any physical technique at ordinary temperatures. If it can be separated, the sample is classified as a mixture.

Mixtures: Heterogeneous and Homogeneous A mixture consists of two or more pure substances that can be separated by physical techniques. In a heterogeneous mixture, the composition of the mixture is not uniform. The uneven distribution of the substances in a heterogeneous mixture can often be detected by the naked eye (Figure 1.8). However, keep in mind there are heterogeneous mixtures that may appear completely uniform but upon closer examination are not. Milk, for example, appears smooth in texture to the unaided eye, but magnification reveals fat and protein globules within the liquid. MATTER (may be solid, liquid, or gas) Anything that occupies space and has mass

HETEROGENEOUS MIXTURE Nonuniform composition

MIXTURES More than one pure substance present. Composition can be varied. HOMOGENEOUS MIXTURE Uniform composition throughout

Physically separable into...

COMPOUNDS Elements united in fixed ratios PURE SUBSTANCES

Fixed composition; cannot be further purified

Chemically separable into...

Combine chemically to form...

ELEMENTS Cannot be subdivided by chemical or physical processes

Figure 1.7  Classifying matter.

10

Chapter 1 / Basic Concepts of Chemistry

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A solution of salt in water. The model shows that salt in water consists of separate, electrically charged particles (ions), but the particles cannot be seen with an optical microscope.

−

+

John C. Kotz

−

A heterogeneous mixture.

+ −

+

A homogeneous mixture.

© Charles D. Winters/Cengage

Granite, an igneous rock, has grains large enough to see with the naked eye.

Figure 1.8  Heterogeneous and homogeneous mixtures.

A homogeneous mixture consists of two or more substances that are uniformly distributed down to the molecular level (Figure 1.8). No amount of optical ­magnification will reveal different properties in different regions of a homogeneous mixture. Homogeneous mixtures are often called solutions. Common examples include air (mostly a mixture of nitrogen and oxygen gases), gasoline (a mixture of carbon- and hydrogen-containing compounds called hydrocarbons), and a soft drink in an unopened container. When a mixture is separated into its pure components, the components are said to be purified. From the earliest days to modern times, the development of methods to separate mixtures is an important goal for chemists. In pharmaceutical ­laboratories, synthesizing a new drug requires pure starting materials. And, the synthesis usually results in a mixture of compounds from which the desired drug must be separated. Outside of a laboratory, purifying compounds may be even more important. For example, your health relies on the removal of harmful bacteria and toxic substances from water before it reaches your kitchen sink. One means of separation, filtration, is used to remove an undissolved solid from a solution. In this process, the mixture is applied to one side of a filter, a porous barrier whose openings are small enough that most of the undissolved solid particles cannot pass through, but the solution can. The solution emerges on the other side of the barrier, leaving the solid particles behind. Efforts at separation are often not completed in a single step, however, and repetition almost always increases purity. For example, soil particles can be separated from water by filtration (Figure 1.9). When the mixture is passed through a filter, many of the particles

© Charles D. Winters/Cengage

A heterogeneous mixture of soil and water When the mixture is poured through the filter paper, the larger soil particles are trapped and the water passes through. The water passing through the filter is more pure than in the mixture.

Figure 1.9  Purifying a hetero­geneous mixture by filtration.

1.3  Classifying Matter

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11

are removed. Repeated filtrations will give water with a higher and higher state of purity. This purification process uses a property of the mixture, its clarity, to measure the extent of purification. When a perfectly clear sample of water is obtained, all of the soil particles are assumed to have been removed.

1.4 Elements Goals for Section 1.4

Hydrogen—gas

© Charles D. Winters/Cengage

Oxygen—gas

Water—liquid

Figure 1.10  Decomposing water to yield hydrogen and oxygen gases.

• Identify the name or symbol for an element, given its symbol or name, respectively. • Use the terms atom, element, and molecule correctly. Passing an electric current through water can decompose it into gaseous ­hydrogen and oxygen (Figure 1.10). Substances like hydrogen and oxygen that are composed of only one type of atom are classified as elements. Several elements are shown in ­Figure 1.11. Though a matter of some debate, 94–98 of the currently known 118 ­elements have been confirmed to exist in nature, though some occur only in trace quantities. The remainder have been created by scientists and have not yet been found in any naturally occurring samples. The names of the elements and their chemical symbols, one- or two-letter abbreviations used in place of the names, are listed in the tables inside the front and back covers of this book. Carbon (C), sulfur (S), iron (Fe), copper (Cu), silver (Ag), tin (Sn), gold (Au), mercury (Hg), and lead (Pb) were known to the early Greeks and Romans as well as the alchemists of ancient China, the medieval Islamic world, and medieval Europe. However, many other elements—such as aluminum (Al), silicon (Si), iodine (I), and helium (He)— were not discovered until the eighteenth and nineteenth centuries. Finally, scientists in the twentieth and twenty-first centuries have made elements that do not exist in nature, such as technetium (Tc) and plutonium (Pu). The stories behind some of the names of the elements are fascinating. Many elements have names and symbols with Latin or Greek origins. Examples include helium (He), from the Greek word helios meaning “sun,” and lead, whose symbol, Pb, comes from the Latin word for “heavy,” plumbum. More recently discovered ­elements have been named for their place of discovery or place of significance, such as americium (Am), californium (Cf), scandium (Sc), europium (Eu), francium (Fr), and polonium (Po). Some elements are named for their discoverers or famous scientists. For example, curium (Cm), einsteinium (Es), fermium (Fm), mendelevium (Md), nobelium (No), and meitnerium (Mt) were named after Marie and Pierre Curie, Albert Einstein, Enrico Fermi, Dmitri Mendeleev, Alfred Nobel, and Lise Meitner, respectively.

Figure 1.11 Elements. ​

© Charles D. Winters/Cengage

Chemical elements can often be distinguished by their color and their state at room temperature.

12

Mercury— liquid

Powdered sulfur—solid

Copper wire— solid

Iron chips— solid

Aluminum— solid

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© Charles D. Winters/Cengage

When writing the symbol for an element, notice that the first letter (but not the second) of an element’s symbol is capitalized. For example, cobalt is Co, not co or CO. The notation co has no chemical meaning, whereas CO represents the chemical compound carbon monoxide. Also note that the element name is not capitalized, except at the beginning of a sentence. The table inside the front cover of this book, in which the symbol and other information for the elements are enclosed in a box, is called the periodic table. This important tool of chemistry is described in more detail beginning in Chapter 2. An atom is the smallest particle of an element that retains the characteristic chemical properties of that element. Some elements, such as neon and argon, are found in nature as isolated atoms. Others are found as molecules, particles consisting of more than one atom in which the atoms are held together by chemical bonds. Examples of molecular elements are the colorless gases of the air, nitrogen (N2) and oxygen (O2) as well as deep purple iodine (I2) and orange liquid bromine (Br2). For each of these cases, the subscript “2” following the element symbol indicates that two atoms of the element exist in a single molecule. Yet other elements consist of infinite networks of atoms; an example of this is diamond, one form of the element carbon.

Diamond.

Diamond.  A diamond consists of a network of carbon atoms linked by chemical bonds.

1.5 Compounds Goals for Section 1.5

•

Sodium is a shiny metal that reacts violently with water (Figure 1.4). Its solidstate structure has sodium atoms tightly packed together.

•

Chlorine is a light yellow-green gas that has a distinctive, suffocating odor and is a powerful irritant to lungs and other tissues. The element is composed of Cl2 molecules in which two chlorine atoms are tightly bound together.

•

Sodium chloride, or common salt (NaCl), is a colorless, crystalline solid composed of sodium and chlorine ions bound tightly together. Its properties are completely unlike those of the two elements from which it is made.

It is important to distinguish between a mixture of elements and a chemical c­ ompound of two or more elements. Pure metallic iron and yellow, powdered sulfur can be mixed in varying proportions. In the chemical compound iron pyrite, ­however, there is no variation in composition. Not only does iron pyrite exhibit properties unique to itself and different from those of iron, sulfur, or a mixture of these two elements, it also has a definite percentage composition by mass (46.55% Fe and 53.45% S). That a compound has a definite composition (by mass) of its combining elements is a basic principle of chemistry and is often referred to as the law of definite proportions or the law of constant composition. Thus, two major differences exist between a mixture and a pure compound: A compound has ­distinctly different characteristics from its parent elements, and it has a definite ­percentage composition (by mass) of its combining elements.

Mixture The material in the dish is a mixture of elements, iron and sulfur. The iron can be separated easily from the sulfur by using a magnet.

© Charles D. Winters/Cengage

A pure substance like sugar, salt, or water, which is composed of two or more ­different elements held together by chemical bonds, is referred to as a chemical compound. Even though only 118 elements are known, there appears to be no limit to the number of compounds that can be made from those elements. As of 2021, over 193 million different compounds were identified in CAS, a ­database created by the American Chemical Society. The properties of a compound, such as its color, hardness, and melting point, are different than those of its constituent elements. Consider common table salt (sodium chloride), which is composed of two elements, sodium and chlorine (­Figure 1.2):

© Charles D. Winters/Cengage

• Use the term compound correctly. • Understand the law of definite proportions (law of constant composition).

Chemical compound Iron pyrite is a chemical compound composed of iron and sulfur. It is often found in nature as perfect, golden cubes. 1.5 Compounds

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13

Name Formula

Water

Methane

Ammonia

Carbon dioxide

H2O

CH4

NH3

CO2

Model

Figure 1.12  Names, formulas, and models of some common molecular compounds.  A common color scheme

used in molecular models is C atoms are gray or black, H atoms are white, N atoms are blue, and O atoms are red.

Compounds with Indefinite Proportions  Though the vast

majority of compounds have definite proportions, a few break this law. For example, a class of high temperature superconductors have the formula La2-xBaxCuO4, in which x varies from 0 to 0.5. (See “Applying Chemical Principles 3.1: Superconductors,” page 177).

Some compounds—such as table salt, NaCl—are composed of ions, which are electrically charged atoms or groups of atoms (Section 2.5). Other compounds— such as water and sugar—consist of molecules. The composition of any compound is represented by its chemical formula. In the formula for water, H2O, for example, the symbol for hydrogen, H, is followed by a subscript “2” that indicates that two atoms of hydrogen occur in a single water molecule. The symbol for oxygen appears without a subscript, indicating that one oxygen atom occurs in the molecule. As you shall see throughout this text, molecules can be represented with models that depict their composition and structure. Figure 1.12 illustrates the names, formulas, and models of the structures of four common molecular compounds.

E xamp le 1.1

Elements and Compounds Problem  Identify whether each of the following is an element or compound: bromine, Br2, and hydrogen peroxide, H2O2.

What Do You Know?   Elements and compounds are both pure substances. An element is composed of only one type of atom. A compound is composed of more than one type of atom, where the atoms are connected by chemical bonds and occur in a definite proportion by mass. Strategy   Look at each formula given and use the guidelines in the “What Do You Know?” section to determine whether the formula is that for an element or a compound.

Solution   Bromine, Br2, is an element because both of the atoms present in the molecule are the same type, bromine atoms. H2O2 is a compound. In H2O2, there are two different types of atoms present, hydrogen atoms and oxygen atoms. They are bonded together in hydrogen peroxide molecules that have a definite composition by mass; each molecule of H2O2 has two atoms of hydrogen and two atoms of oxygen.

Think about Your Answer   If all of the atoms in a molecule are the same type, such as in Br2, then it is a molecule of an element.

Check Your Understanding Identify whether white phosphorus (P 4) and carbon monoxide (CO) are elements or compounds.

14

Chapter 1 / Basic Concepts of Chemistry

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1.6 Properties and Changes Goals for Section 1.6 • Identify several physical properties of common substances. • Relate density to the volume and mass of a substance. • Understand the difference between extensive and intensive properties and give examples.

• Explain the difference between chemical and physical changes. • Identify several chemical properties of common substances. You recognize your friends by their physical appearance: their height and weight and the color of their eyes and hair. The same is true of chemical substances. You can tell the difference between an ice cube and a cube of lead of the same size by their appearance (one is clear and colorless, the other is a lustrous metal), and because one is heavier (lead) than the other (ice). Properties such as these, which can be observed and measured without changing the composition of a substance, are called ­physical properties. The chemical elements in Figure 1.11, for example, differ in terms of their color, appearance, and state (solid, liquid, or gas). Substances can often be classified and identified by their physical properties. Table 1.1 lists several physical properties of matter that chemists commonly use. Density, the ratio of the mass of an object to its volume, is a physical property useful for identifying substances. Density 5



mass volume

(1.1)

For example, you can tell the difference between an ice cube and a cube of lead of identical size because lead has a high density, 11.35 g/cm3 (11.35 grams per cubic centimeter), whereas ice has a density slightly less than 0.917 g/cm3. An ice cube with a volume of 16.0 cm3 has a mass of 14.7 g, whereas a cube of lead with the same volume has a mass of 182 g. The temperature of a sample of matter often affects the numerical values of its ­properties. Density is a particularly important example. Although the change in water density with temperature seems small (Table 1.2), it affects the environment profoundly. Table 1.1



Ice

© Charles D. Winters/Cengage

Lead

Units of Density  The decimal system of units in the sciences is called Le Système International d’Unités, often referred to as SI units. The SI unit of density is kg/m3. In chemistry, the more commonly used unit is g/cm3. To convert from kg/m3 to g/cm3, divide by 1000.

Some Physical Properties

Property

Using the Property to Distinguish Substances

Color

Is the substance colored or colorless? What is the color, and what is its intensity?

State of matter

Is it a solid, liquid, or gas? If it is a solid, what is the shape of the particles?

Melting point

At what temperature does a solid melt?

Boiling point

At what temperature does a liquid boil (at 1 atmosphere pressure)?

Density

What is the substance’s density (mass per unit volume)?

Solubility

What mass of substance can dissolve in a given volume of water or other solvent?

Electric conductivity

Does the substance conduct electricity?

Malleability

Table 1.2 Temperature Dependence of Water Density

Temperature (° C)

Density of Water (g/cm3)

  0 (ice)

0.917

   0 (liq water)

0.99984

  2

0.99994

  4

0.99997

How easily can a solid be deformed?

 10

0.99970

Ductility

How easily can a solid be drawn into a wire?

 25

0.99707

Viscosity

How easily will a liquid flow?

100

0.95836

1.6  Properties and Changes

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15

For example, as the water in a lake cools, the density of the water increases and the denser water sinks as shown in Figure 1.13. This continues until the water temperature reaches 3.98 °C, the point at which water has its maximum density (0.999973 g/cm3). If the water temperature drops farther, the density decreases slightly and the colder water floats on top of water at 3.98 °C. If water is cooled below about 0 °C, solid ice forms. Water has a rare property: Its solid form is less dense than its liquid form, so ice floats on water. The volume of a given mass of liquid changes with temperature, so its density does also. This is why laboratory glassware used to measure precise volumes of solutions always specifies the temperature at which it was calibrated (Figure 1.14).

E xamp le 1.2

Density Problem   A piece of a polypropylene rope (used for water skiing) floats on water, whereas a terephthalate polymer from a soda bottle sinks in water. Place polypropylene, the terephthalate polymer, and water in order of increasing density. What Do You Know?   Density is given by Equation 1.1. An object with a higher density sinks in a liquid of lower density, whereas an object with a lower density floats in a liquid of higher density.

Strategy   Use the observations as to whether a material sinks or floats in water to determine the order of the densities. Solution  The polypropylene rope floats in water, therefore polypropylene is less dense than water. The soda bottle plastic sinks in water; therefore the soda bottle plastic is more dense than the water. This gives the overall order of densities: polypropylene < water < soda bottle plastic.

Think about Your Answer   In a material with a greater density, the matter is more tightly packed for a given mass than in materials of lower density.

Check Your Understanding

© Charles D. Winters/Cengage

Blue dye was added to the left side of the water-filled tank, and ice cubes were placed in the right side.

The water beneath the ice is cooler and denser than the surrounding water, so it sinks. The convection current created by this movement of water is traced by the dye movement as the denser, cooler water sinks.

Figure 1.13  Temperature dependence of physical properties.  The density of water and other substances changes with temperature.

16

© Charles D. Winters/Cengage

Some salad dressings are made from a mixture of olive oil and vinegar. These two liquids do not dissolve significantly in each other. If this mixture is allowed to sit, the two liquids separate from each other with the olive oil floating on top of the vinegar. Which liquid has the greater density?

Figure 1.14  Dependence of density on temperature.  Water and other substances change in density with temperature so laboratory glassware is calibrated for a particular temperature.

Chapter 1 / Basic Concepts of Chemistry

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Extensive and Intensive Properties The physical properties of matter can be classified as either extensive properties or intensive properties. Extensive properties depend on the amount of a substance present. The mass and volume of samples of elements or compounds or the amount of energy transferred as heat from burning gasoline are extensive properties, for example. In contrast, intensive properties do not depend on the amount of substance. A sample of ice will melt at 0 °C, no matter whether it is an ice cube or an iceberg. Although mass and volume are extensive properties, it is interesting that density (the quotient of these two quantities) is an intensive property. The density of gold, for example, is the same (19.3 g/cm3 at 20 °C) whether it is a flake of pure gold or a solid gold ring. Intensive properties are often useful in identifying a material. For example, the temperature at which a material melts (its melting point) is often so characteristic that it can be used to identify the solid (Figure 1.15).

Exam p le 1.3

Extensive and Intensive Properties Problem  A sample of liquid mercury has a shiny surface, melts at 234 K, has a mass of 27.2 g, has a volume of 2.00 cm3, and has a density of 13.6 g/cm3. Which of these properties are extensive properties and which are intensive properties?

What Do You Know?  Extensive properties depend on the amount of a substance present. Intensive properties do not depend on the amount of substance present. Strategy  Determine which of the properties listed depend on the amount of material present and which do not. Solution  The mass and volume of the sample each depend on the amount of material present; the greater the amount of material present, the greater will be the mass and the volume. Mass and volume are extensive properties. The shininess of the surface, the melting point, and the density are properties that are the same regardless of the amount of material present, so they are intensive properties.

Think about Your Answer  Mass and volume are both extensive properties, but their quotient, density, is an intensive property.

Check Your Understanding Identify whether each of the following properties is extensive or intensive: boiling point, hardness, volume of a solution, number of atoms, number of atoms dissolved per volume of solution.

© Charles D. Winters/Cengage

Figure 1.15  An intensive property used to distinguish compounds. 

Naphthalene, a white solid at room temperature, melts at 80.2 °C and so is molten at the temperature of boiling water.



Aspirin, a white solid at room temperature, melts at 135 °C and so remains a solid at the temperature of boiling water.

1.6  Properties and Changes

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17

Physical and Chemical Changes Changes in physical properties are called physical changes. In a physical change the identity of a substance is preserved even though it may have changed its physical state or the gross size and shape of its pieces. A physical change does not result in a new chemical substance being produced. The particles (atoms, molecules, or ions) present before and after the change are the same. An example of a physical change is the melting of a solid (Figure 1.15) or the evaporation of a liquid (Figure 1.16). In either case, the same molecules are present both before and after the change. Their chemical identities have not changed. Physical change • The same molecules are present both before and after the change. O2 molecules in the gas phase

© Charles D. Winters/Cengage

Liquid oxygen (boiling point, –183 °C) is a pale blue liquid.

O2 molecules in the liquid phase

Chemical change • One or more substances (reactants) are transformed into one or more different substances (products) A balloon filled with molecules of hydrogen gas and surrounded by molecules of oxygen in the air. (The balloon floats in air because gaseous hydrogen is less dense than air.)

© Charles D. Winters/Cengage

When ignited with a burning candle, H2 and O2 react to form water, H2O.

O2 (gas)

2 H2 (gas)

2 H2O(gas)

A symbolic and particulate view • The reaction of O2 and H2 Symbolic view

2 H2(gas)

O2(gas)

2 H2O(gas)

Particulate view

Reactants

Products

Figure 1.16  Physical and chemical change.

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Chapter 1 / Basic Concepts of Chemistry

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A physical property of hydrogen gas (H2) is its low density, so a balloon filled with H2 floats in air. Suppose, however, that a lighted candle is brought up to the balloon. When the heat causes the skin of the balloon to rupture, the hydrogen mixes with the oxygen (O2) in the air, and the heat of the burning candle sets off a chemical reaction that produces water, H2O (Figure 1.16). This reaction is an example of a chemical change, in which one or more substances (the reactants) are transformed into one or more different substances (the products). The chemical change in Figure 1.16 is also shown at the particulate level: hydrogen molecules and oxygen molecules react to form water molecules. The representation of the change using chemical formulas is called a chemical equation. It shows that the substances on the left (the reactants) produce the substances on the right (the products). This equation illustrates an important principle of chemical reactions: matter is conserved. The number and identity of the atoms found in the reactants are the same as in the products. Here, there are four atoms of H and two atoms of O before and after the reaction, but the molecules before the reaction are different from those after the reaction. The term chemical property refers to chemical reactions that a substance might undergo. For example, a chemical property of hydrogen gas is that it reacts with oxygen gas, and quite vigorously so.

Exam p le 1.4

Physical and Chemical Changes Problem  Identify each of the following as being either a physical or a chemical change: boiling water and rusting of an iron nail (which converts iron (Fe) to Fe2O3).

What Do You Know?  In a physical change, the chemical identities of the materials do not change, whereas in a chemical change, they do.

Strategy   Examine each of the changes to determine if the chemical identities of the materials change. Solution  In liquid water, the chemical species present is H2O molecules. When water boils, molecules move to the gaseous state. The chemical species is still H2O molecules; the molecules have merely changed state. Boiling water is thus a physical change. Rusting of an iron nail is a chemical change because you begin with iron and oxygen and end up with rust, which is predominantly the chemical compound iron(III) oxide, Fe 2O3. The substance at the end of the process is a different chemical species than the ones with which you started.

Think about Your Answer  Students sometimes confuse changes of state with chemical changes; changes of state are physical changes.

Check Your Understanding Identify whether each of the following is a physical change or a chemical change: (a) melting butter, (b) burning wood, (c) dissolving sugar in water.



1.6  Properties and Changes

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1.7 Energy: Some Basic Principles Goals for Section 1.7 • Identify types of potential and kinetic energy. • Recognize and apply the law of conservation of energy. Units of Energy  Energy in

chemistry is measured in units of joules. (See Chapter 1R and Chapter 5 for calculations involving energy units.)

Energy, a crucial part of many chemical and physical changes, is defined as the capacity to do work. You do work against the force of gravity when carrying yourself and hiking equipment up a mountain. The energy to do this is provided by the food you have eaten. Food is a source of chemical energy—energy stored in chemical compounds and released when the compounds undergo the chemical reactions of metabolism in your body. Chemical reactions almost always either release or absorb energy. Energy can be classified as kinetic or potential. Kinetic energy is energy associated with motion, such as: •

The motion of atoms, molecules, or ions at the submicroscopic (particulate) level (thermal energy). All matter has thermal energy.

•

The motion of macroscopic objects such as a moving tennis ball or automobile (mechanical energy).

•

The movement of electrons in a conductor (electrical energy).

•

The compression and expansion of the spaces between molecules in the transmission of sound (acoustic energy).

Potential energy results from an object’s position or state and includes: •

Energy possessed by a ball held above the floor and by water at the top of a water wheel (gravitational energy) (Figure 1.17a).

•

Energy stored in an extended spring.

•

Energy stored in fuels (chemical energy) (Figure 1.17b).

•

Energy associated with the separation of electrical charges (electrostatic energy) (Figure 1.17c).

(a) Potential energy is converted into mechanical energy.

(b) Chemical potential energy of a fuel and oxygen is converted into thermal and mechanical energy in the launch of a rocket.

Aleksandr Gogolin/Shutterstock.com

Salajean/Shutterstock.com

Handout/Getty Images News/Getty Images

Potential energy and kinetic energy can be interconverted. For example, as water falls over a waterfall, its potential energy is converted into kinetic energy. S­ imilarly, kinetic energy can be converted into potential energy: The kinetic energy of f­ alling water can turn a turbine to produce electricity, which can then be used to c­ onvert

(c) Electrostatic energy is converted to radiant and thermal energy.

Figure 1.17  Energy and its conversion.

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Chapter 1 / Basic Concepts of Chemistry

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Kinetic energy (energy of motion) Diego Barbieri /Shutterstock.com

The diver has potential energy when standing a distance above the water surface.

Heat and work (thermal and mechanical energy)

AkosHorvath /Shutterstock.com

Paolo Bona/Shutterstock.com

Potential energy (energy of position)

Figure 1.18  The law of energy conservation.

The diver’s potential energy is first converted to kinetic energy, which is then transferred to the water.

water into H2 and O2 by electrolysis. Hydrogen gas contains stored chemical ­potential energy because it can be burned to produce heat and light or electricity.

Conservation of Energy Standing on a diving board, you have considerable potential energy because of your position above the water. Once you dive off the board, some of that potential energy is converted into kinetic energy (Figure 1.18). During the dive, the force of gravity accelerates your body so that it moves faster and faster. Your kinetic energy increases and your potential energy decreases. At the moment you hit the water, your velocity is abruptly reduced and much of your kinetic energy is transferred to the water as your body moves it aside. Eventually you float to the surface, and the water becomes still again. If you could see the water molecules, however, you would find that they are moving a little faster in the vicinity of your entry into the water; that is, the kinetic energy of the water molecules is slightly higher. This series of energy conversions illustrates the law of conservation of energy, which states that energy can neither be created nor destroyed. Or, to state this law differently, the total energy of the universe is constant. The law of conservation of energy summarizes the results of many experiments in which the amounts of energy transferred were measured and the total energy content was found to be the same before and after an event. This law can also be examined in the case of a chemical reaction, the reaction of hydrogen and oxygen to form water (Figure 1.16). In this reaction, the reactants (­hydrogen and oxygen) have a certain amount of energy associated with them. When they react, some of this energy is released to their surroundings. If you add up all the energy present before the reaction and all the energy present after the reaction, you will find that the energy was only redistributed; the total amount of energy in the universe has remained constant. Energy has been conserved.

Applying Chemical Principles 1.1 CO2 in the Oceans A 2019 report on the absorption of carbon in the world’s ocean stated that “The global ocean absorbed 34 billion metric tons of carbon from the burning of fossil fuels from 1994 to 2007—a fourfold increase of 2.6 billion metric tons per year when compared to the period starting from the Industrial Revolution in 1800 to 1994. The amount of carbon in the form of CO2 dissolved in the oceans is of great concern and interest because it affects the pH

of the water, that is, its level of acidity. This in turn can affect the growth of marine organisms such as corals, sea urchins, and microscopic coccolithophores (single-cell phytoplankton). Studies have indicated that, in water with a high CO2 content, the spines of sea urchins are greatly impaired, the larvae of orange clown fish lose their homing ability, and the concentrations of calcium, copper, manganese, and iron in sea water are affected, sometimes drastically. Applying Chemical Principles

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and magnitude of ocean geochemical changes potentially unparalleled in at least the last 300 million years of Earth history, raising the possibility that we are entering an unknown territory of marine ecosystem change.”

Cbpix/Shutterstock.com

Questions

Clown fish.  The larvae of the clown fish are affected by increased levels of CO2 in the ocean.

An investigation of the history of ocean acidification ended with the statement that “the current rate of (mainly fossil fuel) CO2 release stands out as capable of driving a combination

1. Much has been written about CO2. What is its name? 2. Give the symbols for the four metals mentioned in this article. 3. Of the four metals mentioned here, which is the most dense? The least dense? (Use an Internet tool such as www.ptable.com to find this information.) 4. The spines of sea urchins, corals, and coccolithophores all are built of the compound CaCO3. Which elements are involved in this compound? Do you know this compound’s name? If not, you might be able to find it using an Internet search. References 1. https://www.ncei.noaa.gov/news/global-ocean-absorbing-more-carbon 2. “The Geological Record of Ocean Acidification,” B. Hönisch, et al., Science, 2012, 335, 1058–1063.

Think–Pair–Share 1. Hypotheses, Laws, and Theories (a) What are the differences between a hypothesis and a theory? (b) Why is the law of constant composition a law and not a theory? 2. Carbon dioxide consists of molecules containing one carbon atom and two oxygen atoms. In each molecule the three atoms are arranged in a line with the carbon in between the two oxygen atoms. Make a drawing, based on the kineticmolecular theory, of the arrangement of carbon dioxide molecules in (a) the solid state, (b) the liquid state, and (c) the gas state. In your drawing, represent the carbon atoms with a solid circle and the oxygen atoms with an open circle. For each case, draw 10 molecules. It is acceptable for your diagram to be two dimensional. 3. Some students come to a chemistry class thinking that all samples of elements consist of isolated atoms and all samples of compounds consist of molecules. (a) Are all elements found in nature as isolated atoms? If not, how else can the atoms be arranged? (b) Are all compounds composed of molecules? If not, what other types of particles are found in some compounds? (c) Is N2 an element or a compound? Explain your answer.

4. Answer the following. (a) Zinc is a shiny silver-colored solid that reacts vigorously with hydrochloric acid. Identify each of these properties as being either a physical property or a chemical property. Explain your reasoning. (b) The mass of a piece of aluminum is 10.0 g, its volume is 3.70 cm3, and its density is 2.70 g/cm3. Identify each of these properties as being either an intensive property or an ­extensive property. Explain your reasoning. (c) Identify each of the following as being either a physical change or a chemical change. Explain your reasoning. (i) Bubbles of carbon dioxide gas form when baking soda is added to vinegar. (ii) Paper burns. (iii) Sugar dissolves in water. (iv) Water boils. 5. Give one or more examples of each of the following types of energy conversions. (a) Chemical potential energy into electrical energy. (b) Chemical potential energy into thermal energy. (c) Thermal energy into mechanical energy. (d) Chemical potential energy into sound and/or light energy.

Chapter Goals Revisited The goals for this chapter are keyed to specific Study Questions to help you organize your review.

1.1  Chemistry and Its Methods • Recognize the difference between a hypothesis and a theory and understand how laws are established. 1, 2.

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Chapter 1 / Basic Concepts of Chemistry

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1.2  Sustainability and Green Chemistry • Understand the principles of green chemistry. 3–6.

1.3  Classifying Matter • Understand the basic ideas of kinetic-molecular theory. 43, 44. • Recognize the importance of representing matter at the macroscopic, submicroscopic, and symbolic levels. 37, 38.

• Recognize the different states of matter (solids, liquids, and gases) and know their characteristics. 31, 43, 53.

• Recognize the difference between pure substances and mixtures and the

difference between homogeneous and heterogeneous mixtures. 33, 34, 44.

1.4 Elements • Identify the name or symbol for an element, given its symbol or name, respectively. 7–10, 31, 32.

• Use the terms atom, element, and molecule correctly. 11, 12, 41, 42.

1.5 Compounds • Use the term compound correctly. 11, 12, 41, 42. • Understand the law of definite proportions (law of constant composition). 13, 14.

1.6  Properties and Changes • Identify several physical properties of common substances. 15, 16, 19, 20, 32, 46, 48.

• Relate density to the volume and mass of a substance. 27, 28, 39, 40, 45, 47, 49, 50, 51, 54, 55, 58.

• Understand the difference between extensive and intensive properties and give examples. 27, 28.

• Explain the difference between chemical and physical changes. 17, 18, 35, 36, 53, 57.

• Identify several chemical properties of common substances. 15, 16, 19, 20, 29, 30.

1.7  Energy: Some Basic Principles • Identify types of potential and kinetic energy. 21–24. • Recognize and apply the law of conservation of energy. 25, 26.

Key Equation Equation 1.1  (page 15)  Density: In chemistry the common unit of density is g/cm3, whereas kg/m3 is commonly used in geology and oceanography.



Density 5



mass volume

Key Equation

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Study Questions ▲ denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully worked solutions in the ­Student Solutions Manual.

Practicing Skills Nature of Science (See Section 1.1.) 1. In the following scenario, identify which of the statements represents a theory, law, or hypothesis. (a) A student exploring the properties of gases proposes that if she decreases the volume of a sample of gas then the pressure exerted by the sample will increase. (b) Many scientists over time have conducted similar experiments and have concluded that pressure and volume are inversely proportional. (c) She proposes that the reason this occurs is that if the volume is decreased, more molecules will collide with a given area of the container walls, causing the pressure to be greater. 2. State whether the following is a hypothesis, theory, or law of science. Global climate change is occurring because of human-generated carbon dioxide. Explain.

Green Chemistry (See Section 1.2.) 3. What is meant by the phrase “sustainable development”? 4. What is meant by the phrase “green chemistry”? 5. Chronic cough is a condition estimated to affect as many as 5% to 10% of adults. Merck and Co., Inc. earned a Presidential Green Chemistry Challenge Award in 2021 for developing a new method of synthesizing gefapixant citrate, a drug to treat chronic cough. The process allows for a six-fold reduction in raw material costs, requires fewer synthetic steps, and achieves a higher yield of the final product than the previous synthetic method. A synthetic step that involved highly hazardous chemicals was replaced, resulting in a safer production process. Finally, the synthesis produces less carbon dioxide and carbon monoxide emissions. Which principles of green chemistry are being followed by this new process compared to the old process? 6. Surfactants are substances used in detergents to reduce the surface tension of liquids in which they are dissolved. Most surfactants are petroleumbased, and their production requires considerable energy and produces toxic waste products. Colonial Chemical was a winner of a 2021 Presidential Green Chemistry Award for the development

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of Suga®Boost surfactants that use more environmentally-friendly chemicals than traditional surfactants. These surfactants are produced from renewable plant-based materials, require less energy to create, require only water as a solvent, and biodegrade into nontoxic substances. Which principles of green chemistry are being followed by this new process compared to the old process?

Matter: Elements and Atoms, Compounds, and Molecules (See Example 1.1.) 7. Give the name of each of the following elements: (a) N (d) I (b) Ca (e) Li (c) Br (f) Fe 8. Give the name of each of the following elements: (a) Cr (d) Cl (b) Ni (e) Ar (c) Mg (f) Ti 9. Give the symbol for each of the following elements: (a) strontium (d) mercury (b) cadmium (e) selenium (c) cobalt (f) bismuth 10. Give the symbol for each of the following elements: (a) platinum (d) thallium (b) gallium (e) tungsten (c) uranium (f) xenon 11. In each of the following pairs, decide which is an element and which is a compound. (a) Na or NaCl (b) sugar or carbon (c) gold or gold(III) chloride 12. In each of the following pairs, decide which is an element and which is a compound. (a) Pt(NH3)2Cl2 or Pt (b) copper or copper(II) oxide (c) silicon or SiO2 13. An 18 g sample of water decomposes into 2 g of hydrogen gas and 16 g of oxygen gas. What masses of hydrogen and oxygen gases would be prepared from 27 g of water? What law of chemistry is used in solving this problem?

Chapter 1 / Basic Concepts of Chemistry

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(See Section 1.6) 15. In each case, decide if the underlined property is a physical or chemical property. (a) The color of elemental bromine is orange-red. (b) Iron turns to rust in the presence of air and water. (c) Hydrogen can explode when ignited in air (Figure 1.16). (d) The density of titanium metal is 4.5 g/cm3. (e) Tin metal melts at 505 K. (f) Chlorophyll, a plant pigment, is green. 16. In each case, decide if the underlined property is a physical or chemical property. (a) Copper wire is used for the transmission of electricity because it is a good electrical conductor. (b) Olive oil pours more slowly from a container than water due to its greater viscosity. (c) The density of liquid water is 1.00 g/mL at 4°C. (d) In wine making, grapes ferment to produce wine, an alcoholic beverage. (e) A pigment used in white paint is titanium(IV) oxide. (f) Gasoline burns when ignited in air. 17. In each case, decide if the change is a chemical or physical change. (a) A puddle of water evaporates after a rain shower ends. (b) Milk turns sour after it is left in a warm place for too long. (c) When baking soda and vinegar are combined, the mixture produces bubbles of carbon dioxide gas. (d) Table sugar, or sucrose, dissolves in water. (e) A tomato turns red as it ripens. (f) A firework explodes filling the sky with bright flashes of light. 18. In each case, decide if the change is a chemical or physical change. (a) A cup of household bleach changes the color of your favorite T-shirt from purple to pink. (b) Water vapor in your exhaled breath condenses in the air on a cold day.



19. Which part of the description of a compound or element refers to its physical properties and which to its chemical properties? (a) Chlorine, a yellow-green gas, reacts with sodium metal to produce sodium chloride. (b) Sodium bicarbonate is a white powdery solid that reacts with an acid to produce carbon dioxide. 20. Which part of the description of a compound or element refers to its physical properties and which to its chemical properties? (a) Copper(II) sulfide is a black solid with a density of 2.71 g/cm3. It reacts readily with an acid to produce gaseous hydrogen sulfide. (b) Magnesium, a silvery metal, reacts with oxygen gas to produce a white compound.

Energy (See Section 1.7.) 21. The flashlight in the photo does not use batteries. Instead, you move a lever, which turns a geared mechanism that results in light from the bulb. What type of energy is used to move the lever? What type or types of energy are produced?

A hand-operated flashlight

22. A solar panel is pictured in the photo. When light shines on the panel, it generates an electric current that can be used to recharge the batteries in an electric car. What types of energy are involved in this setup?

John C. Kotz

Physical and Chemical Properties

(c) Plants use carbon dioxide from the air to make sugar. (d) Butter melts when placed in the sun.

© Charles D. Winters/Cengage

14. A sample of the compound magnesium oxide is synthesized as follows: 60. g of magnesium is burned and produces 100. g of magnesium oxide, indicating that the magnesium combined with 40. g of oxygen in the air. If 30. g of magnesium were used, what mass of oxygen would combine with it? What law of chemistry is used in solving this problem?

A solar panel

Study Questions

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23. Determine which of the following represent potential energy and which represent kinetic energy. (a) acoustic energy (b) thermal energy (c) gravitational energy (d) chemical energy (e) electrostatic energy 24. Determine whether kinetic energy is being converted to potential energy, or vice versa, in the following processes. (a) Water cascades downward in a waterfall. (b) A furnace burns natural gas to heat a house. (c) An electric current is used to recharge a battery. (d) A piston moves when a gas is produced in a chemical reaction. 25. A hot metal block is plunged into water in a well-insulated container. The temperature of the metal block goes down, and the temperature of the water goes up until their temperatures are the same. A total of 1500 J of energy is lost by the metal object. How much energy was absorbed by the water? What law of science is illustrated by this problem?

29. Which observations below describe chemical properties? (a) Mixing vinegar and sodium bicarbonate produces bubbles of carbon dioxide gas. (b) Sugar is soluble in water. (c) Water boils at 100 °C. (d) Ultraviolet light converts O3 (ozone) to O2 (oxygen). (e) Ice is less dense than liquid water. 30. Which observations below describe chemical properties? (a) A reddish-brown coating of rust appears on the surface of a piece of iron. (b) Sodium metal reacts violently with water. (c) The combustion of octane (a compound in gasoline) gives CO2 and H2O. (d) Chlorine is a yellow-green gas. (e) Heat is required to melt ice. 31. The mineral fluorite contains the elements calcium and fluorine and can have various colors, including blue, violet, green, and yellow.

These questions are not designated as to type or location in the chapter. They may combine several concepts. 27. A piece of turquoise is a blue-green solid; it has a density of 2.65 g/cm3 and a mass of 2.5 g. (a) Which of these observations are qualitative and which are quantitative? (b) Which of the observations are extensive and which are intensive? (c) What is the volume of the piece of turquoise? 28. Iron pyrite (fool’s gold, page 13) has a shiny golden metallic appearance. Crystals are often in the form of perfect cubes. A cube 0.40 cm on each side has a mass of 0.32 g. (a) Which of these observations are qualitative and which are quantitative? (b) Which of the observations are extensive and which are intensive? (c) What is the density of the sample of iron pyrite?

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The mineral fluorite, calcium fluoride

(a) What are the symbols of these elements? (b) How would you describe the shape of the fluorite crystals in the photo? What can this tell you about the arrangement of the particles (ions) inside the crystal? 32. Azurite, a blue, crystalline mineral, is composed of copper, carbon, and oxygen.

© Charles D. Winters/Cengage

General Questions

© Charles D. Winters/Cengage

26. A book is held at a height above the floor. It has a certain amount of potential energy. When the book is released, its potential energy is converted into kinetic energy as it falls to the floor. The book hits the floor and comes to rest. According to the law of conservation of energy, the amount of energy in the universe is constant. Where has the energy gone?

Azurite is a deep blue crystalline mineral.  It is surrounded by copper pellets and powdered carbon (in the dish).

Chapter 1 / Basic Concepts of Chemistry

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(a) What are the symbols of the three elements that combine to make the mineral azurite? (b) Based on the photo, describe some of the physical properties of the elements and the mineral. Are any the same? Are any properties different? 33. You have a solution of NaCl dissolved in water. Describe a method by which these two compounds can be separated.

39. Carbon tetrachloride, CCl4, a liquid compound, has a density of 1.58 g/cm3. If you place a piece of a plastic soda bottle (d 5 1.37 g/cm3) and a piece of aluminum (d 5 2.70 g/cm3) in liquid CCl4, will the plastic and aluminum float or sink? 40. The following photo shows copper balls, immersed in water, floating on top of mercury. What are the liquids and solids in this photo? Which substance is most dense? Which is least dense?

Chips of iron mixed with sand

35. Identify the following as either physical changes or chemical changes. (a) Dry ice (solid CO2) sublimes (converts directly from solid to gaseous CO2). (b) The density of mercury metal decreases as the temperature increases. (c) Energy is given off as heat when natural gas (mostly methane, CH4) burns. (d) NaCl dissolves in water. 36. Identify the following as either physical changes or chemical changes. (a) The desalination of sea water (separation of pure water from dissolved salts). (b) The formation of SO2 (an air pollutant) when coal containing sulfur is burned. (c) Silver tarnishes. (d) Iron is heated to red heat. 37. In Figure 1.2 you see a piece of salt and a representation of its internal structure. Which is the macroscopic view and which is the particulate view? How are the macroscopic and particulate views related? 38. In Figure 1.5 you see macroscopic and particulate views of the element bromine. Which are the macroscopic views and which are the particulate views? Describe how the particulate views explain properties of this element related to the state of matter.



© Charles D. Winters/Cengage

© Charles D. Winters/Cengage

34. Small chips of iron are mixed with sand (see photo). Is this a homogeneous or heterogeneous mixture? Suggest a way to separate the iron from the sand.

Water, copper, and mercury

41. Categorize each of the following as an element, a compound, or a mixture. (a) iron pyrite (also known as fool’s gold) (b) carbonated mineral water (c) molybdenum (d) sucrose (also known as table sugar) 42. Categorize each of the following as an element, a compound, or a mixture. (a) sea water (c) bronze (b) sodium chloride (d) 24-carat gold 43. ▲ Make a drawing of the arrangement of particles in each of the following cases based on the kineticmolecular theory and the ideas about atoms and molecules presented in this chapter. For each case, draw 10 particles of each substance. It is acceptable for your diagram to be two dimensional. Represent each atom as a circle, and distinguish each different kind of atom by shading. (a) A sample of solid iron (which consists of iron atoms). (b) A sample of liquid water (which consists of H2O molecules). (c) A sample of water vapor. 44. ▲ Make a drawing of the arrangement of particles in each of the following cases based on the kineticmolecular theory and the ideas about atoms and molecules presented in this chapter. For each case, draw 10 particles of each substance. It is acceptable for your diagram to be two dimensional. Represent each atom as a circle, and distinguish each different kind of atom by shading.

Study Questions

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45. Hexane (C6H14, density 5 0.766 g/cm3), perfluorohexane (C6F14, density 5 1.669 g/cm3), and water are immiscible liquids; that is, they do not dissolve in one another. You place 10 mL of each in a graduated cylinder, along with pieces of high-density polyethylene (HDPE, density 5 0.97 g/cm3), polyvinyl chloride (PVC, density 5 1.36 g/cm3), and Teflon™ (density 5 2.3 g/cm3). None of these common plastics dissolves in these liquids. Describe what you expect to see. 46. ▲ You have a sample of a white crystalline substance from your kitchen. You know that it is either salt or sugar. Although you could decide by taste, suggest another property that you could use to decide. (Hint: You may use the Internet or a handbook of chemistry in the library to find some information.) 47. You can figure out whether a solid floats or sinks if you know its density and the density of the liquid. In which of the liquids listed below will high-density polyethylene (HDPE) float? (HDPE, a common plastic, has a density of 0.97 g/cm3. It does not dissolve in any of these liquids.) Density (g/cm3)

50. Describe an experimental method that can be used to determine the density of an irregularly shaped piece of metal. 51. Diabetes can alter the density of urine, so urine density can be used as a diagnostic tool. Diabetics can excrete too much sugar or excrete too much water. What do you predict will happen to the density of urine under each of these conditions? (Hint: Water containing dissolved sugar is more dense than pure water.) 52. Suggest a way to determine if the colorless liquid in a beaker is water. How could you discover if there is salt dissolved in the water? 53. The following photo shows the element potassium reacting with water to form the element hydrogen, a gas, and a solution of the compound potassium hydroxide.

Properties, Uses

Ethylene glycol

1.1088

Toxic; major component of automobile antifreeze

Water

0.9997

Ethanol

0.7893

Alcohol in alcoholic beverages

Methanol

0.7914

Toxic; gasoline additive to prevent gas line freezing

Acetic acid

1.0492

Component of vinegar

Glycerol

1.2613

Solvent used in home care products

48. You are given a sample of a silvery metal. What information could you use to prove the metal is silver? 49. Milk in a glass bottle was placed in the ­freezing compartment of a refrigerator overnight.

28

Frozen milk in a glass bottle

© Charles D. Winters/Cengage

Substance

By morning, a column of frozen milk emerged from the bottle. Explain this observation.

© Charles D. Winters/Cengage

(a) A homogeneous mixture of water vapor and helium gas (which consists of helium atoms). (b) A heterogeneous mixture consisting of liquid water and solid aluminum; show a region of the sample that includes both substance. (c) A sample of brass (which is a homogeneous solid mixture of copper and zinc).

Potassium reacting with water to produce hydrogen gas and potassium hydroxide.

(a) What states of matter are involved in the reaction? (b) Is the observed change chemical or physical? (c) What are the reactants in this reaction, and what are the products? (d) What qualitative observations can be made concerning this reaction?

Chapter 1 / Basic Concepts of Chemistry

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54. Three liquids of different densities are mixed. Because they are not miscible (do not form a homogeneous solution with one another), they form discrete layers, one on top of the other. Sketch the result of mixing carbon tetrachloride (CCl4, d 5 1.58 g/cm3), mercury (d 5 13.546 g/cm3), and water (d 5 1.00 g/cm3). 55. Four balloons are each filled with a different gas, each having a different density: chlorine, d 5 2.897 g/L

propane, d 5 1.802 g/L

methane, d 5 0.656 g/L

hydrogen, d 5 0.082 g/L

If the density of dry air is 1.12 g/L, which balloon(s) will float in air? 56. A copper-colored metal is found to conduct an electric current. Can you say with certainty that it is copper? Why or why not? Suggest additional information that could provide additional evidence that it is copper.

© Charles D. Winters/Cengage

57. The photo below shows elemental iodine dissolving in ethanol to give a solution. Is this a physical or chemical change?

Elemental iodine dissolving in ethanol.

58. ▲ You want to determine the density of a compound but have only a tiny crystal, and it would be difficult to measure mass and volume accurately. There is another way to determine density, however, called the flotation method. If you place the crystal in a liquid whose density is precisely that of the substance, it will be suspended in the liquid, neither sinking to the bottom of the beaker nor floating to the surface. However, for such an experiment you need a liquid with the precise density of the crystal. You can accomplish this by



mixing two liquids of different densities to obtain a liquid of the desired density. (a) Consider the following: you mix 10.0 mL of CHCl3 (d 5 1.492 g/mL) and 5.0 mL of CHBr3 (d 5 2.890 g/mL), giving 15.0 mL of solution. What is the density of this mixture? (b) Suppose now that you want to determine the density of a small yellow crystal to confirm that it is sulfur. From the literature, you know that sulfur has a density of 2.07 g/cm3. How would you prepare 20.0 mL of the liquid mixture with that density from pure samples of CHCl3 and CHBr3? (Note: 1 mL 5 1 cm3.) 59. A young chemist in Vienna, Austria, wanted to see just how permanent the gold was in his wedding band. The ring was 18-carat gold. (18-carat gold is 75% gold with the remainder copper and silver.) One week after his wedding day he took off the ring, cleaned it carefully, and weighed it. It had a mass of 5.58387 g. He weighed it weekly from then on, and after 1 year it had lost 6.15 mg just from normal wear and tear. He found that the activities that took the greatest toll on the gold were vacationing on a sandy beach and gardening. (a) What are the symbols of the elements that make up 18-carat gold? (b) The density of gold is 19.3 g/cm3. Use one of the periodic tables on the Internet (such as www.ptable.com) to find out if gold is the most dense of all of the known elements. If it is not gold, then what element is the most dense [considering only the elements from hydrogen (H) through uranium (U)]? (c) If a wedding band is 18-carat gold and has a mass of 5.58 g, what mass of gold is contained within the ring? (d) Assume there are 56 million married couples in the United States, and each person has an 18-carat gold ring. What mass of gold is lost by all the wedding rings in the United States in 1 year (in units of grams) if each ring loses 6.15 mg of mass per year? Assuming gold is $1815 per troy ounce (where 1 troy ounce 5 31.1 g), what is the lost gold worth?

Study Questions

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© Charles D. Winters/Cengage

1R

Let’s Review: The Tools of Quantitative Chemistry

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C hapt e r O ut li n e 1R.1 Units of Measurement 1R.2 Making Measurements: Precision, Accuracy, Experimental Error, and Standard Deviation

1R.3 Mathematics of Chemistry 1R.4 Problem Solving by Dimensional Analysis 1R.5 Graphs and Graphing 1R.6 Problem Solving and Chemical Arithmetic

Doing chemistry requires observing chemical reactions and physical changes. You will make qualitative observations—such as changes in color or the evolution of heat—and quantitative measurements of temperature, time, volume, mass, and length or size. To be successful in chemistry, you will need to be familiar with the units used by scientists to make measurements and be able to use quantitative data properly to reach conclusions. These issues are very important to scientists and engineers, and a failure to pay attention to them can have disastrous results. For example, on September 23, 1999, nine months after its launch, the Mars Climate Orbiter, which cost 125 million dollars, reached Mars and began its maneuvers to enter orbit around the planet. The Orbiter failed to reestablish contact with Earth after passing behind Mars. It was determined that the Orbiter failed to enter orbit and fell into the atmosphere of Mars, where it disintegrated. What went wrong? The problem was eventually traced back to a problem with units. Navigational signals were sent from Earth to the Orbiter using English units of pound-seconds, but the Orbiter was designed to use metric units of newton-seconds. As you take measurements in the lab and perform calculations, it is very important that you keep track of all the units used so that you can interpret the results properly.

1R.1 Units of Measurement Goal for Section 1R.1 • Use the common units for measurements in chemistry and make unit conversions (such as from liters to milliliters).

To record and report measurements, the scientific community has chosen a ­modified version of the metric system. This decimal system is called the Système ­International d’Unités (International System of Units), abbreviated SI.

◀ Scientific Instruments and Glassware.  Chemistry is a quantitative science. Many different

instruments and pieces of glassware have been invented to measure the properties of matter.

31

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Table 1R.1

The Seven SI Base Units

Measured Property

0.82 nm

Structure of the aspirin molecule

Name of Unit

Abbreviation

Mass

kilogram

kg

Length

meter

m

Time

second

s

Temperature

kelvin

K

Amount of substance

mole

mol

Electric current

ampere

A

Luminous intensity

candela

cd

All SI units are derived from base units, listed in Table 1R.1. Larger and smaller quantities are expressed by using appropriate prefixes with the base unit (Table 1R.2). The nanometer (nm), for example, is 1 billionth of a meter, that is, 1 3 1029 m (meter). Dimensions on the nanometer scale are common in chemistry and biology. For example, a typical molecule (such as aspirin) is about 1 nm in length and a bacterium is about 1000 nm in length. Indeed, the prefix nano- is also used in the name for a whole area of science, nanotechnology, which involves the synthesis and study of materials having this tiny scale.

Temperature Scales Two temperature scales are commonly used in scientific work: Celsius and Kelvin (Figure 1R.1). The Celsius scale is generally used worldwide for measurements in the laboratory. When calculations incorporate temperature data, however, the ­Kelvin scale is almost always used.

Table 1R.2 Prefix

Common Conversion Factors 

1000 g 5 1 kg 1 3 109 nm 5 1 m 10 mm 5 1 cm 100 cm 5 10 dm 5 1 m 1000 m 5 1 km Conversion factors for SI units are given in Appendix C and inside the back cover of this book.

32

Selected Prefixes Used in the Metric System

Abbreviation

Meaning

Example

Giga-

G

109 (billion)

1 gigahertz 5 1 3 109 Hz

Mega-

M

106 (million)

1 megaton 5 1 3 106 tons

Kilo-

k

103 (thousand)

1 kilogram (kg) 5 1 3 103 g

Deci-

d

1021 (tenth)

1 decimeter (dm) 5 1 3 1021 m

Centi-

c

1022 (one hundredth)

1 centimeter (cm) 5 1 3 1022 m

Milli-

m

1023 (one thousandth)

1 millimeter (mm) 5 1 3 1023 m

Micro-

μ

1026 (one millionth)

1 micrometer (μm) 5 1 3 1026 m

Nano-

n

1029 (one billionth)

1 nanometer (nm) 5 1 3 1029 m

Pico-

p

10212

1 picometer (pm) 5 1 3 10212 m

Femto-

f

10215

1 femtometer (fm) 5 1 3 10215 m

Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry

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Boiling point 212 ° F of water

100 ° C

373.15 K

180 ° F

100 ° C

100 K

0 °C

273.15 K

Freezing point 32 ° F of water

iStock.com/Magnascan

Kelvin (or absolute)

Celsius

Figure 1R.1  Comparison of Fahrenheit, Celsius, and Kelvin scales.  Note that the degree sign (°) is not used with the Kelvin scale.

iStock.com/ValentynVolkov

Fahrenheit

The Celsius Temperature Scale The size of the Celsius degree is defined by assigning zero as the freezing point of pure water (0 °C) and 100 as its boiling point (100 °C). A comfortable room temperature is around 20 °C, and your normal body temperature is 37 °C. The warmest water you can stand to immerse a finger in is probably about 60 °C.

The Kelvin Temperature Scale William Thomson, known as Lord Kelvin (1824–1907), first suggested the temperature scale that now bears his name. The Kelvin scale assigns zero as the lowest temperature that can be achieved, a point called absolute zero. Many experiments have found that this limiting temperature is 2273.15 °C. Kelvin units and Celsius degrees are the same size. Thus, the freezing point of water is reached at 273.15 K; that is, 0 °C 5 273.15 K. The normal boiling point of pure water is 373.15 K. Temperatures in Celsius degrees are converted to kelvins, and vice versa, using the relation T (K) 5



1K (T °C 1 273.15 °C) 1 °C

(1R.1)

Lord Kelvin  William Thomson (1824–1907), known as Lord Kelvin, was a professor of natural philosophy at the University in Glasgow, Scotland, from 1846 to 1899. He was best known for his study of heat and work, from which came the concept of the absolute temperature scale.

Using this equation, you can show that a common room temperature of 23.5 °C is equivalent to 296.7 K. T (K) 5

1K (23.5  C 1 273.15  C ) 5 296.7 K 1 C

Three things to notice about the Kelvin scale are: •

The degree symbol (°) is not used with Kelvin temperatures.

•

The name of the unit is the kelvin (not capitalized).

•

Temperatures are designated with a capital K.



1R.1  Units of Measurement

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33

In 1790, during the French Revolution, the National Assembly of France asked the French Academy of Sciences to come up with a new system of weights and measures. This was the beginning of the metric system. The meter was initially intended to be one ten millionth the distance between the equator and the North Pole along a meridian that crossed Dunkirk, France and Barcelona, Spain. Based on measurements of this value (which were later shown to be in error), a platinum bar was constructed with the distance between its ends at a specified temperature defined as a meter. Since then, the definition of a meter has been revised at various times to allow its value to be determined more accurately and precisely. In 1875, the Treaty of the Meter was signed by 17 countries, and the Bureau Intérnational des Poids et Mésures (BIPM, International Bureau of Weights and Measures) was established. In 1889, a new International Prototype Meter was constructed of a platinum-iridium alloy and adopted by the BIPM. In 1960, a revised system of units, called the Système International d’Unités (SI, International System of Units) was adopted by the BIPM and with it a new definition of the meter. This definition was based on something that could be universally examined in the laboratory. It related the meter to a fraction of the wavelength of light emitted by a type of atom of the element krypton when it undergoes a particular type of electronic transition. Yet another change in the definition occurred in 1983. At this point, the BIPM based the length of a meter on one of the fundamental constants of nature: the speed of light. They defined the speed of light in a

vacuum to be exactly 299,792,458 meters per second. This means that 1 meter is now defined as the distance that light travels in one 299,792,458th of a second. You might note that this definition also ­depends on the definition of the second, another base unit in the SI. On May 20, 2019, revisions to the definitions of four SI base units went into effect. All seven of the base units are now tied to values for seven physical constants of the universe. You will learn about a number of these constants as you proceed through this course. Two of the base units whose definitions were changed are the ­kilogram (the standard unit of mass) and the kelvin (the standard unit of temperature). The kilogram was previously defined as the mass of a cylinder of a particular platinumiridium alloy nicknamed Le Grand K, that was housed in Paris, France. Unfortunately, this block was shown to be mysteriously ­losing mass, clearly not desirable for a standard for mass. Following the revision, the value of a kilogram is now tied to the value of Planck’s constant, which you will learn about in Chapter 6. Planck’s constant is assigned a value of 6.62607015 3 10234 kg m2/s2 (or J· s). The kilogram is thus the mass that makes this value true, given the definitions of the second and the meter. To define the kelvin, the value of Boltzmann’s constant, which you will learn about in Chapter 18, is assigned a value of 1.380649 3 10223 kg m2/s2 · K (or J/K). The kelvin can thus be determined using this value and the previously defined values for the second, meter, and kilogram. What consequences do these changes have for you in your study of general ­chemistry? Not many. The values of the universal constants were selected so that

the new definitions and the previous definitions give as close to the same answer as possible. Thus, using a laboratory balance, 10.00 g using the old definition will still be 10.00 g using the new definition. The new definitions, however, relate the base units to fundamental constants of nature, freeing them from physical standards such as a block of a particular alloy for the kilogram, and will increase the ­accuracy and precision of measurements in the future.

SOTK2011/Alamy Stock Photo

A Closer Look

The SI Base Units

Figure  Le Grand K.  The International

Prototype of the Kilogram was a cylinder of a platinum-iridium alloy that was the standard for a kilogram from 1889 to 2019. Since 2019, the kilogram and all the other SI base units are defined based on their relationship to fundamental constants of nature, whose values are now defined.

Length, Volume, and Mass The meter is the standard unit of length, but objects observed in chemistry are frequently smaller than 1  meter. Measurements are often reported in units of centi­ meters (cm), millimeters (mm), or micrometers (μm), and objects on the atomic and molecular scale have dimensions of nanometers (nm; 1 nm 5 1 3 1029 m) or picometers (pm; 1 pm 5 1 3 10212 m) (Figure 1R.2).

34

Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry

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Ångstrom Units  An older but often-

used non-SI unit for molecular distances is the Ångstrom unit (Å), where 1 Å 5 1.0 3 10210 m. The distance between two carbon atoms in diamond (Figure 1R.2) is 1.54 Å.

© Charles D. Winters/Cengage

0.154 nm

3.0 mm

Figure 1R.2  Dimensions in the molecular world.  Dimensions on the molecular scale are often given in terms of nanometers (1 nm 5 1 3 1029 m) or picometers (1 pm 5 1 3 10212 m). Here, the distance between C atoms in diamond is 0.154 nm.

Exam p le 1R.1

Distances on the Molecular Scale Problem The distance between an O atom and an H atom in a water molecule is 95.8 pm. What is this distance in nanometers (nm)?

What Do You Know?  You are given the interatomic O–H distance. You will need to know (or look up) the relationships of the metric units.

Strategy  You can solve this problem by knowing the conversion

95.8 pm

factor between the units in the information you are given (picometers) and the desired units (meters or nanometers). (For more about conversion factors and their use in problem solving, see page 47.) There is no conversion factor given in Table 1R.2 to change nanometers to picometers directly, but relationships are listed between meters and picometers and between meters and nanometers. Therefore, you will first convert picometers to meters, and then convert ­meters to nanometers.

picometers

x m pm

y nm m

→ meters

→ nanometers

Solution  First, use the conversion factor 1 pm 5 1 3 10212 m to convert 95.8 ­picometers to meters.

95.8 pm 3

1 3 10212 m 5 9.58 3 10211 m 1 pm

Then use the conversion factor 1 nm 5 1 3 1029 m to convert from meters to nanometers.

9.58 3 10211 m 3

1 nm 5 9.58 3 1022 nm or 0.0958 nm 1 3 1029 m

Think about Your Answer  A nanometer is a larger unit than a picometer, so the same distance expressed in nanometers will have a smaller numerical value. The answer agrees with this. Notice how the units cancel in the calculation to leave an answer whose unit is that of the numerator of the conversion factor. The process of using units to guide a calculation is called dimensional analysis. It is explored further on pages 47–48.

Check Your Understanding  The distance between two carbon atoms in diamond (Figure 1R.2) is 0.154 nm. What is this distance in picometers (pm)? In centimeters (cm)?



1R.1  Units of Measurement

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35

Figure 1R.3  Dimensions in chemistry and biology.  These photos are from the research

1 cm

of Professor Joanna Aizenberg of Harvard University.

(c) Scanning electron microscope (SEM) image of a single strand showing its ceramic-composite structure. Scale bar = 20 μm.

5 mm

20 μm

© Charles D. Winters/Cengage

(b) Fragment of the structure showing the square grid of the lattice with diagonal supports. Scale bar = 5 mm.

Figure 1R.4  Some common laboratory glassware.  Volumes

are marked in units of milliliters (mL). Remember that 1 mL is equivalent to 1 cm3.

36

Photos courtesy of Joanna Aizenberg, Bell Laboratories. Reference: J. Aizenberg, et al., Science, Vol. 309, pages 275-278, 2005

(a) Photograph of the glassy sea sponge Euplectella. Scale bar = 1 cm.

The glassy skeleton of a sea sponge (Figure 1R.3) can give you an idea of the range of dimensions used in science. The sea sponge is about 20 cm long and a few centimeters in diameter. A closer look shows more detail of the lattice-like structure. Scientists at Bell Laboratories found that each strand of the lattice is a ceramic-fiber composite of silica (SiO2) and protein less than 100 μm in diameter. Each strand is composed of spicules, which consist of silica nanoparticles 50 to 200 nanometers in diameter. Chemists often use glassware such as beakers, flasks, pipets, graduated cylinders, and burets, which are marked in volume units (Figure 1R.4). Because the SI unit of volume [the cubic meter (m3)] is too large for everyday laboratory use, chemists usually use the liter (L) or the milliliter (mL) for volume measurements. One liter is equivalent to the volume of a cube with sides equal to 10 cm [V 5 (0.1 m)3 5 0.001 m3]. 1 liter (L) 5 1000 cm3 5 1000 mL 5 0.001 m3

Because there are exactly 1000 mL (5 1000 cm3) in a liter, this means that 1 mL 5 0.001 L 5 1 cm3

The units milliliter and cubic centimeter (sometimes abbreviated cc by medical ­professionals) are interchangeable. Therefore, a flask that contains exactly 125 mL has a volume of 125 cm3. Although not widely used in the United States, the cubic decimeter (dm3) is a common unit of volume in the rest of the world. A length of 10 cm is called a ­decimeter (dm), a tenth of a meter. Because a cube with a 10-cm side defines a volume of 1 liter, a liter is equivalent to a cubic decimeter: 1 L 5 1 dm3. Products in Europe, Africa, and other parts of the world are often sold by the cubic decimeter. The deciliter, dL, which is exactly equivalent to 1/10 of a liter (0.100 L) or 100 mL, is widely used in medicine. Standards for concentrations of environmental contaminants are often set as a certain mass per deciliter. For example, the U.S. ­Centers for Disease Control and Prevention recommends that children with more than 3.5 micrograms (3.5 3 1026 g) of lead per deciliter of blood undergo further testing for lead poisoning. Finally, when chemists prepare chemicals for reactions, they often measure the mass of each material. Mass is the fundamental measure of the quantity of matter,

Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry

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A Closer Look

Energy and Food

The U.S. Food and Drug Administration (FDA) mandates that nutritional data, including energy content in Calories (where 1 Cal 5 1 kilocalorie), be included on almost all packaged food. The Nutrition Labeling and Education Act of 1990 requires that the total ­energy from protein, carbohydrates, fat, and alcohol be specified. How is this determined? Initially, the method used was calorimetry. In this method (described in Chapter 5), a food product is burned, and  the energy transferred as heat  in the combustion is

measured. Now, however, energy contents are estimated using the Atwater system. This specifies the following average values for energy sources in foods: 1 1 1 1

g g g g

protein 5 4 kcal (17 kJ) carbohydrate 5 4 kcal (17 kJ) fat 5 9 kcal (38 kJ) alcohol 5 7 kcal (29 kJ)

Because carbohydrates may include some indigestible fiber, the mass of fiber is subtracted from the mass of carbohydrate when calculating the energy from carbohydrates.

As an example, one serving of cashew nuts (about 28 g) has 14 g fat 5 126 kcal 6 g protein 5 24 kcal 7 g carbohydrates 2 1 g fiber 5 24 kcal Total 5 174 kcal (728 kJ) A value of 170 Calories (5 170 kcal) is reported on the package. The calculated and reported values agree to two s­ ignificant figures (see Section 1R.3).

and the SI unit of mass is the kilogram (kg). Smaller masses are expressed in grams (g) or milligrams (mg).

Energy Units When expressing energy quantities, most chemists (and much of the world outside the United States) use the joule (J), the SI unit. The joule is related directly to the units used for mechanical energy: 1 J equals 1 kg · m2/s2. Because the joule is inconveniently small for most uses in chemistry, the kilojoule (kJ), equivalent to 1000 joules, is often the unit of choice. To give you some feeling for joules, suppose you drop a six-pack of soft-drink cans, each full of liquid, on your foot. Although you probably will not take time to calculate it, the kinetic energy at the moment of impact is 10 or more joules. The calorie (cal) is an older energy unit. It is defined as the energy transferred as heat that is required to raise the temperature of 1.00 g of pure liquid water from 14.5 °C to 15.5 °C. A kilocalorie (kcal) is equivalent to 1000 calories. The conversion factor relating joules and calories is 1 calorie (cal) 5 4.184 joules (J)

The dietary Calorie (with a capital C) is often used in the United States to represent the energy content of foods. The dietary Calorie (Cal) is equivalent to a kilocalorie or 1000 calories. Thus, a breakfast cereal that gives you 100.0 Calories of nutritional energy per serving provides 100.0 kcal or 418.4 kJ.

Oesper Collection in the History of Chemistry/ University of Cincinnati

1 kg 5 1000 g and 1 g 5 1000 mg

James Joule  The joule is named for James P. Joule (1818–1889), the son of a wealthy brewer in Manchester, England. The family wealth and a workshop in the brewery gave Joule the opportunity to pursue scientific studies. Among the topics that Joule studied was whether heat was a massless fluid. Scientists at that time referred to this idea as the caloric hypothesis. Joule’s careful experiments showed that heat and mechanical work are related, providing evidence that heat is not a fluid.

1R.2 Making Measurements: Precision, Accuracy, Experimental Error, and Standard Deviation Goal for Section 1R.2 • Recognize and express uncertainties in measurements. The precision of a measurement indicates how well several determinations of the same quantity agree. This is illustrated by the results of throwing darts at a target. In Figures 1R.5a and 1R.5b, the dart thrower was not consistent, and the precision of the darts’ placement on the target is low. In Figures 1R.5c and 1R.5d, the darts are clustered together, indicating much better consistency on the part of the thrower—that is, greater precision.

1R.2  Making Measurements: Precision, Accuracy, Experimental Error, and Standard Deviation

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37

Chemistry in Your Career

Thalia I. Navedo

Thalia I. Navedo

Accuracy and NIST  The National

Institute of Standards and Technology (NIST) is an important resource for standards and data used in science. Comparison with NIST data is a test of the accuracy of the measurement (see www.nist.gov).

At first glance, a knowledge of chemistry may not seem essential to Thalia I. Navedo’s job as a project accountant in construction design. ­Navedo (she/ her/hers) spends her workday on finance and ­accounting tasks such as project set-up, budgeting, subcontracts, billing, safety and compliance, and much more. Yet Navedo, who identifies as ­Hispanic/ Latinx/Black, says that her study of chemistry during college provided skills that she uses daily in her work. “Working in finance involves a good amount of math and problem solving, which were prevalent

all throughout my study of chemistry. Be it dimensional analysis or problem solving, learning chemistry helped me a lot on my analytical skills. . . . Chemistry taught me how to apply the mathematical knowledge into solving real problems and training my brain muscles.” In addition to analytical skills, Navedo’s study of chemistry informs the safety and compliance aspects of her job as she understands and manages the impact of chemicals used in her workplace.

Accuracy is the agreement of a measurement with the accepted value of the quantity. In Figures 1R.5a and 1R.5c, the thrower was not accurate because the average location of the darts is not the bull’s eye. In Figure 1R.5b, the average location of the darts is the bull’s eye, so the thrower was accurate, even though the results were not precise. Figure 1R.5d shows the optimal case where the thrower was accurate as well as precise—the average of all shots is close to the targeted position, the bull’s eye, and the darts are all clustered around this value. Notice that, as shown by Figure 1R.5c, it is possible to be precise without being accurate—the thrower has consistently missed the bull’s eye, although all the darts are clustered precisely around one point on the target. This is analogous to an experiment with some flaw (either in design or in a measuring device) that causes all results to differ from the correct value by the same amount. The accuracy of a result in the laboratory is often expressed in terms of percent error relative to a standard or accepted value, whereas the precision is expressed as a standard deviation.

Experimental Error If you measure a quantity in the laboratory, you may be required to report the error in the result, the difference between your result and the accepted value,

Error in measurement 5 experimentally determined value − accepted value (1R.2)

or the percent error.

(a) Poor precision and poor accuracy

(b) Poor precision and good accuracy

(c) Good precision and poor accuracy

(d) Good precision and good accuracy

Figure 1R.5 ​Precision and accuracy.

38

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error in measurement 3 100% accepted value experimentally determined value 2 accepted value 5 accepted value

Percent error 5

3 100% (1R.3)

Exam p le 1R.2

Accuracy and Error Problem  Suppose a coin has a diameter of 28.054 mm. In an experiment, two students measure this diameter. Student A makes four measurements of the diameter of the coin using a precision tool called a micrometer. Student B measures the same coin using a simple plastic ruler. The two students report the following results: Student A

Student B

28.246 mm

27.9 mm

28.244

28.0

28.246

27.8

28.248

28.1

What is the average diameter and percent error obtained in each case? Which student’s data are more accurate?

What Do You Know?  You know the data collected by the two students and want to compare them with the actual value by calculating the percent error.

Strategy  Step 1. For each set of values, calculate the average of the four measurements. Step 2. Calculate the percent error.

Solution  Step 1. Obtain the average for each set of data by summing the four values and dividing by four, the number of measurements

Average value for Student A 5

28.246 mm 1 28.244 mm 1 28.246 mm 1 28.248 mm 4

5 28.246 mm

Average value for Student B 5

27.9 mm 1 28.0 mm 1 27.8 mm 1 28.1 mm 4

5 27.95 mm 5 28.0 mm Step 2. Use Equation 1R.3 to calculate the percent error for each student.

Percent error for Student A 5

28.246 mm 2 28.054 mm 3 100% 5 0.684% 28.054 mm

Percent error for Student B 5

27.95 mm 2 28.054 mm 3 100% 5 20.4% 28.054 mm

Student B’s average is more accurate because it is closer to the accepted value, and thus it has a smaller percent error.



Percent Error  Percent error can be positive or negative, indicating whether the experimental value is too high or too low compared to the accepted value. In Example 1R.2, Student B’s percent error is 20.4%, indicating it is 0.4% lower than the accepted value.

1R.2  Making Measurements: Precision, Accuracy, Experimental Error, and Standard Deviation

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39

Think about Your Answer  Student A’s results were less accurate, as shown in this example. But Student A’s results were more closely grouped around their average and so were more precise. Possible reasons for the error in Student A’s result are incorrect use of the micrometer or a flaw in the instrument. The rules for determining how many digits should be retained in the calculations and answers will be provided later in the chapter.

Check Your Understanding  A student checked the accuracy of two standard top-loading balances by testing them with a standard 5.000-g mass. The results were as follows: Balance 1:  4.99 g, 5.04 g, 5.03 g, 5.01 g Balance 2:  4.97 g, 4.99 g, 4.95 g, 4.96 g Calculate the average values for balances 1 and 2 and calculate the percent error for each. Which balance is more accurate?

Standard Deviation Laboratory measurements can be in error for two basic reasons. First, determinate errors are errors that can be identified and theoretically avoided or corrected like those caused by faulty instruments, human errors such as incorrect record keeping, and errors inherent in the methods used in the experiment such as other reaction products being formed in a chemical reaction in addition to the desired product. Second, indeterminate (or random) errors arise from uncertainties in a m ­ easurement. One way to judge the indeterminate error in a result is to calculate the standard deviation. The standard deviation of a series of measurements is equal to the square root of the sum of the squares of the deviations for each measurement from the average, divided by one less than the number of measurements. ∑ ( xi 2 x ) N 21 2

Standard deviation 5



(1R.4)

In this equation, the Greek capital letter sigma, , means to add up, xi is each individual measurement, x is the average value, and N is the number of measurements. The use of this equation is shown in Example 1R.3. The standard deviation has a precise statistical significance: Assuming a large number of measurements is used to calculate the average, slightly more than 68% of the values collected are expected to be within one standard deviation of the value determined, and 95% are within two standard deviations.

E xamp le 1R.3

Precision and Standard Deviation Problem  Suppose you carefully measure the mass of water delivered by a 10-mL pipet. (A pipet containing a green solution is shown on the right in Figure 1R.4.) Your results for five measurements are 9.990 g, 9.993 g, 9.975 g, 9.980 g, and 9.982 g. Calculate the ­standard deviation in the mass for this series of measurements.

What Do You Know?  You know the masses obtained for the five samples and the equation for determining the standard deviation. Strategy Step 1. Calculate the average of the measurements. Step 2. Determine the deviation of each individual measurement from the average by subtracting the average from each of the measurements.

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Step 3. Square the individual deviations, and then add these squared values together. Step 4. Finally, calculate the standard deviation by dividing the sum of the squares by one less than the number of measurements and taking the square root.

Solution Step 1. Calculate the average of the measurements. The average of the measurements is calculated by adding together the measured masses and dividing by five, the number of measurements.

Average ( x ) 5

9.990 g 1 9.993 g 1 9.975 g 1 9.980 g 1 9.982 g 5 9.984 g 5

Steps 2-3. Determine the deviation of each individual measurement (xi) from the aver_ age (x ), square the individual deviations, and then add together these squared values. The deviations (difference between each measurement and the average) and squares of the deviations for each measurement are summarized in the following table.

Determination

Measured Mass (g)

Deviation 5 xi2 x (g)

Square of Deviation (g2)

1

9.990

0.006

0.00004

2

9.993

0.009

0.00008

3

9.975

20.009

0.00008

4

9.980

20.004

0.00002

5

9.982

20.002

0.000004

Add together the squares of the deviations.

Sum of the squares of the deviations 5 (0.00004 g2 ) 1 (0.00008 g2 ) 1 (0.00008 g2) 1 (0.00002 g2 ) 1 (0.000004 g2) 5 0.00022 g2 Step 4. Use Equation 1R.4 to calculate the standard deviation.

Standard deviation 5

0.00022 g2 5 521

0.007 g

Think about Your Answer  The standard deviation tells you that if this e­ xperiment were repeated, most of the values would fall in the range from 9.977 g to 9.991 g (±0.007 g from the average value). Even though standard deviations are often reported, it is rare for people today to carry out the detailed calculations shown in this example because many calculators and spreadsheet programs (such as Microsoft Excel or Apple’s Numbers) have methods to perform the calculation using just a few keystrokes.

Check Your Understanding A student obtained the following masses for the mass of an object: 4.99 g, 5.04 g, 5.03 g, and 5.01 g. Determine the standard deviation for these data.

1R.3 Mathematics of Chemistry Goals for Section 1R.3 • Express and use numbers in exponential or scientific notation. • Report the answer of a calculation to the correct number of significant figures.

1R.3  Mathematics of Chemistry

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41

Exponential or Scientific Notation The Eiffel Tower, built in 1889, is the tallest building in Paris. It was designed by the French architect Gustave Eiffel to mark the centennial of the French Revolution. The Tower, constructed of very pure iron, is as tall as an 81-story building. It was supposed to be dismantled in 1909, but the building still stands as a symbol of Paris. Some quantitative information on the structure is given in the following table:

RomanSlavik.com/Shutterstock.com

Eiffel Tower Characteristics

Eiffel Tower (Paris, France)

Quantitative Information

Height

324 meters (m)

Mass of iron

7.3 3 106 kilograms (kg)

Volume of iron

930 cubic meters (m3)

Number of iron pieces

1.8 3 104 pieces

Approximate number of visitors annually

7 3 106 people

Some of the data on the Tower are expressed in exponential notation, or scientific notation, (for example, 7.3 3 106 kilograms). Scientific notation is a way of presenting very large or very small numbers in a compact and consistent form that simplifies calculations. Because of its convenience, scientific notation is widely used in the sciences. In scientific notation a number is expressed as the product of two numbers: N 3 10n. N is the digit term and is a number between 1 and 9.9999. . . . The second number, 10n, the exponential term, is some integer power of 10. For example, 1234 is written in scientific notation as 1.234 3 103, or 1.234 multiplied by 10 three times: 1234 5 1.234 3 101 3 101 3 101 5 1.234 3 103

Conversely, a number less than 1, such as 0.01234, is written as 1.234 3 1022. This notation tells you that 1.234 should be divided twice by 10 to obtain 0.01234: 0.01234 5

1.234 5 1.234 3 1021 3 1021 5 1.234 3 1022 101 3 101

When converting a positive number to scientific notation, notice that the exponent n is positive if the number is greater than 1 and negative if the number is less than 1. The value of n is the number of places by which the decimal is shifted to obtain the number in scientific notation; if the decimal is shifted to the left, n is positive, and if the decimal is shifted to the right, n is negative:

1 2 3 4 5. = 1.2345 × 104 (a) Decimal shifted four places to the left. Therefore, n is positive 4. Comparing Earth and a Plant Cell— Powers of 10 

Earth

  5 12,760,000 meters wide   5 1.276 3 107 meters

Plant cell   5 0.00001276 meters wide   5 1.276 3 1025 meters

0.0 0 1 2 = 1.2 × 10–3 (b) Decimal shifted three places to the right. Therefore, n is negative 3.

If you wish to convert a number in the opposite direction, from scientific notation to one that does not use scientific notation, the procedure is reversed:

6 . 2 7 3 × 102 = 627.3 (a) Decimal point moved two places to the right because n is positive 2.

0 0 6.273 × 10–3 = 0.006273 (b) Decimal point shifted three places to the left because n is negative 3.

42

Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry

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Significant Figures In most experiments, several kinds of measurements must be made, and some can be made more precisely than others. A result calculated from experimental data ­cannot be more precise than the least precise piece of information used in the calculation. This is where the rules for significant figures come in. Significant figures (also called significant digits) are the digits in a measured quantity that are known exactly plus one inexact digit.

Suppose you place a new U.S. dime on the pan of an analytical laboratory balance such as the one pictured in Figure 1R.6 and observe a mass of 2.2653 g. This number has five significant figures or digits because all five numbers are observed. However, you will learn from experience that the final digit on the right (3) is somewhat uncertain because you may notice the balance readings can change slightly and give masses of 2.2652, 2.2653, and 2.2654, with the mass of 2.2653 observed most of the time. In general, in a number representing a scientific measurement, the last digit to the right is taken to be inexact. The uncertainty in the last digit is best determined by taking multiple measurements and calculating the standard deviation of the measurements. However, when the standard deviation is not determined experimentally, it is common practice to assign an uncertainty of ±1 to the last significant figure. Suppose you want to calculate the density of a piece of metal (Figure 1R.7). The mass and dimensions were determined by standard laboratory techniques. Most of these data have two digits to the right of the decimal, but they have different numbers of significant figures. Data Collected

Mass of metal

13.56 g

4

Length

6.45 cm

3

Width

2.50 cm

3

Thickness

3.1 mm 5 0.31 cm

2

balances can determine the mass of an object to the nearest tenth of a milligram.

Significant Figures

The quantity 0.31 cm has two significant figures. The 3 in 0.31 is exactly known, but the 1 is uncertain. That is, the thickness of the metal piece may have been as small as 0.30 cm or as large as 0.32 cm. The width of the piece is reported as 2.50 cm, where 2.5 is known with certainty, but the final 0 is uncertain. There are three significant figures in 2.50. When you read a number in a problem or collect data in the laboratory (­ Figure 1R.8), how do you determine which digits are significant? First, is the number an exact number or a measured quantity? If it is an exact number, you don’t have to worry about the number of significant figures. For example, there are exactly 100 cm in 1 m. You could add as many zeros after the decimal place as you want, and the expression would still be true. Using this relationship in a calculation does not affect how many significant figures you report in your answer. If, however, the number is a measured value, you must take into account ­significant figures. In the data given in Figure 1R.7, the values 13.56 g and 6.45 cm contain only nonzero digits, which are always significant in a reported measurement. Thus, 13.56 g has four significant figures, and 6.45 mm has three. But how many significant figures are in the values 0.31 cm and 2.50 cm? Are the zeroes significant? 1. Zeroes between two other significant digits are significant. For example, the zero in 103 is significant. 2. Zeroes to the right of a nonzero number, and also to the right of a decimal place, are significant. For example, in the number 2.50 cm, the zero is significant.

Figure 1R.6 Analytical laboratory balance and significant figures.  Such

2.50 cm

13.56 g 6.45 cm © Charles D. Winters/Cengage

Measurement

© Charles D. Winters/Cengage

Determining Significant Figures

3.1 mm

Figure 1R.7  Data used to determine the density of a metal.

1R.3  Mathematics of Chemistry

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43

250-mL flask contains 250.0 ± 0.1 mL when full to the mark

50-mL buret marked in 0.10-mL increments

20-mL pipet volume known to the nearest 0.02 mL

Photos: © © Charles D. Winters/Cengage

10-mL graduated cylinder marked in 0.1-mL increments

The 10-mL graduated cylinder is marked in 0.1-mL increments; its contents would normally be estimated to 0.01 mL. However, graduated cylinders are not precision glassware. You can expect no more than 2 significant figures when reading a volume with this cylinder.

A 50-mL buret is marked in 0.10-mL increments, but it may be read with greater precision (0.01 mL).

A volumetric flask is meant to be filled to the mark on the neck. For a 250-mL flask, the volume is known to the nearest 0.1 mL, so the flask contains 250.0 ± 0.1 mL when full to the mark (four significant figures).

A pipet is like a volumetric flask in that it is filled to the mark on its neck. For a 20-mL pipet the volume is known to the nearest 0.02 mL.

Figure 1R.8  Glassware and significant figures.

Zeroes and Common Laboratory Mistakes  Students often find

the mass of a chemical on a balance and fail to write down zeroes that are significant. For example, if a balance displays a mass of 2.340 g, the final zero is significant and must be reported as part of the measured value. The number 2.34 g has only three significant figures and implies the 4 is uncertain, when in fact the balance reading indicated the 4 is certain.

3. Zeroes that are placeholders are not significant. There are two types of numbers that fall under this rule. (a) The first are decimal numbers with zeroes that occur to the left of the first nonzero digit. For example, in 0.0013, only the 1 and the 3 are significant; the zeroes are not. This number has two significant figures. Similarly, 0.31 cm has only two significant figures. (b) The second are numbers with trailing zeroes that must be there to indicate the magnitude of the number. For example, the zeroes in the number 13,000 may or may not be significant; it depends on whether they were measured or not. To avoid confusion with regard to such numbers, assume in this book that trailing zeroes are significant when there is a decimal point to the right of the last zero. Thus, you would say that 13,000 has only two significant figures but that 13,000. has five. The best way to be unambiguous when writing numbers with trailing zeroes is to use scientific notation. For example 1.300 3 104 indicates four significant figures, whereas 1.3 3 104 indicates two.

Using Significant Figures in Calculations When doing calculations using measured quantities, you need to follow some basic rules so that the results reflect the precision of all the measurements that go into the calculations. The rules used for significant figures in this book are as follows: Rule 1.  When adding or subtracting numbers, the number of decimal places in the answer is equal to the number of decimal places in the number with the fewest digits after the decimal. 0.12

2 decimal places

11.9

1 decimal place

110.925

3 decimal places

12.945

answer on calculator

The sum should be reported as 12.9, a number with one decimal place, because 1.9 has only one decimal place.

44

Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry

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Problem Solving Tip 1R.1 Using Your Calculator You will be performing a number of calculations in general chemistry, most of them using a calculator. Many different types of ­calculators are available, so be sure to consult your calculator manual for ­specific instructions to enter scientific ­notation and to find powers and

roots of numbers. You do not want to get questions incorrect because you ­performed the wrong operations on your calculator! To make sure you are using your calculator correctly, try these sample calculations and verify that you obtain the correct answers:

1. (6.02 3 1023)(2.26 3 1025)/367 (Answer 5 3.71 3 1016) 2. (4.32 3 1023)3 (Answer 5 8.06 3 1028) 3. (4.32 3 1023)1/3 (Answer 5 0.163)

Rule 2.  In multiplication or division, the number of significant figures in the answer is determined by the value with the fewest significant figures. 0.01208 5 0.511864. . . (answer on calculator) 0.0236

Because 0.0236 has only three significant digits, while 0.01208 has four, the answer should have three significant digits and be reported as 0.512, or in scientific notation as 5.12 3 1021. Rule 3.  When a number is rounded off, the last digit retained is increased by one (rounded up) only if the following digit is 5 or greater. Full Number

Number Rounded to Three Significant Digits

12.696

12.7

16.349

16.3

18.35

18.4

18.351

18.4

Now you can apply these rules to calculate the density of the piece of metal in Figure 1R.7. Length 3 width 3 thickness 5 volume mass (g) volume (cm3) 13.56 g 5 5 2.7 g/cm3 6.45 cm 3 2.50 cm 3 0.31 cm

Density 5

The calculated density has two significant figures because the thickness has only two significant figures. Remember that a result calculated using only multiplication and division is reported to the same number of significant figures as the value with the fewest significant figures. One last word on significant figures and calculations: When working problems, you should do the calculation with all the digits allowed by your calculator and round off only at the end of the calculation. Rounding off in the middle of a calculation can introduce errors. In Example problems in this book, the answer to each intermediate step is given to the correct number of significant figures plus one extra digit for that step, so that round-off errors do not propagate in the significant figures. The final answers to numerical problems result from retaining several digits more than the number required by the rules of significant figures and rounding to the correct number of significant figures only at the end.



Who Is Right—You or the Book? 

If your answer to a problem in this book does not quite agree with the answer in Appendix N, check if you rounded the answer after each step and then used that rounded answer in the next step. If so, performing the calculation without rounding between steps should resolve the discrepancy.

1R.3  Mathematics of Chemistry

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45

E xamp le 1R.4

Using Significant Figures Problem  An example of a calculation you will do later in the book (Chapter 10) is

( 0.120 ) ( 0.08206 ) ( 273.15 1 5 )

Volume of gas (L) 5

( 230/760.0 )

Calculate the final answer to the correct number of significant figures.

What Do You Know?  You know the rules for determining the number of significant figures for each number in the equation. Strategy  First decide on the number of significant figures represented by each n­ umber and then apply Rules 123 (pages 44–45).

Solution

Number

Number of Significant Figures

Comments

0.120

3

The final 0 on the right is significant.

0.08206

4

Neither the 0 to the left of the decimal place nor the first 0 to the immediate right of the decimal is significant.

273.15 1 5 5 278

3

5 has no decimal places, so the sum cannot either.

230/760.0 5 0.30

2

230 has two significant figures because the trailing zero is not significant. In contrast, there is a decimal point in 760.0, so there are four significant digits. The quotient will have only two significant digits.

The calculation gives 9.0506. . . L. However, an analysis of significant figures shows that one of the pieces of information involved in multiplication and division is known to only two significant figures. Therefore, you should report the volume of gas as  9.1 L , a number with two significant figures.

Think about Your Answer  Be especially careful when you add or subtract two numbers because it is easy to make significant figure errors when doing so. Notice that in the addition portion of this calculation (273.15 1 5 5 278) the sum has three significant figures.

Check Your Understanding  What is the result of the following calculation?

x5

46

(110.7 2 64) (0.056)(0.00216)

Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry

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1R.4 Problem Solving by Dimensional Analysis Goal for Section 1R.4 • Solve problems using dimensional analysis. Figure1R.7 illustrated the data that were collected to determine the density of a piece of metal. The thickness was measured in millimeters, whereas the length and width were measured in centimeters. To find the volume of the sample in cubic centimeters, the length, width, and thickness must all be in the same unit, centimeters, so the thickness was converted to centimeters. 3.1 mm 3

1 cm 5 0.31 cm 10 mm

Here, the thickness in millimeters (3.1 mm) was multiplied by a conversion factor (1 cm/10 mm) to produce the result in the desired unit (0.31 cm). Notice that units are treated like numbers. Because the unit mm is in both the numerator and the denominator, these units are said to cancel. This leaves the answer in centimeters, the desired unit. This approach to problem solving is often called dimensional analysis (or sometimes the factor-label method). It is a general problem-solving approach that uses the dimensions or units of each value as a guide for calculations. A conversion factor expresses the equivalence of a measurement in two different units (1 cm 5 10 mm; 1 g 5 1000 mg; 12 eggs 5 1 dozen; 12 inches 5 1 foot). Because the numerator and the denominator describe the same quantity, the conversion factor is equivalent to the number 1. Therefore, multiplication by this factor does not change the measured quantity, only its units. A conversion factor is always written so that the original units cancel, leaving the answer expressed in the new units. Number in original unit Quantity to express in new units

new unit = new number in new unit original unit

Conversion factor

Quantity now expressed in new units

Using Conversion Factors and Doing Calculations  As you

work problems in this book and read Example problems, notice that proceeding from given information to an answer very often involves a series of multiplications. That is, you multiply the given data by a conversion factor, multiply the answer of that step by another factor, and so on, to get the answer.

Exam p le 1R.5

Using Conversion Factors and Dimensional Analysis Problem  Oceanographers often express the density of sea water in units of kilograms per cubic meter. If the density of sea water is 1.025 g/cm3 at 15 ° C, what is its density in kilograms per cubic meter?

What Do You Know?  You know the density in a unit involving mass in grams and volume in cubic centimeters. These have to be changed to their equivalents in kilograms and cubic meters, respectively. Strategy  To simplify this problem, break it into three steps. Step 1. Convert the mass in grams to kilograms. Step 2. Convert the volume in cubic centimeters to cubic meters. Step 3. Calculate the density by dividing the mass in kilograms by the volume in cubic meters.

1R.4  Problem Solving by Dimensional Analysis

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47

Solution  Step 1. Convert the mass in grams to a mass in kilograms.

1.025 g 3

1 kg 5 1.025 3 1023 kg 1000 g

The given information is known to four significant figures. The conversion factor is an exact number, so using it will not affect the number of significant figures. Step 2. Convert the volume in cm3 to m3. A conversion factor to directly change units of cubic centimeters to cubic meters is not available in the tables in this book. You can find one, however, by cubing (raising to the third power) the relation between the meter and the centimeter: 3

   1m  1 m3 1 cm3 3  5 1 cm3 3  5 1 3 1026 m3  6 3  100 cm   1 3 10 cm  This conversion involves only numbers that are known exactly, so you don’t need to worry about significant figures for this step. You now know that 1 cm3 is equivalent to 1 3 1026 m3. Step 3. Calculate the density of sea water.

Density 5

1.025 3 1023 kg 5  1025 kg/m3  1 3 1026 m3

Think about Your Answer  The number of significant figures reported for the final answer is determined by the given information, 1.025 g, which has four significant figures. The final answer therefore has four significant figures. Densities in units of kg/m3 can often be large numbers. For example, the density of platinum is 21,450 kg/m3, and dry air has a density of 1.204 kg/m3.

Check Your Understanding  The density of gold is 19,320 kg/m3. What is this density in g/cm3?

1R.5 Graphs and Graphing Goals for Section 1R.5 • Read information from graphs. • Prepare and interpret graphs of numerical information, and, if a graph produces a straight line, find the slope and equation of the line.

In a number of instances in this text, graphs are used when analyzing experimental data to obtain a mathematical equation that may help predict new results. The procedure used will often result in a straight line, which has the equation y 5 mx 1 b

In this equation, y is usually referred to as the dependent variable; its value is determined from (that is, is dependent on) the values of x, m, and b. In this equation, x is called the independent variable, and m is the slope of the line. The parameter b is the y-intercept—that is, the value of y when x 5 0. The following example will show two things: (1) how to construct a graph from a set of data points and (2) how to derive an equation for the line generated by the data. Figure 1R.9 includes the set of data points to be graphed. First, mark off each axis in increments of the values of x and y. Here, the x-data are within the range from

48

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3 Experimental data

2.5

y-data

2

x 3.35 2.59 1.08 −1.19

y-intercept, where x = 0 and y = 1.87

y 0.0565 0.520 1.38 2.45

Figure 1R.9  Plotting data.  Using Microsoft Excel with these data and doing a linear regression analysis gives the equation for the line: y 5 20.525x 1 1.87.

1.5

1 x = 2.00, y = 0.82 0.5

Using the points marked with a square, the slope of the line is: Slope =

∆y 0.82 − 1.87 = = −0.525 ∆x 2.00 − 0.00

0 −2

−1

0

1 x-data

2

3

4

22 to 4, so the x-axis is marked off in increments of 1 unit. The y-data fall within the range from 0 to 2.5, so the y-axis is marked off in increments of 0.5. Each data point is marked as a circle on the graph. After plotting the points on the graph (round circles), draw a straight line that comes as close as possible to representing the trend in the data. (Do not just connect the dots!) Because there is always some inaccuracy in experimental data, the straight line that should be drawn is unlikely to pass exactly through every point. To identify the specific equation corresponding to the data, the y-intercept (b) and slope (m) for the equation y 5 mx 1 b must be determined. The y-intercept is the point at which x 5 0 and thus is the point where the line intersects the y-axis. The slope is determined by selecting two points on the line (marked with squares in Figure  1R.9) and calculating the difference in values of y (∆y 5 y2 2 y1) and ­ x (∆x 5 x 2 2 x 1). The slope of the line is then the ratio of these differences, m 5 ∆y/∆x. With the slope and intercept now known, you can write the equation for the line y 5 20.525x 1 1.87

Determining the Slope and y-Intercept with a Computer Program—Least-Squares Analysis  Generally, the easiest

method of determining the slope and y-intercept of a straight line (and thus the line’s equation) is to use a program such as Microsoft Excel or Apple’s Numbers. These programs perform a least-squares or linear regression analysis and give the best straight line based on the data. (This line is referred to in Excel and Numbers as a trendline.)

and you can use this equation to calculate y-values for points that are not part of the  original set of x2y data. For example, when x 5 1.50, you can calculate that y 5 20.525(1.50) 1 1.87 5 1.08.

1R.6 Problem Solving and Chemical Arithmetic Goals for Section 1R.6 • Solve problems using a systematic approach. • Incorporate quantitative information into an algebraic expression and solve that expression.

Many aspects of chemistry involve analyzing quantitative information, so problem solving will be important in your success. However, as in anything you do, careful planning is important, and students usually find it helpful to follow a definite plan as illustrated in all of the examples in the book.



1R.6  Problem Solving and Chemical Arithmetic

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Step 1 State the Problem. Read it carefully—and then read it again. Step 2 What Do You Know? Determine specifically what you are trying to calculate or conclude and what information you are given. What key principles are involved?

What information is known or not known? What information might be present just for context? Organize the information to see what is required and discover the relationships among the data given. Try writing the information down in table form. If the information is numerical, be sure to include units. Strategy Maps  A number of the Example problems in this book are accompanied by a Strategy Map (for instance, Example 1R.6) that outlines a route to a solution.

Step 3 Strategy. One of the greatest difficulties for a student in introductory chemistry is picturing what is being asked for. Try sketching a picture of the situation involved. For example, we sketched a picture of the piece of metal whose density we wanted to calculate and put the dimensions on the drawing (Figure 1R.7). Develop a plan. Have you done a problem of this type before? If not, perhaps the problem is a combination of several simpler ones you have seen before. Break it down into those simpler components. Try reasoning backward from the units of the answer. What data do you need to find an answer in those units? Drawing a strategy map may help you plan how to solve the problem. Step 4 Solution. Execute the plan. Carefully write down each step of the problem, being sure to keep track of the units on each number. (Do the units cancel to give you the answer in the desired units?) Don’t skip steps. Don’t do anything except the simplest steps in your head. Students often say they got a problem wrong because they “made a stupid mistake.” Your instructors—and book authors—also make them, and it is usually because they don’t take the time to write down the steps of the problem clearly. Step 5 Think about Your Answer. Ask yourself whether the answer is reasonable and if you obtained an answer in the correct units. Step 6 Check Your Understanding. In this text, each Example is followed by another

problem for you to try. (The solutions to those questions are given by chapter in Appendix N.) When doing homework Study Questions, try the Practicing Skills questions to see if you understand the basic ideas. These steps for problem solving are ones that many students have found to be successful, so try to conscientiously follow this scheme. But also be flexible. The “What Do You Know?” and “Strategy” steps often blend into a single set of ideas.

Exam pl e 1 R.6 Strategy Map Problem How thick will an oil layer be when a given mass covers a given area?

Problem Solving Problem  A mineral oil has a density of 0.875 g/cm3. Suppose you spread 0.75 g of this oil over the surface of water in a large circular dish with an inner diameter of 21.6 cm. How thick is the oil layer? Express the thickness in centimeters.

What Do You Know?  You know the mass and density of the oil and the diameter of the surface to be covered.

Strategy  It is often useful to begin solving such problems by sketching a picture of the situation. 21.6 cm

50

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This helps you recognize the relationships between the key quantities (volume, area, and thickness) in this problem. If you know the volume and the area covered by the oil, then you can find the thickness because Volume of oil layer 5 (thickness of layer) 3 (area of oil layer) So, you need two things: (1) the volume of the oil layer and (2) the area of the layer. Step 1. Calculate the volume of the oil using the mass and density of the oil. Data/Information Mass and density of the oil and diameter of the circular surface to be covered.

Step 2. Calculate the area covered by the oil. The area can be found because the oil forms a circle, which has an area equal to π 3 r2 (where r is the radius of the dish). Step 3. Calculate the thickness by dividing the volume by the surface area.

Solution  Step 1

Calculate the volume of the oil. The mass of the oil layer is known, so combining the mass of oil with its density gives the volume of the oil used:

0.75 g 3 Step 2

1 cm3 5 0.857 cm3 0.875 g

Calculate the area covered by the oil. The oil is spread over a circular surface, whose area is given by Area 5 π 3 (radius)2 The diameter of the oil layer is 21.6 cm. The radius is half the diameter, 10.8 cm, so Area of oil layer 5 (π)(10.8 cm)2 5 366.4 cm2

Step 3

Calculate the thickness of the oil layer by dividing the volume by the area.

Thickness 5

0.857 cm3 volume 5 5 0.0023 cm area 366.4 cm2  

Think about Your Answer In the volume calculation, the calculator shows 0.857143. . . . The quotient should have two significant figures because 0.75 has two significant figures, so the result of this step is reported as 0.857 cm3, containing one extra digit. In the area calculation, the calculator shows 366.435. . . . The answer to this step should have three significant figures because 10.8 has three; again, this value is reported to one extra digit. When these interim results are combined to calculate the thickness, the final result can have only two significant figures. Remember that premature rounding can lead to errors.

Check Your Understanding  A particular paint has a density of 0.914 g/cm3. You need to cover a wall that is 7.6 m long and 2.74 m high with a paint layer 0.13 mm thick. What volume of paint (in liters) is required? What is the mass (in grams) of the paint layer?

Applying Chemical Principles 1R.1  Out of Gas! On July 23, 1983, a new Boeing 767 jet aircraft was flying at 26,000 feet from Montreal to Edmonton as Air Canada Flight 143. Warning buzzers sounded in the cockpit. One of the world’s largest planes was now a glider—the plane had run out of fuel!

How did this happen? A simple mistake had been made in calculating the amount of fuel required for the flight because of a mixup of units of measurement! Applying Chemical Principles

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51

© Wayne Glowacki/Winnipeg Free Press

The Gimli glider.  After running out of fuel, Air Canada Flight 143 glided 29 minutes before landing on an abandoned airstrip at Gimli, Manitoba, near Winnipeg.

Like all Boeing 767s, this plane had a sophisticated fuel gauge, but it was not working properly. The plane was still allowed to fly, however, because there is an alternative method of determining the quantity of fuel in the tanks. Mechanics can use a stick, much like the oil dipstick in an automobile engine, to measure the fuel level in each of the three tanks. The mechanics in Montreal read the dipsticks, which were calibrated in centimeters, and translated those readings to a volume in liters. According to this, the plane had a total of 7682 L of fuel. Pilots always calculate fuel quantities in units of mass because they need to know the total mass of the plane before take-off. Air Canada pilots had always calculated the quantity of fuel in pounds, but the new 767’s fuel consumption was given in kilograms. The pilots knew that 22,300 kg of fuel was

required for the trip. If 7682 L of fuel remained in the tanks, how much had to be added? This involved using the fuel’s density to convert 7682 L to a mass in kilograms. The mass of fuel to be added could then be calculated, and that mass converted to a volume of fuel to be added. The First Officer of the plane asked a mechanic for the conversion factor to do the volume-to-mass conversion, and the mechanic replied “1.77.’’ Using that number, the First Officer and the mechanics calculated that 4917 L of fuel should be added. But later calculations showed that this is only about one fourth of the required amount of fuel! Why? Because no one thought about the units of the number 1.77. They realized later that 1.77 has units of pounds per liter and not kilograms per liter. Out of fuel, the plane could not make it to Winnipeg, so controllers directed them to the town of Gimli and to a small airport abandoned by the Royal Canadian Air Force. After gliding for almost 30 minutes, the plane approached the Gimli runway. The runway, however, had been converted to a race course for cars, and a race was underway. Furthermore, a steel barrier had been erected across the runway. Nonetheless, the pilot managed to touch down very near the end of the runway. The plane sped down the concrete strip; the nose wheel collapsed; several tires blew—and the plane skidded safely to a stop just before the barrier. The Gimli glider had made it! And somewhere an aircraft mechanic is paying more attention to units on numbers.

Questions

1. What is the fuel density in units of kg/L? (1 pound 5 453.6 g) 2. What mass and what volume of fuel should have been loaded?

Have you ever noticed that there are many ties in swimming competitions? For example, in the 2016 Summer Olympics, there was a two-way tie for the gold medal in the women’s 100-m freestyle and a three-way tie for the silver medal in the men’s 100-m butterfly. Olympic competitions are timed to one hundredth of a second. You might wonder why the officials don’t simply time the events out to a thousandth of a second, something that is technologically feasible and done in some sports, and eliminate most of these ties. The reason relates to the topic of how many digits in a swimming competition are really significant. Consider a 50-m Olympic-sized swimming pool and a 50-m freestyle swimming contest. The current world record of 20.91 seconds for this event was set by César Cielo of Brazil in 2009. Assuming a person is swimming at this rate, the maximum distance traveled in one thousandth of a second is 2.4 mm. The problem arises with the necessary specifications in the dimensions of the pool. There will always be some variation in the lengths of the different lanes due to limitations in the construction of pools. Current specifications allow a lane to be up to 3 cm longer than the stated length of 50.00 m. It would thus not be fair to penalize a swimmer in a lane that could be 3 cm longer for a difference in time that would amount to at most 2.4 mm, and so timing out to thousandths of a second is not done.

52

Richard Heathcote/Getty Images Sport/Getty Images

1R.2  Ties in Swimming and Significant Figures

A Tie for Gold.  Simone Manuel and Penny Oleksiak tie for gold in the 100-m freestyle event at the 2016 Summer Olympics held in Rio de Janeiro, Brazil.

Questions

1. Confirm that a person swimming at the world record rate for the 50-m freestyle would travel 2.4 mm in one thousandth of a second. 2. At this world record rate, how long would it take for a swimmer to travel 3.0 cm? 3. Consider a lane that is 3 cm longer than the stated 50.00 m. What is the percent error in this lane’s length?

Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry

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Think–Pair–Share 1. An antacid tablet is known to contain 1.000 g of calcium carbonate. After analyzing four tablets each, three students report the following results: Student A: average = 0.984 g, standard deviation = 0.005 g Student B: average = 0.975 g, standard deviation = 0.003 g Student C: average = 0.996 g, standard deviation = 0.008 g Which student had the most accurate results? The least accurate results? The most precise results? The least precise results? Explain your answers. 2. Consider the following three measurements: 150 mL, 150. mL, and 150.0 mL. (a) How many significant figures are present in each measurement? (b) Given the general guideline that in many measurements, there is an uncertainty of ±1 in the last significant digit, give the range of values represented by each of the measurements. 3. Carry out the following analysis. (a) Write down the measurements 4.8 m, 4.578 m, and 3.24 m. Underline the inexact digit in each case.

(b) Add the three measurements and report the answer, keeping all digits that would be displayed on a calculator. Underline any digits that have some uncertainty. (c) Based on the principle that a reported number should contain all digits that are known exactly and one digit that is inexact, what answer should be reported for this calculation? How does this answer correspond to the answer required by the addition/subtraction rule for reporting significant figures requires? 4. The density of silver is 10.5 g/cm3. You have a sample of silver that has a mass of 38.7 g. (a) Solve the equation d = m/V for volume (V). Substitute the given data into that equation and calculate the volume in cm3. (b) In an alternative approach, construct a conversion factor that could be used to convert from the mass to the volume and carry out this calculation. (c) Which method do you prefer and why? (Note that some students may prefer one method and other students may prefer the other.)

Chapter Goals Revisited The goals for this chapter are keyed to specific Study Questions to help you organize your review.

1R.1  Units of Measurement • Use the common units for measurements in chemistry and make unit conversions (such as from liters to milliliters). 1–12, 19–22, 39–41.

1R.2 Making Measurements: Precision, Accuracy, Experimental Error, and Standard Deviation • Recognize and express uncertainties in measurements. 23, 24, 50, 62, 64, 68, 73.

1R.3  Mathematics of Chemistry • Express and use numbers in exponential or scientific notation. 25–28. • Report the answer of a calculation to the correct number of significant figures. 29, 30.

1R.4  Problem Solving by Dimensional Analysis • Solve problems using dimensional analysis. 13–18, 43, 44, 59.

1R.5  Graphs and Graphing • Read information from graphs. 32, 33. • Prepare and interpret graphs of numerical information, and, if a graph

produces a straight line, find the slope and equation of the line. 31, 34, 71, 72.

1R.6  Problem Solving and Chemical Arithmetic • Solve problems using a systematic approach. 42, 48, 51, 55–58, 60, 65, 67. • Incorporate quantitative information into an algebraic expression and solve that expression. 35–38.



Chapter Goals Revisited

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53

Key Equations Equation 1R.1  (page 33)  ​Converting a temperature from ° C to K. T (K) 5

1K (T °C 1 273.15 °C) 1 °C

Equation 1R.2  (page 38)  Error in measurement. Error in measurement 5 experimentally determined value 2 accepted value

Equation 1R.3  (page 39)  Percent error. error in measurement 3 100% accepted value experimentally determined value 2 accepted value 5 accepted value

Percent error 5

3 100%

Equation 1R.4  (page 40)  Standard deviation. ∑ ( xi 2 x ) N 21

2

Standard deviation 5

Study Questions denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully worked solutions in the Student Solutions Manual.

▲

Practicing Skills Temperature Scales 1. Many laboratories use 25 °C as a standard temperature. What is this temperature in kelvins? 2. The temperature on the surface of the Sun is 5.5 3 103 °C. What is this temperature in kelvins? 3. Make the following temperature conversions: °C K (a) 16        (b)        370 (c) 40        4. Make the following temperature conversions: °C K (a)        77 (b) 63        (c)        1450

Length, Volume, Mass, and Density (See Example 1R.1.) 5. A marathon distance race covers a distance of 42.195 km. What is this distance in meters? In miles? 6. The average pencil, new and unused, is 19 cm long. What is its length in millimeters? In meters?

54

7. A standard U.S. postage stamp is 2.5 cm long and 2.1 cm wide. What is the area of the stamp in square centimeters? In square meters? 8. A particular cat food can’s lid has a diameter of 8.3 cm. What is the surface area of this lid in square centimeters? In square meters? [Area of a circle 5 (π)(radius)2] 9. A typical laboratory beaker has a volume of 250. mL. What is its volume in cubic centimeters? In liters? In cubic meters? In cubic decimeters? 10. Some soft drinks are sold in bottles with a volume of 1.5 L. What is this volume in milliliters? In cubic centimeters? In cubic decimeters? 11. A book has a mass of 2.52 kg. What is this mass in grams? 12. A new U.S. dime has a mass of 2.265 g. What is its mass in kilograms? In milligrams? 13. Ethylene glycol, C2H6O2, is an ingredient of automobile antifreeze. Its density is 1.11 g/cm3 at 20 °C. If you need 500. mL of this liquid, what mass of the compound, in grams, is required? 14. Acetone is a compound used in many nail polish removers. Its density is 0.7845 g/cm3 at 25 °C. What mass of acetone is present in a sample with a volume of 100. mL?

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15. A piece of silver metal has a mass of 2.365 g. If the density of silver is 10.5 g/cm3, what is the volume of the silver? 16. Lead has a density of 11.35 g/cm3 at 20 °C. What volume of lead is present in a sample with a mass of 50.0 g? 17. You can identify a metal by carefully determining its density (d). An unknown piece of metal, with a mass of 2.361 g, is 2.35 cm long, 1.34 cm wide, and 1.05 mm thick. Which of the following is the element? (a) nickel, d 5 8.91 g/cm3 (b) titanium, d 5 4.50 g/cm3 (c) zinc, d 5 7.14 g/cm3 (d) tin, d 5 7.23 g/cm3 18. Which occupies a larger volume, 600 g of water (with a density of 0.995 g/cm3) or 600 g of lead (with a density of 11.35 g/cm3)?

Energy Units 19. You are on a diet that calls for eating no more than 1200 Cal/day. What is this energy in joules? 20. A 2-inch piece of chocolate cake with frosting provides 1670 kJ of energy. What is this in dietary Calories (Cal)? 21. One food product has an energy content of 170 kcal per serving, and another has 280 kJ per serving. Which food provides the greater energy per serving? 22. A can of soft drink (335 mL) provides 130 Calories. A bottle of mixed berry juice (295 mL) provides 630 kJ. Which provides the greater total energy? Which provides the greater energy per milliliter?

Accuracy, Precision, Error, and Standard Deviation (See Examples 1R.2 and 1R.3.) 23. You and your lab partner are asked to determine the density of an aluminum bar. The mass is known accurately (to four significant figures). You use a simple metric ruler to measure its dimensions and obtain the results for Method A. Your partner uses a precision micrometer and obtains the results for Method B. Method A ( g/cm3)

Method B (g/cm3)

2.2

2.703

2.3

2.701

2.7

2.705

2.4

2.703

The accepted density of aluminum is 2.702 g/cm3. (a) Calculate the average density for each method. (b) Calculate the percent error for each method’s average value.

(c) Calculate the standard deviation for each set of data. (d) Which method’s average value is more precise? Explain your answer. Which method is more accurate? Explain your answer. 24. The accepted value of the mass of aspirin in a tablet of aspirin is 325 mg. Trying to verify that value, you obtain 321 mg, 326 mg, 324 mg, and 329 mg in four separate trials. Your partner measures 327 mg, 329 mg, 328 mg, and 326 mg. (a) Calculate the average value and percent error for your data and your partner’s data. (b) Which of you is more accurate? Explain your answer. (c) Without doing any calculations, predict whose set of data is more precise? Explain your answer. (d) Calculate the standard deviation for your data and your partner’s data. Do these values confirm or disprove your prediction in part (c)?

Exponential Notation and Significant Figures (See Example 1R.4.) 25. Express the following numbers in exponential or scientific notation, and give the number of significant figures in each. (a) 0.0830 g (c) 0.00602 g (b) 136 g (d) 3000 mL 26. Express the following numbers in exponential or scientific notation, and give the number of significant figures in each. (a) 1356 mL (c) 250.0 g (b) 0.03042 L (d) 120 g 27. Express the following numbers without using ­scientific notation (for example 1.23 3 102 5 123), and give the number of significant figures in each. (a) 5.43 3 102 m (c) 6.20 3 1024 L (b) 4.306 3 1022 L (d) 8.42 3 103 mL 28. Express the following numbers without using scientific notation (for example, 1.23 3 102 5 123), and give the number of significant figures in each. (a) 3.25 3 1022 m (c) 4.2 3 103 mL (b) 4.02 3 1023 mL (d) 9.305 3 104 g 29. Carry out the following operations. Provide the answer with the correct number of significant figures. (a) (1.52)(6.21 3 1023) (b) (6.217 3 103)2(5.23 3 102) (c) (6.217 3 103) ÷ (5.23 3 102)  7.779  (d) (0.0546)(16.0000)   55.85  Study Questions

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55

30. Carry out the following operations. Provide the answer with the correct number of significant figures. (a) (6.25 3 102)3 (b) 2.35 3 1023 (c) (2.35 3 1023)1/3 23.56 2 2.3  (d) (1.68)   1.248 3 103   

33. Use the graph below to answer the following questions. (a) Derive the equation for the straight line, y 5 mx 1 b. (b) What is the value of y when x 5 6.0? 25.00

20.00

Graphing (See Section 1R.5)

Number of Kernels

Mass (g)

 5

0.836

12

2.162

35

5.801

15.00 y values

31. To determine the average mass of a popcorn kernel, you collect the following data:

10.00

5.00

Plot the data with number of kernels on the x-axis and mass on the y-axis. Draw the best straight line using the points on the graph (or do a leastsquares or linear regression analysis using a computer program), and then write the equation for the resulting straight line. What is the slope of the line? What does the slope of the line signify about the mass of a popcorn kernel? What is the mass of 20 popcorn kernels? How many kernels are there in a handful of popcorn with a mass of 20.88 g?

0

6.00

y values

5.00 4.00

3.00

4.00

5.00

34. The following data were collected in an experiment to determine how an enzyme works in a biochemical reaction.

Amount of H2O2

Reaction Speed (amount/second)

1.96

4.75 3 1025

1.31

4.03 3 1025

0.98

3.51 3 1025

0.65

2.52 3 1025

0.33

1.44 3 1025

0.16

0.585 3 1025

Solving Equations

3.00

35. Solve the following equation for the unknown value, C.

2.00

(0.502)(123) 5 (750.)C

1.00

0

0.10

0.20

0.30

x values

56

2.00

(a) Plot these data as 1/amount on the x-axis and 1/speed on the y-axis. Draw the best straight line to fit these data points. (b) Determine the equation for the data, and give the values of the y-intercept and the slope. (Note: In biochemistry this is known as a ­Lineweaver–Burk plot, and the y-intercept is related to the maximum speed of the reaction.)

7.00

0

1.00

x values

32. Use the following graph to answer the following questions: (a) What is the value of x when y 5 4.0? (b) What is the value of y when x 5 0.30? (c) What are the slope and the y-intercept of the line? (d) What is the value of y when x 5 1.0? 8.00

0

0.40

0.50

36. Solve the following equation for the unknown value, n. (1.0)(22.4) 5 n(0.082057)(273.15)

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37. Solve the following equation for the unknown value, T. (4.184)(244)(T 2 292.0)  1 (0.449)(88.5)(T 2 369.0) 5 0 38. Solve the following equation for the unknown value, n. 1 1 2246.0 5 1312  2 2 2  n  2

General Questions These questions are not designated as to type or location in the chapter. They may combine several concepts.

43. The anesthetic procaine hydrochloride is often used to deaden pain during dental surgery. The compound is packaged as a 10.% solution (by mass; d 5 1.0 g/mL) in water. If your dentist injects 0.50 mL of the solution, what mass of procaine hydrochloride (in milligrams) is injected? 44. You need a cube of aluminum with a mass of 7.6 g. What must be the length of the cube’s edge (in cm)? (The density of aluminum is 2.698 g/cm3.) 45. You have a 250.0-mL graduated cylinder containing some water. You drop three marbles with a total mass of 95.2 g into the water. What is the average density of a marble?

H3N

NH3 Pt

1.97Å Cl

© Charles D. Winters/Cengage

39. Molecular distances are usually given in nanometers (1 nm 5 1 3 1029 m) or in picometers (1 pm 5 1 3 10212 m). However, the angstrom (Å) unit is sometimes used, where 1 Å 5 1 3 10210 m. (The angstrom unit is not an SI unit.) If the distance between the Pt atom and the N atom in the cancer chemotherapy drug cisplatin is 1.97 Å, what is this distance in nanometers? In picometers?

Cl

Cisplatin

40. The separation between carbon atoms in diamond is 0.154 nm. What is their separation in meters? In picometers (pm)? In Angstroms (Å)? 0.154 nm

A portion of the diamond structure

(a)

(b)

Determining density.  (a) A graduated cylinder with 61 mL of water. (b) Three marbles are added to the cylinder.

46. You have a white crystalline solid, known to be one of the potassium compounds listed below. To determine which, you measure its density. You measure out 18.82 g and transfer it to a graduated cylinder containing kerosene (in which these compounds will not dissolve). The level of liquid kerosene rises from 8.5 mL to 15.3 mL. Calculate the density of the solid, and identify the compound from the following list. (a) KF, d 5 2.48 g/cm3 (b) KCl, d 5 1.98 g/cm3 (c) KBr, d 5 2.75 g/cm3 (d) KI, d 5 3.13 g/cm3 47. ▲ The smallest repeating unit of a crystal of common salt is a cube (called a unit cell) with an edge length of 0.563 nm.

41. A red blood cell has a diameter of 7.5 μm (micrometers). What is this dimension in (a) meters, (b) nanometers, and (c) picometers? 42. The platinum-containing cancer drug cisplatin (Study Question 39) contains 65.0 mass-percent of the metal. If you have 1.53 g of the compound, what mass of platinum (in grams) is contained in this sample?

0.563 nm

Sodium chloride, NaCl Study Questions

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57

(a) What is the volume of this cube in cubic nanometers? In cubic centimeters? (b) The density of NaCl is 2.17 g/cm3. What is the mass of this smallest repeating unit (unit cell)? (c) Each repeating unit is composed of four NaCl units. What is the mass of one NaCl formula unit? 48. Diamond has a density of 3.513 g/cm3. The mass of diamonds is often measured in carats, where 1 carat equals 0.200 g. What is the volume (in cubic centimeters) of a 1.50-carat diamond?

© Charles D. Winters/Cengage

49. The element gallium has a melting point of 29.8 °C. If you hold a sample of gallium in your hand, should it melt? Explain briefly.

Gallium metal

50. ▲ The density of pure water at various temperatures is given below. T(° C)

d (g/cm3)

 4

0.99997

15

0.99913

25

0.99707

35

0.99406

Suppose your laboratory partner tells you the density of water at 20 °C is 0.99910 g/cm3. Is this a reasonable number? Why or why not? 51. When you heat popcorn, it pops because it loses water explosively. Assume a kernel of corn, with a mass of 0.125 g, has a mass of only 0.106 g after popping. (a) What percentage of its mass did the kernel lose on popping? (b) Popcorn is sold by the pound in the United States. Using 0.125 g as the average mass of a popcorn kernel, how many kernels are there in a pound of popcorn? (1 lb 5 453.6 g)

58

52. ▲ The aluminum in a package containing 75 ft2 of kitchen foil weighs approximately 12 ounces. Aluminum has a density of 2.70 g/cm3. What is the approximate thickness of the aluminum foil in millimeters? (1 ounce 5 28.4 g; 12 in 5 1 ft; and 1 in 5 2.54 cm) 53. ▲ Fluoridation of city water supplies has been practiced in the United States for several decades. It is done by continuously adding sodium fluoride to water as it comes from a reservoir. Assume you live in a medium-sized city of 150,000 people and that 660 L (170 gal) of water is used per person per day. What mass of sodium fluoride (in kilograms) must be added to the water supply each year (365 days) to have the required fluoride concentration of 1 ppm (part per million)—that is, 1 kilogram of fluoride per 1 million kilograms of water? (Sodium fluoride is 45.0% fluoride, and water has a density of 1.00 g/cm3.) 54. ▲ Over two centuries ago, Benjamin Franklin showed that 1 teaspoon of oil would cover about 0.5 acre of still water. If you know that 1.0 3 104 m2 5 2.47 acres and that there is approximately 5 cm3 in a teaspoon, what is the thickness of the 0.5-acre layer of oil? How might this thickness be related to the sizes of molecules? 55. ▲ Automobile batteries are filled with an aqueous solution of sulfuric acid. What is the mass of the acid (in grams) in 750. mL of the battery acid solution if the density of the solution is 1.285 g/cm3 and the solution is 38.08% sulfuric acid by mass? 56. ▲ Hydrochloric acid is a solution of the gas hydrogen chloride (HCl) dissolved in water. What mass of HCl is present in 50. mL of a concentrated hydrochloric acid solution that is 37% HCl by mass and has a density of 1.2 g/mL? 57. ▲ Household bleach is often a solution of the compound sodium hypochlorite (NaClO) in water. What volume of a bleach solution that is 5.25% NaClO is needed to provide 8.0 g of NaClO? Assume that the density of the bleach solution is 1.1 g/mL. 58. A 26-meter-tall statue of the Buddha in Tibet is covered with 279 kg of gold. If the gold was applied to a thickness of 0.0015 mm, what surface area is covered (in square meters)? (Gold density 5 19.3 g/cm3) 59. At 25 °C, the density of water is 0.997 g/cm3, whereas the density of ice at 210 °C is 0.917 g/cm3. (a) If a soft-drink can (volume 5 250. mL) is filled completely with pure water at 25 °C and then frozen at 210 °C, what volume does the ice occupy? (b) Can the ice be contained within the can?

Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry

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60. Suppose your bedroom is 18 feet long and 15 feet wide, and the distance from floor to ceiling is 8 feet 6 inches. You need to know the volume of the room in metric units for some scientific calculations. (a) What is the room’s volume in cubic meters? In liters? (b) What is the mass of air in the room in kilograms? In pounds? (Assume the density of air is 1.2 g/L and that the room is empty of furniture.) 61. A spherical steel ball has a mass of 3.475 g and a diameter of 9.40 mm. What is the density of the steel? [The volume of a sphere 5 (4/3)πr3 where r 5 radius.] 62. ▲ You are asked to identify an unknown liquid that is one of the liquids listed below. You pipet a 3.50-mL sample into a beaker. The empty beaker had a mass of 12.20 g, and the beaker plus the liquid weighed 16.08 g.

(b) The unknown is one of the seven metals listed in the following table. Is it possible to identify the metal based on the density you have calculated? Explain. Density (g/cm3)

Metal

Density (g/cm3)

Zinc

7.13

Nickel

 8.90

Iron

7.87

Copper

 8.96

Cadmium

8.65

Silver

10.50

Cobalt

8.90

Metal

64. ▲ There are five hydrocarbon compounds (compounds of C and H) that have the formula C6H14. (Compounds that have the same formula but differ in the way the atoms are attached are called isomers.) All are liquids at room temperature but have slightly different densities. Hydrocarbon

Density at 25 ° C (g/cm3)

Hexane

0.6600

Ethylene glycol

1.1088 (major component of automobile antifreeze)

2,3-Dimethylbutane

0.6616

Water

0.9971

1-Methylpentane

0.6532

Ethanol

0.7893 (alcohol in alcoholic beverages)

2,2-Dimethylbutane

0.6485

2-Methylpentane

0.6645

Acetic acid

1.0492 (active component of vinegar)

Glycerol

1.2613 (solvent used in home care products)

(a) Calculate the density and identify the unknown. (b) If you were able to measure the volume to only two significant figures (that is, 3.5 mL, not 3.50 mL), will the results be sufficiently precise to identify the unknown? Explain. 63. ▲ You have an irregularly shaped piece of an unknown metal. To identify it, you determine its density and then compare this value with known values that you look up in the chemistry library. The mass of the metal is 74.122 g. Because of the irregular shape, you measure the volume by submerging the metal in water in a graduated cylinder. When you do this, the water level in the cylinder rises from 28.2 mL to 36.7 mL. (a) What is the density of the metal? (Use the correct number of significant figures in your answer.)



Density (g/mL)

Substance

(a) You have a pure sample of one of these hydrocarbons, and to identify it you decide to measure its density. You determine that a 5.0-mL sample (measured in a graduated cylinder) has a mass of 3.2745 g (measured on an analytical balance). Assume that the accuracy of the values for mass and volume is plus or minus one (61) in the last significant figure. What is the density of the liquid? (b) Can you identify the unknown hydrocarbon based on your experiment? (c) Can you eliminate any of the five possibilities based on the data? If so, which one(s)? (d) You need a more precise volume measurement to solve this problem, and you redetermine the volume to be 4.93 mL. Based on this new information, what is the unknown compound? 65. ▲ Suppose you have a cylindrical glass tube with a thin capillary opening, and you wish to determine the diameter of the opening. You can do this experimentally by weighing a piece of the tubing before and after filling a portion of the capillary with

Study Questions

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59

mercury. Using the following information, calculate the diameter of the opening. Mass of tube before adding mercury 5 3.263 g Mass of tube after adding mercury 5 3.416 g Length of capillary filled with mercury 5 16.75 mm Density of mercury 5 13.546 g/cm3 Volume of cylindrical capillary filled with mercury 5 (π)(radius)2(length) 66. Copper: Copper has a density of 8.96 g/cm3. An ingot of copper with a mass of 57 kg (126 lb) is drawn into wire with a diameter of 9.50 mm. What length of wire (in meters) can be produced? [Volume of wire 5 (π)(radius)2(length)] 67. ▲ Copper: (a) Suppose you have a cube of copper metal that is 0.236 cm on a side with a mass of 0.1206 g. If you know that each copper atom (radius 5 128 pm) has a mass of 1.055 3 10222 g, how many atoms are there in this cube? What percentage of the volume occupied by the cube is filled with atoms? How much of the lattice is empty space? Why is there empty space in the lattice? (b) Now look at the smallest, repeating unit of the crystal lattice of copper.

Cube of copper atoms

Smallest repeating unit

Knowing that an edge of this cube is 361.47 pm and the density of copper is 8.960 g/cm3, calculate the number of copper atoms in this smallest, repeating unit.

In the Laboratory 69. A sample of unknown metal is placed in a graduated cylinder containing water. The mass of the sample is 37.5 g, and the water levels before and after adding the sample to the cylinder are as shown in the figure. Which metal in the following list is most likely the sample? (d is the density of the metal.) (a) Mg, d 5 1.74 g/cm3 (d) Al, d 5 2.70 g/cm3 (b) Fe, d 5 7.87 g/cm3 (e) Cu, d 5 8.96 g/cm3 (c) Ag, d 5 10.5 g/cm3 (f) Pb, d 5 11.3 g/cm3 25

20

20

15

15

10

10

5

5

Graduated cylinders with unknown metal (right)

70. Iron pyrite is often called fool’s gold because it looks like gold (see page 13). Suppose you have a solid that looks like gold, but you believe it to be fool’s gold. The sample has a mass of 23.5 g. When the sample is lowered into the water in a graduated cylinder (Study Question 69), the water level rises from 47.5 mL to 52.2 mL. Is the sample fool’s gold (d 5 5.00 g/cm3) or pure gold (d 5 19.3 g/cm3)? 71. You can analyze for a copper compound in water using an instrument called a spectrophotometer. [A spectrophotometer is a scientific instrument that measures the absorbance of light (of a given wavelength) by the solution.] The absorbance at a given wavelength of light (A) depends directly on the mass of compound per liter of solution. To calibrate the spectrophotometer, you collect the following data:

68. You set out to determine the density of lead in the laboratory. Using a top loading balance to determine the mass and the water displacement method (Study Question 45) to determine the volume of a variety of pieces of lead, you calculate the following densities: 11.6 g/cm3, 11.8 g/cm3, 11.5 g/cm3, and 12.0 g/cm3. You consult a reference book and find that the accepted value for the density of lead is 11.3 g/cm3. Calculate your average value, percent error, and standard deviation of your results.

60

25

Absorbance (A)

Mass of Copper Compound per Liter (g/L)

0.257

1.029 3 1023

0.518

2.058 3 1023

0.771

3.087 3 1023

1.021

4.116 3 1023

Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry

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Plot the absorbance (A) against the mass of copper compound per liter (g/L), and find the slope (m) and intercept (b) (assuming that A is y and the mass of copper compound per liter of solution is x in the equation for a straight line, y 5 mx 1 b). What is the mass of copper compound in the solution in g/L and mg/mL when the absorbance is 0.635? 72. A gas chromatograph is calibrated for the analysis of isooctane (a major gasoline component) using the following data:



Percent Isooctane (x-data)

Instrument Response (y-data)

0.352

1.09

0.803

1.78

1.08

2.60

1.38

3.03

1.75

4.01

If the instrument response is 2.75, what percentage of isooctane is present? (Data are taken from D. A. Skoog, D. M. West, F. J. Holler, and S. R. Crouch, Analytical Chemistry, An Introduction, 7th ed., Belmont, CA: Brooks/Cole, 2000.) 73. A general chemistry class carried out an experiment to determine the percentage (by mass) of acetic acid in vinegar. Ten students reported the following values: 5.22%, 5.28%, 5.22%, 5.30%, 5.19%, 5.23%, 5.33%, 5.26%, 5.15%, and 5.22%. Determine the average value and the standard deviation from these data. How many of these results fell within 6 one standard deviation of this average value?

Study Questions

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61

2 Atoms, Molecules, and Ions Transition Metals Group 2B (12)

Group 2A (2)

Magnesium—Mg

Titanium—Ti

Vanadium—V

Chromium—Cr

Manganese—Mn

Iron—Fe

Cobalt—Co

Nickel—Ni

Copper—Cu

Mercury—Hg

Group 1A (1) 8A (18)

1A (1)

H

2

Li Be

3

Lithium—Li

2A (2)

1

4

Na Mg K

3A 4A 5A 6A 7A O (17) (13) (14) (15) (16)

B 3B (3)

4B (4)

Ca Sc Ti Y

7B (7)

C

N

Al Si

P

F

Ne

6B (6)

V

Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

S

Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te

5

Rb Sr

6

Cs Ba La Hf Ta

7

Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv

W Re Os Ir

Group 8A (18), Noble Gases

He

5B (5)

(8)

8B 1B 2B (9) (10) (11) (12)

Zinc—Zn

Cl Ar

I

Xe

Pt Au Hg Tl Pb Bi Po At Rn Ts Og

Neon—Ne

Potassium—K Group 4A (14)

Group 3A (13)

Boron—B

Carbon—C

Group 5A (15)

Tin—Sn

Group 6A (16)

Group 7A (17)

Sulfur—S Nitrogen—N2

Bromine—Br Aluminum—Al

Silicon—Si

Lead—Pb

Selenium—Se

Phosphorus—P

© Charles D. Winters/Cengage

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C hapt e r O ut li n e 2.1 Atomic Structure, Atomic Number, and Atomic Mass 2.2 Atomic Weight 2.3 The Periodic Table 2.4 Molecules: Formulas, Models, and Names 2.5 Ions 2.6 Ionic Compounds: Formulas, Names, and Properties 2.7 Atoms, Molecules, and the Mole 2.8 Chemical Analysis: Determining Compound Formulas 2.9 Instrumental Analysis: Determining Compound Formulas

This chapter begins an exploration of the chemistry of the elements—the building blocks of chemistry—and the compounds they form. There are currently 118 known elements, most of which combine to form the millions of compounds known to chemists. You will learn about the submicroscopic particles that make up these pure substances as well as some of the properties you can observe on the macroscopic scale. Some important skills you will learn are how to determine the names and formulas of many compounds and how to perform calculations that will allow you to connect the macroscopic measurement of mass to the particulate understanding of these substances. Throughout this chapter, you will see how the periodic table of the elements is used to help organize much of this information and as a tool to help you understand quantitative relationships in chemistry.

2.1 Atomic Structure, Atomic Number, and Atomic Mass Goals for Section 2.1 • Describe electrons, protons, and neutrons, and the general structure of the atom. • Define the terms atomic number and mass number. • Define isotopes and give the atomic symbol for a specific isotope.

Atomic Structure Around 1900, a series of experiments done by scientists in England, including Sir Joseph John Thomson (1856–1940) and Ernest Rutherford (1871–1937), established a model of the atom that is still the basis of modern atomic theory. ◀ Some of the 118 known elements.

63

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Nucleus with protons (positive electric charge) and neutrons (no electric charge).

Atoms are made of subatomic particles: electrically positive protons, electrically negative electrons, and, in all except one type of hydrogen atom, electrically neutral neutrons. The model places the more massive protons and neutrons in a very small nucleus (Figure 2.1), which contains all the positive charge and almost all the mass of an atom. Electrons, with a much smaller mass than protons or neutrons, surround the nucleus and occupy most of the volume. In an electrically neutral atom, the number of electrons equals the number of protons. As you will see, the chemical properties of elements and compounds depend largely on the electrons in their atoms.

Atomic Number

Electrons (negative electric charge). The number of electrons and protons is equal in an electrically neutral atom.

Figure 2.1  The structure of the atom.  This figure is not drawn to scale. If the nucleus were really the size depicted here, the electron cloud would extend over 200 m. The atom is mostly empty space! In this illustration, the electrons are depicted as a cloud around the nucleus. The most accurate model of the atom represents electrons as waves, not particles.

Historical Perspective on the Development of Our Understanding of Atomic Structure  A brief

history of important experiments and the scientists involved in developing the modern view of the atom is given on pages 72–74. Unified Atomic Mass Units Masses of fundamental atomic particles are sometimes expressed using a non-SI unit called the unified atomic mass unit (u), sometimes referred to as the dalton (Da). One unified atomic mass unit, 1 u, is one-twelfth the mass of an atom of carbon with six protons and six neutrons. Such a carbon atom has a mass of exactly 12 u. 1 atomic mass unit (u) = 1.66054 × 10−24 g.

64

All atoms of a given element have the same number of protons in the nucleus. ­Hydrogen is the simplest element, with one nuclear proton. All helium atoms have two protons, all lithium atoms have three protons, and all beryllium atoms have four protons. The number of protons in the nucleus of an element is given by its atomic number, which is generally indicated by the symbol Z. The 118 known elements are listed in the periodic table inside the front cover of this book and on the list inside the back cover. The integer number at the top of the box for each element in the periodic table is its atomic number. A copper atom (Cu), for example, has an atomic number of 29, so its nucleus contains 29  protons. A ­uranium atom (U) has 92 nuclear protons and Z = 92.

Copper 29

Cu

Atomic number Symbol

Relative Atomic Mass As eighteenth- and nineteenth-century chemists tried to understand how the elements combine, they carried out increasingly quantitative studies aimed at learning, for example, how much of one element combines with another. Based on this work, they learned that substances have a constant composition, which led to the ­conclusion that chemists can define the relative masses of elements that combine to produce a new substance. At the beginning of the nineteenth century, John Dalton (1766–1844) suggested that the combinations of elements involve atoms, and he proposed a relative scale of atom masses. Dalton based this scale on hydrogen having a mass of 1. Later, oxygen atoms were chosen as the standard, and oxygen was assigned a mass of exactly 16, but there were disagreements between chemists and physicists about the details of this definition. In 1959–1960, the International Union of Pure and Applied Chemistry (IUPAC) and the International Union of Pure and Applied Physics (IUPAP) agreed to a new unified standard: a carbon atom having 6 protons and 6 neutrons in the nucleus is assigned a relative mass of exactly 12. The masses of other atoms are determined relative to the mass for this type of carbon atom. For example, chemical experiments and physical measurements show that the mass of an oxygen atom with 8 protons and 8 neutrons is 1.33291 times the mass of a carbon atom with 6 protons and 6 neutrons. So, the oxygen atom has a relative mass of 1.33291 × 12 = 15.9949.

Mass Number Protons and neutrons have relative atomic masses close to 1, while the mass of an electron is only about 1/2000 of this value (Table  2.1). You can estimate the ­approximate mass of an atom if you know the number of neutrons and protons in that atom. The sum of the number of protons and neutrons for an atom is called its mass number and is given the symbol A.

Chapter 2 / Atoms, Molecules, and Ions

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Table 2.1

Properties of Subatomic Particles* mass

Particle

Grams

Relative Atomic Mass

Charge

Symbol −01e or e−

−28

Electron

9.109384 × 10

0.0005485799

1−

Proton

1.672622 × 10−24

1.007276

1+

Neutron

1.674927 × 10−24

1.008665

0

1 1p 1 0n

or p+ or n

How Small Is an Atom?  The radius of the typical atom is between 30 and 300 pm (3 × 10−11 m to 3 × 10−10 m). To get a feeling for the incredible smallness of an atom, consider that 1 cm3 of water contains about three times as many atoms as the Atlantic Ocean contains teaspoons of water.

*These values and others in the book are taken from the National Institute of Standards and Technology website at http://physics.nist.gov/cuu/Constants/index.html

A = mass number = number of protons + number of neutrons

For example, a sodium atom, which has 11 protons and 12 neutrons in its nucleus, has a mass number of 23 (A = 11 p + 12 n). The most common atom of uranium has 92 protons and 146 neutrons, and a mass number of A = 238. Atoms are commonly symbolized with the following notation: Mass number Atomic number

A ZX

Element symbol

The subscript Z is optional because each atomic number represents a unique element. For example, the atoms described previously have the symbols 1213Na and 238 23 Na and 238U. In words, these are referred to as “sodium-23” and 92U, or just “uranium-238.”

Exam p le 2.1

Atomic Composition Problem  How many protons and how many electrons are in an atom of phosphorus with 16  neutrons? What is its mass number? What is the complete symbol for such an atom?

What Do You Know?  You know the name of the element and the number of neutrons. The number of protons and electrons are equal in a neutral atom. Strategy  The number of protons in an atom is given by the atomic number shown on the periodic table. The mass number is the sum of the number of protons and neutrons. Solution  A phosphorus atom has  15 protons and 15 electrons.  A phosphorus atom with 16 neutrons has a  mass number of 31.  Mass number = number of protons + number of neutrons = 15 + 16 = 31 The atom’s complete  symbol is 1315P. 

Think about Your Answer  Once you know the identity of an element, you can determine the number of protons in the nucleus from its atomic number shown on the periodic table. You can determine the mass number if you also know the number of neutrons in the atom.

Check Your Understanding



1.

What is the mass number of an iron atom with 30 neutrons?

2.

How many protons, neutrons, and electrons are in a 64Zn atom?

2.1  Atomic Structure, Atomic Number, and Atomic Mass

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65

Chemistry in Your Career

Tiffany J. Carey

Tiffany J. Carey Tiffany Carey’s first interest was biology, she reports, “But when I learned all life as we know it would not exist if the angle between hydrogen atoms in a ­water molecule was just 5 degrees different, I was intrigued.” Carey went on to earn a B.S. in biology with a chemistry minor and then an M.S. in environmental science and environmental engineering. Today Carey is a quality assurance officer for a major U.S. water utility that delivers 1 billion gallons per day of high-quality drinking water, where she ensures that data generated by the labs are of

the highest integrity and accuracy. Carey enjoys the mutual respect she shares with coworkers, saying that “we all have a part to play in the big picture of enriching the environment and protecting public health for millions of people.” Carey emphasizes the importance of chemistry in the environmental sector. “No one ever told me that the deciding factor in getting a job was going to be having enough chemistry credits, but it is, especially for government or public utility jobs.”

Isotopes

Solid H2O d = 0.917 g/cm3

Solid D2O d = 1.11 g/cm3

Figure 2.2  Ice made from “heavy water” sinks in “ordinary” water.  Water containing ordinary hydrogen (11H, protium) forms a solid that is less dense than liquid H2O, so it floats in the liquid. D2O-ice is denser than liquid H2O, so solid D2O sinks in liquid H2O.

© Charles D. Winters/Cengage

Liquid H2O d = 0.9998 g/cm3

All the atoms in a naturally occurring sample of a given element have the same mass in only a few instances (for example, aluminum, fluorine, and phosphorus). Most elements consist of atoms with several different mass numbers. For example, there are two kinds of boron atoms, one with a mass of about 10 (10B) and a second with a mass of about 11 (11B). Atoms of tin can have any of 10 different masses ranging from 112 to 124. Atoms with the same atomic number but different mass numbers are called isotopes. Scientists often refer to a particular isotope by giving its mass number. For example, an atom of iron with a mass number of 58 is referred to as iron–58 or symbolized as 58Fe. All atoms of an element have the same number of protons. To have different masses, isotopes must have different numbers of neutrons. The nucleus of a 10 B atom (Z = 5) contains five protons and five neutrons, whereas the nucleus of a 11 B atom contains five protons and six neutrons. The isotopes of hydrogen are so important that they have special names and symbols. All hydrogen atoms have one proton. When that is the only nuclear particle, the isotope is called protium, or just hydrogen. The isotope of hydrogen with one neutron, 12H, is called deuterium, or heavy hydrogen (symbol = D). The nucleus of radioactive hydrogen-3, 13H, or tritium (symbol = T), contains one proton and two neutrons. The substitution of one isotope of an element for another isotope of the same element in a compound usually has only a small, almost negligible, effect on chemical and physical properties. An exception is the substitution of deuterium for hydrogen because the mass of deuterium is double that of hydrogen (Figure 2.2).

E xamp le 2.2

Isotope Composition Problem Two naturally occurring isotopes of oxygen are 16O and 18O. How many ­protons, neutrons, and electrons are in the atoms of each of these isotopes?

66

Chapter 2 / Atoms, Molecules, and Ions

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What Do You Know?  You know that 16O and 18O are isotopes of oxygen. You also know that the atomic number for each element is shown on the periodic table and that in a neutral atom the number of protons and number of electrons are the same. Mass number is the sum of the protons and neutrons in an atom. Isotopes have the same number of protons but different numbers of neutrons, which results in different ­ mass numbers.

Strategy Step 1. Determine the number of protons in the atoms from the atomic number of the element shown on the periodic table. Step 2. Determine the number of neutrons in each atom by subtracting the atomic number from the mass number. Number of neutrons = mass number (number of protons and neutrons) − atomic number (number of protons) Step 3. Determine the number of electrons from the number of protons and the charge on the atom. In a neutral atom, the number of electrons equals the number of protons.

Solution  Step 1. The atomic number of oxygen shown on the periodic table is eight, so there are eight protons in the atoms of both 16O and 18O. Step 2. The number of neutrons is the difference between the mass number (protons and neutrons) and the atomic number (protons). Isotopes of the same element have different numbers of neutrons. For 16O: mass number − atomic number = 16 − 8 = 8 neutrons in 16O For 18O: mass number − atomic number = 18 − 8 = 10 neutrons in 18O Step 3. Because there are 8 protons in each isotope, 16O and 18O each contain 8 electrons.

Think about Your Answer  Isotopes of the same element have the same number of protons and the same number of electrons. They differ only in the number of neutrons. Check Your Understanding  Two naturally occurring isotopes of uranium are U and 238U. How many protons, neutrons, and electrons are in the atoms of each of these isotopes? 235

2.2 Atomic Weight Goal for Section 2.2 • Perform calculations that relate the atomic weight (atomic mass) of an element and isotopic abundances and masses.



2.2  Atomic Weight

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67

Determining Atomic Mass and Isotopic Abundance The masses of isotopes and their abundances are determined experimentally by mass spectrometry. In a mass spectrometer, the atoms or molecules of a sample are converted into gaseous, charged species that are then separated based on their mass and charge (Figure 2.3). Modern mass spectrometers can measure isotopic masses to as many as nine significant figures. Except for carbon-12, whose mass is defined to be exactly 12, isotopic masses do not have integer values. Isotopic masses are, however, always close to the mass  numbers for the isotope. For example, the atomic mass of boron-11 (11B, 5 protons and 6 neutrons) is 11.0093, and the atomic mass of iron-58 (58Fe, 26 protons and 32 neutrons) is 57.9333. A sample of water from a lake consists almost entirely of H2O where the H atoms are the 1H isotope. A few molecules, however, have deuterium (2H) substituted for 1H. This is because 99.989% of all hydrogen atoms on Earth are 1H atoms. That is, the abundance of 1H atoms is 99.989%.

Isotopic Masses and the Mass Defect  Actual masses of atoms

are always less than the sum of the masses of the subatomic particles composing that atom. This is called the mass defect, and the reason for it is discussed in Chapter 20.

Percent abundance 5

number of atoms of a given isotope 3 100% (2.1) total number of atoms of all isotopes of that element

The remainder of naturally occurring hydrogen is deuterium, whose abundance is only 0.012%. Tritium, the radioactive 3H isotope, occurs naturally in only trace amounts. Consider the two isotopes of chlorine. The chlorine-35 isotope has an abundance of 75.76%; the abundance of chlorine-37 is 24.24%. Among 10,000 chlorine atoms from an average natural sample, 7576 are chlorine-35 atoms and 2424 of them are chlorine-37 atoms.

Acceleration

Deflection

Heavy ions are deflected too little.

e−e−e− e−e−e− e−e−e−

Gas inlet 1

−

Repeller Electron trap plate 2

Analysis

Magnet

Electron gun

+

Detection

Accelerating plates 3

20Ne+

Magnet 4

Light ions are deflected too much.

1 A sample is introduced as a 4 This chamber is in a magnetic vapor into the ionization chamber. field, which is perpendicular to the direction of the beam of 2 There it is bombarded with highenergy electrons that strip charged particles. 5 electrons from the atoms or The magnetic field causes the molecules of the sample. beam to curve. The radius of curvature depends on the mass 3 The resulting positive particles are accelerated by a series of and charge of the particles (as well negatively charged accelerator as the accelerating voltage and plates into an analyzing chamber. strength of the magnetic field).

To mass analyzer

22Ne+

To vacuum pump 5

21Ne+

Detector

Here, particles of 21Ne+ are focused on the detector, whereas beams of ions of 20Ne+ and 22Ne+ (of lighter or heavier mass) experience greater and lesser curvature, respectively, and so fail to be detected.

A mass spectrum is a plot of the relative abundance of the charged particles versus the ratio of mass/charge (m/Z). Relative Abundance

Vaporization Ionization

100 80 60 40 20 0

20

21

22

m/Z

By changing the magnetic field, charged particles of different masses can be focused on the detector to generate the observed spectrum.

Figure 2.3  Mass spectrometer.  A mass spectrometer separates ions of different mass and charge in a gaseous sample of ions. The instrument allows the researcher to determine the accurate mass of each ion.

68

Chapter 2 / Atoms, Molecules, and Ions

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Atomic Weight Every sample of chlorine has some atoms with an atomic mass of 34.96885 and others with an atomic mass of 36.96590. The atomic weight of the element, the average mass of a representative sample of chlorine atoms, is somewhere between these values. For chlorine for example, the atomic weight is 35.45. If isotope masses and abundances are known, the atomic weight of an element can be calculated using Equation 2.2.  % abundance isotope 1  Atomic weight 5   (mass of isotope 1)  100



 % abundance isotope 2  1  (mass of isotope 2) 1 . . .  100

For chlorine with two isotopes (35Cl, abundance = 75.76%; 24.24%),



(2.2)

Cl, abundance =

37

 75.76   24.24  Atomic weight 5  × 34.96885 1  × 36.96590 5 35.45  100   100 

Equation 2.2 gives an average mass, weighted by the abundance of each isotope for the element. As illustrated by the data in Table 2.2, the atomic weight of an element is often close to the mass of the most abundant isotope or isotopes. For each stable element the atomic weight is given in the periodic table. The atomic weight is shown for a couple of unstable (radioactive) elements, but more often, the mass number of the most stable isotope is given in parentheses. For ­example, the isotopic masses and abundances of the radioactive element uranium (U) are known, so an atomic weight of 238.03 is listed on the periodic table. On the other hand, no atomic weight is shown for neptunium (Np), but the mass number of its most ­stable isotope, 237, is given in parentheses.

Table 2.2 Element Hydrogen

Boron Neon**

Magnesium

Isotope Abundance and Atomic Weight

Symbol H

Atomic Weight

Mass Number

Isotopic Mass

Natural Abundance (%)

 1.008

 1

 1.0078

99.989

D*

 2

 2.0141

 0.012

T†

 3

 3.0160

 0

B Ne

Mg

10.81 20.180

24.305

10

10.0129

19.9

11

11.0093

80.1

20

19.9924

90.48

21

20.9938

 0.27

22

21.9914

 9.25

24

23.9850

78.99

25

24.9858

10.00

26

25.9826

11.01

*D = deuterium; †T = tritium, radioactive; **See the mass spectrum of Ne in Figure 2.3.



2.2  Atomic Weight

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69

A Closer Look

Isotopic Abundances and Atomic Weights Accurate atomic weights for elements are of great importance for quantitative chemistry. The international body that establishes the values for atomic weights is the ­Commission on Isotopic Abundances and Atomic Weights (CIAAW) of the International Union of Pure and Applied Chemistry (IUPAC). Modern mass spectrometers can determine isotopic atomic masses out to nine or more decimal places. For elements with only one naturally occurring isotope, the atomic weights are therefore known to nine or more decimal places. For example, fluorine has only one naturally occurring isotope, 19F, and has an atomic weight of 18.998403162. Most elements, however, have more than one isotope. Their atomic weights depend on both the isotopic masses and the isotopic abundances. In introductory chemistry courses, it is often assumed that the different isotopes of an element are uniformly distributed on the Earth. However, this is not strictly true. Samples of an element’s atoms from different sources may have different isotopic abundances. These differences result in the isotopic abundances being known to fewer significant figures than isotopic masses, which limits the precision of atomic weights. For example, the current value for the atomic weight of copper is 63.546, a value that only goes to the third decimal place. This result follows the general rule for measured values that the final decimal place has some uncertainty. For some elements, the CIAAW has determined that the variation in the abundances of the elements is so large that the atomic weight is best represented as a range of values (Table). In such cases, the atomic weight of the element calculated for normal materials would vary, depending upon the source of the material, but would be somewhere within that range.

Using ranges for atomic weights, however, can make learning chemistry more complicated than desired. To avoid this, the periodic tables in this text show the conventional value for these elements that is often used for samples from an unspecified source.

Elements with Defined Atomic Weight Ranges

Element

Atomic Weight Range

Conventional Value

hydrogen

1.00784–1.00811

1.008

lithium

6.938–6.997

6.94

boron

10.806–10.821

10.81

carbon

12.0096–12.0116

12.011

nitrogen

14.00643–14.00728

14.007

oxygen

15.99903–15.99977

15.999

magnesium

24.304–24.307

24.305

silicon

28.084–28.086

28.085

sulfur

32.059–32.076

32.06

chlorine

35.446–35.457

35.45

argon

39.792–39.963

39.95

bromine

79.901–79.907

79.904

thallium

204.382–204.385

204.38

lead

206.14–207.94

207.2

E xamp le 2.3

Calculating Atomic Weight from Isotope Abundance Problem  Bromine has two naturally occurring isotopes. One has a mass of 78.918 and © Charles D. Winters/Cengage

an abundance of 50.7%. The other isotope has a mass of 80.916 and an abundance of 49.3%. Calculate the atomic weight of bromine. Br2 vapor Br2 liquid

Elemental bromine.  Bromine is a deep orange-red, volatile liquid at room temperature. It consists of Br2 molecules in which two bromine atoms are chemically bonded together. There are two, stable, naturally occurring isotopes of bromine atoms: 79 Br (50.7% abundance) and 81 Br (49.3% abundance).

70

What Do You Know?  You know the mass and abundance of each of the two isotopes.

Strategy  The atomic weight of any element is the weighted average of the masses of the isotopes in a representative sample. Use Equation 2.2 to calculate the atomic weight. Solution Atomic weight of bromine = (50.7/100)(78.918) + (49.3/100)(80.916) =  79.9 

Think about Your Answer  You can also estimate the atomic weight from the data given. There are two isotopes, mass numbers of 79 and 81, in approximately equal abundance. From this, you would expect the average mass to be about 80, midway between the two mass numbers. The calculation bears this out.

Chapter 2 / Atoms, Molecules, and Ions

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Check Your Understanding Verify that the atomic weight of boron is 10.8, given the following information: 10

B mass = 10.0129; percent abundance = 19.9%

11

B mass = 11.0093; percent abundance = 80.1%

Calculating Isotopic Abundances Problem  Antimony, Sb, has two stable isotopes: 121Sb, atomic mass = 120.904, and Sb, atomic mass = 122.904. What are the relative abundances of these isotopes?

123

What Do You Know?  You know the masses of the two isotopes of the element and know that their weighted average, the atomic weight, is 121.76 (see the periodic table). Strategy  To calculate the abundances recognize there are two unknown but related quantities, the fractional abundances of 121Sb and 123Sb (where the fractional abundance of an isotope is the percent abundance of the isotope divided by 100). These are related by Equation 2.2 and by an equation that shows the sum of the two fractional abundances must equal 1. These two equations can be solved for the two unknown fractional abundances.

© Charles D. Winters/Cengage

Exam p le 2.4

A sample of the metalloid antimony.  The element has two stable isotopes, 121Sb and 123Sb.

Solution  Equation 2.2 can be written for antimony as follows. Atomic weight = 121.76 = ( fractional abundance of 121Sb)(120.904) +   (fractional abundance of 123Sb)(122.904) or 121.76 = x(120.904) + y(122.904) where x = fractional abundance of 121Sb and y = fractional abundance of 123Sb. You know that the sum of fractional abundances of the isotopes must equal 1 (x + y = 1). Because y = fractional abundance of

123

Sb = 1 − x, you can make a substitution for y.

121.76 = x(120.904) + (1 − x)(122.904) Expanding this equation, you have 121.76 = 120.904x + 122.904 − 122.904x Finally, solving for x, you find 121.76 − 122.904 = (120.904 − 122.904)x x = 0.572 The fractional abundance of 121Sb is 0.572 and its  percent abundance is 57.2%.  This means that the  percent abundance of 123Sb must be 42.8%. 

Think about Your Answer   You might have predicted that the lighter isotope (121Sb) must be more abundant because the atomic weight is closer to 121 than to 123.



2.2  Atomic Weight

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71

Check Your Understanding Neon has three stable isotopes, one with a small abundance. What are the abundances of the other two isotopes? 20

Ne, mass = 19.992440; percent abundance = ?

21

Ne, mass = 20.993846; percent abundance = 0.27%

22

Ne, mass = 21.991385; percent abundance = ?

Key Experiments The Nature of the Atom and Its Components The idea that atoms are the building blocks for matter was set on the right track by the English chemist John Dalton in the early 1800s, but little was known about atoms at that time and for a long time after. Dalton proposed that an atom was a “solid, massy, hard, impenetrable, moveable particle,” far

+ Slits to focus a

–

narrow beam of rays

Electrically charged deflection plates

+

–

+

+

Negative electrode

Positive electrodes accelerate electrons

A beam of electrons (cathode rays) is accelerated through two focusing slits.

–

Fluorescent sensitized – screen Magnetic field coil To vacuum pump perpendicular to electric field

When passing through an electric field the beam of electrons is deflected.

The experiment is arranged so that the electric field causes the beam of electrons to be deflected in one direction. The magnetic field deflects the beam in the opposite direction.

Figure 1  Cathode rays: Thomson’s experiment to measure the electron’s charge-to-mass ratio.  The second half of the ­nineteenth century saw a series of experiments involving cathode ray tubes. First described in 1869 by William Crookes (1832–1919), a ­cathode ray tube is an evacuated container with two electrodes. When a high voltage is applied, particles (cathode rays) flow from the negative electrode (the cathode) to the anode. These particles were deflected by electric and magnetic fields, and by balancing these effects, it was possible to determine their charge-to-mass

72

from the modern description of a nuclear atom with protons, neutrons, and electrons. Our current understanding required ingenious experiments, carried out in the late 1800s and early 1900s. This section describes the main ideas for a few of these experiments.

Electrically deflected electron beam Undeflected electron beam Magnetically deflected electron beam By balancing the effects of the electrical and magnetic fields, the charge-to-mass ratio of the electron can be determined.

­ niversity ratio (e/m). In 1897, J. J. Thomson (1856–1940) at the U of Cambridge in England estimated that these particles had about three orders of magnitude less mass than a hydrogen atom. They became known as electrons, a term already used to describe the smallest particle of electricity. Thomson reasoned that electrons must originate from the atoms of the cathode, and he speculated that an atom was a uniform sphere of positively charged matter in which negative electrons were embedded, a model that is now known to be incorrect.

Chapter 2 / Atoms, Molecules, and Ions

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particles rays Photographic film or phosphor screen

+

Lead block shield

particles, attracted to + plate

particles particles, attracted to – plate

–

Slit

Charged plates

Radioactive element Emitted radiation passes through oppositely charged plates.

Positive particles deflect toward the – plate, negative particles deflect toward the + plate, and neutral rays continue undeflected.

Identification of the radiation emanating from radioactive substances soon followed. Three types of radiation were observed and given the labels alpha, beta, and gamma. Charge-to-mass studies revealed that alpha rays are helium nuclei (He2+) and beta rays are electrons. Gamma rays have neither mass nor charge; they are a highly energetic form of electromagnetic radiation.

Figure 2 Radioactivity.  Evidence that atoms were made up from smaller particles was also inferred from the discovery of radioactivity. In 1896, Henri Becquerel (1852–1908) found that uranium emitted invisible rays that caused a covered photographic plate to darken. Marie Curie (1867–1934) invented the term radioactivity to describe this new phenomenon. She and her husband Pierre Curie (1859–1906) concluded that the observed radiation is the result of the disintegration of atoms.

Oil atomizer Light source to illuminate drops for viewing X-ray source A fine mist of oil drops is introduced into one chamber. The droplets fall one by one into the lower chamber under the force of gravity.

Positively charged plate

+

+

Telescope

–

– Negatively charged plate

Gas molecules in the bottom chamber are ionized (split into electrons and a positive fragment) by a beam of X-rays. The electrons adhere to the oil drops, some droplets having one electron, some two, and so on.

Figure 3  Millikan’s experiment to determine the electron charge.  Cathode ray experiments allow measurement of the charge-to-mass ratio of a charged particle, but not the charge or mass individually. In 1908, the U.S. physicist Robert Millikan (1868–1953), at the California Institute of Technology, conducted an experiment to measure the charge on the electron. In his experiment, tiny oil droplets were sprayed into a chamber and then subjected to X-rays, causing them to take on a negative charge. The drops could be



Voltage applied to plates

Oil droplets under observation

These negatively charged droplets continue to fall due to gravity. By carefully adjusting the voltage on the plates, the force of gravity on the droplet is exactly counterbalanced

by the attraction of the negative droplet to the upper, positively charged plate. Analysis of these forces leads to a value for the charge on the electron.

suspended in air if the force of gravity was balanced against an electric field, and from an analysis of these forces on the droplet the charge could be calculated. Millikan determined that the electronic charge was 1.592  × 10−19 coulombs (C), not far from ­today’s accepted value of 1.602  × 10−19 C. Millikan correctly assumed this was the fundamental unit of charge. Knowing this value and the charge-to-mass ratio determined by Thomson, the mass of an electron could be calculated.

2.2  Atomic Weight

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73

Beam of particles

Nucleus of gold atoms

Atoms in Electrons occupy gold foil space outside nucleus. Undeflected particles

Gold foil Deflected particles

particles

Some particles are deflected considerably.

A few particles collide head-on with nuclei and are deflected back toward the source.

Most particles pass straight through or are deflected very little.

Figure 4  Rutherford’s experiment to determine the structure of the atom.  Although scientists recognized that atoms were made up of smaller particles, it was not clear how these particles fit together. Around 1910, Ernest Rutherford (1871–1937) established the model accepted today. Rutherford interpreted an experiment conducted by two colleagues, Hans Geiger (1882–1945) and Ernest Marsden (1889–1970), in which they bombarded thin gold foil with α particles. Almost all the particles passed straight through the gold foil as if there was nothing there. However, a few α particles were deflected sideways and some even bounced right back. This experiment proved that an atom of gold is mostly empty space with a tiny nucleus at its center. The electrons surround the nucleus and account for most of the volume of the atom. Rutherford

Source of narrow beam of fast-moving particles

ZnS fluorescent screen

calculated that the central nucleus of an atom occupied only 1/10,000th of its volume. He also estimated that a gold nucleus had a positive charge of around 100 units and a radius of about 10−12 cm. (The values are now known to be 79+ for atomic charge and 10−13 cm for the radius.) The final piece of the picture of atomic structure was not established for another decade. Scientists knew that there had to be something else in the nucleus, and it had to be a heavy particle to account for the mass of the element. In 1932, the British physicist James Chadwick (1891–1974) found the missing particle. These particles, now known as neutrons, have no electric charge and a mass of 1.675  × 10−24  g, slightly greater than the mass of a proton.

2.3 The Periodic Table Goals for Section 2.3 • Know the terminology of the periodic table (periods, groups) and know how to use the information given in the periodic table.

• Recognize similarities and differences in properties of some of the common elements of a group.

Features of the Periodic Table The main organizational features of the periodic table are the following (Figure 2.4): •

74

Elements with similar chemical and physical properties lie in vertical columns called groups or families. The periodic table commonly used in the United States has groups numbered 1 through 8, with each number followed by the letter A or B. The A groups are often called the main group elements and the B groups are the transition elements. In other parts of the world, the groups are numbered 1–18. In this book, groups are referenced using the system commonly used in the United States, followed by the group number using the 1–18 system in parentheses. For example, the elements carbon, silicon, germanium, tin, lead, and flerovium comprise Group 4A (14).

Chapter 2 / Atoms, Molecules, and Ions

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A

Figure 2.4  Periods and groups in the periodic table.  ​An alternative to this labeling system numbers the groups from 1 to 18 going from left to right. This notation is generally used outside the United States.

A

1 2

3 4 5 6 7 8

B 3 4 5 6 7

8

1 2

Groups or Families

1 2 3 4 5 6 7

Main Group Metals Transition Metals

Metalloids Nonmetals

Periods

The horizontal rows of the table are called periods, and they are numbered beginning with 1 for the period containing only H and He. Currently, 118 elements are known, filling periods 1 through 7.

The periodic table can be divided into several regions according to the properties of the elements. In Figures 2.4 and 2.5 (and the table on the inside cover of this book), metals are on the left and are indicated in shades of blue. Nonmetals, on the right (with the exception of hydrogen), are indicated in orange. Metalloids, along the metal-nonmetal boundary, appear in green. Elements gradually become less metallic from left to right across a period. You are probably familiar with many properties of metals from your own experience. At room temperature and normal atmospheric pressure, metals are solids (except for mercury), can conduct electricity, are usually ductile (can be drawn into wires) and malleable (can be rolled into sheets), and can form alloys (mixtures of one or more metals with another metal). Iron (Fe) and aluminum (Al) are used in automobile parts because of their ductility, malleability, and low cost relative to other metals. Copper (Cu) is used in electric wiring because it conducts electricity better than most other metals. The nonmetals, which lie to the right of a diagonal line that stretches from B to Te in the periodic table, have a wide variety of properties. Some are solids (carbon, sulfur, phosphorus, and iodine). Ten elements are gases at room temperature (­hydrogen, oxygen, nitrogen, fluorine, chlorine, helium, neon, argon, krypton, and xenon). One nonmetal, bromine, is a liquid at room temperature. With the exception of carbon in the form of graphite, nonmetals do not conduct electricity, which is one of the main features that distinguishes them from metals. The elements along the diagonal line from boron (B) to tellurium (Te) have properties that make them difficult to classify as metals or nonmetals. Chemists call them metalloids or, sometimes, semimetals. Metalloids are elements that have some of the physical characteristics of a metal but some of the chemical characteristics of a nonmetal. This definition reflects the ambiguity in the behavior of these elements. Antimony (Sb), for example, conducts electricity just as well as many metals. Its chemistry, however, resembles that of phosphorus, a nonmetallic element. We include only B, Si, Ge, As, Sb, and Te in this category, but some chemists include one or more other elements. Some elements can exist in several different and distinct forms called allotropes, each of which has its own properties. For example, oxygen occurs as two allotropes in nature: O2 (usually referred to as oxygen) and O3 (called ozone). Humans will die without a sufficient source of O2 molecules to breathe. On the other hand, O3 is poisonous. Nonetheless, O3 plays an important role in the environment by blocking out some of the radiation that reaches the Earth from the sun.

Placing H in the Periodic Table  Where to place H? Hydrogen is a nonmetal, but periodic tables often show it in Group 1A (1) on the left side of the periodic table even though it is not a metal. However, in some of its reactions, it forms a 1+ ion like other members of Group 1A (1). For this reason, H is often placed in this group.

Forms of silicon

© Charles D. Winters/Cengage

•

Silicon—a metalloid.  Only six elements are generally classified as metalloids or semimetals. This photograph shows solid silicon in various forms, including a wafer that holds printed electronic circuits. Metalloids  For more information on which elements are considered metalloids, see the article “Which Elements are Metalloids?” by René E. Vernon, Journal of Chemical Education, 2013, 90, 1703–1707.

2.3  The Periodic Table

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75

76

Chapter 2 / Atoms, Molecules, and Ions

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H

20

(223)

Fr

87

132.91

Cs

55

85.468

Rb

37

39.098

21

Sc

89

(226)

Ra (227)

Ac

137.33 138.91

88

57

La

Ba

56

Y

88.906

Sr

39

87.62

38

40.078 44.956

Ca

19

Mg

24.305

K

5B (5)

6B (6)

7B (7) (8)

(10)

1B (11)

5

B

3A (13)

V

23

Zr

Nb

41

73

Ta

(267)

Rf

104 (268)

105

Db

178.49 180.95

72

Hf

91.224 92.906

40

47.867 50.942

22

Ti

Figure 2.5  The periodic table of the elements.

Th

Ce

25

Mn

Mo

Tc

75

Re

(98)

43

Pa 238.03

U

(237)

Np

93

92

Pm

61

(277)

108

Hs

91

231.04

Rh

77

Ir

Sm

Pu

(244)

94

29

Cu

Pd

Ag

47

79

Au

Eu

63

(281)

Ds

110

Am

(243)

95

Gd

31

81

80

113 112

Tb

65

(285)

Cn

Cm

(247)

96

(247)

Bk

97

Dy

Cf

(251)

98

7

N

5A (15)

P

15

33

As

83

82

115

Ho

67

(289)

Es

(252)

99

Er

Fm

(257)

100

9

F

7A (17)

35

Br

35.45

17

Cl

Te

I

53

85

Tm

69

(293)

Md

(258)

101

Yb

He

Xe

Lu

71

(294)

118

Og

(222)

86

Rn

131.29

54

83.798

36

Kr

39.95

18

Ar

20.180

10

Ne

4.0026

2

8A (18)

No

(259)

102

Lr

(262)

103

173.05 174.97

70

(294)

Ts

117

116

Lv

(210)

At (209)

84

Po

127.60 126.90

52

78.971 79.904

34

Se

32.06

S

16

15.999 18.998

8

O

6A (16)

167.26 168.93

68

(290)

Mc

114

Fl

208.98

Bi 207.2

Pb

118.71 121.76

Sb

51

Sn

50

72.630 74.922

32

Ge

28.085 30.974

14

Si

12.011 14.007

6

C

4A (14)

162.50 164.93

66

(286)

Nh

200.59 204.38

Tl

Hg

112.41 114.82

In

49

48

Cd

69.723

65.38

Ga

30

Zn

26.982

2B (12)

157.25 158.93

64

(282)

Rg

111

195.08 196.97

78

Pt

106.42 107.87

46

58.693 63.546

28

Ni

150.36 151.96

62

(276)

Mt

109

190.23 192.22

76

Os

(145)

Nd

Ru

45

101.07 102.91

44

144.24

Pr

60

(270)

107

Bh

27

Co

55.845 58.933

26

Fe

140.91

59

(269)

Sg

106

183.84 186.21

74

W

95.95

42

51.996 54.938

24

Cr

(9)

Al

13

4B (4)

Atomic weight

Symbol

Atomic number

12 8B

238.03

U

92

10.81 3B (3)

NONMETALS

METALLOIDS

TRANSITION METALS

MAIN GROUP METALS

9.0122

22.990

Na

11

6.94

Be

4

3

Li

2A (2)

1A (1)

1.008

1

Note: Atomic weights are 58 IUPAC values. For elements Lanthanides for which IUPAC recommends 140.12 ranges of atomic weights, conventional values are shown. 90 Numbers in parentheses are Actinides mass numbers of the most 232.04 stable isotope of an element.

7

6

5

4

3

2

1

Martyn F. Chillmaid/Science Source

Group 1A (1) metals are soft and some, like sodium and potassium, can be cut with a knife.

© Charles D. Winters/Cengage

Group 1A (1) metals react vigorously with water to give hydrogen gas and an alkaline solution of the metal hydroxide.

Figure 2.6  Properties of the alkali metals.

A Brief Overview of the Periodic Table and the Chemical Elements Elements in the leftmost column, Group 1A (1), are known as the alkali metals (except H). The word alkali comes from Arabic. Ancient Arabian chemists discovered that the ashes of certain plants, which they called al-qali, gave water solutions that felt slippery and burned the skin. Chemists now know these ashes contain compounds of Group 1A (1) elements that produce alkaline (basic) solutions. All the alkali metals are solids at room temperature and reactive. For example, they react with water to produce hydrogen and alkaline solutions (Figure  2.6). ­Because of their reactivity, these metals are only found in nature combined in compounds (such as NaCl), never as free elements. Group 2A (2) is also composed entirely of metals that occur naturally only in compounds. Except for beryllium (Be), these elements react with water to produce alkaline solutions, and most of their oxides (such as lime, CaO) form alkaline solutions; hence, they are known as the alkaline earth metals. Magnesium (Mg) and calcium (Ca) are the seventh and fifth most abundant elements in the Earth’s crust, respectively (Table 2.3). Calcium, one of the important elements in teeth and bones, occurs naturally in vast limestone deposits. Calcium carbonate (CaCO3) is the chief constituent of limestone as well as corals, sea shells, marble, and chalk. Radium (Ra), the heaviest alkaline earth element, is radioactive.

Table 2.3

The 10 Most Abundant Elements in the Earth’s Crust

Rank

Element

Abundance (ppm)*

 1

Oxygen

474,000

 2

Silicon

277,000

 3

Aluminum

 82,000

 4

Iron

 41,000

 5

Calcium

 41,000

 6

Sodium

 23,000

 7

Magnesium

 23,000

 8

Potassium

 21,000

 9

Titanium

  5600

10

Hydrogen

  1520

*ppm = parts per million = g per 1000 kg.

2.3  The Periodic Table

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77

A Closer Look

Mendeleev and the Periodic Table

John C. Kotz

Statue of Dmitri Mendeleev and a periodic table.  This statue and mural are at the Institute of Metrology in St. Petersburg, Russia.

78

column where he believed an unknown element should exist. He deduced that these spaces would be filled by undiscovered elements. For example, he left a space between Si (silicon) and Sn (tin) in Group 4A (14) for an element he called eka-silicon. Based on the progression of properties in this group, Mendeleev was able to predict the properties of the missing element. With the discovery of germanium (Ge) in 1886, Mendeleev’s prediction was confirmed. In Mendeleev’s table the elements were ordered by increasing atomic weight. A glance at a modern table, however, shows that, if listed in order of increasing atomic weight, three pairs of elements (Ni and Co, Ar and K, and Te and I) would be out of order. Mendeleev assumed the atomic weights known at that time were i­naccurate—not a bad assumption based on the analytical methods then in use. In fact, his order is correct and his assumption that element

“Key Experiments”) and examined the X-rays emitted in the process. Moseley ­realized the wavelength of the X-rays emitted by a given element was related in a precise manner to the positive charge in the nucleus of the element and that this provided a way to experimentally determine the atomic number of a given element. Indeed, once atomic numbers could be determined, chemists recognized that organizing the elements in a table by increasing atomic number corrected the ­ inconsistencies in Mendeleev’s table. The law of chemical periodicity now states that the properties of the elements are periodic functions of atomic number. References For more on the periodic table, see: 1. J. Emsley: Nature’s Building Blocks—An A–Z Guide to the Elements, Second Edition, New York, Oxford University Press, 2011. 2. E. Scerri, The Periodic Table, Second Edition, New York, Oxford University Press, 2020.

Chapter 2 / Atoms, Molecules, and Ions

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Photos: © Charles D. Winters/Cengage

RE IH E N

Although the arrangement of elSodium Germanium Iodine Copper ements in the periodic table is now understood on the basis of atomic structure, the table was originally developed from many experimental observations of the chemical and physical properties of elements and is the result of the ideas of a number TABELLE I I . of chemists in the eighteenth GRUPPE I. G RUPPE II. G RUPPE III. G RUPPE IV. G RUPPE V. G RUPPE VI. G RUPPE VII. GRU PPE VI I I . and nineteenth centuries. — — — RH 4 RH 3 RH 2 RH — In 1869, at the University of R2O RO R2O3 R 2O 7 RO 4 RO 2 R 2O 5 RO 3 St.  Petersburg in Russia, Dmitri 1 H=1 ­Ivanovich Mendeleev (1834–1907) 2 Li = 7 Be = 9,4 B = 11 C = 12 N = 14 O = 16 F = 19 3 Na = 23 Mg = 24 Al = 27,3 Si = 28 P = 31 S = 32 Cl = 35,5 wrote a textbook on chemistry. As 4 K = 39 Fe = 56, Co = 59, Ca = 40 — = 44 Ti = 48 V = 51 Cr = 52 Mn = 55 he pondered the chemical and Ni = 59, Cu = 63. physical properties of the elements, 5 (Cu = 63) Zn = 65 — = 68 — = 72 As = 75 Se = 78 Br = 80 6 Rb = 85 Ru = 104, Rh = 104, Sr = 87 ?Yt = 88 Zr = 90 Nb = 94 Mo = 96 — = 100 he realized that, if the elements Pd = 106, Ag = 108. were arranged in order of increasing 7 (Ag = 108) Cd = 112 In = 113 Sn = 118 Sb = 122 Te = 125 J = 127 atomic weight, elements with simi8 Cs = 133 ———— Ba = 137 ?Di = 138 ?Ce = 140 — — — lar properties appeared in a regular 9 ( —) — — — — — — 10 — Os = 195, Ir = 197, — ?Er = 178 ?La = 180 Ta = 182 W = 184 — pattern. That is, he saw a ­periodicity Pt = 198, Au = 199. or periodic repetition of the proper11 (Au = 199) Hg = 200 Tl = 204 Pb = 207 Bi = 208 — — ties of elements. Mendeleev orga12 — ———— — — Th = 231 — U = 240 — nized the known elements into a table by lining them up in horizonThe original Mendeleev table showing the places he left for as yet-undiscovered elements. tal rows in order of increasing atomic weight. When he came to an ele- started a new row. As more elements were properties were a function of their atomic ment with properties similar to one already added to the table, new rows were begun, weight was wrong. in the row, he started a new row. For ex- and elements with similar properties (such In 1913, H. G. J. Moseley (1887–1915), ample, the elements Li, Be, B, C, N, O, and as Li, Na, and K) were placed in the same a young English scientist working with F were in a row. Sodium was the next ele- vertical column. Ernest Rutherford (1871–1937), bomment then known; because its properties An important feature of Mendeleev’s barded many different metals with elecclosely resembled those of Li, Mendeleev table was that he left an empty space in a trons in a cathode-ray tube (see page 72,

© Charles D. Winters/Cengage

Figure 2.7  Liquid gallium.  Bromine and mercury are the only elements that are liquids at room temperature and usual atmospheric pressures. Gallium and cesium melt slightly above room temperature.

Gallium melts (melting point = 29.8 °C) when held in the hand.

Group 3A (13) includes a metalloid—boron—and metals—aluminum, gallium (Figure 2.7), indium, and thallium. As a metalloid, boron (B) has a different chemistry than the other elements in its group. Nonetheless, all form compounds with analogous formulas, such as BCl3 and AlCl3, and this similarity marks them as members of the same periodic group. Boron occurs in the mineral borax, a compound used as a cleaning agent, antiseptic, and flux for metal work. Aluminum (Al) is the most abundant metal in the Earth’s crust at 8.2% by mass. It is exceeded in abundance only by the nonmetal oxygen and metalloid silicon, and is usually found in minerals and clays. In Group 4A (14), there is a nonmetal, carbon (C), two metalloids, silicon (Si) and germanium (Ge), and two metals, tin (Sn) and lead (Pb). Because of the change from nonmetallic to metallic behavior, the properties of the elements of this group have more variation than in most others. Nonetheless, there are similarities. For example, these elements form compounds with analogous formulas such as CO2, SiO2, GeO2, and PbO2. Carbon has many allotropes, the best known of which are graphite and diamond. Graphite consists of flat sheets in which each carbon atom is connected to three others (Figure 2.8a). Because the sheets of carbon atoms cling only weakly to one another, the layers slip easily over each other. This explains why graphite is soft, a good lubricant, and used in pencil lead. [Pencil “lead” is not the element lead (Pb) but a composite of clay and graphite that leaves a trail of graphite on the page as you write.] In diamond, each carbon atom is connected to four others at the corners of a tetrahedron, and this extends throughout the solid (Figure  2.8b). This structure causes diamonds to be extremely hard, denser than graphite (d = 3.51 g/cm3 for diamond versus d = 2.22 g/cm3 for graphite), and chemically less reactive. Because diamonds are both hard and excellent conductors of heat, they are used on the tips of metal- and rock-cutting tools. In the late 1980s, another form of carbon was identified as a component of black soot, the stuff that collects when carbon-containing materials are burned in a deficiency of oxygen. This substance is made up of molecules with 60 carbon atoms arranged as a spherical cage (Figure 2.8c). The surface is made up of five- and sixmember rings and resembles a hollow soccer ball. The shape also reminded its discoverers of an architectural dome conceived by the U.S. philosopher and engineer, R. Buckminster Fuller (1895–1983). This led to the official name of the allotrope, buckminsterfullerene, although chemists often call these molecules “buckyballs.” Carbon’s chemistry is unique because of its ability to bond to other carbon ­atoms to form chains and rings to which atoms of other elements can be attached. Most of the millions of chemical compounds known contain carbon.

Elements 113–118.  The heaviest elements in Groups 3A–8A (13–18) with atomic numbers 113–118 are all laboratorycreated and have very short existences. Their properties and reactions are not well known so they have been omitted from the overview of group properties in this chapter.

2.3  The Periodic Table

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79

© Charles D. Winters/Cengage

© Charles D. Winters/Cengage

Mark A. Schneider/Science Source

(a) Graphite. Graphite consists of layers of carbon atoms.

Each six-member ring shares an edge with three other six-member rings and three five-member rings.

Each C atom is connected tetrahedrally to four other C atoms.

Each carbon atom is linked to three others to form a sheet of six-member, hexagonal rings.

(b) Diamond. In diamond the carbon atoms are also arranged in six-member rings, but the rings are not planar.

Figure 2.8  The allotropes of carbon.

(c) Buckyballs. A member of the family called buckminsterfullerenes, C60 is an allotrope of carbon. Sixty carbon atoms are arranged in a spherical cage that resembles a hollow soccer ball. Chemists call this molecule a buckyball. C60 is a black powder; it is shown here in the tip of a pointed glass tube.

Oxides of silicon are the basis of many minerals such as clay, quartz, and beautiful gemstones like amethyst. Tin and lead have been known for centuries because they are easily smelted from their ores. Tin alloyed with copper makes bronze, which was used in ancient times in utensils and weapons. Lead has been used in water pipes and paint, even though the element is toxic to humans. Like Group 4A (14), Group 5A (15) contains nonmetals (N and P), metalloids (As and Sb), and a metal (Bi). In spite of these variations, they form analogous compounds such as the oxides N2O5, P4O10, and As2O5. Nitrogen occurs naturally in the form of the diatomic molecule N2 (Figure 2.9) and makes up about threefourths of Earth’s atmosphere. It is also found in biochemically important ­substances such as chlorophyll, proteins, and DNA. Scientists have long studied ways to make compounds from atmospheric nitrogen, a process called nitrogen fixation. Nature accomplishes this easily in some prokaryotic organisms, but extreme conditions (high temperatures, for example) must be used in the laboratory and industry for N2 to react with other elements (such as H2 to make ammonia, NH3, which is widely used as a fertilizer). Phosphorus is also essential to life. It is an important constituent in bones, teeth, and DNA. The element glows in the dark if it is in the air (owing to its reaction with O2), and its name is based on Greek words meaning “light-bearing.” This  element has several allotropes, the most important being white and red

H2

N2

O2 O3

F2 Cl 2 Br2 I2

Figure 2.9  Elements that exist as diatomic or triatomic molecules.  Seven of the known

elements exist as diatomic, or two-atom, molecules. Oxygen has an additional allotrope, ozone, with three O atoms in each molecule.

80

Chapter 2 / Atoms, Molecules, and Ions

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Iodine, I2 © Charles D. Winters/Cengage

Bromine, Br2



© Charles D. Winters/Cengage

phosphorus. White phosphorus (composed of P4 molecules) ignites spontaneously in air, so it is normally stored under water. When it reacts with air, it forms P4O10, which can react with water to form phosphoric acid (H3PO4), a compound used in food products such as soft drinks. Red phosphorus is used in the striking strips on match books. When a match is struck, potassium chlorate in the match head mixes with some red phosphorus on the striking strip, and the friction is enough to ignite the mixture. In Group 6A (16) there is again a variation of properties. Oxygen, sulfur, and ­selenium are nonmetals, tellurium is a metalloid, and polonium, a radioactive element discovered in 1898 by Marie and Pierre Curie (see “A Closer Look: ­ ­Marie ­Curie”), is a metal. Nonetheless, there is a family resemblance in their chemistries. All of oxygen’s fellow group members form oxygen-containing compounds (SO2, SeO2, and TeO2), and all form sodium-containing compounds (Na2O, Na2S, Na2Se, and Na2Te). Oxygen, which constitutes about 20% of Earth’s atmosphere and which combines readily with most other elements, is at the top of Group 6A (16). Most of the energy that powers life on Earth is derived from reactions in which oxygen combines with other substances. Sulfur has been known in elemental form since ancient times as brimstone or “burning stone” (Figure 2.10). Sulfur, selenium, and tellurium are often referred to collectively as chalcogens (from the Greek word, khalkos, for copper) because most copper ores contain these elements. Their compounds can be foul-smelling and poisonous; nevertheless, sulfur and selenium are essential components of the human diet. By far the most important compound of sulfur is sulfuric acid (H2SO4), which is manufactured in larger amounts than any other compound. As in Group 5A (15), the second- and third-period elements of Group 6A (16) have different structures. Like nitrogen, oxygen is also a diatomic molecule (see Figure 2.9). Unlike nitrogen, however, oxygen has an allotrope, the triatomic molecule ozone, O3. Sulfur, which can be found in nature as a yellow solid, has many allotropes, the most common of which consists of eight-member, crown-shaped rings of sulfur atoms (see Figure 2.10). The Group 7A (17) elements—fluorine, chlorine, bromine, iodine, and radioactive astatine—are all nonmetals and all exist as diatomic molecules. At room temperature fluorine (F2) and chlorine (Cl2) are gases. Bromine (Br2) is a liquid and iodine (I2) is a solid, but bromine and iodine vapor are clearly visible over the liquid or solid (Figure 2.11).

Figure 2.10 Sulfur. The most

common allotrope of sulfur consists of S atoms arranged in eightmember, crown-shaped rings.

Special Group Names Some groups have widely used common names.

Group 1A Group 2A metals Group 7A Group 8A

(1): Alkali metals (2): Alkaline earth (17): Halogens (18): Noble gases

Figure 2.11 Bromine and iodine.  These and other

Group 7A (17) elements are commonly called halogens.

2.3  The Periodic Table

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81

Marie Curie, originally known as ­Maria Skłodowska, was born in Russiancontrolled Warsaw, Poland in 1867. The daughter of school teachers, she hoped to attend college, but at that time the universities in Poland (and many other European countries) refused to admit women. The Sorbonne in Paris, France did admit women, but she did not have enough money to attend. She and her sister Bronya, however, worked out a plan to help each other. Maria would work as a tutor and governess to support Bronya while Bronya studied medicine in Paris, and then Bronya would work to support Maria. Maria also attended what was called a flying, or floating university in Warsaw — an illegal underground movement in which classes were taught informally in private homes and other locations. Of particular importance to Maria was a secret Polish laboratory at the Museum of Industry and Agriculture, where she learned techniques of chemical analysis. In 1891, Bronya fulfilled her promise, and Maria enrolled at the Sorbonne as Marie. She lived frugally, spending almost nothing on luxuries or sometimes even on food. On several occasions, she fainted because she worked constantly and had such a meager diet. She finished first in her physics masters program in the summer of 1893 and second in her mathematics masters the following year. She then decided to do advanced work in physics. In 1896, Henri Becquerel accidentally discovered radioactivity when he placed a uranium compound in a drawer with photographic plates that were covered with black paper. When he developed the plates, an image was produced even though the plates had not been exposed to light. Becquerel correctly surmised that the uranium was emitting some type of radiation. (Abel Niepce de Saint Victor carried out similar experiments in 1857 and came to the same conclusion, but his results were largely ignored.) Marie Curie was considering possible topics for her doctoral thesis at the Sorbonne and had read about Becquerel’s work. She was interested in whether other elements would emit such radiation. Because her French physicist husband, Pierre Curie (1859–1906), was working at the École Supérieure de Physique et de Chimie

82

Hulton Archive/Stringer/Getty Images

A Closer Look

Marie Curie (1867–1934)

Marie and Pierre Curie in their laboratory. Industrielles de la Ville de Paris (ESPCI), she was allowed to perform her research in a laboratory there. After Pierre developed a highly sensitive method to detect radiation from a radioactive source, Marie tested every substance she could find for radioactivity. She determined that the degree to which uranium ore was radioactive—a term she invented to describe the p ­ henomenon— depended on the percent of uranium ­present, which confirmed ­Becquerel’s hypothesis that uranium metal itself was radioactive. When she tested pitchblende, a common ore containing uranium and other metals, she was astonished to find that it was even more radioactive than pure uranium. There was only one explanation: pitchblende contained at least one element more radioactive than uranium. In 1898, using chemical analysis techniques, the Curies separated pitchblende into fractions containing different materials. One fraction, which contained mainly the non-radioactive element bismuth, was nonetheless very radioactive; another contained mostly barium and was also very radioactive. The Curies proposed that each of these fractions contained a new element. They named the element in the bismuth fraction polonium after Marie’s homeland of Poland and the element in the barium fraction radium. Working in a run-down shed at the ESPCI, the Curies started with tons of pitchblende and

performed thousands of steps to successfully isolate about a tenth of a gram of pure radium chloride in 1902. In 1903, Marie earned her doctorate in physics at the Sorbonne, becoming the first women in France to earn this degree. Also in 1903, Henri Becquerel and Marie and Pierre Curie shared the Nobel Prize in physics for their discovery of “spontaneous radioactivity.” Initially, the Nobel committee nominated only Becquerel and Pierre for this honor, but Pierre declared that he would not accept the prize unless Marie was also included. Marie was the first woman to win a Nobel Prize. In 1906, Pierre was killed when he fell beneath the wheels of a horse-drawn vehicle, and Marie was hired to take his place at the Sorbonne, where he had moved in 1900. She was the first woman ever hired to teach at that university. In 1910, working with ­André-Louis Debierne, Marie isolated pure metallic radium. Despite her accomplishments, the French Academy of Sciences chose to elect Édouard Branly instead of Marie Curie to fill a vacancy in the organization in 1911. This was, in part, because of her Polish ancestry, a false rumor that she was Jewish, and being a woman. In 1911–the same year she was rejected by the Academy–Marie was awarded a second Nobel Prize, this time in chemistry for the discovery of radium and polonium. As of 2021, only four people and two organizations have ever won multiple Nobel Prizes. The Radium Institute in Paris, established in 1909, opened in 1914 with ­Marie as the Director of the Physics and Chemistry Laboratory. During World War I, Marie learned that the entire French Army had only one X-ray station. She proposed and helped ­establish mobile X-ray units and then taught herself to drive and operated one of the 20 mobile units, nicknamed “petite Curies,” used during the war. Approximately one  ­million soldiers were treated with these m ­ ­ obile units. ­Marie Curie died at the age of 66 in 1934 from aplastic pernicious a­nemia, a condition she developed from years of ­radiation exposure. A unit of radioactivity ­(curie, Ci) and an element (curium, Cm) are named in honor of Marie and Pierre Curie. One of their daughters, Irène, married ­Frédéric Joliot, and they shared the 1935 Nobel Prize in chemistry for their discovery of induced radioactivity.

Chapter 2 / Atoms, Molecules, and Ions

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The Group 7A (17) elements are among the most reactive of all elements, and all combine violently with alkali metals to form salts such as table salt, NaCl. The name for this group, the halogens, comes from the Greek words hals, meaning “salt,” and genes, for “forming.” The Group 8A (18) elements—helium, neon, argon, krypton, xenon, and radioactive radon—are all nonmetals and are the least reactive elements. All are gases, and none is abundant on Earth or in the Earth’s atmosphere (although argon is the third most abundant gas in dry air at 0.9%). Because of this, they were not discovered until the end of the nineteenth century (see page 118). Their general lack of reactivity is reflected in the common name for this group, the noble gases. Just as the Western nobility class at this time was considered to be generally aloof and hesitant to mix with commoners, these elements are very unreactive toward other elements. Helium, the second most abundant element in the universe after hydrogen, was detected in the sun in 1868 by analysis of the solar spectrum but was not found on Earth until 1895. It is now widely used, with a worldwide production of the gas in 2019 of about 160 billion liters. The biggest single use of helium is to cool the magnets found in magnetic resonance imaging (MRI) units in hospitals and nuclear magnetic resonance (NMR) spectrometers in research laboratories (Figure 2.12). These magnets must be cooled with liquid helium to 4 K because, at this extremely low temperature, the magnets are superconductors of electricity. They can then generate the high magnetic fields needed to produce an image of your body. In addition, helium gas is used to fill weather balloons (and party balloons) and in the semiconductor industry. The United States supplies most of the helium, but there are periodic shortages that seriously disrupt commerce and research. Stretching between Groups 2A and 3A (2 and 13) in the periodic table is a series of elements called the transition elements. These fill the B-groups (3–12) in the fourth through the seventh periods in the center of the periodic table. All are metals, and 13 of them are in the top 30 elements in terms of abundance in the Earth’s crust. Most occur naturally in combination with other elements, but a few—copper (Cu), silver (Ag), gold (Au), and platinum (Pt)—can be found in nature as pure elements. Virtually all of the transition elements have commercial uses. They are used as structural materials (iron, titanium, chromium, copper); in paints (titanium, ­chromium); in the catalytic converters in automobile exhaust systems (platinum and rhodium); in coins (copper, nickel, zinc); and in batteries (manganese, nickel, zinc, cadmium, mercury). Two rows at the bottom of the table accommodate the lanthanides [the series of elements between lanthanum (Z = 57) and hafnium (Z = 72)] and the actinides [the series of elements between actinium (Z = 89) and rutherfordium (Z = 104)]. The lanthanides are often referred to as rare earth elements (Figure 2.13). In fact, they are not very rare but are geologically widely dispersed. In spite of the difficulty in mining rare earth–containing minerals, they have become very important commercially. They are used in magnets, LCD screens, smartphones, computers, hybrid car batteries, wind turbines, and powders used to polish glass. Minerals containing rare earth elements are mined largely in China, and there is concern that a worldwide shortage looms.

Figure 2.12  Helium, a noble gas, and MRI units.  The magnets

of MRI units need to be cooled to 4 K with liquid helium to generate the high magnetic field required. This is the largest use of this noble gas.

2.3  The Periodic Table

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(a) (b)

Figure 2.13  The rare earth elements.  ​(a) A sample of the rare earth element terbium. (b) A mine for ores containing rare earth elements. The rare earth elements are essential for many modern devices. Obtaining sufficient amounts has environmental, economic, and geopolitical consequences.

2.4 Molecules: Formulas, Models, and Names Goals for Section 2.4 • Recognize and interpret molecular formulas, condensed formulas, and structural formulas.

• Recognize and interpret different types of molecular models. • Remember formulas and names of common molecular compounds. • Name and write formulas for binary molecular compounds. In Chapter 1, you learned that some elements (such as oxygen, O2) and many compounds (such as water, H2O) are composed of particles called molecules. A molecule is the smallest identifiable unit into which some pure substances like sugar and water can be divided and still retain the composition and chemical properties of the substance. Such substances are composed of identical molecules consisting of two or more atoms bound firmly together. In the reaction below (and in Figure 2.14), molecules of sulfur, S8, combine with molecules of oxygen, O2, to produce molecules of the compound sulfur dioxide, SO2. S8(s) + 8 O2(g) → 8 SO2(g) sulfur + oxygen → sulfur dioxide

The composition of each element and compound in the chemical change (or chemical reaction) is represented by a formula that indicates the types and numbers of atoms present in one molecule of the substance. For example, one molecule of SO2 is composed of one S atom and two O atoms.

Formulas There is often more than one way to write the formula of a molecular compound. For example, the formula of ethanol (also called ethyl alcohol) can be represented as C2H6O (Figure 2.15). This molecular formula describes the composition of ethanol molecules—two carbon atoms, six hydrogen atoms, and one oxygen atom per

84

Chapter 2 / Atoms, Molecules, and Ions

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Sulfur dioxide, SO2 (g)

Oxygen, O2 (g)

Photos: © Charles D. Winters/Cengage

Sulfur, S8 (s)

Figure 2.14  The reaction of the molecular elements sulfur and oxygen to give the molecular compound sulfur dioxide.

molecule—but it gives no structural information. Structural information—how the atoms are connected and how the molecule fills space—is important because it helps you to understand how a molecule can interact with other molecules. To provide some structural information, it is useful to write a condensed ­formula, which indicates how certain atoms are grouped together. For example, the condensed formula of ethanol, CH3CH2OH (Figure 2.15), indicates that the molecule consists of three groups: a CH3 group, a CH2 group, and an OH group. Writing the formula as CH3CH2OH also shows that the compound is not dimethyl ether, CH3OCH3, a compound with the same molecular formula but with a different ­structure and distinctly different properties. That ethanol and dimethyl ether are different molecules is also clear from their structural formulas (Figure 2.15). This type of formula gives an even higher level of structural detail, showing how all of the atoms are attached within a molecule. The lines between atoms represent the chemical bonds that hold atoms together in the molecule.

Writing Formulas  When writing

molecular formulas of organic compounds (compounds with C, H, and other elements) the convention is to write C first, then H, and finally other elements in alphabetical order. For example, acrylonitrile, a compound used to make consumer plastics, has the condensed formula CH2CHCN. Its molecular formula would be C3H3N.

Molecular Models The physical and chemical properties of compounds are often closely related to their structures (which is why you will see so many molecular models in this book). For example, two well-known features of ice are related to its underlying molecular structure (Figure 2.16). The first is the shape of ice crystals: The sixfold symmetry of macroscopic ice crystals also appears at the particulate level in the form of six-sided rings of hydrogen and oxygen atoms. The second is water’s unusual property of

NAME

Ethanol

MOLECULAR FORMULA C2H6O

CONDENSED FORMULA

STRUCTURAL FORMULA

MOLECULAR MODEL

H H

CH3CH2OH

H

C

C

O

H

H H Dimethyl ether

C2H6O

CH3OCH3

C H



Here the two compounds have the same molecular formula. Condensed or structural formulas, or a molecular model, clearly show these compounds are different.

H

H H

Figure 2.15  Four approaches to showing molecular formulas.  ​

O

C

H

H 2.4  Molecules: Formulas, Models, and Names

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85

The six-sided structure of a snowflake is a reflection of the underlying molecular structure of ice.

Standard Colors for Atoms in Molecular Models  The colors

listed here are used for molecular models in this book and are generally used by chemists. carbon atoms hydrogen atoms oxygen atoms nitrogen atoms

Alexey Kljatov/Shutterstock.com

Ice consists of six-sided rings formed by water molecules, in which each side of a ring consists of two O atoms and an H atom.

Figure 2.16 Ice. ​ Snowflakes reflect the underlying structure of ice.

being less dense as a solid than as a liquid. The lower density of ice, which has enormous consequences for Earth’s climate, results from the fact that molecules of water do not pack together tightly in ice. Because molecules are three dimensional, it is often difficult to represent their structures on paper. Certain conventions have been developed, however, that help represent three-dimensional structures on two-dimensional surfaces. Perspective drawings are often used (Figure 2.17). Molecular models are very useful for visualizing structures. These models show how atoms are attached to one another and show the molecule’s overall three-dimensional structure. In ball-and-stick models, spheres of different colors represent the atoms, and sticks represent the bonds holding them together. ­Molecules can also be represented using space-filling models, representations of molecules in which the atoms are partial spheres that have diameters proportional to those of the atoms and are joined directly to one another. These models are a better representation of relative sizes of atoms and their proximity to each other. A disadvantage of pictures of space-filling models is that atoms can often be hidden from view.

chlorine atoms

Naming Molecular Compounds There are many molecular compounds you will encounter often. You should understand how to name many of them and, in many cases, be able to write their formulas. First consider molecules formed from combinations of two nonmetals, known as binary compounds of nonmetals. Although there are exceptions, most binary molecular compounds are a combination of nonmetallic elements from Groups 4A–7A (14–17) with one another or with hydrogen. The formula is generally

H

H

Bonds in plane of paper

C H

H H

H

John C. Kotz

Bond going away from observer

Bond coming toward observer

Perspective drawing

Plastic model

Ball-and-stick model

Space-filling model

C H

H

The three representations in a single drawing.

Figure 2.17  Ways of depicting a molecule, here the methane (CH4) molecule.

86

Chapter 2 / Atoms, Molecules, and Ions

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written by putting the elements in order of increasing group number. If the two elements are from the same group, then the element further down in the group is usually written first. Hydrogen forms binary compounds with all of the nonmetals except the noble gases. For compounds of oxygen, sulfur, and the halogens, the H atom is generally written first in the formula and is named first. The other nonmetal is named by ­adding -ide to the stem of the name. Compound

Name

HF

Hydrogen fluoride

HCl

Hydrogen chloride

H2S

Hydrogen sulfide

Formulas of Binary Nonmetal Compounds Containing Hydrogen  Simple hydrocarbons

(compounds of C and H) such as methane (CH4) and ethane (C2H6) have formulas written with H following C, and the formulas of ammonia and hydrazine have H following N. Water and the hydrogen halides, however, have the H atom preceding O or the halogen atom. Tradition is the only explanation for such irregularities in writing formulas.

When naming most other binary molecular compounds, the number of ­atoms of a given type in the compound is designated with a prefix. Number of Atoms

Prefix

 1

mono-

 2

di-

 3

tri-

 4

tetra-

 5

penta-

 6

hexa-

 7

hepta-

 8

octa-

 9

nona-

10

deca-

If there is only one atom of a given type and it is the second element listed in the binary compound formula, the prefix “mono-” is used. However, “mono-” is omitted if it is the first element in the formula.

Exam p le 2.5

Binary Molecular Compounds Problem  Provide the missing name or formula for the following binary compounds. (a) What is the name of N2O5? (b) What is the name of CO? (c) What is the formula of tetraphosphorus decaoxide?

What Do You Know?  You know the formulas for N2O5 and CO and the name for tetraphosphorus decaoxide. You know that binary molecular compounds are usually c­ omposed of two nonmetals, and you know the prefixes used in naming binary molecular compounds.



2.4  Molecules: Formulas, Models, and Names

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87

Strategy  For parts a and b, first determine whether the binary compound is likely to be molecular by noting if both elements are nonmetals. The prefix system is used in naming most binary molecular compounds. For part c, the prefixes indicate the number of atoms for each element present in a binary molecular compound. Solution  (a) The two elements present in the compound are nitrogen and oxygen. These are both nonmetals, so the compound is molecular. There are two atoms of nitrogen, so the prefix di- is used. There are five atoms of oxygen, so the prefix penta- is used. The ending of the second element is changed to -ide. The name is dinitrogen pentaoxide. (b) The two elements present are carbon and oxygen, which are both nonmetals, so the compound is molecular. There is only one atom of each element present. The prefix mono- is not used for the first element in a compound’s name, so no prefix is needed for the carbon. The prefix mono- is used with the second element, and the ending of the element’s name needs to be changed to -ide. Rather than say monooxide, however, the word is simplified to monoxide. The name is carbon monoxide. (c) The prefixes indicate the number of atoms for each element present in the binary molecular compound. Tetra- indicates that four atoms of phosphorus are present, and deca- indicates that ten atoms of oxygen are present. The order for writing the element symbols is the same as that in the name: the element further left on the periodic table is written first. The formula is P4O10.

Think about Your Answer  All the compounds in this example were binary molecular compounds, so the appropriate prefixes were used in the names. In the same way that monoxide is used rather than monooxide, some people also drop a final “a” in a prefix used before oxide. Thus, some would have given the name of N2O5 as dinitrogen pentoxide.

Check Your Understanding (1) What are the names of (a) SO2, (b) NI3, and (c) N2O. (2) Write the formulas for (a) dinitrogen tetraoxide, (b) sulfur hexafluoride, and (c) disulfur decafluoride. Hydrocarbons  Compounds such as methane, ethane, propane, and butane belong to a class of hydrocarbons called alkanes.

methane, CH4

ethane, C2H6

88

propane, C3H8

butane, C4H10

Finally, some binary compounds of nonmetals were discovered many years ago and have common names. Compound

Common Name

Compound

Common Name

CH4

Methane

N2H4

Hydrazine

C2H6

Ethane

PH3

Phosphine

C3H8

Propane

NO

Nitric oxide

C4H10

Butane

N2O

Nitrous oxide (laughing gas)

NH3

Ammonia

H2O

Water

Chapter 2 / Atoms, Molecules, and Ions

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2.5 Ions Goals for Section 2.5 • Recognize that metal atoms commonly lose one or more electrons to form positive ions, called cations, and nonmetal atoms often gain electrons to form negative ions, called anions.

• Predict the charge on monatomic cations and anions based on Group number. • Name and write the formulas of ions. The compounds you have encountered so far in this chapter are molecular ­ compounds, that is, compounds that consist of discrete molecules at the particulate level. Ionic compounds make up another major class of compounds, and many may be familiar to you (Figure 2.18). Table salt, or sodium chloride (NaCl), and lime (CaO) are just two. Ionic compounds consist of ions, that is, atoms or groups of atoms that bear a positive or negative electric charge.

Monatomic Ions Atoms of many elements can lose or gain electrons to form monatomic ions, ions consisting of a single atom bearing a charge. Some commonly-encountered ions are listed in Figure 2.19. How do you know if an atom is likely to gain or lose electrons? It depends on whether the element is a metal or nonmetal. In reactions, •

Metals generally lose one or more electrons.

•

Nonmetals frequently gain one or more electrons.

Monatomic Cations If an atom loses an electron (which is transferred to an atom of another element in the course of a reaction), the atom now has one less negative electron than it has positive protons in the nucleus. The result is a positively charged ion called a cation (Figure 2.20a). (The name is pronounced “cat′-i-on.”) For example, the elements of Group 1A (1), such as lithium, lose one electron per atom, resulting in the formation of cations with a 1+ charge.

Figure 2.18  Some common ionic compounds.

Gypsum, CaSO4 ⋅ 2 H2O



Common Name

Calcite, CaCO3

Fluorite, CaF2

Orpiment, As2S3

© Charles D. Winters/Cengage

Hematite, Fe2O3

Name

Formula

Ions Involved

Calcite

Calcium carbonate

CaCO3

Ca2+, CO32−

Fluorite

Calcium fluoride

CaF2

Ca2+, F−

Gypsum

Calcium sulfate dihydrate

CaSO4 ∙ 2 H2O

Ca2+, SO42−

Hematite

Iron(III) oxide

Fe2O3

Fe3+, O2−

Orpiment

Arsenic(III) sulfide

As2S3

As3+, S2−

2.5 Ions

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89

1A (1)

Figure 2.19  Charges on some common monatomic cations and anions.  Metals usually form

cations, and nonmetals usually form anions. Ions with the same magnitude charge (for example, 2+ and 2–) are highlighted with the same color in this figure. NOTE: It is important to recognize that transition metals (and a few main group metals) form cations of several charges. Examples include Cr2+ and Cr3+, Fe2+ and Fe3+, as well as Cu+ and Cu2+. As explained in the text, their names must reflect this.

1+

and 2+ and 2− 3+ and 3− 4+

2A (2)

H+

7A 8A (17) (18)

1−

Li+ 3B Na+ Mg2+ (3) K+ Ca2+

4B (4)

5B (5)

Ti4+

3A 4A 5A 6A − (13) (14) (15) (16) H N3− O2−

6B (6)

7B (7)

(8)

8B (9)

Cr2+

Mn2+

Fe2+

Co2+

Cr3+

1B 2B 3+ (10) (11) (12) Al

Fe3+ Co3+

Ni2+

Ag+ Cd2+

Cs+ Ba2+

Hg22+ Hg2+

→

e−

+

(3 protons and 3 electrons)

Writing Ion Formulas  When writing the formula of an ion, the charge on the ion must be included.

S2− Cl− Se2− Br−

Cu2+ Zn2+

Rb+ Sr2+

Li atom

P3−

Cu+

F−

Sn2+

Te2− I−

Pb2+ Bi3+

Li+ cation (3 protons and 2 electrons)

Elements of Group 2A (2) lose two electrons in reactions, forming cations with a charge of 2+, →

Ca atom

2 e−

+

(20 protons and 20 electrons)

Ca2+ cation (20 protons and 18 electrons)

and elements of Group 3A (13) lose three electrons, forming cations with a charge of 3+. →

Al atom

Group 3A (13) cations Like

aluminum, other metals in Group 3A (13) also lose three electrons to form cations with a 3+ charge. But they also, on occasion, form M+ cations with a single positive charge.

3 e−

(13 protons and 13 electrons)

+

Al3+ cation (13 protons and 10 electrons)

How can you predict the number of electrons gained or lost in reactions of elements in Groups 1A, 2A, and 3A (1, 2, and 13)? •

Metals of Groups 1A, 2A, and 3A lose one or more electrons to form positive ions having a charge equal to the group number of the metal.

(a) Formation of a lithium (Li+) cation from a neutral Li atom.

3e –

e–

2e – 3p 3n

3p 3n

Li

Li +

3p 3n 3e –

3p 3n 2e –

Lithium ion, Li +

A lithium-6 atom is electrically neutral because the number of positive charges (three protons) and negative charges (three electrons) are the same. When it loses one electron, it has one more positive charge than negative charge, so it has a net charge of 1+. The resulting lithium cation is symbolized as Li +.

Lithium, Li

(b) Formation of a fluoride (F –) anion from a neutral F atom.

9e – 9p 10n

10e –

e–

9p 10n

Fluorine, F

F

F–

9p 10n 9e –

9p 10n 10e –

A fluorine-19 atom is also electrically neutral, having nine protons and nine electrons. A fluorine atom can acquire an electron to produce an F− anion. This anion has one more electron than it has protons, so it has a net charge of 1−.

Fluoride ion, F –

Figure 2.20 Ions.

90

Chapter 2 / Atoms, Molecules, and Ions

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•

The number of electrons remaining on the cations formed from Group 1A (1) elements, Group 2A (2) elements, and aluminum is the same as the number of electrons in an atom of the noble gas that precedes it in the periodic table.

Transition metals (B-group elements) also form cations, but unlike the A-group metals, there is no easily predictable pattern of behavior. In addition, transition metals often form several different ions. Iron, for example, may form either Fe2+ or Fe3+ ions in its reactions. Copper may form 1+ or 2+ ions, but silver forms only a 1+ ion.

Monatomic Anions Nonmetals can gain electrons to form negatively charged ions. If an atom gains one or more electrons, there will now be more negatively charged electrons than protons (Figure  2.20b). A negatively charged ion is called an anion (pronounced “an′i-on”). An oxygen atom, for example, can gain two electrons in a reaction to form an ion with the formula O2−: O atom

+

2 e−

→

(8 protons and 8 electrons)

O2− anion (8 protons and 10 electrons)

A chlorine atom can add a single electron to form Cl−. Cl atom (17 protons and 17 electrons)

+

e−

→

Cl− anion (17 protons and 18 electrons)

Two general observations can be made concerning the formation of anions from nonmetals. •

Nonmetals of Groups  5A–7A form negative ions with a charge equal to the group number of the nonmetal minus 8.

•

The number of electrons on the anion is the same as the number of electrons in an atom of the noble gas that follows it in the periodic table.

Notice that the nonmetal hydrogen appears at two locations in Figure 2.19. The H  atom can either lose or gain electrons, depending on the other atoms it encounters. Electron lost:

H (1 proton, 1 electron)  → H+ (1 proton, 0 electrons) + e−

Electron gained:

H (1 proton, 1 electron) + e−  → H− (1 proton, 2 electrons)

Monatomic anion charges Using the 1–18 system of group numbers, the charge on a monatomic anion from Groups 15–17 is equal to the group number minus 18.

Finally, the noble gases very rarely form monatomic cations and never form monatomic anions in chemical reactions.

Naming Monatomic Ions Monatomic ions are named using the following rules: 1. For a monatomic positive ion (that is, a metal cation) from Group 1A (1), Group 2A (2), or aluminum, the name is that of the metal plus the word “­cation.” For example, Al31 is the aluminum cation. 2. Transition metals often form more than one type of positive ion. To specify which ion is involved, the charge of transition metal cations is indicated by a Roman numeral in parentheses immediately following the ion’s name. For example, Co21 is the cobalt(II) cation, and Co31 is the cobalt(III) cation. 3. A monatomic negative ion (that is, a nonmetal anion) is named by adding -­ide to the stem of the name of the nonmetal element from which the ion is derived (Figure 2.21). For example, the anions of the Group 7A (17) elements, the halogens, are known as fluoride, chloride, bromide, and iodide ions and as a group are called halide ions.

Elements with Multiple Ion Charges  These occur especially

in the transition metals. However, some main group metals such as tin (Sn21 and Sn41) and lead (Pb21 and Pb41) can also have multiple ion charges. It is our practice to always indicate the ion charge with a Roman numeral when naming compounds of the transition metal elements and in other cases when multiple charges are possible. 2.5 Ions

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91

1– 3–

H−

2–

hydride ion

F−

N3− O2− nitride ion

oxide ion

P3−

S2−

phosphide sulfide ion ion

fluoride ion

Cl− chloride ion

Se2− Br− selenide bromide ion ion

Te2−

I−

telluride ion

iodide ion

Figure 2.21  Names and charges of some common monatomic anions. 

Polyatomic Ions A polyatomic ion is an electrically-charged particle that consists of two or more atoms bonded together (Figure  2.22 and Table  2.4). For example, carbonate ion, CO32−, a common polyatomic anion, consists of one C atom and three O atoms. The ion has two units of negative charge because there are two more electrons (a total of 32) in the ion than there are protons (a total of 30) in the nuclei of one C atom and three O atoms. The ammonium ion, NH4+, is a common polyatomic cation. In this case, four H atoms surround an N atom, and the ion has a 1+ electric charge. This ion has 10 electrons, but there are 11 positively charged protons in the nuclei of the N and H atoms (7 for N, 1 for each H).

Naming Polyatomic Ions The polyatomic cation you will encounter most in this book and the laboratory is the ammonium ion, NH4+. Do not confuse the ammonium cation with the ammonia molecule, NH3, which has no electric charge and one less H atom. Polyatomic anions are common, especially those containing oxygen (called oxoanions). The names of some of the most common oxoanions are given in T ­ able 2.4. Although most of these names must be memorized, some guidelines can help. 1. For a series of oxoanions with two members, the oxoanion with the greater number of oxygen atoms is given the suffix -ate, and the oxoanion with the smaller number of oxygen atoms has the suffix -ite. For example, consider the following pairs of ions: NO3− is the nitrate ion, whereas NO2− is the nitrite ion. SO42− is the sulfate ion, whereas SO32− is the sulfite ion.

Photos: © Charles D. Winters/Cengage

2. For a series of oxoanions with more than two members, the ion with the largest number of oxygen atoms has the prefix per- and the suffix -ate. The ion with the smallest number of oxygen atoms has the prefix hypo- and the suffix -ite. The chlorine oxoanions are the most commonly encountered example. Perchlorate ion

ClO3−

Chlorate ion

ClO2−

Chlorite ion

ClO−

Hypochlorite ion

CO32– Calcite, CaCO3 Calcium carbonate

per . . . ate increasing oxygen content

ClO4−

. . . ate . . . ite hypo . . . ite

PO43– Apatite, Ca5F(PO4)3 Calcium fluorophosphate

SO42– Celestite, SrSO4 Strontium sulfate

Figure 2.22  Common ionic compounds containing polyatomic ions.

92

Chapter 2 / Atoms, Molecules, and Ions

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Table 2.4 Formula

Formulas and Names of Some Common Polyatomic Ions

Name

Formula

Name

Cation: Positive Ion NH4+

Polyatomic ions  To be successful

Ammonium ion

Anions: Negative Ions Based

on a

CN− −

−

C2O42− on a

Based

on a

ClO−

Acetate ion Hydrogen carbonate ion (or bicarbonate ion)

Chlorite ion

−

Chlorate ion

−

Perchlorate ion

ClO3 ClO4

Oxalate ion Group 5A (15)

element

Based

Nitrite ion

CrO4

Nitrate ion

Cr2O7

on a transition metal

−

NO3

HPO4

2−

H2PO4 Based

2−

Phosphate ion

PO4

3−

−

on a

element

Hypochlorite ion

−

NO2

Group 7A (17)

−

ClO2

Carbonate ion

CO3

2−

Based

element

Cyanide ion

CH3CO2 HCO3

Group 4A (14)

in your study of chemistry, you must know the names and formulas (including the ion charges) of the common ions listed in this table.

Chromate ion

2−

Dichromate ion

−

Permanganate ion

MnO4

Hydrogen phosphate ion Dihydrogen phosphate ion Group 6A (16)

element

OH−

Hydroxide ion

SO32−

Sulfite ion

SO42−

Sulfate ion

HSO4−

Hydrogen sulfate ion (or bisulfate ion)

3. Oxoanions that contain hydrogen are named by adding the word “hydrogen” before the name of the oxoanion. Two hydrogens in an anion are indicated by “dihydrogen.” Some hydrogen-containing oxoanions also have common names. For example, the hydrogen carbonate ion, HCO3−, is called the bicarbonate ion. Ion

Systematic Name

HPO4

2− −

H2PO4

Common Name

Hydrogen phosphate ion Dihydrogen phosphate ion

−

Hydrogen carbonate ion

Bicarbonate ion

−

Hydrogen sulfate ion

Bisulfate ion

−

Hydrogen sulfite ion

Bisulfite ion

HCO3 HSO4 HSO3

2.6 Ionic Compounds: Formulas, Names, and Properties Goals for Section 2.6 • Write formulas for ionic compounds by combining ions in the proper ratio to give a zero overall charge. • Give the names and write the formulas of ionic compounds. • Understand the importance of Coulomb’s law in chemistry, which describes the electrostatic forces of attraction and repulsion of ions.

2.6  Ionic Compounds: Formulas, Names, and Properties

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93

© Charles D. Winters/Cengage

Ionic compounds are electrically neutral combinations of cations and anions. How can you recognize that a compound is likely to be ionic? One indicator is the presence of a metal as the first element in a compound’s formula or the presence of a polyatomic cation such as the ammonium ion (NH4+). Thus, compounds such as NaCl, CuCl2, AgNO3, and (NH4)2CO3 are all classified as being ionic. It is important for you to be able to name ionic compounds and to write their formulas. Because compounds are electrically neutral, they have no net electric charge. Thus, in an ionic compound the numbers of positive and negative ions must be such that the positive and negative charges balance. In sodium chloride, the sodium ion has a 1+ charge (Na+) and the chloride ion has a 1− charge (Cl−). These ions must be present in a 1∶1 ratio, and so the formula is NaCl. The gemstone ruby is largely the compound formed from aluminum ions (Al3+) and oxide ions (O2−). To have a compound with the same number of positive and negative charges, two Al3+ ions must combine with three O2− ions to give a formula of Al2O3. 2 Al3+ ions charge = 2 × (3+) = 6+ 3 O2− ions charge = 3 × (2−) = 6−

The Color of Rubies  The beautiful

red color of a ruby comes from a trace of Cr3+ ions that take the place of a few of the Al3+ ions in the solid.

Al2O3 overall charge = (6+) + (6−) = 0

Calcium, a Group 2A (2) metal, forms a cation having a 2+ charge. It can combine with a variety of anions to form ionic compounds such as those in the following table: Compound

Ion Combination

Overall Charge on Compound

CaCl2

−

Ca  + 2 Cl

(2+) + 2 × (1−) = 0

CaCO3

Ca  + CO3

2−

(2+) + (2−) = 0

Ca3(PO4)2

3 Ca2+ + 2 PO43−

3 × (2+) + 2 × (3−) = 0

2+ 2+

In writing formulas of ionic compounds, the convention is that the symbol of the cation is given first, followed by the anion symbol. Also notice the use of parentheses when more than one polyatomic ion of a given kind is present [as in Ca3(PO4)2]. (None, however, are used when only one polyatomic ion is present, as in CaCO3.)

Problem Solving Tip 4.1 Formulas for Ions and Ionic Compounds Writing formulas for ionic compounds requires that you know the formulas and charges of the most common ions. The charges on monatomic ions are often evident from the position of the element in the periodic table, but you simply have to remember the formulas and charges of polyatomic ions, especially the most common ones such as nitrate, sulfate, carbonate, phosphate, and acetate.

94

If you cannot remember the formula of a polyatomic ion or if you encounter an ion you have not seen before, you may be able to figure out its formula. For example, suppose you are told that the formula for sodium formate is NaCHO2. You know that the sodium ion is Na+, so the formate ion must be the remaining portion of the compound; it must have a charge of 1− to balance the 1+ charge on

the sodium ion. Thus, the formate ion must be CHO2−. Finally, when writing the formulas of ions, you must include the charge on the ion (except in the formula of an ionic compound). Writing Na when you mean sodium ion is incorrect. There is a vast difference in the properties of the element sodium (Na) and those of its ion (Na+).

Chapter 2 / Atoms, Molecules, and Ions

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Exam p le 2.6

Ionic Compound Formulas Problem  For each of the following ionic compounds, write the symbols for the ions present and give the relative number of each: (a) Li2CO3 and (b) Fe2(SO4)3.

What Do You Know?  You know the formulas of the ionic compounds, the predicted charges on monatomic ions (see Figure  2.19), and the formulas and charges of polyatomic ions (see Table 2.4).

Strategy  Divide the formula of the compound into the cations and anions. To accomplish this you will have to recognize, and remember, the formulas of common monatomic and polyatomic ions. Solution (a)  Li2C O3 is composed of two lithium ions, Li+, for each carbonate ion, CO32−.  Li is a Group 1A (1) element and always has a 1+ charge in its compounds. Because the two 1+ charges balance the negative charge of the carbonate ion, the latter must be 2−. (b)  Fe2(SO4)3 contains two iron(III) ions, Fe3+, for every three sulfate ions, SO42−.  The way to recognize this is to recall that sulfate has a 2− charge. Because three sulfate ions are present (with a total charge of 6−), the two iron cations must have a total charge of 6+. This is possible only if each iron cation has a charge of 3+.

Think about Your Answer  Remember that the formula for an ion must include its composition and its charge. Formulas for ionic compounds are always written with the cation first and then the anion, but ion charges are not included.

Identifying Charges on Transition Metal Cations  Because ionic

compounds are electrically neutral, the charges on transition metal cations can be determined if anion charges are known.

Check Your Understanding Give the number and identity of the constituent ions in each of the following ionic compounds: NaF, Cu(NO3)2, and NaCH3CO2.

Exam p le 2.7

Writing Ionic Compound Formulas Problem  Write formulas for ionic compounds composed of an aluminum cation and each of the following anions: (a) fluoride ion, (b) sulfide ion, and (c) nitrate ion.

What Do You Know?  You know the names of the ions involved, the predicted charges on monatomic ions (see Figure  2.19), and the names, formulas, and charges of polyatomic ions (see Table 2.4). Strategy  First decide on the formula of the Al cation and the formula of each anion. Combine the Al cation with each type of anion to form electrically neutral compounds.



2.6  Ionic Compounds: Formulas, Names, and Properties

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95

Solution  Aluminum is a Group 3A (13) metal and therefore has a charge of 3+. (a) Fluorine is a Group 7A (17) element. The charge of the fluoride ion is predicted to be 1− (from 7 − 8 = 1−). Therefore, 3 F− ions are needed to combine with one Al3+. The formula of the compound is  AlF3.  (b) Sulfur is a nonmetal in Group 6A (16), so it forms a 2− anion. Thus, two Al3+ ions [total charge is 6+ = 2 × (3+)] and three S2− ions [total charge is 6− = 3 × (2−)] are needed. The compound has the formula  Al2S3.  (c) The nitrate ion has the formula NO3− (see Table 2.4). The answer is therefore similar to the AlF3 case, and the compound has the formula  Al(NO3)3.  Here parentheses are placed around NO3 to show that three polyatomic NO3− ions are involved.

Think about Your Answer  Compounds are electrically neutral; they have an overall charge of zero. In an ionic compound, the sum of the positive charges must equal the sum of the negative charges. The most common error students make is not knowing the correct charge on an ion.

Check Your Understanding (a) Write the formulas of all electrically neutral ionic compounds that can be formed by combining the cations Na+ and Ba2+ with the anions S2− and PO43−. (b) Iron forms ions having 2+ and 3+ charges. Write the formulas of the compounds formed between chloride ions and these two different iron cations.

Names of Compounds Containing Transition Metal Cations  Be sure

to remember that the charge on a transition metal cation is indicated by a Roman numeral and is included in the name.

Names of Ionic Compounds The name of an ionic compound is built from the names of the positive and negative ions in the compound. The name of the positive cation is given first, followed by the name of the negative anion.

E xamp le 2.8

Naming Ionic Compounds Problem  What is the name for each of the following compounds? (a) CaBr2

(b) (NH4)2CO3

(c) Co2O3

What Do You Know?  You know the formulas of the given compounds. You also know the names and charges of the monatomic and polyatomic ions.

Strategy  First, determine if the compound should be named as an ionic compound. If so, the name consists of the names of the ions in the compound. If a transition metal (or other metal with multiple possible charges) is the cation, you will need to determine its charge and write the charge in Roman numerals after the metal’s name.

96

Chapter 2 / Atoms, Molecules, and Ions

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Solution  (a) The first element listed in the formula (calcium) is a metal, so the compound will be named as an ionic compound. There is one calcium cation (Ca2+), and two bromide anions (Br–) in the formula. The name of the compound is  calcium bromide . Calcium is not a transition metal or other metal that can have multiple charges, so no Roman numerals are needed. (b) This compound contains the ammonium (NH4+) cation, so the compound will be named as an ionic compound. There are two ammonium cations and one carbonate anion (CO32–) in the formula. Its name is  ammonium carbonate . (c) The first element listed in the formula (cobalt) is a transition metal. The compound will be named as an ionic compound, and the charge of the cobalt must be determined.  Keep in mind that the overall compound is electrically neutral, that is, the sum of the ion charges must be zero. There are three oxide ions (O2–) in the formula. This gives an overall negative charge of 3 × (2–) = 6–. The total positive charge must be 6+. There are two cobalt ions whose charges must add up to 6+, so each must be 3+. The compound is  cobalt(III) oxide .

Think about Your Answer  Recall that different rules are used to name binary molecular compounds (see Example 2.5).

Check Your Understanding What is the name for each of the following compounds? (a) NaHSO4

(b) Mg(OH)2

(c) TiCl2

Properties of Ionic Compounds When a particle with a negative electric charge is brought near another particle with a positive electric charge, there is a force of attraction between them (Figure 2.23). In contrast, there is a repulsive force when two particles with the same charge—both positive or both negative—are brought together. These forces are called electrostatic forces, and the force of attraction (or repulsion) between ions is given by Coulomb’s law (Equation 2.3): charge on + and − ions



Force = −k proportionality constant

charge on electron

(n+e)(n−e) d2



(2.3)

distance between ions

where, for example, n+ is +3 for Al3+ and n− is −2 for O2−. Based on Coulomb’s law, the force of attraction (Figure 2.23) between oppositely charged ions increases •

as the ion charges (n+ and n−) increase. Thus, the attraction between ions having charges of 2+ and 2− is greater than that between ions having 1+ and 1− charges.

•

as the distance between the ions becomes smaller.



The Importance of Coulomb’s Law  Coulomb’s law is the

basis for understanding many fundamental concepts of chemistry. Among the chapters where this is important are: Chapter 3: dissolving compounds in water. Chapters 6 & 7: the interaction of electrons and the atomic nucleus. Chapters 8 & 9: the interaction of atoms to form molecules. Chapter 11: the interactions between molecules (intermolecular forces). Chapter 12: the formation of ionic solids. Chapter 13: the solution process.

2.6  Ionic Compounds: Formulas, Names, and Properties

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97

+1 n+ = 1

+

Li+

d small

d

n– = –1

−

+

−1

+2

d large

−2

LiF

F−

Ions such as Li+ and F– are held together by a coulombic force of attraction. Here a lithium ion is attracted to a fluoride ion, and the distance between the nuclei of the two ions is d. (a)

As ion charge increases, force of attraction increases

As distance increases, force of attraction decreases

(b)

Figure 2.23  Coulomb’s law and electrostatic forces.

The simplest ratio of cations to anions in an ionic compound is represented by its formula. However, ionic compounds do not consist of simple pairs or small groups of positive and negative ions. Instead, an ionic solid consists of millions upon millions of ions arranged in an extended three-dimensional network called a crystal ­lattice. A portion of the lattice for NaCl, illustrated in Figure  2.24, represents a ­common way of arranging ions for compounds that have a 1∶1 ratio of cations to anions.

If ionic compounds are prepared in water solution and then isolated as solids, the crystals often retain molecules of water. These compounds are called hydrated compounds. For example, crystals of the red cobalt(II) compound in the Figure have six water molecules per CoCl2. By convention, the formula for this compound is written as CoCl2  ∙  6 H2O. The dot between CoCl2 and 6 H2O ­ indicates that 6  ­molecules of water are ­associated with every CoCl2. The name of the compound is cobalt(II) chloride hexahydrate. Hydrated cobalt(II) chloride, the red solid in the Figure, turns purple and then deep blue as it is heated and loses water to form anhydrous CoCl2; anhydrous means without water. Upon on exposure to moist air, anhydrous CoCl2 takes up water and is converted back into the red hydrated compound. It is this property that allows crystals of the blue compound to be used as a humidity indicator. You may have seen them in small bags packed with a piece of electronic equipment. Hydrated compounds are common. The walls of your home are probably covered with drywall, or plaster board, which contains hydrated calcium sulfate (gypsum,

98

Photos: © Charles D. Winters/Cengage

A Closer Look

Hydrated Ionic Compounds

Cobalt(II) chloride hexahydrate [CoCl2 • 6 H2O] is deep red.

When it is heated, the compound loses some of the water of hydration.

CaSO4  ∙  2 H2O) and anhydrous CaSO4, sandwiched between paper. If gypsum is heated between 120 and 180  °C, the water is partly driven off to give CaSO4  ∙  1/2 H2O, a compound known as plaster of Paris. If you have ever broken an arm or leg and had to have a cast, the cast may have been made of this compound. It is an effective casting material because, when added to water, it

Heating ultimately leads to the deep blue compounds CoCl2 • 2 H2O and CoCl2. Experiments show that some also decomposes to black CoO and HCl.

forms a thick slurry that can be poured into a mold or spread out over a part of the body. As it takes on more water, the material increases in volume and forms a hard, inflexible solid. Plaster of Paris is also a useful material for artists, because the expanding compound fills a mold completely and makes a high-quality reproduction.

Chapter 2 / Atoms, Molecules, and Ions

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© Charles D. Winters/Cengage

Ionic compounds have characteristic properties that can be understood in terms of the charges of the ions and their arrangement in the lattice. Because each ion is surrounded by oppositely charged nearest neighbors, it is held tightly in its allotted location. At room temperature, each ion can move just a bit around its average position, but considerable energy must be added before an ion can escape the attraction of its neighboring ions. Only if enough energy is added will the lattice structure collapse and the substance melt. Greater attractive forces mean that ever more energy—and higher and higher temperatures—is required to cause melting. Thus, Al2O3, composed of Al3+ and O2− ions, melts at a much higher temperature (2072 °C) than NaCl (801 °C), composed of Na+ and Cl− ions. Most ionic compounds are hard solids. That is, the solids are not pliable or soft. This is again related to the lattice of ions. The nearest neighbors of a cation in a lattice are anions, and the force of attraction makes the lattice rigid. However, a blow with a hammer can cause the lattice to break cleanly along a sharp boundary. The hammer blow displaces layers of ions just enough to cause ions of like charge to become nearest neighbors, and the repulsion between these like-charged ions forces the lattice apart (Figure 2.25).

Figure 2.24 Sodium chloride.  A crystal of NaCl

2.7 Atoms, Molecules, and the Mole Goals for Section 2.7 • Understand the mole concept and molar mass and their application. • Use the molar mass of an element and Avogadro’s number in calculations. • Calculate the molar mass of a compound from its formula and a table of atomic

consists of an extended lattice of sodium ions and chloride ions in a 1∶1 ratio. When melted, the crystal lattice collapses and the ions move freely and can conduct an electrical current.

weights.

• Calculate the amount (5 number of moles) of a compound represented by a given mass, and vice versa.

• Use Avogadro’s number to calculate the number of atoms or ions in a compound.

© Charles D. Winters/Cengage

When two chemicals react with each other, chemists want to know how many atoms or molecules of each are used so that formulas can be established for the reaction’s

An ionic solid is rigid owing to the forces of attraction between oppositely charged ions. When struck sharply, however, the crystal can cleave cleanly.

Figure 2.25  Ionic solids.

When a crystal is struck, layers of ions move slightly, and ions of like charge become nearest neighbors.

Repulsions between ions of similar charge cause the crystal to cleave.

2.7  Atoms, Molecules, and the Mole

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Amedeo Avogadro, conte di Quaregna (1776–1856), was an Italian nobleman and a lawyer. In about 1800, he turned to science and was the first professor of mathematical physics in Italy. Avogadro did not propose the notion of a fixed number of particles in a chemical unit. Rather, the number was named in his honor because he performed experiments in the ­nineteenth century that laid the groundwork for the concept. Just how large is Avogadro’s number? One mole of unpopped popcorn kernels would cover the continental United States to a depth of about 9 miles. When the mole was first adopted as an SI base unit, it was defined as the number of elementary entities as there are atoms in exactly 12 g of carbon-12. Using that definition, the mass of 12 g for a mole of carbon-12 was a defined quantity (with an infinite number of significant figures). Avogadro’s number, on the other hand, was a measured quantity that was determined experimentally and changed over time as experiments yielded more accurate and precise values.

The Mole  The term mole was introduced about 1895 by Wilhelm Ostwald (1853–1932), who derived the term from the Latin word moles, meaning a “heap” or a “pile.”

On May 20, 2019, the definition of the mole was officially changed by the Bureau International des Poids et Mesures that oversees the SI base units. In the new definition, the number of particles in a mole is defined to be exactly 6.02214076 × 1023. The value of Avogadro’s number is now a defined quantity. The mass of carbon-12 that corresponds to this number is now an experimentally Amedeo Avogadro measured value with limited accuracy and precision. The new definition of the mole is a fundamental change and has various implications, but most of these are beyond the level of an  introductory chemistry course. The crucial things for you to ­remember are that 1. one mole of any substance contains Avogadro’s number of elementary entities (atoms, molecules, ...) of that substance, and 2. one mole of any substance contains the same number of particles as one mole of any other substance.

Edgar Fahs Smith Collection

A Closer Look

Amedeo Avogadro and His Number

products. To do this, a method of counting atoms and molecules is needed. Is there a way to connect the macroscopic world, the world you can see with your eyes or with ordinary scientific instruments, with the particulate world of atoms, molecules, and ions? The solution to this problem is to define a unit of matter that contains a known number of particles. That chemical unit is the mole. The mole (abbreviated mol) is the SI base unit for measuring an amount of a substance (Table 1R.1, page 32) and is defined as follows: A mole is the amount of a substance that contains exactly 6.02214076 × 1023 elementary entities (atoms, molecules, or other particles). 1 mole = 6.02214076 × 1023 particles

This value is known as Avogadro’s number (symbolized by NA) in honor of Amedeo Avogadro, an Italian lawyer and physicist (1776–1856) who conceived the basic idea (but never determined the number). The key to understanding the concept of the mole is recognizing that one mole always contains the same number of particles, no matter what the substance. One mole of sodium contains the same number of atoms as one mole of iron and the same number of atoms as the number of molecules in one mole of water. An Important Difference Between the Terms Amount and Quantity in Chemistry  The amount of a

substance is the number of moles of that substance. In contrast, quantity refers, for example, to the mass or volume of the substance.

100

Atoms and Molar Mass The mass in grams of one mole of any element is the molar mass of that element. Molar mass is abbreviated with a capital italicized M and has units of grams per mole (g/mol). An element’s molar mass is the quantity in grams numerically equal to its atomic weight. Using copper as an example, Molar mass of copper (Cu) =   mass of 1.000 mol of Cu atoms = 63.55 g∙mol = mass of 6.022 × 1023 Cu atoms

Chapter 2 / Atoms, Molecules, and Ions

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© Charles D. Winters/Cengage

Copper 63.55 g

Sulfur 32.06 g

Magnesium 24.31 g

Silicon 28.09 g

Tin 118.7 g

Figure 2.26  One mole of common elements.  (Left to right) Sulfur powder, magnesium chips, tin, and silicon. (Above) Copper beads.

Figure  2.26 shows one mole of some common elements. Although each of these “piles of atoms” has a different volume and different mass, each contains 6.022 × 1023 atoms. The mole concept is the cornerstone of quantitative chemistry. It is essential to be able to convert from moles to mass and from mass to moles. Dimensional analysis, which is described in Chapter 1R, page 47, shows that this can be done in the following way: MASS Moles to Mass grams Moles × = grams 1 mol molar mass

MOLES CONVERSION Mass to Moles Grams ×

1 mol = moles grams

1/molar mass

For example, what mass, in grams, is represented by 0.35  mol of aluminum? Using the molar mass of aluminum (27.0 g/mol), you can determine that 0.35 mol of Al has a mass of 9.5 g. 0.35 mol Al 3

27.0 g Al 5 9.5 g Al 1 mol Al

Molar masses of the elements are usually known to four or more significant figures. To avoid having the molar mass affect the precision of a calculation, the convention usually followed in this book is to use a value of the molar mass with at least one more significant figure than in any other number in the calculation. For example, if you weigh out 16.5 g of carbon, you use 12.01 g/mol for the molar mass of C to find the amount of carbon present.



2.7  Atoms, Molecules, and the Mole

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101

16.5 g C ×

1 mol C 12.01 g C

= 1.37 mol C

Note that four significant figures are used in the molar mass, but there are three in the sample mass.

E xamp le 2.9

Mass, Moles, and Atoms Problem  Consider two elements in the same vertical column of the periodic table, lead and tin. (a) What mass of lead, in grams, is equivalent to 2.50 mol of lead (Pb)?

© Charles D. Winters/Cengage

(b) What amount of tin, in moles, is represented by 36.6 g of tin (Sn)? How many atoms of tin are in the sample?

Lead.  A 150-mL beaker containing 2.50 mol or 518 g of lead.

(c) What is the mass of an average atom of each of these elements?

What Do You Know?  You know the amount of lead and the mass of tin. You also know, from the periodic tables in this book, the molar masses of lead (207.2 g/mol) and tin (118.7 g/mol). For parts (b) and (c) Avogadro’s number is needed. Strategy Part (a)  Multiply the amount of Pb by the molar mass. Part (b)  Multiply the mass of tin by (1/molar mass). To determine the number of atoms, multiply the amount of tin by Avogadro’s number.

© Charles D. Winters/Cengage

Part (c) Multiply the molar mass of each element by (1/Avogadro’s number).

Solution (a) Convert the amount of lead in moles to mass in grams. 2.50 mol Pb 3

(b) Convert the mass of tin to the amount in moles,

Tin.  A sample of tin having a mass of 36.6 g (or 1.86 × 1023 atoms).

 36.6 g Sn 3

102

1 mol Sn 5 0.3083 mol Sn 5  0.308 mol Sn  118.7 g Sn

and then use Avogadro’s number to find the number of atoms in the sample.

Number of digits  In the first step of

part (b), the highlighted answer (0.308 mol Sn) is reported to the proper number of significant figures. The intermediate value (0.3083 mol Sn) with one extra digit is also shown and carried forward to the next step so that round-off error does not accumulate (see page 45).

207.2 g 5  518 g Pb  1 mol Pb

 0.3083 mol Sn 3

6.022 3 1023 atoms Sn 5  1.86 × 1023 atoms Sn  1 mol Sn

(c) Convert the molar mass of each element to grams per atom. 1 mol Sn 118.7 g Sn 3 5  1.971 × 1022 g Sn/atoms mol Sn 6.022 3 1023 atoms 1 mol Pb 207.2 g Pb 3 5  3.441 × 1022 g Pb/atoms mol Pb 6.022 3 1023 atoms

Chapter 2 / Atoms, Molecules, and Ions

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Think about Your Answer  Parts (a) and (b) were solved using g/mol or mol/g as conversion factors. To be sure you have used them correctly, you should keep track of the units of each term (page 47) and think about the numerical values of your answers. For example, in part (b), if you had inverted the conversion factor (mol/atoms instead of ­atoms/mol), the units would not have canceled properly and you would have calculated that there was less than one atom in 0.308 mol of Sn, an unreasonable answer. In part (c), notice the extremely small values obtained for the average masses of the atoms. Atoms are tiny! Finally, note that these are average masses per atom because of the presence of more than one isotope for each element.

Check Your Understanding What mass of gold, Au, contains 2.6 × 1024 atoms?

Molecules, Compounds, and Molar Mass The formula of a compound tells you the type of atoms or ions in the compound and the relative number of each. For example, one molecule of methane, CH4, is made up of one atom of C and four atoms of H. But suppose you have Avogadro’s number of C  atoms (6.022 × 1023) combined with the proper number of H ­atoms. (For CH4 this means there are 4 × 6.022 × 1023 H  atoms per mole of C  atoms.) What masses of atoms are combined, and what is the mass of this many CH4 molecules? C

+

4H

n

weight is sometimes encountered. This is the sum of the atomic weights of the constituent elements.

CH4

6.022 × 1023 C atoms

4 × 6.022 × 1023 H atoms

6.022 × 1023 CH4 molecules

 =  1.000 mol of C

 =  4.000 mol of H atoms

 =  1.000 mol of CH4 molecules

 =  12.01 g of C atoms  =  4.032 g of H atoms

Molecular Weight and Molar Mass  The old term molecular

 =  16.04 g of CH4 molecules

Students studying chemistry for the first time are often perplexed by the idea of the mole. But it is just a counting unit with an odd name. Pairs and dozens are two other common counting units. For example, a pair of objects has two of the same things (two shoes or two gloves), and there are 12 eggs or apples in a dozen. In the same way, a mole of atoms or a mole of jelly beans has 6.022 × 1023 objects. The great advantage of counting units is that if you know the number of units you also know the number of objects. If you know there are 3.5 dozen apples in



a box, you know there are 42 apples. And, if you have 0.308 mol of tin (36.6 g), you know you have 1.86 × 1023 atoms of tin. When you carry out a chemical reaction in the lab, you need to know how many “chemical units” of an element are involved. Atoms obviously cannot be counted out one by one. Instead, you can weigh a given mass of the element, and, from the molar mass, determine the number of “chemical units” or moles. And, if you really wanted to know the information, you could calculate the number of atoms involved.

© Charles D. Winters/Cengage

A Closer Look

The Mole, a Counting Unit

Counting Units.  The unit dozen, which refers to 12 objects, is a common counting unit. Similarly, the mole is a chemical counting unit. Just as a dozen always has 12 objects, a mole always has 6.022 × 1023 objects.

2.7  Atoms, Molecules, and the Mole

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103

Because you know the number of moles of C and H atoms in 1 mol of CH4, you can calculate the masses of carbon and hydrogen that must be combined. It follows that the mass of CH4 is the sum of these masses. That is, 1 mol of CH4 has a mass equal to the mass of 1 mol of C atoms (12.01 g) plus 4 mol of H atoms (4.032 g). Thus, the molar mass, M, of CH4 is 16.04 g/mol. Because you know the mass of one mole of methane, 16.04 g, and that there are 6.022 × 1023 molecules present in one mole, you can also calculate the mass of an average molecule of CH4. (This is an average mass because there are several isotopes of carbon and hydrogen; the mass of a given molecule is determined by the isotopes making up that molecule.) Average molecular mass 5

C

O

O

C OH C

H

Notice the dual meaning of the subscripts in the molecular formula of a compound. On the particulate scale, the subscripts represent the number of atoms of each element present in one molecule of the compound. Thus, one molecule of CH4 consists of one carbon atom and four hydrogen atoms. On the macroscopic scale, the subscripts refer to the number of moles of each element present in one mole of the compound. One mole of CH4 consists of 1 mole of carbon atoms and 4 moles of hydrogen atoms. Figure 2.27 illustrates 1-mol quantities of several common compounds. To find the molar mass of any compound, you need to add up the atomic weights for each element in the compound, taking into account any subscripts on elements. For example, the molecular formula of aspirin is C9H8O4. In one mole of aspirin there are 9  mol of carbon atoms, 8  mol of hydrogen atoms, and 4  mol of oxygen atoms, which add up to 180.15 g/mol of aspirin.

O

CH3 C

C

C C

H

16.04 g 1 mol ? 5 2.664 3 1023 g/molecule mol 6.022 × 1023 molecule

H

C

Mass of C in 1 mol C9H8O4 5 9 mol C 3

12.01 g C 5 108.09 g C 1 mol C

Mass of H in 1 mol C9H8O4 5 8 mol H 3

1.008 g H 5 8.064 g H 1 mol H

Mass of O in 1 mol C9H8O4 5 4 mol O 3

16.00 g O 5 64.00 g O 1 mol O

H

Aspirin Formula  Aspirin has the

molecular formula C9H8O4 and a molar mass of 180.15 g/mol. Aspirin is the common name of the compound acetylsalicylic acid.

Total mass of 1 mol of C9H8O4 5 molar mass of C9H8O4 5 180.15 g

Aspirin, C9H8O4 180.2 g/mol

Copper(II) chloride dihydrate, CuCl2 ∙ 2 H2O 170.5 g/mol

Iron(III) oxide, Fe2O3 159.7 g/mol

© Charles D. Winters/Cengage

H2O 18.02 g/mol

Figure 2.27  One mole of some compounds.  In the second compound, CuCl2 ∙ 2 H2O, one

formula unit has one Cu2+ ion, two Cl− ions, and two water molecules. The molar mass is the sum of the mass of 1 mol of Cu, 2 mol of Cl, and 2 mol of H2O.

104

Chapter 2 / Atoms, Molecules, and Ions

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As was the case with elements, it is important to be able to convert between amounts (moles) and mass (grams). For example, if you take 325 mg (0.325 g) of aspirin in one tablet, what amount of the compound have you ingested? Based on a molar mass of 180.15 g/mol, there is 0.00180 mol of aspirin per tablet. 0.325 g aspirin 3

1 mol aspirin 5 0.001804 mol aspirin 5 0.00180 mol aspirin 180.15 g aspirin

Using the molar mass of a compound it is possible to determine the number of molecules in any sample from the sample mass and to determine the mass of one molecule. For example, the number of aspirin molecules in one tablet is 0.001804 mol aspirin 3

6.022 3 1023 molecules 5 1.09 3 1021 molecules 1 mol aspirin

and the average mass of one molecule is 1 mol aspirin 180.15 g aspirin 3 5 2.992 3 1022 g/molecule 1 mol aspirin 6.022 3 1023 molecules

Ex am p le 2.10 Strategy Map

Molar Mass and Moles

Problem Calculate the molar mass of oxalic acid. Find the amount of oxalic acid in a given mass. Then find the number of molecules and the number of C atoms in the sample.

Problem  What is the molar mass of oxalic acid (H2C2O4)? What amount of oxalic acid is

Data/Information •­  Mass of sample •­  Formula of compound •­  Avogadro’s number

present in 16.5 g of oxalic acid? How many molecules of oxalic acid are in 16.5 g of the acid? How many atoms of carbon are in 16.5 g of oxalic acid?

What Do You Know?  You know the mass and formula of oxalic acid. The molar mass of the compound can be calculated based on the formula. Strategy Step 1. Calculate the molar mass of the compound. The molar mass is the sum of the masses of the component atoms. Step 2.  Use the molar mass to convert mass to amount (moles). Step 3.  Use Avogadro’s number to calculate the number of molecules from the amount. Step 4.  Determine the number of atoms of carbon.

Solution Step 1

Calculate the molar mass of the compound. The formula for oxalic acid, H2C2O4, tells you that there are 2 mol H, 2 mol C, and 4 mol O in 1 mol of oxalic acid. The molar masses for these elements are on the periodic table. 2 mol H per mol H2C2O4 3

1.008 g H = 2.016 g H per mol H2C2O4 1 mol H

2 mol C per mol H2C2O4 3

12.01 g C = 24.02 g C per mol H2C2O4 1 mol C

4 mol O per mol H2C2O4 3

16.00 g O = 64.00 g O per mol H2C2O4 1 mol O

Molar mass of H2C2O4 =  90.04 g per mol H2C2O4

2.7  Atoms, Molecules, and the Mole

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105

Step 2

Use the molar mass to convert mass to amount (moles). The molar mass (expressed here in units of 1  mol/90.04  g) is used in all mass–mole conversions. 1 mol

 16.5 g H2C2O4 3 90.04 g H2C2O4 5 0.1833 g H2C2O4 5 0.183 mol H2C2O4 Step 3

Use Avogadro’s number to calculate the number of molecules from the amount. 0.1833 mol 3  

6.022 3 1023 molecules 5 1.104 3 1023 molecules 1 mol = 1.10 × 1023 molecules

Step 4

Determine the number of carbon atoms. From the formula, you know that there are two atoms of carbon in each molecule of oxalic acid. 2 C atoms

23 23  1.104 3 10 molecules 3 1 molecule 5 2.21 × 10 C atoms

Think about Your Answer  The mass of oxalic acid is 16.5 g, much less than the mass of a mole (90.04 g), so make sure your answer reflects this. The number of acid molecules should be much less than in one mole of molecules.

Check Your Understanding If you have 454 g of citric acid (H3C6H5O7), what amount (moles) does this represent? How many molecules? How many atoms of carbon?

Ionic compounds such as NaCl do not exist as individual molecules. Thus, for ionic compounds chemists write the simplest formula that shows the relative number of each kind of atom in a formula unit of the compound, and the molar mass is calculated from this formula (M for NaCl = 58.44 g/mol). To differentiate substances like NaCl that do not contain molecules, chemists sometimes refer to their formula mass instead of their molar mass.

2.8 Chemical Analysis: Determining Compound Formulas Goals for Section 2.8 • Express the composition of a compound in terms of percent composition. • Determine the empirical and molecular formula of a compound using percent composition or other experimental data.

Given a sample of an unknown compound, how can its formula be determined? The answer lies in chemical analysis, a major branch of chemistry that deals with the determination of formulas and structures.

Percent Composition A central principle of chemistry is that any sample of a pure compound always consists of the same elements combined in the same proportion by mass. Suppose you have 1.0000 mol of NH3 or 17.031 g. This mass of NH3 is composed of 14.007 g of N (1.0000 mol) and 3.024 g of H (3.000 mol). You can calculate the percent composition by mass for each element in the compound by dividing the mass of each element in the compound by the total mass of the compound and then multiplying by 100.

106

Chapter 2 / Atoms, Molecules, and Ions

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mass of N in 1 mol NH3 3 100% mass of 1 mol NH3

Mass percent N in NH3 5 5

14.007 g N 3 100% 17.031 g NH3

5

composition can be expressed as a percent (mass of an element in a 100-g sample). 82.244% of NH3 mass is nitrogen.

5 82.244% mass of H in 1 mol NH3 3 100% mass of 1 mol NH3

Mass percent H in NH3 5

Molecular Composition Molecular

3.024 g H 3 100% 17.031 g NH3

5 17.76%

H

N H

H

17.76% of NH3 mass is hydrogen.

These values tell you that in a 100.00-g sample there are 82.24 g of N and 17.76 g of H.

Ex am p le 2.11

Using Percent Composition Problem  (a) What is the mass percent of each element in propane, C3H8? (b) What mass of carbon is contained in 454 g of propane?

What Do You Know?  You know the formula of propane. You will need the atomic weights of C and H to calculate the mass percent of each element.

Strategy (a) Mass percent of each element in propane:

Step 1.  Calculate the molar mass of propane.



Step 2. The mass percent of each element is the mass of the element in one mole of the compound divided by the molar mass of the compound and multiplied by 100.

(b) The mass of C in 454 g of C3H8 is obtained by multiplying this mass by the % C and dividing by 100.

Solution (a) Mass percent of each element in C3H8:

Step 1.  The molar mass of C3H8 is 44.10 g/mol.



Step 2.  Mass percent of C and H in C3H8: 3 mol C 1 mol C3H8

3

12.01 g C 5 36.03 g C/1 mol C3H8 1 mol C

Mass percent of C in C3H8 5 8 mol H 1 mol C3H8

3

1.008 g H 5 8.064 g H/1 mol C3H8 1 mol H

Mass percent of H in C3H8 5



36.03 g C 3 100% 5 81.70% C 44.10 g C3H8

8.064 g H 3 100% 5 18.29% H 44.10 g C3H8  

2.8  Chemical Analysis: Determining Compound Formulas

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107

(b) Mass of C in 454 g of C3H8: 454 g C3H8 3

 

81.70 g C 5 371 g C 100.0 g C3H8

Think about Your Answer  Once you know the percent C in the sample, you could calculate the percent H from it knowing that %H = 100% − %C.

Check Your Understanding 1.

Express the composition of ammonium carbonate, (NH4)2CO3, in terms of the mass of each element in 1.00 mol of compound and the mass percent of each element.

2.

What is the mass of carbon in 454 g of octane, C8H18?

Empirical and Molecular Formulas from Percent Composition Now consider the reverse of the procedure just described. That is, use relative mass or percent composition data to find a molecular formula. Suppose you know the identity of the elements in a sample and have determined the mass of each element in a given mass of compound by chemical analysis. You can then calculate the relative amount (moles) of each element, which is also the relative number of atoms of each element in the formula of the compound. For example, for a compound composed of atoms of A and B, the steps from percent composition to a formula are as follows: S T EP 1.

ST EP 2.

ST EP 3.

Convert mass percent to mass

Convert mass to amount

Find mole ratio

%A %B

gA

x mol A

gB

y mol B

ST EP 4 .

Determine the whole-number ratio of A to B x mol A y mol B

AaBb

For example, hydrazine is a compound used to remove oxygen from water in heating and cooling systems and is a close relative of ammonia. Chemical analysis shows that hydrazine is composed of 87.42% N and 12.58% H. What is the molecular formula for hydrazine? Deriving a Formula Percent

composition gives the mass of an element in 100 g of a sample. However, in deriving a formula, any amount of sample is appropriate if you know the mass of each element in that sample mass.

108

Step 1. Convert mass percent to mass. The mass percentages of N and H in hydrazine tell you there are 87.42 g of N and 12.58 g of H in a 100.00-g sample. Step 2. Convert the mass of each element to amount (moles). The amount of each

element in the 100.00-g sample is 87.42 g N 3

1 mol N 5 6.2412 mol N 14.007 g N

12.58 g H 3

1 mol H 5 12.480 mol H 1.008 g H

Chapter 2 / Atoms, Molecules, and Ions

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Problem Solving Tip 2.2 Finding Empirical and Molecular Formulas significant figures can give a misleading result.

• The experimental data available to find a formula may be in the form of percent composition or the masses of elements combined in some mass of compound. No matter what the starting point, the first step is always to convert masses of elements to moles.

• Determining the molecular formula of a compound after calculating the empirical formula requires knowing the molar mass of the compound.

• When finding atom ratios, always divide the larger number of moles by the smaller one.

• Empirical and molecular

• When both the percent composi-

f­ormulas can differ for molecular ­compounds. In contrast, there is no molecular formula for an ionic compound; all that can be recorded is the empirical formula.

• Be sure to use at least three significant figures when calculating empirical formulas. Using fewer

tion and the molar mass are known for a compound, the alternative method mentioned in Think about Your Answer in Example 2.12 can be used.

Step 3. Find the mole ratio of elements. Use the amount (moles) of each element in the 100.00-g sample to find the amount of one element relative to the other. (To do this it is usually best to divide the larger amount by the smaller amount.)

12.480 mol H 2.000 mol H 5 6.2412 mol N 1.000 mol N Step 4. Determine the whole-number ratio. The mole ratio shows that there are 2 mol of H atoms for every 1 mol of N atoms in hydrazine. This is already a wholenumber ratio, so no further adjustment is needed. Thus, in one molecule, two atoms of H occur for every atom of N; that is, the formula is

NH2

This simplest, whole-number atom ratio of atoms in a formula is called the empirical formula. The molecular formula, however, must convey not only this information but also the total number of atoms in the molecule. For example, the empirical formula NH2 for hydrazine tells you the relative numbers of atoms for each element in the compound: Hydrazine has twice as many H atoms as N atoms. But the molecular formula could be NH2, N2H4, N3H6, N4H8, or any other formula with a 1:2 ratio of N to H. To determine the molecular formula from the empirical formula, you need to know the molar mass of the compound. For example, experiments show that the molar mass of hydrazine is 32.0 g/mol, twice the formula mass of NH2, which is 16.0 g/mol. Thus, the molecular formula of hydrazine is two times the empirical formula of NH2, that is, N2H4.

Ex am p le 2.12

Calculating a Formula from Percent Composition Problem  Many soft drinks contain sodium benzoate as a preservative. When you consume the sodium benzoate, it reacts with the amino acid glycine in your body to form hippuric acid, which is then excreted in the urine. Hippuric acid has a molar mass of 179.17 g/mol and is 60.33% C, 5.06% H, and 7.82% N; the remainder is oxygen. What are the empirical and molecular formulas of hippuric acid?

What Do You Know?  You know the mass percent of C, H, and N. The mass percent of oxygen is not known but is obtained by difference. You know the molar mass of ­hippuric acid but will need the atomic weights of C, H, N, and O for the calculation.

2.8  Chemical Analysis: Determining Compound Formulas

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109

Strategy Map

Strategy 

Problem Determine empirical and molecular formulas based on known composition and known molar mass.

Step 1. Convert each mass percent to mass.

Data/Information •­  Molar mass •­  Percent composition Step 1

Step 2. Convert the mass of each element to amount (moles). Step 3. Find the mole ratios of the elements. Step 4. Determine the whole-number ratios to find the empirical formula. Step 5. Divide the known molar mass by the empirical formula mass to determine the molecular formula.

Solution  Convert each mass percent to mass. Assume that each mass percent is equivalent to the mass in grams in a 100-g sample. Thus, there are 60.33 g C, 5.06 g H, and 7.82 g of N in 100 g of hippuric acid. The mass of oxygen in the sample is 100.00 g = 60.33 g C + 5.06 g H + 7.82 g N + mass of O Mass of O = 26.79 g O

Step 2

Convert the mass of each element to amount (moles). Use the molar mass of each element to determine the amount (moles) of each element present in the 100-g sample.

Step 3

60.33 g C 3

1 mol C 5 5.0229 mol C 12.011 g C

5.06 g H 3

1 mol H 5 5.020 mol H 1.008 g H

7.82 g N 3

1 mol N 5 0.5582 mol N 14.01 g N

26.79 g O 3

1 mol O 5 1.6745 mol O 15.999 g O

Find the mole ratios of the elements. To find the mole ratios, the best approach is to base the ratios on the smallest number of moles present—in this case, nitrogen. mol C 5.0229 mol C 9.00 mol C 5 5 5 9 mol C/1 mol N mol N 0.5582 mol N 1.00 mol N mol H 5.020 mol H 8.99 mol H 5 5 5 9 mol H/1 mol N mol N 0.5582 mol N 1.00 mol N mol O 1.6745 mol O 3.00 mol O 5 5 5 3 mol O/1 mol N mol N 0.5582 mol N 1.00 mol N

Step 4

Determine the whole-number ratios to find the empirical formula. These mole ratios are already whole-number ratios, so no further adjustment is needed. There are 9 mol of C, 9 mol of H, and 3 mol of O for each mol of N. Thus, the empirical formula is C9H9NO3.

110

Chapter 2 / Atoms, Molecules, and Ions

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Step 5

Divide the known molar mass by the empirical formula mass to determine the molecular formula. The formula mass of the empirical formula (C9H9NO3) is (9 3 12.011 g/mol) 1 (9 3 1.008 g/mol) 1 (1 × 14.007 g/mol) 1 (3 3 15.999 g/mol) 5 179.175 g/mol The experimentally determined molar mass of hippuric acid is 179.17 g/mol.

molar mass 179.17 g/mol 5 5 0.99997 5 1 empirical formula mass 179.175 g/mol The ratio of the molar mass to the empirical formula mass is 1, so the molecular formula is the same as the empirical formula. The molecular formula is C9H9NO3.

Think about Your Answer  There is another approach to finding the molecular formula if you know both the percent composition and the molar mass of a compound. Multiply the molar mass of the compound by the percent composition of each element divided by 100. This gives the mass of the element in 1 mol of the compound, which can then be converted to amount (moles). For example, for carbon in hippuric acid, 179.17 g hippuric acid 3 108.09 g C 3 Hippuric Acid, C9H9NO3 This substance, which can be isolated as white crystals, is found in the urine of humans and herbivorous animals.

60.33 g C 5 108.09 g C 100 g hippuric acid

1 mol C 5 9.000 mol C 12.011 g C

Likewise, the following values are obtained for the other elements in hippuric acid: 9.07 g of H (8.99 mol), 14.01 g N (1.000 mol), and 48.00 g of O (3.000 mol). This gives a molecular formula of C9H9NO3. However, this approach can only be used when you know both the percent composition and the molar mass.

Check Your Understanding 1.

What is the empirical formula of naphthalene, C10H8?

2.

The empirical formula of acetic acid is CH2O. If its molar mass is 60.05 g/mol, what is the molecular formula of acetic acid?

3.

Isoprene is a liquid compound that can be polymerized to form natural rubber. It is composed of 88.17% carbon and 11.83% hydrogen. Its molar mass is 68.11 g/mol. What are its empirical and molecular formulas?

4.

Camphor is found in camphor wood, much prized for its distinctive odor. It is composed of 78.90% carbon and 10.59% ­hydrogen. The remainder is oxygen. What is its empirical formula?

Determining a Formula from Mass Data The composition of a compound in terms of mass percent gives the mass of each element in a 100.0-g sample. Sometimes, other information about the composition of compounds is collected in the laboratory and used to determine the formulas of compounds. Two ways of doing this are: •



Combining known masses of elements to give a sample of the compound of known mass. Element masses can be converted to amounts (moles), and the ratio of amounts gives the combining ratio of atoms—that is, the empirical formula. This approach is described in Example 2.13.

2.8  Chemical Analysis: Determining Compound Formulas

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•

Decomposing a known mass of an unknown compound into “pieces” of known composition. If the masses of the “pieces” can be determined, the ratio of moles of the “pieces” gives the formula. An example is a decomposition such as Ni(CO)4(ℓ) → Ni(s) + 4 CO(g)

The masses of Ni and CO can be converted to moles, whose 1∶4 ratio would reveal the formula of the compound. This approach is described in Example 2.14.

E xamp le 2.13

Formula of a Compound from Combining Masses Problem  Oxides of virtually every element are known. Bromine, for example, forms several oxides when treated with ozone (O3). Suppose you allow 1.250 g of bromine, Br2, to react with ozone and obtain 1.876  g of BrxOy. What is the empirical formula of the product?

What Do You Know?  You began with a given mass of bromine and all of the bromine became part of bromine oxide of unknown formula. You also know the mass of the product, and because you know the mass of Br in this product, you can determine the mass of O in the product.

Strategy Step 1. The mass of oxygen is determined as the difference between the product mass and the mass of bromine used. Step 2. Calculate the amounts of Br and O from the masses of each element. Step 3. Find the mole ratio of the elements. Step 4. Determine the whole-number ratio to find the empirical formula.

Solution  Step 1. You already know the mass of bromine in the compound, so you can calculate the mass of oxygen in the compound. 1.876 g product − 1.250 g Br2 = 0.626 g O Step 2. Next, calculate the amount of each reactant. Notice that, although Br2 was the reactant, you need to know the amount of Br atoms in the product. 1.250 g Br2 3

1 mol Br2 5 0.0078218 mol Br2 159.81 g Br2

0.0078218 mol Br2 3 0.626 g O 3

2 mol Br 5 0.015644 mol Br 1 mol Br2

1 mol O 5 0.03913 mol O 16.00 g O

Step 3. Find the ratio of moles of O to moles of Br: Mole ratio 5

0.03913 mol O 2.50 mol O 5 0.015644 mol Br 1.00 mol Br

Step 4. T he mole ratio is 2.5 mol O/1.0 mol Br. However, atoms combine in the ratio of small whole numbers, so you should double the mole ratio found in Step 3 to give a whole-number ratio of 5 mol O to 2 mol Br. Thus,  the product is Br2O5 (dibromine pentaoxide). 

112

Chapter 2 / Atoms, Molecules, and Ions

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Think about Your Answer  The whole–number ratio of 5∶2 was found by realizing that 2.5 = 2 1/2 = 5/2. The calculation gave the empirical formula for this compound. To determine whether this is also the molecular formula, the molar mass of the compound would have to be determined.

Check Your Understanding Gallium oxide, GaxOy, forms when gallium is combined with oxygen. Suppose you allow 1.25 g of gallium (Ga) to react with oxygen and obtain 1.68 g of GaxOy. What is the formula of the product?

Ex am p le 2.14

Determining the Formula of a Hydrated Compound hydrated copper(II) sulfate, CuSO4 ∙ x H2O, that is, the number of water molecules for each unit of CuSO4. In the laboratory you weigh out 1.023 g of the solid. After heating the solid thoroughly in a porcelain crucible ­(Figure), 0.654  g of nearly white, anhydrous copper(II) sulfate, CuSO4, remains.

White CuSO4

Blue CuSO4 ∙ x H2O

© Charles D. Winters/Cengage

Problem  You want to know the value of x in blue,

1.023 g CuSO4 ∙ x H2O + heat →   0.654 g CuSO4 + ? g H2O

Strategy Map Problem Determine formula of hydrated salt based on masses of water and dehydrated salt.

What Do You Know?  You know the mass of the copper(II) sulfate sample including water (before heating) and with no water (after heating). Therefore, you know the mass of CuSO4 and can determine the mass of water in the sample.

Strategy  To find x you need to know the amount of H2O per mole CuSO4. Data/Information •­ Mass of sample before and after heating to dehydrate

Step 1. Determine the mass of water released upon heating the hydrated compound. Step 2. Calculate the amount (moles) of CuSO4 and H2O from their masses and molar masses. Step 3.  Determine the smallest whole-number ratio (amount H2O/amount CuSO4).

Solution  Step 1

Determine the mass of water released upon heating the hydrated compound. Mass of hydrated compound − Mass of anhydrous compound, CuSO4 Mass of water

Step 2

 −0.654  0.369 g

Calculate the amount (moles) of CuSO4 and H2O from their masses and molar masses. 0.369 g H2O 3

1 mol H2O 5 0.02048 mol H2O 18.02 g H2O

0.654 g CuSO4 3



1.023 g

1 mol CuSO4 5 0.004098 mol CuSO4 159.6 g CuSO4

2.8  Chemical Analysis: Determining Compound Formulas

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113

Step 3

Determine the mole ratio (amount H2O relative to amount CuSO4). 0.02048 mol H2O 5.00 mol H2O 5 0.004098 mol CuSO4 1.00 mol CuSO4 The water-to-CuSO4 ratio is 5∶1, so the   formula of the hydrated compound is CuSO4 ∙ 5 H2O.  Its name is copper(II) sulfate pentahydrate.

Think about Your Answer  The ratio of the amount of water to the amount of CuSO4 is a whole number. This is almost always the case with hydrated compounds.

Check Your Understanding Hydrated nickel(II) chloride is a beautiful green, crystalline compound. When heated strongly, the compound is dehydrated. If 0.235 g of NiCl2 ∙ x H2O gives 0.128 g of NiCl2 on heating, what is the value of x?

2.9 Instrumental Analysis: Determining Compound Formulas Goals for Section 2.9 • Determine a molecular formula from a mass spectrum. • Identify isotopes using mass spectrometry.

Determining a Formula by Mass Spectrometry In addition to the more traditional chemical methods of determining a molecular formula described in previous sections of this chapter, there are many instrumental methods as well. One of them is mass spectrometry, a technique that was introduced earlier when discussing the existence of isotopes and their relative abundance (see Figure 2.3). If a compound can be vaporized, the vapor can be passed through an electron beam in a mass spectrometer where high-energy electrons collide with the gas-phase molecules. These high-energy collisions cause molecules to lose electrons and become positive ions, which usually fragment into smaller pieces. Figure 2.28 shows the mass spectrum for the compound ethanol, CH3CH2OH. The horizontal axis shows the mass-to-charge ratio (m/Z) of a given ion. Because almost all observed ions have a charge of Z = 1+, the value observed is the mass of the ion. As illustrated in this figure, the cation created from ethanol itself (CH3CH2OH+, m∙Z = 46), called the parent ion or molecular ion, fragments (losing an H atom) to give another cation (CH3CH2O+, m∙Z = 45), which further fragments. Analysis of the spectrum can help identify a compound and can give an accurate molar mass.

Molar Mass and Isotopes in Mass Spectrometry Bromobenzene, C6H5Br, has a molar mass of 157.010 g∙mol. Why, then, are there two prominent lines at mass-to-charge ratios (m∙Z) of 156 and 158 in the mass spectrum of the compound (Figure 2.29)? The answer shows the influence of isotopes on molar mass. Bromine has two naturally occurring isotopes, 79Br and 81Br. They are 50.7% and 49.3% abundant, respectively. Using the most abundant isotopes of C and H (12C and 1H), the mass of the molecule having the 79Br isotope, C6H579Br, is 156. The mass of the molecule containing the 81Br isotope, C6H581Br, is 158. The relative size of these two peaks in the spectrum reflects the relative abundances of the two ­bromine isotopes.

114

Chapter 2 / Atoms, Molecules, and Ions

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Relative abundance of ions

100

CH2OH+ (m/Z = 31)

80

60

C2H5+ (m/Z = 29)

40

CH3CH2OH+ (m/Z = 46)

CH3+ (m/Z = 15)

20

0

CH3CH2O+ (m/Z = 45)

10

One of many fragment ion peaks Parent ion peak

20

30

40

50

Mass-to-charge ratio (m/Z) Figure 2.28  Mass spectrum of ethanol, CH3CH2OH.  The peak (or line) in the spectrum at mass 46 is from an ion of ethanol that has not undergone decomposition (CH3CH2OH+). This ion is referred to as the parent ion or molecular ion. The mass designated by the peak for the parent ion confirms the formula of the mole­cule. Other peaks are for fragment ions. This pattern of lines can provide further, unambiguous evidence of the formula of the compound.

The calculated molar mass of bromobenzene (157.010 g/mol) reflects the abundances of all of the isotopes. In contrast, the mass spectrum has a line for each possible combination of isotopes. This also explains why there are also small lines at the mass-to-charge ratios of 157 and 159. They arise from various combinations of 1 H, 12C, 13C, 79Br, and 81Br atoms. In fact, careful analysis of such patterns can identify a molecule unambiguously. 100

158 = (12C)6(1H)581Br+

Relative abundance of ions

80

156 = (12C)6(1H)579Br+ 60

40

20

0

0

40

80

120

160

Mass-to-charge ratio (m/Z) Figure 2.29  Mass spectrum of bromobenzene, C6H5Br.  Two parent ion peaks are present at

m/Z ratios of 156 and 158. The similar peak heights reflect the near equal abundances of the two isotopes of bromine, 79Br and 81Br.



2.9  Instrumental Analysis: Determining Compound Formulas

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115

E xamp le 2.15

Isotopic Abundance by Mass Spectrometry Problem  The mass spectrum of phosphorus trichloride is illustrated here.

Relative abundance of ions

100

101

80 103 60 136 138

40

20 105

66 68 0 60

80

100 120 Mass-to-charge ratio (m/Z)

140

140

Phosphorus, 31P, has one stable isotope. Chlorine has two stable isotopes, 35Cl and 37Cl. (a) What molecular species give rise to the parent ion peaks at m/Z ratios of 136, 138, and 140? (b) What species give rise to the peaks at m/Z ratios of 101, 103, and 105? (c) Predict the structural formula (see page 85) of phosphorus trichloride from its mass spectrum.

What Do You Know?  You know that PCl3 molecules ionize to form positive ions. Some of the (parent) ions fragment into smaller ions. You also know the mass numbers of each atom. The mass spectrum shows you the mass of each ion divided by its charge (m/Z).

Strategy  Try to generate the m/Z ratios observed in the mass spectrum by combining the mass numbers of the elements (35Cl, 37Cl, and 31P) in various ways. Solution  (a) The parent ion peaks correspond to ions that have not fragmented. A parent ion formed from one 31P atom and three 35Cl atoms has a m/Z ratio of 136 (if the ion charge is 1+). One 31P atom combined with two 35Cl atoms and one 37Cl atom has a m/Z ratio of 138. Finally, a 31P atom combined with one 35Cl and two 37Cl atoms has a m/Z ratio of 140. Thus, the molecular species are P35Cl3 (m/Z = 136), P35Cl237Cl (m/Z = 138), and P35Cl37Cl2 (m/Z = 140). (b) The ions with m/Z ratios of 101, 103, and 105 have the formula PCl2+. The species are P35Cl2 (m/Z = 101), P35Cl37Cl (m/Z = 103), and P37Cl2 (m/Z = 105). (c) The probable structure is below.

Cl Cl

P

Cl

The mass spectrum shows fragment ions of PCl+ (m/Z = 66 and 68) and PCl2+, but no fragment ions of Cl2+ or Cl3+. The absence of Cl2+ or Cl3+ ions is evidence that the chlorine atoms are attached to the phosphorus atom, not each other.

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Chapter 2 / Atoms, Molecules, and Ions

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Think about Your Answer  When identifying ions in a mass spectrum, it is important to use the masses of each isotope of an element, rather than the average atomic mass of the element.

Check Your Understanding The mass spectrum of phosphorus tribromide is illustrated below. Bromine has two stable isotopes, 79Br and 81Br with abundances of 50.7% and 49.3%, respectively.

Relative abundance of ions

100

191

80

60 270 272

189 193 40

20

0 180

268

200

220 240 Mass-to-charge ratio (m/Z)

260

274

280

(a) What molecular species give rise to the parent ion peaks at m/Z ratios of 270 and 272? (b) Explain why the relative abundances of the ions at m/Z ratios of 268 and 274 are ­approximately one-third of those at 270 and 272.

Applying Chemical Principles

In 1991, a hiker in the Alps on the Austrian-Italian border found the well-preserved remains of an approximately 46-year-old man, now nicknamed “The Iceman,” who lived about 5300 years ago. Studies using isotopes of oxygen, strontium, lead, and argon, among others, have helped scientists paint a d ­ etailed picture of the man and his life. The abundance of the 18O isotope of oxygen in a person is related to the latitude and altitude at which the person was born and raised. Oxygen in biominerals such as teeth and bones comes primarily from ingested water. The lakes and rivers on the northern side of the Alps are known to have a lower 18O content than those on the southern side of the mountains. The 18 O content of the teeth and bones of the Iceman was found to be relatively high and characteristic of the watershed south of the Alps. He had clearly been born and raised in that area.



Reuters/Alamy Stock Photo

2.1  Using Isotopes: Ötzi, the Iceman of the Alps

Ötzi the Iceman.  A well-preserved mummy of a man who lived in northern Italy about 5300 years ago.

Applying Chemical Principles

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117

The relative abundance of isotopes of many other elements also varies slightly from place to place and in their incorporation into different minerals. Strontium, a member of the same periodic group as calcium, is incorporated into teeth and bones. The ratio of strontium isotopes, 87Sr/86Sr, and of lead isotopes, 206 Pb/204Pb, in the Iceman’s teeth and bones was characteristic of soils from a narrow region of Italy south of the Alps, which established more clearly where he was born and lived most of his life. The investigators also looked for food residues in the Iceman’s intestines. Although a few grains of cereal were found, they located tiny flakes of mica believed to have broken off stones used to grind grain and that were therefore eaten when the man ate the grain. They analyzed these flakes using argon isotopes, 40Ar and 39Ar, and

found their signature was like that of mica in an area south of the Alps, thus establishing where he lived in his later years. The many isotope studies show that the Iceman lived thousands of years ago in a small area about 10–20 kilometers west of Merano in northern Italy. For details of the isotope studies, see W. Müller, et al., ­Science, 2003, 302, 862–866.

Questions

1. How many neutrons are there in atoms of 18O? In each of the two isotopes of lead? 2. There are three stable isotopes of oxygen (16O, mass 15.9949, 99.757%, 17O, mass 16.9991, 0.038%, and 18O, 17.9992, 0.205%). Use these data to calculate the atomic weight of oxygen.

The practice in medicine for some centuries is to find compounds that are toxic to certain organisms but not so toxic that the patient is harmed. In the early part of the twentieth century, Paul Ehrlich set out to find just such a compound that would cure syphilis, a sexually transmitted disease that was rampant at the time. He screened hundreds of compounds, and found that his 606th compound was effective: an arseniccontaining drug now called salvarsan. It was used for some years for syphilis treatment until penicillin was discovered in the 1930s. Salvarsan was a forerunner of the modern drug industry. Interestingly, what chemists long thought to be a single compound was in fact discovered to be a mixture of compounds. Question 2 will lead you to the molecular formula for each of them.

Questions

1. Arsenic is found widely in the environment and is a major problem in the ground water supply in Bangladesh. Orpiment is one arsenic-containing mineral and enargite is another. The latter has 19.024% As, 48.407% Cu, and 32.569% S. What is the empirical formula of the mineral? 2. Salvarsan was long thought to be a single substance. Recently, however, a mass spectrometry study of the compound shows it to be a mixture of two molecules with the

John C. Kotz

2.2  Arsenic, Medicine, and the Formula of Compound 606

A sample of orpiment, a common arsenic-containing mineral (As2S3).  The name of the element is thought to come from the Greek word for this mineral, which was long favored by seventeenth century Dutch painters as a pigment.

same empirical formula. Each has the composition 39.37% C, 3.304% H, 8.741% O, 7.652% N, and 40.932% As. One has a molar mass of 549  g/mol and the other has a molar mass of 915 g/mol. What are the molecular formulas of the compounds?

2.3  Argon—An Amazing Discovery

118

Wellcome Images CC/DIOMEDIA

Sir William Ramsay (1852–1916).  Ramsay was a Scottish chemist who discovered several of the noble gases (for which he received the Nobel Prize in Chemistry in 1904). Lord Rayleigh received the Nobel Prize in Physics, also in 1904, for the discovery of argon.

Wellcome Images CC/Diomedia

The noble gas argon was discovered by Sir William Ramsay and John William Strutt (the third Lord ­Rayleigh) in England and reported in scientific journals in 1895. In making this discovery, Ramsay and Lord Rayleigh made highly accurate measurements of gas densities. They found that gaseous nitrogen (N2) formed by thermal decomposition of ammonia had a density that was slightly lower than the density of the gas that remained after O2, CO2, and H2O were removed from air. The reason for the difference is that the sample derived from air contained a very small amount of other gases. After removing N2 from the sample by reacting it with red hot magnesium (to form Chapter 2 / Atoms, Molecules, and Ions

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Mg3N2), a small quantity of gas remained that was more dense than air. This was identified as argon. Lord Rayleigh’s experimentally determined densities for oxygen, nitrogen, and air are given below:

from the isotopic masses and fractional abundances.) Assume dry air with CO2 removed is 20.96% (by volume) oxygen, 78.11% nitrogen, and 0.930% argon. Determine the density of argon. 3. Atmospheric argon is a mixture of three stable isotopes, 36Ar, 38 Ar, and 40Ar. Use the information in the table below to determine the atomic mass and natural abundance of 40Ar.

Gas

Density (g/L)

Oxygen

1.42952

Nitrogen, derived from air

1.25718

36

Nitrogen, derived from ammonia

1.25092

38

Air, with water and CO2 removed

1.29327

40

Questions

1. To determine the density of atmospheric nitrogen, Lord Rayleigh removed the oxygen, water, and carbon dioxide from air, then filled an evacuated glass globe with the remaining gas. He determined that a mass of 0.20389  g of nitrogen has a density of 1.25718 g/L under standard conditions of temperature and pressure. What is the volume of the globe (in cm3)? 2. The density of a mixture of gases may be calculated by summing the products of the density of each gas and the fractional volume of space occupied by that gas. (Note the similarity to the calculation of the molar mass of an element

Isotope

Atomic Mass

Abundance (%)

Ar

35.967545

0.334

Ar

37.962732

0.063

Ar

?

?

4. Given that the density of argon is 1.78 g/L under standard conditions of temperature and pressure, how many argon atoms are present in a room with dimensions 4.0  m  × 5.0  m  × 2.4  m that is filled with pure argon under these conditions of temperature and pressure? References 1. Lord Rayleigh and William Ramsay, Proceedings of the Royal Society of London, 1894, 57, 265–287. 2. The Gases of the Atmosphere and Their History, 4th ed., William Ramsay, MacMillan and Co., Limited, London, 1915.

Think–Pair–Share 1. Consider particles compositions:

with

the

following

subatomic

Particle

Protons

Neutrons

Electrons

1

16

16

18

2

18

22

18

3

19

20

19

4

19

21

19

5

20

20

20

6

20

20

18

(a) For each particle, write the complete chemical symbol including the element symbol, atomic number, mass number, and any nonzero charge. (b) Which particles are neutral atoms? Which are isotopes? Which is a cation? Which is an anion? Explain how you arrived at these answers. 2. The figure shows a molecular model of ribose, a component of RNA. The black spheres represent carbon atoms, the red spheres represent oxygen atoms, and the white spheres represent hydrogen atoms.

Ribose

(a) What is the molecular formula? (b) What is the empirical formula? 3. Write the formulas and names for the ionic compounds that can be made by combining the cations: NH41, Mg21, and Fe31 with the anions Cl2, SO422, and PO432. 4. Increasing amounts of carbon dioxide are implicated in climate change. (a) What is the formula of carbon dioxide? (b) How many atoms of carbon and how many atoms of ­oxygen are present in one molecule of carbon dioxide? (c) How many moles of carbon and how many moles of ­oxygen are present in one mole of carbon dioxide? (d) Is there a relationship between the answers for parts b and c? Explain in a few sentences why there should or should not be a relationship.

Think–Pair–Share

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119

5. Empirical Formulas and Mole Ratios (a) What empirical formula (XmYn) is indicated by each of the following mole ratios of the elements X and Y? (i) 1.50 mol Y/mol X (ii) 1.33 mol Y/mol X (iii) 1.67 mol Y/mol X (iv) 2.50 mol Y/mol X (v) 1.25 mol Y/mol X (b) Most textbooks, including this one, instruct students to obtain the mole ratios in an empirical formula problem by dividing the larger number of moles by the smaller number of moles. If you did the opposite and divided the smaller number of moles by the larger number, could you interpret the results to obtain the correct empirical formula? Why

or why not? Why do you think textbooks say to divide the larger number of moles by the smaller number? 6. Two compounds have the molecular formula C2H4Cl2: 1,1-dichloroethane and 1,2-dichloroethane. These are their structural formulas:

H H Cl

C

C

H H H

H

C

C

H

Cl H

Cl Cl

1,1-dichloroethane

1,2-dichloroethane

 ass spectrometry was performed on samples of each comM pound, and the following spectra were obtained.

Mass Spectrum

100

Relative Intensity

80

60

40

20

0

0

20

40

60

80

100

120

80

100

120

m/Z

Mass Spectrum

100

Relative Intensity

80

60

40

20

0

0

20

40

60 m/Z

120

Chapter 2 / Atoms, Molecules, and Ions

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(a) Both spectra have peaks at m/Z 5 98, 100, and 102. Given that carbon is predominantly carbon-12 (99% abundance) and that the element chlorine has two stable isotopes: 35 Cl (76% abundance) and 37Cl (24% abundance), what are the molecular formulas for these ions? Do these ions help identify which spectrum corresponds to which compound?

(b) Do the two spectra provide evidence that the two samples are different compounds? Which spectrum corresponds to which compound? How do you know?

Chapter Goals Revisited The goals for this chapter are keyed to specific Study Questions to help you organize your review.

2.1  Atomic Structure, Atomic Number, and Atomic Mass • Describe electrons, protons, and neutrons, and the general structure of the atom. 1, 3.

• Define the terms atomic number and mass number. 2, 5–7. • Define isotopes and give the atomic symbol for a specific isotope. 8–11, 111.

2.2 Atomic Weight • Perform calculations that relate the atomic weight (atomic mass) of an element and isotopic abundances and masses. 17–22, 166b, 168.

2.3 The Periodic Table • Know the terminology of the periodic table (periods, groups) and know

how to use the information given in the periodic table. 23, 24, 27, 29, 113.

• Recognize similarities and differences in properties of some of the common elements of a group. 26, 30.

2.4 Molecules: Formulas, Models, and Names • Recognize and interpret molecular formulas, condensed formulas, and structural formulas. 31, 32.

• Recognize and interpret different types of molecular models. 31, 32, 85, 129, 134, 138, 139.

• Remember formulas and names of common molecular compounds. 36. • Name and write formulas for binary molecular compounds. 33–36.

2.5 Ions • Recognize that metal atoms commonly lose one or more electrons to form positive ions, called cations, and nonmetal atoms often gain electrons to form negative ions, called anions. 43, 44, 126.

• Predict the charge on monatomic cations and anions based on Group number. 37–40.

• Name and write the formulas of ions. 41, 42.



Chapter Goals Revisited

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121

2.6  Ionic Compounds: Formulas, Names, and Properties • Write formulas for ionic compounds by combining ions in the proper ratio to give a zero overall charge. 47–54.

• Give the names and write the formulas of ionic compounds. 55–62. • Understand the importance of Coulomb’s law in chemistry, which

describes the electrostatic forces of attraction and repulsion of ions. 63–64.

2.7 Atoms, Molecules, and the Mole • Understand the mole concept and molar mass and their application. 65–68, 72, 73.

• Use the molar mass of an element and Avogadro’s number in calculations. 69–71, 115, 116.

• Calculate the molar mass of a compound from its formula and a table of atomic weights. 77–80.

• Calculate the amount (= number of moles) of a compound represented by a given mass, and vice versa. 81, 82.

• Use Avogadro’s number to calculate the number of atoms or ions in a compound. 83–86, 128.

2.8 Chemical Analysis: Determining Compound Formulas • Express the composition of a compound in terms of percent composition. 87–89.

• Determine the empirical and molecular formula of a compound using

percent composition or other experimental data. 95–106, 137, 143, 145.

2.9 Instrumental Analysis: Determining Compound Formulas • Determine a molecular formula from a mass spectrum. 107, 108. • Identify isotopes using mass spectrometry. 109, 110.

Key Equations Equation 2.1 (page 68) Percent abundance of an isotope. Percent abundance 5

number of atoms of a given isotope 3 100% total number of atoms of all isotopes of that element

Equation 2.2 (page 69) Calculate the atomic weight from isotope abundances and the atomic mass of each isotope of an element.  % abundance isotope 1  Atomic weight 5   (mass of isotope 1)  100  % abundance isotope 2  1  (mass of isotope 2) 1 . . .  100

122

Chapter 2 / Atoms, Molecules, and Ions

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Equation 2.3 (page  97) Coulomb’s Law, the force of attraction between oppositely charged ions. charge on + and − ions

Force = −k proportionality constant

charge on electron

(n+e)(n−e) d2 distance between ions

Study Questions ▲  denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully worked solutions in the Student Solutions Manual.

Practicing Skills Atoms: Their Composition and Structure (See Section 2.1 and Example 2.1.) 1. What are the three fundamental particles from which atoms are built? What are their electric charges? Which of these particles constitute the nucleus of an atom? Which is the least massive particle of the three? 2. Define mass number. What is the difference between the mass number and atomic mass of an atom? 3. An atom has a very small nucleus surrounded by an electron cloud. Figure 2.1 represents the nucleus with a diameter of about 2 mm and describes the electron cloud as extending over 200 m. If the diameter of an atom is 1 × 10−8 cm, what is the approximate diameter of its nucleus? 4. A gold atom has a radius of 145 pm. If you could string gold atoms like beads on a thread, how many atoms would you need to have a necklace 36 cm long? 5. Give the complete symbol (AZX), including atomic number and mass number, for each of the following atoms: (a) potassium with 22 neutrons, (b) vanadium with 28 neutrons, and (c) gallium with 38 neutrons. 6. Give the complete symbol (AZX), including atomic number and mass number, for each of the following atoms: (a) iron with 30 neutrons, (b) uranium with 146 neutrons, and (c) lead with 125 neutrons.



7. How many electrons, protons, and neutrons are there in each of the following atoms? (a) magnesium-24, 24Mg (b) tin-119, 119Sn (c) thorium-232, 232Th (d) carbon-13, 13C (e) copper-63, 63Cu (f) bismuth-205, 205Bi 8. Atomic structure. (a) The synthetic radioactive element technetium is used in many medical studies. Give the number of electrons, protons, and neutrons in an atom of technetium-99. (b) Radioactive americium-241 is used in household smoke detectors and in bone mineral analysis. Give the number of electrons, protons, and neutrons in an atom of americium-241. 9. Cobalt has three radioactive isotopes used in medical studies. Atoms of these isotopes have 30, 31, and 33 neutrons, respectively. Give the complete symbol for each of these isotopes. 10. Naturally occurring silver exists as two isotopes having mass numbers 107 and 109. How many protons, neutrons, and electrons are there in each of these isotopes? 11. Name and describe the composition of the three hydrogen isotopes. 12. Which of the following are isotopes of element X, which has an atomic number of 16: 32 36 36 16 30 16 X , 16 X , 18 X , 8 X , and 14 X

Study Questions

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123

Key Experiments Developing Atomic Structure (See pages 72–74.) 13. From cathode ray experiments, J. J. Thomson estimated that the mass of an electron was “about a thousandth” of the mass of a proton. How accurate is that estimate? Calculate the ratio of the mass of an electron to the mass of a proton. 14. In 1886, Eugene Goldstein observed positively charged particles called canal rays moving in the opposite direction to electrons in a cathode ray tube (illustrated below). From their mass, he concluded that these particles were formed from residual gas in the tube. For example, if the cathode ray tube contained helium, the canal rays consisted of He+ ions. Describe a process that could lead to these ions. Cathode rays

Anode

+

–

–

Positive (canal) rays Cathode with holes (pierced disk)

+

+

–

–

+

16. Early in the 1800s, John Dalton proposed that an atom was a “solid, massy, hard, impenetrable, moveable particle.” Critique this description. How does this description misrepresent atomic structure?

Atomic Weight (See Examples 2.3 and 2.4.) 17. Thallium has two stable isotopes, 203Tl and 205Tl. Knowing that the atomic weight of thallium is 204.4, which isotope is the more abundant of the two? 18. Strontium has four stable isotopes. Strontium-84 has a very low natural abundance, but 86Sr, 87Sr, and 88Sr are all reasonably abundant. Knowing that the atomic weight of strontium is 87.62, which of the more abundant isotopes predominates? 19. Verify that the atomic weight of lithium is 6.94, given the following information: Li, mass = 6.015123; percent abundance = 7.6% Li, mass = 7.016003; percent abundance = 92.4%

6

+

7

Electron Gas molecules To vacuum pump

20. Verify that the atomic weight of magnesium is 24.31, given the following information:

Positive ion

Canal rays.  In 1886, Eugene Goldstein detected a stream of particles traveling in the direction opposite to that of the negatively charged cathode rays (electrons). He called this stream of positive particles “canal rays.”

15. Marie and Pierre Curie and others observed that a radioactive substance could emit three types of radiation: alpha (α), beta (β), and gamma (γ). If the radiation from a radioactive source is passed between electrically charged plates, some particles are attracted to the positive plate, some to the negative plate, and others feel no attraction. Which particles are positively charged, which are negatively charged, and which have no charge? Of the two charged particles, which has the most mass? particles rays Photographic film or phosphor screen

+

Lead block shield

particles, attracted to + plate

–

Slit

particles particles, attracted to – plate

Charged plates

Radioactive element

Radioactivity.  Alpha (α), beta (β), and gamma (γ) rays from a radioactive element are separated by passing them between electrically charged plates.

124

Mg, mass = 23.985042; percent abundance = 78.99% Mg, mass = 24.985837; percent abundance = 10.00% 26 Mg, mass = 25.982593; percent abundance = 11.01% 24 25

21. Gallium has two naturally occurring isotopes, 69 Ga and 71Ga, with masses of 68.9256 and 70.9247, respectively. Calculate the percent abundances of these isotopes of gallium. 22. Europium has two stable isotopes, 151Eu and 153 Eu, with masses of 150.9199 and 152.9212, respectively. Calculate the percent abundances of these isotopes of europium.

The Periodic Table (See Section 2.3.) 23. Titanium and thallium have symbols that are easily confused with each other. Give the symbol, atomic number, atomic weight, and group and period number of each element. Are they metals, metalloids, or nonmetals? 24. In Groups 4A–6A (14–16), there are several elements whose symbols begin with S. Name these elements, and for each one give its symbol, atomic number, group number, and period. Describe each as a metal, metalloid, or nonmetal. 25. How many periods of the periodic table have 8 elements, how many have 18 elements, and how many have 32 elements?

Chapter 2 / Atoms, Molecules, and Ions

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26. How many elements occur in the seventh period? What is the name given to the majority of these elements, and what well-known property characterizes them? 27. Select answers to the questions listed below from the following list of elements whose symbols start with the letter C: C, Ca, Cr, Co, Cd, Cl, Cs, Ce, Cm, Cu, and Cf. (You should expect to use some symbols more than once.) (a) Which are nonmetals? (b) Which are main group elements? (c) Which are lanthanides? (d) Which are transition elements? (e) Which are actinides? (f) Which are gases? 28. Give the name and chemical symbol for the following. (a) a nonmetal in the second period (b) an alkali metal in the fifth period (c) the third-period halogen (d) an element that is a gas at 20 °C and 1 atmosphere pressure 29. Classify the following elements as metals, metalloids, or nonmetals: N, Na, Ni, Ne, and Np. 30. Here are symbols for five of the seven elements whose names begin with the letter B: B, Ba, Bk, Bi, and Br. Match each symbol with one of the descriptions below. (a) a radioactive element (b) a liquid at room temperature (c) a metalloid (d) an alkaline earth element (e) a group 5A (15) element

Molecules: Formulas, Models, and Names (See Section 2.4 and Example 2.5) 31. A model of nitric acid is illustrated here. Write the molecular formula for nitric acid, and draw the structural formula. Describe the structure of the molecule. Is it flat? That is, are all the atoms in the plane of the paper? (Color code: nitrogen atoms are blue; oxygen atoms are red; and hydrogen atoms are white.)

32. A model of the amino acid asparagine is illustrated here. Write the molecular formula for the compound, and draw its structural formula.

Asparagine, an an amino Asparagine, aminoacid acid

33. Name each of the following binary molecular compounds: (a) NF3 (c) BI3 (b) HI (d) PF5 34. Name each of the following binary molecular compounds: (a) N2O5 (c) OF2 (b) P4S3 (d) XeF4 35. Give the formula for each of the following compounds: (a) sulfur dichloride (b) dinitrogen pentaoxide (c) silicon tetrachloride (d) diboron trioxide (commonly called boric oxide) 36. Give the formula for each of the following compounds: (a) bromine trifluoride (b) xenon difluoride (c) hydrazine (d) diphosphorus tetrafluoride (e) butane

Ions and Ion Charges (See Figure 2.19 and Table 2.4.) 37. What is the charge on the common monatomic ions of the following elements? (a) magnesium (c) nickel (b) zinc (d) gallium 38. Identify the common charges on monatomic ions of the following elements. (a) sodium (c) iron (b) aluminum (d) copper

Nitric acid

39. What is the charge on the common monatomic ions of the following elements? (a) selenium (c) oxygen (b) fluorine (d) nitrogen Study Questions

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125

40. What is the charge on the common monatomic ions of the following elements? (a) bromine (c) phosphorus (b) sulfur (d) chlorine 41. Give the symbol, including the correct charge, for each of the following ions: (a) barium ion (b) titanium(IV) ion (c) phosphate ion (d) hydrogen carbonate ion (e) sulfide ion (f) perchlorate ion (g) cobalt(II) ion (h) sulfate ion 42. Give the symbol, including the correct charge, for each of the following ions: (a) permanganate ion (b) nitrite ion (c) dihydrogen phosphate ion (d) ammonium ion (e) phosphate ion (f) sulfite ion 43. When a potassium atom becomes a monatomic ion, how many electrons does it lose or gain? What noble gas atom has the same number of electrons as a potassium ion? 44. When oxygen and sulfur atoms become monatomic ions, how many electrons does each lose or gain? Which noble gas atom has the same number of electrons as an oxide ion? Which noble gas atom has the same number of electrons as a sulfide ion? 45. How many protons and electrons are present in each of the following ions? (a) Na1 (c) F− 21 (b) Mg (d) O2− 46. How many protons and electrons are present in each of the following ions? (c) Cu21 (a) Ca21 (b) Cu1 (d) Cl−

Ionic Compounds (See Examples 2.6–2.8) 47. What are the charges on the ions in an ionic compound containing the elements barium and bromine? Write the formula for the compound. 48. What are the charges of the ions in an ionic compound containing cobalt(III) and fluoride ions? Write the formula for the compound.

126

49. Give the name, formula, and the number of each ion that makes up each of the following compounds: (a) K2S (d) (NH4)3PO4 (b) CoSO4 (e) Ca(ClO)2 (c) KMnO4 (f) NaCH3CO2 50. Give the name, formula, and the number of each ion that makes up each of the following compounds: (a) Mg(CH3CO2)2 (d) Ti(SO4)2 (b) Al(OH)3 (e) KH2PO4 (c) CuCO3 (f) CaHPO4 51. Cobalt forms Co21 and Co31 ions. Write the formulas for the two cobalt oxides formed by these transition metal ions. 52. Platinum is a transition element and forms Pt21 and Pt41 ions. Write the formulas for the compounds of each of these ions with (a) chloride ions and (b) sulfide ions. 53. Which of the following are correct formulas for ionic compounds? For those that are not, give the correct formula. (c) Ga2O3 (a) AlCl2 (b) KF2 (d) MgS 54. Which of the following are correct formulas for ionic compounds? For those that are not, give the correct formula. (a) Ca2O (c) Fe2O5 (b) SrBr2 (d) Li2O 55. Name each of the following ionic compounds: (a) K2S (c) (NH4)3PO4 (b) CoSO4 (d) Ca(ClO)2 56. Name each of the following ionic compounds: (a) Ca(CH3CO2)2 (c) Al(OH)3 (b) Ni3(PO4)2 (d) KH2PO4 57. Name each of the following ionic compounds. (a) Na2S (c) NaHSO4 (b) Na2SO3 (d) Na2SO4 58. Name each of the following ionic compounds. (a) Li3N (b) LiNO2 (c) LiNO3 59. Give the formula for each of the following ionic compounds: (a) ammonium carbonate (b) calcium iodide (c) copper(II) bromide (d) aluminum phosphate (e) silver(I) acetate

Chapter 2 / Atoms, Molecules, and Ions

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60. Give the formula for each of the following ionic compounds: (a) calcium hydrogen carbonate (b) potassium permanganate (c) magnesium perchlorate (d) potassium hydrogen phosphate (e) sodium sulfite 61. Write the formulas for the four ionic compounds that can be made by combining each of the cations Na1 and Ba21 with the anions CO322 and I2. Name each of the compounds. 62. Write the formulas for the four ionic compounds that can be made by combining the cations Mg21 and Fe31 with the anions PO432 and NO32. Name each compound formed.

Coulomb’s Law (See Equation 2.3 and Figure 2.23.) 63. Sodium ions, Na1, form ionic compounds with fluoride ions, F2, and iodide ions, I2. The radii of these ions are as follows: Na1 5 116 pm; F2 5 119 pm; and I2 5 206 pm. In which ionic compound, NaF or NaI, are the forces of attraction between cation and anion stronger? Explain your answer. 64. Consider the two ionic compounds NaCl and CaO. In which compound are the cation–anion attractive forces stronger? Explain your answer.

Atoms and the Mole (See Example 2.9.) 65. Calculate the mass, in grams, of each the following: (a) 3.2 mol of aluminum (b) 2.35 × 1023 mol of iron (c) 0.12 mol of calcium (d) 23.0 mol of neon 66. Calculate the mass, in grams, of each the following: (a) 1.24 mol of gold (b) 14.8 mol of He (c) 0.43 mol of platinum (d) 2.42 × 1024 mol of Rh 67. Calculate the amount (moles) represented by each of the following: (a) 87.21 g of Cu (b) 0.024 g of sodium (c) 2.0 mg of iridium (d) 6.75 g of Au 68. Calculate the amount (moles) represented by each of the following: (a) 9.4 g of Li (c) 0.037 g of platinum (b) 0.942 g of tin (d) 1.84 g of Xe

69. Calculate the average mass of an atom (in grams) for each of the following elements. (a) helium (c) scandium (b) fluorine (d) bismuth 70. Calculate the average mass of an atom (in grams) for each of the following elements. (a) beryllium (c) krypton (b) aluminum (d) mercury 71. You are given 1.0-g samples of He, Fe, Li, Si, and C. Which sample contains the largest number of atoms? Which contains the smallest? 72. You are given 0.10-g samples of K, Mo, Cr, and Al. List the samples in order of the amount (moles), from smallest to largest. 73. Analysis of a 10.0-g sample of apatite (a major component of tooth enamel) showed that it was made up of 3.99 g Ca, 1.85 g P, 4.14 g O, and 0.020 g H. List these elements based on relative amounts (moles), from smallest to largest. 74. A semiconducting material is composed of 52 g of Ga, 9.5 g of Al, and 112 g of As. Which element has the largest number of atoms in this material?

Molecules, Compounds, and the Mole (See Example 2.10.) 75. The molecular formula of glucose is C6H12O6. What amount (moles) of carbon is present in 1.0 mol of glucose? In 0.20 mol of glucose? 76. The formula of water is H2O. What amounts (moles) of hydrogen and oxygen are present in 1.0 mol of water? In 4.0 mol of water? 77. Calculate the molar mass of each of the following compounds: (a) Fe2O3, iron(III) oxide (b) BCl3, boron trichloride (c) C6H8O6, ascorbic acid (vitamin C) (d) Mg(NO3)2, magnesium nitrate 78. Calculate the molar mass of each of the following compounds: (a) CaCO3, calcium carbonate, a compound used as an antacid (b) Fe(C6H11O7)2, iron(II) gluconate, a dietary supplement (c) CH3CH2CH2CH2SH, butanethiol, has a skunklike odor (d) C20H24N2O2, quinine, used as an antimalarial drug

Study Questions

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127

79. Calculate the molar mass of each hydrated compound. Note that the water of hydration is included in the molar mass. (See page 104.) (a) Ni(NO3)2 ∙ 6 H2O (b) CuSO4 ∙ 5 H2O 80. Calculate the molar mass of each hydrated compound. Note that the water of hydration is included in the molar mass. (See page 104.) (a) H2C2O4 ∙ 2 H2O (b) MgSO4 ∙ 7 H2O, Epsom salt

84. How many ammonium ions and how many sulfate ions are present in a 0.20 mol sample of (NH4)2SO4? How many atoms of N, H, S and O are contained in this sample? 85. Acetaminophen, whose structure is drawn below, is the active ingredient in some nonprescription pain killers. The recommended dose for an adult is two 500‑mg caplets. How many molecules make up one dose of this drug?

81. What mass is represented by 0.0255 mol of each of the following compounds? (a) C3H7OH, 2-propanol, rubbing alcohol (b) C11H16O2, an antioxidant in foods, also known as BHA (butylated hydroxyanisole) (c) C9H8O4, aspirin (d) (CH3)2CO, acetone, an important industrial solvent 82. Assume you have 0.123 mol of each of the following compounds. What mass of each is present? (a) C14H10O4, benzoyl peroxide, used in acne medications (b) Dimethylglyoxime, used in the laboratory to test for nickel(II) ions CH3 C

N

OH

C

N

OH

CH3

CH3 H

H

C

C

CH3

S

H

H

(d) DEET, a mosquito repellent

HC HC

H C

C

C CH

O

CH2

C

N CH2

CH3

CH3

CH3 83. Sulfur trioxide, SO3, is made industrially in enormous quantities by combining oxygen and sulfur dioxide, SO2. What amount (moles) of SO3 is represented by 1.00 kg of sulfur trioxide? How many molecules? How many sulfur atoms? How many oxygen atoms?

128

86. An Alka-Seltzer tablet contains 325 mg of aspirin (C9H8O4), 1916 mg of NaHCO3, and 1000. mg of citric acid (H3C6H5O7). (The last two compounds react with each other to provide the fizz, bubbles of CO2, when the tablet is put into water.) (a) Calculate the amount (moles) of each substance in the tablet. (b) If you take one tablet, how many molecules of aspirin are you consuming?

Percent Composition

(c) The compound below, responsible for the skunky taste in beer exposed to light.

C

Acetaminophen

(See Example 2.11.) 87. Calculate the mass percent of each element in the following compounds: (a) PbS, lead(II) sulfide, galena (b) C3H8, propane (c) C10H14O, carvone, found in caraway seed oil 88. Calculate the mass percent of each element in the following compounds: (a) C8H10N2O2, caffeine (b) C10H20O, menthol (c) CoCl2 ∙ 6 H2O 89. Calculate the mass percent of copper in CuS, copper(II) sulfide. If you wish to obtain 10.0 g of copper metal from copper(II) sulfide, what mass of CuS (in grams) must you use? 90. Calculate the mass percent of titanium in the mineral ilmenite, FeTiO3. What mass of ilmenite (in grams) is required if you wish to obtain 750 g of titanium?

Chapter 2 / Atoms, Molecules, and Ions

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Empirical and Molecular Formulas (See Example 2.12.) 91. Succinic acid occurs in fungi and lichens. Its empirical formula is C2H3O2, and its molar mass is 118.1 g/mol. What is its molecular formula? 92. An organic compound has the empirical formula C2H4NO. If its molar mass is 116.1 g/mol, what is the molecular formula of the compound? 93. Complete the following table: Empirical Formula

Molar Mass (g/mol)

(a) CH

 26.0

(b) CHO

116.1

(c)

Molecular Formula

C8H16

94. Complete the following table: Empirical Formula

Molar Mass (g/mol)

(a) C2H3O3

150.0

(b) C3H8

 44.1

(c)

Molecular Formula

B4H10

95. Acetylene is a colorless gas used as a fuel in welding torches, among other things. It is 92.26% C and 7.74% H. Its molar mass is 26.02 g/mol. What are the empirical and molecular formulas of acetylene? 96. A large family of boron-hydrogen compounds has the general formula BxHy . One member of this family contains 88.5% B; the remainder is hydrogen. What is its empirical formula? 97. Cumene, a hydrocarbon, is a compound composed only of C and H. It is 89.94% carbon, and its molar mass is 120.2 g/mol. What are the empirical and molecular formulas of cumene? 98. In 2006, a Russian team discovered an interesting molecule they called sulflower because of its shape and because it was based on sulfur. It is composed of 57.17% S and 42.83% C and has a molar mass of 448.70 g/mol. Determine the empirical and molecular formulas of sulflower. 99. Mandelic acid is an organic acid composed of carbon (63.15%), hydrogen (5.30%), and oxygen (31.55%). Its molar mass is 152.14 g/mol. Determine the empirical and molecular formulas of the acid.



100. Nicotine, a poisonous compound found in tobacco leaves, is 74.0% C, 8.65% H, and 17.35% N. Its molar mass is 162 g/mol. What are the empirical and molecular formulas of nicotine?

Determining Formulas from Mass Data (See Examples 2.13 and 2.14.) 101. A compound containing xenon and fluorine was prepared by shining sunlight on a mixture of Xe (0.674 g) and excess F2 gas. If you isolate 0.869 g of the new compound, what is its empirical formula? 102. Elemental sulfur (1.47 g) is combined with fluorine, F2, to give a compound with the formula SFx , a very stable, colorless gas. If you isolate 6.70 g of SFx , what is the value of x? 103. You weigh out 1.523 g of solid BaCl2 ∙ x H2O and heat this compound to remove the water. The mass of the remaining anhydrous compound is 1.298 g. What is the name and formula of the hydrated compound? 104. Glauber’s salt, which has laxative properties, is a hydrate of sodium sulfate. You weigh out 2.343 g of Glauber’s salt (Na2SO4 ∙ x H2O) and heat it to remove the water. The mass of the anhydrous compound that remains is 1.033 g. What is the chemical name and formula of Glauber’s salt? 105. Epsom salt is used in tanning leather and in medicine. It is hydrated magnesium sulfate, MgSO4 ∙ 7 H2O. The water of hydration is lost on heating, with the number lost depending on the temperature. Suppose you heat a 1.394-g sample at 100 °C and obtain 0.885 g of a partially hydrated sample, MgSO4 ∙ x H2O. What is the value of x? 106. You combine 1.25 g of germanium, Ge, with excess chlorine, Cl2. The mass of product, GexCly, is 3.69 g. What is the formula of the product, GexCly?

Mass Spectrometry (See Section 2.9.) 107. The mass spectrum of nitrogen dioxide is illustrated in the following figure. (a) Identify the cations present for each of the four peaks in the mass spectrum. (b) Does the mass spectrum provide evidence that the two oxygen atoms are attached to a central nitrogen atom (ONO), or that an oxygen atom is at the center (NOO)? Explain.

Study Questions

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129

100

30

80

80

NO2

Relative Abundance

Relative abundance of ions

100

60

46

40

20

30 40 Mass-to-charge ratio (m/Z)

100

50

104

30

POF3

60

40

60

General Questions These questions are not designated as to type or location in the chapter. They may combine several concepts.

58

Ni

20

69 47 50 50

66

33

S

Number of protons

10

Number of neutrons

10

Number of electrons    in the neutral atom

0 30

50

110. The highest mass peaks in the mass spectrum of Br2 occur at m/Z 158, 160, and 162. The ratio of intensities of these peaks is approximately 1:2:1. Bromine has two stable isotopes, 79Br (50.7% abundance) and 81Br (49.3% abundance). (a) What molecular species gives rise to each of these peaks? (b) Explain the relative intensities of these peaks. (Hint: Consider the probabilities of each atom combination.)

Symbol

30 25

Name of element 88

70 90 Mass-to-charge ratio (m/Z)

110

109. The mass spectrum of CH3Cl is illustrated here. You know that carbon has two stable isotopes, 12 C and 13C with relative abundances of 98.9% and 1.1%, respectively, and chlorine has two isotopes, 35Cl and 37Cl with abundances of 75.76% and 24.24%, respectively. (a) What molecular species gives rise to the lines at m/Z of 50 and 52? Why is the line at 52 about 1/3 the height of the line at 50? (b) What species might be responsible for the line at m/Z 5 51?

130

40

111. Fill in the blanks in the table (one column per element).

85 Relative abundance of ions

20

(m/Z)

108. The mass spectrum of phosphoryl chloride, POF3, is illustrated here. (a) Identify the cation fragment at a m/Z ratio of 85. (b) Identify the cation fragment at a m/Z ratio of 69. (c) Which two peaks in the mass spectrum provide evidence that the oxygen atom is connected to the phosphorus atom and is not connected to any of the three fluorine atoms?

80

40

0 10

14 0 10

60

20

16

20

CHCl3

112. Potassium has three naturally occurring isotopes (39K, 40K, and 41K), but 40K has a very low natural abundance. Which of the other two isotopes is more abundant? Briefly explain your answer. 113. Crossword Puzzle: In the 2 × 2 box shown here, each answer must be correct four ways: horizontally, vertically, diagonally, and by itself. Instead of words, use symbols of elements. When the puzzle is complete, the four spaces will contain the overlapping symbols of 10 elements. There is only one correct solution. 1

2

3

4

Chapter 2 / Atoms, Molecules, and Ions

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Horizontal 1–2: two-letter symbol for a metal used in ancient times 3–4: two-letter symbol for a metal that burns in air and is found in Group 5A (15) Vertical 1–3: two-letter symbol for a metalloid 2–4: two-letter symbol for a metal used in U.S. coins Single squares: All one-letter symbols 1: a colorful nonmetal 2: colorless, gaseous nonmetal 3: an element that makes fireworks green 4: an element that has medicinal uses Diagonal 1–4: two-letter symbol for an element used in electronics 2–3: two-letter symbol for a metal used with Zr to make wires for superconducting magnets This puzzle first appeared in Chemical & Engineering News, p. 86, December 14, 1987 (submitted by S. J. Cyvin) and in Chem Matters, October 1988. 114. The following chart shows a general decline in abundance in the solar system with increasing mass among the first 30 elements. The decline continues beyond zinc. Notice that the scale on the vertical axis is logarithmic, that is, it progresses in powers of 10. The abundance of nitrogen, for example, is 1/10,000 (1/104) of the abundance of hydrogen. All abundances are plotted as the number of atoms per 1012 atoms of H. (The fact that the abundances of Li, Be, and B, as well as those of the elements near Fe, do not follow the general decline is a consequence of the way that elements are synthesized in stars.) 1014 1012

Relative abundance

1010 108 106

102

H He Li Be B C N O F NeNaMg Al Si P S Cl Ar K Ca Sc Ti V Cr MnFe Co Ni Cu Zn Element

The abundance of the elements in the solar system from H to Zn



115. Copper atoms. (a) What is the average mass of one copper atom? (b) Students in a college computer science class once sued the college because they were asked to calculate the cost of one atom and could not do it. But you are in a chemistry course, and you can do this. (See E. Felsenthal, Wall Street Journal, May 9, 1995.) If the cost of 2.0-mm diameter copper wire (99.999% pure) is currently $80.10 for 7.0 g, what is the cost of one copper atom? 116. Which of the following is impossible? (a) silver foil that is 1.2 × 1024 m thick (b) a sample of potassium that contains 1.784 × 1024 atoms (c) a gold coin of mass 1.23 × 1023 kg (d) 3.43 × 10227 mol of S8 molecules 117. Reviewing the periodic table. (a) Name the element in Group 2A (2) and the fifth period. (b) Name the element in the fifth period and Group 4B (4). (c) Which element is in the second period in Group 4A (14)? (d) Which element is in the fourth period in Group 5A (15)? (e) Which halogen is in the fifth period? (f) Which alkaline earth element is in the third period? (g) Which noble gas element is in the fourth period? (h) Name the nonmetal in Group 6A (16) and the third period. (i) Name a metalloid in the fourth period. 118. Identify two nonmetallic elements that have allotropes and describe the allotropes of each.

104

0

(a) What is the most abundant main group metal? (b) What is the most abundant nonmetal? (c) What is the most abundant metalloid? (d) Which of the transition elements is most abundant? (e) Which halogens are included on this plot, and which is the most abundant?

119. In each case, decide which represents more mass: (a) 0.5 mol of Na, 0.5 mol of Si, or 0.25 mol of U (b) 9.0 g of Na, 0.50 mol of Na, or 1.2 × 1022 atoms of Na (c) 10 atoms of Fe or 10 atoms of K

Study Questions

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131

120. The recommended daily allowance (RDA) of iron for women 19–50 years old is 18 mg. How many moles is this? How many atoms? 121. Order the following samples from smallest to largest mass: (a) 3.79 × 1024 (d) 7.40 mol Si atoms Fe (e) 9.221 mol Na (b) 19.92 mol H2 (f) 4.07 × 1024 atoms Al (c) 8.576 mol C (g) 9.2 mol Cl2 122. ▲  When a sample of phosphorus burns in air, the compound P4O10 forms. One experiment showed that 0.744 g of phosphorus formed 1.704 g of P4O10. Use this information to determine the ratio of the atomic weights of phosphorus and oxygen (mass P/mass O). If the atomic weight of oxygen is assumed to be 16.00, calculate the atomic weight of phosphorus. 123. ▲  When a sample of carbon burns completely in air, the compound CO2 forms. One experiment showed that 0.876 g of carbon formed 3.210 g of CO2. Use this information to determine the ratio of the atomic weights of oxygen and carbon (mass O/mass C). If the atomic weight of carbon is assumed to be 12.01, calculate the atomic weight of oxygen. 124. A reagent occasionally used in chemical synthesis is sodium–potassium alloy. (Alloys are mixtures of metals, and Na-K has the interesting property that it is a liquid.) One formulation of the alloy (the one that melts at the lowest temperature) contains 68 atom percent K; that is, out of every 100 atoms, 68 are K and 32 are Na. What is the mass percent of potassium in sodium–potassium alloy? 125. Write formulas for all of the compounds that can be made by combining the cations NH41 and Ni21 with the anions CO322 and PO432. 126. How many electrons are in a strontium atom (Sr)? Does an atom of Sr gain or lose electrons when forming an ion? How many electrons are gained or lost by the atom? When Sr forms an ion, the ion has the same number of electrons as which one of the noble gases? 127. Which of the following compounds has the highest mass percent of chlorine? (a) BCl3 (d) AlCl3 (b) AsCl3 (e) PCl3 (c) GaCl3

128. Which of the following samples has the largest number of ions? (a) 1.0 g of CaCl2 (d) 1.0 g of SrCO3 (b) 1.0 g of MgCl2 (e) 1.0 g of BaSO4 (c) 1.0 g of CaS 129. The structure of one of the bases in DNA, adenine, is shown here. Which represents the greater mass: 40.0 g of adenine or 3.0 × 1023 molecules of the compound?

Adenine

130. Ionic and molecular compounds of the halogens. (a) What are the names of BaF2, SiCl4, and NiBr2? (b) Which of the compounds in part (a) are ionic, and which are molecular? (c) Which has the largest mass, 0.50 mol of BaF2, 0.50 mol of SiCl4, or 1.0 mol of NiBr2? 131. A drop of water has a volume of about 0.050 mL. How many molecules of water are in a drop of water? (Assume water has a density of 1.00 g/cm3.) 132. Capsaicin, the spicy compound in chili peppers, has the formula C18H27NO3. (a) Calculate its molar mass. (b) If you eat 55 mg of capsaicin, what amount (moles) have you consumed? (c) Calculate the mass percent of each element in the compound. (d) What mass of carbon (in milligrams) is there in 55 mg of capsaicin? 133. Calculate the molar mass and the mass percent of each element in the blue solid compound Cu(NH3)4SO4 ∙ H2O. What is the mass of copper and the mass of water in 10.5 g of the compound? 134. Write the molecular formula and calculate the molar mass for each of the molecules shown here. Which has the largest mass percent of carbon? Of oxygen? (a) ethylene glycol (used in antifreeze)

H H H

O

C

C

O H

H H Ethylene glycol

132

Chapter 2 / Atoms, Molecules, and Ions

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(b) dihydroxyacetone (used in artificial tanning lotions)

H

O

H

O H

C

C

C

H

O H

H

Dihydroxyacetone

(c) ascorbic acid, commonly known as vitamin C Ephedrine

HO

H

H

H

C

C

C

H

OH

O C

OH

C

O

C OH

Ascorbic acid, vitamin C

135. Malic acid, an organic acid found in apples, contains C, H, and O in the following ratios: C1H1.50O1.25. What is the empirical formula of malic acid? 136. Your doctor has diagnosed you as being anemic— that is, as having too little iron in your blood. At the drugstore, you find two iron-containing dietary supplements: one with iron(II) sulfate, FeSO4, and the other with iron(II) gluconate, Fe(C6H11O7)2. If you take 100. mg of each compound, which will deliver more atoms of iron? 137. A compound composed of iron and carbon monoxide, Fex(CO)y , is 30.70% iron. What is the empirical formula for the compound? 138. Ma huang, an extract from the ephedra species of plants, contains ephedrine. The Chinese have used this herb for more than 5000 years to treat asthma. More recently, ephedrine has been used in diet pills that can be purchased over the counter in herbal medicine shops. However, very serious concerns were raised regarding these pills following reports that their use led to serious heart problems. (a) From the following molecular model of ephedrine, determine the molecular formula for ephedrine and calculate its molar mass. (b) What is the weight percent of carbon in ephedrine? (c) Calculate the amount (moles) of ephedrine in a 0.125 g sample. (d) How many molecules of ephedrine are there in 0.125 g? How many C atoms?



139. Saccharin, a molecular model of which is shown below, is more than 300 times sweeter than sugar. It was first made in 1897, when it was common practice for chemists to record the taste of any new substances they synthesized. (a) Write the molecular formula for the compound, and draw its structural formula. (S atoms are yellow.) (b) If you ingest 125 mg of saccharin, what amount (moles) of saccharin have you ingested? (c) What mass of sulfur is contained in 125 mg of saccharin?

Saccharin

140. Name each of the following compounds and indicate which ones are best described as ionic: (a) ClF3 (f) OF2 (b) NCl3 (g) KI (c) SrSO4 (h) Al2S3 (d) Ca(NO3)2 (i) PCl3 (e) XeF4 (j) K3PO4 141. Write the formula for each of the following compounds and indicate which ones are best described as ionic: (a) sodium hypochlorite (b) boron triiodide (c) aluminum perchlorate (d) calcium acetate (e) potassium permanganate (f) ammonium sulfite (g) potassium dihydrogen phosphate (h) disulfur dichloride (i) chlorine trifluoride (j) phosphorus trifluoride Study Questions

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142. Complete the table by placing symbols, formulas, and names in the blanks. Cation

Anion

Name

Formula

ammonium bromide Ba2+

BaS Cl

2

iron(II) chloride

F

PbF2

2

Al31

CO322 iron(III) oxide

143. Empirical and molecular formulas. (a) Fluorocarbonyl hypofluorite is composed of 14.6% C, 39.0% O, and 46.3% F. The molar mass of the compound is 82 g/mol. ­Determine the empirical and molecular formulas of the compound. (b) Azulene, a beautiful blue hydrocarbon, is 93.71% C and has a molar mass of 128.16 g/ mol. What are the empirical and molecular formulas of azulene? 144. Cacodyl, a compound containing arsenic, was reported in 1842 by the German chemist Robert Wilhelm Bunsen. It has an almost intolerable garlic-like odor. Its molar mass is 210 g/mol, and it is 22.88% C, 5.76% H, and 71.36% As. Determine its empirical and molecular formulas. 145. The action of bacteria on meat and fish produces a compound called cadaverine. As its name and origin imply, it stinks! (It is also present in bad breath and adds to the odor of urine.) It is 58.77% C, 13.81% H, and 27.40% N. Its molar mass is 102.2 g/mol. Determine the molecular formula of cadaverine. 146. ▲  In the laboratory you combine 0.125 g of nickel with CO and isolate 0.364 g of Ni(CO)x . What is the value of x? 147. ▲  A compound called MMT was once used to boost the octane rating of gasoline. What is the empirical formula of MMT if it is 49.5% C, 3.2% H, 22.0% O, and 25.2% Mn? 148. ▲  Elemental phosphorus is made by heating calcium phosphate with carbon and sand in an electric furnace. What is the mass percent of phosphorus in calcium phosphate? Use this value to calculate the mass of calcium phosphate (in kilograms) that must be used to produce 15.0 kg of phosphorus. 149. ▲  Chromium is obtained by heating chromium(III) oxide with carbon. Calculate the

134

mass percent of chromium in the oxide, and then use this value to calculate the quantity of Cr2O3 required to produce 850 kg of chromium metal. 150. ▲  Stibnite, Sb2S3, is a dark gray mineral from which antimony metal is obtained. What is the mass percent of antimony in the sulfide? If you have 1.00 kg of an ore that contains 10.6% antimony, what mass of Sb2S3 (in grams) is in the ore? 151. ▲  Direct reaction of iodine (I2) and chlorine (Cl2) produces an iodine chloride, Ix Cly , a bright yellow solid. If you completely consume 0.678 g of I2 in a reaction with excess Cl2 and produce 1.246 g of Ix Cly , what is the empirical formula of the compound? A later experiment showed that the molar mass of Ix Cly is 467 g/mol. What is the molecular formula of the compound? 152. ▲  In a reaction, 2.04 g of vanadium combined with 1.93 g of sulfur to give a pure compound. What is the empirical formula of the product? 153. ▲  Iron pyrite, often called fool’s gold, has the formula FeS2. If you could convert 15.8 kg of iron pyrite to iron metal, what mass of the metal would you obtain? 154. Which of the following statements about 57.1 g of octane, C8H18, is (are) not true? (a) 57.1 g is 0.500 mol of octane. (b) The compound is 84.1% C by weight. (c) The empirical formula of the compound is C4H9. (d) 57.1 g of octane contains 28.0 g of hydrogen atoms. 155. The formula of barium molybdate is BaMoO4. Which of the following is the formula of sodium molybdate? (a) Na4MoO (d) Na2MoO4 (b) NaMoO (e) Na4MoO4 (c) Na2MoO3 156. ▲  A metal M forms a compound with the formula MCl4. If the compound is 74.75% chlorine, what is the identity of M? 157. Pepto-Bismol, which can help provide relief for an upset stomach, contains 262 mg of bismuth subsalicylate, C21H15Bi3O12, per tablet. If you take two tablets for your stomach distress, what amount (in moles) of this active ingredient are you taking? What mass of Bi are you consuming in two tablets? 158. ▲  The weight percent of oxygen in an oxide that has the formula MO2 is 15.2%. What is the molar mass of this compound? What element or elements are possible for M?

Chapter 2 / Atoms, Molecules, and Ions

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159. The mass of 2.50 mol of a compound with the formula ECl4, in which E is a nonmetallic element, is 385 g. What is the molar mass of ECl4? What is the identity of E? 160. ▲  The elements A and Z combine to produce two different compounds: A2Z3 and AZ2. If 0.15 mol of A2Z3 has a mass of 22.48 g and 0.15 mol of AZ2 has a mass of 12.44 g, what are the atomic weights of A and Z? 161. ▲  Polystyrene can be prepared by heating styrene with tribromobenzoyl peroxide in the absence of air. A sample of polystyrene prepared by this method has the empirical formula Br3C6H3(C8H8)n, where the value of n can vary from sample to sample. If one sample has 0.105% Br, what is the value of n? 162. A sample of hemoglobin is found to be 0.335% iron. What is the molar mass of hemoglobin if there are four iron atoms per molecule? 163. ▲  Consider an atom of 64Zn. (a) Calculate the density of the nucleus in grams per cubic centimeter, knowing that the nuclear radius is 4.8 × 1026 nm and the mass of the 64 Zn atom is 1.06 × 10222 g. (The volume of a sphere is [4/3]πr3.) (b) Calculate the density of the space occupied by the electrons in the zinc atom, given that the atomic radius is 0.125 nm and the electron mass is 9.11 × 10228 g. (c) Having calculated these densities, what statement can you make about the relative densities of the parts of the atom? 164. ▲  Estimating the radius of a lead atom. (a) You are given a cube of lead that is 1.000 cm on each side. The density of lead is 11.35 g/cm3. How many atoms of lead are in the sample? (b) Atoms are spherical; therefore, the lead atoms in this sample cannot fill all the available space. As an approximation, assume that 60.% of the space of the cube is filled with spherical lead atoms. Calculate the volume of one lead atom from this information. From the calculated volume (V) and the formula (4/3)πr3 for the volume of a sphere, estimate the radius (r) of a lead atom. 165. A piece of nickel foil, 0.550 mm thick and 1.25 cm square, is allowed to react with fluorine, F2, to give nickel fluoride. (a) How many moles of nickel foil were used? (The density of nickel is 8.902 g/cm3.) (b) If you isolate 1.261 g of the nickel fluoride, what is its formula? (c) What is its complete name?

166. ▲  Uranium is used as a fuel, primarily in the form of uranium(IV) oxide, in nuclear power plants. This question considers some uranium chemistry. (a) A small sample of uranium metal (0.169 g) is heated to between 800 and 900 °C in air to give 0.199 g of a dark green oxide, Ux Oy. How many moles of uranium metal were used? What is the empirical formula of the oxide, Ux Oy? How many moles of Ux Oy must have been obtained? (b) The naturally occurring isotopes of uranium are 234U, 235U, and 238U. Knowing that uranium’s atomic weight is 238.02 g/mol, which isotope must be the most abundant? (c) If the hydrated compound UO2(NO3)2 ∙ z H2O is heated gently, the water of hydration is lost. If you have 0.865 g of the hydrated compound and obtain 0.679 g of UO2(NO3)2 upon heating, how many waters of hydration are in each formula unit of the original compound? (The oxide Ux Oy is obtained if the hydrate is heated to temperatures over 800 °C in the air.) 167. In an experiment, you need 0.125 mol of sodium metal. Sodium can be cut easily with a knife (Figure 2.6), so if you cut out a block of sodium, what should the volume of the block be in cubic centimeters? If you cut a perfect cube, what is the length of the edge of the cube? (The density of sodium is 0.97 g/cm3.) 168. Mass spectrometric analysis showed that there are four isotopes of an unknown element having the following masses and abundances:

Isotope

Mass Number

Isotope Mass

Abundance (%)

1

136

135.9071

 0.185

2

138

137.9060

 0.251

3

140

139.9054

88.45

4

142

141.9093

11.11

Three elements in the periodic table that have atomic weights near these values are lanthanum (La), atomic number 57, atomic weight 138.91; cerium (Ce), atomic number 58, atomic weight 140.12; and praseodymium (Pr), atomic number 59, atomic weight 140.91. Using the data above, calculate the atomic weight, and identify the element if possible.

Study Questions

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135

169. If Epsom salt, MgSO4 ∙ x H2O, is heated to 250 °C, all the water of hydration is lost. Upon heating a 1.687-g sample of the hydrate, 0.824 g of MgSO4 remains. How many molecules of water occur per formula unit of MgSO4? 170. The alum used in cooking is potassium aluminum sulfate hydrate, KAl(SO4)2 ∙ x H2O. To find the value of x, you can heat a sample of the compound to drive off all of the water and leave only KAl(SO4)2. Assume you heat 4.74 g of the hydrated compound and that the sample loses 2.16 g of water. What is the value of x? 171. Tin metal (Sn) and purple iodine (I2) combine to form orange, solid tin iodide with an unknown formula. Sn metal + solid I2  n  solid SnxIy

Weighed quantities of Sn and I2 are combined, where the quantity of Sn is more than is needed to react with all of the iodine. After SnxIy has been formed, it is isolated by filtration. The mass of excess tin is also determined. The following data were collected: Mass of tin (Sn) in the original mixture

1.056 g

Mass of iodine (I2) in the original mixture 1.947 g Mass of tin (Sn) recovered after reaction

0.601 g

What is the empirical formula of the tin iodide obtained? 172. ▲  When analyzed, an unknown compound gave these experimental results: C, 54.0%; H, 6.00%; and O, 40.0%. Four different students used these values to calculate the empirical formulas shown here. Which answer is correct? Why did some students not get the correct answer? (a) C4H5O2 (c) C7H10O4 (b) C5H7O3 (d) C9H12O5 173. ▲  Two general chemistry students working together in the lab weigh out 0.832 g of CaCl2 ∙ 2 H2O into a crucible. After heating the sample for a short time and allowing the crucible to cool, the students determine that the sample has a mass of 0.739 g. They then do a quick calculation. On the basis of this calculation, what should they do next? (a) Congratulate themselves on a job well done. (b) Assume the bottle of CaCl2 ∙ 2 H2O was mislabeled; it actually contained something different. (c) Heat the crucible again, and then reweigh it. 174. To find the empirical formula of tin oxide, you first react tin metal with nitric acid in a porcelain crucible. The metal is converted to tin nitrate, but, upon heating the nitrate strongly, brown nitrogen

136

dioxide gas is evolved and tin oxide is formed. In the laboratory you collect the following data: Mass of crucible

13.457 g

Mass of crucible plus tin

14.710 g

Mass of crucible after heating 15.048 g What is the empirical formula of tin oxide?

Summary and Conceptual Questions The following questions may use concepts from this and the previous chapter. 175. ▲  Identify, from the list below, the information needed to calculate the number of atoms in 1.00 cm3 of iron. Outline the procedure used in this calculation. (a) the structure of solid iron (b) the molar mass of iron (c) Avogadro’s number (d) the density of iron (e) the temperature (f) iron’s atomic number (g) the number of iron isotopes 176. Consider the plot of relative element abundances on page 131. Is there a relationship between abundance and atomic number? Is there any difference between the relative abundance of an element of even atomic number and the relative abundance of an element of odd atomic number? 177. The photo here depicts what happens when a coil of magnesium ribbon and a few calcium chips are placed in water. (a) Based on these observations, what might you expect to see when barium, another Group 2A (2) element, is placed in water? (b) Give the period in which each element (Mg, Ca, and Ba) is found. What correlation do you think you might find between the reactivity of these elements and their positions in the periodic table?

© Charles D. Winters/Cengage

In the Laboratory

Magnesium (left ) and calcium (right ) in water

Chapter 2 / Atoms, Molecules, and Ions

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© Charles D. Winters/Cengage

178. A jar contains some number of jelly beans. To find out precisely how many are in the jar, you could dump them out and count them. How could you estimate their number without counting each one? (Chemists do just this kind of “bean counting” when they work with atoms and molecules. Atoms and molecules are too small to count one by one, so chemists have worked out other methods to determine the number of atoms in a sample.)

How many jelly beans are in the jar?



Study Questions

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137

3 Chemical Reactions N

Precipitation

Adding a solution of K2CrO4 to a solution of Pb(NO3)2 results in formation of a yellow solid, PbCrO4. Gaseous NH3 and HCl in open containers diffuse in air, and when they come into contact, a cloud of solid NH4Cl forms. E

Acid-Base

R

A C

T

I

O

K2CrO4(aq)

E R

A C

NH4Cl(s)

T I O N

PbCrO4(s)

NH3(aq)

HCl(aq)

Pb(NO3)2(aq)

R

Redox

E A C T I O

KOH(aq)

N

K(s)

Potassium reacts vigorously with water to form gaseous H2 and a solution of KOH. © Charles D. Winters/Cengage

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C hapt e r O ut li n e 3.1 Introduction to Chemical Equations 3.2 Balancing Chemical Equations 3.3 Introduction to Chemical Equilibrium 3.4 Aqueous Solutions 3.5 Precipitation Reactions 3.6 Acids and Bases 3.7 Acid–Base Reactions 3.8 Oxidation–Reduction Reactions 3.9 Classifying Reactions in Aqueous Solution

Chemical reactions occur all around us, as well as within us. There are almost limitless numbers of ways that elements and compounds can combine to produce new compounds. Fortunately, many chemical reactions are similar, and they can be grouped or classified. This chapter will introduce three common types of reactions: precipitation reactions, acid–base reactions, and oxidation–reduction reactions. To describe chemical reactions, chemists use symbolic representations (see Figure 1.6) that indicate both the formulas and the amount of each substance involved in the reactions. This chapter begins with instructions for properly using symbolic representations to describe chemical reactions.

3.1 Introduction to Chemical Equations Goals for Section 3.1 • Understand the information conveyed by a balanced chemical equation including the terminology used (reactants, products, stoichiometry, stoichiometric coefficients).

• Recognize that a balanced chemical equation is required by the law of conservation of matter.

Chemistry is about atoms, molecules, and ions and the reactions they undergo. An important way that chemists describe reactions is with balanced chemical equations. A chemical equation shows the reacting elements and compounds and the products of the reaction. A balanced equation has equal numbers of each type of atom on both sides of the equation. For example, when a stream of chlorine gas, Cl2, is directed onto solid phosphorus, P4, the mixture bursts into flame, and a chemical reaction

◀ Chemical reactions are at the heart of chemistry.  Pictured here are three general types of reactions: precipitation, acid–base, and oxidation–reduction (or redox).

139

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produces liquid phosphorus trichloride, PCl3 (Figure 3.1). The balanced chemical equation for this reaction is P4(s) + 6 Cl2(g)

4 PCl3(ℓ)

reactants

product

In a chemical equation, the formulas for the reactants (the substances combined in the reaction) are written to the left of the arrow and the formulas of the products (the substances produced) are written to the right of the arrow. The physical states of reactants and products are usually indicated. The symbol (s) indicates a solid, (g) a gas, and (ℓ) a liquid. A substance dissolved in water, that is, in an aqueous ­solution, is indicated by (aq). In the eighteenth century, the French scientist Antoine Lavoisier (1743–1794) introduced the law of conservation of matter, which states that matter can ­neither be created nor destroyed. This means that if the total mass of reactants is 10 g, and if the reaction completely converts reactants to products, the total mass of the ­products must be 10 g. This also means that if 1000 atoms of a particular e­ lement are contained in the reactants, then those 1000 atoms must also appear in the ­products. Atoms, and thus mass, are conserved in chemical reactions. When applied to the reaction of phosphorus and chlorine, the law of conservation of matter requires that 1 molecule of phosphorus, P4 (with 4 phosphorus atoms), and 6 molecules of Cl2 (with 12 atoms of Cl) will produce 4 molecules of PCl3. Because each PCl3 molecule contains 1 P atom and 3 Cl atoms, 4 PCl3 molecules are needed to account for 4 P atoms and 12 Cl atoms in the product. The equation is balanced; the same number of P and Cl atoms appear on each side of the equation. 6×2= 12 Cl atoms

4×3= 12 Cl atoms

P4(s) + 6 Cl2(g) 4 P atoms

4 PCl3(ℓ) 4 P atoms

In a chemical reaction, the relationship between the amounts of chemical reactants and products is called stoichiometry (pronounced “stoy-key-AHM-uh-tree”). The coefficients in a balanced equation are called stoichiometric coefficients. (In the P4 and Cl2 equation, these are 1, 6, and 4.) They can be interpreted as a number of atoms or molecules: 1 molecule of P4 and 6 molecules of Cl2 produce 4 molecules of PCl3. They can also refer to amounts of reactants and products: 1 mole of P4 combines with 6 moles of Cl2 to produce 4 moles of PCl3.

PCl3

P4

P4(s) + 6 Cl 2(g)

4 PCl 3(ℓ)

R E AC TA N T S

PRODUCT

Photos: © Charles D. Winters/Cengage

Cl2

Figure 3.1  Reaction of solid white phosphorus with chlorine gas.  The product is liquid phosphorus trichloride.

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Chapter 3 / Chemical Reactions

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Chemistry in Your Career Although his educational background is in environmental engineering, Nandan Prabhune uses chemistry daily in his job as Innovation and Technology Head (for Water and Waste Solutions Division) in Pune, India. In this capacity, Prabhune advises industries from mining to manufacturing on how to treat, reuse, and recycle the wastewater from their industrial processes. One of the things he likes best about his job is witnessing how his contributions help industries conserve water. Prabhune recognizes two key components to his success in generating creative solutions to wastewater

Antoine Laurent Lavoisier (1743–1794) On Monday, August 7, 1774, the ­English chemist Joseph Priestley (1733– 1804) isolated oxygen. (The Swedish chemist Carl Scheele [1742–1786] also discovered the element, perhaps in 1773, but did not publish his results until later.) To obtain oxygen, Priestley heated mercury(II) oxide, HgO, causing it to decompose to mercury and oxygen.

2 HgO(s) n 2 Hg(ℓ) + O2(g)

© Charles D. Winters/Cengage

Priestley did not immediately understand the significance of the discovery, but he mentioned it to the French chemist Antoine Lavoisier in October 1774. One of Lavoisier’s contributions to science was his recognition of the importance of exact scientific measurements and of carefully planned experiments, and he applied these methods to the study of oxygen. From this work, Lavoisier proposed that oxygen was an element, that it was

The decomposition of red mercury(II) oxide.  The decomposition reaction yields mercury metal and oxygen gas. The mercury is seen as a film on the surface of the test tube.

challenges: applying fundamental ­principles and i­ntegrating knowledge from across disciplines, including chemistry, microbiology, and physics. Prabhune draws on his knowledge of chemistry when determining the pH and chemical dosing required to precipitate metal ions, fluoride, and silica from industrial wastewaters. He uses acid–base equilibria calculations to determine how to neutralize wastewaters depending on which ions are present and how they interact. “These calculations are used to design wastewater treatment processes [and] equipment and [to] operate the engineered systems,” Prabhune explains.

one of the constituents of the compound water, and that burning involved a reaction with oxygen. He also mistakenly came to believe Priestley’s gas was present in all acids, so he named it oxygen from the Greek words meaning “to form an acid.” In other experiments, Lavoisier observed that the heat produced by a guinea pig exhaling a given amount of carbon dioxide is similar to the quantity of heat produced by burning carbon to give the same amount of carbon dioxide. From these and other experiments he concluded that, “Respiration is a combustion, slow it is true, but otherwise perfectly similar to that of charcoal.” Although he did not understand the details of the process, this was an important step in the development of biochemistry. Lavoisier was a prodigious scientist, and the principles of naming chemical substances that he introduced are still in use today. Furthermore, he wrote a textbook in which he applied the principles of the conservation of matter to chemistry, and he used the idea to write early versions of chemical equations. Because Lavoisier was an aristocrat, he came under suspicion during the Reign of Terror of the French Revolution in 1794. He was an investor in the Ferme Générale, the infamous tax-collecting organization in eighteenth-century France. Tobacco was a monopoly product of the Ferme Générale, and it was common to cheat the purchaser by adding water to the tobacco, a practice that Lavoisier opposed. Nonetheless, because of his involvement with the Ferme, his career was cut short by the guillotine on May 8, 1794, on the charge of “adding water to the people’s tobacco.”

Madame Lavoisier, Marie-Anne Paulze (1758–1836), was a crucial research partner to Antoine. Although not quite 14 when they married, she became a key collaborator in Antoine’s work. She worked in the laboratory, kept careful records of their experiments, and used her artistic talents to illustrate his texts. (She was trained by the painter of the accompanying portrait.) Her ability to translate chemistry texts in English gave the team knowledge of other scientists’ published works. She also ­ ublication of played a major role in the p Antoine’s revolutionary E ­ lementary Treatise on Chemistry in 1789. Finally, in 1804, she married another man of s­ cience, ­Benjamin Thompson (Count Rumford; see page 257); the union was short lived.

Purchase, Mr. and Mrs. Charles Wrightsman Gift, in honor of Everett Fahy, 1977/The Metropolitan Museum of Art

A Closer Look

Nandan Prabhune

Nandan Prabhune

Lavoisier and his wife, Marie-Anne Pierrette Paulze Lavoisier, as painted in 1788 by Jacques-Louis David. Lavoisier was then 45, and Marie-Anne was 30.

3.1  Introduction to Chemical Equations

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141

Photos:© Charles D. Winters/Cengage

(a) Reaction of iron and oxygen to give iron(III) oxide, Fe2O3.

(b) Reaction of sulfur (in the spoon) with oxygen to give sulfur dioxide, SO2.

(c) Reaction of phosphorus and oxygen to give tetraphosphorus decaoxide, P4O10.

Figure 3.2  Reactions of a metal and two nonmetals with oxygen.

3.2 Balancing Chemical Equations Goal for Section 3.2

© Charles D. Winters/Cengage

• Balance simple chemical equations. A balanced equation has the same number of atoms of each element on each side of the equation. Many chemical equations can be balanced by trial and error, and this method is often used. However, more systematic methods exist and are especially useful if reactions are complicated. When balancing chemical equations, there are two important things to remember: • Formulas for reactants and products must be correct. Once the correct formulas for

Figure 3.3  A combustion reaction.  Here, propane, C3H8,

burns to give CO2 and H2O. These simple oxides are always the products of the complete combustion of a hydrocarbon (that is, compounds containing only hydrogen and carbon atoms).

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the reactants and products have been determined, the subscripts in their formulas cannot be changed to balance an equation. Changing the subscripts changes the identity of the substance. For example, you cannot change CO2 to CO to balance an equation; carbon monoxide, CO, and carbon dioxide, CO2, are different compounds.

• Chemical equations are balanced using stoichiometric coefficients. The entire chemical formula of a substance is multiplied by the stoichiometric coefficient.

One general class of chemical reactions is the reaction of metals or nonmetals with oxygen to give oxides of the general formula MxOy. For example, iron reacts with ­oxygen to give iron(III) oxide (Figure 3.2a). 4 Fe(s) + 3 O2(g)  n  2 Fe2O3(s)

The nonmetals sulfur and oxygen react to form sulfur dioxide (Figure 3.2b), S(s) + O2(g)  n SO2(g)

and phosphorus, P4, reacts vigorously with oxygen to give tetraphosphorus decaoxide, P4O10 (Figure 3.2c). P4(s) + 5 O2(g)  n P4O10(s)

Chapter 3 / Chemical Reactions

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The equations written above are balanced. The same number of iron, sulfur, or phosphorus atoms and oxygen atoms occurs on each side of these equations. Every day, you encounter combustion reactions, the burning of a fuel in oxygen accompanied by the evolution of energy as heat (Figure 3.3). The combustion of octane, C8H18, a component of gasoline is an example. 2 C8H18(ℓ) + 25 O2(g)  n  16 CO2(g) + 18 H2O(g)

In all combustion reactions, some or all of the elements in the reactants end up as oxides, compounds containing oxygen. When the reactant is a hydrocarbon (a ­compound that contains only C and H, such as octane), the products of complete combustion are carbon dioxide and water. To illustrate equation balancing, consider the process used to write the ­balanced equation for the complete combustion of propane, C3H8, a common fuel. Step 1  Write correct formulas for the reactants and products. Here propane and oxygen are the reactants, and car-

unbalanced equation

C3H8(g) + O2(g)  88888888888888n CO2(g) + H2O(g)

bon dioxide and water are the products. Step 2 Balance the C atoms. In combustion reactions such as this, it is usually best to balance the C atoms first and leave the O atoms until the end (because O atoms are often found in more than one product). In this case, there are three C atoms in the reactants, so three must occur in the products. Three CO2 molecules are therefore required on the right side.

unbalanced equation

C3H8(g) + O2(g)  88888888888888n 3 CO2(g) + H2O(g)

Step 3  Balance the H atoms. A molecule of propane con-

tains eight H atoms. Each molecule of water has two H atoms, so four molecules of water account for the required eight H atoms on the right side. Step 4 Balance the O atoms. Ten O atoms are on the right side (3 × 2 = 6 in CO2 plus 4 × 1 = 4 in H2O). Five O2 molecules are needed to supply the required ten O atoms.

unbalanced equation

C3H8(g) + O2(g)  88888888888888n  3 CO2(g) + 4 H2O(g)

balanced equation

C3H8(g) + 5 O2(g)  88888888888888n  3 CO2(g) + 4 H2O(g)

Step 5 Verify that the number of atoms of each element is balanced. There are three C atoms, eight H atoms, and ten

O atoms on each side of the equation.

3 C atoms

3 C atoms

8 H atoms

8 H atoms

10 O atoms

10 O atoms

Exam p le 3.1 Strategy Map Problem Balance the equation for the reaction of NH3 and O2. Data/Information The formulas of the reactants and products are given.



Balancing an Equation for a Combustion Reaction Problem   Write the balanced equation for the combustion of ammonia gas (NH3) to give water vapor and nitrogen monoxide gas (NO).

What Do You Know?  You know the correct formulas and/or names for the reactants (NH3 and oxygen, O2) and the products (H2O and nitrogen monoxide, NO). You also know their states. Strategy   First write the unbalanced equation. Next balance the N atoms, then the H atoms, and finally the O atoms.

3.2  Balancing Chemical Equations

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143

Solution Step 1

Write the equation using the correct formulas for the reactants and products. The reactants are NH3(g) and O2(g), and the products are NO(g) and H2O(g). unbalanced equation

NH3(g) + O2(g)  88888888888888n NO(g) + H2O(g) Step 2

Balance the N atoms. There is one N atom on each side of the equation. The N atoms are in balance, at least for the moment. unbalanced equation

NH3(g) + O2(g)  88888888888888n  NO(g) + H2O(g) Step 3

Balance the H atoms. There are three H atoms on the left and two on the right. To have the same number on each side (six), use two molecules of NH3 on the left and three molecules of H2O on the right (which gives six H atoms on each side). unbalanced equation

2 NH3(g) + O2(g)  88888888888888n NO(g) + 3 H2O(g) Notice that after balancing the H atoms, the N atoms are no longer balanced. To bring them into balance, use two NO molecules on the right. unbalanced equation

2 NH3(g) + O2(g)  88888888888888n 2 NO(g) + 3 H2O(g) Step 4

Balance the O atoms. This is best left to the final step. After Step 3, there is an even number of O atoms (two) on the left and an odd number (five) on the right. Because there cannot be an odd number of O atoms on the left (O atoms are paired in O2 molecules), multiply each coefficient on the product side of the equation by two so that an even number of oxygen atoms (ten) now occurs on the right side. Next, multiply the coefficient of NH3 by two so that the number of nitrogen and hydrogen atoms remain balanced with the right side: unbalanced equation

4 NH3(g) + O2(g)  88888888888888n  4 NO(g) + 6 H2O(g) Now the oxygen atoms can be balanced with five O2 molecules on the left side of the equation: balanced equation

4 NH3(g) + 5 O2(g)  88888888888888n  4 NO(g) + 6 H2O(g) Four N atoms, 12 H atoms, and 10 O atoms are on each side of the equation.

Think about Your Answer  An alternative way to write this equation is 2 NH3(g) + 5/2 O2(g)  n  2 NO(g) + 3 H2O(g) where a fractional coefficient has been used. This equation is correctly balanced and has some uses, but chemical equations are usually written with whole-number coefficients.

Check Your Understanding (a) Butane gas, C4H10, can burn completely in air [use O2(g) as the other reactant] to give carbon dioxide gas and water vapor. Write a balanced equation for this combustion reaction. (b) Write a balanced chemical equation for the complete combustion of C3H7BO3(ℓ), a gasoline additive. The products of combustion are CO2(g), H2O(g), and B2O3(s).

144

Chapter 3 / Chemical Reactions

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3.3 Introduction to Chemical Equilibrium Goals for Section 3.3 • Recognize that all chemical reactions are reversible and that reactions eventually reach a dynamic equilibrium.

• Recognize the difference between reactant-favored and product-favored reactions at Up to this point, we have treated chemical reactions as proceeding in one direction only, with reactants being converted completely to products. Nature, however, is more complex than this. All chemical reactions are reversible, in principle, and many reactions lead to incomplete conversion of reactants to products. The formation of stalactites and stalagmites in a limestone cave is an example of a system that depends on the reversibility of a chemical reaction (Figure 3.4). Stalactites and stalagmites are made chiefly of calcium carbonate, a mineral found in underground deposits in the form of limestone, a leftover from ancient oceans. If water seeping through the limestone contains dissolved CO2, a reaction occurs in which the mineral dissolves, giving an aqueous solution of Ca(HCO3)2. CaCO3(s) + CO2(aq) + H2O(ℓ)  n Ca(HCO3)2(aq)

When the mineral-laden water reaches a cave, the reverse reaction occurs; CO2 is released into the cave and solid CaCO3 is deposited. Ca(HCO3)2(aq)  n CaCO3(s) + CO2(g) + H2O(ℓ)

As illustrated in Figure 3.5, these reactions can also be done in a laboratory. A

Reactants: Solutions of CaCl2 (left) and NaHCO3 (right).

© A. N. Palmer

equilibrium.

Figure 3.4  Cave chemistry.  ​ Calcium carbonate stalactites cling to the roof of a cave, and stalagmites grow up from the cave floor. The chemistry producing these formations is a good example of the reversibility of chemical reactions. Figure 3.5 The reversibility of chemical reactions.  The experiments

B

here demonstrate the reversibility of chemical reactions. The system is described by the following balanced chemical equation:

FORWARD REACTION The solutions are mixed, forming H2O, CO2 gas, and CaCO3 solid.

Ca2+(aq) + 2 HCO3−(aq) uv CaCO3(s) + CO2(g) + H2O(ℓ)

Solutions of CaCl2 (a source of Ca2+ ions) and NaHCO3 (a source of HCO3– ions) are mixed and produce solid CaCO3 and CO2 gas.

Photos: © Charles D. Winters/Cengage

C

The reaction can be reversed by bubbling CO2 gas into the CaCO3 suspension.

D

REVERSE REACTION

The CaCO3 dissolves when the solution has been saturated with CO2. Elapsing time...

If CO2 gas is bubbled into a suspension of CaCO3, solid CaCO3 and gaseous CO2 react to produce Ca2+ and HCO3– ions in solution.

3.3  Introduction to Chemical Equilibrium

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145

Another example of a reversible reaction is the reaction of nitrogen with hydrogen to form ammonia gas, a compound made industrially in enormous quantities and used directly as a fertilizer and in the production of other chemicals. N2(g) + 3 H2(g)  n  2 NH3(g)

Nitrogen and hydrogen react to form ammonia, but, under the conditions of the reaction, ammonia also breaks down into nitrogen and hydrogen in the reverse reaction. 2 NH3(g)  n N2(g) + 3 H2(g)

Consider what would happen if you mixed nitrogen and hydrogen in a closed container under the proper conditions for the reaction to occur. At first, N2 and H2 react to produce some ammonia. As the ammonia is produced, however, some NH3 molecules decompose to re-form nitrogen and hydrogen in the reverse reaction (Figure 3.6). At the beginning of the process, the forward reaction to give NH3 predominates, but, as the reactants are consumed, the rate (or speed) of the forward reaction progressively slows. At the same time, the reverse reaction speeds up as the amount of ammonia increases. Eventually, the rate of the forward reaction equals the rate of the reverse reaction. At this point, no further macroscopic change is observed; the amounts of nitrogen, hydrogen, and ammonia stop changing (although the forward and reverse reactions continue). The system has reached chemical equilibrium—the state at which concentrations of reactants and products do not undergo any change. The reaction vessel contains all three substances—nitrogen, hydrogen, and ammonia. Because the forward and reverse processes are still occurring, this state is referred to as a dynamic equilibrium. Systems in dynamic equilibrium are represented by writing a double arrow symbol (uv) connecting the reactants and products. N2(g) + 3 H2(g)  uv  2 NH3(g)

An important principle in chemistry is that chemical reactions always proceed spontaneously toward equilibrium. A reaction will never proceed spontaneously in a direction that takes a system further from equilibrium. A key question is “When a reaction reaches equilibrium, will the reactants be converted largely to products or will most of the reactants still be present?” For now it is useful to define product-favored reactions as reactions in which reactants are completely or largely converted to products when equilibrium is reached. The combustion reactions covered earlier are examples of reactions that are product-favored at equilibrium. In fact, most of the reactions you will study in the rest of this chapter are product-favored reactions at equilibrium. The equations for reactions that are very product-favored are often written using a single arrow (n) connecting reactants and products.

N2(g) + 3H2(g)

Amounts of products and reactants

Reaction begins with 3:1 mixture of H2 to N2.

2 NH3(g)

Equilibrium achieved

H2

Eventually, the amounts of N2, H2, and NH3 no longer change. At this point, the reaction has reached equilibrium. Nonetheless, the forward reaction to produce NH3 continues, as does the reverse reaction (the decomposition of NH3 ).

NH3 N2

As reaction proceeds H2 and N2 produce NH3, but the NH3 also begins to decompose back to H2 and N2.

146

Reaction proceeding toward equilibrium Time

Figure 3.6  The reaction of N2 and H2 to produce NH3.

Chapter 3 / Chemical Reactions

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The opposite of a product-favored reaction is one that is reactant-favored at equilibrium. Such reactions lead to the conversion of only a small amount of the reactants to products. An example of a reactant-favored reaction is the ionization of acetic acid in water where only a tiny fraction of the acid produces ions. CH3CO2H(aq) + H2O(ℓ)  uv CH3CO2−(aq) + H3O+(aq)

Acetic acid is an example of a large number of acids called weak acids because the reaction with water is reactant-favored at equilibrium, and only a few percent of the molecules react with water to form ionic products.

3.4 Aqueous Solutions Goals for Section 3.4 • Explain the difference between electrolytes and nonelectrolytes and recognize examples of each.

• Predict the solubility of ionic compounds in water. Many of the reactions you will study in your chemistry course and almost all of the reactions that occur in living things are carried out in solutions in which the r­ eacting substances are dissolved in water. A solution is a homogeneous mixture of two or more substances. One substance is generally considered the solvent, the medium in which another substance—the solute—is dissolved. The remainder of this chapter is an introduction to some of the types of reactions that occur in ­aqueous ­solutions, solutions in which water is the solvent. First, it is important to understand some basic concepts about the behavior of compounds dissolved in water.

Ions and Molecules in Aqueous Solutions Dissolving an ionic solid requires separating each ion from the oppositely charged ions that surround it in the solid state (Figure 3.7); this process is called dissociation. Water is especially good at dissolving ionic compounds because each water molecule

(−)

(+)

A water molecule is electrically positive on one side (the H atoms) and electrically negative on the other (the O atom). These charges enable water to interact with negative and positive ions in aqueous solution.

Water molecules are attracted to both positive cations and negative anions in aqueous solution.

+

−

Water surrounding a cation

Water surrounding an anion

Figure 3.7  Water as a solvent for ionic substances.  For the sake of simplicity and clarity, the ions in this and other figures are shown surrounded by four or five water molecules. However, experiments show it is often six.

When an ionic substance dissolves in water, each ion is surrounded by up to six water molecules. 2+

− 2+

© Charles D. Winters/Cengage

2+

Copper(II) chloride is added to water. Interactions between water and the Cu2+ and Cl– ions allow the solid to dissolve.

2+

− −

The ions are now sheathed in water molecules. 3.4  Aqueous Solutions

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147

has a positively charged end and a negatively charged end. When an ionic compound dissolves in water, each negative ion becomes surrounded by water molecules with the positive ends of water molecules pointing toward it, and each positive ion becomes surrounded by the negative ends of several water molecules. The forces involved are described by Coulomb’s law (Equation 2.3, page 97). The water-encased ions produced by dissolving an ionic compound are free to move about in solution. Under normal conditions, the movement of ions is random, and the cations and anions from a dissolved ionic compound are dispersed uniformly throughout the solution. However, if two electrodes (conductors of electricity such as copper wire) are placed in the solution and connected to a battery, positive cations are drawn toward the negative electrode and negative anions are drawn toward the positive electrode (Figure 3.8). Conduction of electricity in the solution is a consequence of the movement of charged particles in solution. Compounds whose aqueous solutions conduct electricity are called electrolytes. All ionic compounds that are soluble in water are electrolytes. The extent to which a solution conducts electricity, its conductivity, depends on the ion concentration. You can test the conductivity of a solution by trying to pass electricity through the solution. The greater the ion concentration, the greater the conductivity. For example, for every mole of NaCl that dissolves, 1 mol of Na+ and 1 mol of Cl− ions enter the solution. NaCl(s)  n Na+(aq) + Cl−(aq)

100% Dissociation  n  strong electrolyte

There is a significant concentration of sodium ions and chloride ions in the solution, and the solution is a good conductor of electricity. Substances whose solutions are good electrical conductors are called strong electrolytes (Figure 3.8a). The ions into which an ionic compound dissociates are given by the compound’s name, and the relative amounts of these ions are given by its formula. For example, sodium chloride yields sodium ions (Na+) and chloride ions (Cl−) in solution in a 1:1 ratio.

Strong Electrolyte

Nonelectrolyte

Weak Electrolyte

Bulb is lit, showing solution conducts electricity well.

Bulb is not lit, showing solution does not conduct.

Bulb is dimly lit, showing solution conducts electricity poorly.

Ethanol

2+ Cu2+

−

−

Cl

© Charles D. Winters/Cengage

© Charles D. Winters/Cengage

CuCl2

Acetic acid

−

© Charles D. Winters/Cengage

Figure 3.8

Acetate ion

+ H+

−

+ 2+

2+

2+

− −

(a) A strong electrolyte conducts electricity. ​CuCl2 is completely dissociated into Cu2+ and Cl− ions.

148

−

(b) A nonelectro­lyte does not conduct electricity because no ions are present in solution.

(c) A weak electrolyte conducts electricity poorly because only a few ions are present in solution.

Chapter 3 / Chemical Reactions

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The ionic compound copper(II) chloride, CuCl2, is also a strong electrolyte. In this case, there are two chloride ions for each copper ion in solution. CuCl2(s)  n Cu2+(aq) + 2 Cl−(aq)

Notice that the two chloride ions per formula unit are present as two separate particles in solution. In yet another example, the ionic compound barium nitrate yields barium ions and nitrate ions in solution. For each Ba2+ ion in solution, there are two NO3− ions. Ba(NO3)2(s)  n Ba2+(aq) + 2 NO3−(aq)

Notice that NO3−, a polyatomic ion, does not dissociate further; the ion exists as one unit in aqueous solution. Compounds whose aqueous solutions do not conduct electricity are called nonelectrolytes (Figure 3.8b). The solute particles present in these aqueous solutions are molecules, not ions. For example, when the molecular compound ethanol (C2H5OH) dissolves in water, each molecule of ethanol stays intact as a single unit. Ions are not produced in the solution. Other examples of nonelectrolytes are sucrose (C12H22O11) and antifreeze (ethylene glycol, HOCH2CH2OH). Some molecular compounds (strong acids, weak acids, and weak bases) react with water to produce ions in aqueous solutions and are thus electrolytes. One example is gaseous hydrogen chloride, a molecular compound, which reacts with water to form ions. The aqueous solution is referred to as hydrochloric acid. HCl(g) + H2O(ℓ)  n H3O+(aq) + Cl−(aq)

This reaction is very product-favored. Almost every molecule of HCl ionizes in solution, so hydrochloric acid is a strong electrolyte. Some molecular compounds are weak electrolytes (Figure 3.8c). When these compounds dissolve in water only a small fraction of the molecules ionize to form ions. These aqueous solutions are poor conductors of electricity. Acetic acid is a weak electrolyte. In vinegar, an aqueous solution of acetic acid, only about 0.5% of the molecules of acetic acid are ionized to form acetate (CH3CO2−) and hydronium (H3O+) ions. CH3CO2H(aq) + H2O(ℓ)  uv CH3CO2−(aq) + H3O+(aq)

Figure 3.9 summarizes whether a given type of solute will be present in aqueous solution as ions, molecules, or a combination of ions and molecules.

Solubility of Ionic Compounds in Water Many ionic compounds are quite soluble in water. Others dissolve only to a small extent, while many are essentially insoluble. Fortunately, it is possible to predict Figure 3.9  Predicting the species present in aqueous solution.  When compounds

Solute in an Aqueous Solution

Ionic Compound

Molecular Compound

Acids and Weak Bases

Strong Acids

Strong Electrolyte IONS



Most Molecular Compounds

dissolve, ions may result from ionic or molecular compounds. Some molecular compounds may remain intact as molecules in solution. (Note that hydroxidecontaining strong bases are ionic compounds.)

Weak Acids and Weak Bases

Weak Electrolyte MOLECULES and IONS

Nonelectrolyte MOLECULES 3.4  Aqueous Solutions

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149

Water-soluble compounds Almost all salts of Na+, K+, NH4+ Salts of nitrate, NO3− chlorate, ClO3− perchlorate, ClO4− acetate, CH3CO2−

Almost all salts of Cl−, Br−, I−

Exceptions (not soluble) Halides of Ag+, Hg22+, Pb2+

Water-insoluble compounds Salts containing F−

Salts of sulfate, SO42−

Exceptions (not soluble)

Exceptions (not soluble)

Fluorides of Mg2+, Ca2+, Sr2+, Ba2+, Pb2+

Sulfates of Ca2+, Sr2+, Ba2+, Pb2+, Ag+

Silver compounds

Most metal hydroxides and oxides

Most salts of carbonate, CO32− phosphate, PO43− oxalate, C2O42− chromate, CrO42− sulfide, S2−

Exceptions (soluble) Alkali metal hydroxides and Ba(OH)2 and Sr(OH)2

Hydroxides

Photos: © Charles D. Winters/Cengage

Sulfides

Exceptions (soluble) Salts of NH4+ and the alkali metal cations, and BaS

AgNO3

AgCl

AgOH

(NH4)2S

(a) Nitrates are generally soluble, as are chlorides (exceptions include AgCl). Hydroxides are generally not soluble.

CdS

Sb2S3

NaOH Ca(OH)2 Fe(OH)3 Ni(OH)2

PbS

(b) Sulfides are generally not soluble (exceptions include salts with NH4+ and Na+).

(c) Hydroxides are generally not soluble, except when the cation is a Group 1A (1) metal (or Sr2+ or Ba2+).

Figure 3.10  Guidelines to predict the aqueous solubility of ionic compounds.  If a compound contains one of the ions in the columns on the left side of the chart above, it is predicted to be at least moderately soluble in water. Exceptions to the guidelines are noted.

whether many compounds are soluble. For now, solubility will be considered an either-or question. Compounds that visibly dissolve to a certain extent are soluble, whereas those that show no visible signs of dissolving are insoluble. Figure 3.10 gives broad guidelines that can help you to predict whether an ionic compound is soluble in water based on the ions that make up the compound. For example, sodium nitrate, NaNO 3, contains an alkali metal cation, Na +, and the nitrate anion, NO3−. The presence of either of these ions generally ensures that the compound is soluble in water; at 20 °C, over 90 g of NaNO3 will dissolve in 100 mL of water. In contrast, calcium hydroxide is poorly soluble in water. If a spoonful of solid Ca(OH)2 is added to 100 mL of water, less than 1 g will dissolve at 20 °C. Nearly all of the Ca(OH)2 remains as a solid (Figure 3.10c).

E xamp le 3.2

Solubility Guidelines Problem   Predict whether the following ionic compounds are likely to be watersoluble. For soluble compounds, list the ions present in solution. (a) KCl Solubility Guidelines Observations such as those shown in Figure 3.10 were used to create the solubility guidelines. Note, however, that these are general guidelines. There are exceptions.

150

(b) MgCO3

(c) Fe(OH)3

(d) Cu(NO3)2

What Do You Know?   You know the formulas of the compounds but need to be able to identify the ions that make up each of them in order to use the solubility guidelines in Figure 3.10.

Strategy  Use the solubility guidelines given in Figure 3.10. Soluble ionic compounds will dissociate into their respective ions in solution.

Chapter 3 / Chemical Reactions

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Solution  (a) KCl is composed of K+ and Cl− ions. The presence of either of these ions means that the compound is likely to be soluble in water. The solution contains K+ and Cl− ions  dissolved in water. KCl(s)  n K+(aq) + Cl−(aq) (The solubility of KCl is about 35 g in 100 mL of water at 20 °C.) (b) Magnesium carbonate is composed of Mg 2+ and CO32− ions. Salts containing the carbonate ion usually are insoluble, unless combined with an ion like Na + or NH4+. Therefore, MgCO3 is likely insoluble in water. (The solubility of MgCO3 is less than 0.2 g/100 mL of water.) (c) Iron(III) hydroxide is composed of Fe 3+ and OH− ions. Hydroxides are soluble only when OH− is combined with ions of the alkali metals, strontium, or barium; Fe3+ is a transition metal ion, so Fe(OH)3 is insoluble. (d) Copper(II) nitrate is composed of Cu2+ and NO3− ions. Nitrate salts are soluble, so this compound dissolves in water, giving ions in solution as shown in the equation below. Cu(NO3)2(s) n Cu2+(aq) + 2 NO3−(aq)

Think about Your Answer   For chemists, a set of guidelines like those in Figure 3.10 is useful. If needed, accurate solubility information is available for many compounds in chemical resource books or online databases.

Check Your Understanding  Predict whether each of the following ionic compounds is likely to be soluble in water. If it is soluble, write the formulas of the ions present in aqueous solution.
 (a) LiNO3

(b) CaCl2

(c) Cu(OH)2

(d) NaCH3CO2

3.5 Precipitation Reactions Goals for Section 3.5 • Recognize what ions are formed when an ionic compound dissolves in water. • Recognize exchange reactions in which there is an exchange of anions between the cations of reactants in solution.

• Predict the products of precipitation reactions. • Write net ionic equations for reactions in aqueous solution. With a background on whether compounds will yield ions or molecules when dissolved in water and whether ionic compounds are soluble or insoluble in water, you are ready to examine the types of chemical reactions that occur in aqueous solutions. It is useful to look for patterns that can help you predict the reaction products. Many reactions you will encounter are exchange reactions (sometimes called double displacement, double replacement, or metathesis reactions). In these reactions the ions of the reactants exchange partners.

A+B− + C+D−

A+D− + C+B−

Reactions in which an insoluble solid called a precipitate forms (precipitation reactions) are exchange reactions. For example, aqueous solutions of silver nitrate and potassium chloride react to produce solid silver chloride and aqueous ­potassium nitrate (Figure 3.11).

3.5  Precipitation Reactions

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151

Figure 3.11 ​ Precipitation of silver chloride.

+

−

−

+

+

+

−

© Charles D. Winters/Cengage

−

+

+

+

− −

−

+ +

+

−

(b) Initially, the Ag+ ions (silver) and Cl− ions (yellow) are widely separated.

(c) Ag+ and Cl− ions approach and form ion pairs.

(d) As more and more Ag+ and Cl− ions come together, a precipitate of solid AgCl forms.

(a) Mixing aqueous solutions of silver nitrate and potassium chloride produces white, insoluble silver chloride, AgCl.

Photos: © Charles D. Winters/Cengage

AgNO3(aq) + KCl(aq)  n AgCl(s) + KNO3(aq)

The dark red precipitate PbS forms from the reaction of Pb(NO3)2 and (NH4)2S.

Reactants

Products

Ag+(aq) + NO3−(aq)

Insoluble AgCl(s)

K+(aq) + Cl−(aq)

K+(aq) + NO3−(aq)

The solubility guidelines (Figure 3.10) predict that almost all metal sulfides are insoluble in water. If a solution of a soluble metal compound comes in contact with a source of sulfide ions, the metal sulfide precipitates. Pb(NO3)2(aq) + (NH4)2S(aq)  n PbS(s) + 2 NH4NO3(aq) Reactants

Products

Pb2+(aq) + 2 NO3−(aq)

Insoluble PbS(s)

2 NH4+(aq) + S2−(aq)

2 NH4+(aq) + 2 NO3−(aq)

In yet another example, the solubility guidelines indicate that with the exception of the alkali metal cations (and Sr2+ and Ba2+), metal cations form insoluble hydroxides. Thus, water-soluble iron(III) chloride and sodium hydroxide react to give insoluble iron(III) hydroxide.

Photos: © Charles D. Winters/Cengage

FeCl3(aq) + 3 NaOH(aq)  n Fe(OH)3(s) + 3 NaCl(aq)

The orange precipitate Fe(OH)3 forms from the reaction of FeCl3 and NaOH.

152

Reactants

Products

Fe3+(aq) + 3 Cl−(aq)

Insoluble Fe(OH)3(s)

3 Na+(aq) + 3 OH−(aq)

3 Na+(aq) + 3 Cl−(aq)

E xamp le 3.3

Writing the Equation for a Precipitation Reaction Problem  Does a precipitate form when aqueous solutions of potassium chromate and silver nitrate are mixed? If so, write the balanced equation.

What Do You Know?   The names of the two reactants are given. Precipitation reactions are exchange reactions in which the reactant ions switch partners and form an insoluble solid. You will need information on molecule solubility in Figure 3.10 to determine if any of the possible exchange reaction products are insoluble in water.

Chapter 3 / Chemical Reactions

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Strategy Step 1. Determine the formulas from the reactant names and identify the ions that make up these compounds. Step 3. Determine if the possible product(s) is/are insoluble using information from Figure 3.10. Step 4. Write and balance the equation.

Solution Step 1. Determine the formulas of the reactants and identify the ions. The formulas for silver nitrate and potassium chromate are AgNO3 and K2CrO4, respectively. AgNO3 is composed of Ag+ ions and NO3− ions. K2CrO4 is composed of K+ ions and CrO42− ions. Step 2. Write the formulas for the possible products in this reaction by exchanging cations and anions.

Photos: © Charles D. Winters/Cengage

Step 2. Write formulas for the possible products by exchanging cations and anions.

The red precipitate Ag2CrO4 forms from the reaction of AgNO3 and K2CrO4.

Ag2CrO4 can be formed by combining Ag+ ions and CrO42− ions in a 2:1 ratio. KNO3 can be formed by combining K+ and NO3− ions in a 1:1 ratio. Step 3. Determine whether either possible product is insoluble using information from Figure 3.10. Based on the solubility guidelines, silver chromate is an insoluble compound (chromates are insoluble except for those with Group 1A (1) cations or NH4+), and potassium nitrate is soluble in water. A precipitate of silver chromate is predicted if the reactants are mixed. Step 4. Write and balance the equation. 2 AgNO3(aq) + K2CrO4(aq) n Ag2CrO4(s) + 2 KNO3(aq)

Think about Your Answer  You can figure out that chromate ion (CrO42−) has a

charge of 2− because potassium chromate consists of two potassium ions (K+) for each chromate ion. Likewise, the silver ion must have a 1+ charge because it was initially paired with nitrate ion (NO3−).

Check Your Understanding  In each of the following cases, does a precipitation reaction occur when solutions of the two water-soluble reactants are mixed? Give the formula of any precipitate that forms, and write a balanced chemical equation for the precipitation reactions that occur. (a) sodium carbonate and copper(II) chloride (b) potassium carbonate and sodium nitrate (c) nickel(II) chloride and potassium hydroxide

Net Ionic Equations When aqueous solutions of silver nitrate and potassium chloride are mixed, insoluble silver chloride forms, leaving potassium nitrate in solution (see Figure 3.11). The balanced chemical equation for this process is AgNO3(aq) + KCl(aq)  n AgCl(s) + KNO3(aq)

Another way to represent this reaction is by writing an equation that shows the soluble ionic compounds present in solution as dissociated ions. An aqueous solution of silver nitrate contains Ag+ and NO3− ions, and an aqueous solution of potassium chloride contains K+ and Cl− ions. In the products, potassium nitrate is present in

3.5  Precipitation Reactions

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153

solution as K+ and NO3− ions. However, silver chloride is insoluble and thus is not present in the solution as dissociated ions. It is shown in the equation as AgCl(s). Ag+(aq) + NO3−(aq) + K+(aq) + Cl−(aq)

AgCl(s) + K+(aq) + NO3−(aq) after reaction

before reaction

Net Ionic Equations  All chemical

equations, including net ionic equations, must be balanced. The same number of atoms of each kind must appear on both the product and reactant sides. In addition, the sum of positive and negative charges must be the same on both sides of the equation.

This type of equation is called a complete ionic equation. It represents any soluble ionic compounds (and any strong acids and strong bases, Section 3.6) as aqueous ions. The K+ and NO3− ions are present in solution both before and after reaction, so they appear on both the reactant and product sides of the complete ionic equation. Such ions are often called spectator ions because they do not participate in the net reaction; they only “look on” from the sidelines. Little chemical information is lost if the equation is written without them, so you can simplify the equation to Ag+(aq) + Cl−(aq)  n AgCl(s)

The balanced equation that results from leaving out spectator ions is the net ionic equation for the reaction. The significance of net ionic equations, and the reason that net ionic equations are commonly used, is that they focus attention on the reaction that takes place. Leaving out spectator ions does not mean that K+ and NO3− ions are unimportant in the AgNO3 + KCl reaction. Indeed, Ag+ ions cannot exist alone in solution; a negative ion, in this case NO3−, must be present to balance the positive charge of Ag+. Any anion will do, however, as long as it forms a water-soluble compound with Ag+. Thus, you could use AgClO4 instead of AgNO3. Similarly, there must be a positive ion present to balance the negative charge of Cl−. In this case, the positive ion present is K+ in KCl, but you could use NaCl instead of KCl. The net ionic equation would be the same. Finally, notice that there must always be a charge balance as well as a mass ­balance in a balanced equation. In the Ag+ + Cl− net ionic equation, the cation and anion charges on the left add together to give a net charge of zero, the same as the zero charge on AgCl(s) on the right.

Exampl e 3 .4

Writing and Balancing Net Ionic Equations

Strategy Map Problem Write the balanced net ionic equation for the reaction of BaCl2 + Na2SO4.

Data/Information The formulas of the reactants are given.

Problem  Write a balanced, net ionic equation for the reaction of aqueous solutions of BaCl2 and Na2SO4. What Do You Know?  The formulas for the reactants are given. You should recognize that this is an exchange reaction, and that you will need the information on solubilities in Figure 3.10 (page 150). Strategy Step 1. Determine the formulas for the reaction products by exchanging cations and anions and write a balanced equation. Step 2. Determine the state of each reactant and product (s, ℓ, g, or aq). Step 3. Write the complete ionic equation, which should show any soluble ionic compounds as aqueous cations and anions. Step 4. Write the net ionic equation by eliminating the spectator ions.

Solution  Step 1

154

Determine the products and then write the complete balanced equation. In this exchange reaction, the Ba2+ and Na+ cations exchange anions (Cl− and SO42−) to give BaSO4 and NaCl. Now that the reactants and products are known, you can write an equation for the reaction. To balance the equation, you need to place a two in front of the NaCl.

Chapter 3 / Chemical Reactions

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BaCl2 + Na2SO4 n BaSO4 + 2 NaCl Step 2

Determine the state of each reactant and product. Decide on the solubility of each compound (see Figure 3.10). Compounds containing sodium ions are generally water-soluble, as are those containing chloride ions (with some important exceptions). Sulfate salts are also usually soluble, but one important exception is BaSO4. You can use this information to add the state to each reactant and product. BaCl2(aq) + Na2SO4(aq) n BaSO4(s) + 2 NaCl(aq)

Step 3

Write the complete ionic equation. All soluble ionic compounds dissociate to form ions in aqueous solution. Writing the soluble substances as ions in solution results in the following complete ionic equation. Ba2+(aq) + 2 Cl−(aq) + 2 Na+(aq) + SO42−(aq) n BaSO4(s) + 2 Na+(aq) + 2 Cl−(aq)

Step 4

Write the net ionic equation. Eliminate the spectator ions Na+ and Cl− to give the net ionic equation. Ba2+(aq) + SO42−(aq) n BaSO4(s)

Think about Your Answer  Notice that the sum of ion charges is the same on both sides of the equation. On the left, 2+ and 2− sum to zero; on the right the charge on BaSO4 is also zero.

Check Your Understanding  In each of the following cases, aqueous solutions containing the compounds indicated are mixed. Write balanced net ionic ­equations for the reactions that occur. (a) CaCl2 + Na3PO4 (b) iron(III) chloride and potassium hydroxide (c) lead(II) nitrate and potassium chloride

One application for precipitation reactions is the separation of mixtures of ionic compounds. For example, with the appropriate choice of anions, one metal cation may be selectively precipitated from a mixture.

Exam p le 3.5

Problem   Which of the following soluble ionic compounds can be used to separate a mixture of Cu(NO3)2(aq) and Mg(NO3)2(aq) by selectively precipitating only one of the two metal cations? (a) KCl

(b) NaOH

(c) Na2SO4

(d) KNO3

What Do You Know?  Cu(NO3)2 and Mg(NO3)2 are water soluble and provide Cu2+(aq) and Mg2+(aq) ions, respectively. The anions to be added (Cl −, OH−, SO42−, and NO3−) are all provided as water-soluble ionic compounds. You will need information on solubility from Figure 3.10. You do not need to worry about the nitrate ion forming a precipitate with the added metal cations because nitrate salts are generally soluble.

Strategy  Use Figure 3.10 to predict which combination of cations and anions are insoluble

© Charles D. Winters/Cengage

Separating a Mixture by Selective Precipitation

Precipitation reaction.  The reaction of barium chloride and sodium sulfate described in Example 3.4 produces insoluble barium sulfate and water-soluble sodium chloride.

in water. Determine which, if any, of the anions will precipitate one of the two metal cations.

Solution (a) KCl dissolves in water to produce Cl− ions. Neither Cu2+ nor Mg2+ will precipitate with Cl− because both CuCl2 and MgCl2 are water-soluble compounds. KCl cannot be used to separate the cations in an aqueous solution containing both Cu2+ and Mg2+.

3.5  Precipitation Reactions

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155

(b) NaOH dissolves in water to produce OH− ions. Both Cu2+ and Mg2+ will precipitate as insoluble hydroxides, solid Cu(OH)2 and Mg(OH)2. NaOH will not selectively precipitate one of the two metal cations. (c) Na2SO4 dissolves in water to produce SO42− ions. CuSO4 is soluble in water while MgSO4 is insoluble. Na2SO4 will selectively separate Cu2+ from Mg2+ by precipitating MgSO4. (d) KNO3 dissolves in water to produce NO3− ions. Because both Cu(NO3)2 and Mg(NO3)2 are soluble in water, KNO3 will not selectively precipitate one of the two metal cations.

Think about Your Answer  This example illustrates a laboratory procedure called qualitative chemical analysis. The technique can be used to separate mixtures containing up to dozens of aqueous metal cations. Given a pair of metal cations to be separated, consider what the product might be when combined with a range of anions. You can generally find one anion that will form a precipitate with one cation but not the other, except when trying to separate Group 1A (1) cations, which form soluble compounds with most anions.

Check Your Understanding Is it possible to separate an aqueous mixture of Cl− and SO42− ions with the addition of water-soluble compounds containing each of the following cations? (a) Pb2+

(c) Na+

(b) Ca2+

(d) Cu2+

3.6 Acids and Bases Goals for Section 3.6 • Know the names and formulas of common acids and bases and categorize them as strong or weak.

• Define the Arrhenius and Brønsted-Lowry concepts of acids and bases. • Recognize substances that are amphiprotic and oxides that dissolve in water to give acidic solutions and basic solutions.

Acids and bases are two important classes of compounds. You may already be familiar with some common properties of acids. They produce bubbles of CO2 gas when added to a metal carbonate such as CaCO3 (Figure 3.12a), and they react with many metals to produce hydrogen gas (H2) (Figure 3.12b). Although tasting substances is never done in a chemistry laboratory, you have probably experienced the sour taste of acids such as acetic acid in vinegar and citric acid (commonly found in fruits and added to candies and soft drinks).

Extract of rose petals in alcohol and water

Add base

Photos: © Charles D. Winters/Cengage

Add acid

(a) A piece of coral (mostly CaCO3) dissolves in acid to give CO2 gas.

(b) Zinc reacts with hydrochloric acid to produce zinc chloride and hydrogen gas.

(c) An extract of red rose petals turns deep red on adding acid but turns green on adding base.

Figure 3.12  Some properties of acids and bases.

156

Chapter 3 / Chemical Reactions

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Table 3.1

Oxalic acid H2C2O4

Common Acids and Bases*

Strong Acids (Strong Electrolytes)

Soluble Strong Bases (Strong Electrolytes)**

HCl

Hydrochloric acid

LiOH

Lithium hydroxide

HBr

Hydrobromic acid

NaOH

Sodium hydroxide

HI

Hydroiodic acid

KOH

Potassium hydroxide

HNO3

Nitric acid

Ba(OH)2

Barium hydroxide

HClO3

Chloric acid

Sr(OH)2

Strontium hydroxide

HClO4

Perchloric acid

H2SO4

Sulfuric acid

Weak Acids (Weak Electrolytes)

Weak Base (Weak Electrolyte)

HF

Hydrofluoric acid

NH3

H3PO4

Phosphoric acid

H2CO3

Carbonic acid

CH3CO2H

Acetic acid

H2C2O4

Oxalic acid

H2C4H4O6

Tartaric acid

H3C6H5O7

Citric acid

HC9H7O4

Aspirin

Ammonia

Carboxyl group

Acetic acid CH3CO2H Weak Acids  Common acids and

bases are listed in Table 3.1. There are numerous other weak acids and bases, and many are natural substances. Many naturally occurring weak acids, such as oxalic and acetic acids, contain CO2H or carboxyl groups. (The H of this group is lost as H+.)

*The electrolytic behavior refers to aqueous solutions of these acids and bases. **Ca(OH)2 is often listed as a strong base, although it is poorly soluble.

Acids and bases have some related properties. Solutions of acids or bases, for example, can change the colors of natural pigments (Figure 3.12c). For example, acids change the color of litmus, a dye derived from certain lichens, from blue to red. Adding a base reverses the effect, making the litmus blue again. Thus, acids and bases seem to be opposites. A base can neutralize the effect of an acid, and an acid can neutralize the effect of a base. Table 3.1 lists some common acids and bases.

Naming Common Acids The molecular compound HCl is called hydrogen chloride when it is in the pure, gaseous state. However, an aqueous solution of HCl is acidic, and this solution is given the name hydrochloric acid. The same pattern, adding hydro– at the beginning and an –ic ending, applies to other acids where the anion has an –ide ending. For example, HF(aq) is hydrofluoric acid and H2S(aq) is hydrosulfuric acid. Refer to Table 3.1 and notice that other common acids, such as sulfuric acid (H2SO4) and nitric acid (HNO3), also have names ending in –ic. When anions have names ending in –ate (such as nitrate, sulfate, chlorate, perchlorate, and acetate), the acid associated with that anion has a name ending in –ic. Examples are nitric, sulfuric, chloric, perchloric, and acetic acids. You learned in Chapter 2 that there are series of anions based on chlorine, sulfur, and nitrogen. Among them are the hypochlorite (ClO−) and chlorite (ClO2–) ions as well as the sulfite (SO32−) and nitrite (NO2−) ions. Acids based on ions ending in –ite have names ending in –ous. These acids are named hypochlorous, chlorous, sulfurous, and nitrous acids. These naming conventions are illustrated in Table 3.2.

3.6  Acids and Bases

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157

Table 3.2

Names of Common Ions and Their Corresponding Acids

Common Ions

Corresponding Acids

Cl−, chloride ion

HCl, hydrochloric acid

ClO−, hypochlorite ion

HClO, hypochlorous acid

−

HClO2, chlorous acid

−

HClO3, chloric acid

−

ClO4 , perchlorate ion

HClO4, perchloric acid

S , sulfide ion

H2S, hydrosulfuric acid

SO3 , sulfite ion

H2SO3, sulfurous acid

SO42−, sulfate ion

H2SO4, sulfuric acid

NO2−, nitrite ion

HNO2, nitrous acid

NO3−, nitrate ion

HNO3, nitric acid

ClO2 , chlorite ion Names of Common Ions. Review

pages 92 and 93 for the system for naming common polyatomic ions.

ClO3 , chlorate ion 2−

2−

Acids and Bases: The Arrhenius Definition Over the years, chemists have examined the properties, chemical structures, and reactions of acids and bases and have proposed different definitions of the terms acid and base. The two most commonly used definitions are the one proposed by Svante Arrhenius (1859–1927) and another proposed by Johannes N. Brønsted (1879–1947) and Thomas M. Lowry (1874–1936). In the late 1800s, the Swedish chemist Svante Arrhenius proposed that acids and bases dissolve in water and ultimately form ions. This theory predated any knowledge of the composition and structure of atoms and was not well accepted initially. With a knowledge of atomic structure, however, the presence of ions in solution is almost taken for granted. The Arrhenius definition for acids and bases focuses on formation of H+ and − OH ions in aqueous solutions. •

An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions, H+, in solution. HCl(g) n H+(aq) + Cl−(aq)

•

A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions, OH−, in the solution. NaOH(s) n Na+(aq) + OH−(aq)

•

The reaction of an acid and a base produces a salt and water. Because the characteristic properties of an acid are lost when a base is added, and vice versa, acid–base reactions were logically described as resulting from the combination of H+ and OH− to form water. HCl(aq) + NaOH(aq) n NaCl(aq) + H2O(ℓ)

Arrhenius further proposed that acid strength was related to the extent to which the acid ionized. Some acids such as hydrochloric acid (HCl) and nitric acid (HNO3) ionize almost completely in water; they are strong electrolytes, and so are called strong acids. Other acids such as acetic acid and hydrofluoric acid are incompletely ionized; they are weak electrolytes and are weak acids. Weak acids exist in solution primarily as molecules, and only a small fraction of these molecules ionize to produce H+(aq) ions along with the appropriate anion. Water-soluble compounds that contain hydroxide ions, such as sodium hydroxide (NaOH) and potassium hydroxide (KOH), are strong electrolytes and strong bases.

158

Chapter 3 / Chemical Reactions

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Aqueous NH3 produces a very small number of NH4+ and OH− ions per mole of ammonia molecules

+

© Charles D. Winters/Cengage

−

OH− ions NH3 molecules NH4+ ions

Figure 3.13  Ammonia, a weak electrolyte.  The

name on the bottle, ammonium hydroxide, is misleading. The solution consists almost entirely of NH3 molecules dissolved in water. It is better referred to as aqueous ammonia.

Aqueous ammonia, NH3(aq), is a weak electrolyte. Even though OH− ions are not part of its formula, it does produce ammonium ions and hydroxide ions from its reaction with water and so is a base (Figure 3.13). The fact that this is a weak electrolyte indicates that this reaction with water to form ions is reactant-favored at equilibrium. Most of the ammonia remains in solution in molecular form. NH3(aq) + H2O(ℓ) uv NH4+(aq) + OH−(aq)

Although the Arrhenius theory is still used to some extent and is interesting in a historical context, modern concepts of acid–base chemistry such as the BrønstedLowry theory have gained preference among chemists.

Acids and Bases: The Brønsted-Lowry Definition In 1923, Johannes Brønsted (1879–1947) in Copenhagen, Denmark, and Thomas Lowry (1874–1936) in Cambridge, England, independently suggested a new concept of acid and base behavior. They viewed acids and bases in terms of the transfer of a proton (H+) from one species to another, and they described all acid–base reactions in terms of equilibria. The Brønsted-Lowry theory expanded the scope of the definition of acids and bases and helped chemists make predictions of product- or reactant-favorability based on acid and base strength. The main concepts of the Brønsted-Lowry theory are the following: +

•

An acid is a proton (H ) donor.

•

A base is a proton (H+) acceptor. This definition includes the OH− ion but it also broadens the number and type of bases to include anions derived from acids as well as neutral compounds such as water.

•

An acid–base reaction involves the transfer of a proton from an acid to a base to form a new acid and a new base.

According to Brønsted-Lowry theory, the behavior of acids such as HCl or CH3CO2H in water is better considered as an acid–base reaction. Both species (both Brønsted acids) donate a proton to water (a Brønsted base) forming H3O+(aq), the hydronium ion.

Protons  The major isotope of hydrogen has no neutrons and consists of one proton and one electron. To form H+, the electron is lost, leaving only the proton. Because of this, H+ ions are often referred to as protons.

H3O+ versus H+  The formula for

the hydronium ion, H3O+ is a fairly accurate description and is often used to represent the hydrogen ion in solution. However, there are instances in this textbook when, for simplicity, the hydrogen ion is represented as H+(aq).   Experiments show that other forms of the ion also exist in water, one example being [H3O(H2O)3]+.

3.6  Acids and Bases

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159

Hydrochloric acid, HCl(aq), is a strong electrolyte that ionizes almost completely in aqueous solution. It is classified as a strong acid. Hydrochloric acid, a strong acid, nearly 100% ionized. Equilibrium strongly f­avors products. +

HCl(aq)

hydrochloric acid strong electrolyte ≈100% ionized

H2O(𝓵)

H3O+(aq)

water

hydronium ion

Cl–(aq)

+

chloride ion

In contrast, CH3CO2H, a weak electrolyte, ionizes only to a small extent. It is classified as a weak acid. Acetic acid, a weak acid,  100% ionized. Equilibrium favors reactants. CH3CO2H(aq)

+

acetic acid

H2O(𝓵)

H3O+(aq)

water

hydronium ion

CH3CO2−(aq)

+

acetate ion

Sulfuric acid, a diprotic acid (an acid capable of transferring two H+ ions), reacts with water in two steps. The first step strongly favors products, whereas the second step is reactant-favored. Strong acid: H2SO4(aq) + H2O(𝓵) sulfuric acid ≈100% ionized

Weak acid:

HSO4–(aq) + H2O(𝓵) hydrogen sulfate ion 750 °C). (a) Write a balanced chemical equation for the syn­ thesis of BP. (Hint: Hydrogen is a reducing agent.) (b) Boron phosphide crystallizes in a zinc-blende structure, formed from boron atoms in a facecentered cubic lattice and phosphorus atoms in tetrahedral holes. How many tetrahedral holes are filled with P atoms in each unit cell? (c) The length of a unit cell of BP is 478 pm. What is the density of the solid in g/cm3? (d) Calculate the closest distance between a B and a P atom in the unit cell. (Assume the B atoms do not touch along the cell edge. The B atoms in the faces touch the B atoms at the corners of the unit cell.)

Chapter 12 / The Solid State

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67. There is an old expression “he is so poor he does not have two nickels to rub together.” Nickels in this case are U.S. coins, which are only 25% nickel, the rest being copper. Nickel is important in modern economies, largely because it is used in many important alloys with other metals. (See nitinol in Study Question 12.43.) This question explores nickel aluminide, which is used in high temperature applications. Nickel aluminide, NixAly, has a simple solid-state structure. Its unit cell consists of Al atoms at the corners of a cube with Ni atoms in the cube faces. See Figure 12.23. (a) What are the values of x and y in NixAly? (b) The density of the alloy is 7.71 g/cm3. What is the length of a side of the unit cell? 68. Spinels are described in Study Question 53. Consider two normal spinels, CoAl2O4 and SnCo2O4. What metal ions are involved in each? What are their electron configurations? Are the metal ions paramagnetic, and if so, how many unpaired electrons are involved? 69. Outline a procedure to calculate the percent of space occupied by the atoms in an fcc arrangement. 70. The sample of nephrite jade shown on page 603 has the formula Ca2(Mg, Fe)5(Si4O11)2(OH)2. (Iron in this formula is in the +2 oxidation state.) (a) What is the charge on the (Si4O11)n2 ion in this compound? (b) What is the oxidation state of Si in this compound? (c) What is the percent of iron in a sample of jade that has the formula Ca2(Mg0.35Fe0.65)5(Si4O11)2 (OH)2? (d) The iron ions in the formula for nephrite have the same degree of paramagnetism as seen for Fe21(g). How many unpaired electrons per iron ion does this represent?



71. Phase diagrams for materials that have allotropes can be more complicated than those shown in the chapter. Use the phase diagram for carbon given here to answer the following questions. 1000 Diamond

100 10 P (GPa)

66. ▲ Why is it not possible for a salt with the formula M3X (Na3PO4, for example) to have a facecentered cubic lattice of X anions with M cations in octahedral holes?

Liquid

1 0.1 Graphite

0.01 0.001

0

1

2

3

Vapor

4

5

6

7

8

9

10

T/1000 (K)

(a) How many triple points are present and what phases are in equilibrium for each? (b) Is there a single point where all four phases are in equilibrium? (c) Which is more stable at high pressures, diamond or graphite? (d) Which is the stable phase of carbon at room temperature and 1 atm? 72. Prepare a graph of lattice enthalpy for lithium, sodium, and potassium halides (Table 12.1) versus the sum of the ionic radii for the component ions (Figure 7.12). Evaluate the results and comment on the relationship between these quantities. 73. Two identical swimming pools are filled with uniform spheres of ice packed as closely as possible. The spheres in the first pool are the size of grains of sand; those in the second pool are the size of oranges. The ice in both pools melts. In which pool, if either, will the water level be higher? (Ignore any differences in filling space at the planes next to the walls and bottom.)

Study Questions

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13 Solutions and Their Behavior

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C hapt e r O ut li n e 13.1 Units of Concentration 13.2 The Solution Process 13.3 Factors Affecting Solubility: Pressure and Temperature 13.4 Colligative Properties 13.5 Colloids

In the last few chapters, you have learned about the states of matter: gases (­Chapter 10), liquids (Chapter 11), and solids (Chapter 12). In this chapter, you will learn more about solutions, homogeneous mixtures of two or more substances in a single phase. By convention, the component present in the largest amount is the solvent and the remaining components are solutes (Section 3.4). This chapter explores factors that affect the solubility of solutes in liquid solvents. You will learn that large quantities of sodium chloride dissolve in water, but a mixture of oil and water, perhaps in a salad dressing, readily separates into two phases. For solutions of gases in water, such as dissolved oxygen in a lake, you will learn why the solubility of a gas depends on temperature. Adding a solute to a pure liquid changes the properties of the liquid. In fact, this is why some solutions are made. For instance, a mixture of antifreeze and water in a car’s radiator prevents the coolant from boiling in the summer and freezing in the winter. Adding or removing solutes from your blood affects the rate at which water moves in and out of your cells. For this reason, intravenous fluids must have exact concentrations of solutes such as NaCl and glucose. The freezing point, boiling point, osmotic pressure, and solvent vapor pressure for solutions are examples of colligative properties, which depend on the number of solute particles per solvent molecule, not the identity of the solute.

13.1 Units of Concentration Goals for Section 13.1 • Calculate and use the concentration units: molality, mole fraction, weight percent, and parts per million (ppm).

• Recognize the difference between molarity and molality. To analyze the colligative properties of solutions, you need ways to define solute concentrations that reflect the number of molecules or ions of solute per m ­ olecule of solvent. The concentration unit called molarity, amount of solute (mol)/ volume of solution (L), is used extensively in stoichiometry calculations (see Section 4.5). ◀ Scuba diving: The solubility of gases in your bodily fluids is important to consider when diving.

627

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Unfortunately, this unit is not useful when dealing with most colligative properties because, when preparing a solution of given molarity, the amount of solvent is not known. Four concentration units are described here that do reflect the number of molecules or ions of solute per molecule of solvent: molality, mole fraction, weight percent, and parts per million (ppm). The molality, m, of a solution is defined as the amount of solute (mol) per kilogram of solvent.

Mole Fraction and Gases Mole

fraction was introduced when mixtures of gases were discussed (Section 10.5).

Concentration (c, mol/kg)  molality of solute 

amount of solute (mol) mass of solvent (kg)

(13.1)

To prepare a 1.00 molal solution, for example, you would add 1.00 mol of solute to 1.00 kg of water. The difference between molality (m) and molarity (M) is illustrated in ­Figure 13.1. The flask on the right contains a 0.100 M aqueous solution of potassium chromate. It was made by adding enough water to 0.100  mol of K2CrO4 to make 1.000 L of solution. If 1.00 L (1.00  kg) of water is added to 0.100  mol of K2CrO4 to make a 0.100  molal solution, the volume of solution is greater than 1.000 L (­Figure 13.1, left). The mole fraction, X, of a solution component is defined as the amount of that component (nA) divided by the total amount of all of the components of the mixture (nA + nB + nC + . . .). Mathematically it is represented as Mole fraction of A (X A ) 



nA nA  nB  nC  …

(13.2)

Consider a solution that contains 1.00  mol (46.1  g) of ethanol, C2H5OH, in 9.00 mol (162 g) of water. The mole fraction of alcohol is 0.100, and that of water is 0.900. X ethanol 

1.00 mol ethanol  0.100 1.00 mol ethanol  9.00 mol water

X water 

9.00 mol water  0.900 1.00 mol ethanol  9.00 mol water

Figure 13.1 Preparing 0.100 molal and 0.100 molar solutions.

Note the difference in water levels in the two flasks

© Charles D. Winters/Cengage

Notice that the sum of the mole fractions of the components in the solution equals 1.000, a relationship that must be true, based on how mole fraction is defined. Weight percent is the mass of one component divided by the total mass of the mixture, multiplied by 100%:

Left: 1.00 kg of water was added to 0.100 mol of K2CrO4.

Right: 0.100 mol (19.4 g) of K2CrO4 was mixed with enough water to make 1.000 L of solution.

0.100 molal solution. Water added to flask = 1.00 kg Volume of solution > 1.00 L

0.100 molar solution. Water added to flask < 1.00 kg Volume of solution = 1.00 L 0.100 mol of K2CrO4.

628

Chapter 13 / Solutions and Their Behavior

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Chemistry in Your Career

Uraina Gray-Scully

Uraina Gray-Scully



Weight % A 

Chemists have a role in every sector of the workforce, including the military. Uraina Gray-Scully works for the U.S. Department of Defense, where she applies her chemistry education (B.S. in chemi­ cal engineering and M.S. in mathematical s­ ciences) to ensure that military equipment meets quality standards. “One major focus is Military Color ­Science, where we ensure that all u ­ niforms—from berets, trousers, shirts, and coats, to flags, banners, and insignia—meet specific shade and color ­requirements as set by the Services.” In fact, color is so important to the job that Gray-Scully had to

mass of A  100% mass of A  mass of B  mass of C  …

pass a vision test to demonstrate her superior color acuity. Gray-Scully notes that molecular orbital theory and electrons moving through energy levels explain how dyes can be specially formulated to create camouflaged clothing and other equipment. Gray-Scully uses a spectrophotometer to measure the amount of energy returned in a sample. This allows her to determine where the sample falls in the visible or near-infrared electromagnetic spectrum, a property that corresponds to the perception of color.

(13.3) ppb and ppt  The units ppb (parts

The alcohol–water mixture mentioned previously has 46.1 g of ethanol and 162 g of water, so the total mass of solution is 208 g, and the weight % of alcohol is 46.1 g ethanol  100%  22.2% 46.1 g ethanol  162 g water

Weight percent is a common unit in consumer products (Figure 13.2). Vinegar, for example, is an aqueous solution containing approximately 5% acetic acid and 95% water. The label on a common disinfecting bleach lists its active ingredient as 7.5% sodium hypochlorite (NaClO) and 92.5% inert ingredients. The unit ppm (parts per million) also refers to relative quantities of solute and solvent by mass; 1.0 ppm represents a solution containing 1.0 g of a substance in a sample with a total mass of 1.0 million g (equivalent to 1 mg in 1000 g). This unit is typically used to identify concentrations of solutes in very dilute solutions. It is useful in chemistry (especially in environmental chemistry) and in disciplines such as biology, geology, and oceanography.

Ex am p le 13.1

Calculating Mole Fraction, Molality, and Weight Percent Problem  Suppose you add 1.2 kg of ethylene glycol, HOCH2CH2OH, as an antifreeze to 4.0 kg of water in the radiator of your car. What are the mole fraction, molality, and weight percent of the ethylene glycol?

Editorial Image, LLC/Alamy Stock Photo

Weight % ethanol 

per billion) and ppt (parts per trillion) are related to ppm. Sophisticated analytic techniques are now available that can detect trace impurities in solution at the ppb and ppt range.

Figure 13.2  Weight percent.  ​ The composition of many common products is given in terms of weight percent.

What Do You Know?  You know the identity and masses of solute and solvent. Strategy The amount of solute and solvent can be calculated using the mass and molar mass of each material. The masses and amounts of solute and solvent can then be combined to calculate the concentration in each of the desired concentration units using Equations 13.1–13.3.

13.1  Units of Concentration

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629

Solution The 1.2  kg of ethylene glycol (molar mass = 62.1  g/mol) is equivalent to 19.3 mol, and 4.0 kg of water represents 222 mol. Mole fraction: X glycol 

19.3 mol ethylene glycol   0.080  19.3 mol ethylene glycol  222 mol water

Molality: c glycol 

19.3 mol ethylene glycol  4.88 mol/kg   4.8 m  4.0 kg water

Weight percent:

John C. Kotz

Weight % 

Commercial antifreeze. This solution contains ethylene glycol, HOCH2CH2OH, an organic alcohol that is readily soluble in water.

1.2  103 g ethylene glycol  100%    23%  1.2  10 g ethylene glycol  4.0  103 g water 3

Think about Your Answer  Although the numerical values are very different, the information contained in these values is similar; each relates the relative numbers of solvent and solute particles.

Check Your Understanding (a) If you dissolve 10.0 g (about one heaping teaspoonful) of sugar (sucrose, C12H22O11) in a cup of water (250. g), what are the mole fraction, molality, and weight percent of sugar? (b) Seawater has a sodium ion concentration of 1.08 × 104 ppm. If the sodium is present in the form of dissolved sodium chloride, what mass of NaCl is in each liter of seawater? Seawater is denser than pure water because of dissolved salts. Its density is 1.05 g/mL.

13.2 The Solution Process Goals for Section 13.2 • Understand the process of dissolving a solute in a solvent and recognize the

terminology used (saturated, unsaturated, supersaturated, solubility, miscible).

• Understand the thermodynamics associated with the solution process and calculate the enthalpy of solution from thermodynamic data.

Unsaturated  The term unsaturated is used when referring to solutions with solute concentrations that are less than that of a saturated solution.

If solid CuCl2 is added to a beaker of water, the salt will begin to dissolve (­Figure 13.3). The amount of solid diminishes, and the concentrations of Cu2+(aq) and Cl−(aq) in the solution increase. If you continue to add CuCl2, however, you will eventually reach a point where no additional CuCl2 seems to dissolve. The concentrations of Cu2+(aq) and Cl−(aq) will not increase further, and any additional solid CuCl2 added after this point will remain as a solid at the bottom of the beaker. Such a solution is said to be saturated. Although no change is observed on the macroscopic level when a solution is saturated, it is a different matter on the particulate level. The process of dissolving continues, with Cu2+ and Cl− ions leaving the solid state and entering solution. However, at the same time, solid CuCl2(s) is being formed from Cu2+(aq) and Cl−‌(aq). The rates at which CuCl2 is dissolving and reprecipitating are equal in a saturated solution, so that no net change in the concentration of ions is observed on the macroscopic level. This process is another example of a dynamic equilibrium (Section  3.3) and can  be represented by an equation with substances linked by a set of double arrows (uv): CuCl2(s) uv Cu2+(aq) + 2 Cl−(aq)

630

Chapter 13 / Solutions and Their Behavior

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2+

−

Photos: © Charles D. Winters/Cengage

2+

2+

−

2+

(a) Copper(II) chloride, the solute, is

−

(b) Interactions between water molecules and Cu2+ and Cl− ions allow the solid to dissolve. The ions are now sheathed with water molecules.

added to water, the solvent.

Figure 13.3  Making a solution of copper(II) chloride (the solute) in water (the solvent).  ​When ionic compounds dissolve in water, each ion is surrounded by water molecules. The number of water molecules depends on ion size and charge.

Supersaturated Solutions



evolution of thermal energy. In fact, super­ saturated solutions are used in some heat packs to apply heat to injured muscles.

When crystallization of sodium acetate (NaCH3CO2) from a supersaturated solu­ tion in a heat pack is initiated, the tem­ perature of the heat pack rises to about 50 °C and crystals of solid sodium acetate are detected inside the bag.

Supersaturated solutions.  When a supersaturated solution is disturbed, the excess dissolved salt (here sodium acetate, NaCH3CO2) rapidly crystallizes.

John C. Kotz

Although it may seem a contradiction, it is possible for a solution to hold more dissolved solute than the amount in a saturated solution. Such s­olutions are supersaturated. Supersaturated solu­ tions are unstable, and the excess solid eventually crystallizes from the solu­ tion until the equilibrium concentra­ tion of the solute is reached. The solubility of substances often decreases if the temperature is lowered. Supersaturated solutions are usually made by preparing a saturated solution at an el­ evated temperature and then carefully cooling it. If the rate of crystallization is slow, the solid may not precipitate when the solubility is exceeded. The result is a solution that has more solute than the amount defined by equilibrium conditions. When disturbed in some manner, a supersaturated solution moves toward equilibrium by precipitating solute. This change can occur rapidly, often with the

© Charles D. Winters/Cengage

A Closer Look

The solubility of a solute is the concentration of solute in a saturated solution in equilibrium with undissolved solute. The solubility of CuCl2, for example, is 70.6 g in 100 mL of water at 0 °C. If you add 100.0 g of CuCl2 to 100 mL of water at 0 °C, you can expect 70.6 g to dissolve, and 29.4 g of solid to remain.

Heat of crystallization.  A heat pack relies on the heat evolved by the crystallization of sodium acetate. This photo shows the crystals beginning to form at the bottom left.

13.2  The Solution Process

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631

After mixing Photos: © Charles D. Winters/Cengage

Before mixing

Less dense layer of nonpolar octane, C8H18.

Solution of CuSO4 moves to the top.

Solution of CuSO4 in water.

Homogeneous mixture of nonpolar CCl4 and C8H18 moves to bottom.

More dense layer of nonpolar carbon tetrachloride, CCl4.

Figure 13.4 ​ Miscibility. 

(a) Before mixing. The colorless, denser bottom layer is nonpolar carbon tetrachloride, CCl4. The blue middle layer is a solution of CuSO4 in water, and the colorless, less dense top layer is nonpolar octane, C8H18. This mixture was prepared by carefully layering one liquid on top of another, without mixing.

(b) After mixing. After stirring the mixture, the two nonpolar liquids form a homogeneous mixture. The mixture of CCl4 and C8H18 has a greater density than water.

Liquids Dissolving in Liquids If two liquids can be mixed in any proportion to form a homogeneous mixture they are said to be miscible. In contrast, immiscible liquids do not fully mix in all proportions; they exist in contact with each other as separate layers (Figure  13.4). ­Ethanol (C2H5OH) and water, both polar compounds, are miscible, as are the nonpolar liquids octane (C8H18) and carbon tetrachloride (CCl4). Many similar observations like these have led to a rule of thumb:  Like dissolves like. Nonpolar liquids frequently are miscible, as are polar liquids. On the other hand, nonpolar solvents such as C8H18 and CCl4 are not miscible with water. As a general rule, polar solvents are likely not to be miscible with nonpolar solvents.

Solids Dissolving in Liquids The like-dissolves-like guideline is also often followed when molecular solids dissolve in liquids. Nonpolar solids such as naphthalene, C10H8, dissolve readily in nonpolar solvents such as benzene, C6H6, and hexane, C6H14. Iodine, I2, a nonpolar inorganic solid, dissolves in water to some extent, but, given a choice, it dissolves to a larger extent in a nonpolar liquid such as CCl4 (Figure 13.5). Polar solids are often

Nonpolar I2 Polar H2O

Polar H2O

© Charles D. Winters/ Cengage

Nonpolar CCl4

© Charles D. Winters/ Cengage

Shake the test tube

Nonpolar CCl4 and I2

Figure 13.5  Solubility of nonpolar iodine in polar water and nonpolar carbon tetrachloride. ​ When a solution of nonpolar I2 in water (the brown layer on top in the left test tube) is shaken with nonpolar CCl4 (the colorless bottom layer in the left test tube), the I2 transfers preferentially to the nonpolar solvent. Evidence for this is the purple color of the bottom CCl4 layer in the test tube on the right.

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Chapter 13 / Solutions and Their Behavior

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soluble in polar liquids such as water. Sucrose (sugar), a polar molecular solid with multiple OOH groups, is readily soluble in water, a property important for its use in sweetening beverages. As was seen with nonpolar liquids, most nonpolar solids are likely to be insoluble or have low solubilities in polar solvents such as water. The reverse is also true; polar solutes, sugar for example, are likely to have low solubility in nonpolar solvents such as gasoline. Ionic compounds can be considered extreme examples of polar compounds, and they do not dissolve in nonpolar solvents. Sodium chloride, for example, does not dissolve in liquids such as hexane or CCl4. However, the solubility of ionic compounds in the polar solvent water is not as easily predicted. Recall the solubility guidelines (Figure 3.10), which predict that some ionic compounds are soluble in water while others are not. Network solids, including graphite, diamond, and quartz sand (SiO2), do not dissolve in water. (Where would all the beaches be if sand dissolved in water?) The covalent chemical bonding in network solids is simply too strong to be broken; the lattice remains intact when in contact with water.

OH group

Like dissolves like.  Sucrose has eight OOH groups on each molecule, groups that allow it to form hydrogen bonds with water molecules. As a result, sucrose dissolves readily in water.

Enthalpy of Solution What is the basis for the like-dissolves-like guideline? There are two factors that determine whether any process will occur. One factor is the enthalpy change for the process, here the enthalpy of solution. The second factor is the entropy change for the process. Entropy is a thermodynamic function that is a measure of the dispersal of  energy in a system. In the case of solution formation, the entropy change ­compares the dispersal of energy in the particles in the mixture relative to the pure ­materials (Figure 13.6). The greater the energy dispersal in forming a solution, the greater is the entropy change, and the more favorable the process. In solution ­formation, the entropy effect is often dominant over the enthalpy effect. Entropy will be discussed more fully in Chapter 18. In the current chapter, the enthalpy of solution will be considered. While less important in determining solubility than entropy, it is nonetheless an important factor, and it can be interpreted readily based on attractive forces between particles. It is possible to assess the enthalpy change that occurs when a solution forms by evaluating the intermolecular forces between particles. In pure water and pure methanol (CH3OH), the major force between molecules is hydrogen bonding involving OOH groups. When the two liquids are mixed, hydrogen bonding between methanol and water molecules also occurs. Because solute-solvent and solute-solute forces of attraction are of similar magnitude, you can predict that the enthalpy change for the solution process will be small.

+ H2O

CH3OH

Mixture

Separate liquids

Figure 13.6  Driving the solution process—entropy.  ​When two similar liquids—here, water and methanol—are mixed, the molecules intermingle, and the energy of the system is more dispersed than in the two, separate pure liquids. A measure of this energy dispersal is entropy, a thermodynamic function described in more detail in Chapter 18.

13.2  The Solution Process

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633

Entropy and the Solution Process ​ It is generally accepted that entropy is a more important contributor to the solution process. (See Chapter 18 and T. P. Silverstein, The real reason why oil and water don’t mix. Journal of Chemical Education, 1998, 75, 116–118.)

Similarly, the enthalpy change when dissolving a nonpolar solute in a nonpolar solvent is expected to be near zero. Molecules of pure octane or pure CCl4, both of which are nonpolar, are held together in the liquid phase by London dispersion forces (Section 11.4). When these nonpolar liquids are mixed, the energy associated with the forces of attraction between solute and solvent is similar to the forces of attraction between octane and CCl4 molecules in the pure liquids. As with solutions of polar solutes in polar solvents, little or no energy change occurs; the solution process is expected to be nearly energy neutral. For ionic compounds dissolving in water, a favorable enthalpy change (a negative ΔsolnH) generally leads to a compound being soluble. For example, sodium hydroxide is very soluble, dissolving in water with significant heat evolution (Figure 13.7a). An unfavorable enthalpy factor, however, does not guarantee that an ionic compound will not dissolve. Sodium chloride is quite soluble in water despite the fact that the solution process is slightly endothermic. And, ammonium nitrate is quite soluble in water in a process that is also endothermic, evidenced by the fact that the solution becomes noticeably colder (Figure 13.7b). In these cases, entropy must be the driving force in the solution process. To further understand the energetics of the solution process, consider the process of dissolving KF in water. The energy level diagram in Figure 13.8 will help you follow the changes. Solid potassium fluoride has an ionic crystal lattice with alternating K+ and F− ions held in place by attractive forces due to their opposite charges. In water, these ions are separated from each other and hydrated; that is, they are surrounded by water molecules. Ion–dipole forces of attraction bind water molecules strongly to each ion. The energy change occurring on going from the reactant, KF(s), to the products, K+(aq) and F−(aq), is the sum of the energies of two individual steps:

Photos: © Charles D. Winters/Cengage

1. Energy must be supplied to separate the ions in the lattice against their attractive forces. This is the reverse of the process defining the lattice enthalpy of an ionic compound (Section 12.3), and its value will be equal to −∆latticeH. Separating the ions from one another is highly endothermic because the attractive forces between ions are strong. 2. Energy is evolved when the individual ions dissolve in water, where each ion becomes surrounded by water molecules. Strong forces of attraction (ion– dipole forces) are involved (Section 11.2). This process, referred to as hydration when water is the solvent, is strongly exothermic.

(a) Dissolving NaOH in water is a strongly exothermic process.

Figure 13.7 ​Dissolving ionic solids and enthalpy of solution. 

634

(b) A cold pack contains solid ammonium nitrate, NH4NO3, and a package of water. When the water and NH4NO3 are mixed and the salt dissolves, the temperature of the system drops, owing to the endothermic enthalpy of solution of ammonium nitrate (∆solnH° = +25.7 kJ/mol).

Chapter 13 / Solutions and Their Behavior

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K+(g) + F−(g)

ENERGY

−∆latticeH = +821 kJ/mol ∆hydrationH = −837 kJ/mol

KF(s) ∆solnH = −16 kJ/mol

K+(aq) + F−(aq)

Figure 13.8  Model for energy changes on dissolving KF.  ​An estimate of the magnitude of

the energy change on dissolving an ionic compound in water is obtained by imagining it as occurring in two steps at the particulate level. Here, KF is first separated into cations and anions in the gas phase with an expenditure of 821 kJ per mol of KF. These ions are then hydrated, with ∆hydrationH estimated to be −837 kJ. The net energy change is −16 kJ, a slightly exothermic enthalpy of solution.

We can represent the process of dissolving KF in terms of these chemical equations:

Step 1 KF(s) 8n K+(g) + F−(g)

−∆latticeH

Step 2 K+(g) + F−(g) 8n K+(aq) + F−(aq)

∆hydrationH

The enthalpy of the overall reaction, called the enthalpy of solution (∆solnH), is the sum of the two enthalpy changes. Overall:  KF(s) 8n K+(aq) + F−(aq)  ∆solnH = −∆latticeH + ∆hydrationH

You can use this equation to calculate ∆hydrationH. From the lattice enthalpy for KF (−821 kJ/mol, calculated using a Born-Haber cycle; Section 12.3), and the value of ∆solnH (−16.4  kJ/mol, from calorimetry) you can determine ∆hydrationH to be −837 kJ/mol. As a general rule, to be water-soluble an ionic compound will have an ­enthalpy of solution that is exothermic or only slightly endothermic (Figure 13.9). In the latter instance, it is assumed that the enthalpy-disfavored solution process is overcome by a favorable entropy of solution. If the enthalpy of solution is very endothermic (because of a low hydration energy), then the compound is unlikely to be soluble. Similarly, you can conclude that dissolving ionic compounds in nonpolar solvents, which do not solvate ions strongly, is ­ ­energetically unfavorable. An ionic compound, such as copper(II) sulfate, is not very soluble in nonpolar solvents such as carbon tetrachloride and octane (Figure 13.4). It is also important to recognize that the enthalpy of solution is the difference between two very large numbers. Small variations in either lattice enthalpy or hydration enthalpy can determine whether the solution process is endo- or ­ exothermic.



13.2  The Solution Process

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635

Dipole–dipole attraction and hydrogen bonding

δ+

δ− δ+

δ−

Na+

Ion–dipole attraction; related to enthalpy of hydration, ΔhydrationH

Cl−

Ion–ion attraction; defined by the lattice enthalpy, ΔlatticeH

Figure 13.9  Dissolving an ionic solid in water.  ​This process is a balance of forces. There are intermolecular forces between water molecules, and ion–ion forces are at work in the ionic crystal lattice. To dissolve, the ion–dipole forces between water and the ions (as measured by ∆hydrationH ) must overcome the ion–ion forces (as measured by ∆latticeH ) and the intermolecular forces in water.

Enthalpy of Solution: Thermodynamic Data Tables of thermodynamic values often include values for the enthalpies of ­formation of aqueous solutions of salts. For example, ∆fH° for NaCl(aq) (−407.3  kJ/mol, ­Table 13.1 and Appendix L) refers to the formation of a 1 m solution of NaCl from the elements. It may be considered to involve the enthalpy changes for two steps: (i) the formation of NaCl(s) from the elements Na(s) and Cl2(g) in their standard states (∆fH°), and (ii) the formation of a 1 m solution by dissolving solid NaCl in water (∆solnH°): Formation of NaCl(s): Na(s) + 1⁄2 Cl2(g) 8n NaCl(s)

∆fH° = −411.1 kJ/mol

Dissolving NaCl(s):

NaCl(s) 8n NaCl(aq, 1 m)

∆solnH° = +3.8 kJ/mol

Net process:

Na(s) + 1⁄2 Cl2(g) 8n NaCl(aq, 1 m) ∆fH° = −407.3 kJ/mol

By combining the enthalpies of formation of a solid and of its aqueous solution, you can calculate the enthalpy of solution, as shown in Example 13.2.

636

Table 13.1

Data for Calculating Enthalpy of Solution

Compound

𝚫f  H°(s) (kJ/mol)

𝚫f  H°(aq, 1 m) (kJ/mol)

LiF

−616.9

−611.1

NaF

−573.6

−572.8

KF

−568.6

−585.0

RbF

−557.7

−583.8

LiCl

−408.7

−445.6

NaCl

−411.1

−407.3

KCl

−436.7

−419.5

RbCl

−435.4

−418.3

NaOH

−425.9

−469.2

NH4NO3

−365.6

−339.9

Chapter 13 / Solutions and Their Behavior

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Ex am p le 13.2

Calculating an Enthalpy of Solution Problem  Determine the enthalpy of solution for NH4NO3, the compound used in cold packs (Figure 13.7b).

What Do You Know?  The solution process for NH4NO3 is represented by the equation NH4NO3(s) 8n NH4NO3(aq) The enthalpies of formation of NH4NO3 in the solid state (−365.6 kJ/mol) and in solution (−339.9 kJ/mol) are given in Table 13.1.

Strategy  Equation 5.7 (Section 5.7) states that the enthalpy change for a process is the difference between the enthalpy of formation, ∆f H, of the final state (here the compound in solution) and the initial state (the solid). Solution The enthalpy change for this process is calculated using Equation 5.7 as follows: solnH° 

[n f H°(product)]  [n f H°(reactant)]

  f H° [ NH4NO3(aq) ]   f H° [ NH4NO3(s) ] (where n  1 for both)  339.9 kJ  (  365.6 kJ)   +25.7 kJ 

Think about Your Answer  The process is endothermic, as indicated by the fact that ∆solnH° has a positive value and as verified by the experiment in Figure 13.7b.

Check Your Understanding  Use the data in Table 13.1 to calculate the enthalpy of solution for NaOH.

13.3 Factors Affecting Solubility: Pressure and Temperature Goals for Section 13.3

Table 13.2

• Describe the effects of pressure and temperature on the solubility of a solute. • Use Henry’s law to calculate the solubility of a gas in a solvent. • Apply Le Chatelier’s principle to predict the change in solubility of gases with

Henry’s Law Constants for Gases in Water (25 °C)*

temperature changes.

Gas

Pressure and temperature are two external factors that influence solubility. Both affect the solubility of gases in liquids, whereas only temperature is a factor in the solubility of solids in liquids.

Dissolving Gases in Liquids: Henry’s Law Henry’s law states that the solubility of a gas in a liquid is directly proportional to the gas pressure. Mathematically, this can be represented as

Sg = kHPg 

(13.4)

where Sg is the gas solubility (in mol/kg), Pg is the partial pressure of the gaseous solute, and kH is Henry’s law constant (Table 13.2), a constant characteristic of the solute and solvent.

kH (mol/kg · bar)

He

3.8 × 10−4

H2

7.8 × 10−4

N2

6.0 × 10−4

O2

1.3 × 10−3

CO2

0.034

CH4

1.4 × 10−3

C2H6

1.9 × 10−3

*From https://webbook.nist.gov /chemistry/. Note: 1 bar = ​ 0.9869 atm.

13.3  Factors Affecting Solubility: Pressure and Temperature

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637

Limitations of Henry’s Law ​Henry’s law holds quantitatively only for dilute solutions and for gases that do not interact chemically with the solvent.

© Charles D. Winters/Cengage

Figure 13.10  Gas solubility and pressure.  ​Carbonated beverages are bottled under high CO2 pressure. When the bottle is opened, the pressure is released, and bubbles of CO2 form within the liquid and rise to the surface. After some time, an equilibrium between dissolved CO2 and atmospheric CO2 is reached. The beverage tastes flat when most of its dissolved CO2 is lost.

Carbonated soft drinks illustrate how Henry’s law works. These beverages are packed under pressure in a chamber filled with carbon dioxide gas, and some of this gas dissolves in the beverage. When the can or bottle is opened, the partial pressure of CO2 above the solution drops, causing the solubility of CO2 to drop. Gas bubbles out of the solution (Figure 13.10). You can better understand the effect of pressure on solubility by examining the system at the particulate level. The solubility of a gas is defined as the concentration of the dissolved gas in equilibrium with the substance in the gaseous state. At equilibrium, the rate at which solute gas molecules escape the solution and enter the gaseous state equals the rate at which gas molecules re-enter the solution. An increase in pressure results in more molecules of gas striking the surface of the liquid and entering the solution in a given time. The solution eventually reaches a new equilibrium when the concentration of gas dissolved in the solvent is high enough that the rate of gas molecules escaping the solution again equals the rate of gas molecules entering the solution.

E xamp le 13.3

Using Henry’s Law Problem  What is the concentration of O2 in a freshwater stream in equilibrium with air at 25 °C and at a pressure of 1.0 bar? Express the answer in grams of O2 per kg of water. The mole fraction of O2 in air is 0.21. What Do You Know?  You know the total air pressure, mole fraction of O2 in air, and temperature. You can look up the Henry’s law constant (Table 13.2, 1.3 × 10−3 mol/kg ∙ bar). (Note that 1 bar is approximately 1 atm; Section 10.1.)

Strategy  Step 1.  Use Equation 10.8 (page 515) to calculate the partial pressure of O2 in air. Step 2.  Use Henry’s law to calculate the molar solubility of O2. Step 3.  Convert the molar solubility to the desired units.

Solution Step 1. Use Equation 10.8 to calculate the partial pressure of O2 in air. The mole fraction of O2 in air is 0.21, and the total pressure is 1.0 bar.

P O2  X O2 Ptotal  (0.21) (1.0 bar)  0.21 bar Step 2. Use Henry’s law to calculate the molar solubility of O2. Using the O2 partial ­pressure for Pg in Henry’s law, and the value for the Henry’s law constant from Table 13.2, you obtain:  1.3  103 mol  4 Solubility of O2  kHPg    (0.21 bar)  2.73  10 mol/kg kg  bar 

638

Chapter 13 / Solutions and Their Behavior

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Step 3. Calculate the concentration in g/kg from the concentration in mol/kg and the molar mass of O2:  2.73  104 mol   32.0 g  Solubility of O2     mol    0.0087 g/kg  kg 

Think about Your Answer  The concentration of O2 is 8.7 ppm (8.7 mg/1000 g). This concentration is quite low, but it is sufficient to provide the oxygen required by aquatic life.

Check Your Understanding  What is the concentration of CO2 in water at 25 °C when the partial pressure is 0.33 bar? (Although CO2 reacts with water to give traces of H3O+ and HCO3−, the reaction occurs to such a small extent that Henry’s law is obeyed at low CO2 partial pressures.)

Temperature Effects on Solubility: Le Chatelier’s Principle The solubility of all gases in water decreases with increasing temperature. You may realize this from everyday observations such as the appearance of air bubbles as water is heated below the boiling point. As you will see, it is possible to predict this effect of temperature on the solubility of a gas from the sign of its enthalpy of solution. When gases dissolve in water, they usually do so in an exothermic process. That is, energy is released as heat, and the value of ∆solnH is negative. ∆solnH < O

Gas + liquid solvent u::::::::v saturated solution + energy

The reverse process, loss of dissolved gas molecules from a solution, requires energy as heat. At equilibrium, the rates of the two processes are the same. The effect of temperature on the solubility of a gas in a liquid can be predicted using Le Chatelier’s principle, which states that a change in any of the factors determining an equili­brium causes the system to adjust by shifting in the direction that reduces or counteracts the effect of the change. If a solution of a gas in a liquid is heated, for example, the equilibrium will shift to absorb some of the added energy. Thus, for the case of a gas dissolved in a liquid, where the enthalpy change is exothermic (with a negative ∆solnH), the reaction Exothermic process ∆solnH is negative.

Gas + liquid solvent u:::::::::v saturated solution + energy m88888888888888888888888888888888 Add energy. Equilibrium shifts left.

shifts to the left if the temperature is raised because energy is absorbed in the process that produces free gas molecules and pure solvent. This shift corresponds to less gas dissolved and a lower solubility at higher temperature—the observed result. The solubility of solids in water is also affected by temperature, but, unlike the situation involving solutions of gases, no general pattern of behavior is observed. In Figure 13.11, the solubilities of several salts are plotted versus temperature. The solubility of many salts increases with increasing temperature, but there are notable exceptions. Predictions based on whether the enthalpy of solution is positive or negative are valid most of the time, but exceptions do occur. Chemists take advantage of the variation of solubility with temperature to purify compounds. An impure sample of a compound that is more soluble at higher temperatures is dissolved by heating the solution. The solution is then cooled to decrease the solubility (Figure 13.11c). When the limit of solubility is reached at the lower temperature, crystals of the pure compound form.

13.3  Factors Affecting Solubility: Pressure and Temperature

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639

180

CsCl

NaNO3

140

RbCl

120

LiCl

100 80

NH4Cl

60

KCl

40

NaCl Li2SO4

20 20

Photos: © Charles D. Winters/Cengage

g salt/100 g H2O

160

40 60 80 100 Temperature (°C)

(a) Temperature dependence of the solubility of some ionic compounds.

(b) NH4Cl dissolved in water.

(c) NH4Cl precipitates when the solution is cooled in ice.

Figure 13.11  The temperature dependence of the solubility of some ionic compounds in water.  ​The solubility of most ionic compounds increases with increasing temperature. This is illustrated using NH4Cl (parts b and c).

13.4 Colligative Properties Goals for Section 13.4 • Using Raoult’s law, calculate the effect of dissolved solutes on solvent vapor pressure (Psolvent).

• • • •

Calculate the effect of a solute on the boiling point and freezing point of a solvent. Calculate the osmotic pressure (∏) for solutions. Use colligative properties to determine the molar mass of a solute. Use the van’t Hoff factor in colligative property calculations involving ionic solutes.

Changes in Vapor Pressure: Raoult’s Law Equilibrium Vapor Pressure ​Recall

the equilibrium vapor pressure of a liquid is defined as the pressure of the vapor when the liquid and vapor are in equilibrium (Section 11.6).

When a non-volatile solute dissolves in a liquid, the vapor pressure of the solvent is lowered. Experiments show that the vapor pressure of the solvent over the solution, Psolvent, is proportional to the mole fraction (X) of the solvent. Raoult’s law expresses this relationship:

Psolvent = Xsolvent P°solvent

(13.5)

Raoult’s law tells you that the vapor pressure of solvent over a solution (Psolvent) is lower than that of the pure solvent (P°solvent); it depends on the fraction of molecules in the solution that are solvent molecules. For example, if 95% of the molecules in a solution are solvent molecules (Xsolvent = 0.95), then the vapor pressure of the solvent (Psolvent) is 95% of P°solvent. An ideal solution is defined as one that obeys Raoult’s law precisely. However, just as no gas is truly ideal, no solution is truly ideal. Nonetheless, Raoult’s law is a good approximation of solution behavior in most instances, especially at low solute concentration.

640

Chapter 13 / Solutions and Their Behavior

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The very large crystal of potassium dihydrogen phosphate you see in the photo was grown at the Lawrence Livermore Laboratory in California. It weighs 318  kg and measures 66 ×  53 × 58  cm. The crystal was grown by suspending a thumbnail-sized seed crystal in a 6-foot tank of satu­ rated KH2PO4. The temperature of the solution was gradually reduced from 65 °C over a period of about 50 days. The large crystals are sliced into thin plates, which are used to convert light from a laser from infrared to ultraviolet. You can grow crystals from some ­chemicals found in the supermarket. Alum

(potassium aluminum sulfate) is a particu­ larly good choice. Heat about 100 mL of water and then add the alum until no more will dissolve. Allow the mixture to cool un­ disturbed overnight. The next day small seed crystals will have formed. Remove one small crystal that appears well formed, and then pour the saturated solution into another clean jar. Carefully tie a thread or a piece of nylon fishing line around the seed crystal and suspend it in the satu­ rated solution. (Don’t let it touch the sides or the bottom.) Cover the jar lightly and put it aside in a place where it can sit ­undisturbed for a few days or weeks, and a well-formed crystal will form.

Courtesy of Lawrence Livermore National Laboratory

A Closer Look

Growing Crystals

Giant crystal of potassium dihydrogen phosphate.

For Raoult’s law to apply, the forces of attraction between solute and solvent molecules must be the same as those between solvent molecules in the pure solvent. This is frequently the case when molecules with similar structures are involved. Thus, solutions of one hydrocarbon in another (hexane, C6H14, dissolved in octane, C8H18, for example) follow Raoult’s law quite closely. However, with other solutions, deviations from Raoult’s law are observed. If solvent–solute interactions are stronger than solvent–solvent interactions, the measured vapor pressure will be lower than the value calculated by Raoult’s law. If the solvent–solute interactions are weaker than solvent–solvent interactions, the vapor pressure will be higher. Raoult’s law can be modified to give another useful equation that allows you to calculate the lowering of the vapor pressure of the solvent, ∆Psolvent, as a function of the mole fraction of the solute. ∆Psolvent = Psolvent − P°solvent

Substituting Raoult’s law for Psolvent gives ∆Psolvent = (Xsolvent P°solvent) − P°solvent = −(1 − Xsolvent)P°solvent

In a solution with only the volatile solvent and one nonvolatile solute, the sum of the mole fraction of solvent and solute must be 1: Xsolvent + Xsolute = 1

Therefore, 1 − Xsolvent = Xsolute, and the equation for ∆Psolvent can be rewritten as

∆Psolvent = −Xsolute P°solvent

(13.6)

Thus, the decrease in the vapor pressure of the solvent is proportional to the mole fraction (the relative number of particles) of solute.



13.4  Colligative Properties

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641

E xamp le 13.4

Using Raoult’s Law Problem You dissolve 651  g of ethylene glycol, HOCH2CH2OH, in 1.50  kg of water. What is the vapor pressure of the water over the solution at 90 °C? Assume ideal behavior for the solution. Appendix G lists P°water at various temperatures.

What Do You Know?  This is a Raoult’s law problem where you want to know Psolvent. To calculate Psolvent you need the mole fraction of solvent and the vapor pressure of the pure solvent. You can calculate the mole fraction of solvent using the given masses of solute and solvent and the molar masses of these species. You can look up the vapor pressure of pure water at 90 °C in Appendix G. Strategy  Step 1.  Calculate the mole fraction of the solvent (water). Step 2. Use Raoult’s law (Equation 13.5) to calculate the vapor pressure of the solvent over the solution.

Solution Step 1.  Calculate the mole fraction of the solvent (water).  1 mol  Amount of water  1.50  103 g   83.24 mol water  18.02 g   1 mol   10.49 mol glycol Amount of ethylene glycol  651 g   62.07 g  X water 

83.24 mol water  0.8881 83.24 mol water  10.49 mol glycol

Step 2. Use Raoult’s law (Equation 13.5) to calculate the vapor pressure of the solvent over the solution. The vapor pressure of pure water at 90 °C, P°water, is 525.8 mm Hg (Appendix G). Pwater  X water P  water  (0.8881)(525.8 mm Hg)   467 mm Hg 

Think about Your Answer  Although a substantial mass of ethylene glycol was added to the water, the decrease in the vapor pressure of the solvent was only 59 mm Hg, or about 11%: ∆Pwater = Pwater − P°water = 467 mm Hg − 525.8 mm Hg = −59 mm Hg Notice that the decrease in the vapor pressure can also be calculated using Equation 13.6 using the mole fraction of the solute. Xsolute = 1 − Xsolvent = 1 − 0.8881 = 0.1119 ∆Psolvent = −XsoluteP°solvent = −(0.1119)(525.8 mm Hg) = −58.8 mm Hg

Check Your Understanding  Assume you dissolve 10.0 g of sucrose (C12H22O11) in 225 mL (225 g) of water and warm the water to 60 °C. What is the vapor pressure of the water over this solution? (Appendix G lists P°water at various temperatures.)

642

Chapter 13 / Solutions and Their Behavior

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Boiling Point Elevation Suppose you have a solution of a nonvolatile solute in the volatile solvent benzene. If the solute concentration is 0.200 mol in 100. g of benzene (C6H6) (= 2.00 mol/kg), this means that Xbenzene = 0.865. Using Xbenzene and applying Raoult’s law, you can calculate that the vapor pressure of the solvent at 60 °C will drop from 400. mm Hg for the pure solvent to 346 mm Hg for the solution: Pbenzene = Xbenzene P°benzene = (0.865)(400. mm Hg) = 346 mm Hg

This point is marked on the vapor pressure graph in Figure 13.12. Additional points calculated in the same way for other temperatures define the vapor pressure curve for the solution (the lower curve in Figure 13.12). An important observation you can make in Figure 13.12 is that the vapor pressure lowering caused by the nonvolatile solute leads to an increase in the boiling point. The normal boiling point of a liquid is the temperature at which its vapor pressure is equal to 1 atm or 760 mm Hg (Section 11.6). In Figure 13.12, you see that the normal boiling point of pure benzene (at 760  mm Hg) is about 80  °C. Tracing the vapor pressure curve for the solution, you can see that the vapor pressure reaches 760 mm Hg at a temperature about 5 °C higher than this value. An important question is how the boiling point of the solution varies with solute concentration. The boiling point elevation, ∆Tbp, is directly proportional to the molality of the solute. Elevation in boiling point = ∆Tbp = Kbp msolute



(13.7)

In this equation, Kbp is a proportionality constant called the molal boiling point elevation constant. It has the units of degrees Celsius/molal (°C/m). Values for Kbp are determined experimentally, and different solvents have different values (­Table 13.3). The value of Kbp corresponds to the elevation in boiling point for a 1 m solution. 800

760

700

Vapor pressure (mm)

600

500 ∆P at 60 °C = 54 mm for Xsolute = 0.135

LIQUID

400 300

VAPOR

Pure benzene

Figure 13.12  Lowering the vapor pressure of benzene by addition of a nonvolatile solute.  ​The upper (red) curve represents the vapor pressure of pure benzene, and the lower (blue) curve represents the vapor pressure of a solution containing 0.200 mol of a solute dissolved in 0.100 kg of solvent (2.00 m). This graph was created by doing a series of calculations such as the one shown in the text. As an alternative, the graph could be created by measuring various vapor pressures for the solution in a laboratory experiment.

200 Benzene + solute

100 0 20

BP pure benzene 30

40

50

60

70

∆T 5.1° BP solution 80

Temperature (°C)



13.4  Colligative Properties

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Table 13.3

Some Boiling Point Elevation and Freezing Point Depression Constants

Normal Boiling Point (°C) Pure Solvent

Kbp (°C/m)

Water

100.00

+0.5121

  0.0

  −1.86

Benzene (C6H6)

 80.10

+2.53

  5.50

  −5.12

Camphor (C10H16O)

207.4

+5.611

179.75

−39.7

Chloroform (CHCl3)

 61.70

+3.63

—

—

Solvent

Normal Freezing Point (°C) Pure Solvent

Kfp (°C/m)

V Exampl e 1 3 .5

Boiling Point Elevation Problem  Eugenol, a compound found in nutmeg and cloves, has the formula C10H12O2.

Strategy Map Problem Calculate boiling point of solution.

What is the boiling point of a solution containing 0.144 g of this compound dissolved in 10.0 g of benzene?

What Do You Know?  You know the identity and mass of both the solute and the solvent. You will need to look up the boiling point for pure benzene and the value of Kbp. Strategy

Data/Information • Mass of compound and solvent • Kbp and T(bp solvent) from Table 13.3

Step 1. Calculate the solution concentration (molality, m, in mol/kg) from the amount (moles) of eugenol and the mass of solvent (kg). Step 2.  Calculate the change in boiling point using Equation 13.7. Step 3. Obtain the boiling point of benzene in the solution by adding the change in boiling point to the boiling point of pure benzene.

Solution Step 1

Calculate the solution concentration in molality (mol solute/kg solvent). Convert the mass of eugenol to the amount using the molar mass of eugenol.  1 mol eugenol  4 0.144 g eugenol    8.770  10 mol eugenol 164.2 g  Convert the mass of benzene from grams to kilograms. 10.0 g benzene 

1 kg  0.0100 kg benzene 1000 g

Calculate the molality. c eugenol  Step 2

8.770  104 mol eugenol  8.770  102 m 0.0100 kg benzene

Calculate the change in boiling point using Equation 13.7. The value of Kbp for benzene is listed in Table 13.3 and has a value of 2.53 °C/m. Use this and the concentration of eugenol to calculate the change in the boiling point. ∆Tbp = (2.53 °C/m)(0.08770 m) = 0.222 °C

644

Chapter 13 / Solutions and Their Behavior

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Step 3

Obtain the boiling point of benzene in the solution by adding the change in boiling point to the boiling point of pure benzene. Because the boiling point rises relative to that of the pure solvent, the change in boiling point is added to the boiling point of the pure solvent. 80.10 °C + 0.222 °C =  80.32 °C 

Think about Your Answer  Boiling point elevation is proportional to the solute concentration so sizable increases in boiling point are possible at high concentrations. Eugenol, C10H12O2, is an important component in oil of cloves, a commonly used spice.

Check Your Understanding  What quantity of ethylene glycol, HOCH2CH2OH, must be added to 125 g of water to raise the boiling point by 1.0 °C? Express the answer in grams.

The elevation of the boiling point of a solvent on adding a solute has many practical consequences. One of them is the protection your car’s engine receives in summer using all-season antifreeze. The main ingredient of commercial antifreeze is ethylene glycol, HOCH2CH2OH. The car’s radiator and cooling system are sealed to keep the coolant under pressure, ensuring that it will not vaporize at normal engine temperatures. When the air temperature is high in the summer, however, the radiator could boil over if it were not protected with antifreeze. By adding this nonvolatile liquid, the solution in the radiator has a higher boiling point than that of pure water.

Why Is the Boiling Point of a Solution Elevated and Its Freezing Point Depressed?  The answer to

this question is related to entropy, a thermodynamic function discussed in Chapter 18.

Freezing Point Depression Another consequence of dissolving a solute in a solvent is that the freezing point of the solution is lower than that of the pure solvent (Figure  13.13). For an ideal ­solution, the depression of the freezing point is given by an equation similar to that for the elevation of the boiling point:

Freezing point depression = ∆Tfp = Kfp msolute



(13.8)

where Kfp is the molal freezing point depression constant in degrees Celsius per molal (°C/m). Values of Kfp for a few common solvents are given in Table 13.3. The values are negative quantities, so the result of the calculation is a negative value for ∆Tfp, signifying a decrease in temperature. The practical aspects of freezing point changes from pure solvent to solution are similar to those for boiling point elevation. The very name of the liquid you add to the radiator in your car, antifreeze, indicates its purpose (Figure 13.13a). The label on the container of antifreeze tells you, for example, to add 6 quarts (5.7 L) of antifreeze to a 12-quart (11.4-L) cooling system to lower the freezing point to −34 °C and to raise the boiling point to +109 °C.



13.4  Colligative Properties

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645

Hope Jahren is a botanist who wrote a delightful book, Lab Girl (New York: Alfred Knopf, 2016), about her life and her career in teaching and ­research in universities as well as her passion for understanding the biology of trees. In one chapter, she describes why trees do not freeze in the winter, a topic directly related to the subject of this chapter. “In order to prepare for their long winter journey, trees undergo a process known as ‘hardening.’ First the permeability of the cell walls increases drastically, allowing pure water to flow out while concentrating the sugars, proteins, and acids left behind. These chemicals act as a potent antifreeze, such that the cell can now dip well below freezing and the fluid inside of it will still persist in a syrupy liquid form. The spaces between the cells are now filled with an ul­ trapure distillate of cell water, so pure that there are no stray atoms upon which an ice crystal could nucleate and grow. Ice is a

three-dimensional crystal of molecules, and freezing requires a nucleation spot— some chemical aberration upon which the pattern may start to build. Pure water de­ void of any such site may be ‘supercooled’ to forty degrees below zero and still remain an ice-free liquid. It is in this ‘hardened’ state, with some cells packed full of chemicals and others sectioned off for purity, that a tree embarks on its win­ ter journey, standing unmoved through the frost, sleet, and blizzards of the sea­ son. These trees do not grow during the winter; they merely stand and ride planet Earth to the other side of the sun, where the North Pole will finally be tilted toward the heat source and the tree will experi­ ence summer.” “The vast majority of northern trees pre­ pare well for their wintertime journey, and death due to frost damage is extremely rare. A chilly autumn brings on the same hardening as a balmy one, because trees do not take their cue from the changing

Dugdax/Shutterstock.com

A Closer Look

Hardening of Trees

Freezing point depression.  High concentrations of sugars, proteins, and acids within cells act as an antifreeze during the winter months. temperature. It is the gradual shortening of the days, sensed as a steady decrease of light during the twenty-four-hour cycle, that triggers hardening. Unlike the overall character of winter, which may be mild one winter and punishing the next, the pattern of how light changes through the autumn is exactly the same every year.”

Photos: © Charles D. Winters/Cengage

The dye stayed in solution.

Pure water

Water with antifreeze

(a) Adding antifreeze to water prevents the water from freezing. Here, a jar of pure water (left) and a jar of water to which automobile antifreeze had been added (right) were kept overnight in the freezing compartment of a home refrigerator.

Figure 13.13  Freezing a solution. 

646

Pure solvent formed into ice along the walls of the tube

(b) When a solution freezes, it is pure solvent that solidifies. Here, water containing a purple dye was frozen slowly. Pure ice formed along the walls of the tube, and the dye stayed in solution. The dye concentration increased as more and more solvent was frozen out, and the resulting solution had a lower and lower freezing point. Eventually, the system contained pure, colorless ice along the walls of the tube and a concentrated solution of dye in the center of the tube.

Chapter 13 / Solutions and Their Behavior

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Ex am p le 13.6

Freezing Point Depression Problem  What mass of ethylene glycol, HOCH2CH2OH, must be added to 5.50  kg of water to lower the freezing point of the water from 0.0 °C to −10.0 °C?

What Do You Know?  You know the freezing point of pure water and the desired freezing point for water in the solution. You also know the mass of water and can look up the value for Kfp in Table 13.3.

Strategy  In some ways this is the reverse of Example 13.5. Here you know the change in freezing point, but wish to know how much solute is needed to produce that value of ∆T. The solution concentration (molality, m) can be calculated from ∆Tfp and Kfp (­Table 13.3) using Equation 13.8. Combine the concentration with the mass of solvent to obtain the amount of solute and then its mass. Solution Step 1. Calculate the solute concentration in a solution with a freezing point depression of −10.0 °C. Solute concentration (m) 

Tfp 10.0 °C  5.376 m  K fp 1.86 °C/m

Step 2.  Calculate the amount of solute from the concentration and solvent mass.  5.376 mol glycol   1.00 kg water  (5.50 kg water)   29.57 mol glycol  Step 3.  Calculate the mass of the solute, ethylene glycol.  62.07 g    1840 g glycol  29.57 mol glycol    1 mol 

Think about Your Answer  The value of Kfp tells you that the freezing point of water is lower by 1.86 °C for a 1 molal solution. In this particular problem, the freezing point depression was −10.0 °C, so a molality around 5 m is reasonable. The density of ethylene glycol is 1.11 kg/L, so the volume of glycol to be added is (1.84 kg)(1 L/1.11 kg) = 1.65 L.

Check Your Understanding  In the northern United States, summer cottages are usually closed up for the winter. When doing so, the owners winterize the plumbing by putting antifreeze in the toilet tanks, for example. Will adding 525 g of HOCH2CH2OH to 3.00 kg of water ensure that the water will not freeze at −25 °C?

Osmotic Pressure Osmosis is the movement of solvent molecules through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration. This movement can be demonstrated with a simple experiment. The beaker in Figure 13.14 contains pure water, and the bag and tube hold a concentrated sugar solution. The liquids are separated by a semipermeable membrane, a thin sheet of material (such as a vegetable tissue or cellophane) through which only certain types of molecules can pass. Here, water molecules can pass through the membrane, but larger sugar molecules (or hydrated ions) cannot (Figure  13.15). When the experiment is begun, the liquid levels in the beaker and the tube are the same. Over time, however, the level of the sugar solution inside the tube rises, the level of pure water in the beaker falls, and the sugar solution becomes more dilute. Eventually, no further net change occurs; equilibrium is reached.

13.4  Colligative Properties

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647

(a) The bag attached to the tube contains a solution that is 5% sugar and 95% water. The beaker contains pure water. The bag is made of a material that is semipermeable, meaning that it allows water, but not sugar molecules, to pass through.

The height of the column of solution is a measure of the osmotic pressure. Height of solution column time

Pure water

(b) Over time, water flows from the region of low solute concentration (pure water) to the region of higher solute concentration (the sugar solution). Flow continues until the pressure exerted by the column of solution in the tube above the water level in the beaker is great enough to result in equal rates of passage of water molecules in both directions. The height of the column of solution is a measure of the osmotic pressure.

Water flows from the region of low solute concentration (pure water) to the region of higher solute concentration (the sugar solution).

5% sugar 95% water Semipermeable membrane

Figure 13.14  The process of osmosis. ​

From a molecular point of view, the semipermeable membrane does not pre­ sent a barrier to the movement of water molecules, so they move through the membrane in both directions. Over time, more water molecules pass through the membrane from the pure water side to the solution side than in the opposite direction. The net effect is that water molecules move from regions of low solute concentration to regions of high solute concentration. Why does the system eventually reach equilibrium? The solution in the tube in Figure 13.14 can never reach zero sugar concentration, which would be required to equalize the concentrations of solute on both sides of the membrane. The answer lies in the fact that the solution moves higher and higher in the tube as water moves into the sugar solution. Eventually, the pressure exerted by this column of solution counterbalances the pressure exerted by the water moving through the membrane from the pure water side, and no further net movement of water occurs. An equilibrium of forces is achieved. The pressure created by the column of solution for the system at equilibrium is a measure of the osmotic pressure, ∏. That is, the osmotic Figure 13.15  Osmosis at the particulate level.  ​Osmotic flow through a membrane that is selectively permeable (semipermeable) to water. Dissolved substances such as hydrated ions or large sugar molecules cannot diffuse through the membrane.

semipermeable membrane

+

large molecule

H2O

hydrated ions

−

648

Chapter 13 / Solutions and Their Behavior

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inding sources of freshwater for F humans and agriculture has been a ­ constant battle for centuries, and, if we continue using Earth’s water at the present rate, those problems may ­increase. Although Earth has a­ bundant water, 97% of it is too salty to drink or use on crops. A large portion of the ­remaining 3% is locked in the form of ice in the polar regions and is not eas­ ily obtained. One of the oldest ways to obtain freshwa­ ter from seawater is by evaporation. This is, however, very energy-intensive, and the salt and other materials left behind may not be useful. Reverse osmosis is another method of obtaining freshwater from seawater or groundwater. In this technique, a pressure greater than the osmotic pressure of the impure water is applied to force water through a semipermeable membrane from a region of high solute concentration to one of lower solute concentration, that is, in the reverse direction that the water would move by osmosis.

65.5 million cubic meters of water per day. Smaller reverse osmosis devices are also available for home use.

Although reverse osmosis has been known for over 200 years, only in the last few decades has it been exploited. Now some municipalities obtain drinking water this way, and pharmaceutical companies use it to obtain highly purified water. More than 13,000 reverse osmosis plants are in  operation worldwide, producing about

Pressure

Seawater

Water flow (more concentrated solution)

Concentrate flow

ETAJOE/Shutterstock.com

A Closer Look

Reverse Osmosis for Pure Water

A reverse osmosis plant.

Freshwater

Semipermeable membrane

Reverse osmosis.  Drinking water can be produced from seawater by reverse osmosis. The osmotic pressure of seawater is approximately 27 atm. To obtain freshwater at a reasonable rate, reverse osmosis requires a pressure of about 50 atm. For comparison, bicycle tires usually have an air pressure of 2 to 3 atm.

pressure is the difference between the height of the solution in the tube and the level of pure water in the beaker. From experimental measurements on dilute solutions, it is known that osmotic pressure and concentration (c) are related by the equation

∏ = cRT

(13.9)

In this equation, c is the molar concentration (in moles per liter); R is the gas constant; and T is the absolute temperature (in kelvins). Using a value for the gas law constant of 0.082057 L ∙ atm/K ∙ mol allows calculation of the osmotic pressure ∏ in atmospheres. This equation is analogous to the ideal gas law (PV = nRT), with ∏ taking the place of P and c being equivalent to n/V. Even solutions with low solute concentrations have a significant osmotic pressure. For example, the osmotic pressure of a 1.00 × 10−3 M solution at 298 K is 18.6  mm Hg. Low pressures such as this are easily and accurately measured, so concentrations of very dilute solutions can be determined by this technique. (Compare this to the effect of a solute on freezing point, for example. A dilute aqueous solution of similar concentration, 1.00 × 10−3 mol/kg solvent, would be expected to lower the freezing point by about 0.002 °C, too small to measure accurately.) For this reason determining a molar mass by measuring the osmotic pressure of a ­solution is a particularly useful technique when dealing with compounds having a very high molar mass such as polymers and large biomolecules. Other examples of osmosis are shown in Figure  13.16. In this case, the egg’s membrane serves as the semipermeable membrane. Osmosis occurs in one direction if the concentration of solute is greater inside the egg than in the exterior solution and occurs in the other direction if the concentration of the solution is less inside the egg than in the exterior solution. In both cases, solvent flows from the region of low solute concentration to the region of high solute concentration.

13.4  Colligative Properties

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649

Photos: © Charles D. Winters/Cengage

(b) If the egg, with its shell removed, is placed in pure water, the egg swells.

(a) A fresh egg is placed in dilute acetic acid. The acid reacts with the CaCO3 of the shell but leaves the egg membrane intact.

(c) If the egg, with its shell removed, is placed in a concentrated sugar solution, the egg shrivels.

Figure 13.16  An experiment to observe osmosis.  ​You can try this experiment in your kitchen. In the first step, use vinegar as a source of acetic acid.

E xamp le 13.7

Determining the Osmotic Pressure of a Solution of a Polymer Problem  Polyvinyl alcohol is a water-soluble polymer with an average molar mass of 28,000 g/mol. You dissolve 1.844 g of this polymer in water to give 150. mL of solution. What is the osmotic pressure, measured at 27 °C?

What Do You Know?  Osmotic pressure is calculated using Equation 13.9, ∏ = cRT. You are given the mass and molar mass of the polymer and the temperature and will need the gas law constant.

Strategy Step 1. Calculate the concentration (c) of the polymer in mol/L. Step 2. Calculate the osmotic pressure using Equation 13.9. In this equation, the temperature must be in kelvins, and the value of the gas constant is 0.082057 L ? atm/mol ? K.

Solution Step 1.  Calculate the concentration (c) of the polymer in mol/L. c = 1.844 g(1 mol/28,000 g)/(0.150 L) = 4.39 × 10−4 mol/L Step 2.  Calculate the osmotic pressure. ∏ = cRT

∏ = (4.39 × 10−4 mol/L)(0.082057 L ? atm/mol ? K)(300. K)



∏ =  0.011 atm  (= 8.2 mm Hg)

Think about Your Answer  If the osmotic pressure is measured in a device similar to that shown in Figure 13.14, the height of the column of aqueous solution supported is about 110 mm (8.2 mm Hg × 13.6 mm H2O/mm Hg = 110 mm H2O).

Check Your Understanding  Bradykinin is a small peptide (nine amino acids; 1060 g/mol) that lowers blood pressure by causing blood vessels to dilate. What is the osmotic pressure of a solution of this protein at 20. °C if 0.033 g of the peptide is dissolved in water to give 50.0 mL of solution?

650

Chapter 13 / Solutions and Their Behavior

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A Closer Look

Osmosis and Medicine must be isotonic (Figure B, left). If pure wa­ ter was used, the inside of a blood cell would have a higher solute concentration (lower water concentration), and ­water would flow into the cell. This h ­ ypotonic situation would cause the red blood cells to burst (lyse) ­(Figure B, middle). The opposite situation, hypertonicity, ­ occurs if the intravenous

s­olution is more concentrated than the ­contents of the blood cell (Figure B, right). In this case, the cell would lose water and shrivel up (crenate). To c­ ombat this, a dehy­ drated ­patient is ­rehydrated in the hospital with a sterile ­saline solution that is 0.16 M NaCl, a ­solution that is isotonic with the cells of the body.

Photos: David M. Phillips/Science Source

Osmosis has practical significance in medicine. Patients who become dehy­ drated through illness often need to be given water and nutrients intravenously (Figure A). Water cannot simply be dripped into a patient’s vein, however. Rather, the intravenous solution must have the same overall solute concentra­ tion as the patient’s blood: the solution

John C. Kotz

Figure B  Osmosis and living cells. (left) A cell placed in an isotonic solution. The net

Figure A  An isotonic saline solution. This solution has the same molality as body fluids.

movement of water into and out of the cell is zero because the concentration of solutes inside and outside the cell is the same. (middle) In a hypotonic solution, the concentration of solutes outside the cell is less than that inside. There is a net flow of water into the cell, causing the cell to swell and perhaps to burst (or lyse). (right) In a hypertonic solution, the concentration of solutes outside the cell is greater than that inside. There is a net flow of water out of the cell, causing the cell to dehydrate, shrink, and perhaps die.

Colligative Properties and Molar Mass Determination Earlier in this book you learned how to calculate a molecular formula from an empirical formula when given the molar mass. But how do you know the molar mass of an unknown compound? An experiment must be conducted to find this crucial piece of information, and one way to do so is to use a colligative property of a solution of the compound. The strategy map used with Example 13.8 presents the basic approach.

V Ex am p le 13.8

Determining Molar Mass from Boiling Point Elevation Problem  A solution prepared from 1.25 g of oil of wintergreen (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31 °C. Determine the molar mass of methyl salicylate.

What Do You Know?  You know the masses of the oil of wintergreen and benzene in the solution, as well as the boiling point of the solution. The boiling point of pure benzene and Kbp for benzene are given in Table 13.3.

13.4  Colligative Properties

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651

Strategy Map

Strategy   The molar mass of a compound is the quotient of the mass of a sam-

Problem Calculate molar mass of unknown.

ple (g) and the amount (mol) represented by that sample. Here you know the mass of the sample (1.25 g), so you need to find the amount (mol) that corresponds to this mass.

Data/Information • Mass of unknown and solvent • Boiling point of solution (80.31 C) • Boiling point of pure solvent • Kbp for solvent

Step 1. Determine ∆Tbp from the given boiling point of the solution and the boiling point of pure solvent (benzene). Step 2. Calculate the solution concentration in units of mol/kg using Equation 13.7 (∆Tbp = Kbpm). Step 3. Calculate the amount (mol) of solute (oil of wintergreen) from the solution concentration and the mass of the solvent (kg). Step 4. Determine the molar mass by dividing the mass of the solute by the amount of solute [molar mass = mass (g)/amount (mol)].

Solution Step 1

Determine ∆Tbp. Boiling point elevation (∆Tbp) = 80.31 °C − 80.10 °C = 0.21 °C

Step 2

Calculate the solution concentration (mol/kg) using Equation 13.7. ∆Tbp = Kbpm Concentration (mol/kg) 

Step 3

Tbp 0.21 °C   0.0830 m 2.53 °C/m K bp

Calculate the amount (mol) of solute from the solution concentration and the mass of the solvent.  0.0830 mol  Amount of solute   (0.0990 kg solvent)  0.00822 mol solute  1.00 kg 

Step 4

Determine the molar mass of the solute.

Molar mass 

mass (g) 1.25 g    150 g/mol  0.00822 mol amount (mol)

Think about Your Answer  Methyl salicylate has the formula C8H8O3 and a molar mass of 152.15 g/mol. Given that ∆T has only two significant figures, the calculated value is acceptably close to the actual value.

Check Your Understanding  An aluminum-containing compound has the empirical formula (C2H5)2AlF. Find the molecular formula if 0.448 g of the compound dissolved in 23.46 g of benzene has a freezing point of 5.265 °C.

E xamp le 13.9

Osmotic Pressure and Molar Mass Problem  Beta-carotene is the most important of the A vitamins. Calculate the molar mass of β-carotene if 10.0 mL of a solution in chloroform containing 7.68 mg of β-carotene has an osmotic pressure of 26.57 mm Hg at 25.0 °C.

652

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What Do You Know?  The molar mass of a compound is the quotient of the mass of a sample and the amount represented by that sample. Here you know the mass of the sample (7.68 mg), so you need to find the amount (mol) equivalent to this mass.

Strategy  Step 1. Calculate the solution concentration (mol/L) from the osmotic pressure using Equation 13.9. Step 2. Calculate the amount (mol) of solute from the solution concentration and the volume (L) of the solution. Step 3. Determine the molar mass of the solute from its mass and amount.

Solution Step 1. Calculate the solution concentration (mol/L) using Equation 13.9. ∏ = cRT  1 atm  (26.57 mm Hg)   760 mm Hg  ∏ Concentration (mol/L)   RT (0.082057 L  atm/K  mol)(298.2 K)  1.4287  103 mol/L Step 2. Calculate the amount (mol) of solute from the solution concentration and the volume (L) of the solution. (1.4287 × 10−3 mol/L)(0.0100 L) = 1.429 × 10−5 mol Step 3. Determine the molar mass of the solute.

Molar mass 

7.68  103 g mass (g)   538 g/mol   1.429  105 mol amount (mol)

Think about Your Answer  Beta-carotene is a hydrocarbon with the formula C40H56 (molar mass = 536.9 g/mol).

Check Your Understanding A 1.40-g sample of polyethylene, a common plastic, is dissolved in enough organic solvent to give 100.0 mL of solution. What is the average molar mass of the polymer if the measured osmotic pressure of the solution is 1.86 mm Hg at 25 °C?

Colligative Properties of Solutions Containing Ions In the northern United States and Canada, it is common practice to scatter salt on snowy or icy roads or sidewalks in the winter. When the sun shines on the snow or patch of ice, a small amount melts, and some salt dissolves in the water. As a result of the dissolved solute, the freezing point of the solution is lower than 0 °C. The solution “eats” its way through the ice, breaking it up, and the icy patch is no longer dangerous for drivers or walkers. Salt (NaCl) is the most common substance used on roads because it is inexpensive and dissolves readily in water. Its relatively low molar mass means that the effect per gram is large. In addition, salt is especially effective because it is an electrolyte. That is, it dissolves to give ions in solution: NaCl(s) n Na+(aq) + Cl−(aq)

Recall that  colligative properties depend not on what is dissolved but on the number of particles of solute per solvent particle. When 1 mol of NaCl dissolves, 2 mol of ions

13.4  Colligative Properties

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653

form, which means that the effect on the freezing point of water should be twice as large as that expected for a mole of sugar. A 0.100 m solution of NaCl really contains two solutes, 0.100 m Na+ and 0.100 m Cl−. What should be used to estimate the freezing point depression is the total molality of solute particles: mtotal = m(Na+) + m(Cl−) = (0.100 + 0.100) mol/kg = 0.200 mol/kg ∆Tfp = (−1.86 °C/m)(0.200 m) = −0.372 °C

To estimate the freezing point depression for an ionic compound, first find the molality of solute from the mass and molar mass of the compound and the mass of the solvent. Then, multiply the molality by the number of ions in the formula: two for NaCl, three for Na2SO4, four for LaCl3, five for Al2(SO4)3, and so on. Finally, this is multiplied by the molal freezing point depression constant. Table 13.4 shows that as the concentration of NaCl decreases, ∆Tfp for NaCl approaches but does not quite reach a value that is two times larger than the value determined assuming no dissociation. Likewise, ∆Tfp for Na2SO4 approaches but does not reach a value that is three times larger. The ratio of the experimentally observed value of ∆Tfp to the value calculated assuming no dissociation is called the van’t Hoff factor after Jacobus Henricus van’t Hoff (1852–1911), who studied this phenomenon. The van’t Hoff factor is represented by i. i

T , measured Tfp , measured  fp K fp m Tfp , calculated

or Tfp measured  K fp mi



(13.10)

van’t Hoff factors can be used in calculations of any colligative property. Vapor ­pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure are all larger for electrolytes than for nonelectrolytes of the same molality. The van’t Hoff factor approaches a whole number (2, 3, and so on) only with very dilute solutions. In more concentrated solutions, the experimental freezing point depressions indicate that there are fewer ions in solution than expected. This behavior, which is typical of all ionic compounds, is a consequence of the strong attractions between ions. The result is as if some of the positive and negative ions are paired, decreasing the total molality of particles. Indeed, in more concentrated solutions, and especially in solvents less polar than water, ions are extensively associated in ion pairs and in even larger clusters.

Table 13.4 Mass %

Freezing Point Depressions of Some Ionic Solutions

m (mol/kg)

𝚫Tfp (measured, °C)

𝚫Tfp (calculated, °C)

Tfp, measured Tfp, calculated

NaCl 0.00700

0.0120

−0.0433

−0.0223

1.94

0.500

0.0860

−0.299

−0.160

1.87

1.00

0.173

−0.593

−0.322

1.84

2.00

0.349

−1.186

−0.649

1.83

0.00700

0.00493

−0.0257

−0.00917

2.80

0.500

0.0354

−0.165

−0.0658

2.51

1.00

0.0711

−0.320

−0.132

2.42

2.00

0.144

−0.606

−0.268

2.26

Na2SO4

654

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Ex am p le 13.10

Freezing Point and Ionic Solutions Problem  A 0.0200  m aqueous solution of an ionic compound, Co(NH3)4Cl3, freezes at −0.0640 °C. How many ions does each formula unit of Co(NH3)4Cl3 produce upon being dissolved in water?

What Do You Know?  You know the freezing point depression (−0.0640 °C), the solution concentration, and Kfp. Strategy  The van’t Hoff factor, i, is the ratio of the measured ∆Tfp to the calculated freezing point depression. First, calculate ∆Tfp expected for a solution in which no ions are produced. Compare this value with the actual value of ∆Tfp. The ratio (= i ) reflects the number of ions produced. Solution (a) Calculate the freezing point depression expected for a 0.0200 m solution assuming that the salt does not dissociate into ions. ∆Tfp calculated = Kfpm = (−1.86 °C)(0.0200 m) = −0.0372 °C (b) Compare the calculated freezing point depression with the measured depression. This gives the van’t Hoff factor: i 

Tfp , measured 0.0640 °C  1.72  Tfp , calculated 0.0372 °C

The i value is much greater than 1 and is approaching 2. Therefore, you can conclude that each formula unit of Co(NH3)4Cl3 produces  2 moles of ions in solution. 

Think about Your Answer  The value of i is close to 2, meaning that the complex dissociates into two ions: [Co(NH3)4Cl2]+ and Cl−. As you will see in Chapter 22, the cation is a Co3+ ion surrounded octahedrally by four NH3 molecules and two Cl− ions.

Check Your Understanding 

13.5 Colloids Goal for Section 13.5 • Recognize the properties and importance of colloids. Earlier in this chapter, we defined a solution broadly as a homogeneous mixture of two or more substances in a single phase. To this definition we should add that, in a true solution, no settling of the solute should be observed and the solute particles should be in the form of ions or relatively small molecules. Thus, NaCl and sugar form true solutions in water. You may also be familiar with suspensions, which result, for example, if a handful of fine sand is added to water and shaken vigorously. Sand particles are still visible and gradually settle to the bottom of the beaker or bottle. Colloidal dispersions, also called colloids, represent a state intermediate between a solution and a suspension (Figure  13.17). Colloids include many of the foods you eat and materials around you; among them are JELL-O, milk, fog, and porcelain (Table 13.5).

© Charles D. Winters/Cengage

Calculate the freezing point of 525 g of water that contains 25.0 g of NaCl. Assume i, the van’t Hoff factor, is 1.85 for NaCl.

Figure 13.17  Gold colloid.  ​A water-soluble salt of [AuCl4]− is reduced to give colloidal gold metal. The colloidal gold gives the dispersion its red color (when the particles have a length or diameter of less than 100 nm). Similarly, colloidal gold is used to give a beautiful red color to glass. Since the days of alchemy, some have claimed that drinking a colloidal gold solution was good for the mind. 13.5 Colloids

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655

Table 13.5

Types of Colloids

Type

Dispersing Medium

Dispersed Phase

Examples

Aerosol

Gas

Liquid

Fog, clouds, aerosol sprays

Aerosol

Gas

Solid

Smoke, airborne viruses, automobile exhaust

Foam

Liquid

Gas

Shaving cream, whipped cream

Foam

Solid

Gas

Styrofoam, marshmallow

Emulsion

Liquid

Liquid

Mayonnaise, milk, face cream

Gel

Solid

Liquid

Jelly, JELL-O, cheese, butter

Sol

Liquid

Solid

Gold in water, milk of magnesia, mud

Solid sol

Solid

Solid

Milkglass

Around 1860, the British chemist Thomas Graham (1805–1869) found that substances such as starch, gelatin, glue, and albumin from eggs diffused very slowly in water, compared with sugar or salt. In addition, the former substances differed significantly in their ability to diffuse through a thin membrane: Sugar molecules can diffuse through many membranes, but the very large molecules that make up starch, gelatin, glue, and albumin do not. Moreover, Graham found that he could not crystallize these substances, whereas he could crystallize sugar, salt, and other materials that form true solutions. Graham coined the word colloid (from the Greek, meaning “glue”) to describe this class of substances that are distinctly different from true solutions and suspensions. Modern chemists have been successful in crystallizing some colloidal substances, albeit with difficulty, so there really is no sharp dividing line between these classes based on this property. Colloids do, however, have several distinguishing characteristics. First, many of the substances that form colloids have high molar masses; this is true of proteins such as hemoglobin that have molar masses in the thousands. Second, the particles of a colloid are relatively large (several hundred nanometers in diameter). As a consequence, they exhibit the Tyndall effect; they scatter visible light when dispersed in a solvent, making the mixture appear cloudy (Figure 13.18). Third, even though colloidal particles are large, they are not so large that they settle out.

The light beam is scattered by the colloidal dispersion, exhibiting the Tyndall effect.

Dust in the air scatters the light coming through the trees in a forest along the Oregon coast.

A narrow beam of light from a laser is passed through an NaCl solution (left) and then a colloidal mixture of gelatin and water (right).

Figure 13.18  The Tyndall effect.  ​Colloidal dispersions scatter light, a phenomenon known as the

Tyndall effect.

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Photos: © Charles D. Winters/Cengage

The light beam is minimally scattered by the solution.

© Charles D. Winters/Cengage

Graham also invented the words sol for a colloidal dispersion of a solid substance in a fluid medium and gel for a colloidal dispersion with a structure that prevents it from being mobile. JELL-O is a sol when the solid is first mixed with boiling water, but it becomes a gel when cooled. Other examples of gels are the gelatinous precipitates of Al(OH)3, Fe(OH)3, and Cu(OH)2 (Figure 13.19). Because of the small size of the particles in a colloidal dispersion, they can have a very high surface area. For example, if you have one millionth of a mole of colloidal particles, each assumed to be a sphere with a diameter of 1000 nm, the total surface area of the particles would be on the order of 2 million cm2. It is not surprising, therefore, that many of the properties of colloids depend on the properties of surfaces.

Figure 13.19 Gelatinous precipitates.  ​(moving left to right) Al(OH)3, Fe(OH)3, and Cu(OH)2.

Colloids are classified according to the state of the dispersed phase and the dispersing medium. Table 13.5 lists several types of colloids and gives examples of each. Colloids with water as the dispersing medium can be classified as hydrophobic (from the Greek, meaning “water-fearing”) or hydrophilic (“water-loving”). A hydrophobic colloid is one in which only weak attractive forces exist between the water and the surfaces of the colloidal particles. Examples include dispersions of metals (see Figure  13.17) and of nearly insoluble salts in water. When compounds like AgCl precipitate, the result is often a colloidal dispersion. The precipitation reaction occurs too rapidly for ions to gather from long distances and make large crystals, so the ions aggregate to form small particles that remain suspended in the liquid. Why don’t the particles come together (coagulate) and form larger particles? The answer is that the colloidal particles carry electric charges. To see how this happens, suppose AgCl ion pairs come together to form a tiny particle. If Ag+ ions are still present in substantial concentration in the solution, these positive ions could be attracted to negative Cl− ions on the surface of the particle. Thus the original clump of AgCl ion pairs becomes positively charged, allowing it to attract a secondary layer of anions. The particles, now surrounded by layers of ions, repel one another and are prevented from coming together to form a precipitate (Figure 13.20). Soil particles are often carried by water in rivers and streams as hydrophobic colloids. When river water carrying large amounts of colloidal particles meets seawater with its high concentration of salts, the particles coagulate to form the silt seen at the mouth of the river (Figure 13.21). Municipal water treatment plants often add salts such as Al2(SO4)3 to clarify water. In aqueous solution, aluminum ions exist as −

− − −

−

−

−

−

−

−

+ + − − − + − − − + − + + − − − − − − −

−

repulsion

− − − − − − + + + − + − − − + − − + + + − − − − − − − − −

− − − − + + − + − −

−

−

repulsion −

repulsion −

− −

−

− −

+

− − − − − − + + + − +

+ − −

+ − −

+ + −

−

− −

−

− −

−

− −

colloidal particle surrounded by positive ions sheathed in negative ions

Figure 13.20  Hydrophobic colloids.  ​A hydrophobic colloid is stabilized by positive ions

adsorbed onto each particle and a secondary layer of negative ions. Because the particles bear similar charges, they repel one another, and precipitation is prevented.



NASA Earth Observatory image by Robert Simmon, using Landsat 5 data from the U.S. Geological Survey Global Visualization Viewer

Types of Colloids

Figure 13.21  Formation of silt.  ​Silt forms at a river delta as colloidal soil particles come in contact with saltwater in the ocean. Here, the Connecticut River empties into the Long Island Sound. The high concentration of ions in seawater causes the colloidal soil particles to coagulate. 13.5 Colloids

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657

[Al(H2O)6]3+ cations, which neutralize the charge on the hydrophobic colloidal soil particles, causing these particles to aggregate and settle out. Hydrophilic colloids are strongly attracted to water molecules. They often have groups such as OOH and ONH2 on their surfaces. These groups form strong hydrogen bonds to water, thereby stabilizing the colloid. Proteins and starch are important examples of hydrophilic colloids, and homogenized milk is the most familiar example. Emulsions are colloidal dispersions of one liquid in another, such as oil or fat in water. Familiar examples include salad dressing, mayonnaise, and milk. If vegetable oil and vinegar are mixed to make a salad dressing, the mixture quickly separates into two layers because the nonpolar oil molecules do not interact with the polar water and acetic acid (CH3CO2H) molecules. So why are milk and mayonnaise apparently homogeneous mixtures that do not separate into layers? The answer is that they contain an emulsifying agent such as soap or a protein. Lecithin, found in egg yolks, can act as an emulsifying agent, so mixing egg yolks with oil and vinegar stabilizes the colloidal dispersion known as mayonnaise. To understand this process further, consider how soaps and detergents, substances known as surfactants, function.

Surfactants Soaps and detergents are emulsifying agents. Soap is made by heating a fat with sodium or potassium hydroxide, which produces anions of long-chain carboxylic acids, sometimes referred to as fatty acids. An example is sodium stearate. O H3C(CH2)16

C

O− Na+

hydrocarbon tail polar head soluble in water soluble in oil sodium stearate, a soap Soaps and Surfactants Sodium

soaps are solid at room temperature, whereas potassium soaps are usually liquids. About 30 million tons of household and toilet soap as well as synthetic and soap-based laundry detergents, are produced annually worldwide.

The fatty-acid anion has both a nonpolar, hydrophobic hydrocarbon tail that is soluble in similar hydrocarbons and a polar, hydrophilic head that is soluble in water. Oil cannot be readily washed away from dishes or clothing with water because oil is nonpolar and thus insoluble in water. To clean away the oil, soap is added to the water. The nonpolar molecules of the oil interact with the nonpolar hydrocarbon tails of the soap molecules, leaving the polar heads of the soap to interact with surrounding water molecules. The oil and water then mix (Figure 13.22), and the oil can be washed away. Substances such as soaps that affect the properties of surfaces, and therefore affect the interaction between two phases, are called surface-active agents, or surfactants, for short. A surfactant used for cleaning is a detergent. One function of a surfactant is to lower the surface tension of water, which enhances the cleansing action of the detergent (Figure 13.23). O

detergent molecules −

−

−

− −

−

H −

water −

− −

− −

−

−

−

− oil

−

−

−

−

− −

C

H O

hydrophilic polar head

−

− −

O

+ −

−

−

fabric

hydrophobic nonpolar tail

Figure 13.22  The cleaning action of soap.  ​Soap molecules interact with water through the charged, hydrophilic end of the molecule. The long, hydrocarbon end of the molecule is hydrophobic, but it can bind through dispersion forces with hydrocarbons and other nonpolar substances.

658

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Figure 13.23  Effect of a detergent on the surface tension of water. 

© Charles D. Winters/Cengage

add surfactant

Sulfur (density = 2.1 g/cm3) is carefully placed on the surface of water (density, 1.0 g/cm3). The surface tension of the water keeps the denser sulfur afloat.

Several drops of detergent are then placed on the surface of the water. The surface tension of the water is reduced, and the sulfur sinks to the bottom of the container.

Many detergents used in the home and industry are synthetic. One example is sodium laurylbenzenesulfonate, a biodegradable compound. CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2

SO3− Na+

sodium laurylbenzenesulfonate

In general, synthetic detergents use the sulfonate group, OSO3−, as the polar head instead of the carboxylate group, OCO2−. The carboxylate anions form an insoluble precipitate with any Ca2+ or Mg2+ ions present in water. Because hard water is characterized by high concentrations of these ions, using soaps that contain carboxy­lates produces bathtub rings and telltale gray clothing. The synthetic sulfonate detergents have the advantage that they do not form such precipitates because their calcium salts are more soluble in water.

Applying Chemical Principles 13.1 Distillation Distillation is a means of separating components of a mixture based upon differences in volatility. Because petroleum con­ sists of a large number of compounds of widely varying volatil­ ity, distillation is a key process used to produce gasoline and related products (Figure A). How does this process work? If a mixture is composed of one volatile component and one or more nonvolatile components, a complete separation of the volatile component is possible. But, when two or more compo­ nents in a mixture have nonzero vapor pressures at the mix­ ture’s boiling point, the result of a simple distillation is an incomplete separation. Fractional distillation uses repetitive evaporation– condensation cycles to improve the separation of volatile com­ pounds. When a mixture is boiled, the vapor above the liquid has a higher concentration of the more volatile components than the liquid. If this vapor is condensed, then boiled again,

the resulting vapor is further enriched in more volatile material. The temperature–composition diagram for a hexane–heptane mixture is shown in Figure B. The lower curved line indicates the boiling point of the mixture as a function of the mole frac­ tion of the more volatile component, hexane. The upper curve indicates the mole fraction of hexane in the vapor phase above the boiling solution. The blue line represents the change in composition of the mixture in the course of the fractional distil­ lation of a mixture with an initial hexane mole fraction of 0.20. The solution may be heated until it reaches its boiling point of 90  °C. At this temperature, the mole fraction of hexane in the vapor phase can be determined by drawing a horizontal line at 90 °C that intersects the top curve. This indicates that the vapor  phase has a mole fraction of hexane of 0.38 at this temperature. Applying Chemical Principles

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659

Figure A  Equipment used for fractional distillation of petroleum in an oil refinery.

Questions

1. The blue line on the diagram illustrates the effect of using fractional distillation to separate a mixture of hexane (C6H14) and heptane (C7H16). If a fractional distillation starts with a 0.20  mole fraction of hexane, what is the approximate mole fraction of hexane in the vapor phase after two evaporation–­condensation cycles? 2. How many theoretical plates are required to produce a solu­ tion with a mole fraction of hexane greater than 0.90?

95 Composition of vapor versus temperature

Temperature (°C)

90 85 80

Boiling point of mixture versus mole fraction

75 70

GE_4530/Shutterstock.com

The efficiency of a frac­ tional distillation column is expressed in terms of theoretical plates. Each evaporation–condensation cycle requires one theo­ retical plate. The greater the number of theoretical plates, the better the sep­ aration for most mixtures.

100

65 60

0

0.2

0.4

0.6

0.8

1

Mole fraction, hexane

Figure B  A temperature–composition diagram for hexane (C6H14)–heptane (C7H16) mixtures. 3. What is the mass percent of hexane in a mixture with heptane if the mole fraction of hexane is 0.20? 4. The vapor pressure of pure heptane is 361.5  mm Hg at 75.0 °C and its normal boiling point is 98.4 °C. Use the modi­ fied Clausius–Clapeyron equation (Equation 11.3, page 564) to determine the enthalpy of vaporization for heptane.

On Thursday, August 21, 1986, people and animals around Lake Nyos in Cameroon, a small nation on the west coast of Africa, suddenly collapsed and died. More than 1700 people and hundreds of animals were dead, but there was no apparent cause. What had brought on this disaster? Some weeks later, the mystery was solved. Lake Nyos is a crater lake, which formed when cooled volcanic craters filled with water. Importantly, Lake Nyos contains an enormous amount of dissolved carbon dioxide generated by volcanic ac­ tivity deep in the Earth. Under the high pressure at the bottom of the lake, a very large amount of CO2 dissolves in the water. But on that evening in 1986, something—perhaps a small earthquake or an underwater landslide—disturbed the lake. The CO2-saturated water at the bottom of the lake was carried to the surface, where, under lower pressure, the gas was much less soluble. About one cubic kilometer of carbon dioxide was released into the atmosphere as a geyser of water and CO2 that shot up about 260 feet into the air. Then, because this gas is more dense than air, it hugged the ground and began to move with the prevailing breeze at about 45 miles per hour displac­ ing vital oxygen. The result was that when it reached villages 12 miles away, both people and animals were asphyxiated. Following this disaster, scientists and engineers began work­ ing on a way to avert future such occurrences in Lake Nyos. Beginning in 1990, a major project was begun to install pipes and use siphoning technology to release the dangerous levels of gases in the lower regions of the lake into the atmosphere safely by bringing water from these lower regions up to the surface. It  is believed that, from 2018 onward, this technology has

660

Louise Gubb/Corbis Historical/Getty Images

13.2  Henry’s Law and Exploding Lakes

Lake Nyos in Cameroon (western Africa), the site of a natural disaster. ​In 1986, a huge bubble of CO2 escaped from the lake and asphyxiated more than 1700 people. A syphoning system shown here has been installed to avert future danger.

successfully averted the danger of a repeat eruption of gas at Lake Nyos. The explosive release of CO2 from Lake Nyos is much like what occurs when you shake a bottle of carbonated soft drink. Carbonated sodas are bottled under a high pressure of CO2. Some of the gas dissolves in the soda, but some also remains in the small space above the liquid (called the headspace). The pressure of the CO2 in the headspace is between 2 and 4 atm.

Chapter 13 / Solutions and Their Behavior

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When the bottle cap is removed, the CO2 in the headspace es­ capes rapidly. Some of the dissolved CO2 also comes out of solution, and you see bubbles of gas rising to the surface. If the bottle remains open, this continues until equilibrium is estab­ lished with CO2 in the atmosphere (where the partial pressure of CO2 is 3.75 × 10−4 atm),

Questions

1. If the headspace of a soda is 25 mL and the pressure of CO2 in the space is 4.0 atm (≈4.0 bar) at 25 °C, what amount of CO2 is contained in the headspace? 2. If the CO2 in the headspace escapes into the atmosphere where the partial pressure of CO2 is 3.7 × 10−4 atm, what volume would the CO2 occupy (at 25 °C)? By what amount did the CO2 expand when it was released? 3. What is the solubility of CO2 in water at 25 °C when the pres­ sure of the gas is 3.7 × 10−4 bar? 4. After opening a 1.0 L soda, what mass of the dissolved CO2 is released to the atmosphere to reach equili­brium? Assume that the CO2 dissolved in the soda in the sealed bottle was in equi­ librium with the CO2 in the headspace at 4.0 bar pressure.

CO2(solution) uv CO2(g) and the soda goes flat. If the newly opened soda bottle is undis­ turbed, however, the loss of CO2 from solution is rather slow be­ cause bubble formation is not rapid, and your soda keeps its fizz.

If you have ever been scuba diving, you have heard about nar­ cosis, also called nitrogen narcosis, the martini effect, or rap­ ture of the deep. And you probably discussed the bends with your diving instructor. Both of these effects are caused by gases dissolving in your blood as you breathe O2 and N2 from the scuba tank under pressure. When you dive, the pressure of the air you breathe must be balanced against the external pressure of the water against your body. The pressure on your body when you are under water is normal atmospheric pressure plus the pressure of water, or about 1 atm for every 10 m of depth. To counter this, modern dive regulators automatically supply the breathing mixture at higher and higher pressures as you descend. But here is the problem: as the gas pressure increases, the partial pressures of O2 and N2 increase, causing more gas to dissolve in your blood. These gases also dissolve in nerve cells in the brain and alter nerve transmission, and they have a nar­ cotic effect. The effect is comparable to drinking a martini on an empty stomach or inhaling laughing gas (nitrous oxide, N2O) at the dentist; it makes you slightly giddy. In severe cases, it can impair your judgment and even cause you to take  the regulator out of your mouth and hand it to a fish! Some people can go as deep as 30 to 40 m with no problem, but others experience narcosis at shallower depths. Another problem with breathing air at depths is oxygen tox­ icity. Our bodies are accustomed to a partial pressure of O2 of 0.21 atm. At a depth of 30 m, using compressed air, the par­ tial pressure of O2 is about one atmosphere, which means it is comparable to breathing 100% oxygen at sea level. These higher partial pressures can harm the lungs and cause central nervous system damage. If you ascend too rapidly from a dive, you can experience the bends, a painful and potentially lethal condition in which nitrogen gas bubbles form in the blood as the solubility of nitrogen de­ creases with decreasing pressure. To diminish the likelihood of getting the bends, divers can use a gas mixture called nitrox. This mixture is usually about 32% to 36% O2 with correspondingly less N2. With the lower N2 pressure, less N2 dissolves in the blood. Nitrox allows a diver to ascend more rapidly because less time is



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13.3  Narcosis and the Bends

needed to expel N2 from the blood. And, because of the higher O2 content of this gas mixture, nitrox allows extended dive times. For deep dives (about 100  m) different gas mixtures are used. A common mixture (Trimix) contains about 10% O2, 70% He, and 20% N2; other mixtures contain only oxygen and ­helium (Heliox). The lower percentage of O2 lowers the danger of O2 toxicity. Helium is substituted for N2 because it is less soluble in blood and tissue, but it introduces an interesting side effect. If there is a voice link to the surface, the diver’s speech sounds like Donald Duck! Speech is altered because the veloc­ ity of sound in helium is different from that in air.

Questions

1. Calculate the external pressure experienced by a diver at a depth of 10.0 m. If a scuba diver is using compressed air at this depth, what are the partial pressures of oxygen and nitrogen? (Assume the density of water is 1.00 g/mL.) 2. At a depth of 20 meters the pressure is about 3.0 atm. Cal­ culate the solubility of O2 in water under 3.0 atm pressure. Then, calculate the mass of oxygen that will dissolve in 1.0 L of water under this pressure. 3. A 1.0 L sample of water is shaken in air to allow it to become saturated with both N2 and O2. What is the total amount of dissolved gases? If the solution was now heated and the dis­ solved gases expelled from solution and collected, what is the mole fraction of oxygen in the gas mixture?

Applying Chemical Principles

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Think–Pair–Share 1. Why is molality used instead of molarity for freezing point depression and boiling point elevation experiments? 2. The mass fraction of solute in a solution can be defined as

Mass fraction of solute 

mass of solute (g) mass of solution (g)

How is this unit related to the concentrations expressed as a weight percent, in parts per million, and in parts per billion? 3. What would you observe if you add additional solute to each of the following types of solutions, given sufficient time and mixing?

(a) an unsaturated solution (b) a saturated solution (c) a supersaturated solution 4. For each part, determine whether the pair of liquids is expected to be miscible. Explain each of your answers.

(a) CH3OH and H2O (b) Br2 and CH3CH2CH2CH2CH2CH3 (c) CH3CH2CH2CH2CH2CH3 and H2O 5. Assume that RbCl is a solid for which predictions relating the  enthalpy of solution and the effect of temperature on solu­bility using Le Chatelier’s principle work. If ∆solnH° = +17.1  kJ/mol for RbCl, predict whether RbCl should be more soluble or less soluble if the temperature is increased. Explain your answer. 6. Arrange the following aqueous solutions in order of (i) increas­ ing boiling point and (ii) increasing vapor pressure of water. (Assume that all the solutes are nonvolatile and that any ionic solutes dissociate completely.) Explain your answers. (a) 0.50 m sucrose (C12H22O11) (b) 0.75 m sodium chloride (c) 0.75 m magnesium nitrate (d) 1.0 m glucose (C6H12O6)

Chapter Goals Revisited The goals for this chapter are keyed to specific Study Questions to help you organize your review.

13.1  Units of Concentration • Calculate and use the concentration units: molality, mole fraction, weight percent, and parts per million (ppm). 1, 3, 5, 7, 11, 12.

• Recognize the difference between molarity and molality. 9, 10.

13.2  The Solution Process • Understand the process of dissolving a solute in a solvent and recognize the terminology used (saturated, unsaturated, supersaturated, solubility, miscible). 13, 14, 17, 18, 104, 105.

• Understand the thermodynamics associated with the solution process and calculate the enthalpy of solution from thermodynamic data. 15, 16, 81.

13.3 Factors Affecting Solubility: Pressure and Temperature • Describe the effects of pressure and temperature on the solubility of a solute. 17, 18.

• Use Henry’s law to calculate the solubility of a gas in a solvent. 19–22, 91, 92. • Apply Le Chatelier’s principle to predict the change in solubility of gases with temperature changes. 23, 24.

13.4  Colligative Properties • Using Raoult’s law, calculate the effect of dissolved solutes on solvent vapor pressure (Psolvent). 25–28, 83, 84.

• Calculate the effect of a solute on the boiling point and freezing point of a solvent. 29–36.

• Calculate the osmotic pressure (Π) for solutions. 39–42, 88, 116.

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Chapter 13 / Solutions and Their Behavior

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• Use colligative properties to determine the molar mass of a solute. 43–50, 74, 75.

• Use the van’t Hoff factor in colligative property calculations involving ionic solutes. 51–56, 71, 85, 103.

13.5 Colloids • Recognize the properties and importance of colloids. 57–60, 93, 107.

Key Equations Equation 13.1 (page 628) Molality is defined as the amount of solute per kilogram of solvent. Concentration (c , mol/kg)  molality of solute 

amount of solute (mol) mass of solvent (kg)

Equation 13.2 (page  628) The mole fraction, X, of a solution component (A) is defined as the number of moles of that component (nA, mol) divided by the total number of moles of all of the components of the mixture. Mole fraction of A (X A ) 

nA nA  nB  nC  …

Equation 13.3 (page 629) Weight percent is the mass of one component divided by the total mass of the mixture (multiplied by 100%). Weight % A 

mass of A  100% mass of A  mass of B  mass of C  …

Equation 13.4 (page  637) Henry’s law: The solubility of a gas, Sg, is equal to the

product of the partial pressure of the gaseous solute (Pg) and a constant (kH) characteristic of the solute and solvent. Sg = kHPg

Equation 13.5 (page 640) Raoult’s law: The equilibrium vapor pressure of a solvent over a solution at a given temperature, Psolvent, is the product of the mole fraction of the solvent (Xsolvent) and the vapor pressure of the pure solvent (P°solvent). Psolvent = Xsolvent P°solvent

Equation 13.6 (page 641) The decrease in the vapor pressure of the solvent over a solution, ∆Psolvent, depends on the mole fraction of the solute (Xsolute) and the vapor pressure of the pure solvent (P°solvent). ∆Psolvent = −Xsolute P°solvent

Equation 13.7 (page 643) The elevation in boiling point of the solvent in a solution, ∆Tbp, is the product of the molality of the solute, msolute, and a constant characteristic of the solvent, Kbp. Elevation in boiling point = ∆Tbp = Kbpmsolute

Equation 13.8 (page 645) The depression of the freezing point of the solvent in a solution, ∆Tfp, is the product of the molality of the solute, msolute, and a constant characteristic of the solvent, Kfp. Freezing point depression = ∆Tfp = Kfpmsolute

Equation 13.9 (page 649) The osmotic pressure, ∏, is the product of the solute con-

centration c (in mol/L), the universal gas constant R (0.082057 L ∙ atm/K ∙ mol), and the temperature T (in kelvins). ∏ = cRT

Key Equations

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Equation 13.10 (page  654) This modified equation for freezing point depression accounts for the possible dissociation of a solute. The van’t Hoff factor, i, the ratio of the measured freezing point depression and the freezing point depression calculated assuming no solute dissociation, is related to the relative number of particles produced by a dissolved solute. Tfp measured  K fp mi

Study Questions ▲ denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully worked solutions in the Student Solutions Manual.

Practicing Skills Concentration (See Section 13.1 and Example 13.1.) 1. You dissolve 1.50 g of succinic acid, C2H4(CO2H)2, in 500. mL of water. Assuming that the density of water is 1.00 g/cm3, calculate the molality, mole fraction, and weight percent of succinic acid in the solution. 2. You dissolve 20.0 g of camphor, C10H16O, in 250. mL of ethanol, C2H5OH. Calculate the molality, mole fraction, and weight percent of camphor in this solution. (The density of ethanol is 0.785 g/mL.) 3. Fill in the blanks in the table. Aqueous solutions are assumed. Compound NaI

Molality

Mole Fraction

0.15

C2H5OH C12H22O11

Weight Percent 5.0

Molality

NaNO3 CH3CO2H HOCH2CH2OH

Weight Percent

Mole Fraction

10.0 0.183 15.0

5. What mass of Na2CO3 must you add to 125 g of water to prepare 0.200 m Na2CO3? What is the mole fraction of Na2CO3 in the resulting solution? 6. What mass of KNO3 must be added to 500. g of water to prepare a solution that is 0.0512 m in KNO3? What is the mole fraction of KNO3 in the solution?

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8. You want to prepare an aqueous solution of ethylene glycol, HOCH2CH2OH, in which the mole fraction of solute is 0.125. What mass of ethylene glycol, in grams, should you combine with 955 g of water? What is the molality of the solution? 9. Hydrochloric acid is sold as a concentrated aqueous solution. If the concentration of commercial HCl is 12.0 M and its density is 1.18 g/cm3, calculate the following: (a) the molality of the solution (b) the weight percent of HCl in the solution (c) the mole fraction of HCl in the solution 10. Concentrated sulfuric acid has a density of 1.84 g/cm3 and is 95.0% by weight H2SO4. What is the molality of this acid? What is its molarity? 11. The average lithium ion concentration in seawater is 0.18 ppm. What is the molality of Li+ in seawater?

0.15

4. Fill in the blanks in the table. Aqueous solutions are assumed. Compound

7. You wish to prepare an aqueous solution of glycerol, C3H5(OH)3, in which the mole fraction of the solute is 0.090. What mass of glycerol must you add to 425 g of water to make this solution? What is the molality of the solution?

12. Silver ion has an average concentration of 28 ppb (parts per billion) in U.S. water supplies. (a) What is the molality of the silver ion? (b) If you wanted 1.0 × 102 g of silver and could recover it chemically from water supplies, what volume of water in liters would you have to treat? (Assume the density of water is 1.0 g/cm3.)

The Solution Process (See Section 13.2 and Example 13.2.) 13. Which pairs of liquids are miscible? (a) H2O and CH3CH2CH2CH2CH3 (b) C6H6 (benzene) and CCl4 (c) H2O and CH3CO2H 14. Acetone (CH3COCH3, structure shown here) is quite soluble in water. Explain why.

Chapter 13 / Solutions and Their Behavior

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21. An unopened soda can has an aqueous CO2 concentration of 0.0506 m at 25 °C. What is the pressure of CO2 gas in the can?

O H3CCCH3 Acetone

15. Use the data in Table 13.1 to calculate the enthalpy of solution for (a) NaF and (b) NH4NO3. 16. Use the following data to calculate the enthalpy of solution for sodium perchlorate, NaClO4: ΔfH°(s) = −382.9 kJ/mol

and

Le Chatelier’s Principle ΔfH°(aq, 1 m) = −369.5 kJ/mol

17. You make a saturated solution of NaCl at 25 °C. No solid is present in the beaker holding the ­solution. What can be done to increase the amount of dissolved NaCl in this solution? (See Figure 13.11.) (a) Add more solid NaCl. (b) Raise the temperature of the solution. (c) Raise the temperature of the solution, and add some NaCl. (d) Lower the temperature of the solution, and add some NaCl. 18. Some lithium chloride, LiCl, is dissolved in 100 mL of water in one beaker, and some Li2SO4 is dissolved in 100 mL of water in another beaker. Both are at 10 °C, and both are saturated solutions; some solid remains undissolved in each beaker. Describe what you would observe as the temperature is raised. The following data are ­available to you from a handbook of chemistry: Solubility (g/100 mL) Compound

10 °C

40 °C

Li2SO4

35.5

33.7

LiCl

74.5

89.8

Henry’s Law (See Section 13.3 and Example 13.3.) 19. The partial pressure of O2 in your lungs varies from 25 mm Hg to 40 mm Hg. What mass of O2 can dissolve in 1.0 L of water at 25 °C if the partial pressure of O2 is 40 mm Hg? 20. The Henry’s law constant for O2 in water at 25 °C is given in Table 13.2. Which of the following is a reasonable constant when the temperature is 50 °C? Explain the reason for your choice. (a) 6.7 × 10−4 mol/kg ∙ bar (b) 2.6 × 10−3 mol/kg ∙ bar (c) 1.3 × 10−3 mol/kg ∙ bar (d) 6.4 × 10−2 mol/kg ∙ bar

22. Hydrogen gas has a Henry’s law constant of 7.8 × 10−4 mol/kg ∙ bar at 25 °C when dissolving in water. If the total pressure of gas (H2 gas plus water vapor) over water is 1.00 bar, what is the concentration of H2 in the water in grams per ­milliliter? (See Appendix G for the vapor pressure of water.) (See Section 13.3.) 23. A sealed flask contains water and oxygen gas at 25 °C. The O2 gas has a partial pressure of 1.5 atm. (a) What is the concentration of O2 in the water? (b) If the pressure of O2 in the flask is raised to 1.7 atm, what happens to the amount of ­dissolved O2? What happens to the amount of dissolved O2 when the pressure of O2 gas drops to 1.0 atm? 24. Butane, C4H10, was suggested as the refrigerant in household compressors such as those found in air conditioners. (a) To what extent is butane soluble in water? Calculate the butane concentration in water if the pressure of the gas is 0.21 atm. (kH = 0.0011 mol/kg ∙ bar at 25 °C) (b) If the pressure of butane is increased to 1.0 atm, does the butane concentration increase or decrease?

Raoult’s Law (See Section 13.4 and Example 13.4.) 25. A 35.0-g sample of ethylene glycol, HOCH2CH2OH, is dissolved in 500.0 g of water. The vapor pressure of water at 32 °C is 35.7 mm Hg. What is the vapor pressure of the water–ethylene glycol solution at 32 °C? (Ethylene glycol is nonvolatile.) 26. Urea, (NH2)2CO, which is widely used in fertilizers and plastics, is quite soluble in water. If you dissolve 9.00 g of urea in 10.0 mL of water, what is the vapor pressure of the solution at 24 °C? Assume the density of water is 1.00 g/mL. 27. Pure ethylene glycol, HOCH2CH2OH, is added to 2.00 kg of water in the cooling system of a car. The vapor pressure of the water in the system when the temperature is 90 °C is 457 mm Hg. What mass of glycol was added? (Assume the solution is ideal. See Appendix G for the vapor pressure of water.) 28. Pure iodine (105 g) is dissolved in 325 g of CCl4 at 65 °C. Given that the vapor pressure of CCl4 Study Questions

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at this temperature is 531 mm Hg, what is the vapor pressure of the CCl4–I2 solution at 65 °C? (Assume that I2 does not contribute to the vapor pressure.)

Boiling Point Elevation (See Section 13.4 and Example 13.5.) 29. What is the boiling point of a solution containing 0.200 mol of a nonvolatile solute in 125 g of benzene (C6H6)? 30. What is the boiling point of a solution composed of 15.0 g of urea, (NH2)2CO, in 0.500 kg of water? 31. What is the boiling point of a solution composed of 15.0 g of CHCl3 and 0.515 g of the nonvolatile solute acenaphthene, C12H10, a component of coal tar? 32. A solution of glycerol, C3H5(OH)3, in 735 g of water has a boiling point of 104.4 °C at a pressure of 760 mm Hg. What is the mass of glycerol in the solution? What is the mole fraction of the solute?

Freezing Point Depression (See Section 13.4 and Example 13.6.) 33. A mixture of ethanol, C2H5OH, and water has a freezing point of −16.0 °C. (a) What is the molality of the alcohol? (b) What is the weight percent of alcohol in the solution? 34. Some ethylene glycol, HOCH2CH2OH, is added to your car’s cooling system along with 5.0 kg of water. If the freezing point of the water–glycol solution is −15.0 °C, what mass of HOCH2CH2OH must have been added? 35. You dissolve 15.0 g of sucrose, C12H22O11, in a cup of water (225 g). What is the freezing point of the solution? 36. A typical bottle of wine consists of an 11% solution (by weight) of ethanol (C2H5OH) in water. If the wine is chilled to −20 °C, will the solution begin to freeze?

Osmosis and Osmotic Pressure (See Section 13.4 and Example 13.7.) 37. An aqueous 0.01 M solution of sucrose (C12H22O11, a nonvolatile, non-ionic solute) is placed on one side of a semipermeable membrane, and an aqueous 0.1 M solution of sucrose is placed on the other side. In which direction will there be a net movement of water: from the 0.01 M solution to the 0.1 M solution or from the 0.1 M solution to the 0.01 M solution? 38. An aqueous solution is placed on one side of a semipermeable membrane, and pure water is placed on the other side. In which direction

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will there be a net movement of water: from the s­ olution to the pure water or from the water to the solution? 39. A solution is prepared by dissolving 5.00 g of ­glycerol (C3H5(OH)3, a nonvolatile, non-ionic solute) in enough water to make a 250. mL solution. What is the osmotic pressure (in atm) of this solution at 25 °C? 40. What is the osmotic pressure (in mm Hg) of an aqueous 0.0010 M solution of sucrose (C12H22O11, a nonvolatile, non-ionic solute) at 25 °C? 41. An aqueous solution contains 3.00% phenylalanine (C9H11NO2) by mass. (Phenylalanine is ­non-ionic and nonvolatile.) Find the following: (a) the freezing point of the solution (b) the boiling point of the solution (c) the osmotic pressure of the solution at 25 °C In your view, which of these values is most easily measurable in the laboratory? 42. A 500. mL aqueous solution of sucrose (C12H22O11, a nonvolatile, non-ionic solute) has an osmotic pressure of 28.7 mm Hg at 25 °C. What mass of sucrose is dissolved in this solution?

Colligative Properties and Molar Mass Determination (See Section 13.4 and Examples 13.8 and 13.9.) 43. You add 0.255 g of an orange, crystalline compound whose empirical formula is C10H8Fe to 11.12 g of benzene. The boiling point of the benzene rises from 80.10 °C to 80.26 °C. What are the molar mass and molecular formula of the compound? 44. Butylated hydroxyanisole (BHA) is used in margarine and other fats and oils. (It is used as an antioxidant and prolongs the shelf life of the food.) What is the molar mass of BHA if 0.640 g of the compound, dissolved in 25.0 g of chloroform, produces a solution whose boiling point is 62.22 °C? 45. Benzyl acetate is one of the active components of oil of jasmine. If 0.125 g of the compound is added to 25.0 g of chloroform (CHCl3), the boiling point of the solution is 61.82 °C. What is the molar mass of benzyl acetate? 46. Anthracene, a hydrocarbon obtained from coal, has an empirical formula of C7H5. To find its molecular formula, you dissolve 0.500 g in 30.0 g of benzene. The boiling point of pure benzene is 80.10 °C, whereas the solution has a boiling point of 80.34 °C. What is the molecular formula of anthracene? 47. An aqueous solution contains 0.180 g of an unknown, non-ionic solute in 50.0 g of water. The solution freezes at −0.040 °C. What is the molar mass of the solute?

Chapter 13 / Solutions and Their Behavior

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48. Aluminon, an organic compound, is used as a reagent to test for the presence of the aluminum ion in aqueous solution. A solution of 2.50 g of aluminon in 50.0 g of water freezes at −0.197 °C. What is the molar mass of aluminon? 49. An aqueous solution containing 1.00 g of bovine insulin (a protein, not ionized) per liter has an osmotic pressure of 3.1 mm Hg at 25 °C. ­Calculate the molar mass of bovine insulin. 50. The breakdown of starch in the body produces the sugar maltose, which then is broken down further into glucose. Maltose is a nonvolatile, non-ionic solute. An aqueous solution was prepared by dissolving 1.25 g of maltose in enough water to make a 500.-mL solution. If this solution has an osmotic pressure of 136 mm Hg at 25 °C, what is the molar mass of maltose?

Colligative Properties of Ionic Compounds (See Section 13.4 and Example 13.10.) 51. If 52.5 g of LiF is dissolved in 306 g of water, what is the expected freezing point of the solution? (Assume the van’t Hoff factor, i, for LiF is 2.) 52. To make homemade ice cream, you cool the milk and cream by immersing the container in ice and a concentrated solution of rock salt (NaCl) in water. If you want to have a water–salt solution that freezes at −10. °C, what mass of NaCl must you add to 3.0 kg of water? (Assume the van’t Hoff factor, i, for NaCl is 1.83.) 53. List the following aqueous solutions in order of increasing melting point. (The last three are all assumed to dissociate completely into ions in water.) (a) 0.1 m sugar (c) 0.08 m CaCl2 (b) 0.1 m NaCl (d) 0.04 m Na2SO4 54. Arrange the following aqueous solutions in order of decreasing freezing point. (The last three are all assumed to dissociate completely into ions in water.) (a) 0.20 m ethylene glycol (nonvolatile, nonelectrolyte) (b) 0.12 m K2SO4 (c) 0.10 m MgCl2 (d) 0.12 m KBr 55. Estimate the osmotic pressure of human blood at 37 °C. Assume blood is isotonic with a 0.154 M NaCl solution, and assume the van’t Hoff factor, i, is 1.85 for NaCl. 56. Calculate the osmotic pressure of a 0.0120 M ­solution of NaCl in water at 0 °C. Assume the van’t Hoff factor, i, is 1.94 for this solution.

Colloids (See Section 13.5.) 57. When AgCl precipitates from a solution, a colloidal dispersion often initially forms. What type of colloidal dispersion is it (see Table 13.5)? 58. Mayonnaise is a colloidal dispersion of oil in water. What type of colloidal dispersion is it (see Table 13.5)? 59. When solutions of BaCl2 and Na2SO4 are mixed, the mixture becomes cloudy. After a few days, a white solid is observed on the bottom of the beaker with a clear liquid above it. (a) Write a balanced equation for the reaction that occurs. (b) Why is the solution cloudy at first? (c) What happens during the few days of waiting? 60. The dispersed phase of a certain colloidal dispersion consists of spheres of diameter 1.0 × 102 nm. (a) What are the volume (V = 4⁄3πr3) and surface area (A = 4πr2) of each sphere? (b) How many spheres are required to give a total volume of 1.0 cm3? What is the total surface area of these spheres in square meters?

General Questions These questions are not designated as to type or location in the chapter. They may combine several concepts. 61. Phenylcarbinol is used in nasal sprays as a preservative. A solution of 0.52 g of the compound in 25.0 g of water has a melting point of −0.36 °C. What is the molar mass of phenylcarbinol? 62. (a) Which aqueous solution is expected to have the higher boiling point: 0.10 m Na2SO4 or 0.15 m sugar? (b) For which aqueous solution is the vapor pressure of water higher: 0.30 m NH4NO3 or 0.15 m Na2SO4? 63. Arrange the following aqueous solutions in order of (i) increasing vapor pressure of water and (ii) increasing boiling point. (a) 0.35 m HOCH2CH2OH (a nonvolatile solute) (b) 0.50 m sugar (c) 0.20 m KBr (a strong electrolyte) (d) 0.20 m Na2SO4 (a strong electrolyte) 64. Making homemade ice cream is one of life’s great pleasures. Fresh milk and cream, sugar, and ­flavorings are churned in a bucket suspended in an ice–water mixture, the freezing point of which has been lowered by adding salt. One manufacturer of home ice-cream freezers recommends Study Questions

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adding 2.50 lb (1130 g) of salt (NaCl) to 16.0 lb of ice (7250 g) in a 4-quart freezer. The ice will melt, forming a solution. Calculate the following for the resulting solution: (a) the weight percent of NaCl (b) the mole fraction of NaCl (c) the molality of the solution

© Charles D. Winters/Cengage

65. Dimethylglyoxime [DMG, (CH3CNOH)2] is used as a reagent to precipitate nickel ion. Assume that 53.0 g of DMG has been dissolved in 525 g of ethanol (C2H5OH).

The red, insoluble compound formed between nickel(II) ion and dimethylglyoxime (DMG) is precipitated when DMG is added to a basic solution of Ni2+(aq).

(a) What is the mole fraction of DMG? (b) What is the molality of the solution? (c) What is the vapor pressure of the ethanol over the solution at ethanol’s normal boiling point of 78.4 °C? (d) What is the boiling point of the solution? (DMG does not produce ions in solution.) (Kbp for ethanol = +1.22 °C/m) 66. A 10.7 m solution of NaOH has a density of 1.33 g/cm3 at 20 °C. Calculate the following: (a) the mole fraction of NaOH (b) the weight percent of NaOH (c) the molarity of the solution 67. Concentrated aqueous ammonia has a molarity of 14.8 mol/L and a density of 0.90 g/cm3. What is the molality of the solution? Calculate the mole fraction and weight percent of NH3. 68. If you dissolve 2.00 g of Ca(NO3)2 in 750 g of water, what is the molality of Ca(NO3)2? What is the total molality of ions in solution? (Assume total dissociation of the ionic solid.) 69. If you want a solution that is 0.100 m in ions, what mass of Na2SO4 must you dissolve in 125 g of water? (Assume total dissociation of the ionic solid.) 70. Consider the following aqueous solutions: (i) 0.20 m HOCH2CH2OH (nonvolatile,

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nonelectrolyte); (ii) 0.10 m CaCl2; (iii) 0.12 m KBr; and (iv) 0.12 m Na2SO4. (a) Which solution has the highest boiling point? (b) Which solution has the lowest freezing point? (c) Which solution has the highest water vapor pressure? 71. (a) Which solution is expected to have the higher boiling point: 0.20 m KBr or 0.30 m sugar? (b) Which aqueous solution has the lower freezing point: 0.12 m NH4NO3 or 0.10 m Na2CO3? 72. The solubility of NaCl in water at 100 °C is 39 g/100. g of water. Calculate the boiling point of this solution. (Assume i = 1.8 for NaCl.) 73. Instead of using NaCl to melt the ice on your sidewalk, you decide to use CaCl2. If you add 35.0 g of CaCl2 to 150. g of water, what is the freezing point of the solution? (Assume i = 2.7 for CaCl2.) 74. The smell of ripe raspberries is due to 4-(p-hydroxy­ phenyl)-2-butanone, which has the empirical formula C5H6O. To find its molecular formula, you dissolve 0.135 g in 25.0 g of chloroform, CHCl3. The boiling point of the solution is 61.82 °C. What is the molecular formula of the solute? 75. Hexachlorophene has been used in germicidal soap. What is its molar mass if 0.640 g of the compound, dissolved in 25.0 g of chloroform, produces a solution whose boiling point is 61.93 °C? 76. The solubility of ammonium formate, NH4CHO2, in 100. g of water is 102 g at 0 °C and 546 g at 80 °C. A solution is prepared by dissolving NH4CHO2 in 200. g of water until no more will dissolve at 80 °C. The solution is then cooled to 0 °C. What mass of NH4CHO2 precipitates? (Assume that no water evaporates and that the solution is not supersaturated.) 77. How much N2 can dissolve in water at 25 °C if the N2 partial pressure is 585 mm Hg? 78. Cigars are best stored in a humidor at 18 °C and 55% relative humidity. This means the pressure of water vapor should be 55% of the vapor pressure of pure water at the same temperature. The proper humidity can be maintained by placing a solution of glycerol [C3H5(OH)3] and water in the humidor. Calculate the percent by mass of glycerol that will lower the vapor pressure of water to the desired value. (The vapor pressure of glycerol is negligible.) 79. An aqueous solution containing 10.0 g of starch per liter has an osmotic pressure of 3.8 mm Hg at 25 °C.

Chapter 13 / Solutions and Their Behavior

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(a) What is the average molar mass of starch? (The result is an average because not all starch molecules are identical.) (b) What is the freezing point of the solution? Would it be easy to determine the molecular weight of starch by measuring the freezing point depression? (Assume that the molarity and molality are the same for this solution.)

(b) The density of the solution is 1.014 g/mL. ­Calculate the osmotic pressure of the solution. 87.

80. Vinegar is a 5% solution (by weight) of acetic acid in water. Determine the mole fraction and molality of acetic acid. What is the concentration of acetic acid in parts per million (ppm)? Explain why it is not possible to calculate the molarity of this solution from the information provided. 81. Calculate the enthalpies of solution for Li2SO4 and K2SO4. Are the solution processes exothermic or endothermic? Compare them with LiCl and KCl. What similarities or differences do you find?

Compound

𝚫fH°(s) (kJ/mol)

𝚫fH°(aq, 1 m) (kJ/mol)

Li2SO4

−1436.4

−1464.4

K2SO4

−1437.7

−1413.0

82.

▲

83.

▲ If a volatile solute is added to a volatile solvent, both substances contribute to the vapor pressure over the solution. Assuming an ideal solution, the vapor pressure of each is given by Raoult’s law, and the total vapor pressure is the sum of the vapor pressures for each component. A solution, assumed to be ideal, is made from 1.0 mol of toluene (C6H5CH3) and 2.0 mol of benzene (C6H6). The vapor pressures of the pure solvents are 22 mm Hg and 75 mm Hg, respectively, at 20 °C. What is the total vapor pressure of the mixture? What is the mole fraction of each component in the liquid and in the vapor?

Water at 25 °C has a density of 0.997 g/cm3. Calculate the molality and molarity of pure water at this temperature.

84. A solution is made by adding 50.0 mL of ethanol (C2H5OH, d = 0.789 g/mL) to 50.0 mL of water (d = 0.998 g/mL). What is the total vapor pressure over the solution at 20 °C? (See Study ­Question 83.) The vapor pressure of ethanol at 20 °C is 43.6 mm Hg. 85. A 2.0% (by mass) aqueous solution of novocainium chloride (C13H21ClN2O2) freezes at −0.237 °C. Calculate the van’t Hoff factor, i. How many moles of ions are in the solution per mole of compound? 86. A solution is 4.00% (by mass) maltose and 96.00% water. It freezes at −0.229 °C. (a) Calculate the molar mass of maltose (which is not an ionic compound).

▲ The following table lists the concentrations of the principal ions in seawater:

Ion

Concentration (ppm)

Cl−

1.95 × 104

Na+

1.08 × 104

Mg2+

1.29 × 103

SO42−

9.05 × 102

Ca2+

4.12 × 102

K+

3.80 × 102

Br−

67

(a) Calculate the freezing point of seawater. (b) Calculate the osmotic pressure of seawater at 25 °C. What is the minimum pressure needed to purify seawater by reverse osmosis? 88.

▲

A tree is 10.0 m tall. (a) What must be the total molarity of the solutes if sap rises to the top of the tree by osmotic pressure at 20 °C? Assume the groundwater outside the tree is pure water and that the density of the sap is 1.0 g/mL. (1 mm Hg = 13.6 mm H2O.) (b) If the only solute in the sap is sucrose, C12H22O11, what is its percent by mass?

89. A 2.00% solution of H2SO4 in water freezes at −0.796 °C. (a) Calculate the van’t Hoff factor, i. (b) Which of the following best represents sulfuric acid in a dilute aqueous solution: H2SO4, H3O+ + HSO4−, or 2 H3O+ + SO42−? 90. A compound is known to be a potassium halide, KX. If 4.00 g of the salt is dissolved in exactly 100 g of water, the solution freezes at −1.28 °C. Identify the halide ion in this formula. 91. Nitrous oxide, N2O, laughing gas, is used as an anesthetic. Its Henry’s law constant is 2.4 × 10−2 mol/ kg ∙ bar. Determine the mass of N2O that will dissolve in 500. mL of water, under an N2O pressure of 1.00 bar. What is the concentration of N2O in this solution, expressed in ppm (d (H2O) = 1.00 g/mL)? 92. If a carbonated beverage is bottled under 1.5 bar CO2 pressure, what will be the concentration of dissolved CO2 in that beverage? (kH for CO2 is 0.034 mol/kg bar.) After the pressure is released, what fraction of the dissolved gas will escape before equilibrium with the CO2 in the atmosphere is reached? 93. You are given a flask filled with a colored liquid. Suggest several tests that would allow you to determine whether this is a solution or a colloid. Study Questions

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94. If one is very careful, it is possible to float a needle on the surface of water. (If the needle is magnetized, it will turn to point north and south and become a makeshift compass.) What would happen to the needle if a drop of liquid soap is added to the solution? Explain the observation.

In the Laboratory 95.

▲ A solution of benzoic acid in benzene has a freezing point of 3.1 °C and a boiling point of 82.6 °C. (The freezing point of pure benzene is 5.50 °C, and its boiling point is 80.1 °C.) The structure of benzoic acid is

O C

OH

Benzoic acid, C6H5CO2H

What can you conclude about the state of the benzoic acid molecules at the two different temperatures? Recall the discussion of hydrogen bonding in Section 11.3. 96.

97.

98.

▲

You dissolve 5.0 mg of iodine, I2, in 25 mL of water. You then add 10.0 mL of CCl4 and shake the mixture. If I2 is 85 times more soluble in CCl4 than in H2O (on a volume basis), what are the masses of I2 in the water and CCl4 layers after shaking? (Figure 13.5.) ▲ A solution of 5.00 g of acetic acid in 100. g of benzene freezes at 3.37 °C. A solution of 5.00 g of acetic acid in 100. g of water freezes at −1.49 °C. Find the molar mass of acetic acid from each of these experiments. What can you conclude about the state of the acetic acid molecules dissolved in each of these solvents? Recall the discussion of hydrogen bonding in Section 11.3, and propose a structure for the species in benzene solution. ▲

In a police forensics lab, you examine a package that may contain heroin. However, you find the white powder is not pure heroin but a mixture of heroin (C21H23O5N) and lactose (C12H22O11). To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volumetric flask. You find that the solution has an osmotic pressure of 539 mm Hg at 25 °C. What is the composition of the mixture?

H by burning the compound and collecting the evolved CO2 and H2O. They might then determine the molar mass by measuring the osmotic pressure of a solution of the compound. Calculate the empirical and molecular formulas of a compound, CxHyCr, given the following information: (a) The compound contains 73.94% C and 8.27% H; the remainder is chromium. (b) At 25 °C, the osmotic pressure of a solution containing 5.00 mg of the unknown dissolved in exactly 100 mL of chloroform solution is 3.17 mm Hg.

Summary and Conceptual Questions The following questions may use concepts from this and previous chapters. 101. When salts of Mg2+, Ca2+, and Be2+ are placed in water, the positive ion is hydrated (as is the negative ion). Which of these three cations is most strongly hydrated? Which one is least strongly hydrated? 102. Explain why a cucumber shrivels up when it is placed in a concentrated solution of salt. 103. If you dissolve equal molar amounts of NaCl and CaCl2 in water, the CaCl2 lowers the freezing point of the water almost 1.5 times as much as the NaCl. Why? 104. A 100.-gram sample of sodium chloride (NaCl) is added to 100. mL of water at 0 °C. After equilibrium is reached, about 64 g of solid remains undissolved. Describe the equilibrium that exists in this system at the particulate level. 105. Which of the following substances is/are likely to dissolve in water, and which is/are likely to ­dissolve in benzene (C6H6)? (a) NaNO3 (b) diethyl ether, CH3CH2OCH2CH3 (c) NH4Cl (d) naphthalene, C10H8 H

H

H

H

H

H H

H

99. An organic compound contains carbon (71.17%), hydrogen (5.12%), and the remainder nitrogen. Dissolving 0.177 g of the compound in 10.0 g of benzene gives a solution with a vapor pressure of 94.16 mm Hg at 25 °C. (The vapor pressure of pure benzene at this temperature is 95.26 mm Hg.) What is the molecular formula for the compound?

106. Account for the fact that alcohols such as methanol (CH3OH) and ethanol (C2H5OH) are quite miscible with water, whereas an alcohol with a long carbon chain, such as octanol (C8H17OH), is poorly soluble in water.

100. Chemists often send newly synthesized compounds to commercial laboratories for analysis. These laboratories may determine the weight percent of C and

107. Starch contains COC, COH, COO, and OOH bonds. Hydrocarbons have only COC and COH bonds. Both starch and hydrocarbons can form

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Naphthalene, C10H8

Chapter 13 / Solutions and Their Behavior

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colloidal dispersions in water. Which dispersion is classified as hydrophobic? Which is hydrophilic? Explain briefly. 108. Which substance would have the greater influence on the vapor pressure of water when added to 1000. g of water: 10.0 g of sucrose (C12H22O11) or 10.0 g of ethylene glycol (HOCH2CH2OH)? 109. You have two aqueous solutions separated by a semipermeable membrane. One contains 5.85 g of NaCl dissolved in 100. mL of solution, and the other contains 8.88 g of KNO3 dissolved in 100. mL of solution. In which direction will solvent flow: from the NaCl solution to the KNO3 solution, or from KNO3 to NaCl? Explain briefly. 110. A protozoan (single-celled animal) that normally lives in the ocean is placed in freshwater. Will it shrivel or burst? Explain briefly. 111. ▲ In the process of distillation, a mixture of two (or more) volatile liquids is first heated to convert the volatile materials to the vapor state. Then the vapor is condensed, reforming the liquid. The net result of this liquid n vapor n liquid conversion is to enrich the fraction of the more volatile component in the mixture in the condensed vapor. This can be explained using Raoult’s law. Imagine that you have a mixture of 12% (by weight) ethanol and water (as formed, for example, by the fermentation of grapes). (a) What are the mole fractions of ethanol and water in this mixture? (b) This mixture is heated to 78.5 °C (the normal boiling point of ethanol). What are the equilibrium vapor pressures of ethanol and water at this temperature, assuming Raoult’s law (ideal) behavior? (You will need to derive the equilibrium vapor pressure of water at 78.5 °C from data in Appendix G.) (c) What are the mole fractions of ethanol and water in the vapor? (d) After this vapor is condensed to a liquid, to what extent has the mole fraction of ethanol been enriched? What is the mass fraction of ethanol in the condensed vapor? 112. Sodium chloride (NaCl) is commonly used to melt ice on roads during the winter. Calcium chloride (CaCl2) is sometimes used for this purpose too. Compare the effectiveness of equal masses of these two compounds in lowering the freezing point of water, by calculating the freezing point depression of solutions containing 200. g of each salt in 1.00 kg of water. (An advantage of CaCl2 is that it acts more quickly because it is hygroscopic, that is, it absorbs moisture from the air to give a solution and begin the process. A disadvantage is that this compound is more costly.)

113. Review the trend in values of the van’t Hoff factor i as a function of concentration (Table 13.4). Use the following data to calculate the van’t Hoff factor for a NaCl concentration of 5.00% (by weight) (for which ΔT = −3.05 °C) and a Na2SO4 concentration of 5.00% (by weight) (for which ΔT = −1.36 °C). Are these values in line with your expectations based on the trend in the values given in Table 13.4? Speculate on why this trend is seen. 114. The table below gives experimentally determined values for freezing points of 1.00% solutions (mass %) of a series of acids. (a) Calculate the molality of each solution, determine the calculated freezing points, and then calculate the values of the van’t Hoff factor i. Fill these values into the table.

Acid (1.00 mass %)

Molality (mol/kg H2O)

Tmeasured (°C)

HNO3

−0.56

CH­3CO2H

−0.32

H2SO4

−0.42

H2C2O4

−0.30

HCO2H

−0.42

CCl3CO2H

−0.21

Tcalculated (°C)

i

(b) Analyze the results, comparing the values of i for the various acids. How do these data relate to acid strengths? (The discussion of strong and weak acids in Section 3.6 will assist you to answer this question.) 115. It is interesting how the Fahrenheit temperature scale was established. One report, given by Fahrenheit in a paper in 1724, stated that the value of 0 °F was established as the freezing temperature of saturated solutions of sea salt. In the chemical literature, it is reported that the freezing point of a 20% by mass solution of NaCl is −16.46 °C. (This is the lowest freezing temperature reported for solutions of NaCl.) Does this value lend credence to this story of how the Fahrenheit scale was established? 116. The osmotic pressure exerted by seawater at 25 °C is about 27 atm. Calculate the concentration of ions dissolved in seawater that is needed to give an osmotic pressure of this magnitude. (Desalinization of seawater is accomplished by reverse osmosis. In this process, an applied pressure forces water through a membrane against a concentration gradient. The minimum external force needed for this process is 27 atm. Actually, to accomplish the process at a reasonable rate, the applied pressure needs to be about twice this value.)

Study Questions

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14

Chemical Kinetics: The Rates of ­Chemical Reactions

Fundamental Photographs

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C hapt e r O ut li n e 14.1

Rates of Chemical Reactions

14.2

Reaction Conditions and Rate

14.3

Effect of Concentration on Reaction Rate

14.4

Concentration–Time Relationships: Integrated Rate Laws

14.5

A Microscopic View of Reaction Rates

14.6 Catalysts 14.7

Reaction Mechanisms

­ acroscopic Chemical kinetics is the study of the rates of chemical reactions. On the m level, the field of chemical kinetics addresses what reaction rate means, how to determine a reaction rate experimentally, and how factors such as temperature and reactant concentrations influence rates. At the particulate level, the concern is with the reaction mechanism, the detailed pathway taken by atoms and molecules as a reaction proceeds. The goal is to use data from the macroscopic world of chemistry to understand how chemical reactions occur at the particulate level—and then to apply this information to control important reactions.

14.1 Rates of Chemical Reactions Goals for Section 14.1 • Calculate the average rate of a reaction from concentration–time data. • Relate the rates for the disappearance of reactants and formation of products for a chemical reaction.

You have encountered the concept of rates in your daily lives. The speed of an automobile is the distance traveled per unit time (for example, kilometers per hour) and the rate of water flowing from a faucet is the volume delivered per unit time (perhaps liters per minute). In each case, a change is measured over an interval of time. Similarly, the rate of a chemical reaction refers to the change in concentration of a reactant or product per unit of time. Reaction rate 

change in concentration change in time

◀ Catalysts speed up chemical reactions.  This chapter is about chemical kinetics, the study of

reaction rates. Factors that can affect reaction rates include concentration, temperature, and the presence of a catalyst. Catalysts are substances that speed up a chemical reaction without being permanently changed in the process. Biological catalysts are called enzymes. In this photo, ground beef liver that contains the enzyme catalase is added to a solution of hydrogen peroxide in a beaker. The enzyme speeds up the decomposition of hydrogen peroxide into water and oxygen gas.

673

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Photos: © Charles D. Winters/Cengage

A few drops of blue food dye are added to water, followed by a solution of bleach.

The color fades as the dye reacts with the bleach.

Figure 14.1  An experiment to measure rate of reaction.  The absorbance of the solution can be measured at various times using a spectrophotometer (see Section 4.9), and these values can be used to determine the concentration of the dye.

Calculating a Rate The average speed of an automobile is the distance traveled divided by the time elapsed, or ∆(distance)/∆(time). If an automobile travels 3.9 km in 4.5 min (0.075 h), its average speed is (3.9 km/0.075 h), or 52 km/h. Average rates of ­chemical reactions can be determined similarly. Two quantities, concentration and time, are measured. Concentrations can be determined several ways, such as by measuring the absorbance of light by a solution, a property that can be related to the concentration of a species in solution (Figure 14.1). The average rate of the ­reaction is the change in the concentration per unit time. Consider the decomposition of N2O5 in a solvent. This reaction occurs according to the following equation: N2O5 n 2 NO2 + 1⁄2 O2

Concentrations and time elapsed for a typical experiment done at 30.0 °C are illustrated by the graph in Figure 14.2. 2 NO2 + ½ O2

N2O5

1.40 1.30

Average rate, 40 min to 55 min 0.12 mol/L = 0.0080 mol/L • min 15 min

15 min to decrease [N2O5] from 1.22 to 1.10

1.20

Rate =

1.10

Instantaneous rate when [N2O5] = 0.34 M −∆[N2O5] Rate = ∆t 0.22 mol/L − 0.42 mol/L = − 380 min – 240 min = 0.0014 mol/L • min

1.00 [N2O5], mol/L

0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20

Average rate, 390 min to 540 min 0.12 mol/L Rate = = 0.00080 mol/L • min 150 min

0.42 ∆[N2O5] ∆t

0.22

0.10 0

0

60

120

180

240

300 360 420 Time (t), minutes

480

540

600

660

Figure 14.2  A plot of reactant concentration versus time for the decomposition of N2O5.

674

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Chemistry in Your Career

Alex Matsche

Alex Matsche Alex Matsche completed his B.S. in chemistry at Gallaudet University, the only university in the world dedicated to Deaf education specifically. He began his career as a medicinal chemist in the healthcare industry but was soon inspired to start his own business selling CBD products. Matsche began with only three products, but through his research and development efforts he now offers 20. He uses chemistry in his work every day. “I basically can do many things in the company such as extraction

from raw plants to collect specific chemical compounds, eliminating some impurities, developing products (ranging from dietary supplements, sport medicine, pet supplements, and skin care products), calculating the correct dosages, analyzing data from third party labs to verify compounds and dosages, and accurate product labeling.” Matsche enjoys working in the lab developing new products and is proud of what he has accomplished as a Deaf chemist and business owner.

The average rate of this reaction for any interval of time can be expressed as the negative of the change in N2O5 concentration divided by the change in time: Rate of decomposition of N2O5  

change in [N2O5 ]  [N2O5 ]  t change in time

The negative sign is required because N 2O5 is a reactant and its concentration decreases with time (that is, ∆[N2O5] = [N2O5]final − [N2O5]initial is negative), but rate is always expressed as a positive quantity. For example, using data from ­Figure 14.2, the average rate of disappearance of N2O5 between 40 and 55 minutes is given by Rate  

0.12 mol/L  [N2O5 ] (1.10 mol/L)  (1.22 mol/L)   15 min t 55 min  40 min

Calculating Changes When

Rate  0.0080 mol N2O5 consumed/L  min

Note the units for reaction rates; if concentration is expressed in mol/L, the units for rate will be mol/(L ∙ time). Also notice that the average rate for the decomposition of N2O5 is the negative of the slope of a line that connects the two points of interest on the concentration–time graph. The rate of the reaction decreases during the course of the reaction. You can verify this by comparing the average rate of N2O5 disappearance calculated previously (when the concentration decreased by 0.12 mol/L in 15 minutes) to the average rate of reaction calculated for the time interval from 390 min to 540 min (when the concentration drops by 0.12 mol/L in 150 minutes). The average rate in this later stage of this reaction is only one-tenth of the previous value. 

calculating a change in a quantity, always subtract the initial quantity from the final quantity: Δc = cfinal − cinitial.

 [N2O5 ] (0.10 mol/L)  (0.22 mol/L) 0.12 mol/L   t 540 min  3900 min 150 min  0.00080 mol/L  min

You might also ask what the instantaneous rate is at a single point in time. In an automobile, the instantaneous rate can be read from the speedometer. For a chemical reaction, you can determine the instantaneous rate from the concentration–time graph by drawing a line tangent to the concentration–time curve at a particular time and determining its slope (Figure 14.2). For example, when [N2O5] = 0.34 mol/L and t = 300 min, the rate is Rate when [N2O5 ] is 0.34 M  

 [N2O5 ] 0.20 mol/L  t 140 min

 0.0014 mol/L  min

At that particular moment in time (t = 300 min), N2O5 is consumed at a rate of 0.0014 mol/L ∙ min.

14.1  Rates of Chemical Reactions

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675

E xamp le 14.1

Rate of Reaction Problem   Data collected on the concentration of a dye as a function of time are given in the following graph. Using these data, estimate the value of (a) the average rate of change in the dye concentration over the first 2.0 minutes, and (b) the average rate of change during the fifth minute (from t = 4.0 min to t = 5.0 min). Dye concentration (×105) (mol/L)

3.50 3.00 2.50 2.00 1.50 1.00 0.50 0.00

0

1

2

3

4

5

6

7

8

Time (min)

What Do You Know?  The concentration of dye as a function of time is presented as a graph. From this curve you can identify the concentration of dye at a specific time. Notice from the label of the y-axis that the concentrations plotted were multiplied by 105. The point plotted at t = 0 min thus corresponds to a concentration of 3.4 × 10–5 mol/L. Strategy   To find the average rate, calculate the difference in concentration at the beginning and end of a time period (∆c = cfinal − cinitial) and divide by the elapsed time. The negative of this value is the average rate.

Solution  (a) Average rate over first 2  min: The concentration of dye decreases from 3.4 × 10−5 mol/L at t = 0 min to 1.7 × 10−5 mol/L at t = 2.0 min. The average rate of the reaction in this interval of time is Average rate = 

 [Dye ] (1.7  105 mol/L)  (3.4  105 mol/L)  2.0 min t =  8.5 × 10−6 mol/L ∙ min 

(b) Average rate during the fifth minute: The concentration of dye decreases from 0.90 × 10−5 mol/L at t = 4.0 min to 0.60 × 10−5 mol/L at t = 5.0 min. The average rate of the reaction in this interval of time is Concentration (mol/L)

0.05

Average rate = 

0.04



0.03

=  3.0  × 10−6 mol/L ∙ min 

Think about Your Answer  Notice that the rate of reaction decreases as the concentration of dye decreases. This tells you that the rate of the reaction is related to the concentration of dye.

0.02 0.01 0.00

 [Dye ] (0.60  105 mol/L)  (0.90  105 mol/L)  1.0 min t

0

2

4 6 Time (hours)

Concentration versus time for the decom­position of sucrose. ​ (“Check Your Understanding” in Example 14.1.)

676

8

Check Your Understanding  Sucrose decomposes to fructose and glucose in acid solution. A plot of the concentration of sucrose as a function of time is given in the margin. What is the rate of change of the sucrose concentration over the first 2.0 hours? What is the rate of change over the last 2.0 hours?

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Relative Rates and Stoichiometry During a chemical reaction, amounts of reactants decrease with time, and amounts of products increase. So far, for the decomposition of N2O5, the rate has been determined from the disappearance of the reactant, −∆[N2O5]/∆t, but you can also express the rate of appearance of products, either as ∆[NO2]/∆t or as ∆[O2]/∆t. Rates based on changes in concentrations of products do not require the negative sign in the ∆(concentration)/∆t expression because the concentration of a product increases, and the expression will already have a positive value. Furthermore, the numerical values of rates defined in terms of NO2 or O2 will be different from the value of –∆[N2O5]/∆t due to the different stoichiometric coefficients for the substances in the chemical equation. For example, the rate of appearance of NO2 will be twice the rate of disappearance of N2O5 because the stoichiometry requires two NO2 molecules to be formed for each N2O5 molecule consumed. To take the stoichiometry into account when defining the rate of a reaction, include a factor for each reactant and product of 1/x where x is the substance’s stoichiometric coefficient. This produces a single value of the reaction rate regardless of the reactant or product being monitored. For the general reaction aA+bBncC+dD

the reaction rate is defined as Reaction rate  

1  [ A] 1  [B ] 1  [C ] 1  [D ]    a t b t c t d t

Note that the reaction rate is now on a per mole of reaction basis. In the example above of the decomposition of N2O5 reaction [N2O5(g) n 2 NO2(g) + 1⁄2 O2(g)], the rate expression that relates all the reactants and products is Reaction rate  

 [ O2 ]  [N2O5 ] 1  [NO2 ]  2  t t 2 t

Ex am p le 14.2

Relative Rates and Stoichiometry Problem   Relate the rates for the disappearance of reactants and formation of prod-

Mole of Reaction  One mole of

reaction is said to occur when the reaction has occurred according to the number of moles given by the coefficients in the chemical equation (Chapter 4, page 200). For the chemical equation 2 H2(g) + O2(g) n 2 H2O(ℓ), 1 mol of reaction has occurred when 2 mol of hydrogen gas react with 1 mol of oxygen gas to form 2 mol of liquid water.

ucts for the following reaction: 4 PH3(g) n P4(g) + 6 H2(g)

What Do You Know?   The stoichiometric coefficients in the balanced equation can be used to evaluate the relative rates for the disappearance of the starting material and formation of the products.

Strategy   In this reaction, PH3 disappears, and P4 and H2 are formed. Consequently, the value of ∆[PH3]/∆t will be negative, whereas ∆[P4]/∆t and ∆[H2]/∆t will be positive. To relate the rates to each other, divide each ∆[reagent]/∆t by its stoichiometric coefficient in the balanced equation. Solution  Because 4 mol of PH3 disappear for every 1 mol of P4 formed, the numerical value of the rate of formation of P4 is one-fourth of the rate of disappearance of PH3. Similarly, P4 is formed at only one-sixth of the rate that H2 is formed. Reaction rate  

1   [PH3 ]  1   [H2 ]   [P4 ]    4  t  t 6  t 

Think about Your Answer  In determining the rate of a chemical reaction, you should take into account (1) whether a substance is a reactant or product and (2) the stoichiometric coefficient of the substance in the balanced chemical equation for the reaction.

14.1  Rates of Chemical Reactions

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677

Check Your Understanding  What are the relative rates of appearance or disappearance of each product and reactant in the decomposition of nitrosyl chloride, NOCl? 2 NOCl(g) n 2 NO(g) + Cl2(g)

14.2 Reaction Conditions and Rate Goal for Section 14.2 • Describe how reaction conditions (reactant concentrations, temperature, presence of a catalyst, and the state of the reactants) affect reaction rate.

Several factors—reactant concentrations, temperature, and presence of catalysts— affect the rate of a reaction. If the reactant is a solid, the surface area available for reaction is also a factor. The iodine clock reaction (Figure 14.3) illustrates the effects of concentration and temperature. The reaction mixture contains hydrogen peroxide (H2O2), iodide ion (I−), ascorbic acid (vitamin C), and starch (an indicator of the presence of iodine, I2). A sequence of reactions begins with the slow oxidation of iodide ion to I2 by H2O2. H2O2(aq) + 2 I−(aq) + 2 H3O+(aq) n 4 H2O(ℓ) + I2(aq)

As I2 is formed in the solution, vitamin C rapidly reduces it back to I−. 2 H2O(ℓ) + I2(aq) + C6H8O6(aq) n C6H6O6(aq) + 2 H3O+(aq) + 2 I−(aq)

When all the vitamin C has been consumed, I2 remains in solution and forms a blueblack complex with starch. The time it takes for the given amount of vitamin C to react is measured. For the first experiment (A in Figure 14.3) the time required is 51 seconds. When the concentration of iodide ion is smaller (B), the time required for the vitamin C to be consumed is longer, 1 minute 33 seconds. Finally, when the concentrations are the

Smaller concentration of I− than in Experiment A.

Same concentration as in Experiment B, but at a higher temperature.

A

B

C

Initial Experiment

Change concentration

Change temperature © Charles D. Winters/Cengage

Hot bath

Solutions containing vitamin C, H2O2, I−, and starch are mixed.

Initial Experiment The blue color of the starchiodine complex develops in 51 seconds.

Change Concentration The blue color of the starchiodine complex develops in 1 minute 33 seconds when the solution is less concentrated than in A.

Change Temperature The blue color of the starchiodine complex develops in 56 seconds when the solution is the same concentration as in B but at a higher temperature.

Figure 14.3  The iodine clock reaction.  This reaction illustrates the effects of concentration and temperature on reaction rate. (You can do these experiments yourself with reagents available in the supermarket. For details, see S. W. Wright, The vitamin C clock reaction, Journal of Chemical Education, 2002, 79, 41–43.)

678

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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© Charles D. Winters/Cengage

The rate of decomposition of hydrogen peroxide is increased by the catalyst MnO2. Here, H2O2 (as a 30% aqueous solution) is poured onto the black solid MnO2 and rapidly decomposes to O2 and H2O.

© Charles D. Winters/Cengage

Bubbles of O2 gas are seen rising from the potato in the solution.

Steam forms because of the high heat of reaction.

© Thomas Eisner with Daniel Aneshansley, Cornell University

The energy involved in the reaction lets the insect eject hot water and irritating chemicals with explosive force.

The enzyme catalase, found in potatoes and other sources, is used to catalyze H2O2 decomposition.

A bombardier beetle uses the catalase catalyzed decomposition of H2O2 as a defense mechanism.

Figure 14.4  Catalyzed decomposition of H2O2.

same as in experiment B but the reaction mixture is heated, the reaction occurs more rapidly (56 seconds). This experiment illustrates two features that are true of most reactions: •

If the concentration of a reactant is increased, the reaction rate also increases.

•

Chemical reactions occur more rapidly at higher temperatures.

Catalysts are substances that accelerate chemical reactions but are not themselves consumed. Consider the effect of a catalyst on the decomposition of h ­ ydrogen peroxide, H2O2, to form water and oxygen. 2 H2O2(aq) n O2(g) + 2 H2O(ℓ)

This decomposition is very slow; a solution of H2O2 can be stored for many months with only minimal change in concentration. Adding manganese(IV) oxide, an iodidecontaining salt, or the enzyme catalase—a biological catalyst—causes this reaction to occur rapidly, as shown by the vigorous bubbling that occurs when the gaseous oxygen escapes from the solution (Figure 14.4 and Chapter Opening Photograph). The surface area of a solid reactant can also affect the reaction rate. Only molecules at the surface of a solid can come in contact with other reactants. The smaller the particles of a solid, the more molecules are found on the solid’s surface. With very small particles, the effect of surface area on rate can be quite dramatic (Figure 14.5). Farmers know that fine dust particles (suspended in the air of an enclosed space such as a silo or feed mill where grain is stored) represent a major explosion hazard.

© Charles D. Winters/Cengage

Figure 14.5  The contrasting rates of combustion of lycopodium powder in bulk and finely powdered form.

The spores of the common lycopodium clubmoss burn only with difficulty when piled in a dish.



If the spores are ground to a fine powder and sprayed into a flame, combustion is rapid.

14.2  Reaction Conditions and Rate

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679

14.3 Effect of Concentration on Reaction Rate Goal for Section 14.3 • Derive a rate equation from experimental information using the method of initial rates. One important goal in studying the kinetics of a reaction is to determine its mechanism; that is, how the reaction occurs at the molecular level. The place to begin is to learn, by experiment, how reactant concentrations affect the reaction rate. The resulting relationship between reactant concentrations and reaction rate is expressed by an equation called a rate equation, or rate law.

Rate Equations The effect of concentration can be determined by evaluating how the rate is affected when the concentrations of the reactants are varied (with the temperature held constant). Consider, for example, the gas-phase decomposition of N2O5 to NO2 and O2. N2O5(g) n 2 NO2(g) + 1⁄2 O2(g)

Figure 14.6 presents data on the concentration of N2O5 as a function of time for reactions beginning with two different concentrations of N 2O5. The initial rate for the first 15 min (0.25 h) of the reaction is calculated in each case. If you start with 0.500 M N2O5, the initial rate is 0.12 mol/ L ∙ h and if you start with 1.00 M N2O5, the initial rate is 0.24 mol/ L ∙ h. That is, doubling the concentration of N2O5 doubles the reaction rate. This result shows that the rate for this reaction must be directly proportional to the N2O5 concentration: N2O5(g) n 2 NO2(g) + 1⁄2 O2(g) Rate of reaction ∝ [N2O5]

where the symbol ∝ means “proportional to.” This proportionality is expressed by the rate equation Rate of reaction = −∆[N2O5]/∆t = k[N2O5]

where the proportionality constant, k, is called the rate constant. This rate equation tells you that this reaction rate is proportional to the concentration of the reactant. Thus, when [N2O5] is doubled, the reaction rate doubles, for example. Figure 14.6 Reactant concentration versus time data for the gas-phase decomposition of N2O5, beginning with two different concentrations of N2O5.

2 NO2 + ½ O2

N2O5

1.00

Average rate = 0.24 mol/L • h

0.90 0.80

[N2O5], mol/L

0.70 0.60 0.50 0.40 0.30 0.20

Average rate = 0.12 mol/L • h

0.10 0

680

0

1.0

2.0

3.0 4.0 Time (t), hours

5.0

6.0

7.0

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Generally, for a reaction such as a A + b B n x X

the rate equation has the form Rate of reaction = k[A]m[B]n

The rate equation expresses the fact that the rate of reaction is proportional to the reactant concentrations, each concentration being raised to some power. The exponents (m and n) in the rate equation must be determined by experiment; they are not necessarily the stoichiometric coefficients (a and b) for the balanced chemical equation. The exponents in this equation are often positive whole numbers, but they can also be negative numbers, fractions, or zero. A catalyst does not appear as a reactant in the balanced, overall equation for a reaction (Section 14.6). But if a homogeneous catalyst, a catalyst in the same phase as the reactants, is present, its concentration might also be included in the rate equation, even though the catalytic species is not a product or reactant in the equation for the reaction. Consider, for example, the decomposition of hydrogen peroxide in the presence of a catalyst such as iodide ion. 

I (aq) H2O2(aq)   → H2O()  1⁄2 O2(g)

Experiments show that this reaction has the following rate equation: Reaction rate = −∆[H2O2]/∆t = k[H2O2][I−]

Here, the concentration of the catalyst, I−, appears in the rate law, even though it is not part of the balanced equation.

The Order of a Reaction

Designation of Catalysts in Chemical Reactions  A common

practice is to identify catalysts by name or symbol above the reaction arrow, as shown in the example. A homogeneous catalyst is one in the same phase as the reactants. For example, in the iodide ion-catalyzed decomposition of H2O2 in water, the I− ion is a homogeneous catalyst.

The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms. For example, the reaction of NO and Cl2: NO(g) + 1⁄2 Cl2(g) n NOCl(g)

has the following experimentally determined rate law: Reaction rate = −∆[NO]/∆t = k[NO]2[Cl2]

This reaction is second-order in NO, first-order in Cl2, and third-order overall. While first- and second-order reactions are common, other reaction orders are also observed, including zero-order reactions. An example of a zero-order reaction is the decomposition of ammonia on a platinum surface at 856 °C. NH3(g) n 1⁄2 N2(g) + 3⁄2 H2(g)

When the concentration of ammonia is high, the reaction rate is independent of NH3 concentration. The rate law for this reaction is Reaction rate = k[NH3]0 = k

Reaction order is important because it gives some insight into how the reaction occurs. This is described further in Section 14.7.

The Rate Constant, k The rate constant, k, is a proportionality constant that relates rate and concentration at a given temperature. It is an important quantity because it enables you to find the reaction rate for a new set of concentrations. To see how to use k, consider

14.3  Effect of Concentration on Reaction Rate

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681

the substitution of Cl− ion by water in the cancer chemotherapy agent cisplatin, Pt(NH3)2Cl2. Pseudo–First Order Reaction ​

Actually, the reaction of cisplatin and water is a secondorder reaction where Rate = k[Pt(NH3)2Cl2][H2O]. If the reaction is done in a great excess of water, however, the concentration of water does not change by a significant amount over the course of the reaction, that is, [H2O] = constant. Under such circumstances, the reaction is said to be run under pseudo– first-order conditions, and the rate law is given by the equation Rate = k ′[Pt(NH3)2Cl2], where k ′ = k[H2O]. Time and Rate Constants The time in a rate constant can be seconds, minutes, hours, days, years, or whatever time unit is appropriate. The fraction 1/time can also be written as time−1. For example, 1/y is equivalent to y−1, and 1/s is equivalent to s−1.

Pt(NH3)2Cl2(aq)

+

[Pt(NH3)2(H2O)Cl]+(aq)

H2O(𝓵)

+ Cl−(aq)

+

+

The rate law for this reaction is Reaction rate = −∆[Pt(NH3)2Cl2]/∆t = k[Pt(NH3)2Cl2]

and the rate constant, k, is 0.27/h at 25 °C. Knowing k allows you to calculate the rate at a particular reactant concentration—for example, when [Pt(NH3)2Cl2] = ​0.018 mol/L: Reaction rate = (0.27/h)(0.018 mol/L) = 0.0049 mol/L ∙ h

As noted earlier, reaction rates have units of mol/L ∙ time when concentrations are given as moles per liter. Rate constants must have units consistent with the units for the other terms in the rate equation. •

First-order reactions: The units of k are 1/time.

•

Second-order reactions: The units of k are L/mol ∙ time.

•

Zero-order reaction: The units of k are mol/L ∙ time.

Finally, you should recognize that reactions can range from excruciatingly slow to lightning fast, and this is reflected in the wide-ranging values of k. The cisplatin reaction above has k = 0.27/h at 25 °C. In contrast, sucrose decomposes in a firstorder reaction to fructose and glucose with k = 0.0036/h at 25 °C. And the very rapid combination of I atoms to form I2 molecules in the gas phase has k = 4 × 1011 L/mol ∙ s at 23 °C.

Determining a Rate Equation One way to determine a rate equation is by using the method of initial rates. The initial rate is the instantaneous reaction rate at the start of the reaction (the rate at t = 0). An approximate value of the initial rate can be obtained by mixing the reactants and determining the reaction rate after 1% to 2% of the limiting reactant has been consumed. Measuring the rate during the initial stage of a reaction is convenient because initial concentrations are known (Figure 14.6). As an example of the determination of a reaction rate law using initial rates, consider the following reaction of nitrogen monoxide with chlorine. NO(g) + 1⁄2 Cl2(g) n NOCl(g)

Reactant concentrations and initial rates for this reaction for several experiments at 50 °C are collected in the table below. initial concentrations

(mol/L)

682

Experiment

[NO]

[Cl2]

Initial Rate, −𝚫[NO]/𝚫t (mol/L ∙ s)

1

0.250

0.250

1.43 × 10−6

g × 2

g No change

g × 4

2

0.500

0.250

5.72 × 10−6

3

0.250

0.500

2.86 × 10−6

4

0.500

0.500

11.4 × 10−6

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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•

Compare Experiments 1 and 2: Here, [Cl2] is held constant, and [NO] is doubled. This change in [NO] by a factor of 2 leads to a reaction rate increase by a factor of 4; that is, the rate is proportional to the square of the NO concentration.

•

Compare Experiments 1 and 3: In experiments 1 and 3, [NO] is held constant, and [Cl2] is doubled, causing the rate to double. That is, the rate is proportional to [Cl2]. The rate law that reflects these experimental observations is Reaction rate = −∆[NO]/∆t = k[NO]2[Cl2]

Using this equation, you can predict that doubling both concentrations at the same time should cause the rate to go up by a factor of 8, as is determined if experiments 1 and 4 are compared [(1.43 × 10−6 mol/L ∙ s) × 8 = 11.4 × 10−6 mol/L ∙ s]. If the rate equation is known, the value of k, the rate constant, can be found by substituting values for the rate and concentration into the rate equation. Using the data for the NO/Cl2 reaction from the first experiment gives Reaction rate  1.43  106 mol/L  s  k(0.250 mol/L)2(0.250 mol/L) k

1.43  106 mol/L  s  9.15  105 L2 /mol 2  s (0.250 mol/L)2(0.250 mol/L)

Ex am p le 14.3

Determining a Rate Equation Problem  The rate of the reaction between CO and NO2 at 540 K CO(g) + NO2(g) n CO2(g) + NO(g) was measured starting with various concentrations of CO and NO2. Determine the rate equation and the value of the rate constant. initial concentrations

(mol/L) Experiment

[CO]

[NO2]

Initial Rate (mol/L ∙ h)

1

5.10 × 10−4

0.350 × 10−4

 3.4 × 10−8

2

5.10 × 10−4

0.700 × 10−4

 6.8 × 10−8

3

5.10 × 10−4

0.175 × 10−4

 1.7 × 10−8

4

1.02 × 10−3

0.350 × 10−4

 6.8 × 10−8

5

1.53 × 10−3

0.350 × 10−4

10.2 × 10−8

What Do You Know?  The table contains concentrations of the two reactants and initial rates for five experiments.

Strategy  For a reaction involving two reactants, the general approach is to keep the concentration of one reactant constant and then decide how the rate of reaction changes as the concentration of the other reactant is varied.

Solution   The initial concentration of CO is the same in the first three experiments. Comparing experiments 1 and 2, you can set up the following ratios: k [ CO ]2 [NO2 ]2 Rate 2  m n Rate 1 k [ CO ]1 [NO2 ]1 m



n

14.3  Effect of Concentration on Reaction Rate

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683

Because k and the concentration of CO are constant in these two cases, they cancel, leaving  [NO2 ]2  Rate 2   Rate 1  [NO2 ]1 

n

Substituting the values from the table, you obtain  0.700  104 mol/L  6.8  108 mol/L  h  8  0.350  104 mol/L  3.4  10 mol/L  h 2.0  (2.00)n n1

n

Thus, doubling the NO2 concentration in experiment 2 relative to experiment 1 leads to a twofold increase in the rate. The reaction is first-order in NO2. This finding is confirmed by experiment 3. Decreasing [NO2] in experiment 3 to half its original value causes the rate to decrease by half. In a similar fashion, the data in experiments 1 and 4 (with constant [NO2]) show that doubling [CO] doubles the rate, and the data from experiments 1 and 5 show that tripling the concentration of CO triples the rate. These results mean that the reaction is first-order in [CO]. Thus, you now know the rate equation is  Reaction rate = k[CO][NO2]  The rate constant, k, can be found by inserting data for one of the experiments into the rate equation. Using data from experiment 1, for example, Rate  3.4  108 mol/L  h  k(5.10  104 mol/L)(0.350  104 mol/L)  k = 1.9 L/mol ∙ h 

Think about Your Answer  To check your answer, calculate k for two or more of the experiments. If these values differ significantly, then you know that you have made a mistake in calculating the reaction orders.

Check Your Understanding  The initial rate (−∆[NO]/∆t) of the reaction of nitrogen monoxide and oxygen NO(g) + 1⁄2 O2(g) n NO2(g) was measured for various initial concentrations of NO and O2 at 25 °C. Determine the rate equation from these data. What is the value of the rate constant, k, and what are its units? initial concentrations

(mol/L)

684

Experiment

[NO]

[O2]

Initial Rate (mol NO/L ∙ s)

1

0.020

0.010

0.028

2

0.020

0.020

0.057

3

0.020

0.040

0.114

4

0.040

0.020

0.227

5

0.010

0.020

0.014

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Problem Solving Tip 14.1 Determining a Rate Equation: A Logarithmic Approach In Example 14.3, the ratio of the rate equations simplified to 2 = 2n, allowing for the determination of the exponent (n = 1) by inspection. What do you do when the ratio of the reactant concentrations or the exponents are not integers? Consider if you had obtained the following ratios.  [R ]  Rate in experiment 2   2 Rate in experiment 1  [R ]1 

n

 5.25  103 mol/L  4.05  108 mol/L  h  8  3.50  103 mol/L  1.80  10 mol/L  h

You can solve this problem by taking the logarithm of both sides of the equation. log(2.25) = log(1.50)n Using a property of logarithms, this equation becomes log(2.25) = n log(1.50) Solving for n, you obtain n

n

2.25  ( 1.50 )

n

log(2.25)  2.000 log(1.50)

Thus, the reaction is second-order with respect to reactant R.

Ex am p le 14.4

Using a Rate Equation to Determine Rates Problem   Using the rate equation and rate constant determined for the reaction of CO and NO2 at 540 K in Example 14.3, determine the initial rate of the reaction when [CO] = 3.8 × 10−4 mol/L and [NO2] = 0.650 × 10−4 mol/L.

What Do You Know?   The rate law and the value for the rate constant (1.9 L/mol ∙ h) are both known. The concentrations of the two reactants are given.

Strategy   A rate equation consists of three parts: a rate, a rate constant (k), and the concentration terms. If two of these parts are known (here k and the concentrations), the third can be calculated. Solution Substitute k (= 1.9 L/mol ∙ h) and the concentration of each reactant into the rate law determined in Example 14.3. Reaction rate  k[CO][NO2]  (1.9 L/mol  h)(3.8  104 mol/L)(0.650  104 mol/L) Reaction rate  4.7  108 mol/L  h

Think about Your Answer   As a check on the calculated result, it is often useful to make an educated guess at the answer before carrying out the mathematical solution. You know the reaction is first-order in both reactants. Compare the concentration values given in this problem with the concentration values in experiment 1 in Example 14.3. In this ­example, [CO] is about three-fourths the value in experiment 1, whereas [NO2] is almost twice the value. The effects do not precisely offset each other, but you might predict that the difference in rates between this experiment and experiment 1 will be fairly small, with the rate in this experiment being just a little greater. The calculated value bears this out.

Check Your Understanding  The rate constant, k, at 25 °C is 0.27/h for the reaction Pt(NH3)2Cl2(aq) + H2O(ℓ) n [Pt(NH3)2(H2O)Cl]+(aq) + Cl−(aq) and the rate equation is Reaction rate = k[Pt(NH3)2Cl2] Calculate the rate of reaction when the concentration of Pt(NH3)2Cl2 is 0.020 M.



14.3  Effect of Concentration on Reaction Rate

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685

14.4 Concentration–Time Relationships: Integrated Rate Laws Goals for Section 14.4 • Use the relationships between reactant concentration and time for zero-order, firstorder, and second-order reactions.

• Apply graphical methods for determining reaction order and the rate constant from experimental data.

• Use the concept of half-life (t1/2), especially for first-order reactions. It is often important for a chemist to know how long a reaction must proceed to reach a particular concentration of some reactant or product, or what the reactant and product concentrations will be after some time has elapsed. For this reason, it would be useful to have a mathematical equation that relates time and c­ oncentration—an equation that describes concentration–time curves like the one shown in Figure 14.2. With such an equation, you could calculate the concentration at any given time or the length of time needed for a given amount of reactant to react.

First-Order Reactions Suppose the reaction R n products is first-order. This means the reaction rate is directly proportional to the concentration of R raised to the first power, or, mathematically, 

[R]  k [R] t

This relationship can be transformed into a very useful equation called an ­integrated rate equation (because integral calculus is used in its derivation). ln



[R]t  kt [R]0

(14.1)

Here, [R]0 and [R]t are concentrations of the reactant at time t = 0 and at a later time, t, respectively. The ratio of concentrations, [R]t /[R]0, is the fraction of reactant that remains after a given time has elapsed. Equation 14.1 can be used to carry out many useful calculations. For example, •

If [R]t /[R]0 is measured in the laboratory after some amount of time has elapsed, then k can be determined.

•

If [R]0 and k are known, then the concentration of material remaining after a given amount of time ([R]t) can be calculated.

•

If k is known, then the time elapsed until a specific fraction ([R]t /[R]0) remains can be determined.

Finally, notice that  k for a first-order reaction is independent of concentration; k has units of time−1 (y−1 or s−1, for example). This means you can choose any convenient unit for [R]t and [R]0: moles per liter, moles, grams, number of atoms, number of molecules, or gas pressure.

686

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Ex am p le 14.5

The First-Order Rate Equation Problem   In the past, cyclopropane, C3H6, was used in a mixture with oxygen as an anesthetic. (This practice has almost ceased today because the compound is flammable.) When heated, cyclopropane rearranges to propene in a first-order process.

H C H

H C H

H C

CH3CH

CH2

H

cyclopropane

propene

Rate = k[cyclopropane]  k = 2.42 h−1 at 500 °C If the initial concentration of cyclopropane is 0.050 mol/L, how much time (in hours) must elapse for its concentration to drop to 0.010 mol/L?

What Do You Know?   The reaction is first-order in cyclopropane. You know the rate constant, k, and the initial and final concentrations of this reactant, [R]0 and [R]t. Strategy  Use Equation 14.1 to calculate the time (t) elapsed to reach a concentration of 0.010 mol/L.

Solution   Values for [cyclopropane]t, [cyclopropane]0, and k are substituted into ­Equation 14.1; t (time) is the unknown: [R]t  kt [R]0 [0.010 mol/L] ln  (2.42 h1)t [0.050 mol/L] ln

t 

ln(0.20) (1.609)    0.665 h  2.42 h1 2.42 h1

Think about Your Answer   Notice that the concentration units cancel and that the unit for the answer is a time unit (h). In addition, the negative sign in the initial equation is cancelled by the negative value of the calculated logarithm, leaving the time as a positive number, which it must be.

Check Your Understanding  Sucrose, a sugar, decomposes in acid solution to give glucose and fructose. The reaction is first-order in sucrose, and the rate constant at 25 °C is k = 0.21 h−1. If the initial concentration of sucrose is 0.010 mol/L, what is its concentration after 5.0 h?



14.4  Concentration–Time Relationships: Integrated Rate Laws

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687

E xamp le 14.6

Using the First-Order Rate Equation Problem   Hydrogen peroxide decomposes in a dilute sodium hydroxide solution at 20 °C in a first-order reaction: H2O2(aq) n H2O(ℓ) + 1⁄2 O2(g) Rate = k[H2O2] with k = 1.06 × 10−3 min−1 What is the fraction of H2O2 remaining after 100. minutes? If the initial concentration of H2O2 is 0.020 mol/L, what is the concentration of H2O2 at this time?

What Do You Know?  This is a first-order reaction. The rate constant k, initial concentration of H2O2, and elapsed time are given. Strategy   Because the reaction is first-order in H2O2, use Equation 14.1. Here, [H2O2]0, k, and t are known, and you are asked to find the value of the fraction remaining, [H2O2]t/ [H2O2]0. Once this value is known, and knowing [H2O2]0, you can calculate [H2O2]t.

Solution  Substitute the rate constant and time into Equation 14.1. ln

[H2O2]t  kt  (1.06  103 min1)(100. min) [H2O2]0 ln

[H2O2]t  0.1060 [H2O2]0

Take the antilogarithm of −0.1060 [that is, the inverse of the natural logarithm of −0.1060 or e−0.1060] to find the fraction of H2O2 remaining. Fraction remaining 

[H2O2]t   0.8994 =  0.899  [H2O2]0

The calculated fraction remaining is 0.90, thus the concentration of H2O2 remaining is 90% of the initial concentration. [H2O2]t =  0.8994 [H2O2]0 = 0.8994 (0.020 mol/L) =  0.018 mol/L 

Think about Your Answer   Although H2O2 is unstable, its rate of decomposition is very slow, particularly in a dilute solution. However, sodium hydroxide catalyzes the decomposition. The rate of the reaction can be studied by measuring the volume of O2 gas evolved as a function of time.

Check Your Understanding  Gaseous azomethane (CH3N2CH3) decomposes to ethane and nitrogen when heated: CH3N2CH3(g) n CH3CH3(g) + N2(g) The decomposition of azomethane is a first-order reaction with k = 3.6 × 10−4 s−1 at 600 K. (a) A sample of gaseous CH 3 N 2 CH 3 is placed in a flask and heated at 600 K for 150 ­seconds. What fraction of the initial sample remains after this time? (b) How long must a sample be heated so that 99% of the sample has decomposed?

Second-Order Reactions Suppose the reaction R n products is second-order. The rate equation is 

688

[R]  k[R]2 t

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Using calculus, this relationship can be transformed into the following equation that relates reactant concentration and time: 1 1   kt [R]t [R]0



(14.2)

Just as for the integrated first-order rate equation, [R]0 is the concentration of reactant at time t = 0, and [R]t is the concentration at a later time, but now k is the second-order rate constant, which has the units of L/mol ∙ time.

Ex am p le 14.7

Using the Second-Order Integrated Rate Equation Problem  The gas-phase decomposition of HI HI(g) n 1⁄2 H2(g) + 1⁄2 I2(g) has the rate equation 

[HI]  k[HI]2 t

where k = 30. L/mol ∙ min at 443 °C. How much time does it take for the concentration of HI to drop from 0.010 mol/L to 0.0050 mol/L at 443 °C?

What Do You Know?  Equation 14.2 is used for a second-order reaction. The rate constant, k, and the initial and final concentrations of HI are given; the elapsed time is the unknown. Strategy  Substitute the values of [HI]0, [HI]t , and k into Equation 14.2, and solve for the unknown, t. Solution   Here, [HI]0 = 0.010 mol/L, [HI]t = 0.0050 mol/L, and k = 30. L/mol • min. ­Substitute these values into Equation 14.2. 1 1   (30. L/mol  min)t 0.0050 mol/L 0.010 mol/L (2.00  102 L/mol)  (1.00  102 L/mol)  (30. L/mol  min)t t =  3.3 minutes 

Think about Your Answer  Keep track of the units for each quantity. Most of the units cancel, leaving only the unit for the elapsed time (minutes).

Check Your Understanding  Using the rate constant for HI decomposition given in this example, calculate the concentration of HI after 12 minutes if [HI]0 = 0.010 mol/L.

Zero-Order Reactions If a reaction (R n products) is zero-order, the rate equation is 

[R]  k[R]0 t

This equation leads to the integrated rate equation where the units of k are mol/L ∙ s.

[R]t  [R]0  kt

(14.3)

Zero-order kinetics are observed, for example, with gas-phase reactions that occur on metal surfaces. Here, the rate of the reaction, such as the decomposition of ammonia on platinum (Section 14.3, page 681), is independent of reactant concentrations so long as the number of gas-phase reactant molecules is much greater than the number of binding

14.4  Concentration–Time Relationships: Integrated Rate Laws

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689

sites on the metal surface. Note, however, that as the reaction proceeds and the number of reactant molecules decreases, the reaction order, and thus the rate law, is likely to change.

Graphical Methods for Determining Reaction Order and the Rate Constant Finding the Slope of a Line ​See

Section 1R.5 (pages 48–49), for a description of methods for finding the slope of a line.

Chemists often determine the order of a reaction and its rate constant using graphical methods. If rearranged slightly, Equations 14.1, 14.2, and 14.3 have the form y = mx + b. This is the equation for a straight line, where m is the slope of the line and b is the y -intercept. In each of these equations, x = t. Zero-order

First-order

Second-order

[R]t = − kt + [R]0

ln [R]t = − kt + ln [R]0

1 1 = + kt + [R]t [R]0

y

mx

b

y

mx

b

y

mx

b

As an example of the graphical method for determining reaction order, consider the decomposition of azomethane. CH3N2CH3(g) n CH3CH3(g) + N2(g)

The decomposition of azomethane was followed at 600 K by observing the decrease in its partial pressure (P(CH3N2CH3)) with time. (Recall from Chapter 10 that pressure is proportional to concentration at a given temperature and volume.) As shown in ­Figure 14.7a, a graph of ln P(CH3N2CH3) versus time produces a straight line, which shows that the reaction is first-order in CH3N2CH3. The slope of the line can be determined, and the negative of the slope equals the rate constant for the reaction, 3.6 × 10−4 s−1. The decomposition of NO2 is a second-order process. NO2(g) n NO(g) + 1⁄2 O2(g) Rate = k[NO2]2

This fact can be verified with a plot of 1/[NO2] versus time, which is a straight line (Figure 14.7b). Here, the slope of the line is equal to k. For a zero-order reaction (Figure 14.7c), a plot of concentration versus time gives a straight line with a slope equal to the negative of the rate constant. Table 14.1 summarizes the relationships between concentration and time for first-, second-, and zero-order processes.

Half-Life and First-Order Reactions The half-life, t1/2 , of a reaction is the time required for the concentration of a reactant ([R]t) to decrease to one-half its initial value ([R]0). It is a convenient way to describe the rate at which a reactant is consumed in a chemical reaction: The longer the half-life, the slower the reaction. [R]t  1⁄2[R]0 or

Table 14.1

690

[R]t  1⁄2 [R]0

Characteristic Properties of Reactions of the Type R 88n Products

Integrated Rate Equation

Order

Rate Equation

0

−Δ[R]/Δt = k[R]0

[R]t − [R]0 = −kt

1

−Δ[R]/Δt = k[R]1

2

−Δ[R]/Δt = k[R]2

StraightLine Plot

Slope

k Units

[R]t vs. t

−k

mol/L ∙ time

ln ([R]t /[R]0) = −kt

ln [R]t vs. t

−k

1/time

(1/[R]t) − (1/[R]0) = kt

1/[R]t vs. t

  k

L/mol ∙ time

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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P × 1O2 atm

ln P

0 1000 2000 3000 4000

8.20 5.72 3.99 2.78 1.94

−2.50 −2.86 −3.22 −3.58 −3.94

−2

Rate = kPCH3N2CH3

−2.5

Slope = −k

−3

=

ln P

Time (s)

[(−3.76) − (−3.04)] (3500 − 1500) s

= −3.6 × 10−4 s−1

−3.5

k = 3.6 × 10−4 s−1

−4 CH3N2CH3(g)

CH3CH3(g) + N2(g)

1000

2000 3000 Time (s)

−4.5 0

4000

(a) First-order reaction. A plot of the natural logarithm of the CH3N2CH3 pressure versus time for the decomposition of azomethane results in a straight line with a negative slope. The rate constant k = −slope. (Recall that for a first-order reaction, k is independent of concentration, so any convenient unit proportional to concentration can be used to determine k. In this case the partial pressure of azomethane in atm was used.)

[NO2] (mol/L)

1/[NO2] (L/mol)

0 30 60 90 120

0.020 0.015 0.012 0.010 0.0087

50 67 83 100 115

120

Rate = k[NO2]2 Slope = k

100 1 (L/mol) [NO2]

Time (s)

=

80

(105 − 61) mol/L (100 − 20) s

k = 0.55 L/mol ∙ s

60 NO(g) + ½ O2(g)

NO2(g) 40 0

30

60

90

120

Time (s) (b) Second-order reaction. A plot of 1/[NO2] versus time for the decomposition of NO2 results in a straight line. The rate constant k = slope.

[NH3] (mmol/L)

200 280 600 750 800 900

1.75 1.65 1.15 0.94 0.85 0.70

2.00 [NH3], mmol/L (1 mmol = 10−3mol)

Time (s)

Rate = k[NH3]0 Slope = −k

1.50

=

1.00

(0.540 − 1.29) mmol/L (1000 − 500) s

= −1.5 × 10−3 mmol/L ∙ s k = 1.5 × 10−3 mmol/L ∙ s

0.50

0

NH3(g) 200

½ N2(g) + ³∕² H2(g) 400

600

800

1000

Time (t), seconds (c) Zero-order reaction. A plot of the concentration of ammonia, [NH3], against time for the decomposition of NH3 on a metal surface at 856 °C is a straight line, indicating that this is a zero-order reaction. The rate constant k = −slope.

Figure 14.7  Graphical methods for determining reaction order.

Half-life is used primarily when dealing with first-order processes. To evaluate t1/2 for a first-order reaction, substitute [R]t /[R]0 = 1⁄2 and t = t1/2 into the integrated first-order rate equation (Equation 14.1), ln (1⁄2)  kt 1⁄2

or

ln 2  kt 1⁄2 14.4  Concentration–Time Relationships: Integrated Rate Laws

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691

Half-Life and Radioactivity ​ Half-

life is a term often encountered when dealing with radioactive elements. Radioactive decay is a first-order process, and half-life is commonly used to describe how rapidly a radioactive element decays. See Chapter 20 and Example 14.9.

Rearranging this equation (and substituting ln 2 = 0.693) provides a useful equation that relates half-life and the first-order rate constant: t 1⁄2 



0.693 k

(14.4)

This equation identifies an important feature of first-order reactions: t 1⁄2 is ­independent of concentration. To illustrate the concept of half-life, consider again the first-order decomposition of azomethane, CH3N2CH3. CH3N2CH3(g) n CH3CH3(g) + N2(g)

Half-Life Equations for Other Reaction Orders ​For a zero-order

reaction, R n products

t 1⁄2 

Given a rate constant of 3.6 × 10−4 s−1, the half-life is 1.9 × 103 s or 32 min.

[R]0 2k

t 1⁄2 

0.693  1.9  103 s (or 32 min) 3.6  104 s1

The partial pressure of azomethane is plotted as a function of time in F­ igure 14.8, and this graph shows that P(azomethane) decreases by half every 32 min. The initial pressure of azomethane was 820 mm Hg, but it dropped to 410 mm Hg in 32 min, and then dropped to 205 mm Hg in another 32 min. That is, after two half-lives (64 min), the partial pressure is (1⁄2) × (1⁄2) = (1⁄2)2 = 1⁄4 or 25% of the initial pressure. After three half-lives, the partial pressure has dropped further to 102 mm Hg or 12.5% of the initial value and is equal to (1⁄2) × (1⁄2) × ​ (1⁄2) = (1⁄2)3 = 1⁄8 of the initial value. It can be hard to visualize whether a reaction is fast or slow from the rate constant value. Can you tell from the rate constant, k = 3.6 × 10−4 s−1, whether the azomethane decomposition will take seconds, hours, or days to reach completion? Probably not, but you can get a better sense of reaction rate from the value of the half-life for the reaction (32 min). This indicates that you would only have to wait a few hours for the reactant to be essentially consumed.

For a second-order reaction, R n products

t 1⁄2 

Rate = k[CH3N2CH3] with k = 3.6 × 10−4 s−1 at 600 K

1 k[R]0

Note that in both cases the half-life depends on the initial concentration.

Partial pressure (mm Hg) of azomethane (10−2)

9 First-order decomposition:

8

CH3N2CH3(g)

CH3CH3(g) + N2(g)

k = 3.6 × 10−4 s−1

7 6 5

1 half-life, 1900 s P = 1⁄ 2 (820 mm Hg)

4

The pressure of CH3N2CH3 is halved every 1900 s (32 min).

2 half-lives, 3800 s P = 1⁄ 4 (820 mm Hg)

3 2 1 0

0

500

1000

1500

2000

2500

3000

3500

4000

4500

Time (s)

Figure 14.8  Half-life of a first-order reaction.  This plot of pressure versus time is similar in shape to plots of concentration versus time for all other first-order reactions.

692

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Ex am p le 14.8

Half-Life and a First-Order Process Problem   Sucrose, C12H22O11, decomposes to fructose and glucose in acid solution with the rate law Rate = k[C12H22O11]  k = 0.216 h−1 at 25 °C (a) What is the half-life of sucrose at this temperature? (b) What amount of time is required for 87.5% of the initial concentration of sucrose to decompose?

What Do You Know?  The decomposition of sucrose is a first-order reaction. The rate constant for the reaction (at 25 °C) is given.

Strategy  (a) Use Equation 14.4 to calculate the half-life from the rate constant. (b) After 87.5% of the C12H22O11 has decomposed, 12.5% (or one-eighth of the sample) remains. To reach this point, three half-lives are required. Half-Life

Fraction Remaining

1

0.5

2

0.25

3

0.125

Therefore, you multiply the half-life calculated in part (a) by 3.

Solution (a) The half-life for the reaction is t1⁄2 = 0.693/k = 0.693/(0.216 h−1) = 3.208 hours =  3.21 hours  (b) Three half-lives must elapse before the fraction remaining is 0.125, so Time elapsed = 3 × 3.208 h =  9.63 hours 

Think about Your Answer  Half-life is a convenient way to describe the speed of a first-order reaction. In this example, you can quickly see that complete decomposition of a sample of sucrose requires many hours.

Check Your Understanding  The catalyzed decomposition of hydrogen peroxide is first-order in [H2O2]. It was found that the concentration of H2O2 decreased from 0.24 M to 0.060 M over a period of 282 minutes. What is the half-life of H2O2? What is the rate constant for this reaction? What is the initial rate of decomposition at the beginning of this experiment (when [H2O2] = 0.24 M)?

Ex am p le 14.9

Half-Life and Radioactive Decay Problem  Radioactive decay is a first-order process. Radioactive radon-222 gas (222Rn) occurs naturally as a product of the uranium decay series. The half-life of 222Rn is 3.8 days. Suppose a flask originally contained 4.0 × 1013 atoms of 222Rn. How many atoms of 222Rn will remain after one month (30. d)?



14.4  Concentration–Time Relationships: Integrated Rate Laws

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693

Strategy Map

What Do You Know?  This process follows first-order kinetics. The half-life of 222Rn

Problem Find the final number of atoms of a radioactive isotope after a given time period.

Strategy 

Data/Information • Half-life for decay • Initial number of atoms • Time elapsed Step 1

and the number of atoms initially present are known.

Step 1.  Calculate the rate constant, k, from the half-life using Equation 14.4. Step 2.  Knowing the rate constant, the number of atoms at the beginning ([R]0), and  the  elapsed time (30. days), use the integrated first-order rate equation (­Equation 14.1) to c­ alculate the number of atoms remaining ([R]t).

Solution  Calculate the rate constant from the half-life. k 

Step 2

0.693 0.693   0.182 d1   t 1⁄2 3.8 d

Use Equation 14.1 to calculate the number of atoms remaining after 30. days. ln

[Rn]t  (0.182 d1)(30. d)  5.47 4.0  1013 atoms [Rn]t  e5.47  0.0042 4.0  1013 atoms [Rn]t =  2 × 1011 atoms 

Think about Your Answer  Thirty days is approximately eight half-lives for this element. This means that the number of atoms present at the end of the month is approximately (1⁄2)8 or 1⁄256th of the original number. In other words, about 99.6% of the Rn atoms decayed during this period of time.

Check Your Understanding  Americium is used in smoke detectors and in medicine for the treatment of certain malignancies. One isotope of americium, 241 Am, has a rate constant, k, for radioactive decay of 0.0016 y−1. In contrast, radioactive iodine-125, which is used for studies of thyroid functioning, has a rate constant for decay of 0.011 d−1. (a) What are the half-lives of these isotopes? (b) Which isotope decays faster? (c) If you are given a dose of iodine-125 containing 1.6 × 1015 atoms, how many atoms remain after 2.0 d?

14.5 A Microscopic View of Reaction Rates Goals for Section 14.5 • Describe the collision theory of reaction rates and use collision theory to describe the effects of reactant concentration, molecular orientation, and temperature on reaction rate.

• Relate activation energy (Ea) to the rate of a reaction. • Understand reaction coordinate diagrams. • Use the Arrhenius equation, or one of its modified forms, to calculate the activation energy from rate constants at different temperatures.

Chemists often turn to the particulate level of chemistry to explain chemical phenomena. Looking at the way reactions occur at the atomic and molecular levels can give some insight into the various influences on rates of reactions.

694

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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A Closer Look

Rate Laws, Rate Constants, and Reaction Stoichiometry A basic definition of the rate of a reaction is that it represents the change in concentration of the reactants and products over time. When you deal with reaction rates quantitatively, however, you need to be specific about the reaction stoichiometry. Consider the first-order decomposition of N2O5, a reaction mentioned earlier.

N2O5(g) n 2 NO2(g) + 1⁄2 O2(g) The rate of the reaction can be expressed (and measured in the lab) as

  [ O2 ]    [N2O5 ]  1   [NO2 ]  Rate       2  2  t   t   t  If you follow the disappearance of N2O5 as a measure of reaction rate and base the definition of rate on the stoichiometry above, you should write the following rate law.

Following the reasoning above, the reaction rate would be written as follows:

  [ O2 ]  1   [N2O5 ]  1   [NO2 ]  Rate          2  t  4  t   t  This leads to the following rate law:

1   [N2O5 ]  Rate     k [N2O5 ] 2  t  Rearranging this to solve for the rate of disappearance of N2O5 yields the following equation:

  [N2O5 ]    2k [N2O5 ]  t  with the following integrated rate equation and half-life equation. Integrated rate equation:

  [N2O5 ]  Rate     k [N2O5 ]  t  From this definition, it also follows that the integrated rate ­equation is

[N O ] ln 2 5 t  kt [N2O5]0 and the half-life equation is

t1⁄2 = 0.693/k You can also, however, write the equation for the decomposition of N2O5 as follows:

2 N2O5(g) n 4 NO2(g) + O2(g)

ln

[N2O5]t  2kt [N2O5]0

Half-life equation:

t1⁄2 = 0.693/2k′ The integrated rate laws derived based on the two different chemical equations have the same form, but the rate constants do not have the same values; instead k = 2k′. Notice, however, that both calculations give the same value for the half-life, as they should; half-life does not depend on how the equation is written. For more on these issues, see K. T. Quisenberry and J. Tellinghuisen, Journal of Chemical Education, 2006, 83, 510–512.

Let us review the macroscopic observations made so far concerning reaction rates. There are wide differences in rates of reactions—from very fast reactions like the explosion that occurs when hydrogen and oxygen are exposed to a spark or flame (Figure 1.16), to slow reactions like the formation of rust that occurs over days, weeks, or years. For a specific reaction, factors that influence reaction rate include the concentrations of the reactants, the temperature of the reaction system, and the presence of catalysts. Each of these influences can be explained using the collision theory of reaction rates, which states that three conditions must be met for a reaction to occur: 1. The reacting molecules must collide with one another. 2. The reacting molecules must collide with sufficient energy to initiate the process of breaking and forming bonds. 3. The molecules must collide in an orientation that can lead to rearrangement of the atoms and the formation of products.

Collision Theory: Concentration and Reaction Rate Consider the gas-phase reaction of nitric oxide and ozone, an environmentally important reaction that contributes to both natural and human-caused ozone decomposition: NO(g) + O3(g) n NO2(g) + O2(g)

14.5  A Microscopic View of Reaction Rates

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Figure 14.9  The effect of concentration on the frequency of molecular collisions.

2

4

1 2

1

3

1 4

2

3

(a) 1 NO : 16 O3 − 2 hits/second.

(b) 2 NO : 16 O3 − 4 hits/second.

(c) 1 NO : 32 O3 − 4 hits/second.

A single NO molecule, moving among 16 O3 molecules, is shown colliding with two of them per second.

If two NO molecules move among 16 O3 molecules, you would predict that four NO—O3 collisions would occur per second.

If the number of O3 molecules is doubled (to 32), the frequency of NO—O3 collisions is also doubled, to four per second.

The rate law for this product-favored reaction is first-order in each reactant: Rate = k[NO][O3]. Why does this reaction have this rate law? Imagine a flask containing a mixture of NO and O3 molecules in the gas phase. Both kinds of molecules are in rapid and random motion within the flask. They strike the walls of the vessel, but they also collide with other molecules. It is reasonable to propose that the rate of their reaction should be related to the number of collisions, which is in turn related to their concentrations (Figure 14.9). ­Doubling the concentration of one reactant in the NO + O3 reaction, say NO, will lead to twice the number of molecular collisions. Figure 14.9a shows a single molecule of one of the reactants (NO) moving randomly among sixteen O3 molecules. In a given time period, it might collide with two O3 molecules. The number of NOOO3 collisions will double, however, if the concentration of NO molecules is doubled (to 2, as shown in Figure 14.9b) or if the number of O3 molecules is doubled (to 32, as in Figure 14.9c). Thus, the dependence of reaction rate on concentration can be explained: The number of collisions between the two reactant molecules is directly proportional to the concentration of each reactant, and the rate of the reaction shows a first-order dependence on each reactant.

Jamie Schwaberow/NCAA Photos/Getty Images

Collision Theory: Activation Energy

Figure 14.10  An analogy to chemical activation energy.  For the volleyball to go over the net, the player must give it sufficient energy.

696

Molecules require some minimum energy to react. Chemists visualize this as an energy barrier that must be surmounted by the reactants for a reaction to occur (­Figure 14.10). The energy required to cross the barrier is called the activation energy, Ea. If the activation energy is low, a high proportion of the molecules in a sample may have sufficient energy to react. The reaction will be fast. If the activation energy is high, only a few reactant molecules in a sample may have enough energy. The reaction will be slow. To further illustrate an activation energy barrier, consider the high temperature conversion of NO2 and CO to NO and CO2. At the molecular level, imagine that the reaction involves the transfer of an O atom from an NO 2 molecule to a CO molecule. NO2(g) + CO(g) uv NO(g) + CO2(g)

The process can be described with an energy diagram called a reaction coordinate diagram (Figure 14.11). The horizontal axis describes the reaction progress as the reaction proceeds, and the vertical axis represents the potential energy of the system during the reaction. When NO2 and CO approach and O atom transfer begins, an NOO bond is being broken, and a CPO bond is forming. Energy input (the activation energy) is required for this to occur. The energy of the system reaches a maximum at the transition state. At the transition state, sufficient energy has been concentrated in the appropriate bonds; bonds in the reactants can now break, and

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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NO2(g) + CO(g) Reactants

Activation energy for forward reaction.

Figure 14.11 Activation energy for the reaction of NO2 and CO to give NO and CO2.

NO(g) + CO2(g) Transition state

Products

Energy

Ea = 132 kJ/mol Ea′ = 358 kJ/mol

Reactants NO2 + CO

Net energy change for forward reaction.

Activation energy for reverse reaction.

Products NO + CO2

∆H = −226 kJ/mol

More about Molecular Orientation and Reaction Coordinate Diagrams

The rate of a chemical reaction depends, among other things, on the proper alignment of the reactants. A reaction will occur only if the ­reactants collide in the proper orientation. A well-studied example of this is the substitution of a chlorine atom of CH3Cl by an ion such as F−. Here, the F− ion attacks the molecule from the side opposite the Cl substituent. As F− begins to form a bond to carbon, the COCl bond weakens, and the CH3 portion of the molecule changes shape. As time progresses, the products CH3F and Cl− are formed.

+ F−

CH3Cl

[F • • • CH3 • • • Cl]

−

+ CH3F

Cl−

But what happens if F− approaches from the side of the molecule with the Cl atom or along one of the COH bonds? There



Energy of intermediate

Energy

A Closer Look

Reaction progress

Ea for step 1

Ea for step 2

Reactants Products Reaction progress

Figure A  A reaction coordinate diagram for a two-step reaction, a process involving an intermediate. will be no reaction, no matter how much energy the reactants have! The misalignment of reactants is one reason that reactions can be slow, particularly when the reactants have complex structures. Assuming the reactants are aligned properly, the CH3Cl/F− reaction would follow a reaction coordinate diagram such as that in Figure 14.11. Another substitution reaction that has been thoroughly studied is the reaction of methanol, CH3OH, with Br− ion in the presence of an acid. This is a two-step reaction, which is described by the reaction coordinate diagram in Figure A. In the first step, an H+ ion attaches to the O of the COOOH group in a rapid, reversible reaction. The energy of this protonated species, CH3OH2+, a reaction

intermediate, is higher than the energies of the reactants. The reaction intermediate is represented by the dip in the curve shown in Figure A. In the second step, a halide ion, say Br−, attacks the intermediate to produce methyl bromide, CH3Br, and water. There is an activation energy barrier in both the first step and second step. Notice in Figure A, as in Figure 14.11, that the energy of the products is lower than the energy of the reactants. The reaction is exothermic.

+ H3O+

CH3OH

+ CH3OH2+

H2O

+ CH3OH2+

Br− +

CH3Br

H2O

14.5  A Microscopic View of Reaction Rates

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new bonds can form to give products. The system is poised to go on to products, but it could also return to the reactants. Because the transition state is at a maximum in potential energy, a molecular species with this structure cannot be isolated. Using computer molecular modeling techniques, however, chemists can describe what the transition state must look like. In the NO 2 + CO reaction, 132 kJ/mol is required to reach the transition state, that is, the top of the energy barrier. As the reaction continues toward the products—as the NOO bond breaks and a CPO bond forms—the reaction evolves energy, 358 kJ/mol. The net energy change involved in this exothermic reaction is −226 kJ/mol. ∆H = +132 kJ/mol + (−358 kJ/mol) = −226 kJ/mol

What happens if the reverse reaction is carried out? That reaction requires 358 kJ/mol to reach the transition state, and 132 kJ/mol is evolved on proceeding to the products, NO2 and CO. The reaction in this direction is endothermic, requiring a net input of +226 kJ/mol.

Collision Theory: Activation Energy and Temperature In a laboratory or in the chemical industry, a chemical reaction is often carried out at elevated temperatures because this allows the reaction to occur more rapidly. Conversely, it is sometimes desirable to lower the temperature to slow down a chemical reaction (to avoid an uncontrollable reaction or a potentially dangerous explosion). A discussion of the effect of temperature on reaction rate begins by recalling there is a distribution of energies for molecules in a sample of a gas or liquid. The molecules in a sample have a wide range of energies, described earlier as a Boltzmann distribution of energies (Figure 10.11). In any sample of a gas or liquid, some molecules have very low energies, others have very high energies, but most have some intermediate energy. As the temperature increases, the average energy of the molecules in the sample increases, as does the fraction having higher energies. For example, in the reaction discussed above the conversion of NO2 and CO to products at room temperature is slow because only a small fraction of the molecules has enough energy to reach the transition state. The rate can be increased by heating the sample. Raising the temperature increases the reaction rate by increasing the fraction of molecules with enough energy to surmount the activation energy barrier (Figure 14.12).

Lower temperature Number of molecules with a given energy

Figure 14.12  Energy distribution curve.  ​Note that this resembles Figure 10.11, the Boltzmann distribution function, for a collection of gas molecules.

Higher temperature T1 T2 At a higher temperature, a larger fraction of the molecules has sufficient energy to react. A lower fraction of molecules has the required energy at lower temperatures. Energy The minimum energy required for a given reaction.

698

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Collision Theory: Effect of Molecular Orientation on Reaction Rate Not only must reacting molecules have sufficient energy to react, but the reactant molecules must also come together in the correct orientation. For the reaction of NO2 and CO, imagine that the transition state structure has one of the O atoms of NO2 beginning to bind to the C atom of CO in preparation for O atom transfer (Figure 14.11). The lower the probability of achieving the proper alignment, the smaller the value of k, and the slower the reaction. Imagine what happens when two or more complicated molecules collide. In only a small fraction of the collisions will the molecules come together in exactly the right orientation. Thus, only a tiny fraction of the collisions can be effective. No wonder some reactions are slow. Conversely, it is amazing that so many are fast!

The Arrhenius Equation The observation that reaction rates depend on the energy and frequency of collisions between reacting molecules, on the temperature, and on whether the ­collisions have the correct geometry is summarized by the Arrhenius equation: k = Ae−Ea /RT

Frequency factor

Fraction of molecules with minimum energy for reaction

(14.5)

In this equation, k is the rate constant, R is the gas constant with a value of 8.31446 × 10−3 kJ/K ∙ mol, and T is the kelvin temperature. The parameter A is called the frequency factor. It is related to the number of collisions and to the ­fraction of collisions that have the correct geometry; A is specific to each reaction and is temperature dependent.  The factor e−Ea/RT represents the fraction of molecules having the minimum energy for reaction; it always has a value less than 1. As the following table shows, this fraction changes significantly with temperature. Temperature (K)

Value of e−Ea/RT for Ea = 40 kJ/mol-rxn

298

9.7 × 10−8

400

6.0 × 10−6

600

3.3 × 10−4

Interpreting the Arrhenius Equation 

(a) The exponential term gives the fraction of molecules having sufficient energy for reaction and is a function of T. (b) Although a complete understanding of the frequency factor goes beyond the level of this text, note that A becomes smaller as the reactants become larger. This reflects the fact that larger molecules have a smaller probability of coming together in the appropriate geometry.

The Arrhenius equation is significant because •

it can be used to calculate Ea from the temperature dependence of the rate ­constant, and

•

it can be used to calculate the rate constant, if Ea, T, and A are known.

If rate constants for a given reaction are measured at several temperatures then you can use graphical techniques to determine the activation energy of a reaction. ­Taking the natural logarithm of each side of Equation 14.5 gives

Ea, Reaction Rates, and Temperature ​ An often-used rule of thumb is that reaction rates double for every 10 °C rise in temperature in the vicinity of room temperature.

 E  ln k  ln A    a   RT 

Rearranging this expression slightly shows that ln k and 1/T are related linearly.





g

Ea  1     ln A  m Arrhenius equation R T g g

y =

mx

ln k  

+ b



(14.6)

m Equation for straight line 14.5  A Microscopic View of Reaction Rates

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699

This means that, if the natural logarithm of k (ln k) is plotted versus 1/T, the result is a downward-sloping straight line with a slope of (−Ea/R). The activation energy, Ea, can be obtained from the slope of this line (Ea = −R × slope).

E xamp le 14.10

Determination of Ea Using the Arrhenius Equation Problem   Using the experimental data shown in the table, calculate the activation energy Ea for the reaction 2 N2O(g) n 2 N2(g) + O2(g)

k (L/mol ∙ s)

Experiment

Temperature (K)

1

1125

11.59

2

1053

 1.67

3

1001

 0.380

4

 838

 0.0011

What Do You Know?  Rate constants are given at several temperatures. Strategy   To solve this problem graphically, first calculate ln k and 1/T for each data point. These data are then plotted, and Ea is calculated from the resulting straight line (slope = −Ea /R).

Solution  First, calculate 1/T and ln k.

2

ln k

1

8.889 × 10−4

   2.4501

2

9.497 × 10−4

   0.513

3

9.990 × 10−4

−0.968

4

11.9 × 10−4

−6.81

Ea R E a  R  Slope  (8.31  103 kJ/K  mol)(3.03  104 K)  250 kJ/mol Slope  

0 ln k

1/T (K−1)

Plotting these data gives the graph shown in Figure 14.13. A linear least squares fit of the line by Microsoft Excel gives the slope as –3.03 × 104 K. The activation energy is evaluated from this slope.

4

−2 −4

Think about Your Answer   Notice the units of the answer. Using the value for R = 8.31 × 10−3 kJ/K ∙ mol will give an answer with the units kJ/mol.

−6 −8 −10

Experiment

Check Your Understanding  7

8

9 10 11 12 ( 1 ) × 104 K−1 T

Figure 14.13  Arrhenius plot.  ​ A plot of ln k versus 1/T for the reaction 2 N2O(g) n 2 N2(g) + O2(g). The slope of the line can be used to calculate Ea. See Example 14.10.

700

Rate constants were determined for the decomposition of acetaldehyde (CH3CHO) in the temperature range 700 to 1000 K. Use these data to determine Ea for the reaction using a graphical method. T (K) k (L/mol ∙ s)

700 0.011

760 0.105

840 2.17

1000  145

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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The activation energy, Ea, for a reaction can be obtained algebraically if k is known at two different temperatures. You can write an equation for each set of these conditions:  E  ln k1    a   ln A  RT1 

and

 E  ln k2    a   ln A  RT2 

Subtracting one of these equations from the other gives ln k2  ln k1  ln



k2 E 1 1  a    k1 R  T2 T1 

(14.7)

Example 14.11 demonstrates the use of this equation.

Ex am p le 14.11

Calculating Ea Numerically Problem   Use values of k determined at two different temperatures to calculate the value of Ea for the decomposition of HI: 2 HI(g) n H2(g) + I2(g) k1 = 2.15 × 10−8 L/(mol ∙ s) at 6.50 × 102 K (T1) k2 = 2.39 × 10−7 L/(mol ∙ s) at 7.00 × 102 K (T2)

What Do You Know?  Values for the rate constant at two temperatures are given. Strategy  Substitute values of k1, k2, T1, and T2 into Equation 14.7 and solve for Ea. Solution ln

Ea 2.39  107 L/(mol  s) 1 1     2.15  108 L/(mol  s) 8.314  103 kJ/K  mol  7.00  102 K 6.50  102 K 

Solving this equation gives  Ea = 180 kJ/mol. 

Think about Your Answer  Another way to write the difference in fractions in brackets is 1 T1  T2 1 T  T   TT  2 1  1 2 This expression is sometimes easier to use.

Check Your Understanding  The colorless gas N2O4 decomposes to the brown gas NO2 in a first-order reaction. N2O4(g) n 2 NO2(g) The rate constant k = 4.5 × 103 s−1 at 274 K and k = 1.00 × 104 s−1 at 283 K. What is the activation energy, Ea?



14.5  A Microscopic View of Reaction Rates

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701

14.6 Catalysts Goals for Section 14.6 • Describe the functioning of a catalyst and its effect on the activation energy of a reaction.

• Understand the lock-and-key and induced-fit models for substrate binding to enzymes.

• Understand the basics of the Michaelis-Menten model of enzyme kinetics. A catalyst is a substance that speeds up the rate of a chemical reaction, and you have seen several examples of catalysts in earlier discussions in this chapter; MnO2, iodide ion, the enzyme catalase, and hydroxide ion all catalyze the decomposition of hydrogen peroxide (Figure 14.4). Catalysts are not consumed in a chemical reaction. They are, however, intimately involved in the details of the reaction at the particulate level. A catalyst’s function is to provide a different pathway with a lower activation energy for the reaction.

Effect of Catalysts on Reaction Rate To illustrate how a catalyst participates in a reaction, consider the isomerization of cis-2-butene to the slightly more stable isomer trans-2-butene. CH3

H3C C H

C

(g)

C

H

H3C

H

cis-2-butene

H Transition state

C

(g) CH3

trans-2-butene

End rotates  bond breaks

A Closer Look

The activation energy for the uncatalyzed conversion is relatively large—264 kJ/ mol—because the π bond must be broken to allow one end of the molecule to rotate into a new position. Because of the high activation energy, this is a slow reaction, and rather high temperatures are required for it to occur at a reasonable rate. The cis- to trans-2-butene reaction is greatly accelerated by a catalyst, iodine, and the reaction can occur in the presence of iodine at a temperature several hundred degrees lower than for the uncatalyzed reaction. Iodine is not consumed (nor is it a

702

Thinking about Kinetics, Catalysis, and Bond Energies  s you will see in Section 14.7, one of A the most important uses of chemical kinetics is to give chemists an idea as to the reaction mechanism, the stepby-step process by which a chemical reaction occurs. These individual steps involve bond breaking and bond formation and guide the chemical reaction from reactants to products. Bond breaking requires energy, while bond formation evolves energy.

Consider the uncatalyzed conversion between cis- and trans-2-butene. The postulated mechanism for this reaction involves rotation around the double bond, which would occur if the two ends of the molecule twist out of alignment by 90° resulting in the π bond breaking. You can estimate the energy of the π bond in this species using data from the table of bonddissociation enthalpies (Table 8.8). The average bond-dissociation enthalpies of

CPC double and COC single bonds are 610 kJ/mol and 346 kJ/mol, respectively; the difference between these two values is 264 kJ/mol-rxn. Note the correspondence between this value and the measured value for the activation energy of this reaction, 264 kJ/molrxn. Matching the estimated bond energy and the activation energy is evidence in support of this mechanism.

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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trans-2-butene

1 2

Step 5: I atoms reform I2

Step 1: I2 dissociation

Step 4: I atom dissociates to form trans isomer

cis-2-butene

Step 2: I atom attaches to cis isomer

Figure 14.14  The mechanism of the iodine-catalyzed isomerization of cis-2-butene to trans-2-butene.  ​In the text and in this figure, the addition of an iodine atom to cis-2-butene to eventually form trans-2-butene is emphasized. The iodine atom, however, can also add to trans2-butene to form cis-2-butene. Both the forward and the reverse reactions are sped up, so that the system reaches equilibrium (Chapter 15) more quickly.

Step 3: C C bond rotation

product), and it does not appear in the overall balanced equation. It does appear in the reaction rate law, however; the rate of the reaction depends on the square root of the iodine concentration: Rate = k[cis-2-butene][I2]1/2

The presence of I2 changes the way the butene isomerization reaction occurs; that is, it changes the mechanism of the reaction (Figure 14.14). The best hypothesis is that iodine molecules first dissociate to form iodine atoms (Step 1). This requires much less energy input than rotation of the CPC double bond. An iodine atom then adds to one of the C atoms of the CPC double bond (Step 2). This converts the double bond between the carbon atoms to a single bond (the π bond is broken) and allows the ends of the molecule to twist freely relative to each other (Step 3). If the iodine atom then dissociates from the intermediate, the double bond can re-form in the trans configuration (Step 4). The iodine atom is now free to add to another molecule of cis-2-butene. The result is a kind of chain reaction, as one molecule of cis-2-butene after another is converted to the trans isomer. The chain is broken if the iodine atom recombines with another iodine atom to re-form molecular iodine. An energy profile for the catalyzed reaction (Figure 14.15) shows that the overall energy barrier is much lower than for the uncatalyzed reaction. Five separate steps are identified for the mechanism in the energy profile. This proposed mechanism also includes a series of reaction intermediates, species formed in one step of the reaction and consumed in a later step. Iodine atoms are intermediates, as are the free radical species formed when an iodine atom adds to cis-2-butene. Five important points are associated with this mechanism: •

Iodine molecules, I2, dissociate to atoms and then re-form. On the macroscopic level, the concentration of I2 is unchanged. Iodine does not appear in the balanced, stoichiometric equation even though it appears in the rate equation. This is generally true of catalysts.

•

Both the catalyst I2 and the reactant cis-2-butene are in the gas phase. If a catalyst is present in the same phase as the reacting substance, it is called a h ­ omogeneous catalyst. On the other hand, if the catalyst is in a different phase than the reacting substance, it is called a heterogeneous catalyst; an example of this is the solid MnO2 catalyst used to decompose aqueous H2O2 in Figure 14.4.



14.6 Catalysts

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703

300 Profile for uncatalyzed reaction. Ea = 264 kJ/mol

Energy (kJ/mol)

200 3

2 100

1

5 ~75 kJ/mol

0

Profile for catalyzed reaction. The shape of the barrier changes and the value of Ea is lower.

4

~118 kJ/mol 4 kJ/mol

(Reactants) cis-C4H8 + I2

Products trans-C4H8 + I2

−100

Reaction progress

Figure 14.15  Energy profile for the iodine-catalyzed reaction of cis-2-butene to form trans-2-butene.  ​A catalyst accelerates a reaction by altering the mechanism so that the activation energy is lowered. With a smaller barrier to overcome, more reacting molecules have sufficient energy to surmount the barrier, and the reaction occurs more rapidly.

•

Iodine atoms and the radical species formed by addition of an iodine atom to a 2-butene molecule are intermediates.

•

The activation-energy barrier to reaction is significantly lower because the mechanism changed. Dropping the activation energy from 264 kJ/mol for the uncatalyzed reaction to about 150 kJ/mol for the catalyzed process makes the catalyzed reaction 1015 times faster!

•

The diagram of energy-versus-reaction progress has five energy barriers (five humps appear in the curve). This feature in the diagram means that the reaction occurs in a series of five steps.

Many reactions are catalyzed by acids and bases. Acid catalysts are often used in the synthesis of organic molecules, and an example is an esterification reaction. An ester is produced from the condensation reaction of a carboxylic acid and an alcohol (Section 23.4). For example, the reaction of acetic acid with methanol produces methyl acetate and water. O

O CH3COH + HOCH3

CH3COCH3 + H2O

Without a catalyst, the system is slow to reach equilibrium, but the rate of the process is greatly increased by the addition of a strong acid (such as HCl or H2SO4). The acid catalyzes two steps in the reaction mechanism. First, the acid protonates the carbonyl oxygen (the O atom of the CPO group) on the acetic acid molecule. +OH

O CH3COH + H+

CH3COH

OH CH3COH +

Protonation weakens the carbon-oxygen double bond and increases the positive ­character of the carbon atom. The carbon atom then bonds to the oxygen atom of methanol, producing a tetrahedral intermediate, and a proton is returned to the solution. OH CH3COH + HOCH3 +

OH CH3COH +

HOCH3

704

OH CH3COH + H+ OCH3

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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In the second catalyzed part of the reaction, one of the hydroxyl groups is first protonated and is then removed as water. +OH

OH

OH CH3COH + H+

+

CH3COCH3 + H2O

CH3COH2 OCH3

OCH3

Finally, a proton is lost, resulting in the formation of methyl acetate. +OH

CH3COCH3

O CH3COCH3 + H+

Remember, catalysts are not consumed in chemical reactions. Although protons appear twice in the mechanism as reactants, they also appear twice as products. Overall, there is no net production or consumption of protons in the reaction. It is important to know that although a catalyst increases the rate of a chemical reaction, it does not influence whether a reaction is product-favored or reactantfavored at equilibrium. A catalyst simply enables a reaction to reach equilibrium more quickly.

Enzymes Most reactions required for life occur too slowly on their own, so organisms have developed biological catalysts to speed them up to the appropriate levels. These biological catalysts are called enzymes, which are typically large proteins (­Section 24.1). Enzyme-catalyzed reactions are often 107 to 1014 times faster than uncatalyzed reactions. For an enzyme to catalyze a reaction, several key steps must occur: 1. The reactant (often called the substrate) must bind to the enzyme. 2. The chemical reaction must take place. 3. The products(s) of the reaction must leave the enzyme so that more substrate can bind and the process can be repeated. The location in the enzyme where the substrate binds and the reaction occurs is called the active site. The active site often consists of a cavity or cleft in the structure of the enzyme into which the substrate or part of the substrate can fit in just the right orientation so that specific bonds can be broken and/or made. How do enzymes bring about such fast and specific chemical reactions? An early model of enzyme binding is the lock-and-key model (Figure 14.16a). The shape of the substrate is pictured as fitting into the enzyme's active site precisely, much like a key (the substrate) fits a lock (the enzyme). This model has been supplanted by the induced-fit model (Figure 14.16b) in which the enzyme is not viewed as a rigid structure like a lock, but instead as a very large molecule that has some threedimensional flexibility. The enzyme begins with a structure that does not match the shape of the substrate exactly. Instead, binding of the substrate causes the structure of the enzyme to change so that it fits the substrate. This model can also incorporate another feature of enzyme action. A catalyst functions by providing a pathway for reaction that has a lower activation energy for the reaction. An enzyme's active site, therefore, should optimally bind to and thus stabilize the transition state for the reaction, as opposed to the substrate. When a substrate binds, it undergoes changes in its shape, referred to as conformational changes, that push it in the direction that leads to the formation of the transition state of the reaction. In the induced-fit model, therefore, both the enzyme and the substrate undergo conformational changes upon binding that lower the activation energy for the reaction, leading to a faster reaction.

14.6 Catalysts

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705

Lock-and-key model

Induced-fit model

Campbell, M. K., Farrell, S. O., & McDougal, O. M. Biochemistry, 9ed, 2018, Cengage Learning.

Substrate

Active site

+ 1

Substrate

3

+ 1

3

2 Enzyme

1 2

2 Enzymesubstrate complex

(a)  In the lock-and-key model, the enzyme’s active site is pictured as being complementary to the structure of the substrate.

Figure 14.16  Models of substrate binding.

3 Enzyme

1

3 2 Enzymesubstrate complex

(b)  In the induced-fit model, the enzyme’s structure is viewed as being flexible so that, upon binding of the substrate, it changes to become complementary to the substrate.

The rate laws of enzyme-catalyzed reactions demonstrate an interesting property. At low substrate concentrations, the rate law of the reaction is often first-order in substrate, S (Figure 14.17). At high substrate concentrations, however, the rate of the reaction levels off and the rate law is zero-order in S. In 1913, Leonor Michaelis and Maud L. Menten proposed a general theory of enzyme action based on these kinetic observations. They assumed that the substrate and the enzyme, E, form a complex, ES. The chemical reaction then occurs, and the enzyme and the product, P, are released. E + S uv ES ES n P + E

Rate (V)

Vmax

[S]

Figure 14.17  Rate of enzymecatalyzed reaction. ​This plot of reaction rate versus substrate concentration [S] is typical of reactions catalyzed by enzymes that follow the Michaelis-Menten model.

At low substrate concentrations, the first-order kinetics indicate that adding more substrate will increase the rate of the reaction. Why then do the kinetics change at high substrate concentration? Because there is only so much enzyme present, the active sites in the available enzyme molecules become saturated at high substrate concentrations, and the rate reaches its maximum value. Michaelis and Menten were further able to show that their mechanism led to the following equation for the rate (or velocity) of the reaction, V. V 

Vmax[S] KM  [S]

where Vmax is the maximum rate of the reaction and K M is a constant called the Michaelis constant. When V is plotted versus [S], this equation fits the curve in ­Figure 14.17 well.

14.7 Reaction Mechanisms Goals for Section 14.7 • Understand the concept of a reaction mechanism and the relation of the mechanism to the overall, stoichiometric equation for a reaction.

• Describe the elementary steps of a mechanism, and give their molecularity. • Identify reaction intermediates in a mechanism. • Predict the rate law for a reaction, given a simple mechanism and the identity of the rate-determining step.

706

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Why study chemical kinetics? Beyond simply telling you what factors influence the rate of a chemical reaction and how you might be able to speed up or slow down a particular reaction, chemical kinetics also gives you insight into how a chemical reaction actually occurs. The reaction mechanism for a chemical reaction is the sequence of bond-making and bond-breaking steps that occurs during the conversion of reactants to products. Based on the rate equation for a reaction, and by applying chemical intuition, chemists can often make an educated guess about the mechanism for the reaction. In some reactions, the conversion of reactants to products in a single step is envisioned as the logical mechanism. For example, the uncatalyzed isomerization of cis-2-butene to trans-2-butene is best described as a single-step reaction (see Figure 14.15). Most chemical reactions occur in a sequence of steps, however. A multiple-step mechanism was proposed for the iodine-catalyzed 2-butene isomerization reaction (Figures 14.14 and 14.15). Another example of a multistep reaction is the reaction of bromine and NO: Br2(g) + 2 NO(g) n 2 BrNO(g)

A single-step reaction would require that three reactant molecules collide simultaneously in just the right orientation. The probability of this occurring is small; thus, it is reasonable to look for a mechanism that occurs in a series of steps, with each step involving only one or two molecules. In one possible mechanism, Br2 and NO might combine in an initial step to produce an intermediate species, Br2NO. +

Step 1

NO

Br2

Br2NO

This intermediate would then react with another NO molecule to give the reaction products. +

Step 2 NO

+ Br2NO

BrNO

BrNO

The equation for the overall reaction is obtained by adding the equations for these two steps:

Step 1

Br2(g) + NO(g) uv Br2NO(g)

Step 2

NO(g) + Br2NO(g) n 2 BrNO(g)

Overall Reaction

Br2(g) + 2 NO(g) n 2 BrNO(g)

Each step in a multistep reaction sequence is an elementary step, defined by a chemical equation that describes a single molecular event such as the formation or rupture of a chemical bond resulting from a molecular collision. Each step has its own activation energy, Ea, and rate constant, k. Adding the equations for each step must give the balanced equation for the overall reaction, and the time required to complete all of the steps defines the overall reaction rate. The series of steps constitutes a possible reaction mechanism.

14.7  Reaction Mechanisms

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707

Molecularity of Elementary Steps Elementary steps are classified by the number of reactant molecules (or ions, atoms, or free radicals) that come together. This whole, positive number is called the molecularity of the elementary step. When one molecule is the only reactant in an elementary step, the reaction is a unimolecular process. A bimolecular elementary process involves two molecules, which may be identical (A + A n products) or different (A + B n products). The mechanism proposed for the decomposition of ozone in the stratosphere illustrates the use of these terms. Step 1

Unimolecular

O3(g) n O2(g) + O(g)

Step 2

Bimolecular

O3(g) + O(g) n 2 O2(g) 2 O3(g) n 3 O2(g)

Overall Reaction

A termolecular elementary step involves three molecules, which could be the same or different (3 A n products; 2 A + B n products; or A + B + C n products). Be aware, however, the simultaneous collision of three molecules has a low probability, unless one of the molecules involved is in high concentration, such as a solvent molecule. In fact, most termolecular processes involve the collision of two reactant molecules and a third, inert molecule. The function of the inert molecule is to absorb the excess energy produced when a new chemical bond is formed by the first two molecules. For example, N2 is unchanged in a termolecular reaction between oxygen molecules and oxygen atoms that produces ozone in the upper atmosphere: O(g) + O2(g) + N2(g) n O3(g) + energetic N2(g)

The probability that four or more molecules will simultaneously collide with sufficient kinetic energy and proper orientation to react is so small that reaction molecularities greater than three are never proposed. How does molecularity fit in with the previous description of collision theory? You can see that bimolecular or termolecular reactions can occur by collisions of reactant molecules with sufficient energy and in the correct orientation to react. But what about a unimolecular reaction? How can collisions be involved? In a unimolecular reaction, the reacting species must acquire enough energy to break apart. This energy can come from collisions of the reacting molecule with other molecules, such as with other molecules of the same kind, with solvent molecules, or with some high-energy background molecule. Atmospheric ozone, for example, disintegrates into O2 molecules and O atoms by colliding with high-energy nitrogen molecules.

Rate Equations for Elementary Steps The rate equation for a reaction must be determined by experiment; it cannot be predicted from the overall stoichiometry.  In contrast, the rate equation for any ­elementary step is defined by its reaction stoichiometry.  The rate equation of an elementary step is given by the product of its rate constant and the concentrations of the reactants in that step.  You can therefore write the rate equation for any elementary step, as shown by examples in the following table:

708

Elementary Step

Molecularity

Rate Equation

A n product

Unimolecular

Rate = k[A]

A + B n product

Bimolecular

Rate = k[A][B]

A + A n product

Bimolecular

Rate = k[A]2

2 A + B n product

Termolecular

Rate = k[A]2[B]

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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A Closer Look

Organic Bimolecular Substitution Reactions

Experimentally determined rate laws can be used to propose mechanisms and to eliminate other mechanisms from consideration. To decide among possible mechanisms, however, additional experimental information and chemical intuition are also required. Consider the reaction between 2-bromooctane and iodide ion (­Figure A) (a reaction similar to the reaction of fluoride ion and chloromethane (page  697)). This reaction follows second-order kinetics; it is first-order in each of the reactants, and from this you know that a single-step mechanism is one possibility. A possible single-step mechanism is shown in Figure B. This involves attack of iodide ion on the side of the carbon atom opposite to Br, which leads to breaking of the carbon–bromine bond

H I− + C6H13

C

H C6H13

CH3

Br

C

CH2 + Br−

I

Figure A  Reaction of 2-bromooctane with iodide ion. this is the correct mechanism, and if you start with a compound having the geometry shown, then the product of the reaction should have the inverted geometry. This is exactly what is found in the laboratory and provides strong evidence in support of this mechanism.

concurrent with formation of a carbon– iodine bond. Notice that the geometry around the carbon atom in 2-bromooctane inverts in the course of this mechanism. (The geometry around the carbon has been inverted like an umbrella inverts in a strong wind.) If

I−

C6H13 H3C C Br + H

I

CH3

C6H13

H3C C

Br

H

I

C

C6H13

+ Br−

H

Figure B  Proposed mechanism for the reaction of 2-bromooctane with iodide ion.

For example, the rate laws for each of the two elementary steps in the decomposition of ozone are Rate for (unimolecular) Step 1 = k[O3] Rate for (bimolecular) Step 2 = k′[O3][O]

The two rate constants (k and k′ in this example) are not expected to have the same value (nor the same units, because the two steps have different molecularities).

Ex am p le 14.12

Elementary Steps Problem   The hypochlorite ion undergoes self-oxidation–reduction to give chlorate ions, ClO3−, and chloride ions. 3 ClO−(aq) n ClO3−(aq) + 2 Cl−(aq) This reaction is thought to occur in two steps: Step 1

ClO−(aq) + ClO−(aq) n ClO2−(aq) + Cl−(aq)

Step 2

ClO2−(aq) + ClO−(aq) n ClO3−(aq) + Cl−(aq)

What is the molecularity of each step? Write the rate equation for each elementary step. Show that the sum of these reactions gives the equation for the net reaction.

What Do You Know?   A two-step mechanism for the reaction of ClO− to form Cl− and ClO3− is proposed.

Strategy  The molecularity is the number of ions or molecules involved in an elementary step. The rate equation for each elementary step involves the concentration of each ion or molecule in the elementary step, raised to the power of its stoichiometric coefficient.

14.7  Reaction Mechanisms

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709

Solution   Because two ions are involved in each elementary step,  each step is bi­molecular.  The rate equation for any elementary step involves the product of the concentrations of the reactants. Thus, in this case, the rate equations are Step 1

 Rate = k1[ClO−]2 

Step 2

 Rate = k2[ClO2−][ClO−] 

The ClO2− ion is an intermediate, a product of the first step and a reactant in the second step. It therefore cancels out, leaving the balanced equation for the overall reaction: Step 1

ClO−(aq) + ClO−(aq) n ClO2 (aq)  + Cl−(aq)

Step 2

ClO2 (aq)  + ClO−(aq) n ClO3−(aq) + Cl−(aq)

Sum of steps

3 ClO−(aq) n ClO3−(aq) + 2 Cl−(aq) 

Think about Your Answer  Other mechanisms are possible. The next question to ask is “What evidence will let you decide between several different mechanisms?”

Check Your Understanding  Nitrogen monoxide is reduced by hydrogen to give nitrogen and water: 2 NO(g) + 2 H2(g) n N2(g) + 2 H2O(g) One possible mechanism for this reaction involves the following reactions: 2 NO(g) n N2O2(g) N2O2(g) + H2(g) n N2O(g) + H2O(g) N2O(g) + H2(g) n N2(g) + H2O(g) What is the molecularity of each of the three steps? What is the rate equation for each step? Identify the intermediates in this reaction; how many different intermediates are there? Show that the sum of these elementary steps gives the equation for the overall reaction.

Reaction Mechanisms and Rate Equations The dependence of rate on concentration is an experimental fact. Mechanisms, in contrast, are constructs of our imagination, intuition, and good chemical sense. When you propose a mechanism, you are making a hypothesis about how the reaction occurs at the particulate level. Several mechanisms can often be proposed that correspond to the observed rate equation. A good mechanism is a worthy goal because it allows chemists to understand the chemistry better and perhaps to predict how to control a reaction better and how to design new experiments. One of the important guidelines of kinetics is that products of a reaction can never be produced at a rate faster than the rate of the slowest step. If one step in a

710

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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multistep reaction is slower than the others, then the rate of the overall reaction is limited by the combined rates of all elementary steps up through the slowest step in the mechanism. Often the overall reaction rate and the rate of the slow step are nearly the same. If the slow step determines the rate of the reaction, it is called the rate-determining step, or rate-limiting step. Imagine that a reaction takes place with a mechanism involving two sequential steps, and assume that you know the rates of both steps. The first step is slow and the second is fast: Elementary Step 1:

k1 A  B  → X M Slow, Ea large

Elementary Step 2:

k2 M  A  → Y Fast, Ea small

Overall Reaction:

2 A  B → X  Y

In the first step, A and B come together and slowly react to form one of the products (X) plus another reactive species, M. Almost as soon as M is formed, however, it is rapidly consumed by reacting with another molecule of A to form the second ­product Y. The rate-determining elementary step in this example is the first step. That is, the rate of the first step is equal to the rate of the overall reaction. This step is bimolecular and so it has the rate equation: Rate = k1[A][B]

where k1 is the rate constant for that step. The overall reaction is expected to have this same second-order rate equation. Now consider how these ideas can be applied to the mechanism of a real ­reaction: the second-order reaction of nitrogen dioxide with fluorine. 2 NO2(g) + F2(g) n 2 FNO2(g) Rate = k[NO2][F2]

The rate equation rules out the possibility that the reaction occurs in a single step. If this was a single-step reaction, the rate law would be Rate = k[NO2]2[F2], with a second-order dependence on [NO2]. Because a single-step reaction is ruled out, the mechanism must include at least two steps. You can also conclude from the rate law that the rate-determining elementary step must involve NO2 and F2 in a 1:1 ratio. One possible mechanism for the NO 2/F 2 reaction proposes that NO 2 and F2 first react to produce one molecule of the product (FNO2) plus one F atom. In a second step, the fluorine atom produced in the first step reacts with additional NO2 to give a second molecule of product. If the first, bimolecular step is rate determining, the rate equation would be Rate = k1[NO2][F2], the same as the experimentally observed rate equation. The experimental rate constant would be the same as k1. Elementary Step 1:

Slow

NO2(g) + F2(g)

Elementary Step 2:

Fast

NO2(g) + F(g)

Overall Reaction:

2 NO2(g) + F2(g)

k1 k2

Can You Derive a Mechanism? ​ At this introductory level, you cannot be expected to derive reaction mechanisms. Given a mechanism, however, you can decide whether it agrees with experimental rate laws.

FNO2(g) + F(g) FNO2(g) 2 FNO2(g)

The fluorine atom formed in the first step of the NO2/F2 reaction is a reaction intermediate. It does not appear in the equation describing the overall reaction. ­Reaction intermediates usually have only a fleeting existence, but in some instances they have long enough lifetimes to be observed. When this is possible, the detection and identification of an intermediate provide strong evidence supporting the proposed mechanism.



14.7  Reaction Mechanisms

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711

E xamp le 14.13

Elementary Steps and Reaction Mechanisms Problem   Oxygen atom transfer from NO2 to CO produces nitrogen monoxide and ­carbon dioxide. NO2(g) + CO(g) n NO(g) + CO2(g) The mechanism for this reaction depends on the temperature, with one mechanism followed at low temperatures (less than 500 K) and another mechanism followed at high temperatures. The rate equation for this reaction at temperatures less than 500 K is Rate = k[NO2]2. Can the low-temperature reaction occur in one bimolecular step?

What Do You Know?  The reaction equation and the rate law for the reaction are given.

Strategy   Write the rate law based on the equation for the NO2 + CO reaction occurring as if it were an elementary step. If this rate law corresponds to the observed rate law, then a one-step mechanism is possible.

Solution   If the reaction occurs by the collision of one NO2 molecule with one CO ­molecule, the rate equation would be Rate = k[NO2][CO] This does not agree with experiment, so  the mechanism must involve more than a single step. 

Think about Your Answer   Because the rate equation is second-order in [NO2], the rate-determining step in this multistep reaction must involve the collision of two NO2 molecules. In one possible mechanism, the reaction occurs in two bimolecular steps, the first one slow and the second one fast: Elementary Step 1:

Slow, ratedetermining

Elementary Step 2:

Fast

Overall Reaction:

2 NO2(g)

NO3(g) + NO(g)

NO3(g) + CO(g)

NO2(g) + CO2(g)

NO2(g) + CO(g)

NO(g) + CO2(g)

The first (rate-determining) step has a rate equation that agrees with experiment, so this is a possible mechanism.

Check Your Understanding  The Raschig reaction produces hydrazine, N2H4, an industrially important reducing agent, from NH3 and OCl− in basic, aqueous solution. A proposed mechanism is Step 1:

Fast

NH3(aq) + OCl−(aq) n NH2Cl(aq) + OH−(aq)

Step 2:

Slow

NH2Cl(aq) + NH3(aq) n N2H5+(aq) + Cl−(aq)

Step 3:

Fast

N2H5+(aq) + OH−(aq) n N2H4(aq) + H2O(ℓ)

(a) What is the overall equation? (b) Which step of the three is rate determining? (c) Write the rate equation for the rate-determining elementary step. (d) What reaction intermediates are involved?

712

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Reaction Mechanisms Involving an Initial Equilibrium Step A common two-step reaction mechanism involves an initial fast reaction that produces an intermediate, followed by a slower second step in which the intermediate is converted to the final product. The rate of the reaction is determined by the second step. The rate of that step, however, depends on the concentration of the intermediate. The concentration of an intermediate should not appear as a term in the overall rate equation because this concentration will probably not be measurable and because such a rate law cannot be compared directly with the experimentally determined rate law. Therefore, the expression for the intermediate must be replaced with another expression in terms of quantities measurable in the lab. The reaction of nitrogen monoxide and oxygen is an example of a two-step reaction where the first step is fast and the second step is rate determining. 2 NO(g) + O2(g) n 2 NO2(g) Rate = k[NO]2[O2]

The experimentally determined rate law shows second-order dependence on NO and first-order dependence on O2. Although this rate law would be correct for a termolecular reaction, experimental evidence indicates that an intermediate is formed in this reaction. A possible two-step mechanism that proceeds through an intermediate is Elementary Step 1:

Fast, equilibrium

k1

NO(g) + O2(g)  uv OONO(g) k−1 intermediate

Elementary Step 2:

Slow, rate-determining

k2

NO(g) + OONO(g)  n  2 NO2(g) 2 NO(g) + O2(g)  n  2 NO2(g)

Overall Reaction:

The second step of this reaction is the slow step and determines the overall rate. You can write a rate law for the second step, but this rate law cannot be compared directly with the experimental rate law because it contains the concentration of an intermediate, OONO: Rate = k2[NO][OONO]

To eliminate the concentration of intermediate from this rate expression, look at the rapid first step, which involves an equilibrium between the intermediate species and the reactants. At the beginning of the reaction, NO and O 2 react rapidly and produce the intermediate OONO. The rate of formation can be defined by a rate law with a rate constant k1: Rate of production of OONO (NO + O2 n OONO) = k1[NO][O2]

Because the intermediate is consumed only very slowly in the second step, it is possible for the OONO to revert to NO and O2 before it reacts further: Rate of reverse reaction (OONO n NO + O2) = k−1[OONO]

As NO and O2 form OONO, their concentrations drop, so the rate of the forward reaction decreases. At the same time, the concentration of OONO builds up, so the rate of the reverse reaction increases. At equilibrium, the rates of the forward and reverse reactions become the same.

Mechanisms with an Initial Equilibrium ​In this mechanism,

the forward and reverse reactions in the first elementary step are so much faster than the second elementary step that equilibrium is established before any significant amount of OONO is consumed by NO to give NO2. The state of equilibrium for the first step remains throughout the lifetime of the overall reaction.

Rate of forward reaction = rate of reverse reaction k1[NO][O2] = k−1[OONO]

Rearranging this equation gives k1 [OONO]  K k1 [NO][O2]

14.7  Reaction Mechanisms

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713

Equilibrium Constant The important concept of chemical equilibrium was introduced in Chapter 3 and will be described in more detail in Chapters 15–18.

Both k1 and k−1 are constants (they will change only if the temperature changes). The ratio of the two rate constants must also be a constant. This constant is defined as the equilibrium constant and is given the symbol K. From this, you can derive an expression for the concentration of OONO: [OONO] = K [NO][O2]

If K[NO][O2] is substituted for [OONO] in the rate law for the rate-determining elementary step, you have Rate = k2[NO][OONO] = k2[NO]{K [NO][O2]} = k2K[NO]2[O2]

Because both k2 and K are constants, their product is another constant k′, and you have Rate = k′[NO]2[O2]

This is exactly the rate law derived from experiment. Thus, the sequence of reactions on which the rate law is based may be a reasonable mechanism for this reaction. It is not the only possible mechanism, however. This rate equation is also consistent with the reaction occurring in a single termolecular step, and another possible mechanism is illustrated in Example 14.14.

E xamp le 14.14

Reaction Mechanism Involving an Equilibrium Step Problem  The NO + O2 reaction described in the text could also occur by the following mechanism: Elementary Step 1: Fast, equilibrium k1

NO(g) + NO(g)  uv N2O2(g) k−1 intermediate

Elementary Step 2: Slow, rate-determining k2

N2O2(g) + O2(g)  n  2 NO2(g) Overall Reaction: 2 NO(g) + O2(g)  n  2 NO2(g) Show that this mechanism leads to the following experimental rate law: Rate = k[NO]2[O2].

What Do You Know?   A possible mechanism for the reaction of NO and O 2 is given.

Strategy  The rate law for the reaction is derived using the stoichiometry of the ratedetermining step. If this step involves an intermediate, an expression for the concentration of this intermediate will need to be found that has only the concentrations of species that can be measured in the lab, such as reactants, products, or homogeneous catalysts. Solution  The rate law for the rate-determining elementary step is Rate = k2[N2O2][O2] The intermediate N2O2 cannot appear in the final derived rate law. However, you can find an expression relating the concentration of the intermediate to the concentrations of the reactants, using the equilibrium constant expression for the first step. [N2O2] and [NO] are related by the equilibrium constant. k1 [N O ]  2 22  K k1 [NO]

714

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Solving this equation for [N2O2] gives [N2O2] = K[NO]2. When this is substituted into the derived rate law  Rate  = k2{K[NO]2}[O2]  the resulting equation is identical with the experimental rate law where k2K = k.

Think about Your Answer   Three mechanisms have been proposed for the NO + O2 reaction. The challenge for chemists is to decide which is correct. In this case, further experimentation detected the species OONO as a short-lived intermediate, thus providing evidence for the mechanism involving this intermediate.

Check Your Understanding  One possible mechanism for the decomposition of nitryl chloride, NO2Cl, is k1

Elementary Step 1:  Fast, equilibrium

NO2Cl(g) uv NO2(g) + Cl(g)

Elementary Step 2:  Slow

NO2Cl(g) + Cl(g)  n NO2(g) + Cl2(g)

k−1

k2

What is the overall reaction? What rate law would be derived from this mechanism? What effect does increasing the concentration of the product NO2 have on the reaction rate?

Reaction Mechanisms Involving a Free-Radical Chain Reaction Chain reactions, in which one of the products of the reaction feeds back into an earlier step of the reaction and thus serves to keep the reaction going in a selfsustaining fashion, are sometimes encountered. These reactions are important in the synthesis of some organic chemicals and in some types of polymerization reactions, but chain reactions can become uncontrolled and can lead to explosions. An example of a chain reaction process is the reaction of methane and chlorine, which results in the formation of chloromethanes. The reaction does not occur when these reactants are mixed at room temperature and in the dark. This exothermic reaction requires either ultraviolet light or temperatures in the range of 300 °C to occur. If the reactants are present in a 1:1 ratio, the products are mainly CH3Cl and HCl, along with smaller amounts of CH2Cl2 and other chlorinated methanes. CH4(g) + Cl2(g) n HCl(g) + CH3Cl(g) (plus lesser amounts of CH2Cl2, CHCl3, and CCl4)

The mechanism is believed to involve a series of steps: 1. Initiation Step. Heat or UV light causes the chlorine molecule to dissociate into two chlorine atoms. Only a few chlorine atoms are needed to initiate the chain reaction. Cl2 n 2 ·Cl    ∆H = D(ClOCl) = 242 kJ/mol-rxn

2. Propagation Steps. The highly energetic chlorine atom is a free radical (­Section 8.6). As the ·Cl atom moves randomly in the system, various molecular collisions with other species (CH4, Cl2, and Cl) are expected to occur. Collision with a Cl2 molecule is not productive, but collision with a methane molecule can result in the formation of a ClOH bond concurrent with COH bond breaking. An estimate of the enthalpy of this process using bond-dissociation enthalpy values (Table 8.8) indicates that this reaction should be slightly exothermic.

D, Bond-Dissociation Enthalpy ​

Recall that D is a commonly used symbol for bond-dissociation enthalpies (Section 8.10).

CH3OH + ·Cl n ·CH3 + HCl    ∆H = D(COH) − D(HOCl) ∆H = 413 kJ/mol–rxn − 432 kJ/mol–rxn = −19 kJ/mol-rxn

14.7  Reaction Mechanisms

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Problem Solving Tip 14.2 Relating Rate Equations and Reaction Mechanisms experimental rate equation, the principles of stoichiometry, molecular structure and bonding, general chemical experience, and intuition.

The connection between an experimental rate equation and the proposed reaction mechanism is important in chemistry. 1. Experiments are performed to determine the experimental rate equation. 2. A mechanism for the reaction is proposed on the basis of the

3. The proposed reaction mechanism is used to derive a rate equation. If the derived and experimental rate equations are the same, the

postulated mechanism may be a reasonable hypothesis. 4. If more than one mechanism can be proposed and they all predict derived rate equations in agreement with the experiment, then more experiments must be done.

The methyl radical, ·CH3, generated in this step is also highly reactive. A collision with a Cl2 molecule can result in the formation of a COCl bond and a chlorine atom. This is also an exothermic reaction. ·CH3 + ClOCl n CH3Cl + ·Cl    ∆H = D(ClOCl) − D(COCl) ∆H = 242 kJ/mol–rxn − 339 kJ/mol–rxn = −97 kJ/mol-rxn

H2 and Cl2 Chain Reaction Another example of a chain reaction is the reaction of H2 and Cl2 gases to produce HCl gas. The reaction occurs when the ClOCl bond is broken by blue or UV light, but not by red or green light. There are many videos of the explosive reaction on the internet.

The chlorine atom formed in this step is now available to react with another molecule of CH4. The propagation steps can repeat over and over, in a self-sustaining fashion, a feature characteristic of a chain reaction. As the reaction progresses, the concentration of CH4 decreases and the concentration of the product, CH3Cl, increases. With these changes, the probability of collisions between CH3Cl and ·Cl atoms increases. These collisions lead to the formation of small amounts of CH2Cl2 and the other more highly chlorinated species. 3. Termination Step. The chain reaction will continue as long as the free radicals (·Cl or ·CH3) are present. It will end if the radicals cease to be present. This will happen if two Cl radicals (or any two radicals) collide and react. 2 ·Cl n Cl2    ∆H = −D(ClOCl) = −242 kJ/mol-rxn

The concentration of the various radical species in these reactions is exceedingly small. Thus, the probability of such a reaction is fairly low and the radical chain reaction will have a long lifetime before it stops. Summing the equations for the individual steps to give CH3Cl gives the overall reaction equation. The enthalpy change for this reaction, ΔH = −116 kJ/mol-rxn, is the sum of the enthalpies of the individual steps.

Applying Chemical Principles 14.1  Enzymes–Nature's Catalysts Many natural systems are controlled by catalysts called enzymes. One example is catalase (chapter opening photograph and Figure 14.4), which speeds up the decomposition of h ­ ydrogen peroxide. In the process of aerobic metabolism, molecular oxygen is reduced predominantly to water. A small amount of the oxygen, however, is reduced only partially, to hydrogen peroxide, which is highly toxic. Catalase ensures that the hydrogen peroxide formed does not build up in the body.

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Carbon dioxide is another product of aerobic metabolism. Carbon dioxide dissolves in water to a small extent to produce carbonic acid, which ionizes to give H3O+ and HCO3− ions. Reaction 1: CO2(g) uv CO2(aq) Reaction 2: CO2(aq) + H2O(ℓ) uv H2CO3(aq) Reaction 3: H2CO3(aq) + H2O(ℓ) uv H3O+(aq) + HCO3−(aq)

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Photos: © Charles D. Winters/Cengage

(a)

t=0s

(b)

A few drops of blood are added to a cold solution of CO2 in water. (You can use commercial soda water.)

t=3s

A few drops of a dye (bromothymol blue) are added to the solution, the yellow color indicating an acidic solution.

(c)

t = 15 s

(d)

A less-than-stoichiometric amount of sodium hydroxide is added, converting the H2CO3 to HCO3− (and CO32−). The blue color of the dye indicates a basic solution.

t = 17 s

(e)

t = 21 s

The blue color begins to fade after some seconds as CO2 forms more H2CO3. The amount of H2CO3 formed is finally sufficient to consume the added NaOH, and the solution is again acidic.

The acid-base chemistry of CO2 in water.

The enzyme carbonic anhydrase speeds up reactions 2 and 3. Many of the H3O+ ions produced by ionization of H2CO3 (reaction 3) are picked up by hemoglobin in the blood as hemoglobin loses O2. The resulting HCO3− ions are transported back to the lungs. When hemoglobin again takes on O2, it releases H3O+ ions. These ions and HCO3− re-form H2CO3, from which CO2 is liberated and exhaled. You can do an experiment that illustrates the effect of carbonic anhydrase. First, add a small amount of NaOH to a cold, aqueous solution of CO2. The solution becomes basic immediately because there is not enough H2CO3 in the solution to use up the NaOH. After some seconds, however, dissolved CO2 slowly produces more H2CO3, which consumes NaOH, and the solution is again acidic. Now try the experiment again, this time adding a few drops of blood to the solution (Figure). The solution becomes acidic very quickly. Carbonic anhydrase in blood speeds up the

reaction of CO2(aq) to HCO3−(aq) by a factor of about 107, as evidenced by the more rapid reaction under these conditions.

Questions

1. Catalase can decompose hydrogen peroxide to O2 and water about 107 times faster than the uncatalyzed reaction. If the latter requires one year, about how much time is required by the enzyme-catalyzed reaction? 2. How many moles of carbonic anhydrase (approximate molar mass = 29,000 g/mol) are present in 100. mL of a solution containing a 2 mg/mL concentration (approximately physiological concentration) of carbonic anhydrase? Carbonic anhydrase can catalyze very fast reactions. Some forms of this enzyme have a turnover number equal to 1 × 106 s–1. This means that they can convert substrate to products at a rate of 1 × 106 moles of CO2 per mole of enzyme per second. What mass of CO2 could be converted to HCO3− in 1 s by the amount of enzyme calculated for the solution in the first part of this question?

14.2  Kinetics and Mechanisms: A 70-Year-Old Mystery Solved Toward the end of the nineteenth century, the gas phase reaction of H2 with I2 was shown to be first-order for each reactant. H2(g) + I2(g) uv 2 HI(g) For approximately 70 years, the accepted mechanism was an elementary bimolecular collision that results in an exchange of atoms. However, in 1967, John H. Sullivan determined that this single-step mechanism is incorrect. Sullivan provided evidence that the reaction actually occurs in two steps. The first step of the mechanism is the dissociation of elemental iodine into iodine atoms. The dissociation of I2 is a fast equilibrium process that produces a relatively constant concentration of iodine atoms. Fast equilibrium

I2(g) uv 2 I(g)

The second step of the mechanism is a slow termolecular reaction between two iodine atoms and elemental hydrogen.

Slow H2(g) + 2 I(g) n 2 HI What was Sullivan’s evidence for the revised mechanism? First, he worked at temperatures too low to allow thermal decomposition of I2. At these temperatures, little or no hydrogen iodide is formed. Second, he used a technique called flash photolysis to create iodine atoms from I2. In this technique, a mixture of hydrogen and iodine was irradiated with a strong pulse of light. Sullivan found that the rate of the reaction was dependent on the square of the concentration of iodine atoms created by flash photolysis, which is consistent with the second elementary step in his reaction mechanism. The story of the mechanism of the hydrogen-iodine reaction is a good lesson: chemists should not trust a proposed mechanism just because it fits with an experimentally determined reaction order. Verifying mechanisms often requires identifying intermediates during the course of a reaction. Sullivan both identified an intermediate and controlled its production. Applying Chemical Principles

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717

Colin Cuthbert/Science Source

Questions

1. Sullivan used 578-nm light to dissociate I 2 molecules to I atoms. (a) What is the energy (in kJ/mol) of 578-nm light? (b) Breaking an I2 bond requires 151 kJ/mol of energy. What is the longest wavelength of light that has enough energy to dissociate I2? 2. Show that the two-step mechanism proposed by Sullivan yields the correct (overall second-order) rate law. 3. Why is a termolecular elementary step likely to be the slowest in a mechanism? 4. Determine the activation energy for the reaction of H2 and I2 to produce HI, given the data in the table.

Temperature (°C)

Rate Constant (M−1 s−1)

144.7

1.40 × 10−12

207.5

1.52 × 10−9

246.9

5.15 × 10−8

An apparatus for carrying out flash photolysis experiments in the study of kinetics.

Reference J. H. Sullivan, Journal of Chemical Physics, 1967, 46, 73−77.

1. When using the initial rates method to determine the rate law for a reaction, the concentration change of a reactant or product is measured over a short period of time. The time interval is chosen such that only a small change in the concentrations occurs, perhaps 1% to 2% of the reactants are consumed. For reactions that are first- or second-order with respect to each reactant, it is important that only small changes in reactant concentrations occur. Why? 2. For the following generalized reaction, A + 2 B n C + D, determine the reaction order for A and B given the tabulated initial rates data. Try to solve this problem without using a calculator. Explain your reasoning.

[A]0 (M)

[B]0 (M)

Rate (M/s)

0.028

0.015

1.8 × 10−3

0.028

0.030

7.2 × 10−3

0.014

0.030

3.6 × 10−3

3. The following graphs were obtained during the study of a first-order decomposition reaction. Identify the data sets collected at the same temperature? Identify the data set(s) collected at the highest temperature. Identify the data sets that started with the same initial concentration of reactant. Explain.

718

ln[A]

Think–Pair–Share

1 2 3

Time 4. The molecularity of the individual steps in a mechanism are almost always either unimolecular or bimolecular. (a) Explain what the terms “unimolecular step” and “bimolecular step” mean. (b) Explain why termolecular steps in a mechanism are rare. (c) The rate law for the following reaction, NO(g) + 1⁄2 Cl2(g) n NOCl(g)  Rate = k[NO]2[Cl2]

is third-order overall. Does this mean that this is one of the rare reactions that occurs in a single termolecular step? If not, explain. 5. Depending upon conditions, the catalyzed decomposition of HI(g) to H2(g) and I2(g) on a platinum surface will have either a first- or zero-order rate law. (a) What factor(s) can cause the rate law for this reaction to change? (b) If conditions change so that the reaction goes from zeroto first-order, does this also mean that the mechanism for the reaction changes? Explain.

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Chapter Goals Revisited The goals for this chapter are keyed to specific Study Questions to help you organize your review.

14.1  Rates of Chemical Reactions • Calculate the average rate of a reaction from concentration–time data. 5, 6. • Relate the rates for the disappearance of reactants and formation of products for a chemical reaction. 1–4.

14.2  Reaction Conditions and Rate • Describe how reaction conditions (reactant concentrations, temperature, presence of a catalyst, and the state of the reactants) affect reaction rate. 10−12, 99, 101.

14.3  Effect of Concentration on Reaction Rate • Derive a rate equation from experimental information using the method of initial rates. 13−16, 64, 72.

14.4 Concentration–Time Relationships: Integrated Rate Laws • Use the relationships between reactant concentration and time for zeroorder, first-order, and second-order reactions. 17−30.

• Apply graphical methods for determining reaction order and the rate constant from experimental data. 39–44, 66, 67, 69.

• Use the concept of half-life (t1/2), especially for first-order reactions. 31−38, 76, 93.

14.5  A Microscopic View of Reaction Rates • Describe the collision theory of reaction rates and use collision theory

to describe the effects of reactant concentration, molecular orientation, and temperature on reaction rate. 104, 109.

• Relate activation energy (Ea) to the rate of a reaction. 45, 46, 48. • Understand reaction coordinate diagrams. 49, 50, 59, 108. • Use the Arrhenius equation, or one of its modified forms, to calculate the activation energy from rate constants at different temperatures. 45, 47, 48, 74, 81, 82.

14.6 Catalysts • Describe the functioning of a catalyst and its effect on the activation energy of a reaction. 102, 108.

• Understand the lock-and-key and induced-fit models for substrate binding to enzymes. 51, 52.

• Understand the basics of the Michaelis-Menten model of enzyme kinetics. 53, 54, 95.



Chapter Goals Revisited

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14.7  Reaction Mechanisms • Understand the concept of a reaction mechanism and the relation of the mechanism to the overall, stoichiometric equation for a reaction. 58, 60.

• Describe the elementary steps of a mechanism, and give their molecularity. 55−58, 96.

• Identify reaction intermediates in a mechanism. 83, 102. • Predict the rate law for a reaction, given a simple mechanism and the identity of the rate-determining step. 58, 60, 71, 79, 83, 92, 97.

Key Equations Equation 14.1 (page 686) Integrated rate equation for a first-order reaction (in

which −∆[R]/∆t = k[R]).

ln

[R ]t [R ]0

 kt

Here, [R]0 and [R]t are concentrations of the reactant at time t = 0 and at a later time, t. The ratio of concentrations, [R]t/[R]0, is the fraction of reactant that remains after a given time has elapsed.

Equation 14.2 (page 689) Integrated rate equation for a second-order reaction (in which −∆[R]/∆t = k[R]2).

1 1   kt [R]t [R]0

Equation 14.3 (page 689) Integrated rate equation for a zero-order reaction (in

which −∆[R]/∆t = k[R]0).

[R]t − [R]0 = –kt

Equation 14.4 (page 692) The relation between the half-life (t1/2) and the rate constant (k) for a first-order reaction.

t 1⁄2 

0.693 k

Equation 14.5 (page 699) Arrhenius equation in exponential form. k = Ae−Ea /RT Frequency factor

Fraction of molecules with minimum energy for reaction

k is the rate constant, A is the frequency factor; Ea is the activation energy; T is the temperature (in kelvins); and R is the gas constant (= 8.31446 × 10−3 kJ/K ∙ mol).

Equation 14.6 (page 699) Expanded Arrhenius equation in logarithmic form. ln k = −

y

720

=

Ea 1 + ln A R T mx

+b

Arrhenius equation

Equation for straight line

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Equation 14.7 (page 701) A version of the Arrhenius equation used to calculate the activation energy for a reaction when you know the values of the rate constant at two temperatures (in kelvins). ln k2  ln k1  ln

k2 E 1 1  a    k1 R  T2 T1 

Study Questions denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully worked solutions in the Student Solutions Manual.

▲

Practicing Skills

6. Phenyl acetate, an ester, reacts with water according to the equation

Reaction Rates (See Section 14.1 and Examples 14.1–14.2.) 1. Give the relative rates of disappearance of reactants and formation of products for each of the following reactions. (a) 2 O3(g) n 3 O2(g) (b) 2 HOF(g) n 2 HF(g) + O2(g) 2. Give the relative rates of disappearance of reactants and formation of products for each of the following reactions. (a) NO(g) + 1/2 O2(g) n NO2(g) (b) 2 NH3(g) n N2(g) + 3 H2(g) 3. In the reaction 2 O3(g) n 3 O2(g), the rate of formation of O2 is 1.5 × 10−3 mol/L ∙ s. What is the rate of decomposition of O3? 4. In the synthesis of ammonia, if Δ[NH3]/Δt = 6.4 × 10−4 mol/L ∙ min, what is −Δ[H2]/Δt? N2(g) + 3 H2(g) n 2 NH3(g)

5. Experimental data are listed here for the reaction A n 2 B. Time (s)

[B] (mol/L)

 0.00

0.000

10.0

0.326

20.0

0.572

30.0

0.750

40.0

0.890

(a) Prepare a graph from these data; connect the points with a smooth line; and calculate the rate of change of [B] for each 10-s interval from 0.0 s to 40.0 s. Does the rate of change decrease from one time interval to the next? Suggest a reason for this result. (b) How is the rate of change of [A] related to the rate of change of [B] in each time interval? Calculate the rate of change of [A] for the time interval from 10.0 s to 20.0 s.

O

O CH3COC6H5 + H2O phenyl acetate

CH3COH + C6H5OH acetic acid

phenol

The data in the table were collected for this reaction at 5 °C. Time (s)

[Phenyl acetate] (mol/L)

 0

0.55

15.0

0.42

30.0

0.31

45.0

0.23

60.0

0.17

75.0

0.12

90.0

0.085

(a) Plot the phenyl acetate concentration versus time, and describe the shape of the curve observed. (b) Calculate the rate of change of the phenyl acetate concentration during the period 15.0 s to 30.0 s and also during the period 75.0 s to 90.0 s. Why is one value smaller than the other?

Concentration and Rate Equations (See Section 14.3 and Examples 14.3–14.4.) 7. Using the rate equation Rate = k[A]2[B], define the order of the reaction with respect to A and B. What is the total order of the reaction? 8. Using the rate equation Rate = k[A]2[B]2, define the order of the reaction with respect to A and B. What is the total order of the reaction? 9. A reaction has the experimental rate equation Rate = k[A][B]2. How will the rate change if the concentrations of both A and B are doubled?

Study Questions

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10. A reaction has the experimental rate equation Rate = k[A]2. How will the rate change if the concentration of A is tripled? If the concentration of A is halved? 11. The reaction between ozone and nitrogen dioxide at 231 K is first-order in both [NO2] and [O3]. 2 NO2(g) + O3(g) n N2O5(g) + O2(g)

(a) Write the rate equation for the reaction. (b) If the concentration of NO2 is tripled (and [O3] is not changed), what is the change in the reaction rate? (c) What is the effect on reaction rate if the concentration of O3 is halved (with no change in [NO2])? 12. Nitrosyl bromide, NOBr, is formed from NO and Br2: 2 NO(g) + Br2(g) n 2 NOBr(g)

Experiments show that this reaction is second-order in NO and first-order in Br2. (a) Write the rate equation for the reaction. (b) How does the initial reaction rate change if the concentration of Br2 is changed from 0.0022 mol/L to 0.0066 mol/L? (c) What is the change in the initial rate if the concentration of NO is changed from 0.0024 mol/L to 0.0012 mol/L? 13. The data in the table are for the reaction of NO and O2 at 660 K. NO(g) + 1⁄2 O2(g) n NO2(g)

Reactant ­Concentration (mol/L) [NO]

[O2]

Rate of ­Disappearance of NO (mol/L ∙ s)

0.010

0.010

2.5 × 10−5

0.020

0.010

1.0 × 10−4

0.010

0.020

5.0 × 10−5

(a) Determine the order of the reaction for each reactant. (b) Write the rate equation for the reaction. (c) Calculate the rate constant. (d) Calculate the rate (in mol/L ∙ s) at the instant when [NO] = 0.015 mol/L and [O2] = 0.0050 mol/L. (e) At the instant when NO is reacting at the rate 1.0 × 10−4 mol/L ∙ s, what is the rate at which O2 is reacting and NO2 is forming?

722

14. The reaction 2 NO(g) + 2 H2(g) n N2(g) + 2 H2O(g)

was studied at 904 °C, and the data in the table were collected. Reactant Concentration (mol/L) [NO]

[H2]

Rate of Appearance of N2 (mol/L ∙ s)

0.420

0.122

0.136

0.210

0.122

0.0339

0.210

0.244

0.0678

0.105

0.488

0.0339

(a) Determine the order of the reaction for each reactant. (b) Write the rate equation for the reaction. (c) Calculate the rate constant for the reaction. (d) Find the rate of appearance of N2 at the instant when [NO] = 0.350 mol/L and [H2] = 0.205 mol/L. 15. Data for the reaction NO(g) + 1⁄2 O2(g) n NO2(g) are given (for a particular temperature) in the table. Concentration (mol/L) Experiment

[NO]

[O2]

Initial Rate (mol NO/L ∙ h)

1

3.6 × 10−4  5.2 × 10−3

3.4 × 10−8

2

3.6 × 10−4 1.04 × 10−2

6.8 × 10−8

3

1.8 × 10−4 1.04 × 10−2

1.7 × 10−8

4

1.8 × 10−4  5.2 × 10−3

?

(a) What is the rate law for this reaction? (b) What is the rate constant for the reaction? (c) What is the initial rate of the reaction in experiment 4? 16. Data for the following reaction are given in the table below. CO(g) + NO2(g) n CO2(g) + NO(g)

Concentration (mol/L) Experiment

[CO]

[NO2]

Initial Rate (mol/L ∙ h)

1

5.0 × 10−4

0.36 × 10−4

3.4 × 10−8

2

5.0 × 10−4

0.18 × 10−4

1.7 × 10−8

3

1.0 × 10−3

0.36 × 10−4

6.8 × 10−8

4

1.5 × 10−3

0.72 × 10−4

?

(a) What is the rate law for this reaction? (b) What is the rate constant for the reaction? (c) What is the initial rate of the reaction in experiment 4?

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Concentration–Time Relationships (See Section 14.4 and Examples 14.5–14.7.) 17. The rate equation for the hydrolysis of sucrose (C12H22O11) to fructose and glucose (which both have the molecular formula C6H12O6) C12H22O11(aq) + H2O(ℓ) n 2 C6H12O6(aq)

is −Δ[sucrose]/Δt = k[C12H22O11]. After 27 minutes at 27 °C, the sucrose concentration decreased from 0.0146 M to 0.0132 M. Find the rate constant, k. 18. The decomposition of N2O5 in CCl4 is a first-order reaction. N2O5(g) n 2 NO2(g) + 1⁄2 O2(g)

If 2.56 mg of N2O5 is present initially and 2.50 mg is present after 4.26 minutes at 55 °C, what is the value of the rate constant, k?

24. The decomposition of nitrogen dioxide at a high temperature NO2(g) n NO(g) + 1⁄2 O2(g)

is second-order in this reactant. The rate constant for this reaction is 3.40 L/mol ∙ min. Determine the time needed for the concentration of NO2 to decrease from 2.00 mol/L to 1.50 mol/L. 25. At a different temperature than in Study Question 24, the decomposition of NO2(g) has a secondorder rate constant k = 1.1 L/mol ∙ s. If a vessel containing NO2(g) has an initial concentration of 1.9 × 10–2 mol/L, how long will it take for 75% of the NO2(g) to decompose? 26. The dimerization of butadiene, C4H6, to form 1,5-cyclo­octadiene is a second-order process that occurs when the diene is heated.

19. The decomposition of SO2Cl2 is a first-order reaction: SO2Cl2(g) n SO2(g) + Cl2(g)

The rate constant for the reaction is 2.8 × 10−3 min−1 at 600 K. If the initial concentration of SO2Cl2 is 1.24 × 10−3 mol/L, how long will it take for the concentration to drop to 0.31 × 10−3 mol/L? 20. The conversion of cyclopropane to propene (Example 14.5) occurs with a first-order rate constant of 2.42 h−1. How long will it take for the concentration of cyclopropane to decrease from an initial concentration of 0.080 mol/L to 0.010 mol/L? 21. Hydrogen peroxide, H2O2(aq), decomposes to H2O(ℓ) and O2(g) in a reaction that is first-order in H2O2. H2O2(aq) n H2O(ℓ) + 1⁄2 O2(g)

This reaction has a rate constant k = 1.06 × 10−3 min−1 at a given temperature. (a) How long will it take for 15% of a sample of H2O2 to decompose? (b) How long will it take for 85% of the sample to decompose?

C4H6(g) n 1⁄2 C8H12(g)

In an experiment, a sample of 0.0087 mol of C4H6 was heated in a 1.0-L flask. After 600. s, 21% of the butadiene had dimerized. Calculate the rate constant for this reaction. 27. The decomposition of ammonia on a metal surface to form N2 and H2 is a zero-order reaction (­Figure 14.7c). At 873 °C, the value of the rate constant is 1.5 × 10−3 mol/L ∙ s. How long it will take to completely decompose 0.16 g of NH3 in a 1.0-L flask? 28. Under certain conditions, the decomposition of HI on a metal surface to form H2 and I2 is a zero-order reaction with a rate constant k = 3.8 × 10−3 mol/L ∙ s. How long will it take for the concentration of HI to decrease from 0.280 mol/L to 0.050 mol/L? 29. Ammonium cyanate, NH4NCO, rearranges in water to yield urea, (NH2)2CO. The rate law for the reaction is −Δ[NH4NCO]/Δt = k[NH4NCO]2 and k = 0.012 L/mol ∙ min. What was the initial concentration of NH4NCO if its concentration 2.0 days after dissolving it in water is 1.5 × 10−3 mol/L?

22. Dinitrogen monoxide, N2O(g), decomposes to N2 and O2 gases in a reaction that is first-order in N2O with a rate constant k = 0.0131 min−1 at a given temperature. (a) How long will it take for 25% of an N2O sample to decompose? (b) How long will it take for 75% of the sample to decompose?

30. Hydrogen iodide decomposes when heated, forming H2(g) and I2(g). The rate law for this reaction is −∆[HI]/∆t = k[HI]2. At 443 °C, k = 30. L/mol ∙ min. If the initial HI(g) concentration is 2.5 × 10−2 mol/L, what concentration of HI(g) will remain after 15. min?

23. The dimerization of tetrafluoroethylene, C2F4(g) to form C4F8(g) is second-order with a rate constant k = 0.052 L/mol∙s at a given temperature. Determine the time needed to lower the concentration of C2F4 from 0.44 mol/L to 0.22 mol/L.

31. In a first-order reaction, 75% of the reactant (R) decomposes in 270 s. What is the half-life for the reaction? What is the rate constant for the reaction R n products?



Half-Life (See Section 14.4 and Examples 14.8 and 14.9.)

Study Questions

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32. In a first-order decomposition reaction, the reactant (R) concentration decreases from 0.480 mol/L to 0.060 mol/L in 12.6 days. What is the half-life for the reaction? What is the rate constant for the reaction R n products? 33. The rate equation for the decomposition of N2O5 (giving NO2 and O2) is Rate = k[N2O5]. The value of k is 6.7 × 10−5 s−1 for the reaction at a particular temperature. (a) Calculate the half-life of N2O5. (b) How long does it take for the N2O5 concentration to drop to one-tenth of its original value? 34. Gaseous azomethane, CH3NPNCH3, decomposes in a first-order reaction when heated:

is first-order in SO2Cl2, and the reaction has a halflife of 245 minutes at 600 K. If you begin with 3.6 × 10−3 mol of SO2Cl2 in a 1.0-L flask, how long will it take for the amount of SO2Cl2 to decrease to 2.00 × 10−4 mol? 36. The compound Xe(CF3)2 decomposes in a firstorder reaction to elemental Xe with a half-life of 30. min. If you place 7.50 mg of Xe(CF3)2 in a flask, how long must you wait until only 0.25 mg of Xe(CF3)2 remains? 37. The radioactive isotope 64Cu is used in the form of copper(II) acetate to study Wilson’s disease. The isotope has a half-life of 12.70 hours. What fraction of radioactive copper(II) acetate remains after 64 h?

 15.0

0.0835

 30.0

0.0680

 80.0

0.0350

120.0

0.0220

NH3(g) n NH2(g) + H(g)

The data in the table for this reaction were collected at a high temperature.

The rate constant for this reaction at 600 K is 0.0216 min−1. If the initial quantity of azomethane in the flask is 2.00 g, how much remains after 0.0500 h? What mass of N2 is formed in this time? SO2Cl2(g) n SO2(g) + Cl2(g)

[N2O] (mol/L)

40. Ammonia decomposes when heated according to the equation

CH3NPNCH3(g) n N2(g) + C2H6(g)

35. The decomposition of SO2Cl2

Time (min)

Time (h)

[NH3] (mol/L)

 0

8.00 × 10−7

25

6.75 × 10−7

50

5.84 × 10−7

75

5.15 × 10−7

Plot ln [NH3] versus time and 1/[NH3] versus time. What is the order of this reaction with respect to NH3? Find the rate constant for the reaction from the slope. 41. Gaseous NO2 decomposes at 573 K. NO2(g) n NO(g) + 1⁄2 O2(g)

The concentration of NO2 was measured as a function of time. A graph of 1/[NO2] versus time gives a straight line with a slope of 1.1 L/mol ∙ s. What is the rate law for this reaction? What is the rate constant? 42. The decomposition of HOF occurs at 25 °C. HOF(g) n HF(g) + 1⁄2 O2(g)

Using the data in the table below, determine the rate law, and then calculate the rate constant. [HOF] (mol/L)

Time (min)

38. Radioactive gold-198 is used in the diagnosis of liver problems. The half-life of this isotope is 2.7 d. If you begin with a 5.6-mg sample of the isotope, how much of this sample remains after 1.0 d?

0.850

 0

0.810

 2.00

0.754

 5.00

Graphical Analysis: Rate Equations and k

0.526

20.0

(See Section 14.4.)

0.243

50.0

39. Data for the decomposition of dinitrogen monoxide N2O(g) n N2(g) + ⁄2 O2(g) 1

on a gold surface at 900 °C are given below. Verify that the reaction is first-order by preparing a graph of ln[N2O] versus time. Derive the rate constant from the slope of the line in this graph. Using the rate law and value of k, determine the decomposition rate at 900 °C when [N2O] = 0.035 mol/L.

724

43. For the reaction C2F4 n 1⁄2 C4F8, a graph of 1/‌[C2F4] versus time gives a straight line with a slope of +0.04 L/mol ∙ s. What is the rate law for this reaction? 44. Butadiene, C4H6(g), dimerizes when heated, forming 1,5-cyclooctadiene, C8H12. The data in the table were collected.

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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H2C

CHCH

CH2

1,3-butadiene

½

H2 C HC

CH2 CH

HC H2 C

CH CH2

(a) Is the reaction exothermic or endothermic? (b) Does the reaction occur in more than one step? If so, how many?

1,5-cyclooctadiene

   0

−3

 200.

−3

 500.

−3

 800.

−3

1200.

1.0 × 10 8.7 × 10

Time (s)

−2

7.7 × 10 6.9 × 10 5.8 × 10

(a) Use a graphical method to verify that this is a second-order reaction. (b) Calculate the rate constant for the reaction.

Kinetics and Energy (See Section 14.5 and Examples 14.10 and 14.11.) 45. Calculate the activation energy, Ea, for the reaction 2 N2O5(g) n 4 NO2(g) + O2(g)

from the observed rate constants: k at 25 °C = 3.46 × 10−5 s−1 and k at 55 °C = 1.5 × 10−3 s−1. 46. If the rate constant for a reaction triples when the temperature rises from 3.15 × 102 K to 3.30 × 102 K, what is the activation energy of the reaction? 47. When heated to a high temperature, cyclobutane, C4H8, decomposes to ethylene: C4H8(g) n 2 C2H4(g)

The activation energy, Ea, for this reaction is 260 kJ/mol. At 800 K, the rate constant k = 0.0315 s−1. Determine the value of k at 850 K. 48. When heated, cyclopropane is converted to propene (Example 14.5). Rate constants for this reaction at 470 °C and 510 °C are k = 1.10 × 10−4 s−1 and k = 1.02 × 10−3 s−1, respectively. Determine the activation energy, Ea, from these data. 49. The reaction of H2 molecules with F atoms H2(g) + F(g) n HF(g) + H(g)

has an activation energy of 8 kJ/mol and an enthalpy change of −133 kJ/mol. Draw a diagram similar to Figure 14.11 for this process. Indicate the activation energy and enthalpy change on this diagram. 50. Answer the following questions based on the diagram below.



Energy

[C4H6] (mol/L)

Products Reactants

Reaction progress

Enzymes (See Section 14.6.) 51. Compare the lock-and-key and induced-fit models for substrate binding to an enzyme. 52. To which species should an enzyme bind best: the substrate, transition state, or product of a reaction? 53. According to the Michaelis-Menten model, if 1/Rate is plotted versus 1/[S], the intercept of the plot (when 1/[S] = 0) is 1/Ratemax. Using the data below at a given temperature, for a given enzyme and its substrate (S), calculate the maximum rate of the reaction, Ratemax. [S], mol/L

Rate, mmol/min

2.500

0.588

1.00

0.500

0.714

0.417

0.526

0.370

0.250

0.256

54. The enzyme carbonic anhydrase catalyzes the transformation of carbon dioxide into hydrogen carbonate ions. This reaction was studied by H. DeVoe and G. B. Kistiakowsky (Journal of the American Chemical Society, 1961, 83, 274–280) and found to obey the Michaelis-Menten model. Use the data below at a given temperature to calculate the maximum rate of the reaction, Ratemax. See Question 53 for the graphical method to use. [CO2] (mol/L)

Reaction Rate (mol/L ∙ s)

−3

 1.3 × 10

2.8 × 10−5

 2.5 × 10−3

5.0 × 10−5

 5.0 × 10−3

8.3 × 10−5

20.0 × 10−3

 17 × 10−5

Study Questions

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Reaction Mechanisms (See Section 14.7 and Examples 14.12–14.14.) 55. What is the rate law for each of the following elementary reactions? (a) NO(g) + NO3(g) n 2 NO2(g) (b) Cl(g) + H2(g) n HCl(g) + H(g) (c) (CH3)3CBr(aq) n (CH3)3C+(aq) + Br−(aq) 56. What is the rate law for each of the following elementary reactions? (a) NO2(g) + CO(g) n NO(g) + CO2(g) (b) N2O4(g) n 2 NO2(g) (c) 2 NO2(g) n NO3(g) + NO(g) 57. Ozone, O3, in the Earth’s upper atmosphere decomposes according to the equation 2 O3(g) n 3 O2(g)

The mechanism of the reaction is thought to proceed through an initial fast, reversible step followed by a slow, second step. Step 1:  Fast, reversible O3(g) uv O2(g) + O(g)

Step 2:  Slow O3(g) + O(g) n 2 O2(g)

(a) Which of the steps is rate-determining? (b) Write the rate equation for the rate-determining step. 58. At temperatures below 500 K, the reaction of NO2(g) and CO(g) is thought to occur in two steps to give NO and CO2: Step 1:  Slow

(a) Identify each of the following as a reactant, product, or intermediate: NO2(g), CO(g), NO3(g), CO2(g), NO(g). (b) Draw a reaction coordinate diagram for this reaction. Indicate on this drawing the activation energy for each step and the overall enthalpy change. 60. The mechanism for the reaction of CH3OH and HBr is believed to involve two steps. The overall reaction is exothermic. Step 1:  Fast, endothermic CH3OH + H+ uv CH3OH2+

Step 2:  Slow CH3OH2+ + Br− n CH3Br + H2O

(a) Write an equation for the overall reaction. (b) Draw a reaction coordinate diagram for this reaction. (c) Show that the rate law for this reaction is Rate = k[CH3OH][H+][Br−].

General Questions These questions are not designated as to type or location in the chapter. They may combine several concepts. 61. A reaction has the following experimental rate equation: Rate = k[A]2[B]. If the concentration of A is doubled and the concentration of B is halved, what happens to the reaction rate? 62. For a first-order reaction, what fraction of reactant remains after six half-lives have elapsed? 63. To determine the concentration dependence of the rate of the reaction

NO2(g) + NO2(g) n NO(g) + NO3(g)

Step 2:  Fast NO3(g) + CO(g) n NO2(g) + CO2(g)

(a) Show that the elementary steps add up to give the overall, stoichiometric equation. (b) What is the molecularity of each step? (c) What must be the experimental rate equation for this mechanism to be possible? (d) Identify any intermediates in this reaction. 59. A proposed mechanism for the reaction of NO2 and CO is

H2PO3−(aq) + OH−(aq) n HPO32−(aq) + H2O(ℓ)

you might measure [OH−] as a function of time using a pH meter. (To do so, you would set up conditions under which [H2PO3−] remains constant by using a large excess of this reactant.) How would you prove a second-order rate dependence for [OH−]? 64. Data for the following reaction are given in the table. 2 NO(g) + Br2(g) n 2 NOBr(g)

Experiment

[NO] (M)

[Br2] (M)

Initial Rate (mol/L ∙ s)

1

2.5 × 10−2

2.0 × 10−2

6.0 × 10−2

2

1.0 × 10−1

2.0 × 10−2

9.6 × 10−1

3

2.5 × 10−2

5.0 × 10−2

1.5 × 10−1

Step 1:  Slow, endothermic 2 NO2(g) n NO(g) + NO3(g)

Step 2:  Fast, exothermic NO3(g) + CO(g) n NO2(g) + CO2(g)

Overall Reaction:  Exothermic NO2(g) + CO(g) n NO(g) + CO2(g)

726

What is the order of the reaction with respect to [NO] and [Br2], and what is the overall order of the reaction? What is the value for the rate constant?

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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65. Formic acid decomposes at 550 °C according to the equation The reaction follows first-order kinetics. In an experiment, it is determined that 75% of a sample of HCO2H has decomposed in 72 seconds. Determine t½ for this reaction. 66. Isomerization of CH3NC occurs slowly when CH3NC is heated.

1/[C2F4] (L/mol)

HCO2H(g) n CO2(g) + H2(g)

50 40 30 20 10 0

0

100

300 500 Time, seconds

CH3NC(g) n CH3CN(g)

To study the rate of this reaction at 488 K, data on [CH3NC] were collected at various times. Analysis led to the following graph. (a) What is the rate law for this reaction? (b) What is the equation for the straight line in this graph? (c) Calculate the rate constant for this reaction. (d) How long does it take for half of the sample to isomerize? (e) What is the concentration of CH3NC after 1.0 × 104 s? −4.0

ln[CH3NC]

−6.0 −7.0 4000 8000 Time, seconds

12,000

67. When heated, tetrafluoroethylene dimerizes to form octafluorocyclobutane. C2F4(g) n 1⁄2 C4F8(g)

To determine the rate of this reaction at 488 K, the data in the table were collected. Analysis was done graphically, as shown in the accompanying graph. [C2F4] (M)

Time (s)

0.100

  0

0.080

 56

0.060

150.

0.040

335

0.030

520.

(a) What is the rate law for this reaction? (b) What is the value of the rate constant? (c) What is the concentration of C2F4 after 600 s? (d) How long will it take until the reaction is 90% complete?

68. Data in the table were collected at 540 K for the following reaction: CO(g) + NO2(g) n CO2(g) + NO(g)

Initial Concentration (mol/L) [CO]

[NO2]

Initial Rate (mol/L ∙ h)

5.1 × 10−4

0.35 × 10−4

 3.4 × 10−8

5.1 × 10−4

0.70 × 10−4

 6.8 × 10−8

5.1 × 10−4

0.18 × 10−4

 1.7 × 10−8

1.0 × 10−3

0.35 × 10−4

 6.8 × 10−8

−3

−4

10.2 × 10−8

1.5 × 10

0.35 × 10

Using the data in the table: (a) Determine the reaction order with respect to each reactant. (b) Derive the rate equation. (c) Calculate the rate constant, giving the correct units for k.

−5.0

0

700

69. Ammonium cyanate, NH4NCO, rearranges in water to give urea, (NH2)2CO. NH4NCO(aq) n (NH2)2CO(aq)

Time (min)

[NH4NCO] (mol/L)

0

0.458

4.50 × 10

1

0.370

1.07 × 10

2

0.292

2.30 × 10

2

0.212

6.00 × 10

2

0.114

Using the data in the table: (a) Decide whether the reaction is first-order or second-order. (b) Calculate k for this reaction. (c) Calculate the half-life of ammonium cyanate under these conditions. (d) Calculate the concentration of NH4NCO after 12.0 h.

Study Questions

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727

70. NOx, a mixture of NO and NO2, plays an essential role in the production of pollutants found in photo­chemical smog. The NOx in the atmosphere is slowly broken down to N2 and O2 in a firstorder reaction. The average half-life of NOx in the smokestack emissions in a large city during daylight is 3.9 h. (a) Starting with 1.50 mg in an experiment, what quantity of NOx remains after 5.25 h? (b) How many hours of daylight must have elapsed to decrease 1.50 mg of NOx to 2.50 × 10−6 mg? 71. At temperatures below 500 K, the reaction between carbon monoxide and nitrogen dioxide

73. The decomposition of dinitrogen pentaoxide N2O5(g) n 2 NO2(g) + 1⁄2 O2(g)

has the following rate equation: Rate = k[N2O5]. It has been found experimentally that the decomposition is 20.5% complete in 13.0 hours at 298 K. Calculate the rate constant and the half-life at 298 K. 74. The data in the table give the temperature dependence of the rate constant for the reaction N2O5(g) n 2 NO2(g) + 1⁄2 O2(g). Plot these data in the appropriate way to derive the activation energy for the reaction and determine its value.

CO(g) + NO2(g) n CO2(g) + NO(g)

has the following rate equation: Rate = k[NO2]2. Which of the three mechanisms suggested here best agrees with the experimentally observed rate equation? Mechanism 1

Single, elementary step NO2 + CO n CO2 + NO

Mechanism 2

Two steps

 Slow

NO2 + NO2 n NO3 + NO

 Fast

NO3 + CO n NO2 + CO2

Mechanism 3

Two steps

 Slow

NO2 n NO + O

 Fast

CO + O n CO2

2 NO2(g) + F2(g) n 2 NO2F(g)

Use the rate data in the table to do the following: (a) Write the rate equation for the reaction. (b) Indicate the order of reaction with respect to each component of the reaction. (c) Find the numerical value of the rate constant, k.

Experiment

[NO2]

[F2]

[NO2F]

Initial Rate (mol F2/L ∙ s)

1

0.001

0.005

0.001

2.0 × 10−4

2

0.002

0.005

0.001

4.0 × 10−4

3

0.006

0.002

0.001

4.8 × 10−4

4

0.006

0.004

0.001

9.6 × 10−4

5

0.001

0.001

0.001

4.0 × 10−5

6

0.001

0.001

0.002

4.0 × 10−5

728

k (s−1)

338

4.87 × 10−3

328

1.50 × 10−3

318

4.98 × 10−4

308

1.35 × 10−4

298

3.46 × 10−5

273

7.87 × 10−7

75. The decomposition of gaseous dimethyl ether at ordinary pressures is first-order. Its half-life is 25.0 min at 500 °C: CH3OCH3(g) n CH4(g) + CO(g) + H2(g)

72. ▲ Nitryl fluoride can be made by treating nitrogen dioxide with fluorine:

Initial Concentrations (mol/L)

T (K)

(a) Starting with 8.00 g of dimethyl ether, what mass remains (in grams) after 125 min and after 145 min? (b) Calculate the time in minutes required to decrease 7.60 ng (nanograms) to 2.25 ng. (c) What fraction of the original dimethyl ether remains after 150 min? 76. The decomposition of phosphine, PH3, proceeds according to the equation PH3(g) n 1⁄4 P4(g) + 3⁄2 H2(g)

It is found that the reaction has the following rate equation: Rate = k[PH3]. The half-life of PH3 is 37.9 s at 120 °C. (a) How much time is required for three-fourths of the PH3 to decompose? (b) What fraction of the original sample of PH3 remains after 2.00 min? 77. The thermal decomposition of diacetylene, C4H2, was studied at 950 °C. Use the following data (K. C. Hou and H. B. Palmer, Journal of Physical Chemistry, 1965, 69, 858–862) to determine the order of the reaction.

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Time (ms)

Concentration of C4H2 (mol/L)

  0

1.02 × 10−4

 50.

5.05 × 10−5

100.

2.59 × 10−5

150.

2.01 × 10−5

200.

1.44 × 10−5

250.

1.30 × 10−5

78. Kinetic experiments were conducted to determine the value of the rate constant, k, for the thermal decomposition of diacetylene, C4H2, at temperatures below 1100 K (K. C. Hou and H. B. Palmer, Journal of Physical Chemistry, 1965, 69, 858–862). Calculate Ea for this reaction from a plot of ln k ­versus 1/T. T (K)

k (L mol−1 min−1)

 973

10.6

1023

29.8

1073

71.1

[Mn(CO)5(CH3CN)]+ + NC5H5  88n [Mn(CO)5(NC5H5)]+ + CH3CN

are given in the table. Calculate Ea from a plot of ln k versus 1/T. T (K)

k (min−1)

298

0.0409

308

0.0818

318

0.157

82. The gas-phase reaction 2 N2O5(g) n 4 NO2(g) + O2(g)

has an activation energy of 103 kJ/mol, and the rate constant is 0.0900 min−1 at 328.0 K. Find the rate constant at 313.0 K.

2 NO(g) + O2(g) n 2 NO2(g)

The mechanism of the reaction is thought to be Step 1:  2 NO(g) uv N2O2(g)

2 O3(g) n 3 O2(g)

The mechanism of the reaction is thought to proceed through an initial fast equilibrium and a slow step: Step 1:  Fast, reversible

O3(g) uv O2(g) + O(g)

Step 2:  Slow

O3(g) + O(g) n 2 O2(g)

Show that the mechanism agrees with this experimental rate law: Rate = −(1/2)Δ[O3]/Δt = k [O3]2/[O2].

80. Hundreds of different reactions occur in the stratosphere, among them reactions that destroy the Earth’s ozone layer. The table below lists several (second-order) reactions of Cl atoms with ozone and organic compounds; each is given with its rate constant.



81. Data for the reaction

83. A reaction that occurs in Earth’s atmosphere is the oxidation of NO to the brown gas NO2.

79. The ozone in the Earth’s ozone layer decomposes according to the equation

Reaction

For equal concentrations of Cl and the other reactant, which is the slowest reaction? Which is the fastest reaction?

Rate Constant (298 K, cm3/molecule ∙ s)

(a) Cl + O3 n ClO + O2

1.2 × 10−11

(b) Cl + CH4 n HCl + CH3

1.0 × 10−13

(c) Cl + C3H8 n HCl + C3H7

1.4 × 10−10

(d) Cl + CH2FCl n HCl + CHFCl

3.0 × 10−18

rapidly established equilibrium

Step 2:  N2O2(g) + O2(g) n 2 NO2(g) slow Which is the rate determining step? Is there an intermediate in the reaction? If this is the correct mechanism for this reaction, what is the experimentally determined rate law? 84. The decomposition of SO2Cl2 to SO2 and Cl2 is first-order in SO2Cl2. SO2Cl2(g) n SO2(g) + Cl2(g) Rate = k[SO2Cl2] where k = 0.17/h

(a) What is the rate of decomposition when [SO2Cl2] = 0.010 M? (b) What is the half-life of the reaction? (c) If the initial pressure of SO2Cl2 in a flask is 0.050 atm, what is the pressure of all gases (i.e., the total pressure) in the flask after the reaction has proceeded for one half-life?

Study Questions

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85. The decomposition of nitrogen dioxide at a high temperature

Time (min)

NO2(g) n NO(g) + 1⁄2 O2(g)

is second-order in this reactant. (a) Determine the rate constant for this reaction if it takes 1.76 min for the concentration of NO2 to fall from 0.250 mol/L to 0.100 mol/L. (b) If the chemical equation is written as 2 NO2(g) n 2 NO(g) + O2(g)

what is the value of the rate constant? 86. Hydrogen peroxide, H2O2(aq), decomposes to H2O(ℓ) and O2(g) 2 H2O2(aq) n 2 H2O(ℓ) + O2(g)

At a particular temperature, the following data were collected for the initial rate of appearance O2. [H2O2] (mol/L)

Initial Reaction Rate (mol O2/L ∙ min)

0.0500

5.30 × 10−5

0.100

1.06 × 10−4

0.200

2.12 × 10−4

(a) What is the rate law for this reaction? (b) Calculate the value of the rate constant for this reaction. (c) If the chemical equation for this reaction is written as H2O2(aq) n H2O(ℓ) + 1⁄2 O2(g)

88. ▲ The compound 1,3-butadiene (C4H6) forms 1,5-cyclooctadiene, C8H12 at higher temperatures. C4H6(g) n 1⁄2 C8H12(g)

Use the following data to determine the order of the reaction and the rate constant, k. (Note that the total pressure is the pressure of the ­unreacted C4H6 at any time plus the pressure of the C8H12.)

 0

436

 3.5

428

11.5

413

18.3

401

25.0

391

32.0

382

41.2

371

89. ▲ Hypofluorous acid, HOF, is very unstable, decomposing in a first-order reaction to give HF and O2, with a half-life of 30. min at room temperature: HOF(g) n HF(g) + 1⁄2 O2(g)

If the partial pressure of HOF in a 1.00-L flask is initially 1.00 × 102 mm Hg at 25 °C, what are the total pressure in the flask and the partial pressure of HOF after exactly 30 min? After 45 min? 90. ▲ The decomposition of SO2Cl2 is first-order in SO2Cl2, SO2Cl2(g) n SO2(g) + Cl2(g)

with a half-life of 245 min at 600 K. If you begin with a partial pressure of SO2Cl2 of 25 mm Hg in a 1.0-L flask, what is the partial pressure of each reactant and product after 245 min? What is the partial pressure of each reactant and product after 12 h? 91. ▲ Nitramide, NO2NH2, decomposes slowly in aqueous solution according to the following reaction:

what is the value of the rate constant? 87. ▲ Egg protein albumin is precipitated when an egg is cooked in boiling (100 °C) water. Ea for this first-order reaction is 52.0 kJ/mol. Estimate the time to prepare a 3-min egg at an altitude at which water boils at 90 °C.

Total Pressure (mm Hg)

NO2NH2(aq) n N2O(g) + H2O(ℓ)

The reaction follows the experimental rate law Rate 

k[NO2NH2] [H3O]

(a) What is the apparent order of the reaction in a pH buffered solution? (In a pH buffered solution, the concentration of H3O+ is a constant.) (b) Which of the following mechanisms is the most appropriate for the interpretation of this rate law? Explain. (Note that when writing the expression for K, the equilibrium constant, [H2O] is not involved. See Chapter 15.) Mechanism 1 k1  NO2NH2 88n N2O + H2O

730

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Mechanism 2

k3  NO2NH3+ 88n N2O + H3O+

© Charles D. Winters/Cengage

k2  NO2NH2 + H3O+ uv NO2NH3+ + H2O k2′ (rapid equilibrium) (rate-limiting step)

Mechanism 3 k4  NO2NH2 + H2O uv NO2NH− + H3O+ k4′ (rapid equilibrium) k5  NO2NH− 88n N2O + OH−

(rate-limiting step)

k6  H3O+ + OH− 88n 2 H2O

(very fast reaction)

(c) Show the relationship between the experimentally observed rate constant, k, and the rate constants in the selected mechanism. (d) Based on the experimental rate law, will the reaction rate increase or decrease if the pH of the solution is increased? 92. Many biochemical reactions are catalyzed by acids. A typical mechanism consistent with the experimental results (in which HA is the acid and X is the reactant) is

Fading of the color of phenolphthalein with time (elapsed time about 3 minutes)

Time (s)

Concentration of Phenol­ phthalein (mol/L)

  0.00

0.0050

 10.5

0.0045

 22.3

0.0040

 35.7

0.0035

 51.1

0.0030

 69.3

0.0025

 91.6

0.0020

120.4

0.0015

160.9

0.0010

−

+

Fast, reversible:

HA uv H + A

230.3

0.00050

Step 2:

Fast, reversible:

X + H uv XH

299.6

0.00025

Step 3:

Slow

XH+ n products

Step 1:

+

+

What rate law is derived from this mechanism? What is the order of the reaction with respect to HA? How would doubling the concentration of HA affect the reaction?

In the Laboratory 93. The color change accompanying the reaction of phenolphthalein with strong base is ­illustrated in the following photographs. The change in concentration of the dye can be followed by ­spectrophotometry (Section 4.9), and some data collected by that approach are given below. The initial concentrations were [phenolphthalein] = 0.0050 mol/L and [OH−] = 0.61 mol/L. (Data are taken from review materials for kinetics at https://chemed.chem.purdue.edu.) (For more details on this reaction see L. Nicholson, Journal of Chemical Education, 1989, 66, 725–726.)

(a) Plot the data above as [phenolphthalein] versus time, and determine the average rate from t = 0 to t = 15 s and from t = 100 s to t = 125 s. Does the rate change? If so, why? (b) Use a graphical method to determine the order of the reaction with respect to phenolphthalein. Write the rate law, and determine the rate constant. (c) What is the half-life for the reaction? 94. ▲ You want to study the hydrolysis of the beautiful green, cobalt-based complex called trans-dichlorobis(ethylenediamine)cobalt(III) ion,

H2C H2C



H2 N

Cl Co

N H2

Cl

+

H2 N CH 2 N CH2 H2

Study Questions

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In this hydrolysis reaction, the green complex ion trans–[Co(en)2Cl2]+ forms the red complex ion [Co(en)2(H2O)Cl]2+ as a Cl− ion is replaced with a water molecule on the Co3+ ion (en = H2NCH2CH2NH2). trans–[Co(en)2Cl2]+(aq) + H2O(ℓ) n  green

[Co(en)2(H2O)Cl]2+(aq) + Cl−(aq) red

The reaction progress is followed by observing the color of the solution. The original solution is green, and the final solution is red, but at some intermediate stage when both the reactant and product are present, the solution is gray.

intermediate solution

Photos: © Charles D. Winters/Cengage

original solution

Slow:

trans–[Co(en)2Cl2]+(aq) n [Co(en)2Cl]2+(aq) + Cl−(aq)

Fast:

[Co(en)2Cl]2+(aq) + H2O(ℓ) n [Co(en)2(H2O)Cl]2+(aq)

(a) Based on the reaction mechanism, what is the predicted rate law? (b) As the reaction proceeds, the color changes from green to red with an intermediate stage where the color is gray. The gray color is reached at the same time, no matter what the concentration of the green starting material (at the same temperature). How does this show the reaction is first-order in the green form? Explain. (c) The activation energy for a reaction can be found by plotting ln k versus 1/T. However, you do not need to measure k directly in this case. Instead, because k = −(1/t)ln([R]/[R]0), the reciprocal of the time needed to achieve the gray color is a measure of k. Use the data below to find the activation energy. Temperature (° C)

Time Needed to Achieve Gray Color (for the Same Initial Concentration)

56

156 s

60

114 s

65

  88 s

75

  47 s

95. The enzyme chymotrypsin catalyzes the hydrolysis of a peptide containing phenylalanine. Using the data below at a given temperature, calculate the maximum rate of the reaction, Ratemax.

final solution

Changes in color with time as Cl− ion is replaced by H2O in a cobalt(III) complex.  The shape in the middle of the beaker is a vortex that arises because the solutions are being stirred using a magnetic stirring bar in the bottom of the beaker.

Reactions such as this have been studied extensively, and experiments suggest that the initial, slow step in the reaction is the breaking of the CoOCl bond to give a five-coordinate intermediate. The intermediate is then attacked rapidly by water.

732

Peptide Concentration (mol/L)

Reaction Rate (mol/L ∙ min)

 2.5 × 10−4

2.2 × 10−6

 5.0 × 10−4

3.8 × 10−6

10.0 × 10−4

5.9 × 10−6

15.0 × 10−4

7.1 × 10−6

96. ▲ The substitution of CO in Ni(CO)4 by another molecule L [where L is an electron-pair donor such as P(CH3)3] was studied some years ago and led to an understanding of some of the general principles that govern the chemistry of compounds having metal–CO bonds. (See J. P. Day, F. Basolo, and R. G. Pearson: Journal of the American Chemical Society, 1968, 90, 6927–6933.) A detailed study of the kinetics of the reaction led to the following mechanism:

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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Ni(CO)4 n Ni(CO)3 + CO

Slow:

98. The acid-catalyzed iodination of acetone

Ni(CO)3 + L n Ni(CO)3L Fast: (a) What is the molecularity of each of the elementary reactions? (b) Doubling the concentration of Ni(CO)4 increased the reaction rate by a factor of 2. Doubling the concentration of L had no effect on the reaction rate. Based on this information, write the rate equation for the reaction. Does this agree with the mechanism described? (c) The experimental rate constant for the reaction, when L = P(C6H5)3, is 9.3 × 10−3 s−1 at 20 °C. If the initial concentration of Ni(CO)4 is 0.025 M, what is the concentration of the product after 5.0 min?

97. ▲ The oxidation of iodide ion by the hypochlorite ion in the presence of hydroxide ions I−(aq) + ClO−(aq) n IO−(aq) + Cl−(aq)

was studied at 25 °C, and the following initial rates data (Y. Chia and R. E. Connick, Journal of Physical Chemistry, 1959, 63, 1518–1519) were collected: Initial Concentrations (mol/L) Experiment

[ClO−]

[I−]

Initial Rate [OH−] (mol IO−/L ∙ s)

1

4.0 × 10−3 2.0 × 10−3

1.0

4.8 × 10−4

2

2.0 × 10−3 4.0 × 10−3

1.0

5.0 × 10−4

3

2.0 × 10−3 2.0 × 10−3

1.0

2.4 × 10−4

4

2.0 × 10−3 2.0 × 10−3

0.50

4.6 × 10−4

(a) Determine the rate law for this reaction. (b) One mechanism that has been proposed for this reaction is the following: Step 1:

ClO− + H2O uv HOCl + OH−

fast, reversible

Step 2:

I− + HOCl n HOI + Cl−

slow

Step 3:

HOI + OH− n IO− + H2O

fast

Show that the rate law predicted by this mechanism matches the experimentally determined rate law in part a. (Note that when writing the expression for K, the equilibrium constant, [H2O] is not involved. See Chapter 15.)



CH3COCH3(aq) + I2(aq) n CH3COCH2I(aq) + HI(aq)

is a common laboratory experiment used in general chemistry courses to teach the method of initial rates. The reaction is followed spectrophotometrically by the disappearance of the color of iodine in the solution. The following data (J. P. Birk and D. L. Walters, Journal of Chemical Education, 1992, 69, 585–587) were collected at 23 °C for this reaction. Initial Concentrations (mol/L) Experiment [CH3COCH3]

[H+]

[I2]

Initial Rate (mol I2/L ∙ s)

1

1.33

0.162

0.00665

8.1 × 10−6

2

1.33

0.323

0.00665

1.7 × 10−5

3

0.667

0.323

0.00665

7.6 × 10−6

4

0.333

0.323

0.00665

3.8 × 10−6

5

0.333

0.323

0.00332

3.6 × 10−6

Determine the rate law for this reaction.

Summary and Conceptual Questions The following questions may use concepts from this and ­previous chapters. 99. Hydrogenation reactions, processes wherein H2 is added to a molecule, are usually catalyzed. An excellent catalyst is a very finely divided metal suspended in the reaction solvent. Explain why finely divided rhodium, for example, is a much more efficient catalyst than a small block of the metal. 100. ▲  Suppose you have 1000 blocks, each of which is 1.0 cm on a side. If all 1000 of these blocks are stacked to give a cube that is 10. cm on a side, what fraction of the 1000 blocks have at least one surface on the outside surface of the cube? Next, divide the 1000 blocks into eight equal piles of blocks and form them into eight cubes, 5.0 cm on a side. What fraction of the blocks now have at least one surface on the outside of the cubes? How does this mathematical model pertain to Study Question 99? 101. The following statements relate to the reaction for the formation of HI: H2(g) + I2(g) n 2 HI(g)   Rate = k[H2][I2]

Determine which of the following statements are true. If a statement is false, indicate why it is incorrect.

Study Questions

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733

(a) The reaction must occur in a single step. (b) This is a second-order reaction overall. (c) Raising the temperature will cause the value of k to decrease. (d) Raising the temperature lowers the activation energy for this reaction. (e) If the concentrations of both reactants are doubled, the rate will double. (f) Adding a catalyst in the reaction will cause the initial rate to increase. 102. Chlorine atoms contribute to the destruction of the Earth’s ozone layer by the following sequence of reactions: Cl + O3 n ClO + O2 ClO + O n Cl + O2

where the O atoms in the second step come from the decomposition of ozone by sunlight:

105. The reaction cyclopropane n propene occurs on a platinum metal surface at 200 °C. (The platinum is a catalyst.) The reaction is first-order in cyclopropane. Indicate how the following quantities change (increase, decrease, or no change) as this reaction progresses, assuming constant temperature. (a) [cyclopropane] (b) [propene] (c) [catalyst] (d) the rate constant, k (e) the order of the reaction (f) the half-life of cyclopropane 106. Isotopes are often used as tracers to follow an atom through a chemical reaction, and the following is an example. Acetic acid reacts with methanol.

O3(g) n O(g) + O2(g)

104. Identify which of the following statements are incorrect. If the statement is incorrect, rewrite it to be correct. (a) Reactions are faster at a higher temperature because activation energies are lower. (b) Rates increase with increasing concentration of reactants because there are more collisions between reactant molecules. (c) At higher temperatures, a larger fraction of molecules have enough energy to get over the activation energy barrier. (d) Catalyzed and uncatalyzed reactions have identical mechanisms.

734

+ CH3CO2H

+

CH3OH

+ CH3CO2CH3

+

H2O

Explain how you could use the isotope 18O to show whether the oxygen atom in the water comes from the OOH of CH3CO2H or the OOH of CH3OH. 107. Examine the reaction coordinate diagram given here.

Energy

What is the net equation on summing these three equations? Why does this lead to ozone loss in the stratosphere? What is the role played by Cl in this sequence of reactions? What name is given to species such as ClO? 103. Describe each of the following statements as true or false. If false, rewrite the sentence to make it correct. (a) The rate-determining elementary step in a reaction is the slowest step in a mechanism. (b) It is possible to change the rate constant by changing the temperature. (c) As a reaction proceeds at constant temperature, the rate remains constant. (d) A reaction that is third-order overall must involve more than one step.

Reactants Products Reaction progress

(a) How many steps are in the mechanism for the reaction described by this diagram? (b) Is the reaction overall exothermic or endothermic?

Chapter 14 / Chemical Kinetics: The Rates of ­C hemical Reactions

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108. Draw a reaction coordinate diagram for an ­exothermic reaction that occurs in a single step. Identify the activation energy and the net energy change for the reaction on this diagram. Draw a second diagram that represents the same reaction in the presence of a catalyst, assuming a single-step reaction is involved here also. Identify the activation energy of this reaction and the energy change. Is the activation energy in the two drawings different? Does the energy evolved in the two reactions differ? 109. Consider the reaction of ozone and nitrogen monoxide to form nitrogen dioxide and oxygen.

Which of the following orientations for the collision between ozone and nitrogen monoxide could perhaps lead to an effective collision between the molecules? (a)

(b)

(c)

O3(g) + NO(g) n NO2(g) + O2(g) (d)



Study Questions

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735

© A. N. Palmer

15

Principles of Chemical Reactivity: Equilibria

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C hapt e r O ut li n e 15.1 Chemical Equilibrium: A Review 15.2 The Equilibrium Constant and Reaction Quotient 15.3 Determining an Equilibrium Constant 15.4 Using Equilibrium Constants in Calculations  15.5 More about Balanced Equations and Equilibrium Constants 15.6 Disturbing a Chemical Equilibrium

In Chapter 3 you were introduced to the concept of chemical equilibrium. This is a dynamic process in which the products of a reaction convert back to reactants at the same rate that they are produced. In this chapter, you will learn that the extent to which a reactant occurs can be described by an equilibrium constant, a number that relates the concentrations of the products and reactants present at equilibrium. A major goal of this chapter and the next two chapters is to describe equilibria in quantitative terms. That is, you will learn to use equilibrium constants to determine the concentrations of products and reactants in systems that have attained equilibrium.

15.1 Chemical Equilibrium: A Review Goal for Section 15.1 • Understand that chemical reactions are reversible and that chemical equilibria are dynamic.

If you mix solutions of CaCl2 and NaHCO3, a chemical reaction is immediately detected: a gas (CO2) bubbles from the mixture, and an insoluble white solid, CaCO3, forms (Figure 15.1a). The reaction occurring is Ca2+(aq) + 2 HCO3−(aq)  n CaCO3(s) + CO2(g) + H2O(ℓ)

If you next add pieces of dry ice (solid CO2) to the CaCO3 suspension (or if you bubble gaseous CO2 into the mixture), the solid CaCO3 dissolves (Figures 15.1b and 15.1c). This happens because a reaction occurs that is the reverse of the reaction that led to precipitation of CaCO3; that is: CaCO3(s) + CO2(g) + H2O(ℓ)  n Ca2+(aq) + 2 HCO3−(aq)

Now imagine what will happen if the solution of Ca2+ and HCO3− ions is in a closed container (unlike the reaction in Figure  15.1). As the reaction begins, Ca2+ and HCO3− react to give products at some rate. As the reactants are depleted, the rate of this reaction slows. At the same time, however, the reaction products (CaCO3, CO2, ◀ Dynamic and reversible.  Calcium carbonate stalactites cling to the roof of the cave, and

stalagmites grow up from the cave floor. The chemistry producing these formations is a good introduction to the concept of chemical equilibrium.

737

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Photos: © Charles D. Winters/Cengage

(a) Combining solutions of NaHCO3 and CaCl2 produces solid CaCO3 and CO2 gas.

(b) Dry ice (the white solid) is added to the slurry of CaCO3 precipitated in (a).

(c) The calcium carbonate dissolves upon adding sufficient dry ice (CO2) to give Ca2+(aq) and HCO3−(aq).

Figure 15.1  The CO2/Ca21/H2O system.  See Figure 3.5, which shows the same reaction and illustrates the chemistry of the system.

and H2O) begin to combine and re-form Ca2+ and HCO3−. Eventually, the rate of the forward reaction, the formation of CaCO3, and the rate of the reverse reaction, the redissolving of CaCO3, become equal. Both the forward and reverse reactions continue to occur, but no further macroscopic change is observed. The system is now at equilibrium. For equilibrium systems, such as the formation and dissolution of calcium ­carbonate reactions discussed previously, the single arrow between the reactants and products is replaced with double arrows (uv) to signify that both the forward and reverse reactions occur to a measurable extent and that the reaction will be studied using the concepts of chemical equilibria. Ca2+(aq) + 2 HCO3−(aq) uv CaCO3(s) + CO2(g) + H2O(ℓ)

The calcium carbonate equilibrium illustrates some important features of chemical reactions. •

Chemical reactions are reversible, at least in theory.

•

Chemical reactions proceed spontaneously in the direction that leads toward equilibrium.

•

In a closed system, a state of equilibrium between reactants and products is achieved eventually and remains unless there is an outside disturbance.

•

Outside forces can disturb an equilibrium.

15.2 The Equilibrium Constant and Reaction Quotient Goals for Section 15.2 • Write the equilibrium constant expression, K, for a chemical reaction. • Recognize that the equilibrium constant can be written relating K to concentrations (Kc) or, if the equilibrating species are gases, to partial pressures (Kp). Convert between Kc and Kp values.

• Predict whether a reaction is reactant-favored or product-favored at equilibrium based on the value of K.

738

Chapter 15 / Principles of Chemical Reactivity: Equilibria

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• Write the reaction quotient expression, Q, for a chemical reaction, and use Q to decide whether a reaction is at equilibrium (Q 5 K) or if there will be a net conversion of reactants to products (Q , K) or products to reactants (Q . K) to attain equilibrium.

The concentrations of reactants and products for a reaction at equilibrium are related by a mathematical equation. For example, for the reaction of hydrogen and iodine to produce hydrogen iodide, experiments have shown that at equilibrium the ratio of the square of the HI concentration to the product of the H2 and I2 concentrations is a constant. H2(g) + I2(g) uv 2 HI(g) [HI]2  constant (K ) at equilibrium [H2][I2]

This constant,  K, the equilibrium constant, is always the same within experimental error for all experiments done  at a given temperature. Suppose, for example, the concentrations of H2 and I2 in a flask are each initially 0.0175 mol/L at 425 °C and no HI is present. Over time, the concentrations of H2 and I2 will decrease, and the concentration of HI will increase until a state of equilibrium is reached (­Figure 15.2). If the gases in the flask are then analyzed, the observed concentrations would be [H2] = [I2] = 0.0037 mol/L and [HI] = 0.0276 mol/L. The ­following table—which is often called an ICE table for initial, change, and equilibrium c­ oncentrations— summarizes these results: Equation

H2(g)

I = Initial concentration (M)

0.0175

0.0175

−0.0138

−0.0138

+0.0276

0.0037

0.0037

0.0276

C = Change in concentration as the reaction proceeds to equilibrium (M) E = Equilibrium concentration (M)

I2(g)

1

uv

2 HI(g)   0

ICE Table: Initial, Change, and Equilibrium  Throughout our

discussions of chemical equilibria, quantitative information for reactions will be summarized in amounts tables (Section 4.1) or ICE tables. These tables show what the initial (I ) concentrations are, how those concentrations change (C  ) on proceeding to equilibrium, and what the concentrations are at equilibrium (E  ).

The second line in the table gives the change in concentration of reactants and products upon proceeding to equilibrium. Changes are always equal to the difference between the equilibrium and initial concentrations. Change in concentration = equilibrium concentration − initial concentration

Putting the equilibrium concentration values from the ICE table into the expression for the constant (K) gives a value of 56. [HI]2 (0.0276)2   56 [H2][I2] (0.0037)(0.0037) H2(g) + I2(g)

Concentration of reactants and products (M)

0.030

[HI]

0.025 Reactant and product concentrations reach equilibrium values in about 130 minutes in this system. No further net change occurs.

0.020 0.015 0.010

The final concentrations of H2, I2, and HI depend on the initial concentrations of H2 and I2. If the reaction begins with a different set of initial concentrations, the equilibrium concentrations will be different, but the quotient [HI]2/[H2][I2] will always be the same at a given temperature.

[H2]

0.005 0



Figure 15.2  The reaction of H2 and I2 reaches equilibrium.  ​

2 HI(g)

[I2] 0

20

40

60

80 100 120 140 160 Time (minutes) Reactants proceeding toward equilibrium

180

200

15.2  The Equilibrium Constant and Reaction Quotient

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Other experiments can be performed using the H2/I2 reaction with different concentrations of reactants or mixtures of reactants and products. Regardless of the initial amounts, when equilibrium is achieved, the ratio [HI]2/[H2][I2] is always the same, 56, at this temperature. The observation that the ratio of the product and reactant concentrations for the H2 and I2 reaction is always the same can be generalized to other reactions. For the general chemical reaction a A + b B uv c C + d D

the equilibrium constant, K, is defined as follows:

Equilibrium constant  K 

[C]c [D]d [A]a[B]b

(15.1)

Equation 15.1 is called the equilibrium constant expression.  If the ratio of products to reactants as defined by Equation 15.1  matches the equilibrium constant value, the system is at equilibrium. Conversely, if the ratio has a different value, the system is not at equilibrium.

Writing Equilibrium Constant Expressions

© Charles D. Winters/Cengage

In an equilibrium constant expression,

Figure 15.3 Burning sulfur.  Elemental sulfur burns

A Closer Look

in oxygen with a beautiful blue flame to give SO2 gas.

•

all concentrations are equilibrium values.

•

product concentrations appear in the numerator, and reactant concentrations appear in the denominator.

•

each concentration is raised to the power of its stoichiometric coefficient in the balanced chemical equation.

•

solids, pure liquids, and solvents do not appear in an equilibrium expression. (However, if a solid or pure liquid is involved in a reaction it must be physically present in the reaction mixture for the system to be at equilibrium.)

•

the value of the constant K depends on the temperature.

•

values of K are dimensionless. (“A Closer Look: Activities and Units of K.”)

Reactions Involving Solids As noted, the concentrations of any solid reactants and products are not included in the equilibrium constant expression. The oxidation of solid, yellow sulfur produces colorless sulfur dioxide gas (Figure 15.3),

Activities and Units of K

In the text we state that “values of K are dimensionless.” After all our care in using units in this book, this might seem strange. However, advanced thermodynamics informs us that equilibrium constants should really be calculated from the activities of reactants and products, not from their concentrations or partial pressures. Activities can be considered “effective” ­concentrations or partial pressures. The activity of a substance in solution is obtained by calculating the ratio of its

740

S(s) + O2(g) uv SO2(g)

concentration, [X], relative to a standard concentration (1 M), and then multiplying this ratio by a correction factor called an activity coefficient. Because it involves a ratio of concentrations, the activity is dimensionless. Likewise, the activity of a gas is obtained from the ratio of its partial pressure, PX, relative to a standard pressure (1  bar) and then multiplying by an activity coefficient. In general chemistry, all the activity coefficients are usually assumed equal to 1 and so the activity of a solute or a gas is numerically the same as

its concentration or partial pressure, respectively. This assumption is best met for solutes in very dilute solutions or gases at low pressures. Regardless of the values of activity coefficients, values of K are properly calculated using dimensionless quantities and so they have no units. Another consequence of using activities is that solids do not appear in the K ­expression. This is because the activity of a solid is 1. Similarly, pure liquids and solvents are not included because their ­ ­activities are also 1.

Chapter 15 / Principles of Chemical Reactivity: Equilibria

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a reaction with the following equilibrium constant expression. K 

[SO2] [O2]

In reactions involving solids, experiments show that the equilibrium concentrations of other reactants or products—here, O2 and SO2—do not depend on the amount of solid present (as long as some solid is present at equilibrium).

Reactions in Solution There are also special considerations for reactions that occur in solution when the solvent (water, for example) is either a reactant or a product. Consider ammonia, which is a weak base owing to its (incomplete) reaction with water. NH3(aq) + H2O(ℓ) uv NH4+(aq) + OH−(aq)

Because the water concentration is very high in a dilute ammonia solution, the concentration of water is essentially unchanged by the reaction.  The general rule for reactions in aqueous solution is that the concentration of water is not included in the equilibrium constant expression. Thus, for the aqueous ammonia equilibrium you should write the equilibrium constant expression as K 

[NH4][OH] [NH3 ]

Reactions Involving Gases: Kc and Kp Concentration data can be used to calculate equilibrium constants for both aqueous and gaseous systems. In these cases, the symbol K is sometimes given the subscript “c” for “concentration,” as in Kc. For gases, however, equilibrium constant expressions can also be written in terms of partial pressures of reactants and products. If reactant and product quantities are given in partial pressures (in atmospheres or in bars), then K is given the subscript “p,” as in Kp. H2(g) + I2(g) uv 2 HI(g) Kp 

Using Partial Pressures in Place of Concentrations ​If you

rearrange the ideal gas law, [PV = nRT ], you find that the gas concentration, (n/V ), is equivalent to P/RT. Thus, the partial pressure of a gas is proportional to its concentration [P = (n/V )RT ].

PHI2 PH2 PI2

Notice that the basic form of the equilibrium constant expression is the same as for Kc. In some cases, the numerical values of Kc and Kp are the same, but they are different when the numbers of moles of gaseous reactants and products are different. “A Closer Look: Equilibrium Constant Expressions for Gases—Kc and Kp” shows how Kc and Kp are related and how to convert from one to the other.

Ex am p le 15.1

Writing Equilibrium Constant Expressions Problem  Write the equilibrium constant expressions (Kc) for the following reactions. (a) N2(g) + 3 H2(g) uv 2 NH3(g) (b) H2CO3(aq) + H2O(ℓ) uv HCO3−(aq) + H3O+(aq)

What Do You Know?  You have balanced chemical equations from which you can write the equilibrium constant expressions.

Strategy  Product concentrations always appear in the numerator and reactant concentrations in the denominator. Each concentration should be raised to a power equal to the stoichiometric coefficient in the balanced equation. In reaction (b), water is the solvent, and its concentration does not appear in the equilibrium constant expression.

15.2  The Equilibrium Constant and Reaction Quotient

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741

Solution (a) K c 

[NH3 ]2 [N2 ][H2 ]3

(b) K c 

[HCO3][H3O] [H2CO3 ]

Think about Your Answer   Always check to ensure you have the products in the numerator and reactants in the denominator. Confusing these is a common source of ­student error.

Check Your Understanding  Write the equilibrium constant expression for each of the following reactions in terms of concentrations. (a) CO2(g) + C(s) uv 2 CO(g) (b) [Cu(NH3)4]2+(aq) uv Cu2+(aq) + 4 NH3(aq) (c) CH3CO2H(aq) + H2O(ℓ) uv CH3CO2−(aq) + H3O+(aq)

The Magnitude of the Equilibrium Constant, K The magnitude of the equilibrium constant can vary widely. A large value for the equilibrium constant means that the concentrations of the products are larger than the concentrations of the reactants at equilibrium. That is, the products are favored over the reactants at equilibrium. K > 1: Reaction is product-favored at equilibrium. The concentrations of products are greater than the concentrations of the reactants at equilibrium.

An example is the reaction of nitrogen monoxide and ozone. NO(g) + O3(g) uv NO2(g) + O2(g)

A Closer Look

Kc 

Equilibrium Constant Expressions for Gases—Kc and Kp Many metal carbonates, such as limestone, decompose upon heating to yield the metal oxide and CO2 gas.

Consider the equilibrium constant for the reaction of N2 and H2 to produce ammonia in terms of partial pressures, Kp.

Looking carefully at these examples and others, you will find that

CaCO3(s) uv CaO(s) + CO2(g)

N2(g) + 3 H2(g) uv 2 NH3(g)

where ∆n is the change in the number of moles of gas that occurs when reactants are transformed into products.

The equilibrium constant for this reaction can be expressed either in terms of the number of moles per liter of CO2, Kc = [CO2], or in terms of the partial pressure of CO2, Kp = PCO2. From the ideal gas law, you know that

P =  (n/V )RT  = (concentration in mol/L) × RT Therefore, for this reaction Kp = PCO2 =  [CO2]RT. Because Kc = [CO2], this leads to the conclusion that Kp = Kc(RT). For the decomposition of calcium carbonate, Kp is the product of Kc and the factor RT. That is, the values of Kp and Kc are not the same.

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[NO2 ][O2 ]  6  1034 at 25 C [NO][O3 ]

(PNH3 )2 Kp  (PN2 )(PH2 )3 Does Kc, the equilibrium constant in terms of concentrations, have the same value as Kp? You can answer this question by substituting for each pressure in Kp the equivalent expression [C  ](RT). That is,

{[NH3 ](RT )}2 {[N2 ](RT )}{[H2 ](RT )}3 [NH3 ]2 1 Kc    3 2 [N2 ][H2 ] (RT ) (RT )2

Kp 

or Kp = Kc(RT)−2 Once again, you see that Kp and Kc are not the same but are related by some function of RT.

Kp = Kc(RT)∆n

∆n = total moles of gaseous products −  total moles of gaseous reactants For the decomposition of CaCO3,

∆n = 1 − 0 = 1 whereas the value of ∆n for the ammonia synthesis is

∆n = 2 − 4 = −2 What about a reaction in which ∆n is zero, such as the oxidation of NO by ozone?

NO(g) + O3(g) uv NO2(g) + O2(g) Now Kp = Kc.

Chapter 15 / Principles of Chemical Reactivity: Equilibria

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Chemistry in Your Career

Dr. Sonali Paul

Dr. Sonali Paul Dr. Sonali Paul (she/her/hers) is a teacher and researcher at the University of Engineering and Management in Kolkata, India. Dr. Paul’s background in botany (B.S., Presidency College, Kolkata) combined with her advanced study of biochemistry (M.S. and Ph.D., University of Calcutta) has informed her research into the remediation of arsenic contamination in agriculture. Arsenic occurs naturally in the Earth's crust, but it poses a serious threat to human health when it enters the food chain. A known carcinogen, arsenic targets enzyme reactions, which

means that arsenic toxicity has a broad range of negative effects on nearly every organ system. Rice plants are found to be particularly susceptible to arsenic because they grow in waterlogged paddies, where there is ample time for the uptake of arsenic from contaminated ground water. Rice  is a staple food for much of Asia, including the area of West Bengal, India, where Dr. Paul focuses much of her research. Dr. Paul’s current project is formulating a fertilizer that reduces arsenic uptake, which could protect the health and livelihood of populations that depend on rice.

The large value of K indicates that, at equilibrium, [NO2][O2] >> [NO][O3]. If stoichiometric amounts of NO and O3 are mixed and allowed to come to equilibrium, virtually none of the reactants will be found. Essentially, all will have been converted to NO2 and O2. A chemist would say that “the reaction has gone to completion.” Conversely, a small value of K means that little of the products exists when equilibrium has been achieved. That is, the reactants are favored over the products at equilibrium. K < 1: Reaction is reactant-favored at equilibrium. Concentrations of reactants are greater than concentrations of products at equilibrium.

This is true for the formation of ozone from oxygen. 3⁄2

Kc 

O2(g) uv O3(g)

[O3 ]  2.5  1029 at 25 C [O2 ] 3 / 2

The small value of K indicates that, at equilibrium, [O3] K: If Q is greater than K, some products must be converted to reactants for the reaction to reach equilibrium. This will increase the reactant concentrations and decrease the product concentrations.

To illustrate these points, consider the transformation of butane to isobutane (2-methylpropane). Butane

Isobutane CH3

CH3CH2CH2CH3

Kc =

CH3CHCH3

[isobutane] = 2.50 at 298 K [butane]

Any mixture of butane and isobutane, whether at equilibrium or not, can be represented by the reaction quotient Q (=[isobutane]/[butane]). Suppose you have a mixture of 0.0030  mol/L of butane and 0.0040  mol/L of isobutane (at 298 K ) (Figure 15.4a). This means that the reaction quotient, Q, is Q

(a) Not at equilibrium. Q 5 4/3 , K. Here, four isobutane molecules and three butane molecules are present. The reaction will proceed to convert butane into isobutane to reach equilibrium.

[ isobutane ]  [ butane ]

(b) At equilibrium. Q 5 5/2 5 K. Here, five isobutane molecules and two butane molecules are present. The reaction is at equilibrium.

0.0040  1.3 0.0030

(c) Not at equilibrium. Q 5 6/1 . K. Here, six isobutane molecules and one butane molecule are present. The reaction will proceed to convert isobutane into butane to reach equilibrium.

Figure 15.4  The interconversion of isobutane and butane.  Only when the concentrations of isobutane and butane are in the ratio [isobutane]/[butane] = 2.5 is the system at equilibrium (b). With any other ratio of concentrations, there will be a net conversion of one compound into the other until equilibrium is achieved.

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Chapter 15 / Principles of Chemical Reactivity: Equilibria

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This set of concentrations does not represent an equilibrium system because Q < Kc. To reach equilibrium, some butane molecules must be converted to molecules of isobutane, thereby lowering [butane] and raising [isobutane]. This transformation will continue until the ratio [isobutane]/[butane] = 2.5; that is, until Q = Kc (Figure 15.4b). What happens when there is too much isobutane in the system relative to the amount of butane? Suppose [isobutane] = 0.0060  mol/L but [butane] is only 0.0010 mol/L (Figure 15.4c). Now the reaction quotient Q is greater than Kc (Q > Kc), and the system is again not at equilibrium. It will proceed to equilibrium by converting isobutane molecules to butane molecules.

Ex am p le 15.2

The Reaction Quotient Problem  The brown gas nitrogen dioxide, NO2, can exist in equilibrium with the colorless gas N2O4. Kc = 170 at 298 K for the reaction 2 NO2(g) uv N2O4(g)  Kc = 170 Suppose that the concentration of NO2 is 0.015 M and the concentration of N2O4 is 0.025 M. Is the system at equilibrium? If not, in which direction will the reaction proceed to achieve equilibrium?

What Do You Know?  You are given the balanced chemical equation, the value of the equilibrium constant Kc , and concentrations of reactant and product. Strategy

• •

Write the expression for the reaction quotient, Q, from the balanced equation.

•

Decide if Q is greater than, less than, or equal to Kc and in which direction the reaction will proceed.

Substitute concentrations of reactant and product into the expression and calculate Q.

Solution Q

[N2O4 ] (0.025)   110 [NO2]2 (0.015)2

The value of Q is less than the value of Kc (Q < K), so  the reaction is not at equilibrium. The  system proceeds to equilibrium by converting NO2 to N2O4, thereby increasing [N2O4], and decreasing [NO2]  until Q = Kc.

Think about Your Answer  When calculating Q, make sure that you raise each concentration to the power corresponding to the stoichiometric coefficient.

Check Your Understanding  Answer the following questions regarding the butane uv isobutane equilibrium (Kc = 2.50 at 298 K). (a) Is the system at equilibrium when [butane] =  0.00097 M and [isobutane] = 0.00218 M? If not, in which direction will the reaction proceed to achieve equilibrium? (b) Is the system at equilibrium when [butane] =  0.00075 M and [isobutane] = 0.00260 M? If not, in which direction will the reaction proceed to achieve equilibrium?



15.2  The Equilibrium Constant and Reaction Quotient

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15.3 Determining an Equilibrium Constant Goals for Section 15.3 • Calculate an equilibrium constant given the reactant and product concentrations at equilibrium.

• Calculate an equilibrium constant given the initial concentrations of the reactants and products and the concentration of one reactant or product at equilibrium.

When the concentrations of all of the reactants and products are known at equilibrium, an equilibrium constant can be calculated by substituting the data into the equilibrium constant expression. Assume that a mixture of SO2, O2, and SO3 is at equilibrium at 852 K. 2 SO2(g) + O2(g) uv 2 SO3(g)

The equilibrium concentrations are found to be [SO2] = 3.61 × 10−3  mol/L, [O2] = 6.11 × 10−4 mol/L, and [SO3] = 1.01 × 10−2 mol/L. Substituting these data into the equilibrium constant expression, you can calculate the value of Kc. Kc 

[SO3 ]2 (1.01  102)2   1.28  104 at 852 K [SO2 ]2[O2 ] (3.61  103)2(6.11  104 )

(Notice that Kc has a large value; at 852 K, the oxidation of sulfur dioxide is productfavored at equilibrium.) More often, an experiment provides information on the initial quantities of reactants and the concentration at equilibrium of only one reactant or one product. The equilibrium concentrations of the rest of the reactants and products must then be calculated based on the reaction stoichiometry. As an example, again consider the oxidation of sulfur dioxide to sulfur trioxide and suppose that 0.00100 mol each of SO2 and O2 are placed in a 1.00-L flask at a different temperature. When equilibrium is achieved, 0.00054 mol of SO3 is formed. This information can be used to calculate the equilibrium constant for the reaction at this temperature. Set up an ICE ­table (Sections  4.1 and 15.2) showing the initial concentrations, the changes in those concentrations upon proceeding to equilibrium, and the concentrations at equilibrium.

Equation

2 SO2(g)

Initial (M)

0.00100 −2x

Change (M) Equilibrium (M)

0.00100 − 2x = 0.00100 − 0.00054 = 0.00046

O2(g)

1

uv

0.00100 −x 0.00100 − x = 0.00100 − 0.00054/2 = 0.00073

2 SO3(g) 0 +2x 2x = 0.00054

The quantities in the ICE table come from the following analysis:

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•

Initial: The concentration of both SO2 and O2 is 0.00100 mol/1.00 L = 0.00100 M. No SO3 was placed in the flask, so its initial concentration is 0 M.

•

Change: The amount of O2 consumed is designated as −x mol/L. It is given a minus sign because O2 is consumed. It then follows from the reaction stoichiometry that the amount of SO2 consumed is −2x, and the amount of SO3 produced is +2x.

Chapter 15 / Principles of Chemical Reactivity: Equilibria

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Equilibrium (a) You know from the experiment that [SO3] at equilibrium is 0.00054 M. Therefore, 2x = 0.00054 M. (b) T  he equilibrium concentration of SO2 is equal to the initial concentration minus what was consumed. Therefore, [SO2] is (0.00100 − 2x) M or 0.00046 M. (c) The amount of O2 consumed is half of the amount of SO3 produced or x (= 0.00027 mol/L). Therefore, the equilibrium concentration of O2 is (0.00100 − x) M, or 0.00073. With the equilibrium concentrations now known, it is possible to calculate Kc. •

Kc 

[SO3]2 (0.00054)2   1.9  103 [SO2]2[O2] (0.00046)2(0.000733)

V Ex am p le 15.3

Calculating an Equilibrium Constant (Kc) Using Concentrations Problem  In aqueous solution, iron(III) ions react with iodide ions to give iron(II) ions and triiodide ions, I3−. Suppose the initial concentration of Fe3+ ions is 0.200 M, the initial I− ion concentration is 0.300 M. When equilibrium is achieved, the concentration of I3− ions is 0.0866 M. What is the value of Kc? 2 Fe3+(aq) + 3 I−(aq) uv 2 Fe2+(aq) + I3−(aq)

What Do You Know?  You are given the balanced equation (from which the equiStrategy Map Problem Calculate Kc for reaction of Fe3+ with I− to give Fe2+ and I3−.

librium constant expression can be written), initial concentrations of the reactants, and the concentration of one product (I3−) after equilibrium is reached.

Strategy Step 1. Write the balanced equation and set up an ICE table. Step 2. Enter the initial concentrations on the initial (I) line of the table.

Data/Information • Balanced equation • Initial concentrations of reactants • Equilibrium concentration of one product

Step 3. Using the variable x, determine the changes in concentrations of reactants and products according to the stoichiometry of the reaction. Enter the changes in concentration on the change (C) line of the table. Step 4. Knowing the equilibrium concentration of I3− is x and that x 5 0.0866 M, ­calculate the equilibrium concentrations for each species. Enter the equilibrium concentrations on the equilibrium (E) line of the table. Step 5. Enter the equilibrium concentrations of reactants and products into the equilibrium constant expression and solve for Kc.

Solution Step 1

Write the balanced equation and set up an ICE table. The strategies for completing the ICE table are given in Steps 2–4.

Equation



2 Fe3+

3 I−

1

uv

2 Fe2+

I3−

1

Initial (M)

0.200

0.300

0

0

Change (M)

−2x

−3x

+2x

+x

Equilibrium (M)

0.200 − 2(0.0866)  = 0.0268

0.300 − 3(0.0866)  = 0.0402

2(0.0866) = 0.1732

0.0866

15.3  Determining an Equilibrium Constant

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747

Step 2

Enter the initial concentrations on the initial (I) line of the table.

Step 3

Using the variable x, determine the changes in concentrations of reactants and products according to the stoichiometry of the reaction. Begin by writing that x mol/L of I3− ions are produced upon proceeding to equilibrium. Based on the reaction stoichiometry, this means that 2x mol/L of Fe2+ ions must also be produced, 2x mol/L of Fe3+ ions are consumed, and 3x mol/L of I− ions are consumed.

Step 4

Calculate the equilibrium concentrations of the reactants and products. The equilibrium concentration of I3− is known; x = 0.0866 M. Calculate the equilibrium concentrations of the other reactants and products by adding the changes in concentrations to their initial concentrations.

Step 5

Solve the equilibrium expression for Kc. The concentration of each substance at equilibrium is now known, and Kc can be calculated. Kc 

[Fe2]2[I3] (0.1732)2(0.0866)   5.6 × 104 3 2  3 [Fe ] [I ] (0.0268)2(0.0402)3

Think about Your Answer  The key to this problem is that the concentration of one product, I3−, was known at equilibrium. The equilibrium concentrations of the other product and of the reactants were derived from this based on the reaction stoichiometry. Also notice that the calculated Kc is much greater than one, so the reaction is product-­ favored at equilibrium. This is consistent with the equilibrium concentrations calculated in the ICE table.

Check Your Understanding  A solution is prepared by dissolving 0.050 mol of diiodocyclohexane, C6H10I2, in the solvent CCl4. The total solution volume is 1.00 L. When the reaction C6H10I2 uv C6H10 + I2 reaches equilibrium at 35 °C, the concentration of I2 is 0.035 mol/L. (a) What are the concentrations of C6H10I2 and C6H10 at equilibrium? (b) Calculate Kc, the equilibrium constant.

15.4 Using Equilibrium Constants in Calculations Goal for Section 15.4 • Use equilibrium constants to calculate the concentration (or pressure) of a reactant or product at equilibrium.

Suppose the value of Kc or Kp and the initial concentrations of reactants are known, and you want to know the equilibrium concentrations. ICE tables will again be used to summarize the initial conditions, the changes that occur upon proceeding to equilibrium, and the final conditions.

748

Chapter 15 / Principles of Chemical Reactivity: Equilibria

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VEx am p le 15.4

Calculating Equilibrium Concentrations Problem  The equilibrium constant Kc (= 55.64) for H2(g) + I2(g) uv 2 HI(g) was determined at 425 °C. If 0.130 mol each of H2 and I2 is placed in a 25.0-L flask at 425 °C, what are the concentrations of H2, I2, and HI when the system reaches equilibrium?

What Do You Know?  You are given the balanced equation (from which the equiStrategy Map Problem What are the concentrations of H2, I2, and HI when the system reaches equilibrium?

Data/Information • Balanced equation • Value of Kc • Initial concentrations of reactants

librium constant expression can be written), the value of Kc, and both the initial amounts of the reactants and the volume of the container (from which initial concentrations of the reactants can be calculated).

Strategy Step 1. Write the equilibrium constant expression and set up an ICE table. Step 2. Enter the initial concentrations of H2 and I2 on the initial (I) line. Step 3. Using the variable x, determine the changes in concentrations of reactants and products according to the stoichiometry of the reaction. Enter the changes in concentration on the change (C) line of the table. Step 4. Enter the expressions for the equilibrium concentrations of all three species on the equilibrium line (E) of the ICE table. Step 5. Insert the expressions for the equilibrium concentrations into the equilibrium constant expression and solve for x. Step 6. Use the calculated value of x to solve for the final concentration of each species.

Solution Step 1

Write the equilibrium constant expression and set up an ICE table. Kc 

[HI]2  55.64 [H2][I2]

The strategies for completing the ICE table are given in Steps 2–4.

Equation

H2(g)

Initial (M)

0.130 mol/25.0 L  = 5.200 × 10−3 M

0.130 mol/25.0 L  = 5.200 × 10−3 M

0

Change (M)

−x

−x

+2x

Equilibrium (M)

−3

5.200 × 10

I2(g)

1

M − x

−3

5.200 × 10

uv

M − x

2 HI(g)

2x

Step 2

Enter the initial concentrations of H2 and I2 on the initial (I) line.

Step 3

Using the variable x, determine the changes in concentrations of reactants and products according to the stoichiometry of the reaction and enter these on the change (C) line. Based on the reaction stoichiometry the change in [H2] and [I2] is −x, and the change in [HI] is +2x.

Step 4

Enter the expressions for the equilibrium concentrations of all three species on the equilibrium line (E). The variable x remains in the concentration terms for all three species. All three concentrations are related through the same unknown variable x.



15.4  Using Equilibrium Constants in Calculations

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749

Step 5

Solve the equilibrium expression for x. Now the expressions for the equilibrium concentrations can be substituted into the equilibrium constant expression. 55.64 

3

(5.200  10

(2 x)2 (2 x)2  3  x)(5.200  10  x) (5.200  103  x)2

In this case, you can solve for the unknown quantity x by taking the square root of both sides of the equation, 2x 5.200  103  x 7.4592 (5.200  103  x)  0.03879  7.4592x  2 x K c  7.4592 

0.03879  9.4592x x  4.101  103 Step 6

Solve for the final concentration of each species. With x known, you can solve for the equilibrium concentrations of the reactants and products.  [H2] = [I2] = 5.200 × 10−3 – x =  1.099 × 10–3 M = 1.10 × 10−3 M [  HI]  = 2x = 8.201 × 10−3 M = 8.20 × 10–3 M 

Think about Your Answer  It is always wise to verify the answer by substituting the value of each concentration back into the equilibrium constant expression to see if your calculated Kc agrees with the one given in the problem. In this case,

(8.201  103)2  55.7 (1.099  103)2 which agrees quite well with the value of Kc .

Check Your Understanding  At some temperature, Kc = 33 for the reaction H2(g) + I2(g) uv 2 HI(g) Assume the initial concentrations of both H2 and I2 are 6.00 × 10−3 mol/L. Find the concentration of each reactant and product at equilibrium.

Calculations Where the Solution Involves a Quadratic Expression Suppose you are studying the decomposition of PCl5 to form PCl3 and Cl2. You know that Kc = 1.20 at a given temperature. PCl5(g) uv PCl3(g) + Cl2(g)

If the initial concentration of PCl5 is 0.0920 M, what will be the concentrations of  reactant and products when the system reaches equilibrium? Following the ­procedures outlined in Example 15.4, you would set up an ICE table to define the ­equilibrium concentrations of reactants and products.

750

Chapter 15 / Principles of Chemical Reactivity: Equilibria

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Reaction

PCl5(g)

uv

PCl3(g)

Cl2(g)

1

Initial (M)

0.0920

0

0

Change (M)

−x

+x

+x

Equilibrium (M)

0.0920 − x

x

x

Substituting into the equilibrium constant expression, you have Kc  1.20 

[PCl 3][Cl 2] ( x)( x)  [PCl 5] 0.0920  x

Expanding the algebraic expression results in a quadratic equation, x2 + 1.20x − 0.1104 = 0

Using the quadratic formula (Appendix A; a = 1, b = 1.20, and c = −0.1104), you find two roots to the equation: x = 0.08586 and −1.286. Because a negative value of x (leads to negative concentrations of PCl3 and Cl2) is not chemically meaningful, the answer is x = 0.08586 M. Therefore, at equilibrium [PCl5] = 0.0920 − 0.08586 = 0.00614 = 0.0061 M [PCl3] = [Cl2] = 0.08586 = 0.0859 M

Although a solution to a quadratic equation can always be obtained using the quadratic formula, in many instances you can find an acceptable answer by using a realistic approximation to simplify the calculation. To illustrate this, consider another equilibrium, the dissociation of I2  molecules to form I  atoms, for which Kc = 5.6 × 10−12 at 500 K.

Solving Quadratic Equations ​

Quad­ratic equations are usually solved using the quadratic formula (Appendix A). An alternative is the method of successive approximations, also outlined in Appendix A. Most equilibrium expressions can be solved quickly by this method, and you are urged to try it. This will remove the uncertainty of whether K expressions need to be solved exactly.

I2(g) uv 2 I(g) Kc 

[I]2  5.6  1012 [I2]

If the initial I2 concentration is 0.45 M the ICE table is Reaction

I2(g)

uv

2 I(g)

Initial (M)

0.45

0

Change (M)

−x

+2x

Equilibrium (M)

0.45 − x

2x

Again, the equilibrium constant expression produces a quadratic equation. Kc  5.6  1012 

(2x)2 (0.45  x)

This could be solved using the quadratic formula, but there is a simpler way to reach an answer. The value of Kc is very small, indicating that the reaction is reactant-­ favored at equilibrium. In fact, Kc is so small that subtracting x from the original reactant concentration (0.45 mol/L) in the denominator of this equation leaves it essentially unchanged. That is, (0.45 − x) is essentially equal to 0.45. Dropping the x from the denominator simplifies the expression.



15.4  Using Equilibrium Constants in Calculations

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751

Kc  5.6  1012 

(2 x)2 (0.45)

Solving this equation gives x = 7.94 × 10−7 mol/L. From this value, you can determine that [I2] = 0.45 − x = 0.45 mol/L and [I] = 2x = 1.6 × 10−6 mol/L. Notice that the answer confirms the assumption that the dissociation of I2 is so small that [I2] at equilibrium is essentially equal to the initial concentration. When is it possible to simplify a quadratic equation in this way? The decision depends on both the value of the initial concentration of the reactant and the value of x, which is in turn related to the value of K. Consider the general reaction A uv B + C

where K = [B][C]/[A]. Assume that there is no B or C initially present, that you know K and the initial concentration of A (= [A]0), and wish to find the equilibrium concentrations of B and C (= x). The equilibrium constant expression now is Kc 

[B][C] ( x)( x)  [A] [A]0  x

When Kc is very small, the value of x will be much less than [A]0, so [A]0 − x ≅ [A]0. Therefore, you can write the expression Kc 



[B][C] ( x)( x)  [A] [A]0

(15.3)

A good guideline to follow is:  If 100 × Kc < [A]0, the approximate expression will give acceptable values of equilibrium concentrations  (to two significant figures). For more about this, see “Problem Solving Tip 15.1: When Do You Need to Use the Quadratic Formula?”

E xamp le 15.5

Calculating Equilibrium Concentrations Using an Equilibrium Constant Problem  The reaction N2(g) + O2(g) uv 2 NO(g) contributes to air pollution whenever a fuel is burned in air at a high temperature, as in a gasoline engine. At 1500 K, Kc = 1.0 × 10−5. Suppose a sample of air has [N2] = 0.080 mol/L and [O2] = 0.020 mol/L before any reaction occurs. Calculate the equilibrium concentrations of reactants and products after the mixture is heated to 1500 K.

What Do You Know?  As in Example 15.4, you know the value of Kc and can write the equilibrium expression from the balanced equation. You also know the initial concentrations of the reactants and can define the equilibrium concentrations in terms of the amounts of N2 and O2 consumed (= x).

Strategy  Set up an ICE table, and then substitute the equilibrium concentrations into the equilibrium constant expression. The result will be a quadratic equation. This expression can be solved using the methods outlined in Appendix A or using the guideline in the text to simplify the calculation.

752

Chapter 15 / Principles of Chemical Reactivity: Equilibria

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Solution  First set up an ICE table where the amounts of N2 and O2 consumed are designated as x.

Equation

N2(g)

1

O2(g)

2 NO(g)

uv

Initial (M)

0.080

0.020

0

Change (M)

−x

−x

+2x

Equilibrium (M)

0.080 − x

0.020 − x

2x

Next, substitute the equilibrium concentrations into the equilibrium constant expression. K c  1.0  105 

[NO]2 (2 x)2  [N2][O2] (0.080  x)(0.020  x)

Refer to the guideline to decide whether an approximate solution (Equation 15.3) is ­acceptable. Here, 100 × Kc (= 1.0 × 10−3) is smaller than either of the initial reactant concentrations (0.080 and 0.020). This means you can use the approximate expression. K c  1.0  105 

[NO]2 (2 x)2  [N2][O2] (0.080)(0.020)

Solve the expression for x. 1.60 × 10−8 = 4x2 x = 6.32 × 10−5 M Therefore, the reactant and product concentrations at equilibrium are  [N2] = 0.080 − 6.32 × 10−5 ≈ 0.080 M [O2] = 0.020 − 6.32 × 10−5 ≈ 0.020 M   [  NO]  = 2x = 1.3 × 10−4 M 

Think about Your Answer  The value of x obtained using the approximation is the same as that obtained from the quadratic formula.

Check Your Understanding  The decomposition of PCl5(g) to form PCl3(g) and Cl2(g) has Kc =33.3 at a high temperature. If the initial concentration of PCl5 is 0.1000 M, what are the equilibrium concentrations of the reactants and products?

Problem Solving Tip 15.1 When Do You Need to Use the Quadratic Formula? In most equilibrium calculations, the quantity x may be neglected in the denominator of the equation K =  x2/([A]0 − x) if x is less than 10% of the concentration of reactant initially present (= [A]0). The guideline presented in the text for making the approximation that [A]0 − x = [A]0 when 100 × K < [A]0 reflects this fact.



In general, when K is about 1 or greater, the approximation cannot be made. If K is much less than 1 and 100 × K < [A]0 (you will see many such cases in Chapter 16), the approximate expression (K = x2/[A]0) gives an acceptable answer. If you are not certain, first make the assumption that the unknown (x) is small,

and solve the approximate expression (K = (x)­2/[A]0). Next, compare the approximate value of x with [A]0. If x has a value equal to or less than 10% of [A]0, there is no need to solve the full equation using the quadratic formula.

15.4  Using Equilibrium Constants in Calculations

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753

15.5 More about Balanced Equations and Equilibrium Constants Goal for Section 15.5 • Know how K changes as different stoichiometric coefficients are used in a balanced equation, if the equation is reversed, or if several equations are added to give a new net equation.

Using Different Stoichiometric Coefficients Chemical equations can be balanced using different sets of stoichiometric coefficients. For example, the equation for the oxidation of carbon to give carbon monoxide can be written C(s) + 1⁄2 O2(g) uv CO(g)

for which the equilibrium constant expression is K1 

[CO]  4.6  1023 at 25 C [O2 ]1⁄2

You can write the chemical equation equally well, however, as 2 C(s) + O2(g) uv 2 CO(g)

and the equilibrium constant expression is now K2 

[CO]2  2.1  1047 at 25 C [O2 ]

When you compare the two equilibrium constant expressions you find that K2  = (K1)2; that is, 2

K2 

 [CO]  [CO]2   K 12 1⁄  [O2]  [O2] 2 

When the stoichiometric coefficients of a balanced equation are multiplied by some factor, the equilibrium constant for the new equation is the old equilibrium constant raised to the power of the multiplication factor.

Reversing a Chemical Equation Consider what happens if a chemical equation is reversed. Here, you will compare the value of Kc for formic acid transferring an H+ ion to water HCO2H(aq) + H2O(ℓ) uv HCO2−(aq) + H3O+(aq) K1 

[HCO2][H3O]  1.8  104 at 25 C [HCO2 H]

with the opposite reaction, the gain of an H+ ion by the formate ion, HCO2−. HCO2−(aq) + H3O+(aq) uv HCO2H(aq) + H2O(ℓ) K2 

[HCO2 H]  5.6  103 at 25 C [HCO2][H3O]

Here, K2 = 1/K1. The equilibrium constants for a reaction and its reverse are reciprocals.

754

Chapter 15 / Principles of Chemical Reactivity: Equilibria

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

+

−

−

+

+

−

− −

+

+

+ © Charles D. Winters/Cengage

+

+

−

+ AgCl(s) in water

−

+

+

+

−

+ + After adding NH3(aq)

Figure 15.5  Dissolving silver chloride in aqueous ammonia.  (left) A precipitate of AgCl(s) is suspended in water. (right) When aqueous ammonia is added, the ammonia reacts with the trace of silver ion in solution, the equilibrium shifts, and the silver chloride dissolves.

Adding Two Chemical Equations It is often useful to add two equations to obtain the equation for a net process. As an example, consider the reactions that occur when silver chloride dissolves in water (to a very small extent) and ammonia is added to the solution. Ammonia reacts with silver ions to form a water-soluble compound, Ag(NH3)2Cl (Figure 15.5). Adding the equation for dissolving solid AgCl to the equation for the reaction of Ag+ ions with ammonia gives the equation for the net reaction, dissolving solid AgCl in aqueous ammonia. (All equilibrium constants are given at 25 °C.) AgCl(s) uv Ag+(aq) + Cl−(aq)

K1 = [Ag+][Cl−] = 1.8 × 10−10

Ag+(aq) + 2 NH3(aq) uv [Ag(NH3)2]+(aq)

K2 

[Ag(NH3)2]  1.1  107 [Ag][NH3]2

Net equation: AgCl(s) + 2 NH3(aq) uv [Ag(NH3)2]+(aq) + Cl−(aq)

To obtain the equilibrium constant for the net reaction, Knet, multiply the equilibrium constants for the two reactions, K1 × K2. Knet  K1  K2  [ Ag ][Cl] 

[Ag(NH3)2 ] [Ag(NH3)2 ][Cl]  [ Ag ][NH3 ]2 [NH3 ]2

Knet = K1 × K2 = 2.0 × 10−3 When two or more chemical equations are added to produce a net equation, the equilibrium constant for the net equation is the product of the equilibrium constants for the added equations.

Problem Solving Tip 15.2 Balanced Equations and Equilibrium Constants You now know: 1. how to write an equilibrium constant expression from the balanced equation, recognizing that the concentrations of solids, pure liquids, and liquids used as solvents do not appear in the expression.



2. that when the stoichiometric co­efficients in a balanced equation are changed by a factor of n, Knew = (Kold)n. 3. that when a balanced equation is reversed, Knew = 1/Kold.

4. that when several balanced equations (each with its own equilibrium constant, K1, K2, etc.) are added to obtain a net, balanced equation, Knet = K1 × K2 × K3 × . . . .

15.5  More about Balanced Equations and Equilibrium Constants

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755

E xamp le 15.6

Balanced Equations and Equilibrium Constants Problem  A mixture of nitrogen, hydrogen, and ammonia is brought to equilibrium. When the equation is written using whole-number coefficients, as follows, the value of Kc is 3.5 × 108 at 25 °C. N2(g) + 3 H2(g) uv 2 NH3(g)

Equation 1:

K1 = 3.5 × 108

However, the equation can also be written as given in Equation 2. What is the value of K2? 1

⁄2 N2(g) + 3⁄2 H2(g) uv NH3(g)

Equation 2:

K2 = ?

The decomposition of ammonia to the elements (Equation 3) is the reverse of its formation (Equation 1). What is the value of K3? 2 NH3(g) uv N2(g) + 3 H2(g)

Equation 3:

K3 = ?

What Do You Know?  You know the value of Kc for a given balanced equation. You want to know how the value of Kc changes as the stoichiometric coefficients change or when the equation is reversed. Strategy  Determine how the desired reaction is related to the given reaction(s). (Was a given chemical reaction multiplied by a factor? Was a reaction reversed? Were two or more reactions added?) Use the relationships discussed to determine the effect(s) of these transformations on the given K value(s). See also Problem Solving Tip 15.2. Solution  Equation 2 is obtained by multiplying Equation 1 by 1/2. Thus K2 is equal to K1 raised to the one-half power, K11/2. To confirm this relationship between K1 and K2, write the equilibrium constant expressions for these two balanced equations. K1 

[NH3]2 [N2][H2]3

K2 

[NH3] 1 3 [N2] ⁄2 [H2] ⁄2

Writing these expressions makes it clear that K2 is the square root of K1. K 2  ( K1 )

1⁄ 2



K1  3.5  108   1.9 × 104 

Equation 3 is the reverse of Equation 1, and its equilibrium constant expression is K3 

[N2][H2]3 [NH3]2

In this case, K3 is the reciprocal of K1. That is, K3 = 1/K1. K3 

1 1   2.9 × 10−9  3.5  108 K1

Think about Your Answer  Notice that the production of ammonia from the elements has a large equilibrium constant and is product-favored at equilibrium (Section  15.2). As expected, the reverse reaction, the decomposition of ammonia to its elements, has a small equilibrium constant and is reactant-favored at equilibrium.

Check Your Understanding The conversion of oxygen to ozone has a very small equilibrium constant. 3⁄2

O2(g) uv O3(g)  K = 2.5 × 10−29

(a) What is the value of K when the equation is written using whole-number coefficients? 3 O2(g) uv 2 O3(g) (b) What is the value of K for the conversion of ozone to oxygen? 2 O3(g) uv 3 O2(g)

756

Chapter 15 / Principles of Chemical Reactivity: Equilibria

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

15.6 Disturbing a Chemical Equilibrium Goal for Section 15.6 • Predict how a system at equilibrium will respond if the reaction conditions are changed.

The equilibrium between reactants and products may be disturbed in three ways: (1) by changing the temperature, (2) by changing the concentration of a reactant or product, or (3) by changing the volume (for systems involving gases). A change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner as to reduce or counteract the effect of the change. This statement is often referred to as Le Chatelier’s principle (Section 13.3). The principle describes how the quantities of reactants and products are adjusted so that equilibrium is restored, that is, so that the reaction quotient is once again equal to the equilibrium constant.

Effect of the Addition or Removal of a Reactant or Product Consider the following experiment: You have a chemical system initially at equilibrium. To this you add (or take away) one or more of the reactants or products. The system will no longer be at equilibrium. When the system returns to equilibrium the new equilibrium concentrations of reactants and products will be different, but the value of the equilibrium constant expression will still equal K. To illustrate this, look again at the butane/isobutane equilibrium. CH3 CH3CH2CH2CH3

CH3CHCH3

butane

isobutane

Kc = 2.5

Suppose the equilibrium mixture consists of two molecules of butane and five molecules of isobutane (Figure 15.6). The reaction quotient, Q, is 5/2 (or 2.5/1), the value of the equilibrium constant for the reaction. Now you add seven more

Seven isobutane are added.

The system returns to equilibrium.

Q = 5/2 = K

Q = 12/2 > K

An equilibrium mixture of five isobutane molecules and two butane molecules.

Seven isobutane molecules are added, so the system is no longer at equilibrium.

Figure 15.6  Addition of more reactant or product to an equilibrium system.

Q = 10/4 = K A net of two isobutane molecules has changed to butane molecules, to once again give an equilibrium mixture where the ratio of isobutane to butane is 5:2 (or 2.5:1).

15.6  Disturbing a Chemical Equilibrium

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757

molecules of isobutane to the mixture to give a ratio of 12 isobutane molecules to two butane molecules. The reaction quotient is now 6 (= 12/2). This means Q is greater than K, and the system will change to reestablish equilibrium. To do so, some molecules of isobutane must be changed into butane molecules, a process that continues until the ratio [isobutane]/[butane] is once again 2.5/1. In this particular case, if two of the 12 isobutane molecules change to butane, the ratio of isobutane to butane is again equal to Kc (= 10/4 = 2.5/1), and equilibrium is reestablished. Le Chatelier’s principle provides a shortcut to determine the direction in which the reaction occurs to restore equilibrium. The initial equilibrium was disturbed by  the addition of the product isobutane. Le Chatelier’s principle indicates that the chemical reaction will occur in the direction that reduces the effect of this change. In this case, that means that the reaction will occur in the direction that reduces the amount of isobutane. That is, the reaction will occur in the reverse direction as ­written, from products (isobutane) to reactants (butane).

VExampl e 1 5 .7

Effect of Concentration Changes on Equilibrium Problem  Assume equilibrium has been established in a 1.00-L flask with [butane] =  0.00500 mol/L and [isobutane] = 0.0125 mol/L. Butane uv Isobutane   Kc = 2.50 Then 0.0150 mol of butane is added. What are the concentrations of butane and isobutane when equilibrium is reestablished? Strategy Map Problem What are the equilibrium ­concentrations of reactant and product after adding excess reactant?

Data/Information • Balanced equation • Value of Kc • Initial concentrations of reactant and product and amount of excess reactant added

What Do You Know?  Here you know the value of Kc , the balanced equation, the original equilibrium concentrations of reactant and product, and the amount of reactant added to the system at equilibrium. Strategy Step 1. Write the balanced equation and set up an ICE table. Step 2. Enter the initial concentrations of butane and isobutane on the initial (I) line of the table. Step 3. Assign the variable x to represent changes in concentration on the change (C) line. Step 4. Enter the expressions for the equilibrium concentrations of butane and isobutane on the third line (E) of the ICE table. Step 5. Insert the expressions for the equilibrium concentrations into the equilibrium constant expression and solve for x. Step 6. Use the calculated value of x to solve for the final concentration of each species.

Solution Step 1

Write the equilibrium constant expression and set up an ICE table. The strategies for completing the ICE table are given in Steps 2–4.

758

Equation

Butane

Initial (M)

0.0200

0.0125

Change in concentration to reestablish equilibrium (M)

−x

+x

Equilibrium (M)

0.0200 − x

0.0125 + x

uv

Isobutane

Chapter 15 / Principles of Chemical Reactivity: Equilibria

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Step 2

Enter the initial concentrations on the initial (I) line of the table. The concentration of butane is the sum of its initial moles (0.00500 mol) and the added butane (0.0125 mol) divided by the volume (1.0 L).

Step 3

Assign the variable x to represent changes in concentration on the change (C) line. After adding excess butane, Q < Kc. The concentration of butane must decrease by an amount x and that of isobutane must increase by an amount x to reestablish equilibrium.

Step 4

Enter the expressions for the equilibrium concentrations on the equilibrium (E) line.

Step 5

Solve the equilibrium expression for x. K c  2.50 

[isobutane] [butane]

0.0125  x 0.0200  x 2.50 (0.0200  x )  0.0125  x 2.50 

x  0.01071 mol/L Solve for the final concentration of each species.

Step 6

 [butane] = 0.0200 − x = 0.00929 M = 0.0093 M [isobutane] = 0.0125 + x = 0.02321 M = 0.0232 M  Answer check: new ratio [isobutane]/[butane] = 0.02321/0.00929 = 2.5

Think about Your Answer  As predicted by Le Chatelier’s principle, the stress on the system from adding butane to the equilibrium mixture is relieved by converting some butane to isobutane to achieve a new equilibrium mixture where Q again equals Kc. In the new equilibrium mixture, the butane concentration lies between its initial value and its value immediately upon the addition of excess butane, and the isobutane concentration is greater than the original value.

Check Your Understanding  Equilibrium exists between butane and isobutane when [butane] = 0.020 M and [isobutane] = 0.050 M. An additional 0.0200 mol/L of isobutane is added to the mixture. What are the concentrations of butane and isobutane after equilibrium is again attained?

Effect of Volume Changes on Gas-Phase Equilibria For an equilibrium system involving gases, what happens to the gas concentrations or pressures if the size of the container is changed? (Such a change occurs, for example, when fuel and air are compressed in an automobile engine.) The answer is that the gas concentrations also change if the container volume changes. And, if the concentrations change, then, depending on the stoichiometry, the equilibrium composition can change. As an example, consider the following equilibrium: 2 NO2(g) uv N2O4(g) brown gas

Kc =

colorless gas

[N2O4] = 170 at 298 K [NO2]2 15.6  Disturbing a Chemical Equilibrium

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What happens if the volume of the flask holding the gases is suddenly halved? The immediate result is that the concentrations of both gases will double. For example, assume equilibrium is initially established when [N2O4] is 0.0280 mol/L and [NO2] is 0.0128 mol/L. When the volume is halved, [N2O4] becomes 0.0560 mol/L, and [NO2] is 0.0256  mol/L. The reaction quotient, Q, under these circumstances is (0.0560)/(0.0256)2 = 85.4. Now, Q is less than K, and to return to equilibrium, the quantity of product must increase at the expense of the reactant. Thus, the new equilibrium composition will have a higher concentration of N2O4 than immediately after the volume change. 2 NO2(g)

N2O4(g)

decrease volume of container NO2 is converted to N2O4 until equilibrium is attained

In returning to equilibrium, the concentration of NO2 decreases twice as much as the concentration of N2O4 increases because one molecule of N2O4 is formed by consuming two molecules of NO2. This occurs until the reaction quotient, Q = [N2O4]/[NO2]2, is once again equal to Kc. The net effect of the volume decrease is to decrease the number of molecules in the gas phase. The conclusions for the NO2/N2O4 equilibrium can be generalized: •

for reactions involving gases, the stress of a volume decrease (a pressure increase) is counterbalanced by a change in the equilibrium composition to one with a smaller number of gas molecules.

•

for a volume increase (a pressure decrease), the equilibrium composition favors the side of the reaction with the larger number of gas molecules.

•

for a reaction in which there is no change in the number of gas molecules when proceeding from reactants to products, such as in the reaction of H2 and I2 to produce HI [H2(g) + I2(g) uv 2 HI(g)], a volume change will have no effect.

Effect of Temperature Changes on Equilibrium Composition Changing the temperature of a system at equilibrium is different from other equilibrium disturbances because  the value of the equilibrium constant changes with temperature. You can make a qualitative prediction about the effect if you know whether the reaction is exothermic or endothermic. As an example, consider the endothermic reaction of N2 with O2 to give NO. N2(g) + O2(g) uv 2 NO(g)   ∆rH° = +180.6 kJ/mol-rxn Kc 

[NO]2 [N2][O2]

Le Chatelier’s principle allows you to predict how the value of K will vary with temperature. The formation of NO from N2 and O2 is endothermic; that is, energy must be provided as heat for the reaction to occur. You might imagine that heat is a reactant. heat + N2(g) + O2(g) uv 2 NO(g)

If the system is at equilibrium and the temperature is increased, thus increasing the energy added as heat, the system will adjust to alleviate this stress. The way to counteract the energy input is to use up some of the energy added as heat by consuming N2 and O2 and producing more NO as the system returns to equilibrium. This raises the value of the numerator ([NO]2) and lowers the value of the denominator ([N2] [O2]) in the reaction quotient, Q, resulting in a higher value of Kc. The prediction regarding the N2/O2/NO reaction is correct as you see in the following table of equilibrium constants at various temperatures. The equilibrium constant and thus the proportion of NO in the equilibrium mixture increases with temperature.

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Equilibrium Constant, Kc

Temperature (K)

−31

 298

6.7 × 10

−10

 900

1.7 × 10

−3

2300

4.5 × 10

As another example, consider the combination of molecules of the brown gas NO2 to form colorless N2O4. An equilibrium between these compounds is readily achieved in a closed system (Figure 15.7). 2 NO2(g) uv N2O4(g)  ∆rH° = −57.1 kJ/mol-rxn Kc 

[N2O4 ] [NO2]2

Equilibrium Constant, Kc

Temperature (K)

1300

273

 170

298

Here, the reaction is exothermic, so you might imagine heat as being a reaction product. By lowering the temperature of the system, as in Figure 15.7, some energy is removed as heat. The removal of energy can be counteracted if the reaction produces energy as heat by the combination of NO2  molecules to give more N2O4. Thus, the equilibrium concentration of NO2 decreases; the concentration of N2O4 increases; and the value of K is larger at lower temperatures. In summary, when the temperature of a system at equilibrium increases, the equilibrium will shift in the direction that absorbs energy as heat (Table 15.1)—that is, in the endothermic direction.

•

if the temperature decreases, the equilibrium will shift in the direction that releases energy as heat—that is, in the exothermic direction.

•

changing the temperature changes the value of K.

Higher Temperature At 50 °C the equilibrium is shifted toward NO2, as indicated by the darker brown color.

© Marna G. Clarke/Cengage

•

2 NO2(g)

N2O4(g)

Brown gas

Colorless gas

Lower Temperature The mixture of NO2 and N2O4 gas at 0 °C is only slightly brown, indicating a smaller concentration of the brown gas NO2 than at 50 °C.

Figure 15.7  Effect of temperature on an equilibrium.  ​The tubes in the photograph both contain gaseous

NO2 (brown) and N2O4 (colorless) at equilibrium. Kc is larger at lower temperatures for this exothermic reaction, so the mixture is dominated by colorless N2O4 at lower temperatures.

15.6  Disturbing a Chemical Equilibrium

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Table 15.1

Effects of Disturbances on Equilibrium Composition

Change as Mixture Returns to Equilibrium

Disturbance

Effect on Equilibrium

Effect on K

Reactions Involving Solids, Liquids, or Gases Addition of reactant*

Some of added reactant is consumed

Product concentration increases

No change

Addition of product*

Some of added product is consumed

Reactant concentration increases

No change

Rise in temperature

Energy is absorbed by system

Shift in endothermic direction

Change

Drop in temperature

Energy is generated by system

Shift in exothermic direction

Change

Reactions Involving Gases Decrease in volume, increase in pressure

Pressure decreases

Composition changes to reduce total number of gas molecules

No change

Increase in volume, decrease in pressure

Pressure increases

Composition changes to increase total number of gas molecules

No change

*Does not apply when an insoluble solid reactant or product is added. Recall that solids do not appear in the reaction quotient.

Applying Chemical Principles 15.1 Applying Equilibrium Concepts—The Haber–Bosch Ammonia Process Nitrogen-containing substances are used around the world to stimulate the growth of field crops. Farmers have used animal waste for centuries as a natural fertilizer. In the nineteenth century, industrialized coun­ tries imported nitrogen-rich marine bird manure from Peru, Bolivia, and Chile, but the supply of this material was limited. In 1898, William Ramsay (the discoverer of the noble gases, “Applying Chemical Principles 2.3: Argon— An Amazing Discovery,” page  118) pointed out that the amount of fixed nitrogen in the world was being depleted and predicted that world food shortages would occur by the midtwentieth century as a result. That Ramsay’s prediction failed to materialize was due in part to the work of Fritz Haber (1868–1934). In about 1908, Haber developed a method to make ammonia directly from the elements,

H2, N2, and ammonia N2 and H2

Heat exchanger

A mixture of H2 and N2 is pumped over Catalyst a catalytic surface.

Uncombined N2 and H2 Recirculating pump

Heating coil

N2(g) + 3 H2(g) uv 2 NH3(g) and, a few years later, Carl Bosch (1874–1940) perfected the industrial scale synthesis. Ammonia is now made for pennies per kilogram and is consistently ranked in the top five chemicals produced in the United States, around 14 billion kg produced annually. In addition to its direct use as a fertilizer, it is also a starting material for making nitric acid and ammonium nitrate, among other things. The manufacture of ammonia is a good example of the role that kinetics and chemical equilibria play in practical chemistry. The N2 + H2 reaction is product-favored at equilibrium at

762

Cooling coil

Liquid ammonia

Unchanged reactants are recycled into the catalytic chamber.

The Haber–Bosch process for ammonia synthesis. The NH3 is collected as a liquid (at −33 °C).

25 °C [Kc (calc’d value) = 3.5 × 108]. In addition, the reaction is exothermic (∆rH ° = −92.2 kJ/mol-rxn). Unfortunately, the rate of production of ammonia at 25  °C is slow, so it is necessary to carry out the reaction at a higher temperature and use a catalyst to increase the reaction rate. An effective catalyst for the Haber–Bosch process is Fe3O4 mixed with KOH, SiO2, and Al2O3 (all inexpensive chemicals). Because the catalyst is not effective below 400 °C, the process is carried out at 450–500 °C. The problem, however,

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is that the equilibrium constant declines with tem­perature, as predicted by Le Chatelier’s principle. At 450  °C, Kc (experimental value) = 0.16. When carrying out the reaction at this high temperature, two things are done to increase ammonia yields. The first is to raise the pressure. This does not change the value of K, but an increase in pressure can be compensated by converting 4 mol of reactants to 2 mol of product, thereby increasing the percent conversion to NH3. The second is to remove the a­ mmonia as it  is produced. Both increasing the pressure and removing ­ammonia take advantage of Le Chatelier’s principle to produce ammonia under reactant-favored conditions.

(b) Urea is formed in the reaction of ammonia and CO2. 2 NH3(g) + CO2(g) uv (NH2)2CO(s) + H2O(g) Would high pressure favor urea production? Would high temperature? (∆f H° for solid urea = −333.1 kJ/mol-rxn). Explain your answers. 2. Hydrogen is used in the Haber–Bosch process, and this is made from natural gas in a process called steam reforming.

CH4(g) + H2O(g) n CO(g) + 3 H2(g) CO(g) + H2O(g) n CO2(g) + H2(g) (a) Are the two reactions above endo- or exothermic? (b) To obtain the H2 necessary to manufacture 15 billion kg of NH3, what mass of CH4 is required, and what mass of CO2 is produced as a by-product (assuming complete conversion of the CH4)?

Questions

1. Anhydrous ammonia is used directly as a fertilizer, but much of it is also converted to other fertilizers such as ammonium nitrate and urea. (a) Write a balanced chemical equation for the reaction of ­ammonia with nitric acid to produce ammonium nitrate.

15.2 Trivalent Carbon The octet rule is a guiding principle in organic chemistry. As a result, when someone discovers a molecule that does not obey the octet rule, organic chemists are always interested. The synthesis of the triphenylmethyl radical, (C6H5)3C, was one such event. The triphenylmethyl radical (compound 2), the first known persistent organic free radical, was discovered over 100 years ago by Moses Gomberg, a chemist at the University of Michigan. Gomberg set out to make hexaphenylethane [(C6H5)3C–C(C6H5)3] (compound 3). When he combined the reactants (C6H5)3CCl (compound 1) and Zn, he obtained a yellow solution that became more intensely yellow when heated and was reactive toward oxygen and halogens. This extreme reactivity led Gomberg to conclude that the yellow color was due to the presence of (C6H5)3C in solution. The existence of the stable free radical was explained by the fact that there are three large phenyl groups around a carbon atom, which prevent the radical from undergoing the expected coupling, or dimerization, to hexaphenylethane. Actually, the triphenylmethyl radical does couple (or dimerize), but not the way Gomberg had expected. A diamagnetic, white crystalline solid (compound 4) can be isolated from the solutions containing the radical. The coupling occurs between a methyl carbon radical on one molecule and the phenyl ring on a second. When redissolved in benzene, the original yellow solution forms. Studies determined that compounds 4 and 2 exist in solution in equilibrium. The value of Kc for the dimer–monomer equilibrium (4 uv 2) in benzene is 4.1 × 10−4 at 20 °C. Interestingly, Gomberg finished his initial publication with the following statement: “This work will be continued and

1

C

2

Cl + Zn

3

C•

C

C

4

C

C H

I  wish to reserve the field for myself.” In the United States, chemical research is competitive and Gomberg’s wish was not respected.

Questions

1. Freezing point depression (see Section 13.4) is one way to determine the molar mass of a compound. The freezing point depression constant of benzene is –5.12 °C/m. (a) When a 0.503 g sample of the white crystalline dimer is dissolved in 10.0 g benzene, the freezing point of benzene is decreased by 0.542 °C. Verify that the molar mass of the dimer is 475 g/mol when determined by freezing point depression. Assume that no dissociation of the dimer occurs.

Applying Chemical Principles

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763

(b) The correct molar mass of the dimer is 487 g/mol. Explain why the dissociation equilibrium causes the freezing point depression calculation to yield a lower molar mass for the dimer. 2. What concentration of the monomer (2) will exist at equilibrium with 0.015 mol/L of the dimer (4) in benzene at 20 °C? 3. A 0.64  g sample of the white crystalline dimer (4) is dissolved in 25.0 mL of benzene at 20 °C. Use the equilibrium constant to calculate the concentrations of monomer (2) and dimer (4) in this solution. 4. Predict whether the dissociation of the dimer to the monomer is exothermic or endothermic, based on the fact that at higher temperatures the yellow color of the solution intensifies.

5. Which of the organic species mentioned in this story is paramagnetic? (a) triphenylmethyl chloride (b) triphenylmethyl radical (c) the triphenylmethyl dimer Reference

M. Gomberg, Journal of the American Chemical Society, 1900, 22, 757–771.

Think–Pair–Share 1. Ionic compounds are often described as either soluble or insoluble in water. Many ionic compounds that are considered insoluble are actually sparingly soluble, and their solubility can be determined using equilibrium constants. An example is the dissolution of AgCl in water, described by the reaction:

3. Water is produced from the combustion of hydrogen gas. The reaction is so product-favored that chemists generally do not write the reaction as an equilibrium expression. However, the equilibrium constants, Kp, at 25 °C to produce both liquid water and gaseous water are known.

AgCl(s) uv Ag+(aq) + Cl−(aq)  K = 1.8 × 10−10 at 25 °C

H2(g) + 1⁄2 O2(g) uv H2O()  Kp = 3.54 × 1041 H2(g) + 1⁄2 O2(g) uv H2O(g)  Kp = 1.12 × 1040

If 5.5 mg AgCl (3.8 × 10−5 mol AgCl) is placed in a flask and diluted to a volume of 1.0 L, will all the AgCl dissolve? Your explanation should use the reaction quotient, Q. 2. The decomposition of solid ammonium chloride is an endothermic reaction.

(a) Manipulate the two chemical expressions so that they add together to give the equilibrium expression for the conversion of liquid water to its vapor: H2O() uv H2O(g)

NH4Cl(s) uv NH3(g) + HCl(g) Predict the effect on the equilibrium for each change listed below. In each case, explain your response. (a) NH4Cl(s) added. (b) NH3 is removed. (c) HCl is added. (d) The reactants and products are transferred to a flask with a larger volume. (e) The temperature is decreased.

and then calculate the equilibrium constant (K p) for this reaction. (b) Write the equilibrium constant expression (Kp) for the conversion of liquid water to gaseous water at 25 °C. (c) Is the conversion of liquid water to gaseous water an endothermic or exothermic process? (d) If an equilibrium system of liquid water and its vapor is heated, does the equilibrium constant change? If so, does Kp get larger or smaller? Explain.

Chapter Goals Revisited The goals for this chapter are keyed to specific Study Questions to help you organize your review:

15.1  Chemical Equilibrium: A Review • Understand that chemical reactions are reversible and that chemical equilibria are dynamic. 77, 78.

15.2  The Equilibrium Constant and Reaction Quotient • Write the equilibrium constant expression, K, for a chemical reaction. 1–4. • Recognize that the equilibrium constant can be written relating K to

concentrations (Kc) or, if the equilibrating species are gases, to partial pressures (Kp). Convert between Kc and Kp values. 29, 30.

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• Predict whether a reaction is reactant-favored or product-favored at equilibrium based on the value of K. 75.

• Write the reaction quotient expression, Q, for a chemical reaction, and use

Q to decide whether a reaction is at equilibrium (Q = K) or if there will be a net conversion of reactants to products (Q < K) or products to reactants (Q > K) to attain equilibrium. 5–8, 41, 42.

15.3  Determining an Equilibrium Constant • Calculate an equilibrium constant given the reactant and product concentrations at equilibrium. 9–11.

• Calculate an equilibrium constant given the initial concentrations of the

reactants and products and the concentration of one reactant or product at equilibrium. 13, 14, 35, 39, 40, 49, 50.

15.4  Using Equilibrium Constants in Calculations • Use equilibrium constants to calculate the concentration (or pressure) of a reactant or product at equilibrium. 15–19, 48, 53, 54, 56, 63.

15.5 More about Balanced Equations and Equilibrium Constants • Know how K changes as different stoichiometric coefficients are used in a balanced equation, if the equation is reversed, or if several equations are added to give a new net equation. 21–28, 36, 37, 43.

15.6  Disturbing a Chemical Equilibrium • Predict how a system at equilibrium will respond if the reaction conditions are changed. 31–34, 47, 60, 70, 71.

Key Equations Equation 15.1 (page 740) The equilibrium constant expression. At equilibrium, the ratio of products to reactants (each raised to the power corresponding to its stoichiometric coefficient) has a constant value, K (at a particular temperature). For the general reaction a A + b B uv c C + d D, Equilibrium constant  K 

[C]c [D]d [A]a[B]b

Equation 15.2 (page 743) For the general reaction a A + b B uv c C + d D, the ratio of product to reactant concentrations at any point in the reaction is the reaction quotient, Q. Reaction quotient  Q 

[C]c [D]d [A]a[B]b

Equation 15.3 (page 752) This approximation is used to solve for the equilibrium concentrations of B and C (= x) in the general reaction A uv B + C when the value of 100 × K is less than the original concentration of A(= [A]0). Kc 



[B][C] ( x)( x)  [A]0 [A] Key Equations

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765

Study Questions ▲ denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully worked solutions in the Student Solutions Manual.

Practicing Skills Writing Equilibrium Constant Expressions (See Section 15.2 and Example 15.1.) 1. Write equilibrium constant expressions for the ­following reactions. For gases, use either partial pressures or concentrations. (a) 2 H2O2(g) uv 2 H2O(g) + O2(g) (b) CO(g) + 1⁄2 O2(g) uv CO2(g) (c) C(s) + CO2(g) uv 2 CO(g) (d) NiO(s) + CO(g) uv Ni(s) + CO2(g) 2. Write equilibrium constant expressions for the ­following reactions. For gases, use either partial pressures or concentrations. (a) 3 O2(g) uv 2 O3(g) (b) Fe(s) + 5 CO(g) uv Fe(CO)5(g) (c) (NH4)2CO3(s) uv  2 NH3(g) + CO2(g) + H2O(g) (d) Ag2SO4(s) uv 2 Ag+(aq) + SO42−(aq) 3. Write balanced chemical equations that correspond to the following equilibrium constant expressions. [SO3]2 (c) K  [Pb2][Cl]2 (a) K  2 [SO2] [O2]

(b) K 

[CO2]2 [CO]2 [O2]

[F][H3O] (d) K  [HF]

4. Write balanced chemical equations that correspond to the following equilibrium constant expressions. [CO]2 [O2] (c) K  [Mg 2][OH]2 (a) K  2 [CO2]

(b) K 

[CH4 ][CCl 4 ] [CH2Cl 2]2

(d) K 

[HF][OH] [F]

The Equilibrium Constant and Reaction Quotient (See Section 15.2 and Example 15.2.) 5. Kc = 5.6 × 10−12 at 500 K for the dissociation of iodine molecules to iodine atoms. I2(g) uv 2 I(g)

A mixture has [I2] = 0.020 mol/L and [I] = 2.0 × 10−8 mol/L. Is the reaction at equilibrium (at 500 K)? If not, which way must the reaction proceed to reach equilibrium? 6. The reaction 2 NO2(g) uv N2O4(g)

766

has an equilibrium constant, Kc, of 170 at 25 °C. If 2.0 × 10−3 mol of NO2 is present in a 10.-L flask along with 1.5 × 10−3 mol of N2O4, is the system at equilibrium? If it is not at equilibrium, does the concentration of NO2 increase or decrease as the system proceeds to equilibrium? 7. A mixture of SO2, O2, and SO3 at 1000 K contains the gases at the following concentrations: [SO2] = 5.0 × 10−3 mol/L, [O2] = 1.9 × 10−3 mol/L, and [SO3] = 6.9 × 10−3 mol/L. Is the reaction at equilibrium? If not, which way will the reaction proceed to reach equilibrium? 2 SO2(g) + O2(g) uv 2 SO3(g) Kc = 279

8. The equilibrium constant, Kc, for the reaction 2 NOCl(g) uv 2 NO(g) + Cl2(g)

is 1.6 × 10−3 at 300 °C. A mixture contains the gases at the following concentrations: [NOCl] = 5.0 × 10−3 mol/L, [NO] = 2.5 × 10−3 mol/L, and [Cl2] = 2.0 × 10−3 mol/L. Is the reaction at equilibrium at 300 °C? If not, in which direction does the reaction proceed to come to equilibrium?

Calculating an Equilibrium Constant (See Section 15.3 and Example 15.3.) 9. The reaction PCl5(g) uv PCl3(g) + Cl2(g)

was examined at 250 °C. At equilibrium, [PCl5] = 4.2 × 10−5 mol/L, [PCl3] = 1.3 × 10−2 mol/L, and [Cl2] = 3.9 × 10−3 mol/L. Calculate Kc for the reaction. 10. An equilibrium mixture of SO2, O2, and SO3 at a high temperature contains the gases at the following concentrations: [SO2] = 3.77 × 10−3 mol/L, [O2] = 4.30 × 10−3 mol/L, and [SO3] = 4.13 × 10−3 mol/L. Calculate the equilibrium constant, Kc, for the reaction. 2 SO2(g) + O2(g) uv 2 SO3(g)

11. The reaction C(s) + CO2(g) uv 2 CO(g)

occurs at high temperatures. At 700 °C, a 200.0-L tank contains 1.0 mol of CO, 0.20 mol of CO2, and 0.40 mol of C at equilibrium. (a) Calculate Kc for the reaction at 700 °C. (b) Calculate Kc for the reaction, also at 700 °C, if the amounts at equilibrium in the 200.0-L tank

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are 1.0 mol of CO, 0.20 mol of CO2, and 0.80 mol of C. (c) Compare the results of (a) and (b). Does the quantity of carbon affect the value of Kc? Explain. 12. Hydrogen and carbon dioxide react at a high temperature to give water and carbon monoxide. H2(g) + CO2(g) uv H2O(g) + CO(g)

(a) Laboratory measurements at 986 °C show that there are 0.11 mol each of CO and H2O vapor and 0.087 mol each of H2 and CO2 at equilibrium in a 50.0-L container. Calculate the equilibrium constant for the reaction at 986 °C. (b) Suppose 0.024 mol each of H2 and CO2 are placed in a 300.0-L container. When equilibrium is achieved at 986 °C, what amounts of CO(g) and H2O(g), in moles, would be present? [Use the value of Kc from part (a).] 13. A mixture of CO and Cl2 is placed in a reaction flask: [CO] = 0.0102 mol/L and [Cl2] = 0.00609 mol/L. When the reaction

cyclohexane

The equilibrium constant is estimated to be 0.12 at 25 °C. If you originally place 0.058 mol of cyclohexane in a 2.8-L flask, what are the concentrations of cyclohexane and methylcyclopentane when equilibrium is established? 17. The equilibrium constant for the dissociation of iodine molecules to iodine atoms I2(g) uv 2 I(g)

is 3.76 × 10−3 at 1000 K. Suppose 0.105 mol of I2 is placed in a 12.3-L flask at 1000 K. What are the concentrations of I2 and I when the system comes to equilibrium? 18. The equilibrium constant, Kc, for the reaction

CO(g) + Cl2(g) uv COCl2(g)

has come to equilibrium at 600 K, [Cl2] = 0.00301 mol/L. (a) Calculate the concentrations of CO and COCl2 at equilibrium. (b) Calculate Kc. 14. You place 0.0300 mol of pure SO3 in an 8.00-L flask at 1150 K. At equilibrium, 0.0058 mol of O2 was formed. Calculate Kc for the reaction at 1150 K. 2 SO3(g) uv 2 SO2(g) + O2(g)

methylcyclopentane

N2O4(g) uv 2 NO2(g)

at 25 °C is 5.9 × 10−3. Suppose 14.8 g of N2O4 is placed in a 5.0-L flask at 25 °C. Calculate the following: (a) the amount of NO2 (mol) present at equilibrium; (b) the percentage of the original N2O4 that is dissociated. 19. Iodine dissolves in water, but its solubility in a nonpolar solvent such as CCl4 is greater.

Using Equilibrium Constants 15. The value of Kc for the interconversion of butane and isobutane is 2.5 at 25 °C.

Nonpolar I2 Polar H2O

Nonpolar CCl4

butane

isobutane

If you place 0.017 mol of butane in a 0.50-L flask at 25 °C and allow equilibrium to establish, what will be the equilibrium concentrations of the two forms of butane? 16. Cyclohexane, C6H12, a hydrocarbon, can isomerize or change into methylcyclopentane, a compound of the same formula (C5H9CH3) but with a different molecular structure.



Polar H2O Shake the test tube Photos: © Charles D. Winters/Cengage

(See Section 15.4 and Examples 15.4 and 15.5.)

Nonpolar CCl4 and I2

Extracting iodine (I2) from water with the nonpolar solvent CCl4.  I2 is more soluble in CCl4 and, after shaking a mixture of water and CCl4, the I2 has accumulated in the more dense CCl4 layer.

The equilibrium constant is 85.0 for the process I2(aq) uv I2(CCl4)

You place 0.0340 g of I2 in 100.0 mL of water. After shaking it with 10.0 mL of CCl4, how much I2 remains in the water layer?

Study Questions

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20. Carbonyl bromide decomposes to carbon monoxide and bromine. COBr2(g) uv CO(g) + Br2(g)

Kc is 0.190 at 73 °C. If you place 0.0500 mol of COBr2 in a 2.00-L flask and heat it to 73 °C, what are the equilibrium concentrations of COBr2, CO, and Br2? What percentage of the original COBr2 decomposed at this temperature?

Manipulating Equilibrium Constant Expressions (See Section 15.5 and Example 15.6.) 21. Which of the following correctly relates the equilibrium constants for the two reactions shown? A + B uv 2 C

K1

2 A + 2 B uv 4 C

K2

(a) K2 = 2K1 (b) K2 = K12

(c) K2 = 1/K1 (d) K2 = 1/K12

22. Which of the following correctly relates the equilibrium constants for the two reactions shown? 2A + B uv 2 C

K1

A + 1⁄2 B uv C

K2

(a) K2 = 2K1 (b) K2 = (1⁄2)K1

(c) K2 = K11⁄2 (d) K2 = 1/K11⁄2

23. Which of the following correctly relates the equilibrium constants for the two reactions shown? A + 2B uv C

K1

2C uv 2A + 4B

K2

(a) K2 = −2K1 (b) K2 = 1/K1

(c) K2 = −K12 (d) K2 = 1/K12

24. Which of the following correctly relates the equilibrium constants for the two reactions shown? A + B uv 2 C

K1

C uv 1⁄2 A + 1⁄2 B

K2

(a) K2 = 1/(K1)1⁄2 (b) K2 = 1/K1

(c) K2 = K12 (d) K2 = −K11⁄2

25. Consider the following equilibria involving SO2(g) and their corresponding equilibrium constants. SO2(g) + 1⁄2 O2(g) uv SO3(g)

K1

2 SO3(g) uv 2 SO2(g) + O2(g)

K2

Which of the following expressions relates K1 to K2? (a) K2 = K12 (d) K2 = 1/K1 2 (b) K2  = K1 (e) K2 = 1/K12 (c) K2 = K1

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26. The equilibrium constant K for the reaction CO2(g) uv CO(g) + 1⁄2 O2(g)

is 6.66 × 10−12 at 1000 K. Calculate K for the reaction 2 CO(g) + O2(g) uv 2 CO2(g)

27. Calculate K for the reaction SnO2(s) + 2 CO(g) uv Sn(s) + 2 CO2(g)

given the following information: SnO2(s) + 2 H2(g) uv Sn(s) + 2 H2O(g)

K = 8.12

H2(g) + CO2(g) uv H2O(g) + CO(g)

K = 0.771

28. Calculate K for the reaction FeO(s) + H2(g) uv Fe(s) + H2O(g)

given the following information: H2O(g) + CO(g) uv H2(g) + CO2(g)

K = 1.6

FeO(s) + CO(g) uv Fe(s) + CO2(g)

K = 0.67

29. Relationship of Kc and Kp: (a) Kp for the following reaction is 0.16 at 25 °C. What is the value of Kc? 2 NOBr(g) uv 2 NO(g) + Br2(g)

(b) The equilibrium constant, Kc, for the following reaction is 1.05 at 350 K. What is the value of Kp? 2 CH2Cl2(g) uv CH4(g) + CCl4(g)

30. Relationship of Kc and Kp: (a) The equilibrium constant, Kc, for the following reaction at 25 °C is 170. What is the value of Kp? N2O4(g) uv 2 NO2(g)

(b) Kc for the decomposition of ammonium hydrogen sulfide is 1.8 × 10−4 at 25 °C. What is the value of Kp? NH4HS(s) uv NH3(g) + H2S(g)

Disturbing a Chemical Equilibrium (See Section 15.6 and Example 15.7.) 31. Dinitrogen trioxide decomposes to NO and NO2 in an endothermic process (ΔrH° = 40.5 kJ/mol-rxn). N2O3(g) uv NO(g) + NO2(g)

Predict the effect of the following changes on the position of the equilibrium; that is, state which way the equilibrium will shift (left, right, or no change) when each of the following changes is made. (a) adding more N2O3(g) (b) adding more NO2(g) (c) increasing the volume of the reaction flask (d) lowering the temperature

Chapter 15 / Principles of Chemical Reactivity: Equilibria

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32. Kp for the following reaction is 0.16 at 25 °C: 2 NOBr(g) uv 2 NO(g) + Br2(g)

The enthalpy change for the reaction at standard conditions is +16.3 kJ/mol-rxn. Predict the effect of the following changes on the position of the equilibrium; that is, state which way the equilibrium will shift (left, right, or no change) when each of the following changes is made. (a) removing some Br2(g) (b) adding more NOBr(g) (c) increasing the temperature (d) decreasing the container volume 33. Consider the isomerization of butane with an ­equilibrium constant of K = 2.5. (See Study ­Question 15.) The system is originally at ­equilibrium with [butane] = 1.0 M and [­isobutane] = 2.5 M. (a) If 0.50 mol/L of isobutane is suddenly added and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas? (b) If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas? 34. The decomposition of NH4HS NH4HS(s) uv NH3(g) + H2S(g)

is an endothermic process. Using Le Chatelier’s principle, explain how increasing the temperature would affect the equilibrium. If more NH4HS is added to a flask in which this equilibrium exists, how is the equilibrium affected? What if some additional NH3 is placed in the flask? What will happen to the pressure of NH3 if some H2S is removed from the flask?

General Questions These questions are not designated as to type or ­location in the chapter. They may combine several concepts. 35. Suppose 0.086 mol of Br2 is placed in a 1.26-L flask and heated to 1756 K, a temperature at which the halogen dissociates to atoms. Br2(g) uv 2 Br(g)

If Br2 is 3.7% dissociated at this temperature, calculate Kc. 36. The equilibrium constant for the reaction N2(g) + O2(g) uv 2 NO(g)

is 1.7 × 10−3 at 2300 K.

(a) What is K for the reaction when written as follows? ⁄2 N2(g) + 1⁄2 O2(g) uv NO(g)

1

(b) What is K for the following reaction? 4 NO(g) uv 2 N2(g) + 2 O2(g)

37. Kp for the formation of phosgene, COCl2, is 6.5 × 1011 at 25 °C. CO(g) + Cl2(g) uv COCl2(g)

What is the value of Kp for the dissociation of phosgene? COCl2(g) uv CO(g) + Cl2(g)

38. The equilibrium constant, Kc, for the following reaction is 1.05 at 350 K. 2 CH2Cl2(g) uv CH4(g) + CCl4(g)

If an equilibrium mixture of the three gases at 350 K contains 0.0206 M CH2Cl2(g) and 0.0163 M CH4, what is the equilibrium concentration of CCl4? 39. Carbon tetrachloride can be produced by the following reaction: CS2(g) + 3 Cl2(g) uv S2Cl2(g) + CCl4(g)

Suppose 0.12 mol of CS2 and 0.36 mol of Cl2 are placed in a 10.0-L flask. After equilibrium has been achieved, the mixture contains 0.090 mol CCl4. Calculate Kc. 40. Equal numbers of moles of H2 gas and I2 vapor are mixed in a flask and heated to 700 °C. The initial concentration of each gas is 0.0088 mol/L, and 78.6% of the I2 is consumed when ­equilibrium is achieved according to the equation H2(g) + I2(g) uv 2 HI(g)

Calculate Kc for this reaction. 41. The equilibrium constant for the butane uv ­isobutane isomerization reaction is 2.5 at 25 °C. If 1.75 mol of butane and 1.25 mol of isobutane are mixed, is the system at equilibrium? If not, when it proceeds to equilibrium, which reagent increases in concentration? Calculate the ­concentrations of the two compounds when the system reaches equilibrium. 42. At 2300 K the equilibrium constant for the formation of NO(g) is 1.7 × 10−3. N2(g) + O2(g) uv 2 NO(g)

(a) Analysis shows that the concentrations of N2 and O2 are both 0.25 M and that of NO is 0.056 M under certain conditions. Is the system at equilibrium? Study Questions

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(b) If the system is not at equilibrium, in which direction does the reaction proceed? (c) When the system is at equilibrium, what are the equilibrium concentrations? 43. Which of the following correctly relates the two equilibrium constants for the two reactions shown? NOCl(g) uv NO(g) + 1⁄2 Cl2(g)

K1

2 NO(g) + Cl2(g) uv 2 NOCl(g)

(a) K2 = −K1 (b) K2 = 1/(K1)1⁄2 2

K2

1/K12

(c) K2 = (d) K2 = 2K1

44. Consider the following equilibrium: COBr2(g) uv CO(g) + Br2(g)

Kc = 0.190 at 73 °C

(a) A 0.50 mol sample of COBr2 is transferred to a 9.50-L flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species. (b) The volume of the container is decreased to 4.5 L and the system allowed to return to equilibrium. Calculate the new equilibrium concentrations. (Hint: The calculation will be easier if you view this as a new problem with 0.50 mol of COBr2 transferred to a 4.5-L flask.) (c) What is the effect of decreasing the container volume from 9.50 L to 4.50 L? 45. Heating a metal carbonate leads to decomposition. BaCO3(s) uv BaO(s) + CO2(g)

Predict the effect on the equilibrium of each change listed below. Answer by choosing (i) no change, (ii) shifts left, or (iii) shifts right. (a) add BaCO3 (c) add BaO (b) add CO2 (d) raise the temperature (e) increase the volume of the flask containing the reaction 46. Carbonyl bromide decomposes to carbon monoxide and bromine. COBr2(g) uv CO(g) + Br2(g)

Kc is 0.190 at 73 °C. Suppose you place 0.500 mol of COBr2 in a 2.00-L flask and heat it to 73 °C. After equilibrium has been achieved, you add an additional 2.00 mol of CO. (a) How is the equilibrium mixture affected by adding more CO? (b) When equilibrium is reestablished, what are the new equilibrium concentrations of COBr2, CO, and Br2? (c) How has the addition of CO affected the percentage of COBr2 that decomposed?

770

47. Phosphorus pentachloride decomposes at elevated temperatures. PCl5(g) uv PCl3(g) + Cl2(g)

An equilibrium mixture at some temperature consists of 3.120 g of PCl5, 3.845 g of PCl3, and 1.787 g of Cl2 in a 10.0-L flask. If you add 1.418 g of Cl2, how will the equilibrium be affected? What will the concentrations of PCl5, PCl3, and Cl2 be when equilibrium is reestablished? 48. Ammonium hydrogen sulfide decomposes on heating. NH4HS(s) uv NH3(g) + H2S(g)

If Kp for this reaction is 0.11 at 25 °C (when the partial pressures are measured in atmospheres), what is the total pressure in the flask at equilibrium? 49. Ammonium iodide dissociates reversibly to ammonia and hydrogen iodide if the salt is heated to a sufficiently high temperature. NH4I(s) uv NH3(g) + HI(g)

Some ammonium iodide is placed in a flask, which is then heated to 400 °C. If the total pressure in the flask when equilibrium has been achieved is 705 mm Hg, what is the value of Kp (when partial pressures are in atmospheres)? 50. When solid ammonium carbamate sublimes, it dissociates completely into ammonia and carbon dioxide according to the following equation: (NH4)(H2NCO2)(s) uv 2 NH3(g) + CO2(g)

At 25 °C, experiment shows that the total pressure of the gases in equilibrium with the solid is 0.116 atm. What is the equilibrium constant, Kp? 51. The equilibrium reaction N2O4(g) uv 2 NO2(g) has been thoroughly studied (Figure 15.7). (a) If the total pressure in a flask containing NO2 and N2O4 gas at 25 °C is 1.50 atm and the value of Kp at this temperature is 0.148, what fraction of the N2O4 has dissociated to NO2? (b) What happens to the fraction dissociated if the volume of the container is increased so that the total equilibrium pressure falls to 1.00 atm? 52. In the gas phase, acetic acid exists as an equilibrium of monomer and dimer molecules. (The dimer consists of two molecules linked through hydrogen bonds.)

Chapter 15 / Principles of Chemical Reactivity: Equilibria

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If 0.050 mol of COCl2 is placed in a 12.5-L flask, what is the total pressure at equilibrium at 600 K? 57. ▲ A 15-L flask at 300 K contains 6.44 g of a mixture of NO2 and N2O4 in equilibrium. What is the total pressure in the flask? (Kp for 2 NO2 (g) uv N2O4(g) is 7.1.) The equilibrium constant, Kc, at 25 °C for the monomer–dimer equilibrium 2 CH3CO2H uv (CH3CO2H)2

was determined to be 3.2 × 104. Assume that acetic acid initially has a concentration of 5.4 × 10−4 mol/L at 25 °C and that no dimer is present initially. (a) What percentage of the acetic acid is converted to dimer? (b) As the temperature increases, in which direction does the equilibrium shift? (Recall that hydrogen-bond formation is an exothermic process.) 53. Assume 3.60 mol of ammonia is placed in a 2.00-L vessel and allowed to decompose to the elements at 723 K. 2 NH3(g) uv N2(g) + 3 H2(g)

If the experimental value of Kc is 6.3 for this reaction at the temperature in the reactor, calculate the equilibrium concentration of each reagent. What is the total pressure in the flask? 54. The total pressure for a mixture of N2O4 and NO2 is 0.36 atm. If Kp = 7.1 (at 25 °C), calculate the partial pressure of each gas in the mixture. 2 NO2(g) uv N2O4(g)

55. Kc for the decomposition of ammonium hydrogen sulfide is 1.8 × 10−4 at 25 °C. NH4HS(s) uv NH3(g) + H2S(g)

(a) When the pure salt decomposes in a flask, what are the equilibrium concentrations of NH3 and H2S? (b) If NH4HS is placed in a flask already containing 0.020 mol/L of NH3 and then the system is allowed to come to equilibrium, what are the equilibrium concentrations of NH3 and H2S? 56. ▲ COCl2 decomposes to CO and Cl2 at high temperatures. Kc at 600 K for the reaction is 0.0071.

58. ▲ Lanthanum oxalate decomposes when heated to lanthanum(III) oxide, CO, and CO2. La2(C2O4)3(s) uv La2O3(s) + 3 CO(g) + 3 CO2(g)

(a) If, at equilibrium and 373 K, the total pressure in a 10.0-L flask is 0.200 atm, what is the value of Kp? (b) Suppose 0.100 mol of La2(C2O4)3 was originally placed in the 10.0-L flask. What quantity of La2(C2O4)3 remains unreacted at equilibrium at 373 K? 59. ▲ The reaction of hydrogen and iodine to give hydrogen iodide has an equilibrium constant, Kc, of 56 at 435 °C. (a) What is the value of Kp? (b) Suppose you mix 0.045 mol of H2 and 0.045 mol of I2 in a 10.0-L flask at 425 °C. What is the total pressure of the mixture before and after equilibrium is achieved? (c) What is the partial pressure of each gas at equilibrium? 60. Sulfuryl chloride, SO2Cl2, is used as a reagent in the synthesis of organic compounds. When heated to a sufficiently high temperature, it decomposes to SO2 and Cl2. SO2Cl2(g) uv SO2(g) + Cl2(g)

Kc = 0.045 at 375 °C

(a) A 12.0-L flask containing 7.80 g of SO2Cl2 is heated to 375 °C. What is the concentration of each compound in the system when equilibrium is achieved? What fraction of SO2Cl2 has dissociated? (b) What are the concentrations of SO2Cl2, SO2, and Cl2 at equilibrium in the 12.0-L flask at 375 °C if you begin with a mixture of SO2Cl2 (7.80 g) and Cl2 (0.10 atm)? What fraction of SO2Cl2 has dissociated? (c) Compare the fractions of SO2Cl2 in parts (a) and (b). Do they agree with your expectations based on Le Chatelier’s principle?

COCl2(g) uv CO(g) + Cl2(g)



Study Questions

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771

61. ▲ Hemoglobin (Hb) can form a complex with both O2 and CO. For the reaction HbO2(aq) + CO(g) uv HbCO(aq) + O2(g)

at body temperature, K is about 200. If the ratio [HbCO]/[HbO2] comes close to 1, death is probable. What partial pressure of CO in the air is likely to be fatal? Assume the partial pressure of O2 is 0.20 atm. 62. ▲ Limestone decomposes at high temperatures. CaCO3(s) uv CaO(s) + CO2(g)

At 1000 °C, Kp = 3.87. If pure CaCO3 is placed in a 5.00-L flask and heated to 1000 °C, what quantity of CaCO3 must decompose to achieve the equilibrium pressure of CO2? 63. At 1800 K, oxygen dissociates very slightly into its atoms. O2(g) uv 2 O(g)   Kp = 1.2 × 10−10

If you place 0.050 mol of O2 in a 10.-L vessel and heat it to 1800 K, how many O atoms are present in the flask? 64. ▲ Nitrosyl bromide, NOBr, dissociates readily at room temperature. NOBr(g) uv NO(g) + 1⁄2 Br2(g)

Some NOBr is placed in a flask at 25 °C and allowed to dissociate. The total pressure at equilibrium is 190 mm Hg and the compound is 34% dissociated. What is the value of Kp?

67. ▲ A sample of N2O4 gas with a pressure of 1.00 atm is placed in a flask. When equilibrium is achieved, 20.0% of the N2O4 was converted to NO2 gas. (a) Calculate Kp. (b) If the original pressure of N2O4 is 0.10 atm, what is the percent dissociation of the gas? Is the result in agreement with Le Chatelier’s principle? 68. ▲ A reaction important in smog formation is O3(g) + NO(g) uv O2(g) + NO2(g)

Kc = 6.0 × 1034

(a) If the initial concentrations are [O3] = 1.0 × 10−6 M, [NO] = 1.0 × 10−5 M, [NO2] = 2.5 × 10−4 M, and [O2] = 8.2 × 10−3 M, is the system at equilibrium? If not, in which direction does the reaction proceed? (b) If the temperature is increased, as on a very warm day, will the concentrations of the products increase or decrease? (Hint: Calculate the enthalpy change for the reaction to find out if it is exothermic or endothermic.)

In the Laboratory 69. ▲ The ammonia complex of trimethylborane, (NH3)B(CH3)3, dissociates at 100 °C to its components with Kp = 4.62 (when the pressures are in atmospheres). (NH3)B(CH3)3(g)

B(CH3)3(g)

+

NH3(g)

65. ▲ Boric acid and glycerin form a complex B(OH)3(aq) + glycerin(aq) uv B(OH)3 ∙ glycerin(aq)

with an equilibrium constant of 0.90. If the concentration of boric acid is 0.10 M, how much glycerin should be added, per liter, so that 60.% of the boric acid is in the form of the complex? 66. ▲ The dissociation of calcium carbonate has an equilibrium constant of Kp = 1.16 at 800 °C. CaCO3(s) uv CaO(s) + CO2(g)

(a) What is Kc for the reaction? (b) If you place 22.5 g of CaCO3 in a 9.56-L container at 800 °C, what is the pressure of CO2 in the container? (c) What percentage of the original 22.5-g sample of CaCO3 remains undecomposed at equilibrium?

772

+

If NH3 is changed to some other molecule, the equilibrium constant is different. For [(CH3)3P]B(CH3)3

Kp = 0.128

For [(CH3)3N]B(CH3)3

Kp = 0.472

(a) If you place 0.010 mol of each complex in separate flasks of equal volume, which flask has the largest partial pressure of B(CH3)3 at 100 °C? (b) If 0.73 g (0.010 mol) of (NH3)B(CH3)3 is placed in a 1.25-L flask and heated to 100 °C, what is the partial pressure of each gas in the equilibrium mixture, and what is the total pressure? What is the percent dissociation of (NH3)B(CH3)3?

Chapter 15 / Principles of Chemical Reactivity: Equilibria

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2 CrO42−(aq)  +  2 H3O+(aq) uv Cr2O72−(aq)  +  3 H2O(ℓ)

Photos: © Charles D. Winters/Cengage

70. The following photograph shows what occurs when a solution of potassium chromate is treated with a few drops of concentrated hydrochloric acid. Some of the bright yellow chromate ion is converted to the orange dichromate ion.

Before

Add HCl

Photos: © Charles D. Winters/Cengage

(a) Adding KSCN to iron(III) nitrate

(a) Explain this experimental observation in terms of Le Chatelier’s principle. (b) What would you observe if you treated the orange solution with sodium hydroxide? Explain. 71. Figure (a) shows what occurs when a solution of iron(III) nitrate is treated with a few drops of aqueous potassium thiocyanate. The nearly ­colorless iron(III) ion is converted to a red [Fe(H2O)5SCN]2+ ion. (This is a classic test for the presence of iron(III) ions in solution.) [Fe(H2O)6]3+(aq) + SCN−(aq) uv [Fe(H2O)5SCN]2+(aq) + H2O(ℓ)

Photos: © Charles D. Winters/Cengage

[Ni(H2O)6]2+

Add ammonia NH3

(b) Adding Ag+

(a) As more KSCN is added to the solution (Figure a), the color becomes even more red. Explain this observation. (b) Silver ions form a white precipitate with SCN− ions. Upon adding a few drops of aqueous silver nitrate to a red solution of [Fe(H2O)5SCN]+ ions, the red color become less intense (Figure b). Explain this observation. 72. ▲ The photographs at the bottom of the page show what occurs when you add ammonia to aqueous nickel(II) nitrate and then add ethylenediamine (NH2CH2CH2NH2) to the intermediate bluepurple solution. [Ni(H2O)6]2+(aq) + 6 NH3(aq) green

uv [Ni(NH3)6]2+(aq) + 6 H2O(ℓ)

K1

blue-purple

[Ni(NH3)6]2+(aq) + 3 NH2CH2CH2 NH2(aq) blue-purple

uv [Ni(NH2CH2CH2NH2)3]2+(aq) + 6 NH3(aq)

K2

violet

(a) Write a chemical equation for the formation of [Ni(NH2CH2CH2NH2)3]2+ from [Ni(H2O)6]2+ and ethylenediamine, and relate the value of K for this reaction to K1 and K2. (b) Which species, [Ni(NH2CH2CH2NH2)3]2+, [Ni(NH3)6]2+, or [Ni(H2O)6]2+ is the most stable? Explain.

[Ni(NH3)6]2+

Add ethylenediamine NH2CH2CH2NH2

[Ni(NH2CH2CH2NH2)3]2+

Aqueous nickel(II) nitrate



Study Questions

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773

73. Decide whether each of the following statements is true or false. If false, change the wording to make it true. (a) When two chemical equations are added to give a net equation, the equilibrium constant for the net equation is the product of the equilibrium constants of the summed equations. (b) The equilibrium constant for a reaction has the same magnitude but is opposite in sign as K for the reverse reaction. (c) Only the concentration of CO2 appears in the equilibrium constant expression for the reaction CaCO3(s) uv CaO(s) + CO2(g). (d) For the reaction CaCO3(s) uv CaO(s) + CO2(g), the value of K is numerically the same, whether the amount of CO2 is expressed as moles/liter or as gas pressure. (e) For the reaction CaCO3(s) uv CaO(s) + CO2(g), the standard enthalpy of reaction at 298 K is +179 kJ/mol-rxn. Increasing the ­temperature increases the value of K. 74. Neither PbCl2 nor PbF2 is appreciably soluble in water. If solid PbCl2 and solid PbF2 are placed in equal amounts of water in separate beakers, in which beaker is the concentration of Pb2+ greater? Equilibrium constants for these solids dissolving in water are as follows: PbCl2(s) uv Pb2+(aq) + 2 Cl−(aq)

Kc = 1.7 × 10−5

PbF2(s) uv Pb2+(aq) + 2 F−(aq)

Kc = 3.7 × 10−8

75. Characterize each of the following as product- or reactant-favored at equilibrium. (a) CO(g) + 1⁄2 O2(g) uv CO2(g) Kp = 1.2 × 1045 (b) H2O(g) uv H2(g) + 1⁄2 O2(g) Kp = 9.1 × 10−41 (c) CO(g) + Cl2(g) uv COCl2(g) Kp = 6.5 × 1011 76. ▲ The size of a flask containing colorless N2O4(g) and brown NO2(g) at equilibrium (Figure 15.7) is rapidly reduced to half the original volume. N2O4(g) uv 2 NO2(g)

77. Describe an experiment that would prove that the system 3 H2(g) + N2(g) uv 2 NH3(g) is a dynamic equilibrium. (Hint: Consider using a stable isotope such as 15N or 2H.) 78. At low temperatures, the cobalt(II) ion in aqueous solution with added chloride ions is deep red and in the form of [Co(H2O)6]2+ ions. As the temperature is increased, chloride ions replace the water molecules, yielding the blue [CoCl4]2− ion. The reaction is reversed upon cooling the solution. [Co(H2O)6]2+(aq) + 4 Cl−(aq) uv  [CoCl4]2− + 6 H2O(ℓ)

  

(a) Is the conversion of the red cation to the blue anion exothermic or endothermic? (b) If hydrochloric acid is added to the violet mixture of cobalt(II) ions shown on the left in the following photographs, the blue [CoCl4]2− ion is favored (middle photograph). If water is then added to the mixture, a red solution ­favoring [Co(H2O)]2+ results (right ­photograph). Explain these observations in terms of Le ­Chatelier’s principle.

Solution of cobalt(II) chloride in dilute hydrochloric acid

Solution after adding more hydrochloric acid

Solution after adding more water

(c) How do these observations prove the reaction is reversible?

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Photos: © Charles D. Winters/Cengage

The following questions may use concepts from this and previous chapters.

(a) What color change (if any) is observed immediately upon halving the flask size? (b) What color change (if any) is observed during the process in which equilibrium is reestablished in the flask?

© Charles D. Winters/Cengage

Summary and Conceptual Questions

79. Suppose a tank initially contains H2S at a pressure of 10.00 atm and a temperature of 800 K. When the reaction attains equilibrium, the partial pressure of S2 vapor is 0.020 atm. Calculate Kp. 2 H2S(g) uv 2 H2(g) + S2(g)



80. Pure PCl5 gas is placed in a 2.00-L flask. After heating to 250 °C the pressure of PCl5 is initially 2.000 atm. However, the gas slowly decomposes to gaseous PCl3 and Cl2 until equilibrium is reached. At that point, the partial pressure of Cl2 is 0.814 atm. Calculate Kp for the decomposition.

Study Questions

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775

Maks Narodenko/Shutterstockd.com

16

Principles of Chemical Reactivity: The Chemistry of Acids and Bases

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C hapt e r O ut li n e .

16.1

The Brønsted-Lowry Concept of Acids and Bases

16.2

Water and the pH Scale

16.3

Equilibrium Constants for Acids and Bases

16.4

Acid–Base Properties of Salts

16.5

Predicting the Direction of Acid–Base Reactions

16.6

Types of Acid–Base Reactions

16.7

Calculations with Equilibrium Constants

16.8

Polyprotic Acids and Bases

16.9

Molecular Structure, Bonding, and Acid–Base Behavior

16.10  The Lewis Concept of Acids and Bases

The focus of this chapter is acid–base chemistry, with an emphasis on BrønstedLowry concepts of acids and bases. Protic acids and bases are among the most common substances in nature. Amino acids are the building blocks of proteins. The repository of genetic information in your cells is DNA, deoxyribonucleic acid. The pH of lakes, rivers, and oceans is affected by dissolved acids and bases, and many bodily functions depend on acids and bases. In addition to the Brønsted-Lowry concept, you will learn another acid–base concept, one that does not require the transfer of a proton in a reaction. Developed in the 1930s by Gilbert N. Lewis, the Lewis acid–base concept is based upon the sharing of an electron pair between an acid and a base.

16.1 The Brønsted-Lowry Concept of Acids and Bases Goals for Section 16.1 • Recognize common monoprotic and polyprotic acids and bases, and write balanced equations for their ionization in water.

• Recognize the Brønsted acid and base in a reaction, and identify the conjugate partner of each.

• Identify the acid and base reactions undergone by amphiprotic substances. In Chapter 3, you were introduced to two definitions of acids and bases: the Arrhenius definition and the Brønsted-Lowry definition. According to the Arrhenius ◀ Acids are present in many fruits.  You can recognize acids in foods from their sour taste. Two of the most sour fruits are lemons and limes. These fruits contain citric acid, H3C6H5O7.

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Brønsted-Lowry Theory This chapter limits the discussion to aqueous solutions. However, the theory applies equally well to non-aqueous systems.

definition, an acid is any substance that, when dissolved in water, increases the concentration of hydrogen ions, H+. An Arrhenius base is any substance that increases the concentration of hydroxide ions, OH−, when dissolved in water. The Brønsted-Lowry definition of acids and bases is more general and views acid–base behavior in terms of proton transfer from one substance to another. A Brønsted-Lowry acid is a proton (H+) donor, and a Brønsted-Lowry base is a proton acceptor. This definition extends the list of acids and bases and the scope of acid–base reactions, and it helps chemists make predictions of product- or reactantfavorability based on acid and base strength. A wide variety of Brønsted-Lowry acids is known. These include molecular compounds such as nitric acid, HNO3(aq) + H2O(ℓ)

NO3−(aq) + H3O+(aq)

acid

− +

cations such as NH4+, the ammonium ion NH4+(aq) + H2O(ℓ)

NH3(aq) + H3O+(aq)

acid + +

anions such as HSO4−, the hydrogen sulfate ion HSO4−(aq)

+

H2O(ℓ)

SO42−(aq) + H3O+(aq)

−

2− +

and hydrated metal cations. [Fe(H2O)6]3+(aq) + H2O(ℓ) uv [Fe(H2O)5(OH)]2+(aq) + H3O+(aq)

Similarly, many different species can act as Brønsted-Lowry bases in their reactions with water. These include molecular compounds, NH3(aq) + H2O(ℓ) base

NH4+(aq) + OH−(aq) + −

and anions. CO32−(aq) + H2O(ℓ)

HCO3−(aq) + OH−(aq)

2−

− −

Acids such as HF, HCl, HNO3, and CH3CO2H (acetic acid) are capable of donating one proton and so are called monoprotic acids. Other acids, called polyprotic acids (Table 16.1), are capable of donating two or more protons. Phosphoric acid, for example, can undergo three ionization reactions in water.

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Table 16.1

Polyprotic Acids and Bases

Acid Form

Amphiprotic Form

Base Form

H2S (hydrosulfuric acid or hydrogen sulfide)

HS− (hydrogen sulfide ion)

S2− (sulfide ion)

H3PO4 (phosphoric acid)

H2PO4− (dihydrogen phosphate ion)

PO43− (phosphate ion)

HPO4

2−

(hydrogen phosphate ion)

−

H2CO3 (carbonic acid)

HCO3 (hydrogen carbonate ion or bicarbonate ion)

CO32− (carbonate ion)

H2C2O4 (oxalic acid)

HC2O4− (hydrogen oxalate ion)

C2O42− (oxalate ion)

H3PO4(aq) + H2O(ℓ) uv H3O+(aq) + H2PO4−(aq) H2PO4−(aq) + H2O(ℓ) uv H3O+(aq) + HPO42−(aq)

Just as there are acids that can donate more than one proton, there are polyprotic bases that can accept more than one proton. The fully deprotonated anions of polyprotic acids are polyprotic bases; examples include S2−, PO43−, CO32−, and C2O42−. The carbonate ion, for example, can accept two protons. CO32−(aq) + H2O(ℓ) uv HCO3−(aq) + OH−(aq)

© Charles D. Winters/Cengage

HPO42−(aq) + H2O(ℓ) uv H3O+(aq) + PO43−(aq)

HCO3−(aq) + H2O(ℓ) uv H2CO3(aq) + OH−(aq)

Some molecules (such as water) and ions can behave either as Brønsted acids or bases and are referred to as being amphiprotic (Section 3.6). An example of an amphiprotic anion is the dihydrogen phosphate ion (Table 16.1). H2PO4−(aq) + H2O(ℓ) uv H3O+(aq) + HPO42−(aq)

Carboxylic acid groups

acid

H2PO4−(aq) + H2O(ℓ) uv H3PO4(aq) + OH−(aq) base

As you will learn in Chapter 17 (“Applying Chemical Principles 17.2: Take a Deep Breath”), amphiprotic ions such as HCO3−and H2PO4− are particularly important in biochemical systems.

Conjugate Acid–Base Pairs The reaction of the hydrogen carbonate ion and water exemplifies a feature of ­Brønsted acid–base chemistry:  The reaction of a Brønsted acid and base produces a new acid and base. 

Tartaric acid, H2C4H4O6, is a naturally occurring diprotic acid.  Tartaric acid and its potassium salt are found in grapes and other fruits. The acidic protons are the H atoms of the carboxylic acid (OCO2H) groups.

conjugate pair 1 conjugate pair 2

HCO3−(aq) + H2O(ℓ) base

acid

OH−(aq) + H2CO3(aq) base

acid

−

+

−

+

In the forward direction, HCO3− is the Brønsted base because it captures H+ from the Brønsted acid, H2O. The products are a new Brønsted acid, H2CO3, and base, OH−.

16.1  The Brønsted-Lowry Concept of Acids and Bases

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A conjugate acid–base pair consists of two species that differ from each other by the presence of one hydrogen ion. Thus, H2CO3 and HCO3− are a conjugate acid–base pair. In this pair, HCO3− is the conjugate base of the acid H2CO3, and H2CO3 is the conjugate acid of the base HCO3−. There is a second conjugate acid– base pair in this reaction: H2O and OH−. In fact,  every reaction between a Brønsted acid and a Brønsted base involves two conjugate acid–base pairs. 

16.2 Water and the pH Scale Goals for Section 16.2 • Use the water ionization constant, Kw. • Use the pH concept. Because chemists often use aqueous solutions of acids and bases and because the acid–base reactions in your body occur in your aqueous interior, it is important to understand the acid–base behavior of water in terms of chemical equilibria.

Water Autoionization and the Water Ionization ­Constant, Kw Kw and Temperature  The equation Kw = [H3O+][OH−] is valid for pure water and for any aqueous solution. However, the numerical value for Kw is temperature dependent. T (°C)

Kw

10

0.29 × 10−14

15

0.45 × 10−14

20

0.68 × 10−14

25

1.01 × 10−14

30

1.47 × 10−14

50

5.48 × 10−14

In pure water, an equilibrium exists between water and the hydronium and hydroxide ions. 2 H2O(ℓ)

H3O+(aq) + OH−(aq)

+

+

+

−

This autoionization reaction of water was demonstrated over a century ago by Friedrich Kohlrausch (1840–1910). He found that, even after water is painstakingly purified, it still conducts electricity to a small extent. We now know this is because autoionization produces very low concentrations of H3O+ and OH− ions. Water autoionization is the cornerstone of modern concepts of aqueous acid–base behavior. The water autoionization equilibrium is very reactant-favored at equilibrium. In fact, in pure water at 25 °C, only about two out of a billion (109) water molecules are ionized at any instant. A quantitative expression of this idea is the equilibrium constant equation for autoionization.

Kw = [H3O+][OH−] = 1.0 × 10−14 at 25 °C

(16.1)

There are several important aspects of this equation.

The Magnitude of [H3O+] in Pure Water  The ionization of two

water molecules out of a billion produces an H3O+ concentration of 1 × 10−7 M. To have a sense of this tiny amount, this would be like identifying just 16 people out of the current population of the Earth, about 7.95 billion.

•

The equilibrium constant is given a special symbol, Kw , and is known as the autoionization constant for water.

•

Electrical conductivity measurements of pure water show that [H 3O +] = ​ [OH−] = 1.0 × 10−7 M at 25 °C, so Kw has a value of 1.0 × 10−14 at 25 °C.

•

Based on the rules for writing equilibrium constants, the concentration of water is not given in the expression for Kw.

In pure water, the hydronium ion and hydroxide ion concentrations are equal, and the water is said to be neutral. If some acid or base is added to pure water, however, the equilibrium 2 H2O(ℓ) uv H3O+(aq) + OH−(aq)

is disturbed. Adding an acid such as HCl raises the concentration of the H3O+ ions. To oppose this increase, Le Chatelier’s principle (Section 15.6) predicts that a small

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fraction of the H3O+ ions will react with OH− ions from water autoionization to form water. This lowers [OH−] until the product of [H3O+] and [OH−] is again equal to 1.0 × 10−14 at 25 °C. Similarly, adding a base to pure water gives a basic solution because the OH− ion concentration has increased. Le Chatelier’s principle predicts that some of the added OH− ions will react with H3O+ ions present in the solution from water autoionization, thereby lowering [H3O+] until the value of the product of [H3O+] and [OH−] equals 1.0 × 10−14 at 25 °C. Thus, for aqueous solutions at 25 °C, • In a neutral solution, [H3O+] = [OH−]. Both equal 1.0 × 10−7 M. • In an acidic solution, [H3O+] > [OH−]. [H3O+] > 1.0 × 10−7 M and [OH−] < 1.0 × 10−7 M. • In a basic solution, [H3O+] < [OH−]. [H3O+] < 1.0 × 10−7 M and [OH−] > 1.0 × 10−7 M.

Ex am p le 16.1

Hydronium and Hydroxide Ion Concentrations in a Solution of a Strong Base Problem  What are the hydroxide ion and hydronium ion concentrations in a 0.0012 M aqueous solution of NaOH at 25 °C?

What Do You Know?  You know the concentration of NaOH and that it is a strong base, 100% dissociated into ions in water. Strategy  Because NaOH is a strong base that has one hydroxide ion per formula unit, the OH− ion concentration is the same as the NaOH concentration. The H3O+ ion concentration can then be calculated using Equation 16.1. Solution  The initial concentration of OH− is 0.0012 M. 0.0012 mol NaOH per liter n 0.0012 M Na+(aq) +  0.0012 M OH−(aq)  Substituting the OH− concentration into Equation 16.1 gives Kw = 1.0 × 10−14 = [H3O+][OH−] = [H3O+](0.0012) and so [H3O] 

1.0  1014  8.3  1012 M 0.0012

Think about Your Answer   Why didn’t we take into account the ions produced by water autoionization when we calculated the concentration of hydroxide ions? ­ utoionization should add OH− ions to the solution. If x is equal to the concentration of A OH− ions generated by the autoionization of water, then, when equilibrium is achieved, [OH−] = (0.0012 M + OH− from water autoionization) = (0.0012 M + x) In pure water, the concentration of OH− ion generated by autoionization is 1.0 × 10−7 M. Le Chatelier’s principle (Section 15.6) suggests that the concentration from autoionization should be even smaller when OH− ions are already present in solution from NaOH; that is, x should be much less than 1.0 × 10−7 M. This means x in the term (0.0012 + x) is insignificant compared with 0.0012. (Following the rules for significant figures, the sum of 0.0012 and a number even smaller than 1.0 × 10−7 is 0.0012.)

16.2  Water and the pH Scale

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781

Check Your Understanding  What are the hydronium ion and hydroxide ion concentrations in 4.0 × 10−3 M HCl(aq) at 25 °C? (Recall that because HCl is a strong acid, it is almost 100% ionized in water.)

The pH Scale The concentration of hydronium ions in an aqueous solution can vary from less than 10−15 M in concentrated strong bases to greater than 10 M in concentrated strong acids—in other words, over 16 orders of magnitude. The pH scale compresses this range of concentrations to values from roughly 15 to −1. The pH of a solution is defined as the negative of the base-10 logarithm (log) of the hydronium ion concentration (Section 4.6). Working with Logarithms See

Appendix A for more on using logarithms.

pH   log[H3O]



(4.4 and 16.2)

In a similar way, the pOH of a solution is defined as the negative of the base-10 logarithm of the hydroxide ion concentration. pOH   log[OH]



(16.3)

In pure water, the hydronium and hydroxide ion concentrations are both 1.0 × 10−7 M. Therefore, for pure water at 25 °C pH = −log (1.0 × 10−7) = 7.00

In the same way, you can show that the pOH of pure water is also 7.00 at 25 °C. If you take the negative logarithms of both sides of the expression K w =  [H3O+][OH−], you obtain another useful equation. K w  1.0  1014  [H3O][OH]  log K w   log (1.0  1014)   log ([H3O][OH]) pK w  14.00   log [H3O]  ( log [OH]) pK w  14.00  pH  pOH



(16.4)

The sum of the pH and pOH of a solution must be equal to 14.00 at 25 °C. As illustrated in Figures 4.9 and 16.1, solutions with pH less than 7.00 (at 25 °C) are acidic, whereas solutions with pH greater than 7.00 are basic. Solutions with pH = 7.00 at 25 °C are neutral.

pH

[H3O+]

[OH−]

pOH

14.00

1.0 × 10−14

1.0 × 100

0.00

10.00

1.0 × 10−10

1.0 × 10−4

4.00

7.00

1.0 × 10−7

1.0 × 10−7

7.00

4.00

1.0 × 10−4

1.0 × 10−10

10.00

0.00

1.0 × 100

1.0 × 10−14

14.00

Basic

Figure 16.1  pH and pOH.  ​

This figure illustrates the relationship of hydronium ion and hydroxide ion concentrations and of pH and pOH at 25 °C.

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Neutral

Acidic

Chapter 16 / Principles of Chemical Reactivity: The Chemistry of Acids and Bases

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Chemistry in Your Career

Dr. Cassandra Callmann

Dr. Cassandra Callmann Dr. Cassandra Callmann is Assistant Professor of Chemistry at the University of Texas at Austin, a position she calls “the best job in the world.” In addition to teaching undergraduate chemistry courses, Dr. Callmann runs her own research program. “My team and I are constantly pushing the boundaries of science and researching new materials to address significant questions in human health.” She credits her education (B.S. in Biochemistry and Ph.D. in Chemistry) for the foundational knowledge she needed to be

successful as both a chemistry researcher and an educator. Dr. Callmann is passionate about mentoring students to help them develop the skills and experiences necessary to be successful in their own careers. She notes the unique perspective she brings to her department as a first-generation academic and woman. “I am honored to be able to serve as a role model for the undergraduates in my class and my graduate students and show them that everyone can be successful scientists!”

16.3 E  quilibrium Constants for Acids and Bases Goals for Section 16.3 • Write equilibrium constant expressions for acids and bases. • Understand the relationship between Ka for an acid and Kb for its conjugate base. • Calculate pKa from Ka (or Ka from pKa), and understand how pKa is correlated with acid strength.

Acids and bases can be divided roughly into two groups: strong electrolytes (such as HCl, HNO3, and NaOH) and weak electrolytes, such as CH3CO2H and NH3 (­Figure 16.2 and Table 3.1). Hydrochloric acid is a strong acid, so almost 100% of the acid ionizes to produce hydronium and chloride ions. In contrast, acetic acid is a weak electrolyte because it ionizes only to a small extent in water. CH3CO2H(aq) + H2O(ℓ) uv H3O+(aq) + CH3CO2−(aq)

The acid, its anion, and the hydronium ion are all present at equilibrium in solution, but the ions are present in low concentration relative to the acid concentration. For example, in a 0.100 M solution of acetic acid, [H3O+] and [CH3CO2−] are each about 0.0013 M whereas the concentration of un-ionized acetic acid, [CH3CO2H], is 0.099 M. Similarly, ammonia is a weak base. NH3(aq) + H2O(ℓ) uv NH4+(aq) + OH−(aq)

Only about 1% of ammonia molecules in a 0.100 M solution react with water to produce the ammonium and hydroxide ions. One way to define the relative strengths of a series of acids is to measure the pH of solutions of acids of equal concentration: the lower the pH, the greater the concentration of hydronium ion, and the stronger the acid. Similarly, for a series of weak bases, [OH−] will increase, and the pH will increase as the bases become stronger. •

For a strong monoprotic acid, [H3O+] in solution is equal to the original acid concentration. Similarly, for a strong monoprotic base, [OH−] is equal to the original base concentration.

•

For a weak acid, [H3O+] is much less than the original acid concentration. That is, [H3O+] is smaller than if the acid were a strong acid of the same concentration. Similarly, a weak base gives a smaller [OH−] than if the base were a strong base of the same concentration.

•

For a series of weak monoprotic acids with the same concentration, [H 3O+] increases (and the pH decreases) as the acids become stronger. Similarly, for a series of weak bases, [OH−] increases (and the pH increases) as the bases become stronger.



The Leveling Effect  The acid

strengths of strong acids are indistinguishable in water. They all ionize almost 100% to produce hydronium ion, H3O+, so these acids appear to have the same acid strength in water. This is known as the leveling effect.

16.3  Equilibrium Constants for Acids and Bases

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783

HCl almost completely ionizes in aqueous solution.

NH3

−

−

−

+

+

+

−

−

−

−

+

CH3CO2H

The weak base ammonia reacts to a small extent with water to give a weakly basic solution.

Photos: © Charles D. Winters/Cengage

−

HCl

Acetic acid, CH3CO2H, ionizes only slightly in water.

+

−

+

−

+

+

+

−

+

−

Strong Acid

Weak Acid

(a) Hydrochloric acid, a strong acid, is sold for household use as muriatic acid.

Weak Base

(b) Vinegar is a solution of acetic acid, a weak acid that ionizes only to a small extent in water.

(c) Ammonia is a weak base, ionizing to a small extent in water.

Figure 16.2  Strong and weak acids and bases.

The relative strength of an acid or base in water can also be expressed quantitatively with an equilibrium constant, often called an ionization constant. For the general acid HA, the ionization reaction and equilibrium constant expression are HA(aq) + H2O(ℓ) uv H3O+(aq) + A−(aq) Ka 

[H3O][A] [HA]



(16.5)

where the equilibrium constant, K, has a subscript “a” to indicate that it is an equilibrium constant for an acid in water. For weak acids, the value of Ka is less than 1 because only a small fraction of the weak acid ionizes. For a series of acids, the acid strength increases as the value of Ka increases. The equilibrium expression for a weak base B in water is treated similarly. Here, the constant K is written with a subscript “b.” Its value is less than 1 for weak bases. B(aq) + H2O(ℓ) uv BH+(aq) + OH−(aq) Kb 

[BH][OH] [B]



(16.6)

Some acids and bases are listed in Table 16.2, each with its value of Ka or Kb. The following are important concepts concerning this table.

784

•

A large value of K indicates that ionization products are strongly favored, whereas a small value of K indicates that reactants are favored.

•

The strongest acids are at the upper left. They have the largest Ka values. Ka values become smaller on descending the chart as acid strength declines.

•

The strongest bases are at the lower right. They have the largest K b values. Kb values become larger on descending the chart as base strength increases.

•

 The weaker the acid, the stronger its conjugate base.  That is, the smaller the value of Ka for an acid, the larger the value of Kb for its conjugate base.

•

Some acids or bases are listed as having Ka or Kb values that are large or very small. Aqueous acids that are stronger than H3O+ are almost completely ionized (HNO3, for example), so their Ka values are large. Their conjugate bases (such as NO3−) do not produce meaningful concentrations of OH− ions, so their Kb values are very small. Similar arguments follow for strong bases and their conjugate acids.

Chapter 16 / Principles of Chemical Reactivity: The Chemistry of Acids and Bases

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Ionization Constants for Some Acids and Their Conjugate Bases at 25 °C

Acid Name

Acid

Ka

Base

Kb

Base Name

Perchloric acid

HClO4

Large

ClO4−

Very small

Perchlorate ion

Sulfuric acid

H2SO4

Large

HSO4−

Very small

Hydrogen sulfate ion

Very small

Chloride ion

NO3

Very small

Nitrate ion

H2O

1.0 × 10

Water

Hydrochloric acid Nitric acid Hydronium ion Sulfurous acid

Large

HNO3

Cl

−

Large

+

−14

H3O

1.0

H2SO3

1.2 × 10

HSO3

8.3 × 10

Hydrogen sulfite ion

−

−2

−

−13

Hydrogen sulfate ion

HSO4

1.2 × 10

SO4

8.3 × 10

Sulfate ion

Phosphoric acid

H3PO4

7.5 × 10−3

H2PO4−

1.3 × 10−12

Dihydrogen phosphate ion

Hexaaquairon(III) ion

[Fe(H2O)6]3+

6.3 × 10−3

[Fe(H2O)5OH]2+

1.6 × 10−12

Pentaaquahydroxoiron(III) ion

Hydrofluoric acid

HF

7.2 × 10−4

F−

1.4 × 10−11

Fluoride ion

HNO2

4.5 × 10

2.2 × 10

Nitrite ion

HCO2H

1.8 × 10

5.6 × 10

Formate ion

C6H5CO2H

6.3 × 10

C6H5CO2

1.6 × 10

Benzoate ion

CH3CO2H

1.8 × 10

−

5.6 × 10

Acetate ion

Propanoic acid

CH3CH2CO2H

1.3 × 10

CH3CH2CO2

7.7 × 10

Propanoate ion

Hexaaquaaluminum ion

[Al(H2O)6]3+

7.9 × 10−6

[Al(H2O)5OH]2+

1.3 × 10−9

Pentaaquahydroxoaluminum ion

Carbonic acid

H2CO3

4.2 × 10−7

HCO3−

2.4 × 10−8

Hydrogen carbonate ion

Hexaaquacopper(II) ion

[Cu(H2O)6]2+

1.6 × 10−7

[Cu(H2O)5OH]+

6.3 × 10−8

Pentaaquahydroxocopper(II) ion

H2S

1 × 10

1 × 10

Hydrogen sulfide ion

Nitrous acid Formic acid Benzoic acid Acetic acid Increasing Acid Strength

HCl

−

Hydrogen sulfide

−2

−4 −4 −5 −5 −5

−7

−13

2−

−

−11

NO2

−

−11

HCO2

−

−10 −10

CH3CO2

−

−10

−

−7

HS

H2PO4

6.2 × 10

HPO4

1.6 × 10

HSO3

−

6.2 × 10

SO3

2−

1.6 × 10

Sulfite ion

Hypochlorous acid

HClO

3.5 × 10

−

2.9 × 10

Hypochlorite ion

Hexaaqualead(II) ion

[Pb(H2O)6]

1.5 × 10

[Pb(H2O)5OH]

6.7 × 10

Pentaaquahydroxolead(II) ion

Hexaaquacobalt(II) ion

[Co(H2O)6]2+

1.3 × 10−9

[Co(H2O)5OH]+

7.7 × 10−6

Pentaaquahydroxocobalt(II) ion

Boric acid

B(OH)3(H2O)

7.3 × 10−10

B(OH)4−

1.4 × 10−5

Tetrahydroxoborate ion

Ammonium ion

NH4+

5.6 × 10−10

NH3

1.8 × 10−5

Ammonia

Hydrocyanic acid

HCN

4.0 × 10

2.5 × 10

Cyanide ion

Hexaaquairon(II) ion

[Fe(H2O)6]

Dihydrogen phosphate ion Hydrogen sulfite ion

−

−8 −8 −8

2+

2+

−

−8

−10

HCO3

Hexaaquanickel(II) ion

[Ni(H2O)6]

Hydrogen phosphate ion

−7 −7 −7

ClO

+

−

−7

−5

CN

Hydrogen phosphate ion

3.2 × 10

[Fe(H2O)5OH]

3.1 × 10

Pentaaquahydroxoiron(II) ion

4.8 × 10

CO3

2.1 × 10

Carbonate ion

−10 −11

Hydrogen carbonate ion

2−

+

−5 −4

2−

−11

2.5 × 10

[Ni(H2O)5OH]

4.0 × 10

Pentaaquahydroxonickel(II) ion

HPO4

3.6 × 10

−13

PO4

2.8 × 10

Phosphate ion

Water

H2O

1.0 × 10−14

OH−

1.0

Hydroxide ion

Hydrogen sulfide ion*

HS−

1 × 10−19

S2−

1 × 105

Sulfide ion

Ethanol

C2H5OH

Very small

C2H5O−

Large

Ethoxide ion

Ammonia Hydrogen

2+

2−

NH3 H2

+

3−

−

−4 −2

Very small

NH2

Large

Amide ion

Very small

−

Large

Hydride ion

H

*The values of Ka for HS− and Kb for S2− are estimates.



16.3  Equilibrium Constants for Acids and Bases

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785

Increasing Base Strength

Table 16.2

Problem Solving Tip 16.1 Strong or Weak? How can you tell whether an acid or a base is weak? The easiest way is to remember those few that are strong. All others are probably weak. Common strong acids include the following: Hydrohalic acids: HCl, HBr, and HI (but not HF)

Nitric acid: HNO3

All Group 1A (1) hydroxides: LiOH, NaOH, KOH, RbOH, CsOH

Sulfuric acid: H2SO4 (for loss of first H+ only)

Group 2A (2) hydroxides: Sr(OH)2 and Ba(OH)2. [Neither Mg(OH)2 nor Ca(OH)2 dissolve appreciably in water. However, some texts consider them strong bases.]

Perchloric and chloric acids: HClO4 and HClO3 Some common strong bases include the following:

A comparison of some common acids and bases will illustrate some of these ideas. For example, HF is a stronger acid than HClO, which is in turn stronger than HCO3−, Increasing acid strength HCO3−

HClO

HF

Ka = 4.8 × 10−11

Ka = 3.5 × 10−8

Ka = 7.2 × 10−4

and their conjugate bases become stronger from F− to ClO− to CO32−. Increasing base strength CO32−

ClO−

F−

Kb = 2.1 × 10−4

Kb = 2.9 × 10−7

Kb = 1.4 × 10−11

Acids and bases are abundant in nature (Figure 16.3). Many naturally occurring acids contain the carboxylic acid group (OCO2H), and a few are illustrated here. Notice that the organic portion of the molecule has an effect on its relative strength (as described further in Section 16.9). Ka increases; acid strength increases

Kb of conjugate base increases

propanoic acid, CH3CH2CO2H Ka = 1.3 × 10−5

acetic acid, CH3CO2H Ka = 1.8 × 10−5

formic acid, HCO2H Ka = 1.8 × 10−4

There are many naturally occurring weak bases (Figure 16.3). Ammonia and its conjugate acid, the ammonium ion, are part of the nitrogen cycle in the environment (Section 25.1). Biological systems reduce nitrate ion to NH3 and NH4+ and incorporate nitrogen into amino acids and proteins. Many bases are derived from NH3 by replacement of the H atoms with organic groups.

H H

N

H H 3C

H

ammonia

H

methylamine −5

Kb = 1.8 × 10

786

N

H

C6H5

N

H

aniline −4

Kb = 5.0 × 10

Kb = 4.0 × 10−10

Chapter 16 / Principles of Chemical Reactivity: The Chemistry of Acids and Bases

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C

H2 OH H2 C C OH C C

O

C OH O

O H3C O

O

CH3 N

N N

N

CH3

The tartness of lemons and oranges comes from the weak acid citric acid. The acid is found widely in nature and in many consumer products.

Caffeine is a stimulant and a weak base.

Photos: © Charles D. Winters/Cengage

HO

Figure 16.3  Natural acids and bases.  Hundreds of acids and bases occur in nature. Our foods contain a wide variety, and biochemically important molecules are often acids and bases.

Ammonia is a weaker base than methylamine (Kb for NH3 < Kb for CH3NH2). This means that the conjugate acid of ammonia, NH4+ (Ka = 5.6 × 10−10), is a stronger acid than the methylammonium ion, the conjugate acid of methylamine, CH3NH3+ (Ka = 2.0 × 10−11).

Ka and Kb Values for Polyprotic Acids Like all polyprotic acids, phosphoric acid ionizes in a series of steps. First ionization step: Ka1 = 7.5 × 10−3 H3PO4(aq) + H2O(ℓ) uv H2PO4−(aq) + H3O+(aq)

Second ionization step: Ka2 = 6.2 × 10−8 H2PO4−(aq) + H2O(ℓ) uv HPO42−(aq) + H3O+(aq)

Third ionization step: Ka3 = 3.6 × 10−13 HPO42−(aq) + H2O(ℓ) uv PO43−(aq) + H3O+(aq)

The Ka value for each successive step becomes smaller because it is more difficult to remove H+ from a negatively charged ion, such as H2PO4−, than from a neutral molecule, such as H3PO4. Similarly, the larger the negative charge of the anionic acid, the more difficult it is to remove H+. For many inorganic polyprotic acids, Ka values become smaller by about 105 for each proton removed. Similarly, successive base equilibrium expressions can be written for phosphate ion, PO43−, in water. First step: Kb1 = 2.8 × 10−2 PO43−(aq) + H2O(ℓ) uv HPO42−(aq) + OH−(aq)

Second step: Kb2 = 1.6 × 10−7 HPO42−(aq) + H2O(ℓ) uv H2PO4−(aq) + OH−(aq)

Third step: Kb3 = 1.3 × 10−12 H2PO4−(aq) + H2O(ℓ) uv H3PO4(aq) + OH−(aq)

In a trend that is similar to polyprotic acid dissociation, the Kb value for each successive step becomes smaller by about 105.

16.3  Equilibrium Constants for Acids and Bases

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787

Logarithmic Scale of Relative Acid Strength, pKa Most chemists and biochemists use a logarithmic scale to report and compare relative acid strengths. pK a   log K a



(16.7)

The pKa of an acid is the negative log of the Ka value (just as pH is the negative log of the hydronium ion concentration). For example, acetic acid has a pKa value of 4.74. pKa = −log (1.8 × 10−5) = 4.74

The pKa value becomes smaller as the acid strength increases. 8888888  Acid strength increases  88888n

Propanoic Acid

Acetic Acid

CH3CH2CO2H

CH3CO2H

Ka = 1.3 × 10

Ka = 1.8 × 10

pKa = 4.89

pKa = 4.74

−5

Formic Acid HCO2H −5

Ka = 1.8 × 10−4 pKa = 3.74

88888888888  pKa decreases  888888888n

Relating the Ionization Constants for an Acid and Its Conjugate Base Look again at Table 16.2. From the top of the table to the bottom, the strengths of the acids decline (Ka becomes smaller), and the strengths of their conjugate bases increase (the values of Kb increase). These observations are connected: the product of Ka for an acid and Kb for its conjugate base is equal to a constant, specifically Kw. K a  Kb  K w



(16.8)

A useful relationship for an acid–conjugate base pair follows from Equation 16.8. (16.9)

pKa + pKbb = pK = pKw pKa + pK w



Consider the specific case of the ionization of a weak acid, say HCN, and the interaction of its conjugate base, CN−, with H2O at 25 °C. Weak acid:

HCN(aq) + H2O(ℓ) uv H3O+(aq) + CN−(aq)

Ka = 4.0 × 10−10

pKa = 9.40

Conj. base:

CN−(aq) + H2O(ℓ) uv HCN(aq) + OH−(aq)

Kb = 2.5 × 10−5

pKb = 4.60

Kw = 1.0 × 10−14

pKw = 14.00

2 H2O(ℓ) uv H3O+(aq) + OH−(aq)

Adding the equations gives the chemical equation for the autoionization of water, and the product of Ka and Kb is indeed 1.0 × 10−14. That is,  [H O][CN ]   [HCN][OH]  K a  Kb   3  [H3O][OH]  K w  [HCN]   [CN ] 

Equation 16.8 is useful because Kb for a base can be calculated from the Ka value for its conjugate acid. The value of Kb for the cyanide ion at 25 °C, for example, is Kb for CN 

1.0  1014 Kw  2.5  105  4.0  1010 K a for HCN

Similarly, the pKb value for a base can be calculated from the pKa value for its conjugate acid. The value of pKb for the cyanide ion at 25 °C is pKb for CN− = pKw − pKa = 14.00 − 9.40 = 4.60.

788

Chapter 16 / Principles of Chemical Reactivity: The Chemistry of Acids and Bases

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16.4 Acid–Base Properties of Salts Goal for Section 16.4 • Describe the acid–base properties of salts. A number of the acids and bases listed in Table 16.2 are cations or anions. As described earlier, anions can act as Brønsted bases because they can accept a proton from an acid to form the conjugate acid of the ion. CO32−(aq) + H2O(ℓ) uv HCO3−(aq) + OH−(aq) Kb = 2.1 × 10−4

You should also notice that many metal cations in water are Brønsted acids. [Al(H2O)6]3+(aq) + H2O(ℓ) uv [Al(H2O)5(OH)]2+(aq) + H3O+(aq) Ka = 7.9 × 10−6

Table 16.3 summarizes the acid–base properties of some common cations and anions in aqueous solution. As you look over this table, notice the following points: •

Anions that are conjugate bases of strong acids (for example, Cl− and NO3−) are such weak bases that they have no effect on solution pH.

•

There are numerous basic anions (such as CH3CO2−). All are the conjugate bases of weak acids.

•

The acid–base behavior of anions of polyprotic acids depends on the extent of deprotonation. A fully deprotonated anion (such as CO32−) is basic. A partially deprotonated anion (such as HCO3−) is amphiprotic and is capable of undergoing both acid and base reactions with water. A solution may be acidic or basic depending on the relative strengths of the partially deprotonated anion as an acid or as a base, which can be assessed by looking at the magnitudes of Ka and Kb of the ion.

•

Alkali metal and alkaline earth cations have no measurable effect on solution pH.

•

Acidic cations fall into two categories: (a) ammonium ion (and its organic derivatives), and (b) metal cations with 2+ and 3+ charges. All metal cations are hydrated in water, forming ions such as [M(H2O)6]n+. However, only when M is a 2+ or 3+ ion does the ion act as an acid.

•

Basic cations such as [Al(H2O)5(OH)]2+ are conjugate bases of acidic cations such as [Al(H2O)6]3+.

Acid and Base Properties of Some Ions in Aqueous Solution

Neutral Anions

Basic

Cl−

NO3−

−

−

CH3CO2− −

Cations

HSO4−

HPO4

H2PO4−

SO32−

HSO3−

Br

ClO3

HCO2

I−

ClO4−

CO32−

HCO3−

Li+

−

S

HS

F−

NO2−

2−

−

OCl

[Al(H2O)5(OH)]2+ (for example)

Na+

Ca2+

+

Ba2+

K



SO42−

PO43−

2−

Acidic

CN−

[Al(H2O)6]3+ and hydrated transition metal cations (such as [Fe(H2O)6]3+) NH4+

say that, when ions interact with water to produce acidic or basic solutions, the ions “hydrolyze” in water, or they undergo “hydrolysis.” Thus, some books refer to the Ka and Kb values of ions as “hydrolysis constants,” Kh.

© Charles D. Winters/Cengage

Table 16.3

Hydrolysis Reactions and Hydrolysis Constants  Chemists often

Many aqueous metal cations in salts are Brønsted acids.  A pH measurement of a dilute solution of copper(II) sulfate shows that the solution is acidic. Among the common cations, Al3+ and transition metal ions form acidic solutions in water.

16.4  Acid–Base Properties of Salts

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789

E xamp le 16.2

Acid–Base Properties of Salts Problem  Determine whether each of the following salts gives rise to an acidic, basic, or neutral solution in water. (a) NaNO3

(d) NaHCO3

(b) K3PO4

(e) NH4F

(c) FeCl2

What Do You Know?  You know the salt formulas and, from Tables 16.2 and 16.3, you know the effect of the constituent ions on pH.

Strategy  First, identify the cation and anion in each salt. Next, use Tables 16.2 and 16.3 to describe the acid–base properties of each ion. Finally, decide on the overall acid–base properties of the salt. Solution (a) NaNO3: Neither the sodium ion, Na+, nor the nitrate ion, NO3− (the very weak conjugate base of a strong acid), affects the solution pH. This salt gives a  neutral, aqueous solution (pH = 7).  (b) K3PO4: The K+ ion, like the Na+ ion, does not affect the solution pH. The phosphate ion, PO43−, is the fully deprotonated anion from the polyprotic acid H3PO4. It acts as a base in water and accepts an H+ ion from water. PO43−(aq) + H2O(ℓ) uv HPO42−(aq) + OH−(aq)

The resulting solution is  basic (pH > 7).

(c) FeCl2: Chloride ion is the very weak conjugate base of the strong acid HCl, so it does not contribute OH− ions to the solution. The Fe2+ ion in water, [Fe(H2O)6]2+ is a ­Brønsted acid and donates a proton to water. [Fe(H2O)6]2+(aq) + H2O(ℓ) uv [Fe(H2O)5OH]+ + H3O+(aq)

Therefore, an aqueous solution of FeCl2 is  acidic, pH < 7. 

(d) NaHCO3: The Na+ ion does not affect the solution pH. Some additional information is needed concerning salts of amphiprotic anions such as HCO3−. Because they have an ionizable hydrogen, they can act as acids, HCO3−(aq) + H2O(ℓ) uv CO32−(aq) + H3O+(aq)

Ka = 4.8 × 10−11

but because they are anions, they can also act as bases and accept an H+ ion from water. HCO3−(aq) + H2O(ℓ) uv H2CO3(aq) + OH−(aq)

Kb = 2.4 × 10−8

Whether the solution is acidic or basic depends on the relative magnitudes of Ka and Kb. In the case of the hydrogen carbonate anion, Kb is larger than Ka, so [OH−] is larger than [H3O+], and  an aqueous solution of NaHCO3 will be basic. 

(e) NH4F: This is a salt based on an acidic cation and a basic anion. Here, the ammonium ion would decrease the pH, and the fluoride ion would increase the pH. NH4+(aq) + H2O(ℓ) uv H3O+(aq) + NH3(aq)

Ka (NH4+) = 5.6 × 10−10

F−(aq) + H2O(ℓ) uv HF(aq) + OH−(aq)

Kb (F−) = 1.4 × 10−11

Because Ka (NH4+) > Kb (F−), the ammonium ion is a slightly stronger acid than the fluoride ion is a base. The resulting solution is  acidic. 

790

Chapter 16 / Principles of Chemical Reactivity: The Chemistry of Acids and Bases

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Think about Your Answer  There are several important points here:

•

Anions that are conjugate bases of strong acids—such as Cl− and NO3−—have no effect on solution pH.

•

To determine whether a salt is acidic, basic, or neutral, you must take into account both the cation and the anion. When a salt has an acidic cation and a basic anion, the pH of the solution will be determined by the ion that is the stronger acid or base.

Check Your Understanding  For each of the following salts in water, predict whether the pH will be greater than, less than, or equal to 7. (a) KBr

(b) NH4NO3

(c) AlCl3

(d) Na2HPO4

16.5 P  redicting the Direction of Acid–Base Reactions Goal for Section 16.5 • Write equations for acid–base reactions, and decide whether they are productor reactant-favored at equilibrium.

According to the Brønsted-Lowry theory, all acid–base reactions can be written as equilibria involving the acid and base and their conjugates. Acid + base uv conjugate base of the acid + conjugate acid of the base

In Sections 16.3 and 16.4, equilibrium constants were used to provide information about the relative strengths of acids and bases. They can also be used to decide whether a particular acid–base reaction is product- or reactant-favored at equilibrium. Hydrochloric acid is a strong acid. Its equilibrium constant for reaction with water is very large, with the equilibrium lying completely to the right. HCl(aq) + H2O(ℓ) uv H3O+(aq) + Cl−(aq) Strong acid (≈ 100% ionized), Ka >> 1 [H3O+] ≈ initial concentration of the acid

There are two acids and two bases in this reaction. The acids are HCl and H3O+, and the bases are H2O and Cl−. The relative strengths of these acids and bases can be assessed using the values for Ka and Kb in Table 16.2. Of the two acids, HCl (large Ka) is stronger than H3O+ (Ka = 1.0). Of the two bases, H2O (Kb = 1.0 × 10−14) is stronger than Cl− (very small Kb) and wins out in the competition for the proton.  At equilibrium, there will be greater concentrations of the species present on the side of the reaction with the weaker acid and base.  conjugate pair 1 conjugate pair 2 HCl(aq) stronger acid than H3O+



+

H2O(ℓ) stronger base than Cl−

K>1

H3O+(aq) weaker acid than HCl

+

Cl−(aq) weaker base than H2O

16.5  Predicting the Direction of Acid–Base Reactions

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791

In contrast to HCl and other strong acids, acetic acid, a weak acid, ionizes to only a very small extent. CH3CO2H(aq) + H2O(ℓ) uv H3O+(aq) + CH3CO2−(aq) Weak acid (< 100% ionized), Ka = 1.8 × 10−5 [H3O+] 100 × Ka; 0.020 M > 6.3 × 10−3). Therefore, you can use the approximate expression. K a  6.3  105 

x2 0.020

Solving for x, you obtain x  K a  (0.020)  0.00112 M and you find that [H3O+] = [C6H5CO2−] = 0.00112 M = 0.0011 M Finally, the pH of the solution is found to be pH = −log (1.12 × 10−3) =  2.95 

Think about Your Answer   The H3O+ concentration was determined using the approximation that (0.020 − x) ≈ 0.020. If the approximation is not made and the exact expression is solved, x = [H3O+] = 0.0011 M. This is the same answer to two significant figures obtained from the approximate expression.

Check Your Understanding  What are the equilibrium concentrations of acetic acid, the acetate ion, and H3O+ for a 0.10 M solution of acetic acid (Ka =  1.8 × 10−5)? What is the pH of the solution?



16.7  Calculations with Equilibrium Constants

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799

E xamp le 16.6

Calculating Equilibrium Concentrations and pH from Ka and Using the Method of Successive Approximations Problem  What is the concentration of H3O+ and the pH of a 0.0010 M solution of formic acid? What is the concentration of formic acid at equilibrium? The acid is moderately weak, with Ka = 1.8 × 10−4. HCO2H(aq) + H2O(ℓ) uv HCO2−(aq) + H3O+(aq)

What Do You Know?  You know the value of Ka for the acid and its initial concentration. You need to find the equilibrium concentration of H3O+ to calculate the pH. Strategy   This is like Example 16.5, where you wanted to find the concentration of a reaction product, except that an approximate solution will not be possible. The strategy is the same. Step 1.  Write the balanced equation and set up an ICE table. Step 2.  Enter the initial concentration of HCO2H on the Initial (I) line of the ICE table. Step 3. Using the variable x, enter the changes in concentrations of the reactants and products on the Change line (C) of the table. Step 4. Enter the equilibrium concentrations on the Equilibrium (E) line of the ICE table. Step 5. Write an equilibrium constant expression, Ka, for the acid-ionization reaction. Enter the expressions for the final equilibrium concentrations of all three species into the equilibrium constant expression. Step 6.  Solve for x (= [H3O+]) and convert to pH.

Solution Step 1–4.  Write the balanced chemical equation and complete the ICE table.

Equilibrium

HCO2H + H2O

Initial (M) Change (M) Equilibrium (M)

uv

HCO2−

+

H3O+

0.0010

 0

 0

−x

+x

+x

(0.0010 − x)

 x

 x

Step 5. Write an equilibrium constant expression, Ka, for the acid ionization reaction and enter the expressions for the final equilibrium concentrations. Ka 

[H3O][HCO2] ( x)( x)  1.8  104  0.0010  x [HCO2H]

Step 6. Solve for x (=[H3O+]) and convert to pH. In this example, [HA]0 (= 0.0010 M) is not greater than 100 × Ka (= 1.8 × 10−2), so the usual approximation is not reasonable. Thus, you have to find the equilibrium concentrations by solving the exact expression using the quadratic formula or by successive approximations (Appendix A). The successive approximations method is used here. To use the successive approximations approach, begin by solving the approximate expression for x. 1.8  104 

( x)( x) 0.0010

Solving this, you find x = 4.24 × 10−4. Put this value into the expression for x in the denominator of the exact expression.

800

Chapter 16 / Principles of Chemical Reactivity: The Chemistry of Acids and Bases

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1.8  104 

( x)( x) ( x)( x)  0.0010  x 0.0010  4.24  104

Solving this equation for x, you now find x = 3.22 × 10−4. Again, put this value into the denominator, and solve for x. 1.8  104 

( x)( x) ( x)( x)  0.0010  x 0.0010  3.22  104

Continue this procedure until the value of x does not change from one cycle to the next. In this case, several more cycles give x = [H3O+] = [HCO2−] = 3.44 × 10−4 M = 3.4 × 10−4 M Thus, [HCO2H] = 0.0010 − x ≈  0.0007 M  and the pH of the formic acid solution is pH =−log (3.44 × 10−4) =  3.46 

Think about Your Answer   If you had used the approximate expression to find the H3O+ concentration, you would have obtained a value of [H3O+] = 4.2 × 10−4 M. The simplifying assumption led to a large error, about 24%. The approximate solution fails in this case because (a) the acid concentration is small and (b) the acid is not very weak.

Check Your Understanding  What are the equilibrium concentrations of HF, F− ion, and H3O+ ion in a 0.00150 M ­solution of HF? What is the pH of the solution?

Just as acids can be molecular species or ions, bases can also be molecular species or ions (Figures 16.3–16.5). Many molecular bases are based on nitrogen, with ammonia being the simplest. Many nitrogen-containing bases occur naturally, such as caffeine and nicotine. The anionic conjugate bases of weak acids make up another group of bases. The following example describes the calculation of the pH for a solution of such a basic salt, sodium acetate.

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Ammonia, NH3 Kb = 1.8 × 10−5

Caffeine, C8H10N4O2 Kb = 2.5 × 10−4 Benzoate ion, C6H5CO2− Kb = 1.6 × 10−10

Phosphate ion, PO43− Kb = 2.8 × 10−2

Figure 16.5  Examples of weak bases.  ​Weak bases in water include molecules with one or more N atoms that can accept an H+ ion and anions of weak acids such as benzoate and phosphate.



16.7  Calculations with Equilibrium Constants

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Exampl e 1 6 .7

The pH of a Solution of a Weakly Basic Salt, Sodium Acetate Problem  What is the pH of a 0.015 M solution of sodium acetate, NaCH3CO2? What Do You Know?   You know sodium acetate is basic in water because the acetate ion, the conjugate base of a weak acid, acetic acid, reacts with water to form OH− (Tables 16.2 and 16.3). (You also know that the sodium ion of sodium acetate does not affect the solution pH.) Finally, you know the value of Kb for the acetate ion (Table 16.2) and its initial concentration. You need to find the equilibrium concentration of H3O+ to calculate the pH. Strategy Map Problem Calculate the pH for a weak base knowing the value of Kb.

Strategy  This is like Examples 16.5 and 16.6, where you wanted to find the concentration of a reaction product. Step 1.  Write the balanced equation and set up an ICE table. Step 2.  Enter the initial concentration of CH3CO2− on the Initial (I) line of the ICE table. Step 3. The variable x represents changes in concentration, so the change in [CH3CO2−] is −x and the changes in product concentrations are +x. Enter these values on the Change (C) line on the table.

Data/Information • Base concentration • Value of Kb

Step 4. Enter the equilibrium concentrations of all three species on the Equilibrium (E) line of the ICE table. Step 5. Write an equilibrium constant expression, Kb, for the base ionization reaction. Enter the expressions for the final equilibrium concentrations of all three species into the equilibrium constant expression. Step 6.  Solve for x (= [OH−]). Step 7. Determine the [H3O+] from the [OH−] using the expression Kw = [H3O+][OH−], then calculate pH from [H3O+].

Solution Step 1

Write the balanced equation and set up an ICE table. The steps for completing the ICE table are given in Steps 2–4.

Equilibrium Initial (M) Change (M) Equilibrium (M) Step 2

CH3CO2− + H2O

uv

CH3CO2H

+

OH−

0.015

 0

 0

−x

+x

+x

(0.015 − x)

 x

 x

Enter the initial concentration of CH3CO22 on the Initial (I) line of the ICE table. The initial concentration of CH3CO2− is 0.015 M. Its conjugate acid, CH3CO2H, is not initially present. The concentration of hydroxide ion, OH−, in pure water from water autoionization is small (10−7 M) and can be ignored.

Step 3

Using the variable x, enter the changes in concentrations of the reactants and products on the Change line (C) of the table. The change in [CH3CO2−] is −x and the changes in [CH3CO2H] and [H3O+] are +x.

Step 4

802

Enter the equilibrium concentrations on the Equilibrium (E) line of the ICE table.

Chapter 16 / Principles of Chemical Reactivity: The Chemistry of Acids and Bases

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Step 5

Write an equilibrium constant expression, Kb, for the base ionization reaction and enter the expressions for the final equilibrium concentrations into the ­equilibrium constant expression. The value of Kb for the acetate ion is 5.6 × 10−10 (Table 16.2). K b  5.6  1010 

Step 6

[CH3CO2H][OH] x2   [CH3CO2 ] 0.015  x

Solve for x (= [OH−]). The acetate ion, a weak base, has a very small value of Kb. Therefore, assume that x, the concentration of hydroxide ion generated by the reaction of acetate ion with water, is very small. Therefore, the approximate expression can be used to solve for x. K b  5.6  1010 

x2 0.015

x  [OH]  [CH3CO2H]  (5.6  1010)(0.015)  2.90  106 M Step 7

Determine the [H3O+] and then calculate pH. To calculate the pH of the solution, you need the hydronium ion concentration. In aqueous solutions, it is always true that, at 25 °C Kw = 1.0 × 10−14 = [H3O+][OH−] [H3O] 

Kw 1.0  1014   3.45  109 M  [OH ] 2.90  106

Therefore, pH   log(3.45  109)  8.46

Think about Your Answer   The hydroxide ion concentration (x) is quite small relative to the initial acetate ion concentration so the approximate expression is appropriate. (Note that 100 × Kb is less than the initial base concentration.)

Check Your Understanding  The weak base, ClO− (hypochlorite ion), is used in the form of NaClO as a disinfectant in swimming pools and water-treatment plants. What are the concentrations of HClO and OH− and the pH of a 0.015 M solution of NaClO?

Ex am p le 16.8

Calculating the pH after the Reaction of a Weak Base with a Strong Acid Problem  What is the pH of the solution that results from mixing 25 mL of 0.016 M NH3 and 25 mL of 0.016 M HCl?

What Do You Know?   You know this is an acid–base reaction (HCl + NH3), and you know the amount of each reactant (calculated from the volume and concentration of each). Because equal volumes and concentrations are involved, neither the acid nor the base is in excess, and none will remain after the reaction. The weak acid NH4+ is produced. You will need to determine its initial concentration following the reaction and look up its acid-ionization equilibrium constant. Determine the pH using the strategy for Examples 16.5 and 16.6.



16.7  Calculations with Equilibrium Constants

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803

Strategy Map

Strategy  This question requires solving a stoichiometry problem, followed by solving

Problem Calculate pH of solution after the reaction of a weak base and a strong acid.

Step 1. Write the balanced chemical equation for the reaction between a weak base (NH3) and a strong acid (HCl). This reaction has a large equilibrium constant and is not treated as an equilibrium.

Data/Information • Reactant concentrations • Reactant volumes

an acid–base equilibrium problem.

Step 2. Treat the reaction of NH3 with HCl as a stoichiometry problem. Determine the amounts (moles) of products and then the concentration of the weak acid (NH4+) produced. Step 3. Write the balanced chemical equation for the acid–base reaction of the weak acid NH4+ with H2O, and set up an ICE table. Solve for the pH using the strategy from Example 16.5 or 16.6.

Solution Step 1

Write the balanced chemical equation for the reaction of NH3 and HCl. The net ionic equation for the reaction that occurs is NH3(aq) + H3O+(aq) n NH4+(aq) + H2O(ℓ)

Step 2

Determine the amount (moles) and then the concentration of the weak acid (NH4+) produced. Amount of HCl and NH3 consumed: (0.025 L HCl)(0.016 mol/L) = 4.00 × 10−4 mol HCl (0.025 L NH3)(0.016 mol/L) = 4.00 × 10−4 mol NH3 Equal amounts of HCl and NH3 were added, and they react in a 1:1 stoichiometric ratio, so neither reactant is limiting. Either reactant can be used to calculate the amount of NH4+ produced in the reaction.  1 mol NH4  4.00  104 mol NH3   4.00  104 mol NH4  1 mol NH3  Combining 25 mL of HCl with 25 mL of NH3 gives a total solution volume of 50. mL. Therefore, the concentration of NH4+ is [NH4] 

Step 3

4.00  104 mol  8.00  103 M 0.050 L

Write the balanced chemical equation for the acid–base reaction of NH4+ and water, set up an ICE table, and solve for the pH. The concentration of NH4+ produced from the reaction of NH3 with HCl is treated as the initial concentration of NH4+. As a weak acid, NH4+ is slightly ionized, creating equal ­concentrations of NH3 and H3O+.

NH4+ + H2O

Equilibrium Initial (M)

uv

0.00800 −x

Change (M) Equilibrium (M)

(0.00800 − x)

NH3

+

H3O+

0

0

+x

+x

x

x

Substitute the values in the table into the Ka expression for the ammonium ion. K a  5.6  1010 

804

[H3O][NH3] ( x)( x)  [NH4] 0.00800  x

Chapter 16 / Principles of Chemical Reactivity: The Chemistry of Acids and Bases

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The ammonium ion is a very weak acid, as reflected by the very small value of Ka. Therefore, x, the concentration of hydronium ion generated by reaction of ammonium ion with water, is assumed to be very small, and the approximate expression is used to solve for x. (Here 100 × Ka is much less than the original acid concentration.) K a  5.6  1010 

x2 0.00800

x  (5.6  1010)(0.00800)  [H3O]  [NH3]  2.12  106 M pH   log(2.12  106)  5.67

Think about Your Answer   As predicted (Table 16.4), the solution after mixing equal amounts of a strong acid and weak base is weakly acidic.

Check Your Understanding  Calculate the pH after mixing 15 mL of 0.12 M acetic acid with 15 mL of 0.12 M NaOH. What are the major species in solution at equilibrium (besides water), and what are their concentrations?

Problem Solving Tip 16.2 What Is the pH After Mixing Equal Amounts (Moles) of an Acid and a Base?

Table 16.4 (page 794) summarizes the outcome of mixing various types of acids and bases. But how do you calculate a numerical value for the pH, particularly in the case of mixing a weak acid with a strong base or a weak base with a strong acid? The strategy (Example 16.8) is to recognize that this involves two calculations: a stoichiometry calculation and an equilibrium calculation. The

key is determining the concentration of the weak acid or weak base produced when the acid and base are mixed. Answering the following questions will guide you to an answer: (a) What amounts of acid and base are used (in moles)? (This is a stoichiometry problem.) (b) What is the total volume of the solution after mixing the acid and base solutions?

(c) What is the concentration of the weak acid or base produced on mixing the acid and base solutions? (d) Using the concentration found in Step (c), what is the hydronium ion concentration in the solution? (This is an equilibrium problem.) Determine the pH from the hydronium ion concentration.

16.8 Polyprotic Acids and Bases Goal for Section 16.8 • Use the equilibrium constant, Ka1 or Kb1, and other information to calculate the pH of a solution of a polyprotic acid or base.

Because polyprotic acids are capable of donating more than one proton, they pre­ sent additional challenges when predicting the pH of their solutions. For many inorganic polyprotic acids, such as phosphoric acid, carbonic acid, and hydrosulfuric acid (H2S), the ionization constant for each successive loss of a proton is about 104 to 106 smaller than the previous ionization step. This means that the first ionization step of a polyprotic acid produces up to about a million times more H3O+ ions than the second step. For this reason,  the pH of many inorganic polyprotic acids depends primarily on the hydronium ion generated in the first ionization

16.8  Polyprotic Acids and Bases

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805

step; the hydronium ion produced in the second step can be neglected.  The same principle applies to the fully deprotonated conjugate bases of polyprotic acids. This is illustrated by the calculation of the pH of a solution of carbonate ion, an important base in the environment.

© Charles D. Winters/Cengage

E xamp le 16.9

A polyprotic acid.  Malic acid, C4H6O5, is a diprotic acid occurring in apples. It is also classified as an alpha-hydroxy acid because it has an OH group on the C atom next to the CO2H (in the alpha position). It is one of a large group of natural alphahydroxy acids such as lactic acid, citric acid, and ascorbic acid. Alpha-hydroxy acids are touted as an ingredient in antiaging skin creams. They work by accelerating the natural process by which skin replaces the outer layer of cells with new cells.

Calculating the pH of the Solution of a Polyprotic Base Problem  The carbonate ion, CO32−, is a base in water, forming the hydrogen carbonate ion, which in turn can form carbonic acid. CO32−(aq) + H2O(ℓ) uv HCO3−(aq) + OH−(aq)

Kb1 = 2.1 × 10−4

HCO3−(aq) + H2O(ℓ) uv H2CO3(aq) + OH−(aq)

Kb2 = 2.4 × 10−8

What is the pH of a 0.10 M solution of Na2CO3?

What Do You Know?  You know the balanced equations and the values of Kb for the ions as well as the concentration of the carbonate ion.

Strategy  The first ionization constant, Kb1, is much larger than the second, Kb2, so the hydroxide ion concentration in the solution results almost entirely from the first step. Therefore, you can calculate the OH− concentration produced by considering only the first ionization step.

Solution  Set up an ICE table for the reaction of the carbonate ion (Equilibrium Table 1). Equilibrium Table 1—Reaction of CO32− Ion

Equilibrium Initial (M) Change Equilibrium (M)

CO32− + H2O

uv

HCO3−

+

OH−

0.10

  0

  0

−x

+x

+x

(0.10 − x)

  x

  x

Based on this table, the equilibrium concentration of OH− (= x) can then be calculated. K b1  2.1  104 

[HCO3][OH] x2  2 [CO3 ] 0.10  x

Because Kb1 is relatively small, it is reasonable to make the approximation that (0.10 − x) ≈ 0.10. Therefore, x  [HCO3]  [OH]  (2.1  104 )(0.10)  4.58  103 M Using this value of [OH−], you can calculate the pOH of the solution, pOH = −log (4.58 × 10−3) = 2.34 and then use the relationship pH + pOH = 14.00 to calculate the pH. pH = 14.00 − pOH =  11.66  The concentration of the carbonate ion is approximately 0.10 M. [CO32−] = 0.10 − 0.00458 ≈ 0.10 M At this point, you have answered the question that was asked in this problem, but you might wonder if the pH predicted is accurate since only the first ionization step was considered. The HCO3− ion produced in the first step could acquire another proton to give H2CO3, and this could affect the pH. But does this occur to a meaningful extent? To test this, set up a second ICE Table.

806

Chapter 16 / Principles of Chemical Reactivity: The Chemistry of Acids and Bases

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Equilibrium Table 2—Reaction of HCO3− Ion

Initial (M) Change (M) Equilibrium (M)

uv

H2CO3

+

OH−

4.58 × 10−3

0

4.58 × 10−3

−y

+y

+y

y

(4.58 × 10−3 + y)

(4.58 × 10

−3

− y)

Because Kb2 is so small, the second step occurs to a much smaller extent than the first step. This means the amount of H2CO3 and OH− produced in the second step (= y) is much smaller than 4.58 × 10−3 M. Therefore, it is reasonable that both [HCO3−] and [OH−] are very close to 4.58 × 10−3 M. K b2  2.4  108 

[H2CO3 ][OH] ( y)(4.58  103)  4.58  103 [HCO3]

Because [HCO3−] and [OH−] have nearly identical values, they cancel from the expression, and you find that [H2CO3] is equal to Kb2. y = [H2CO3] = Kb2 = 2.4 × 10−8 M The concentration of OH− produced in this reaction (= y) is negligible. All of the hydroxide ion is essentially produced in the first equilibrium process.

© Charles D. Winters/Cengage

HCO3− + H2O

Equilibrium

Sodium carbonate, a polyprotic base.  This common substance is a base in aqueous solution. Its primary use is in the glass industry. Although it used to be manufactured, it is now mined as the mineral trona, Na2CO3 ∙ NaHCO3 ∙ 2 H2O.

Think about Your Answer  It is almost always the case that the pH of a solution of a weak inorganic polyprotic acid is due to the hydronium ion generated in the first ionization step. Similarly, the pH of a polyprotic base is due to the OH− ion produced in the first hydrolysis step.

Check Your Understanding  What is the pH of a 0.10 M solution of oxalic acid, H2C2O4? What are the concentrations of H3O+, HC2O4−, and the oxalate ion, C2O42−? (See Appendix H for Ka values.)

16.9 M  olecular Structure, Bonding, and Acid–Base Behavior Goal for Section 16.9 • Appreciate the connection between the structure of a compound and its acidity or basicity.

One of the most interesting aspects of chemistry is the correlation between a molecule’s structure and bonding with its chemical properties. Because acids and bases play such a key role in chemistry, principles relating acid or base character to structure are useful. This correlation allows chemists to predict properties of newly discovered molecules.

Acid Strength of the Hydrogen Halides, HX Aqueous HF is a weak Brønsted acid in water, whereas the other hydrohalic acids—aqueous HCl, HBr, and HI—are all strong acids. Experiments show that the acid strength increases in the order HF Ka (or pH < pKa), then [HInd] > [Ind−]

•

when [H3O+] < Ka (or pH > pKa), then [HInd] < [Ind−]

Now apply these conclusions to, for example, the titration of an acid with a base using an indicator whose pKa value is nearly the same as the pH at the equivalence point (Figure 17.10). At the beginning of the titration, the pH is low and [H 3O+] is high; the acid form of the indicator (HInd) predominates, so its color is the one observed. As the titration progresses and the pH increases ([H3O+] decreases), less HInd and more Ind− exist in solution. Finally, just after the equivalence point is reached, [Ind−] is much larger than [HInd], and the color of [Ind−] is observed. Several questions remain to be answered. If you are trying to analyze for an acid and add an indicator that is a weak acid, won’t this affect the analysis? Recall that you use only a tiny amount of an indicator in a titration. Although the acidic indicator molecules also react with the base as the titration progresses, so little indicator is present that any error is not significant. Another question is whether you could accurately determine the pH by observing the color change of an indicator. In practice, your eyes are not quite that good. Assume, qualitatively, that you see the color of HInd when [HInd]/[Ind−] is about 10/1, and the color of Ind− when [HInd]/[Ind−] is about 1/10. This means the color change is observed over a hydronium ion concentration interval of about 2 pH units. However, as you can see in Figures 17.4–17.7, on passing through the equivalence point of these titrations, the pH changes by as many as 7 units. As Figure 17.11 shows, a variety of indicators is available, each changing color in a different pH range. If you are analyzing an acid or base by titration, you must

4

5

6

7

8

9

10

11

12

13

14

Crystal violet Cresol red Thymol blue Erythrosin B 2, 4-Dinitrophenol Bromophenol blue Methyl orange Bromocresol green Methyl red Eriochrome black T Bromocresol purple Alizarin Bromothymol blue Phenol red m-Nitrophenol o-Cresolphthalein Phenolphthalein Thymolphthalein Alizarin yellow GG

Figure 17.11  Common acid–base indicators.  ​The color changes occur over a range of pH values. Notice that a few indicators have color changes over two different pH ranges.

852

Chapter 17 / Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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choose an indicator that changes color in a range that includes the pH at the equivalence point. For the titration of a strong acid with a strong base, an indicator that changes color in the pH range 7 ± 2 should be used. On the other hand, the pH at the equivalence point in the titration of a weak acid with a strong base is greater than 7, and you should choose an indicator that changes color at a pH near the anticipated equivalence point.

17.3 Solubility of Salts Goals for Section 17.3 • Write the equilibrium constant expression relating concentrations of ions in solution to Ksp for any insoluble salt.

• Calculate the Ksp value for a salt from its solubility and calculate the solubility of a salt from its Ksp.

• Recognize how the presence of a common ion affects the solubility of a salt. • Understand how hydrolysis of basic anions affects the solubility of a salt. • Recognize that salts with anions derived from weak acids have increased solubility in acidic solutions.

Precipitation reactions (Section 3.5) are exchange reactions in which one of the products is a water-insoluble compound such as CaCO3, CaCl2(aq) + Na2CO3(aq) n CaCO3(s) + 2 NaCl(aq)

Photos: John C. Kotz

that is, a compound having a water solubility of less than about 0.01 mol of dissolved material per liter of solution (Figure 17.12). How do you know when to predict an insoluble compound as the product of a reaction? In Chapter 3, some guidelines are listed for predicting solubility (Figure 3.10). Now, by treating the solubility of salts as equilibrium systems, it is possible to quantify the amount of a salt that will dissolve in water. It is also possible to explore conditions under which some ionic compounds precipitate and others do not.

Crocoite, lead(II) chromate, PbCrO4

Rhodochrosite, manganese(II) carbonate, MnCO3

Figure 17.12  Some insoluble substances. ​



Copper(II) minerals: green malachite, CuCO3 ∙ Cu(OH)2, and blue azurite, 2 CuCO3 ∙ Cu(OH)2

17.3  Solubility of Salts

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853

The Solubility Product Constant, Ksp If some AgBr is placed in pure water, a tiny amount of the compound dissolves, and an equilibrium is established. AgBr(s) uv Ag+(aq) + Br−(aq)

When sufficient AgBr has dissolved and equilibrium is attained, the solution is said to be saturated (Section 13.2), and experiments show that the concentrations of the silver and bromide ions in the solution are each about 7.35 × 10−7 M at 25 °C. The extent to which an insoluble salt dissolves can be expressed in terms of the e­ quilibrium constant for the dissolving process. In this case, the appropriate expression is Ksp = [Ag+][Br−] Writing Equilibrium Constant Expressions  Solids are not

included in these equations.

The equilibrium constant that reflects the solubility of a compound is referred to as its solubility product constant. Chemists frequently use the notation Ksp for such constants, the subscript “sp” denoting a solubility product. The water solubility of a compound, and thus its Ksp value, can be estimated by determining the concentration of the cation or anion when the compound dissolves. For example, if you find that AgBr dissolves to give a silver ion concentration of 7.35 × 10−7 mol/L, you know that 7.35 × 10−7 mol of AgBr dissolved per liter of solution (and that the bromide ion concentration also equals 7.35 × 10−7 M). Therefore, the value of the equilibrium constant for AgBr is Ksp = [Ag+][Br−] = (7.35 × 10−7)(7.35 × 10−7) = 5.40 × 10−13 (at 25 °C)

Equilibrium constants for dissolving other insoluble salts can be calculated in the same manner. The solubility product constant, Ksp, for any salt always has the form AxBy(s) uv x Ay+(aq) + y Bx−(aq)      Ksp = [Ay+]x[Bx−]y 

(17.6)

For example, CaF2(s) uv Ca2+(aq) + 2 F−(aq)

Ksp = [Ca2+][F−]2 = 5.3 × 10−11

Ag2SO4(s) uv 2 Ag+(aq) + SO42−(aq)

Ksp = [Ag+]2[SO42−] = 1.2 × 10−5

The numerical values of Ksp for a few salts are given in Table 17.2, and more values are collected in Appendix J. Table 17.2

Some Common Insoluble Compounds and Their Ksp Values*

Formula

Name

Ksp (25 °C)

Common Names/Uses

CaCO3

Calcium carbonate

3.4 × 10−9

Calcite, Iceland spar

MnCO3 FeCO3 CaF2 AgCl

Manganese(II) carbonate Iron(II) carbonate Calcium fluoride Silver chloride

2.3 × 10

−11

Rhodochrosite (forms rose-colored crystals)

3.1 × 10

−11

Siderite

5.3 × 10

−11

Fluorite (source of HF and other inorganic fluorides)

1.8 × 10

−10

Chlorargyrite

−13

Used in photographic film

AgBr

Silver bromide

5.4 × 10

CaSO4

Calcium sulfate

4.9 × 10−5

Hydrated form is commonly called gypsum

BaSO4

Barium sulfate

1.1 × 10−10

Barite (used in “drilling mud” and as a component of paints)

SrSO4

Strontium sulfate

3.4 × 10−7

Celestite

Ca(OH)2

Calcium hydroxide

5.5 × 10

−6

Slaked lime

*The values in this table were taken from Lange’s Handbook of Chemistry, 15th ed., New York: McGraw-Hill Publishers, 1999. A ­ dditional Ksp values are given in Appendix J.

854

Chapter 17 / Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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Do not confuse the solubility of a compound with its solubility product constant. The solubility of a salt is the quantity or amount present in some volume of a saturated solution, expressed in grams per 100 mL, moles per liter, or other units. The solubility product constant is an equilibrium constant. However, there is a connection between them: If one is known, the other can, in principle, be calculated.

Relating Solubility and Ksp Solubility product constants are determined by careful laboratory measurements of the concentrations of ions in solution.

Exa m p le 17.8

Ksp from Solubility Measurements

Strategy Map Problem Calculate Ksp of CaF2 from its solubility.

Problem   Calcium fluoride, the main component of the mineral fluorite (page 580), ­dissolves to a slight extent in water. CaF2(s) uv Ca2+(aq) + 2 F−(aq)  Ksp = [Ca2+][F−]2 Calculate the Ksp value for CaF2 if the calcium ion concentration is determined experimentally to be 2.3 × 10−4 mol/L.

Data/Information • Concentration of Ca2+ • Balanced equation • Ksp expression

What Do You Know?  You know the balanced equation for the process, the Ca2+ concentration, and the Ksp expression.

Strategy Step 1.  Calculate the fluoride ion concentration from [Ca2+] and stoichiometry. Step 2. Insert the values for [F−] and [Ca2+] into the Ksp expression and calculate the value of Ksp.

Solution Step 1

Calculate the F− ion concentration. When CaF2 dissolves in water, the balanced equation shows that the concentration of F− must be twice the Ca2+ concentration. If [Ca2+] = 2.3 × 10−4 M, then [F−] = 2 × [Ca2+] = 4.6 × 10−4 M

Step 2

Insert the equilibrium concentrations into the Ksp expression and calculate the Ksp. Ksp = [Ca2+][F−]2 = (2.3 × 10−4)(4.6 × 10−4)2 =  4.9 × 10−11

Think about Your Answer   A common error is to forget the reaction stoichiometry. Be sure to notice that for every Ca2+ ion in solution there are two F− ions.

Check Your Understanding  The barium ion concentration, [Ba2+], in a saturated solution of barium fluoride is 3.6 × 10−3 M. Calculate the value of the Ksp for BaF2. BaF2(s) uv Ba2+(aq) + 2 F−(aq)

Ksp values for insoluble salts can be used to calculate the solubility of a solid salt (as shown in Examples 17.9 and 17.10) or to determine whether a solid will precipitate when solutions of its anion and cation are mixed.

17.3  Solubility of Salts

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855

(b) Barium sulfate is opaque to X-rays, so it is used by physicians to examine the digestive tract. A patient drinks a “cocktail” containing BaSO4, and the progress of the BaSO4 through the digestive organs can be followed by X-ray ­analysis. This photo is an X-ray image of the gastrointestinal tract after a person ingested barium sulfate.

Radiological imaging/Shutterstock.com

© Charles D. Winters/Cengage

(a) A sample of the mineral barite, which is mostly barium sulfate. Drilling mud, used in drilling oil wells, consists of clay, barite, and calcium carbonate.

Figure 17.13  Barium sulfate.  ​Barium sulfate, a white solid, is quite insoluble in water (Ksp = 1.1 × 10−10) (Example 17.9).

Exam pl e 1 7 .9

Solubility from Ksp Problem  The K sp for the mineral barite (BaSO 4, Figure 17.13) is 1.1 × 10 −10 at 25 °C. Calculate the solubility of barium sulfate in water in moles per liter and grams per liter.

Strategy Map Problem Calculate solubility of BaSO4 from Ksp.

What Do You Know?  You know the formula for the mineral and its Ksp value. Strategy   When BaSO4 dissolves, equimolar amounts of Ba2+ ions and SO42− ions are produced. Thus, the solubility of BaSO4 can be estimated by calculating the equilibrium concentration of either Ba2+ or SO42− from the solubility product constant. Step 1.  Write the balanced equation and the Ksp expression. Step 2. Set up an ICE table, designating the unknown concentrations of Ba2+ and SO42− in solution as x. Step 3.  Using x for [Ba2+] and [SO42−] in the Ksp expression, solve for x. Step 4. Convert the solubility of BaSO4 from mol/L to g/L using the molar mass of BaSO4.

Data/Information • Ksp for BaSO4 • Balanced equation

Solution Step 1

Write the balanced equation and the Ksp expression. The equation for the solubility of BaSO4 and its Ksp expression are BaSO4(s) uv Ba2+(aq) + SO42−(aq)  Ksp = [Ba2+][SO42−] = 1.1 × 10−10

Step 2

Set up an ICE table. Denote the solubility of BaSO4 (in mol/L) by x; that is, x moles of BaSO4 dissolve per liter. Therefore, both [Ba2+] and [SO42−] must also equal x at equilibrium.

Equation

856

BaSO4(s)

uv

Ba2+(aq)

+

SO42−(aq)

Initial (M)

  0

  0

Change (M)

+x

+x

Equilibrium (M)

  x

  x

Chapter 17 / Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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Step 3

Solve the Ksp expression for x. Because Ksp is the product of the barium ion and sulfate ion concentrations, K sp is the square of the solubility, x, Ksp = [Ba2+][SO42−] = 1.1 × 10−10 = (x)(x) = x2 and so the value of x is x  [Ba2]  [SO42] 

K sp  1.1  1010  1.05  105 M  1.0  105 M

The  solubility of BaSO4 in pure water is 1.0 × 10−5 mol/L.  Step 4

Determine the solubility of BaSO4 in units of g/L. To find its solubility in g/L, multiply by the molar mass of BaSO4. Solubility in g/L = (1.05 × 10−5 mol/L)(233 g/mol) =  0.0024 g/L

Think about Your Answer  As noted in Figure 17.13, BaSO4 is used to investigate the digestive tract. It is fortunate the compound is so insoluble, because water- and acidsoluble barium salts are toxic.

Check Your Understanding  Calculate the solubility of AgCN in moles per liter and grams per liter. Ksp for AgCN is 6.0 × 10−17.

Ex am p le 17.10

Solubility from Ksp Problem  Knowing that the Ksp value for MgF2 is 5.2 × 10−11, calculate the solubility of the salt in moles per liter and grams per liter.

What Do You Know?   You know the formula for magnesium fluoride and its Ksp value.

Strategy   The solubility must be defined in terms that will allow you to solve the Ksp expression for the concentration of one ion. From stoichiometry, you can say that if x mol of MgF2 dissolves, then x mol of Mg2+ and 2x mol of F− appear in solution. This means the MgF2 solubility (in moles dissolved per liter) is equivalent to the concentration of Mg2+ ions in the solution. Step 1.  Write the balanced equation and the Ksp expression. Step 2.  S et up an ICE table, designating the unknown concentrations of Mg2+ and F− in solution as x and 2x, respectively. Step 3.  Using x for [Mg2+] and 2x for [F−] in the Ksp expression, solve for x. Step 4.  Convert the solubility of MgF2 from mol/L to g/L using the molar mass of MgF2.

Solution Step 1.  Write the balanced equation and the Ksp expression. MgF2(s) uv Mg2+(aq) + 2 F−(aq)  Ksp = [Mg2+][F−]2 = 5.2 × 10−11 Step 2.  Set up an ICE table. The concentrations of Mg2+ and F− in solution are x and 2x, respectively.



17.3  Solubility of Salts

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857

Equation

MgF2(s)

uv

Mg2+(aq)

+

2 F−(aq)

Initial (M)

  0

  0

Change (M)

+x

+2x

Equilibrium (M)

  x

  2x

Step 3.  Solve the Ksp expression for x. Substituting the equilibrium concentrations for [Mg2+] and [F−] into the Ksp expression, you find Ksp = [Mg2+][F−]2 = (x)(2x)2 = 4x3 Solving the equation for x, x

3

K sp  4

3

5.2  1011  2.35  104  2.4  104 4

you find that  2.4  × 10−4 moles of MgF2 dissolve per liter.  Step 4.  Calculate the solubility in units of g/L. Multiply by the molar mass of MgF2 to convert mol/L to g/L. (2.35 × 10−4 mol/L)(62.3 g/mol) =  0.015 g MgF2/L 

Think about Your Answer   Problems like this one might provoke the question, “Aren’t you counting things twice when you multiply x by 2 and then square it as well?” in the expression Ksp = (x)(2x)2. The answer is no. The 2 in the 2x term is based on the stoichiometry of the compound. The exponent of 2 on the F− ion concentration arises from the rules for writing equilibrium expressions.

Check Your Understanding  Calculate the solubility of Ca(OH)2 in moles per liter and grams per liter (Ksp = 5.5 × 10−6).

The relative solubilities of salts can often be deduced by comparing values of solubility product constants, but be careful! For example, the Ksp value for silver chloride is AgCl(s) uv Ag+(aq) + Cl−(aq)  Ksp = 1.8 × 10−10

and the Ksp for silver iodide is AgI(s) uv Ag+(aq) + I−(aq)     Ksp = 8.5 × 10−17

The calculated solubility of silver chloride is 1.3 × 10−5 mol/L and that for silver iodide is 9.2 × 10−9 mol/L. Thus for these two salts with the same 1:1 ratio of ions, the one with the lower Ksp value has the lower solubility. On the other hand, the Ksp for silver chromate is Ag2CrO4(s) uv 2 Ag+(aq) + CrO42−(aq)  Ksp = 9.0 × 10−12

In spite of the fact that Ag2CrO4 has a smaller numerical Ksp value than AgCl, the chromate salt is about 10 times more soluble than the chloride salt. If you determine solubilities from Ksp values as in the examples above, you would find the solubility of AgCl is 1.3 × 10−5 mol/L, whereas that of Ag2CrO4 is 1.3 × 10−4 mol/L. From this you can draw the following conclusion: Direct comparisons of the solubility of two salts on the basis of their Ksp values can be made only for salts having the same cation-to-anion ratio.

858

Chapter 17 / Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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Minerals and gems are among nature’s most beautiful creations. Many, such as rubies, are metal oxides, and the various types of quartz are based on silicon dioxide. Another large class of gemstones consists largely of metal silicates. These include emerald, topaz, aquamarine, and tourmaline. Carbonates represent another large class of minerals and a few gemstones. Rhodochrosite, one of the most beautiful red stones, is manganese(II) carbonate. And one of the most abundant minerals on Earth is limestone, calcium carbonate, which is also a major component of sea shells and corals. Hydroxides are represented by azurite, which is a mixed carbonate/hydroxide with the

formula Cu3(OH)2(CO3)2. Turquoise is a mixed hydroxide/phosphate based on copper(II), the source of the blue color of turquoise. Among the most common minerals are sulfides such as golden iron pyrite (FeS2), black stibnite (Sb2S3), red cinnabar (HgS), and yellow orpiment (As2S3). Other smaller classes of minerals exist; one of the smallest is the class based on the halides, and the best example is fluorite. Fluorite, CaF2, exhibits a wide range of colors from purple to green to yellow. What do all of these minerals and gems have in common? They are all insoluble or poorly soluble in water. If they were more soluble, they would be dissolved in the world’s lakes and oceans.

John C. Kotz

A Closer Look

Minerals and Gems—The Importance of Solubility

Mineral samples (clockwise from upper right): golden iron pyrite, green fluorite, yellow orpiment, black (needles) stibnite, and (center) blue azurite. ​Formulas are in the text (see also Figure 17.12).

This means, for example, that you can directly compare solubilities of 1:1 salts such as the silver halides by comparing their Ksp values. AgI (Ksp = 8.5 × 10−17) < AgBr (Ksp = 5.4 × 10−13) < AgCl (Ksp = 1.8 × 10−10) increasing Ksp and increasing solubility

Similarly, you could compare 1:2 salts such as the lead halides, PbI2 (Ksp = 9.8 × 10−9) < PbBr2 (Ksp = 6.6 × 10−6) < PbCl2 (Ksp = 1.7 × 10−5) increasing Ksp and increasing solubility

but you cannot directly compare the solubility of a 1:1 salt (AgCl) with a 2:1 salt (Ag2CrO4) by only comparing the Ksp values.

Solubility and the Common Ion Effect The test tube on the left in Figure 17.14 contains a precipitate of silver acetate, AgCH3CO2, in water. The solution is saturated, and the silver ions and acetate ions in the solution are in equilibrium with solid silver acetate. AgCH3CO2(s) uv Ag+(aq) + CH3CO2−(aq) AgNO3 added

Ag+

Ag+

CH CO − 3 2 + Ag

+

Ag −

−

CH3CO2 −

3CO2

Ag+

CH CO − 3 2

Ag+ CH3CO2− Ag+ CH3C −

CH3CO2

Ag+

−

CH3CO2

Ag

+

Ag

+

Ag

−

CH3CO2

+ Ag

NO3 −

CH 3CO 2 −

Ag

−

NO 3

Ag+

−

O2 H 3C

C

CH 3CO 2

−

CH 3CO 2

NO

Ag +

Ag+

Ag+ 3 −

C

CH3CO2− Ag

−

CH3CO2

+

Ag

Ag+

CH3CO2− Ag

−

3CO2

Ag+

Ag+ CH3CO2− Ag+ CH3C 3CO2− Ag+ CH3CO2− Ag+ CH3C CH3CO2− Ag+ CH3CO2− Ag Ag+ CH3CO2− Ag+ CH3CO2− Ag

Photos: © Charles D. Winters/Cengage

−

NO 3

−

O2 H 3C

More solid silver acetate forms

Figure 17.14  The common ion effect.  ​The tube (left) contains a saturated solution of silver acetate, AgCH3CO2. When 1.0 M AgNO3 is added to the tube (right), more solid silver acetate forms.

17.3  Solubility of Salts

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859

A Closer Look

Solubility Calculations The K sp value reported for lead(II) chloride, PbCl2, is 1.7 × 10−5. If you assume the appropriate equilibrium in solution is

PbCl+(aq) + Cl−(aq)

undissociated salt dissolved in water

PbCl2(s) uv Pb2+(aq) + 2 Cl−(aq)

the calculated solubility of PbCl2 is 0.016 M. The experimental value for the solubility of the salt, however, is 0.036 M, more than twice the calculated value! The problem is that  chemical behavior is often much more complicated than the e­ quation defining Ksp. The main problem in the lead(II) chloride case, and in many others, is that the compound dissolves but is not 100% dissociated into its constituent ions. Instead, some dissolves as the undissociated salt or forms ion pairs. Other problems that lead to discrepancies between calculated and

K = 0.63

PbCl2(aq)

ion pairs

K = 0.0011

PbCl2(s)

K = 0.026 Ksp = 1.7 × 10−5

Pb2+(aq) + 2 Cl−(aq)

slightly soluble salt

experimental solubilities are the reactions of ions (particularly anions) with water and complex ion formation. Complex ion formation is illustrated by the fact that lead chloride is more soluble in the presence of excess chloride ion, owing to the formation of the complex ion PbCl42−.

PbCl2(s) + 2 Cl−(aq) uv PbCl42−(aq)

100% dissociated into ions

Hydrolysis and complex ion formation are described in more detail later (page 861). References • S. J. Hawkes, Journal of Chemical Education, 1998, 75, 1179–1181. • R. W. Clark and J. M. Bonicamp, Journal of Chemical Education, 1998, 75, 1182–1185.

But what happens if the silver ion concentration is increased, say by adding silver nitrate? Le Chatelier’s principle (Section 15.6) suggests—and observations c­ onfirm— that more silver acetate precipitate should form because silver ion was added, causing the equilibrium to shift to form more silver acetate. The effect of adding silver ions to a saturated silver acetate solution is another example of the common ion effect. Adding a common ion to a saturated solution of a salt lowers the salt solubility (unless a complex ion can form; see “A Closer Look: Solubility Calculations”).

E xamp le 17.11

The Common Ion Effect and Salt Solubility Problem  If solid AgCl is placed in 1.00 L of 0.55 M NaCl, what mass of AgCl will dissolve? What Do You Know?  You know the formula of the insoluble compound, its Ksp (Table 17.2 and Appendix J), and its molar mass. You also know that the presence of an ion common to the equilibrium (Cl−) suppresses the solubility of AgCl.

Strategy Step 1.  Write the balanced equation and the Ksp expression for AgCl. Step 2.  Set up an ICE table and enter 0.55 M as the initial concentration of the common ion, Cl−. Define the solubility of AgCl as x. This is the increase in the concentration of Ag+ and Cl− in solution as AgCl dissolves. Step 3.  Solve the Ksp expression for x. Step 4.  Determine the solubility of AgCl in units of g/L.

Solution Step 1.  Write a balanced equation and the Ksp expression.

AgCl(s) uv Ag+(aq) + Cl−(aq)  Ksp = [Ag+][Cl−] = 1.8 × 10−10

860

Chapter 17 / Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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Step 2.  Set up an ICE table. Be sure to include the initial concentration of Cl− on the Initial (I) line.

Equation

AgCl(s)

uv

Ag+(aq)

+

Cl−(aq)

Initial (M)

  0

  0.55

Change (M)

+x

+x

Equilibrium (M)

  x

  0.55 + x

Step 3.  Solve the Ksp expression for x. The equilibrium concentrations from the table are substituted into the Ksp expression, Ksp = 1.8 × 10−10 = [Ag+][Cl−] = (x)(0.55 + x) This is a quadratic equation and can be solved by the methods in Appendix A. An easier approach, however, is to make the approximation that x is very small with respect to 0.55 [and so (0.55 + x) ≈ 0.55]. This is a reasonable assumption because you know that the solubility is very small without the common ion Cl− and that it will be even smaller in the presence of added Cl−. Therefore, Ksp = 1.8 × 10−10 = (x)(0.55) x = [Ag+] = 3.27 × 10−10 M As predicted by Le Chatelier’s principle, the solubility of AgCl in the presence of added Cl− is much less (3.3 × 10−10 M) than in pure water (1.3 × 10−5 M). Step 4.  Determine the solubility of AgCl in units of g/L. To find its solubility in g/L, multiply by the molar mass of AgCl. (3.27 × 10−10 mol/L)(143 g/mol) =  4.7 × 10−8 g/L 

Think about Your Answer   The approximation made here is similar to the approximations made in acid–base equilibrium problems. As a final step, you should check its validity by substituting the calculated value of x into the exact expression Ksp = (x) (0.55 + x). If the product (x)(0.55 + x) is the same as the given value of Ksp, the approximation is valid. Ksp = (x)(0.55 + x) = (3.3 × 10−10)(0.55 + 3.3 × 10−10) = 1.8 × 10−10

Check Your Understanding  Calculate the solubility of BaSO4 (a) in pure water and (b) in the presence of 0.010 M Ba(NO3)2. Ksp for BaSO4 is 1.1 × 10−10. 

There are two important general ideas from Example 17.11: •

The solubility of a salt is reduced by the presence of a common ion, in accordance with Le Chatelier’s principle.

•

Assuming that the amount of common ion added to the solution is very large compared to the amount of that ion coming from the insoluble salt simplifies the calculations. This is almost always the case, but you should check to be sure.

The Effect of Basic Anions on Salt Solubility The next time you are tempted to wash a supposedly insoluble salt down the kitchen or laboratory drain, stop and consider the consequences. Many metal ions such as lead, chromium, and mercury are toxic in the environment. Even if a so-called insoluble salt of one of these cations does not appear to dissolve, its solubility in



17.3  Solubility of Salts

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861

© Charles D. Winters/Cengage

Figure 17.15  Lead(II) sulfide (galena).  This and other metal sulfides dissolve in water to a greater extent than expected because the sulfide ion reacts with water to form HS− and OH−. PbS(s) + H2O(ℓ) uv Pb2+(aq) + HS−(aq) + OH−(aq) The model of PbS shows that the unit cell is cubic, a feature reflected by the cubic crystals of the mineral galena.

water may be greater than you think, in part owing to the possibility that the anion of the salt is a weak base or the cation is a weak acid. Lead(II) sulfide, PbS, which is found in nature as the mineral galena (­Figure 17.15), demonstrates the effect of the acid–base properties of an ion on salt solubility. When placed in water, a trace amount dissolves, PbS(s) uv Pb2+(aq) + S2−(aq)

and one product of the reaction is the sulfide ion, which is itself a strong base. S2−(aq) + H2O(ℓ) uv HS−(aq) + OH−(aq)  Kb1 = 1 × 105

The sulfide ion undergoes hydrolysis (reaction with water), which decreases its concentration, and the equilibrium process for dissolving PbS shifts to the right. Thus, the lead ion concentration in solution is greater than expected from the simple dissociation of the salt. The lead(II) sulfide example leads to the following general observation: Any salt containing an anion that is the conjugate base of a weak acid will dissolve in water to a greater extent than given by Ksp.

This means that salts of phosphate, acetate, carbonate, and cyanide, as well as sulfide, can be affected, because all of these anions undergo the general hydrolysis reaction X−(aq) + H2O(ℓ) uv HX(aq) + OH−(aq)

The observation that ions from insoluble salts can undergo hydrolysis is related to another useful, general conclusion: Metal Sulfide Solubility  The true

solubility of a metal sulfide is better represented by a modified solubility product constant, K ′sp, which is defined as:

MS(s) uv M2+(aq) + S2−(aq)  Ksp = [M2+][S2−] S2−(aq) + H2O(ℓ) uv  HS−(aq) + OH−(aq)  Kb = [HS−][OH−]/[S2−]

Net reaction: MS(s) + H2O(ℓ) uv  HS−(aq) + M2+(aq) + OH−(aq)  K ′sp = [M2+][HS−][OH−] = Ksp × Kb

Values for K ′sp for several metal sulfides are included in ­Appendix J (Table J.2).

Insoluble salts in which the anion is the conjugate base of a weak acid dissolve in strong acids.

Insoluble salts containing anions such as acetate, carbonate, hydroxide, phosphate, and sulfide dissolve in strong acids. For example, you know that if a strong acid is added to a water-insoluble metal carbonate such as CaCO3, the salt dissolves (Section 3.7). CaCO3(s) + 2 H3O+(aq) n Ca2+(aq) + 3 H2O(ℓ) + CO2(g)

You can think of this as the result of a series of reactions. CaCO3(s) uv Ca2+(aq) + CO32−(aq)

Ksp = 3.4 × 10−9

CO32−(aq) + H3O+(aq) uv HCO3−(aq) + H2O(ℓ)

1/Ka2 = 1/4.8 × 10−11 = 2.1 × 1010

HCO3−(aq) + H3O+(aq) uv H2CO3(aq) + H2O(ℓ)

1/Ka1 = 1/4.2 × 10−7 = 2.4 × 106

Overall: CaCO3(s) + 2 H3O+(aq) uv Ca2+(aq) + 2 H2O(ℓ) + H2CO3(aq) Knet = (Ksp)(1/Ka2)(1/Ka1) = 1.7 × 108

Carbonic acid, a product of this reaction, is not stable, H2CO3(aq) uv CO2(g) + H2O(ℓ)  K ≈ 105

and you see CO 2 bubbling out of the solution, a process that moves the CaCO3 + H3O+ equilibrium even further to the right. Calcium carbonate dissolves completely in strong acid! Many metal sulfides are also soluble in strong acids FeS(s) + 2 H3O+(aq) uv Fe2+(aq) + H2S(aq) + 2 H2O(ℓ)

as are metal phosphates (Figure 17.16), Ag3PO4(s) + 3 H3O+(aq) uv 3 Ag+(aq) + H3PO4(aq) + 3 H2O(ℓ)

and metal hydroxides. Mg(OH)2(s) + 2 H3O+(aq) uv Mg2+(aq) + 4 H2O(ℓ)

862

Chapter 17 / Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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Photos: © Charles D. Winters/Cengage

Add strong acid.

precipitate of AgCl and Ag3PO4

Figure 17.16  The effect of the anion on salt solubility in acid.  ​(left) A precipitate of AgCl (white) and Ag3PO4 (yellow). (right) Adding a strong acid (HNO3) dissolves Ag3PO4 (and leaves insoluble AgCl). The basic anion PO43− reacts with acid to give H3PO4, whereas Cl− is too weakly basic to form HCl.

precipitate of AgCl

In general, the solubility of a salt containing the conjugate base of a weak acid is increased by addition of a stronger acid to the solution. In contrast, salts are not soluble in strong acid if the anion is the conjugate base of a strong acid. For example, AgCl is not soluble in strong acid AgCl(s) uv Ag+(aq) + Cl−(aq)

Ksp = 1.8 × 10−10

H3O+(aq) + Cl−(aq) uv HCl(aq) + H2O(ℓ)

K Ksp). 3. If Q > Ksp, the solution is oversaturated and precipitation will occur.

If Q > Ksp, then precipitation will occur until Q = Ksp.

E xamp le 17.12

Solubility and the Reaction Quotient Problem   Solid AgCl has been placed in a beaker of water. After some time, the concentrations of Ag+ and Cl− are each 1.2 × 10−5 mol/L. Has the system reached equilibrium? If not, will more AgCl dissolve? What Do You Know?  You know the balanced equation for dissolving the AgCl, its value of Ksp (Appendix J), and the Ag+ and Cl− concentrations.

Strategy Step 1. Write the balanced equation for dissolving the salt and the equilibrium constant expression. The reaction quotient expression, Q, will have the same format but use the concentrations actually present. Step 2. Enter the concentrations of Ag+ and Cl− into the reaction quotient expression and solve for Q. Step 3. Compare Q to Ksp to decide if the system is at equilibrium (that is, if Q = Ksp). If not, determine the direction the reaction proceeds.

864

Chapter 17 / Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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Solution Step 1.  Write the balanced equation and the equilibrium constant expression. AgCl(s) uv Ag+(aq) + Cl−(aq)  Ksp = [Ag+][Cl−] = 1.8 × 10−10 Step 2.  Enter the concentrations in the reaction quotient expression and solve for Q. Q = [Ag+][Cl−] = (1.2 × 10−5)(1.2 × 10−5) = 1.4 × 10−10 Step 3. Decide if the system is in equilibrium. If not, determine the direction the reaction proceeds. Here, Q (1.4 × 10−10) is less than Ksp (1.8 × 10−10). The solution is not yet saturated, and AgCl will continue to dissolve until Q = Ksp.

Think about Your Answer   Dissolving is often a fairly slow process. If you measure the ion concentrations at a particular time, they may not have yet reached an equilibrium value, as is the case here.

Check Your Understanding  Solid PbI2 (Ksp = 9.8 × 10−9) is placed in a beaker of water. After a period of time, the lead(II) concentration is measured and found to be 1.1 × 10−3 M. Has the system reached equilibrium? That is, is the solution saturated? If not, will more PbI2 dissolve?

Ksp, the Reaction Quotient, and Precipitation Reactions Using the reaction quotient and the solubility product constant, you can decide (1) if a precipitate will form when the ion concentrations are known or (2) what concentrations of ions are required to begin the precipitation of an insoluble salt. Suppose the concentration of magnesium ion in an aqueous solution is 1.5 × 10−6 M. If enough NaOH is added to make the solution 1.0 × 10−4 M in hydroxide ion, OH−, will precipitation of Mg(OH)2 occur (Ksp = 5.6 × 10−12)? If not, will it occur if the concentration of OH− is increased to 1.0 × 10−2 M? One strategy is similar to that in Example 17.12. That is, use the ion concentrations to calculate the value of Q and then compare Q with Ksp to decide if the system is at equilibrium. Begin with the equation for the dissolution of Mg(OH)2. Mg(OH)2(s) uv Mg2+(aq) + 2 OH−(aq)

When the concentrations of magnesium and hydroxide ions are the first ones stated above, Q is less than Ksp. Q = [Mg2+][OH−]2 = (1.5 × 10−6)(1.0 × 10−4)2 = 1.5 × 10−14 Q (1.5 × 10−14) < Ksp (5.6 × 10−12)

This means the solution is not yet saturated, and precipitation does not occur. When [OH−] is increased to 1.0 × 10−2 M, the reaction quotient is now larger than Ksp. Q = (1.5 × 10−6)(1.0 × 10−2)2 = 1.5 × 10−10 Q(1.5 × 10−10) > Ksp (5.6 × 10−12)

Precipitation of Mg(OH)2 occurs and will continue until the Mg2+ and OH− ion concentrations have decreased to the point where their product is equal to Ksp. The next example shows a similar problem: deciding how much of the precipitating agent is required to begin the precipitation of an ion at a given concentration level.



17.4  Precipitation Reactions

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Exampl e 1 7 .13

Ion Concentrations Required to Begin Precipitation

Strategy Map Problem What concentration of SO4 22 is required to begin precipitation of BaSO4?

Problem  The concentration of barium ion, Ba2+, in a solution is 0.010 M. (a) What concentration of sulfate ion, SO 42−, is required to begin the precipitation of BaSO4? (b) When enough SO42− has been added so that the concentration of sulfate ion in the solution reaches 0.015 M, what concentration of barium ion will remain in solution?

What Do You Know?   There are three terms in the Ksp expression: Ksp and the

Data/Information • Ksp value for BaSO4 • Initial [Ba2+]

anion and cation concentrations. Here, you know Ksp (1.1 × 10−10) and one of the ion concentrations. You can calculate the other ion concentration.

Strategy Step 1.    Write the balanced equation for dissolving BaSO4 and the Ksp expression. Step 2(a).  Use the Ksp expression to calculate [SO42−] when [Ba2+] = 0.010 M. Step 2(b).  Use the Ksp expression to calculate [Ba2+] when [SO42−] = 0.015 M.

Solution Step 1

Write the balanced equation for dissolving BaSO4 and the Ksp expression. BaSO4(s) uv Ba2+(aq) + SO42−(aq)  Ksp = [Ba2+][SO42−] = 1.1 × 10−10

Step 2(a)

Solve the equilibrium expression for [SO42−] when the [Ba2+] = 0.010 M. When the product of the ion concentrations exceeds the Ksp (= 1.1 × 10−10)—that is, when Q > Ksp—precipitation will occur. The Ba2+ concentration is known (0.010 M), so the SO42− ion concentration necessary for precipitation can be calculated. [SO42] 

K sp 1.1  1010   1.1 × 10−8 M  2 [Ba ] 0.010

If the sulfate ion concentration is slightly greater than 1.1 × 10−8 M, BaSO4 will begin to precipitate. Step 2(b)

Solve the equilibrium expression for [Ba2+] when the [SO42−] = 0.015 M. If the sulfate ion concentration is increased to 0.015 M, the maximum concentration of Ba2+ ion that can be present in solution (in equilibrium with BaSO4) is [Ba2] 

K sp 1.1  1010   7.3 × 10−9 M 2 [SO4 ] 0.015

Think about Your Answer   The fact that the barium ion concentration is so small when [SO42−] = 0.015 M means that the Ba2+ ion has been essentially removed from solution. (It began at 0.010 M and has decreased by a factor of about 1 million.)

Check Your Understanding  What is the minimum concentration of I− that can cause precipitation of PbI2 from a 0.050 M solution of Pb(NO3)2? Ksp for PbI2 is 9.8 × 10−9. What concentration of Pb2+ ions remains in solution when the concentration of I− is 0.0015 M?

866

Chapter 17 / Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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Ex am p le 17.14

Ksp and Precipitations Problem   Suppose you mix 100.0 mL of 0.0200 M BaCl2 with 50.0 mL of 0.0300 M Na2SO4. Will BaSO4 (Ksp = 1.1 × 10−10) precipitate?

What Do You Know?   You know the concentration of two different compounds in solutions of differing volumes. You know that BaCl2 and Na2SO4 will combine to give a precipitate, BaSO4, if they are mixed in sufficient concentration. You also know the Ksp value for BaSO4. Strategy  Here, you mix two solutions, one containing Ba2+ ions and the other SO42− ions, both with known concentration and volume, and insoluble BaSO4 may be formed. Step 1. Find the concentration of each of the ions after mixing (assuming no precipitation). Step 2. Write the balanced equation for dissolving BaSO4 and the Ksp expression. Step 3. Knowing the ion concentrations in the combined solution, calculate Q and ­compare it with the Ksp value for BaSO4 to decide if BaSO4 will precipitate under these circumstances.

Solution Step 1. Find the concentrations of Ba2+ and SO422 after mixing. Use the equation c1V1 = c2V2 (Section 4.5) to calculate c2, the concentration of the Ba2+ and SO42− ions after mixing, to give a new solution with a volume of 150.0 mL (= V2). [Ba2] after mixing  [SO42] after mixing 

(0.0200 mol/L)(0.1000 L)  0.01333 M 0.1500 L (0.0300 mol/L)(0.0500 L)  0.01000 M 0.1500 L

Step 2.  Write the balanced equation for dissolving BaSO4 and the Ksp expression. BaSO4(s) uv Ba2+(aq) + SO42−(aq)  Ksp = [Ba2+][SO42−] = 1.1 × 10−10 Step 3.  Calculate Q and compare it with the Ksp value for BaSO4. Q = [Ba2+][SO42−] = (0.01333)(0.01000) = 1.33 × 10−4  Q is much larger than Ksp, so BaSO4 precipitates. 

Think about Your Answer The Ksp value for BaSO4 is very small, so mixing solutions with Ba2+ and SO42− ions, even in very low concentration, can lead to the precipitation of BaSO4.

Check Your Understanding  You have 100.0 mL of 0.0010 M silver nitrate. Will AgCl precipitate if you add 5.0 mL of 0.025 M HCl?

Complex Ions  Complex ions are

17.5 Equilibria Involving Complex Ions Goal for Section 17.5 • Understand the formation of complex ions in solution and recognize how this can increase the solubility of an insoluble salt.

prevalent in chemistry and are the basis of such biologically important substances as hemoglobin and vitamin B12. They are described in more detail in Chapter 22. See also Section 16.10.

17.5  Equilibria Involving Complex Ions

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© Charles D. Winters/Cengage

Dimethylglyoximate complex of Ni2+ ion

[Ni(NH3)6]2+ [Ni(H2O)6]2+

Figure 17.18  Complex ions.  ​

The green solution contains soluble [Ni(H2O)6]2+ ions in which water molecules are bound to Ni2+ ions by ion–dipole forces. This complex ion gives the solution its green color. The Ni2+ammonia complex ion is purple. The red, insoluble solid is the dimethylglyoximate complex of the Ni2+ ion [Ni(C4H7O2N2)2] (model at top). Formation of this beautiful red insoluble compound is the classical test for the presence of the aqueous Ni2+ ion.

Metal ions exist in aqueous solution as complex ions (Section 16.10). ­Complex ions consist of the metal ion and other molecules or ions bound into a single entity. In water, metal ions are always surrounded by water molecules, with the negative end of the polar water molecule, the oxygen atom, attracted to the positive metal ion. In the case of Ni 2+, the ion exists as [Ni(H 2O) 6] 2+ in water. Upon adding ammonia, water molecules are displaced successively, and in the presence of a high enough concentration of ammonia, the complex ion [Ni(NH3)6]2+ is formed. Many organic molecules also form complex ions with metal ions, one example being the complex with the dimethylglyoximate ion in Figure 17.18. The molecules or ions that bind to metal ions are called ligands (Chapter 22). In aqueous solution, metal ions and ligands exist in equilibrium, and the e­ quilibrium constants for these reactions are referred to as formation constants, Kf (­Appendix K). For example, Cu2+(aq) + NH3(aq) uv [Cu(NH3)]2+(aq)

Kf1 = 2.0 × 104

[Cu(NH3)]2+(aq) + NH3(aq) uv [Cu(NH3)2]2+(aq)

Kf2 = 4.7 × 103

[Cu(NH3)2]2+(aq) + NH3(aq) uv [Cu(NH3)3]2+(aq)

Kf3 = 1.1 × 103

[Cu(NH3)3]2+(aq) + NH3(aq) uv [Cu(NH3)4]2+(aq)

Kf4 = 2.0 × 102

In these reactions, Cu 2+ begins as [Cu(H 2 O) 4 ] 2+ , but ammonia successively displaces the water molecules. Overall, the formation of the tetraammine copper(II) complex ion has an equilibrium constant of 2.1 × 1013 (= Kf1 × Kf2 × Kf3 × Kf4). Cu2+(aq) + 4 NH3(aq) uv [Cu(NH3)4]2+(aq)

Kf = 2.1 × 1013

E xamp le 17.15

Complex Ion Equilibria Problem   What is the concentration of Cu2+ ions in a solution prepared by adding 0.00100 mol of Cu(NO3)2 to 1.00 L of 1.50 M NH3? Kf for the copper-ammonia complex ion [Cu(NH3)4]2+ is 2.1 × 1013.

What Do You Know?  Here you know the concentration of the species that form the complex ion (Cu2+ and NH3), and you know the formation constant for the complex ion.

Strategy   The formation constant for the complex ion is very large, so you start with the assumption that all of the Cu2+ ions react with NH3 to form [Cu(NH3)4]2+. That is, the initial concentration of the complex ion [Cu(NH3)4]2+ is 0.00100 M. This cation then dissociates to produce Cu2+ ions and additional NH3 in solution. The equilibrium constant for the dissociation of [Cu(NH3)4]2+ is the reciprocal of Kf because the dissociation of the ion is the reverse of its formation. Step 1. W  rite a balanced equation for the dissociation of the complex ion that formed in solution and set up an ICE table. Step 2. A  ssume that all of the Cu2+ ions in the solution are in the form of [Cu(NH3)4]2+ (0.00100 M). This means that [NH3] = original concentration − 4 × 0.00100 M. Step 3. A  ssume the concentration of complex ion dissociated at equilibrium is x, so x mol/L of Cu2+ are released to solution as are 4x mol/L of NH3. Step 4. U  se the equilibrium concentrations of the ions in the expression for Kdissociation (= 1/Kf ) and solve for x (which is the concentration of Cu2+ at equilibrium).

868

Chapter 17 / Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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Solution  Set up an ICE table for the dissociation of [Cu(NH3)4]2+. Cu2+(aq) + 4 NH3(aq)

Equation

[Cu(NH3)4]2+(aq)

Initial (M)

  0.00100

  0

  1.50 − 0.00400 M

Change (M)

−x

+x

+4x

Equilibrium (M)

  0.00100 − x ≈ 0.00100

  x

  1.50 − 0.00400  + 4x  ≈ 1.50

uv

Here you assume that x is so small that the concentration of the complex ion is very nearly 0.00100 M and that the NH3 concentration at equilibrium is essentially what was there originally. K dissociation 

1 1 [Cu2][NH3 ]4 ( x)(1.50)4    2.1  1013 0.00100 Kf [[Cu(NH3)4 ]2]

x =  [Cu2+] = 9.4 × 10−18 M 

Think about Your Answer  Make sure to test your assumption that x is so small it can be neglected in determining the equilibrium concentrations of [Cu(NH3)4]2+ and NH3. It certainly is in this case.

Check Your Understanding  Silver nitrate (0.0050 mol) is added to 1.00 L of 1.00 M NH3. What is the concentration of Ag+ ions at equilibrium? Ag+(aq) + 2 NH3(aq) uv [Ag(NH3)2]+(aq)  Kf = 1.1 × 107

Solubility and Complex Ions Silver chloride does not dissolve either in water or in strong acid, but it does dissolve in ammonia because it forms a water-soluble complex ion, [Ag(NH 3) 2] + (Figure 17.19).

AgCl(s), Ksp = 1.8 × 10−10 (a) AgCl precipitates upon adding NaCl(aq) to AgNO3(aq).

NaBr(aq)

[Ag(NH3)2]+(aq)

(b) The precipitate of AgCl dissolves upon adding aqueous NH3 to give water-soluble [Ag(NH3)2]+.

Na2S2O3(aq)

AgBr(s), Ksp = 5.4 × 10−13 (c) The silver-ammonia complex ion is changed to insoluble AgBr upon adding NaBr(aq).

Photos: © Charles D. Winters/Cengage

NH3(aq)

[Ag(S2O3)2]3−(aq)

(d) Solid AgBr is dissolved upon adding Na2S2O3(aq). The product is the water-soluble complex ion [Ag(S2O3)2]3−.

Figure 17.19  Forming and dissolving precipitates.  ​Insoluble compounds often dissolve upon addition of a complexing agent.

17.5  Equilibria Involving Complex Ions

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869

AgCl(s) + 2 NH3(aq) uv [Ag(NH3)2]+(aq) + Cl−(aq)

Dissolving AgCl(s) can be viewed as a two-step process. First, AgCl dissolves minimally in water, giving Ag+(aq) and Cl−(aq) ion. Then, the Ag+(aq) ion combines with NH 3 to give the ammonia complex. Lowering the Ag +(aq) concentration through complexation with NH3 shifts the solubility equilibrium to the right, and more solid AgCl dissolves. AgCl(s) uv Ag+(aq) + Cl−(aq)

Ksp = 1.8 × 10−10

Ag+(aq) + 2 NH3(aq) uv [Ag(NH3)2]+(aq)

Kf = 1.1 × 107

This is an example of combining or coupling two (or more) equilibria where one is a reactant-favored reaction and the other is product-favored. The large value of the formation constant for [Ag(NH3)2]+ means that the equilibrium lies well to the right, and AgCl can dissolve in the presence of NH3. If you combine Kf with Ksp, you obtain the net equilibrium constant for the interaction of AgCl and aqueous ammonia. Knet  =  Ksp × Kf = (1.8 × 10−10)(1.1 × 107) = 2.0 × 10−3 Knet  2.0  103 

[[Ag(NH3)2]][Cl] [NH3]2

Even though the value of Knet seems small, if you use a large concentration of NH3, the concentration of [Ag(NH3)2]+ in solution can be appreciable. Silver chloride is thus much more soluble in the presence of ammonia than in pure water. The stabilities of various complex ions involving Ag+ can be compared using the values of their formation constants. Formation Equilibrium +

−

Kf −

Ag (aq) + 2 Cl (aq) uv [AgCl2] (aq)

1.1 × 105

Ag+(aq) + 2 S2O32−(aq) uv [Ag(S2O3)2]3−(aq)

2.9 × 1013

Ag+(aq) + 2 CN−(aq) uv [Ag(CN)2] −(aq)

1.3 × 1021

The formation of all three silver complexes is strongly product-favored, and the cyanide complex ion [Ag(CN)2]− is the most stable of the three. Figure 17.19 shows that beginning with a precipitate of AgCl, adding aqueous ammonia dissolves the precipitate to give the soluble complex ion [Ag(NH3)2]+. Silver bromide is even more stable than [Ag(NH3)2]+, so AgBr (Ksp = 5.4 × 10−13) forms in preference to the complex ion upon adding bromide ion. If thiosulfate ion, S2O32−, is then added, AgBr dissolves due to the formation of [Ag(S2O3)2]3−, a complex ion with a large formation constant (2.9 × 1013).

E xamp le 17.16

Complex Ions and Solubility Problem   What is the value of the equilibrium constant, Knet, for dissolving AgBr in a solution containing the thiosulfate ion, S2O32− (Figure 17.19)? Does AgBr dissolve upon adding aqueous sodium thiosulfate to the solid?

What Do You Know?   There are two equilibria here. One is for dissolving AgBr in water to give Ag+ and Br− ions, and its equilibrium constant is Ksp. The other is the formation of [Ag(S 2O 3) 2] 3− ions from Ag + and S 2O 32− ions; its equilibrium constant is Kf.

Strategy   Summing several equilibrium processes gives the net chemical equation. Knet is the product of the values of K of the summed chemical equations (Section 15.5).

870

Chapter 17 / Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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Solution   The overall reaction for dissolving AgBr in the presence of the thiosulfate anion is the sum of two equilibrium processes. AgBr(s) uv Ag+(aq) + Br−(aq)

Ksp = 5.4 × 10−13

Ag+(aq) + 2 S2O32−(aq) uv [Ag(S2O3)2]3−(aq)

Kf = 2.9 × 1013

Net chemical equation:

AgBr(s) + 2 S2O32−(aq) uv [Ag(S2O3)2]3−(aq) + Br−(aq)

 Knet = Ksp × Kf = 16 

 AgBr is predicted to dissolve in aqueous Na2S2O3, as observed (Figure 17.19). 

Think about Your Answer   The value of Knet is greater than 1, indicating a ­ roduct-favored reaction at equilibrium. And, although the equilibrium constant is not p large, the majority of the Ag+ ion can be complexed if an excess of S2O32− is added.

Check Your Understanding  Calculate the value of the equilibrium constant, Knet, for dissolving Cu(OH)2 in aqueous ammonia (to form the complex ion [Cu(NH3)4]2+) (Figure 16.9).

Applying Chemical Principles 17.1  Everything that Glitters . . . Gold mining evokes images of miners panning for gold in mountain streams or chopping rocks with pick axes. However, the percentage of gold in most deposits is too low for these extraction methods to be feasible. Today’s miners blast and crush enormous quantities of gold-bearing ore and then dissolve gold from the ore using a chemical process involving cyanide ion and oxygen.

For thousands of years gold has been used in jewelry and for currency. Gold does not tarnish, and it can be drawn into wires and hammered into sheets. In modern society, the most important application of this element may be in electronics, where its high conductivity and resistance to corrosion make it invaluable for wires and connectors.

4 Au(s) + 8 NaCN(aq) + O2(g) + 2 H2O(ℓ) n   4 NaAu(CN)2(aq) + 4 NaOH(aq) The solution containing the complex ion [Au(CN)2]− is filtered to separate it from the solids. Metallic gold is then recovered by reacting the cyanide complex with zinc. The zinc reduces the gold(I) complex to elemental gold and then joins with cyanide ions to form [Zn(CN)4]2−. Large volumes of 0.035% aqueous sodium cyanide are used for gold extraction. Unfortunately, unsafe disposal and accidental discharges have resulted in environmental disasters. In 2000, the collapse of dams in Baia Mare, Romania, resulted in millions of liters of cyanide waste entering the Tisza and Danube rivers. All aquatic life for miles downstream was killed. Although research into safer extraction methods is ongoing, extraction by cyanide ion is still the primary means of obtaining gold.

NASA

Questions Using gold.  The 18 mirror segments and stationary mirror of the Webb Space Telescope are covered with a thin film of gold. (The film is about 100 nm thick and is composed of approximately 48 g of gold.) The vacuum deposited film makes the mirrors highly reflective, which enables the telescope to capture stunning images of deep space.

1. Approximately 0.10 g of sodium cyanide is fatal to humans. What volume (in mL) of 0.035 mass percent NaCN solution contains a fatal dose of sodium cyanide? Assume the density of the solution is 1.0 g/mL. 2. What is the minimum volume of 0.0071 M NaCN(aq) necessary to dissolve the gold from 1.0 metric ton (1000 kg) of ore if the ore contains 0.012% gold? Applying Chemical Principles

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871

3. Use the formation constant of [Au(CN)2]− in Appendix K to determine the equilibrium concentration of Au+(aq) in a solution that is 0.0071 M CN− and 1.1 × 10−4 M [Au(CN)2]−. Is it reasonable to conclude that 100% of the gold in solution is present as the [Au(CN)2]− complex ion? Explain. 4. Silver undergoes similar reactions as those shown for gold. Both metals react with cyanide ion in the presence of oxygen to form soluble complexes, and both are reduced by zinc. The reaction of Ag+ with cyanide ion may be viewed as two sequential steps: Step 1.

Ag+(aq) + CN−(aq) uv AgCN(s)

Step 2.

AgCN(s) + CN−(aq) uv [Ag(CN)2]−(aq) Ag+(aq) + 2 CN−(aq) uv [Ag(CN)2]−(aq) Kf = 1.3 × 1021

(a) Use the solubility product equilibrium constant (Appendix J) of AgCN(s) to determine the equilibrium constant for Step 1. (b) Use the equilibrium constants from Step 1 and the overall reaction to determine the equilibrium constant for Step 2. (c) Excess AgCN(s) is combined with 1.0 L of 0.0071 M CN− (aq) and allowed to equilibrate. Calculate the equilibrium concentrations of CN − and [Ag(CN) 2] − using the equilibrium constant for Step 2. Assume no change in volume occurs. 5. Write a balanced chemical equation for the reaction of NaAu(CN)2(aq) and Zn(s).

17.2  Take a Deep Breath Maintenance of pH is vital to the cells of all living organisms because enzyme activity is influenced by pH. The primary protection against harmful pH changes in cells is provided by buffer systems, which maintain the intracellular pH of most cells between 6.9 and 7.4. Two important biological buffer systems control pH in this range: the bicarbonate/carbonic acid system (HCO3−/H2CO3) and the phosphate system (HPO42−/H2PO4−). The bicarbonate/carbonic acid buffer is important in blood plasma, where three equilibria are important. CO2(g) uv CO2(dissolved) © Charles D. Winters/Cengage

CO2(dissolved) + H2O(ℓ) uv H2CO3(aq) H2CO3(aq) + H2O(ℓ) uv H3O+(aq) + HCO3−(aq) The overall equilibrium for the second and third steps has pKoverall = 6.3 at 37 °C, the temperature of the human body. Thus, 7.4  6.3  log

[HCO3] [CO2(dissolved)]

Although the value of pK overall is about 1 pH unit away from the blood pH, the natural partial pressure of CO2 in the alveoli of the lungs (about 40 mm Hg) is sufficient to keep [CO2(dissolved)] at about 1.2 × 10−3 M and [HCO3−] at about 1.5 × 10−2 M as required to maintain this pH. If blood pH rises above about 7.45, you can suffer from a condition called alkalosis. Respiratory alkalosis can arise from hyperventilation when a person breathes quickly to expel CO2 from the lungs. This has the effect of lowering the CO2 concentration, which in turn leads to a lower H3O+ concentration and a higher pH. This same condition can also arise from severe anxiety or from an oxygen deficiency at high altitude. It can ultimately lead to overexcitability of the central nervous system, muscle spasms, convulsions, and death. One way to treat acute respiratory alkalosis is to breathe into a paper bag. The CO2 you exhale is recycled. This raises the blood CO 2 level and causes the equilibria above to shift to the right, thus raising the hydronium ion concentration and lowering the pH.

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Metabolic alkalosis can occur if you take large amounts of sodium bicarbonate to treat stomach acid (which is mostly HCl at a pH of about 1 to 2). It also commonly occurs when a person vomits profusely. This depletes the body of hydronium ions, which leads to an increase in bicarbonate ion concentration. Athletes can use the H2CO3/HCO3− equilibrium to enhance their performance. Strenuous activity produces high levels of lactic acid, and this can lower blood pH and cause muscle cramps. To counteract this, athletes will prepare before a race by hyperventilating for some seconds to raise blood pH, thereby helping to neutralize the acidity from the lactic acid. Acidosis is the opposite of alkalosis. There was a case of a toddler who came to the hospital with viral gastroenteritis and metabolic acidosis. He had severe diarrhea, was dehydrated, and had a high rate of respiration. One function of the bicarbonate ion is to neutralize stomach acid in the intestines. However, because of his diarrhea, the toddler was losing bicarbonate ions in his stool, and his blood pH was too low. To compensate, the toddler was breathing rapidly and blowing off

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CO2 through the lungs (the effect of which is to lower [H3O+] and raise the pH). Respiratory acidosis results from a build-up of CO2 in the body. This can be caused by pulmonary problems, by head injuries, or by drugs such as anesthetics and sedatives. It can be reversed by breathing rapidly and deeply. Doubling the breathing rate increases the blood pH by about 0.23 units.

Questions

Most importantly, the pKa for the H2PO4− ion is 7.20, which is very close to the normal pH in the body. H2PO4−(aq) + H2O(ℓ) uv H3O+(aq) + HPO42−(aq) 1. What should the ratio [HPO42−]/[H2PO4−] be to control the pH at 7.40? 2. A typical total phosphate concentration in a cell, [HPO42−] + [H2PO4−], is 2.0 × 10−2 M. What are the concentrations of HPO42− and H2PO4− at pH 7.40?

Phosphate ions are abundant in cells, both as the ions themselves and as important substituents on organic molecules.

Think–Pair–Share 1. Buffer capacity is often defined as the moles of strong acid or base required to change the pH of a buffer by 1 pH unit. (a) Estimate the amount of NaOH required to increase the pH of each of the following acetic acid–acetate ion buffers by 1 pH unit and then place the buffers in order from highest to lowest buffer capacity for the addition of 1.0 M NaOH(aq). (i) 1.0 L of a buffer consisting of 1.0 M CH3CO2H and 1.0 M CH3CO2−. (ii) 0.50 L of a buffer consisting of 1.0 M CH3CO2H and 1.0 M CH3CO2−. (iii) 1.0 L of a buffer consisting of 0.5 M CH3CO2H and 1.5 M CH3CO2−. (iv) 1.0 L of a buffer consisting of 1.5 M CH3CO2H and 0.5 M CH3CO2−. (b) Estimate the buffer capacity of the same buffers with respect to the addition of 1.0 M HCl. Place the buffers in order from highest to lowest buffer capacity. 2. The pK a for H 2PO 4− is 7.20. Suggest three methods for ­m aking 1 L of a buffer containing 1 mol of H 2PO 4− and 1 mol of its conjugate base, HPO42−, using only combinations of two or more of the following substances and solutions: NaH2PO4(s), Na2HPO4(s), 10.0 M NaOH, 10.0 M HCl, and distilled H2O. 3. The pKa for a weak acid, HA, is 5.00 (Ka = 1.0 × 10−5). A 25.0 mL sample of 0.10 M HA(aq) is titrated with 0.10 M NaOH(aq) while measuring the pH. Sketch the titration

curve (pH versus mL of OH−). Estimate the pH after adding the following volumes of OH−: 0.0 mL, 12.5 mL, 22.5 mL, 25.0 mL, and 30.0 mL. (If you ignore the volume changes that occur with the addition of NaOH, you may be able to estimate the pH values without using a calculator.) 4. A series of indicators and the pH ranges in which they change color are given below. Which indicator is suitable for the titration of a weak acid with a pKa of 5.0? Explain.

Indicator

pH Range of Color Changes

Methyl red

4.2 (red)−6.2 (yellow)

Bromothymol blue

6.0 (yellow)−7.7 (blue)

Phenolphthalein

8.2 (colorless)−9.8 (pink)

5. You want to separate (by selective precipitation using hydroxide ion) a mixture of 0.10 M Mg2+(aq) and 0.10 M Ca2+(aq). Assume that concentrated NaOH is added with little to no volume change. (Ksp[Ca(OH)2] = 5.5 × 10−6 and Ksp[Mg(OH)2] = 5.6 × 10−12) (a) At what pH will Ca(OH)2 begin precipitating? (b) What concentration of Mg2+ remains in solution at the pH where Ca(OH)2 begins precipitating? (c) What percentage of Mg2+ was removed from solution at the pH when Ca(OH)2 begins precipitating?

Chapter Goals Revisited The goals for this chapter are keyed to specific Study Questions to help you organize your review.

17.1 Buffers • Predict the effect of the addition of a common ion on the pH of the solution of a weak acid or base. 1–6, 9, 10.

• Recognize the composition of a buffer and know how to prepare a buffer solution. 19–24.

• Understand how a buffer solution resists changes to its pH. 19, 20, 29, 30, 106.

Chapter Goals Revisited

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• Calculate the pH of a buffer solution of given composition using either the equation for the equilibrium for a weak acid or the Henderson– Hasselbalch equation. 3−10, 13−18.

• Calculate the pH of a buffered solution after the addition of acid or base. 25–28.

17.2  Acid–Base Titrations • Recognize the differences between a titration curve for a strong acid–

strong base titration and a titration curve in which one of the substances is weak. 35, 36.

• Calculate the pH at various points in acid–base titrations (initial pH, the

pH prior to the equivalence point, at the equivalence point, and after the equivalence point). 31−34, 37, 38, 114, 115.

• Describe how an indicator functions in an acid–base titration. 39, 40, 113−115.

17.3  Solubility of Salts • Write the equilibrium constant expression relating concentrations of ions in solution to Ksp for any insoluble salt. 45, 46.

• Calculate the Ksp value for a salt from its solubility and calculate the solubility of a salt from its Ksp. 47−50, 55, 60.

• Recognize how the presence of a common ion affects the solubility of a salt. 63−66.

• Understand how hydrolysis of basic anions affects the solubility of a salt. 71−72.

• Recognize that salts with anions derived from weak acids have increased solubility in acidic solutions. 67−70, 122.

17.4  Precipitation Reactions • Evaluate the reaction quotient Q to determine if a precipitate will form when the ion concentrations are known. 73−76, 78.

• Calculate the ion concentrations required to begin the precipitation of an insoluble salt. 77, 92, 103.

17.5  Equilibria Involving Complex Ions • Understand the formation of complex ions in solution and recognize how this can increase the solubility of an insoluble salt. 79−84.

Key Equations Equation 17.1 (page 836) Hydronium ion concentration in a buffer solution composed of a weak acid and its conjugate base. [H3O] 

[acid]  Ka [conjugate base]

Equation 17.2 (page 837) Henderson–Hasselbalch equation. To calculate the pH of a buffer solution composed of a weak acid and its conjugate base. pH  pK a  log

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[conjugate base] [acid]

Chapter 17 / Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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Equation 17.3 (page 846) Equation to calculate the hydronium ion concentration before the equivalence point in the titration of a weak acid with a strong base. [H3O] 

[weak acid remaining]  Ka [conjugate base produced]

Equation 17.4 (page 846) Equation to calculate the pH before the equivalence point in the titration of a weak acid with a strong base based on the Henderson– Hasselbalch equation. pH  pK a  log

[conjugate base produced] [weak acid remaining]

Equation 17.5 (page 846) The relationship between the pH of the solution and the

pKa of the weak acid (or [H3O+] and Ka) at the halfway or half-neutralization point in the titration of a weak acid with a strong base (or of a weak base with a strong acid). [H3O+] = Ka and pH = pKa

Equation 17.6 (page 854) The general equilibrium constant expression, Ksp (solubility constant) for dissolving a poorly soluble salt, AxBy.

AxBy(s) uv x Ay+(aq) + y Bx−(aq)  Ksp = [Ay+]x[Bx−]y

Study Questions ▲ denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully worked solutions in the Student Solutions Manual.

Practicing Skills The Common Ion Effect and Buffer Solutions (See Section 17.1 and Examples 17.1 and 17.2.) 1. Does the pH of the solution increase, decrease, or stay the same when you (a) add solid ammonium chloride to a dilute aqueous solution of NH3? (b) add solid sodium acetate to a dilute aqueous solution of acetic acid? (c) add solid NaCl to a dilute aqueous solution of NaOH? 2. Does the pH of the solution increase, decrease, or stay the same when you (a) add solid sodium oxalate, Na2C2O4, to 50.0 mL of 0.015 M oxalic acid, H2C2O4? (b) add solid ammonium chloride to 75 mL of 0.016 M HCl? (c) add 20.0 g of NaCl to 1.0 L of 0.10 M sodium acetate, NaCH3CO2? 3. What is the pH of a solution that consists of 0.20 M ammonia, NH3, and 0.20 M ammonium chloride, NH4Cl? 4. What is the pH of a solution that consists of 0.25 M formic acid, HCO2H, and 0.25 M sodium formate, NaHCO2?

5. What is the pH of 1.0 L of 0.20 M acetic acid to which 16.4 g of sodium acetate, NaCH3CO2, is added? 6. What is the pH of 1.0 L of 0.30 M ammonia to which 16.0 g of ammonium chloride, NH4Cl, is added? 7. What is the pH of the solution that results from adding 30.0 mL of 0.015 M KOH to 50.0 mL of 0.015 M benzoic acid? 8. What is the pH of the solution that results from adding 25.0 mL of 0.12 M HCl to 25.0 mL of 0.43 M NH3? 9. What is the pH of the buffer solution that contains 2.2 g of NH4Cl in 250 mL of 0.12 M NH3? Is the final pH lower or higher than the pH of the 0.12 M ammonia solution? 10. Lactic acid (CH3CHOHCO2H) is found in sour milk, in sauerkraut, and in muscles after activity. (Ka for lactic acid = 1.4 × 10−4.) (a) If 8.40 g of NaCH3CHOHCO2, sodium lactate, is added to 5.00 × 102 mL of 0.100 M lactic acid, what is the pH of the resulting buffer solution? (b) Is the pH of the buffered solution lower or higher than the pH of the lactic acid solution?

Study Questions

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875

11. What mass of sodium acetate, NaCH3CO2, must be added to 1.00 L of 0.10 M acetic acid to give a solution with a pH of 4.50? 12. What mass of ammonium chloride, NH4Cl, must be added to exactly 5.00 × 102 mL of 0.10 M NH3 solution to give a solution with a pH of 9.50?

Using the Henderson–Hasselbalch Equation (See Section 17.1 and Example 17.3.) 13. Calculate the pH of a solution that has an acetic acid concentration of 0.050 M and a sodium acetate concentration of 0.075 M. 14. Calculate the pH of a solution that has an ammonium chloride concentration of 0.050 M and an ammonia concentration of 0.045 M. 15. What must the ratio of acetic acid to acetate ion be to have a buffer with a pH value of 4.50? 16. What must the ratio of H2PO4− to HPO42− be to have a buffer with a pH value of 7.50? 17. A buffer is composed of formic acid and its conjugate base, the formate ion. (a) What is the pH of a solution that has a formic acid concentration of 0.050 M and a sodium formate concentration of 0.035 M? (b) What must the ratio of acid to conjugate base be to have a pH value 0.50 units higher than the value calculated in part (a)? 18. A buffer solution is composed of 1.360 g of KH2PO4 and 5.677 g of Na2HPO4. (a) What is the pH of the buffer solution? (b) What mass of KH2PO4 must be added to decrease the buffer solution pH by 0.50 unit from the value calculated in part (a)?

Preparing a Buffer Solution (See Section 17.1 and Example 17.4.) 19. Which of the following combinations would be the best to buffer the pH of a solution at approximately 5? (a) HCl and NaCl (b) NH3 and NH4Cl (c) CH3CO2H and NaCH3CO2 20. Which of the following combinations would be the best to buffer the pH of a solution at approximately 7? (a) H3PO4 and NaH2PO4 (b) NaH2PO4 and Na2HPO4 (c) Na2HPO4 and Na3PO4 21. Describe how to prepare a buffer solution from NaH2PO4 and Na2HPO4 to have a pH of 7.5.

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22. Describe how to prepare a buffer solution from CH3CO2H and NaCH3CO2 to have a pH of 4.5. 23. Determine the volume (in mL) of 1.00 M NaOH that must be added to 250 mL of 0.50 M CH3CO2H to produce a buffer with a pH of 4.50. 24. Determine the volume (in mL) of 1.00 M HCl that must be added to 750 mL of 0.50 M HPO42− to produce a buffer with a pH of 7.00.

Adding an Acid or a Base to a Buffer Solution (See Section 17.1 and Example 17.5.) 25. A buffer solution was prepared by adding 4.95 g of sodium acetate, NaCH3CO2, to 2.50 × 102 mL of 0.150 M acetic acid, CH3CO2H. (a) What is the pH of the buffer? (b) What is the pH of 1.00 × 102 mL of the buffer solution if you add 82 mg of NaOH to the solution? 26. You dissolve 0.425 g of NaOH in 2.00 L of a buffer solution that has [H2PO4−] = [HPO42−] = 0.132 M. What is the pH of the solution before adding NaOH? After adding NaOH? 27. A buffer solution is prepared by adding 0.125 mol of ammonium chloride to 5.00 × 102 mL of 0.500 M solution of ammonia. (a) What is the pH of the buffer? (b) If 0.0100 mol of HCl gas is bubbled into 5.00 × 102 mL of the buffer, what is the new pH of the solution? 28. What is the pH change when 20.0 mL of 0.100 M NaOH is added to 80.0 mL of a buffer solution consisting of 0.169 M NH3 and 0.183 M NH4Cl? 29. A buffer consists of 0.50 mol of NaH2PO4 and 0.50 mol of Na2HPO4 diluted to a volume of 1.0 L. (a) What is the pH of the buffer? (b) What volume of 0.15 M NaOH is required to raise the pH of the buffer by 1.00? 30. A buffer consists of 0.36 mol of CH3CO2H and 0.36 mol of NaCH3CO2 diluted to a volume of 1.0 L. (a) What is the pH of the buffer? (b) What volume of 0.10 M HCl is required to decrease the pH of the buffer by 1.00?

More about Acid–Base Reactions: Titrations (See Section 17.2 and Examples 17.6 and 17.7.) 31. Phenol, C6H5OH, is a weak organic acid. Suppose 0.515 g of the compound is dissolved in enough water to make 125 mL of solution. The resulting solution is titrated with 0.123 M NaOH.

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C6H5OH(aq) + OH−(aq) uv C6H5O−(aq) + H2O(ℓ)

(a) What is the pH of the original solution of phenol? (b) What are the concentrations of all of the ­following ions at the equivalence point: Na+, H3O+, OH−, and C6H5O−? (c) What is the pH of the solution at the equivalence point? 32. Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 1.00 × 102 mL of solution and then titrate the solution with 0.108 M NaOH. C6H5CO2H(aq) + OH−(aq) uv  C6H5CO2−(aq) + H2O(ℓ)

(a) What was the pH of the original benzoic acid solution? (b) What are the concentrations of all of the following ions at the equivalence point: Na+, H3O+, OH−, and C6H5CO2−? (c) What is the pH of the solution at the equivalence point? 33. You require 36.78 mL of 0.0105 M HCl to reach the equivalence point in the titration of 25.0 mL of aqueous ammonia. (a) What was the concentration of NH3 in the original ammonia solution? (b) What are the concentrations of H3O+, OH−, and NH4+ at the equivalence point? (c) What is the pH of the solution at the equivalence point? 34. A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalence point. C6H5NH2(aq) + H3O+(aq) uv  C6H5NH3+(aq) + H2O(ℓ)

(a) What was the concentration of aniline in the original solution? (b) What are the concentrations of H3O+, OH−, and C6H5NH3+ at the equivalence point? (c) What is the pH of the solution at the equivalence point?

Titration Curves and Indicators (See Section 17.2 and Figures 17.4–17.11.) 35. Without doing detailed calculations, sketch the curve for the titration of 30.0 mL of 0.10 M NaOH with 0.10 M HCl. Indicate the approximate pH at the beginning of the titration and at the equivalence point. What is the total solution volume at the equivalence point?



36. Without doing detailed calculations, sketch the curve for the titration of 50 mL of 0.050 M pyridine, C5H5N (a weak base), with 0.10 M HCl. ­Indicate the approximate pH at the beginning of the titration and at the equivalence point. What is the total solution volume at the equivalence point? 37. You titrate 25.0 mL of 0.10 M NH3 with 0.10 M HCl. (a) What is the pH of the NH3 solution before the titration begins? (b) What is the pH at the equivalence point? (c) What is the pH at the halfway point of the titration? (d) What indicator in Figure 17.11 could be used to detect the equivalence point? (e) Calculate the pH of the solution after adding 5.00, 15.0, 20.0, 22.0, and 30.0 mL of the acid. Combine this information with that in parts (a)–(c) and plot the titration curve. 38. Construct a rough plot of pH versus volume of base for the titration of 25.0 mL of 0.050 M HCN with 0.075 M NaOH. (a) What is the pH before any NaOH is added? (b) What is the pH at the halfway point of the titration? (c) What is the pH when 95% of the required NaOH has been added? (d) What volume of base, in milliliters, is required to reach the equivalence point? (e) What is the pH at the equivalence point? (f) What indicator would be most suitable for this titration? (Figure 17.11.) (g) What is the pH when 105% of the required base has been added? 39. Using Figure 17.11, suggest an indicator to use in each of the following titrations: (a) The weak base pyridine is titrated with HCl. (b) Formic acid is titrated with NaOH. (c) Ethylenediamine, a weak diprotic base, is titrated with HCl. 40. Using Figure 17.11, suggest an indicator to use in each of the following titrations. (a) NaHCO3 is titrated to CO32− with NaOH. (b) Hypochlorous acid is titrated with NaOH. (c) Trimethylamine is titrated with HCl.

Solubility Guidelines (See Sections 3.4 and 3.5, Figure 3.10, and Example 3.2.) 41. Name two insoluble salts of each of the following ions. (a) Cl− (b) Zn2+ (c) Fe2+ Study Questions

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42. Name two insoluble salts of each of the following ions. (a) CO32− (b) Ba2+ (c) I−

51. You add 0.979 g of Pb(OH)2 to 1.00 L of pure water at 25 °C. The pH is 9.15. Estimate the value of Ksp for Pb(OH)2.

43. Using the solubility guidelines (Figure 3.10, page 150), predict whether each of the following is insoluble or soluble in water. (a) (NH4)2CO3 (c) NiS (b) ZnSO4 (d) BaSO4

52. You place 1.234 g of solid Ca(OH)2 in 1.00 L of pure water at 25 °C. The pH of the solution is found to be 12.35. Estimate the value of Ksp for Ca(OH)2.

44. Predict whether each of the following is insoluble or soluble in water. (a) Ni(OH)2 (c) MgCl2 (b) Zn(NO3)2 (d) K2S

Writing Solubility Product Constant Expressions (See Section 17.3.) 45. For each of the following insoluble salts, (i) write a balanced equation showing the equilibrium occurring when the salt is added to water, and (ii) write the Ksp expression. (a) AgCN (b) NiCO3 (c) AuBr3 46. For each of the following insoluble salts, (i) write a balanced equation showing the equilibrium occurring when the salt is added to water, and (ii) write the Ksp expression. (a) CaCO3 (b) PbBr2 (c) Ag2S

Calculating Ksp

(See Section 17.3 and Example 17.8.) 47. When 1.55 g of solid thallium(I) bromide is added to 1.00 L of water, the salt dissolves to a small extent. TlBr(s) uv Tl+(aq) + Br−(aq)

The thallium(I) and bromide ions in equilibrium with TlBr each have a concentration of 1.9 × 10−3 M. What is the value of Ksp for TlBr? 48. At 20 °C, a saturated aqueous solution of silver acetate, AgCH3CO2, contains 1.0 g of the silver compound dissolved in 100.0 mL of solution. ­Calculate Ksp for silver acetate. AgCH3CO2(s) uv Ag+(aq) + CH3CO2−(aq)

49. When 250 mg of SrF2, strontium fluoride, is added to 1.00 L of water, the salt dissolves to a very small extent. SrF2(s) uv Sr2+(aq) + 2 F−(aq)

At equilibrium, the concentration of Sr2+ is found to be 1.03 × 10−3 M. What is the value of Ksp for SrF2? 50. Calcium hydroxide, Ca(OH)2, dissolves in water to the extent of 0.824 g per liter. What is the value of Ksp for Ca(OH)2? Ca(OH)2(s) uv Ca2+(aq) + 2 OH−(aq)

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Estimating Salt Solubility from Ksp

(See Section 17.3 and Examples 17.9 and 17.10.) 53. Estimate the solubility of silver iodide in pure water at 25 °C (a) in moles per liter and (b) in grams per liter. AgI(s) uv Ag+(aq) + I−(aq)

54. What is the molar concentration of Au+(aq) in a saturated solution of AuCl in pure water at 25 °C? AuCl(s) uv Au+(aq) + Cl−(aq)

55. Estimate the solubility of calcium fluoride, CaF2, (a) in moles per liter and (b) in grams per liter of pure water. CaF2(s) uv Ca2+(aq) + 2 F−(aq)

56. Estimate the solubility of lead(II) iodide (a) in moles per liter and (b) in grams per liter of pure water. 57. The Ksp value for radium sulfate, RaSO4, is 4.2 × 10−11. If 25 mg of radium sulfate is placed in 1.00 × 102 mL of water, does all of it dissolve? If not, how much dissolves? 58. If 25 mg of barium fluoride is placed in 250 mL of pure water, does all of it dissolve? If not, how much dissolves? 59. Use Ksp values to decide which compound in each of the following pairs is more soluble (Appendix J). (a) PbCl2 or PbBr2 (b) HgS or FeS (c) Fe(OH)2 or Zn(OH)2 60. Use Ksp values to decide which compound in each of the following pairs is more soluble (Appendix J). (a) AgCl or AgSCN (b) PbCO3 or PbSO4 (c) AgCl or MgF2 (d) SrF2 or PbF2 61. Use Ksp values to decide which compound in each of the following pairs is more soluble (Appendix J). (a) TlBr or PbBr2 (b) AuCl or Hg2Cl2 (c) TlI or Hg2I2

Chapter 17 / Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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62. Use Ksp values to decide which compound in each of the following pairs is more soluble (Appendix J). (a) BaCO3 or Ag2CO3 (b) TlI or PbI2 (c) AgBr or Hg2Br2

The Common Ion Effect and Salt Solubility (See Section 17.3 and Example 17.11.) 63. Calculate the molar solubility of silver thiocyanate, AgSCN, in pure water and in water containing 0.010 M NaSCN. 64. Calculate the solubility of silver bromide, AgBr, in moles per liter, in pure water. Compare this value with the molar solubility of AgBr in 225 mL of water to which 0.15 g of NaBr was added. 65. Compare the solubility, in milligrams per milliliter, of silver iodide, AgI, (a) in pure water and (b) in water that is 0.020 M in AgNO3. 66. What is the solubility, in milligrams per milliliter, of BaF2, (a) in pure water and (b) in water containing 5.0 mg/mL KF? 67. Calculate the solubility, in moles per liter, of iron(II) hydroxide, Fe(OH)2, in a solution buffered to a pH of 7.00. 68. Calculate the solubility, in moles per liter, of ­magnesium hydroxide, Mg(OH)2, in a solution buffered to a pH of 9.50.

The Effect of Basic Anions on Salt Solubility (See Section 17.3) 69. Which insoluble compound in each pair should be more soluble in nitric acid than in pure water? (a) PbCl2 or PbS (b) Ag2CO3 or AgI (c) Al(OH)3 or AgCl

(See Section 17.4 and Examples 17.12–17.14.) 73. You have a solution that has a lead(II) ion concentration of 0.0012 M. If enough soluble chloride-containing salt is added so that the Cl− concentration is 0.010 M, will PbCl2 precipitate? 74. Sodium carbonate is added to a solution in which the concentration of Ni2+ ion is 0.0024 M. Will precipitation of NiCO3 occur (a) when the concentration of the carbonate ion is 1.0 × 10−6 M or (b) when it is 100 times greater (1.0 × 10−4 M)? 75. If the concentration of Zn2+ in 10.0 mL of water is 1.63 × 10−4 M, will zinc hydroxide, Zn(OH)2, precipitate when 4.0 mg of NaOH is added? 76. You have 95 mL of a solution that has a lead(II) concentration of 0.0012 M. Will PbCl2 precipitate when 1.20 g of solid NaCl is added? 77. If the concentration of Mg2+ ion in seawater is 1350 mg/L, what OH− concentration is required to precipitate Mg(OH)2? 78. Will a precipitate of Ca(OH)2 form when 25.0 mL of 0.010 M NaOH is combined with 125.0 mL of a 0.10 M solution of calcium chloride?

Equilibria Involving Complex Ions (See Section 17.5 and Examples 17.15 and 17.16.) 79. Zinc hydroxide is amphoteric (Section 16.10). Use equilibrium constants to show that, given sufficient OH−, Zn(OH)2 can dissolve in NaOH. 80. Solid silver iodide, AgI, can be dissolved by adding aqueous sodium cyanide. Calculate Knet for the following reaction. AgI(s) + 2 CN−(aq) uv [Ag(CN)2]−(aq) + I−(aq)

81.

What amount of ammonia (moles) must be added to dissolve 0.050 mol of AgCl suspended in 1.0 L of water? ▲

70. Which compound in each pair should be more soluble in nitric acid than in pure water? (a) AgCN or AgBr (b) PbF2 or PbI2 (c) CuI or Cu(OH)2

82. Can you dissolve 15.0 mg of AuCl in 100.0 mL of water if you add 15.0 mL of 6.00 M NaCN?

71. Which compound in each pair should be more soluble in water than is predicted by a calculation from Ksp? (a) AgCl or AgCN (b) Ag2C2O4 or AgBr (c) Hg2I2 or Hg2CO3

84. The chemistry of silver cyanide: (a) Calculate the solubility of AgCN(s) in water from the Ksp value. (b) Calculate the value of the equilibrium constant for the reaction

72. Which compound in each pair is more soluble in water than is predicted by a calculation from Ksp? (a) AgI or Ag2CO3 (b) PbCO3 or PbCl2 (c) AgCl or AgCN

Precipitation Reactions

83. What is the solubility of AgCl (a) in pure water and (b) in 1.0 M NH3?

AgCN(s) + CN−(aq) uv [Ag(CN)2]−(aq)

from Ksp and Kf values and predict from this value whether AgCN(s) would dissolve in KCN(aq).

Study Questions

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879

(c) Determine the equilibrium constant for the reaction AgCN(s) + 2 S2O32−(aq) uv  [Ag(S2O3)2]3−(aq) + CN−(aq)

Calculate the solubility of AgCN in a solution containing 0.10 M S2O32− and compare the value to the solubility in water [part (a)].

92. A sample of hard water contains about 2.0 × 10−3 M Ca2+. A soluble fluoridecontaining salt such as NaF is added to ­fluoridate the water (to aid in the ­prevention of dental cavities). What is the maximum concentration of F− that can be present without precipitating CaF2?

General Questions These questions are not designated as to type or location in the chapter. They may combine several concepts.

87. If you mix 48 mL of 0.0012 M BaCl2 with 24 mL of 1.0 × 10−6 M Na2SO4, will a precipitate of BaSO4 form? 88. Calculate the hydronium ion concentration and the pH of the solution that results when 20.0 mL of 0.15 M acetic acid, CH3CO2H, is mixed with 5.0 mL of 0.17 M NaOH.

Household sources of fluoride ion.  Adding fluoride ion to drinking water (or toothpaste) prevents the formation of ­dental cavities.

93. What is the pH of a buffer solution prepared from 5.15 g of NH4NO3 and 100. mL of 0.15 M NH3? What is the new pH if the solution is diluted with pure water to a volume of 500. mL? 94. If you place 5.0 mg of SrSO4 in 1.0 L of pure water, will all of the salt dissolve before equilibrium is established, or will some salt remain undissolved? © Charles D. Winters/ Cengage

86. In each of the following cases, decide whether a precipitate will form when mixing the indicated reagents, and write a balanced equation for the reaction. (a) Na2SO4(aq) + Mg(NO3)2(aq) (b) K3PO4(aq) + FeCl3(aq)

© Charles D. Winters/Cengage

85. In each of the following cases, decide whether a precipitate will form when mixing the indicated reagents, and write a balanced equation for the reaction. (a) NaBr(aq) + AgNO3(aq) (b) KCl(aq) + Pb(NO3)2(aq)

89. Calculate the hydronium ion concentration and the pH of the solution that results when 50.0 mL of 0.40 M NH3 is mixed with 25.0 mL of 0.20 M HCl. 90. For each of the following cases, decide whether the pH is less than 7, equal to 7, or greater than 7. (a) Equal volumes of 0.20 M ammonia, NH3, and 0.20 M HCl are mixed. (b) Equal volumes of 0.10 M acetic acid, CH3CO2H, and 0.10 M KOH are mixed. (c) 25 mL of 0.015 M NH3 is mixed with 12 mL of 0.015 M HCl. (d) 150 mL of 0.20 M HNO3 is mixed with 75 mL of 0.40 M NaOH. (e) 25 mL of 0.45 M H2SO4 is mixed with 25 mL of 0.90 M NaOH. 91. Rank the following compounds in order of increasing solubility in water: Na2CO3, BaCO3, Ag2CO3.

880

SO42–

Celestite, SrSO4 Strontium sulfate

95. Describe the effect on the pH of the following actions or explain why there is no effect: (a) Adding sodium acetate, NaCH3CO2, to 0.100 M CH3CO2H (b) Adding NaNO3 to 0.100 M HNO3 96. What volume of 0.120 M NaOH must be added to 100. mL of 0.100 M NaHC2O4 to reach a pH of 4.70?

Chapter 17 / Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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97.

A buffer solution is prepared by dissolving 1.50 g each of benzoic acid, C6H5CO2H, and sodium benzoate, NaC6H5CO2, in 150.0 mL of solution. (a) What is the pH of this buffer solution? (b) Which buffer component must be added, and in what quantity, to change the pH to 4.00? (c) What quantity of 2.0 M NaOH or 2.0 M HCl must be added to the buffer to change the pH to 4.00? ▲

104. ▲ A solution contains 0.10 M iodide ion, I−, and 0.10 M carbonate ion, CO32−. (a) If solid Pb(NO3)2 is slowly added to the solution, which salt will precipitate first, PbI2 or PbCO3? (b) What will be the concentration of the first ion that precipitates (CO32− or I−) when the second, more soluble salt begins to precipitate?

99. What is the equilibrium constant for the following reaction? AgCl(s) + I−(aq) uv AgI(s) + Cl−(aq)

Does the equilibrium lie predominantly to the left or to the right? Will AgI form if iodide ion, I−, is added to a saturated solution of AgCl? 100. Calculate the equilibrium constant for the following reaction. Zn(OH)2(s) + 2 CN−(aq) uv  Zn(CN)2(s) + 2 OH−(aq)

Does the equilibrium lie predominantly to the left or to the right? 101. Suppose you harvest 28 g of rhubarb leaves with an oxalic acid content of 1.2% by weight. (a) What volume of 0.25 M NaOH is required to titrate completely the oxalic acid in the leaves? (b) What mass of calcium oxalate could be formed from the oxalic acid in these leaves? 102. The solubility product constant for calcium oxalate is estimated to be 4 × 10−9. What is its solubility in grams per liter? 103. ▲ In principle, the ions Ba2+ and Ca2+ can be separated by the difference in solubility of their fluorides, BaF2 and CaF2. If you have a solution that is 0.10 M in both Ba2+ and Ca2+, CaF2 will begin to precipitate first as fluoride ion is added slowly to the solution. (a) What concentration of fluoride ion will precipitate the maximum amount of Ca2+ ion without precipitating BaF2? (b) What concentration of Ca2+ remains in ­solution when BaF2 just begins to precipitate?



© Charles D. Winters/Cengage

98. What volume of 0.200 M HCl must be added to 500.0 mL of 0.250 M NH3 to have a buffer with a pH of 9.00?

Lead(II) iodide (Ksp = 9.8 × 10−9) is a bright yellow solid.

105. ▲ A solution contains Ca2+ and Pb2+ ions, both at a concentration of 0.010 M. You wish to separate the two ions from each other as completely as possible by precipitating one but not the other using aqueous Na2SO4 as the precipitating agent. (a) Which will precipitate first as sodium sulfate is added, CaSO4 or PbSO4? (b) What will be the concentration of the first ion that precipitates (Ca2+ or Pb2+) when the second, more soluble salt begins to precipitate? 106. Buffer capacity is defined as the number of moles of a strong acid or strong base required to change the pH of 1 L of the buffer solution by one unit. What is the buffer capacity of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate? 107. Some photographic film is coated with crystals of AgBr suspended in gelatin. Some of the silver ions are reduced to silver metal upon exposure to light. Unexposed AgBr is then dissolved with sodium thiosulfate in the fixing step. AgBr(s) + 2 S2O32−(aq) uv  [Ag(S2O3)2]3−(aq) + Br−(aq)

(a) What is the equilibrium constant for this reaction? (b) What mass of Na2S2O3 must be added to dissolve 1.00 g of AgBr suspended in 1.00 L of water?

Study Questions

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881

108. The Ca2+ ion in hard water is precipitated as CaCO3 by adding soda ash, Na2CO3. If the calcium ion concentration in hard water is 0.010 M and if the Na2CO3 is added until the ­carbonate ion concentration is 0.050 M, what ­percentage of the calcium ions was removed from the water? (You may neglect carbonate ion hydrolysis.)

112. ▲ You will often work with salts of Fe3+, Pb2+, and Al3+ in the laboratory. (All are found in nature, and all are important economically.) If you have a solution containing these three ions, each at a concentration of 0.10 M, what is the order in which their hydroxides precipitate as aqueous NaOH is slowly added to the solution? 113. Aniline hydrochloride, (C6H5NH3)Cl, is a weak acid. (Its conjugate base is the weak base aniline, C6H5NH2.) The acid can be titrated with a strong base such as NaOH.

© Charles D. Winters/Cengage

C6H5NH3+(aq) + OH−(aq) uv  C6H5NH2(aq) + H2O(ℓ)

This sample of calcium carbonate (Ksp = 3.4 × 10−9) was deposited in a cave formation.

In the Laboratory 109. Each pair of ions below is found together in aqueous solution. Using the table of solubility product constants in Appendix J, devise a way to separate these ions by adding a reagent to precipitate one of the ions as an insoluble salt and leaving the other in solution. (a) Ba2+ and Na+ (b) Ni2+ and Pb2+ 110. Each pair of ions below is found together in aqueous solution. Using the table of solubility product constants in Appendix J, devise a way to separate these ions by adding a reagent to precipitate one of the ions as an insoluble salt and leave the other in solution. (a) Cu2+ and Ag+ (b) Al3+ and Fe3+ 111. ▲ The cations Ba2+ and Sr2+ can be precipitated as very insoluble sulfates. (a) If you add sodium sulfate to a solution containing these metal cations, each with a concentration of 0.10 M, which is precipitated first, BaSO4 or SrSO4? (b) What will be the concentration of the first ion that precipitates (Ba2+ or Sr2+) when the second, more soluble salt begins to precipitate?

882

Assume 50.0 mL of 0.100 M aniline hydrochloride is titrated with 0.185 M NaOH. (Ka for aniline hydrochloride is 2.4 × 10−5.) (a) What is the pH of the (C6H5NH3)Cl solution before the titration begins? (b) What is the pH at the equivalence point? (c) What is the pH at the halfway point of the titration? (d) Which indicator in Figure 17.11 could be used to detect the equivalence point? (e) Calculate the pH of the solution after adding 10.0, 20.0, and 30.0 mL of base. (f) Combine the information in parts (a), (b), (c), and (e), and plot an approximate titration curve. 114. The weak base ethanolamine, HOCH2CH2NH2, can be titrated with HCl. HOCH2CH2NH2(aq) + H3O+(aq) uv  HOCH2CH2NH3+(aq) + H2O(ℓ)

Assume you have 25.0 mL of a 0.010 M solution of ethanolamine and titrate it with 0.0095 M HCl. (Kb for ethanolamine is 3.2 × 10−5.) (a) What is the pH of the ethanolamine solution before the titration begins? (b) What is the pH at the equivalence point? (c) What is the pH at the halfway point of the titration? (d) Which indicator in Figure 17.11 is the best choice to detect the equivalence point? (e) Calculate the pH of the solution after adding 5.00, 10.0, 20.0, and 30.0 mL of the acid. (f) Combine the information in parts (a), (b), (c), and (e), and plot an approximate titration curve.

Chapter 17 / Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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116. A buffer solution with a pH of 12.00 consists of Na3PO4 and Na2HPO4. The volume of solution is 200.0 mL. (a) Which component of the buffer is present in a larger amount? (b) If the concentration of Na3PO4 is 0.400 M, what mass of Na2HPO4 is present? (c) Which component of the buffer must be added to change the pH to 12.25? What mass of that component is required? 117. To have a buffer with a pH of 2.50, what volume of 0.150 M NaOH must be added to 100. mL of 0.230 M H3PO4? 118. ▲ What mass of Na3PO4 must be added to 80.0 mL of 0.200 M HCl to obtain a buffer with a pH of 7.75? 119. You have a solution that contains AgNO3, Pb(NO3)2, and Cu(NO3)2. Devise a separation method that results in having Ag+ in one test tube, Pb2+ in another, and Cu2+ in a third test tube. Use solubility guidelines and Ksp and Kf values. 120. Once you have separated the three salts in Study Question 119 into three test tubes, you need to confirm their presence. (a) For Pb2+ ion, one way to do this is to treat a precipitate of PbCl2 with K2CrO4 to produce the bright yellow insoluble solid, PbCrO4. Using Ksp values, confirm that the chloride salt should be converted to the chromate salt.



K2CrO4 added

© Charles D. Winters/Cengage

115. For the titration of 50.0 mL of 0.150 M ethylamine, C2H5NH2, with 0.100 M HCl, find the pH at each of the following points, and then use that information to sketch the titration curve and decide on an appropriate indicator. (a) At the beginning, before HCl is added (b) At the halfway point in the titration (c) When 75% of the required acid is added (d) At the equivalence point (e) When 10.0 mL more HCl is added than is required (f) Sketch the titration curve. (g) Suggest an appropriate indicator for this titration.

Stirred

PbCl2 precipitate

White PbCl2 is converted to yellow PbCrO4 on adding K2CrO4.

(b) Suggest a method for confirming the presence of Ag+ and Cu2+ ions using complex ions.

Summary and Conceptual Questions The following questions may use concepts from this and ­previous chapters. 121. Suggest a method for separating a precipitate ­consisting of a mixture of solid CuS and solid Cu(OH)2. 122. Which of the following barium salts should dissolve in a strong acid such as HCl: Ba(OH)2, BaSO4, or BaCO3? 123. Explain why the solubility of Ag3PO4 can be greater in water than is calculated from the Ksp value of the salt. 124. Two acids, each approximately 0.01 M in concentration, are titrated separately with a strong base. The acids show the following pH values at the equivalence point: HA, pH = 9.5, and HB, pH = 8.5. (a) Which is the stronger acid, HA or HB? (b) Which of the conjugate bases, A− or B−, is the stronger base? 125. Composition diagrams, commonly known as alpha plots, are often used to visualize the species in a solution of an acid or base as the pH is varied. The diagram for 0.100 M acetic acid is shown here.

Study Questions

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883

1.00

1.00 Fraction of CH3CO2H

0.80 Fraction of CH3CO2−

0.60

Alpha

Alpha

0.80

0.40 0.20

0.60 Fraction of H2CO3

Fraction of CO32−

0.40 0.20

0.00 2.0

4.0

6.0

0.00

3

5

7

pH

The plot shows how the fraction [alpha (α)] of acetic acid in solution, 

[CH3CO2H] [CH3CO2H]  [CH3CO2]

changes as the pH increases (blue curve). (The red curve shows how the fraction of acetate ion, CH3CO2−, changes as the pH increases.) Alpha plots are another way of viewing the relative concentrations of acetic acid and acetate ion as a strong base is added to a solution of acetic acid in the course of a titration. (a) Explain why the fraction of acetic acid declines and that of acetate ion increases as the pH increases. (b) Which species predominates at a pH of 4, acetic acid or acetate ion? What is the situation at a pH of 6? (c) Consider the point where the two lines cross. The fraction of acetic acid in the solution is 0.5, and so is that of acetate ion. That is, the solution is half acid and half conjugate base; their concentrations are equal. At this point, the graph shows the pH is 4.74. Explain why the pH at this point is 4.74. 126. The composition diagram, or alpha plot, for the important acid–base system of carbonic acid, H2CO3, is illustrated. (See Study Question 125 for more information on such diagrams.)

884

Fraction of HCO3−

9 pH

11

13

15

(a) Explain why the fraction of bicarbonate ion, HCO3−, rises and then falls as the pH increases. (b) What is the composition of the solution when the pH is 6.0? When the pH is 10.0? (c) If you wanted to buffer a solution at a pH of 11.0, what should be the ratio of HCO3− to CO32−? 127. The chemical name for aspirin is acetylsalicylic acid. It is believed that the analgesic and other desirable properties of aspirin are due not to the aspirin itself but rather to the simpler compound salicylic acid, C6H4(OH)CO2H, which results from the breakdown of aspirin in the stomach. O C

OH OH

Salicylic acid

(a) Give approximate values for the following bond angles in the acid: (i) COCOC in the ring; (ii) OOCPO; (iii) either of the COOOH angles; and (iv) COCOH. (b) What is the hybridization of the C atoms of the ring? Of the C atom in the OCO2H group?

Chapter 17 / Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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128. Aluminum hydroxide reacts with phosphoric acid to give AlPO4. The substance is used industrially in adhesives, binders, and cements. (a) Write the balanced equation for the preparation of AlPO4 from aluminum hydroxide and phosphoric acid. (b) If you begin with 152 g of aluminum hydroxide and 3.00 L of 0.750 M phosphoric acid, what is the theoretical yield of AlPO4?



(c) If you place 25.0 g of AlPO4 in 1.00 L of water, what are the concentrations of Al3+ and PO43− at equilibrium? (Neglect hydrolysis of aqueous Al3+ and PO43− ions.) Ksp for AlPO4 is 1.3 × 10−20. (d) Does the solubility of AlPO4 increase or decrease on adding HCl? Explain.

© Charles D. Winters/Cengage

(c) Experiment shows that 1.00 g of salicylic acid will dissolve in 460 mL of water. If the pH of this solution is 2.4, what is Ka for the acid? (d) If you have salicylic acid in your stomach and if the pH of gastric juice is 2.0, calculate the percentage of salicylic acid present in the stomach in the form of the salicylate ion, C6H4(OH)CO2−. (e) Assume you have 25.0 mL of a 0.014 M solution of salicylic acid and titrate it with 0.010 M NaOH. What is the pH at the halfway point of the titration? What is the pH at the equivalence point?

A sample of hydrated aluminum phosphate, a mineral known as augelite.

Study Questions

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885

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18

Principles of Chemical Reactivity: Entropy and Free Energy

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C hapt e r O ut li n e 18.1 Spontaneity and Dispersal of Energy: Entropy 18.2 Entropy: A Microscopic Understanding 18.3 Entropy Measurement and Values 18.4 Entropy Changes and Spontaneity 18.5 Gibbs Free Energy 18.6 Calculating and Using Standard Free Energies, ΔrG° 18.7 The Interplay of Kinetics and Thermodynamics

In Chapter 5, you were introduced to the first law of thermodynamics—the internal energy change (∆U) for a system is equal to the sum of the energy transferred as heat (q) and the energy transferred as work (w) between the system and the surroundings. Measuring changes in energy as heat and work, however, are not enough to tell you in which direction a chemical reaction will proceed, or whether a ­reaction will go to completion. The goal of this chapter is to describe a method for p ­ redicting the direction and the degree to which chemical or physical changes occur. This will require a more thorough examination of entropy (S), a concept touched upon in Chapter 13 (page 633) and Chapter 16 (see “A Closer Look: Acid Strengths and Molecular Structure,” page 811).

18.1 Spontaneity and Dispersal of Energy: Entropy Goals for Section 18.1 • Understand the concept of spontaneity: spontaneous processes occur without outside intervention and proceed to equilibrium, a result of energy dispersal.

• Know the second law of thermodynamics: a spontaneous process is one that leads to an increase in entropy in the universe.

• Recognize that an entropy change is the energy transferred as heat for a reversible process divided by the kelvin temperature.

You have encountered many examples of chemical changes (chemical reactions) and physical changes (the formation of mixtures, expansion of gases, and changes of state, to name a few) in this book. Change is central to chemistry, so it is important to understand the factors that determine whether a change will occur. ­Chemists use the term spontaneous to represent a change that occurs without outside ◀ The Combustion of Hydrogen Gas.  The exothermic reaction of hydrogen gas with oxygen to form water. (A hydrogen-filled balloon was ignited with a candle.)

887

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gas-filled flask

evacuated flask open valve

When the valve is opened, the gas expands irreversibly to fill both flasks.

intervention. There is an important observation to be made:  Spontaneous changes occur only in the direction that leads to equilibrium.  Many exothermic reactions are spontaneous. Examples you have seen include hydrogen and oxygen combining to form water, methane burning to give CO2 and H2O, Na and Cl2 reacting to form NaCl, and HCl(aq) and NaOH(aq) reacting to form H2O and NaCl(aq). It would be tempting to conclude that evolution of energy as heat is the criterion that determines whether a reaction or process is spontaneous. Further inspection, however, reveals significant flaws in this reasoning. This is especially evident when you consider some common spontaneous changes that are endothermic or energy neutral: •

Dissolving NH4NO3. The ionic compound NH4NO3 dissolves spontaneously in water. The process is endothermic (∆r H° = +25.7 kJ/mol).

•

Expansion of a gas into a vacuum. A system is set up with two flasks connected by a valve (Figure 18.1). One flask is filled with a gas, and the other is evacuated. When the valve is opened, the gas flows spontaneously from one flask to the other until the pressure is the same throughout. The expansion of an ideal gas is energy neutral (although expansion of most real gases is endothermic).

•

Phase changes. Melting of ice is an endothermic process. Above 0 °C, ice melts spontaneously. Below 0 °C, melting is not spontaneous. At 0 °C, no net change will occur; liquid water and ice coexist at equilibrium. This example illustrates that temperature can have a role in determining spontaneity and that equilibrium is an important aspect of the problem.

•

Energy transfer as heat. The temperature of cold water sitting in a warm environment will rise until it reaches the ambient temperature. The energy required for this endothermic process comes from the surroundings. Energy transfer as heat from a hotter object (the surroundings) to a cooler object (the water) is spontaneous.

•

Chemical Reactions. The reaction of H2 and I2 to form HI is endothermic, and the reverse reaction, the decomposition of HI to form H2 and I2, is exothermic. If H2(g) and I2(g) are mixed, a reaction forming HI will occur [H2(g) + I2(g) uv 2 HI(g)] until equilibrium is reached. Furthermore, if HI(g) is placed in a container, there will also be a reaction, but in the reverse direction, until equilibrium is achieved. Approach to equilibrium occurs spontaneously from either direction. 

Figure 18.1 Spontaneous expansion of a gas.

Further reflection reveals that evolution of heat cannot be a sufficient criterion in determining spontaneity. The first law of thermodynamics tells you that in any

Problem Solving Tip 18.1 A Review of Concepts of Thermodynamics To understand the thermodynamic concepts introduced in this chapter, be sure to review the following ideas from Chapter 5. System: The part of the universe under study. Surroundings: The rest of the universe exclusive of the system, capable of exchanging energy and/or matter with the system. Exothermic: Energy transfers as heat from the system to the surroundings.

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Endothermic: Energy transfers as heat from the surroundings to the system. First law of thermodynamics: The law of conservation of energy; energy cannot be created or destroyed. The change in internal energy of a system is the sum of energy transferred into or out of the system as heat and/or as work, ∆U = q + w. Enthalpy change: The energy transferred as heat under conditions of constant pressure when the only type of work possible is P-V work.

State function: A quantity whose value depends only on the state of the system; changes in a state function can be calculated based on the initial and final states of a system. Standard conditions: Pressure of 1 bar (1 bar = 0.98692 atm) and solution concentration of 1 m. Standard enthalpy of formation, ∆f H°: The enthalpy change occurring when 1 mol of a compound is formed from its elements in their standard states.

Chapter 18 / Principles of Chemical Reactivity: Entropy and Free Energy

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S 



qrev T

© Charles D. Winters/Cengage

process, energy must be conserved. If energy is transferred out of a system, then the same amount of energy must be transferred to the surroundings. Exothermicity of the system is always accompanied by an endothermic change in the ­surroundings. If energy evolution were the only factor determining whether a change is spontaneous, then for every spontaneous process there would be a corresponding ­nonspontaneous change in the surroundings. Adherence to the first law of thermodynamics alone is not enough to determine whether a change is spontaneous. What does determine spontaneity? Consider the following experiment: a piece of hot metal is placed in a beaker of cooler water (Figure 18.2). Energy transfers as heat spontaneously from the metal to the water until the temperature of the metal and the water is the same. In accord with the first law of thermodynamics, the total amount of energy in the combination of the piece of metal and the water is ­conserved, but there is directionality to this process. Energy transfers as heat from a ­hotter object to a colder object; you will never see a net flow of energy in the opposite direction, from a colder object to a hotter object. Is there a way to explain the directionality of this energy transfer? Now consider the initial and final states for this process of adding the hot metal to the cold water. Initially, energy is concentrated in the piece of metal. At the end of the process, energy has been dispersed between the metal and the water. This is the indicator for the directionality of chemical processes.  In a spontaneous process, energy goes from being more concentrated to being more dispersed.  There is a state function called entropy (S) that allows the dispersal of energy to be quantified. The second law of thermodynamics states that  a spontaneous ­process is one that results in an increase of the entropy of the universe. In a spontaneous process ∆S(universe) is greater than zero; this is a measure of the dispersal of energy in the process.  Because thermal energy is the result of the random motion of particles, potential energy is dispersed when it is converted to thermal energy. This conversion occurs when energy is transferred as heat, q. It is therefore not surprising that q is a part of the mathematical definition of ∆S. In addition, the effect of a given quantity of energy transferred as heat on energy dispersal is different at different temperatures. It turns out that a given q has a greater effect on ∆S at a lower temperature than at a higher temperature; that is, the extent of energy dispersal is inversely proportional to the temperature. The definition for ∆S is thus related to the quotient q/T, but it must be a little more specific about q. The value of q used in the calculation of an entropy change must be the energy transferred as heat under what are called reversible conditions, qrev. (This is described more fully in “A Closer Look: Reversible and Irreversible ­Processes.”) The mathematical definition of ∆S is therefore qrev divided by the ­absolute (Kelvin) temperature with units of J/K:

Figure 18.2  A spontaneous process.  ​The heated metal cylinder is placed in water. Energy transfers as heat spontaneously from the metal to water, that is, from the hotter object to the cooler object. Entropy  For a more complete

discussion of entropy, see the papers by F. L. Lambert, such as “Entropy Is Simple, Qualitatively,” Journal of Chemical Education, 2002, 79, 1241–1246, and references therein.

Second Law of Thermodynamics 

For a spontaneous process, ∆S(universe) > 0.

(18.1)

18.2 Entropy: A Microscopic Understanding Goal for Section 18.2 • Understand the statistical basis of entropy, that entropy is proportional to the number of ways that energy can be dispersed, that is, to the number of microstates available to a system.

Entropy is a measure of the extent of energy dispersal. In all spontaneous physical and chemical processes, energy changes from being localized or concentrated to being more dispersed or spread out.  In a spontaneous process, the change in entropy, ∆S, of the universe indicates the extent to which energy is dispersed. 

18.2  Entropy: A Microscopic Understanding

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A Closer Look

Reversible and Irreversible Processes

To determine entropy changes experimentally, the energy transferred by heating and cooling must be measured for a reversible process. But what is a reversible process? The test for reversibility is that after carrying out a change along a given path (for example, energy added as heat), it must be possible to return to the starting point by the same path (energy taken away as heat) without altering the surroundings. Melting of ice and freezing of water at 0 °C are examples of reversible processes. Given a mixture of ice and water at equilibrium, adding energy as heat in small increments will convert ice to water; removing energy

decreases by the amount of work expended by the surroundings. The system can be restored to its original state, but the surroundings will be altered in the process. In summary, there are two important points concerning reversibility: • At every step along a reversible pathway between two states, the system remains at equilibrium. • Spontaneous processes follow irreversible pathways and involve nonequilibrium conditions. To determine the entropy change for a process, it is necessary to identify a reversible pathway. Only then can an entropy change for the process be calculated from qrev and the Kelvin temperature.

as heat in small increments will convert water back to ice. Reversibility is closely associated with equilibrium. Assume a system is at equilibrium. Reversible changes can be made by very slightly perturbing the equilibrium and letting the system readjust. Spontaneous processes are not reversible. Suppose a gas is allowed to expand into a vacuum. No work is done in this process because there is no force to resist expansion. To return the system to its original state, it is necessary to compress the gas. Doing so means doing work on the system, however, because the system will not return to its original state on its own. In this process, the energy of the surroundings

So far, you have a definition of entropy, but no explanation for why dispersal of energy occurs. In order to explain this, you need to consider energy in its quantized form and matter on the atomic level.

Dispersal of Energy

Quantization of Matter and Energy ​ In the examples that follow dealing with the submicroscopic nature of matter, imagine that matter is made up of atoms and that energy is also quantized, coming in packets of energy called quanta.

You can explore the dispersal of energy using a simple example: energy being transferred as heat between hot and cold gaseous atoms. Consider an experiment involving two containers, one with hot atoms and the other with cold atoms. The atoms move randomly in each container and collide with the walls. When the containers are in contact, energy is transferred through the container walls. Eventually, both containers will be at the same temperature; the energy originally localized in the hotter atoms is distributed over a greater number of atoms; and the atoms in each container will have the same distribution of energies. A statistical explanation shows why energy is dispersed in a system. With statistical arguments, systems must include large numbers of particles for the arguments to be accurate. It will be easiest, however, to look first at simple examples with only a few particles to understand the underlying concepts and then extrapolate the ­conclusions to larger systems. Consider the system in Figure 18.3 in which, initially, there is one atom (1) with two discrete packets, or quanta, of energy and three other atoms (2, 3, and 4) with no energy. Collisions among the atoms allow energy to be transferred so that, Possible distribution of energy packets

Figure 18.3  Energy dispersal.  ​Possible

ways of distributing two packets of energy among four atoms. To keep the analysis simple, assume that initially there is one atom with 2 quanta of energy (1) and three atoms (2, 3, and 4) with no energy. There are 10 different ways to distribute the 2 quanta of energy among the four atoms. In these figures, atoms with the quanta of energy are shown in yellow, and those without are shown in blue. In addition, the atoms on which the quanta reside are listed below each figure.

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3

1 4

2 1,1

3

1 2

4

2,4

1

3

1 2

2

4

4

1,2

1,3

3

1

1 2

4

2

2,3

2

4

2 2,2 1

1 3

4,4

3

1 4

1,4 3

4

1

3

3

2 3,4

4

3 4

2 3,3

Chapter 18 / Principles of Chemical Reactivity: Entropy and Free Energy

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Chemistry in Your Career

Kaitlyn Tibbetts

Kaitlyn Tibbetts chemicals react to paper currency is key to identifying counterfeits. Local retail cashiers often use a detection pen to verify the authenticity of larger bills. When the chemicals in the ink react with paper that does not share the same composition of genuine currency paper, it reacts by turning brown. However, chemicals in certain household products can cause counterfeit currency to react like genuine currency paper as well. Tibbetts enjoys the variety and fast pace of her job. “Every request, every situation is unique; it keeps you on your toes and constantly intrigued.”

Kaitlyn Tibbetts is an analyst in the Investigative Support Operations Center of the U.S. Secret Service. “We conduct rapid analysis for ­ongoing investigations and protective threat cases. We also handle the agency’s Wanted Warrant Subjects and the research of individuals and businesses for cases.” Tibbetts uses chemistry in a variety of ways in her work, including detecting fingerprints, determining the compositions of substances, tracing DNA, and identifying counterfeit currency. She explains that understanding how certain

over time, all distributions of the two packets of energy over the four atoms are seen. There are 10 different ways to distribute these 2 quanta of energy over the four atoms. Each of these 10 different ways to distribute the energy is called a ­microstate. In only one of these microstates do the 2 quanta remain on atom 1. In fact, only in 4 of the 10 microstates [1,1; 2,2; 3,3; and 4,4] is the energy concentrated on a single atom. In the majority of cases, 6 out of 10, the energy is distributed to two different atoms. Even in this small sample (four atoms) with only two packets of energy, it is more likely that at any given time the energy will be distributed to two atoms rather than concentrated on a single atom. There is a distinct preference that the energy will be dispersed over a greater number of atoms. Now look at a second example of four atoms but with the quantity of energy increased from 2 quanta to 6 quanta. Assume that two atoms initially have 3 quanta of energy each. The other two atoms initially have zero energy (Figure 18.4a). Number of different ways to achieve this arrangement

6

5

5

4 3 2

ENERGY QUANTA

ENERGY QUANTA

4 6

12

12

24

6

6

4

4 3 2

1

1

0

0

(a) Initially, two particles each have 3 quanta of energy, and the other two have none. A total of 6 quanta of energy will be distributed once the four particles interact.

12

4

Figure 18.4 Energy dispersal.  ​Possible ways of distributing 6 quanta of energy among four atoms. A total of 84 microstates is possible.

(b) Once the particles begin to interact, there are nine ways to distribute the 6 available quanta. Each of these arrangements will have multiple ways of distributing the energy among the four atoms. Part (c) shows how the arrangement on the right can be achieved four ways. 6 ENERGY QUANTA

5 4 3

a

c

b

d

2 1

b c

d

a

c

d

a

b

d

a

b c

0 (c) There are four different ways to have four particles (a, b, c, and d) such that one particle has 3 quanta of energy and the other three each have 1 quantum of energy.

18.2  Entropy: A Microscopic Understanding

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Oesper Collection in the History of Chemistry/University of Cincinnati

Ludwig Boltzmann (1844–1906).  Engraved on his tombstone in Vienna, Austria, is his equation defining entropy. The constant k is now known as Boltzmann’s constant.

How Many Microstates?  To give you a sense of the number of microstates available to a substance, consider a mole of ice at 273 K, where S° = ​41.3 J/K ∙ mol. Using Boltzmann’s equa­tion, W =  101,299,000,000,000,000,000,000,000. That is, there are many, many more microstates for 1 mol of ice than there are atoms in the universe (about 1080).

Statistical Thermodynamics ​ The arguments presented here come from a branch of chemistry called statistical thermodynamics. See H. Jungermann, Journal of Chemical Education, 2002, 83, 1686–1694.

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Through collisions, energy can be transferred to achieve different distributions of energy among the four atoms. In all, there are 84 microstates, falling into nine basic patterns. For example, one possible arrangement has one atom with 3 quanta of energy, and three atoms with 1 quantum each. There are four microstates in which this is true (Figure 18.4c). Increasing the number of quanta from 2 to 6 with the same number of atoms increased the number of possible microstates from 10 to 84. With this increase, the proportion of microstates in which the energy is concentrated on one atom decreased from 40% to under 5%. Statistical analyses for larger aggregates of atoms and energy quanta become increasingly complex, but the conclusions are even more compelling. As the number of particles and/or quanta increases, the number of energy microstates grows rapidly. This larger number of microstates allows the energy to be dispersed to a greater extent. Ludwig Boltzmann proposed that the entropy of a system (the dispersal of energy at a given temperature) results from the number of microstates available.  As the number of microstates increases, so does the entropy of the system.  He expressed this idea in the equation

S = k lnW

(18.2)

which states that the entropy of a system, S, is proportional to the natural ­logarithm of the number of accessible microstates, W, that belong to a given energy of a ­system or substance. (The proportionality constant, k, is now known as Boltzmann’s ­constant and has a value of 1.381 × 10−23 J/K.) Within these microstates, it turns out that those states that disperse energy over the largest number of atoms are vastly more probable than the others.

Dispersal of Matter: Dispersal of Energy Revisited In many processes, it appears that the dispersal of matter also contributes to ­spontaneity, but the reason these processes are spontaneous is really energy dispersal and not matter dispersal. For example, the expansion of a gas into a ­vacuum ­(Figure 18.1) is spontaneous. How can this be related to energy dispersal and entropy? Begin with the premise that all energy is quantized and that this applies to any system, including gas molecules in a room or in a reaction flask. You know from the previous discussion of kinetic-molecular theory that the molecules in a gas sample have a distribution of energies (Figure 10.11) (often referred to as a Boltzmann ­distribution). The molecules are assigned to (or occupy) quantized microstates. Some molecules are in states of high or low energy, but most are in states near the average energy of the system. (For a gas in a laboratory-sized container, the energy levels are so closely spaced that, for most purposes, there is a continuum of energy states.) When the gas expands to fill a larger container, the average energy of the sample and the energy for the particles in a given energy range are constant. However, quantum mechanics show (for now, you will have to take our word for it) that as a consequence of having a larger volume in which the molecules can move in the expanded state, there is an increase in the number of microstates and that those microstates are even more closely spaced than before (Figure 18.5). The result of this greater density of microstates is that the number of microstates available to the gas particles increases when the gas expands. Gas expansion, a dispersal of matter, leads to the dispersal of energy over a larger number of microstates and thus to an increase in entropy. The logic applied to the expansion of a gas into a vacuum can be used to rationalize the mixing of two gases, the mixing of two liquids, or the dissolution of a solid

Chapter 18 / Principles of Chemical Reactivity: Entropy and Free Energy

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Note that for a gas in a container of the size likely to be found in a laboratory, the energy levels are so closely spaced that the system is usually regarded as having a continuum of energy levels, rather than quantized energy levels. ENERGY LEVELS

ENERGY LEVELS

Gas expands into a new container, doubling the volume.

Energy levels for a gas in a container. Shading indicates the total energy available.

Energy levels for a gas in a new container with twice the volume. More energy states are now available with the same total energy. The states are closer together.

Figure 18.5  Energy (and matter) dispersal.  ​As the size of the container for the chemical or physical change increases, the number of microstates accessible to the atoms or molecules of the system increases, and the density of states increases. A consequence of the distribution of molecules over a greater number of microstates is an increase in entropy.

in a liquid (Figure 18.6). For example, if flasks containing O2 and N2 are connected (in an experimental setup like that in Figure 18.5), the two gases diffuse together, eventually leading to a mixture in which O2 and N2 molecules are evenly distributed throughout the total volume. A mixture of O2 and N2 will never separate into samples of each component of its own accord. The gases spontaneously move toward a situation in which each gas and its energy are maximally dispersed. The energy of the system is dispersed over a larger number of microstates, and the entropy of the system ­ increases. Indeed, this is a large part of the explanation for the fact that similar liquids (such as oil and gasoline or water and ethanol) readily form ­homogeneous ­solutions. Recall the rule of thumb that “like dissolves like” (Section 13.2). Entropy Change on Gas Expansion ​

A Summary: Entropy, Entropy Change, and Energy Dispersal According to Boltzmann’s equation (Equation 18.2), entropy is proportional to the number of ways energy can be dispersed in a substance, that is, to the number of microstates available to the system (W). The number of microstates increases with an increased number of particles, with an increase in energy, and with an increase in volume. There will be an increase in entropy, ∆S, if there is an increase in the number of microstates over which energy can be dispersed.

The entropy change for a gas expansion can be calculated from ΔS = nR ln(Vfinal/Vinitial)

At a given temperature, V is pro­ portional to the number of microstates, so the equation is directly related to k ln(Wfinal/Winitial).

∆S = Sfinal − Sinitial = k (lnWfinal − lnWinitial) = k ln(Wfinal/Winitial)

Time



Photos: © Charles D. Winters/Cengage

The focus for the remainder of the chapter will be on the change in entropy (ΔS) that occurs due to the dispersal of energy in a system and its surroundings during a physical or chemical change. Figure 18.6 Dissolving KMnO4 in water.  ​A small quantity of solid, purple KMnO4 is added to water (left). With time, the solid dissolves, and the highly colored MnO4− ions (and the K+ ions) become dispersed throughout the solution. Entropy makes a large contribution to the mixing of liquids and solutions (Section 13.2).

18.2  Entropy: A Microscopic Understanding

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18.3 Entropy Measurement and Values Goals for Section 18.3 • Recognize that assigning a perfect crystal at 0 K to have zero entropy (the third law of thermodynamics) establishes a means of evaluating the entropy of a substance.

• Define standard molar entropies, S°, and evaluate and compare factors affecting S° values (temperature, volume, molecular structure, states of matter).

• Use values of standard molar entropies to calculate changes in entropy, ∆rS°, for a chemical reaction.

•

Raising the temperature of a substance corresponds to adding energy as heat. Thus, the entropy of a substance will increase with an increase in temperature.

•

Conversions from solid to liquid and from liquid to gas typically require large inputs of energy as heat. Consequently, there is a large increase in entropy in conversions involving changes of state (Figure 18.7).

John C. Kotz

© Charles D. Winters/Cengage

Negative Entropy Values  A glance at thermodynamic tables indicates that ions in aqueous solution can and do have negative entropy values listed. However, these are not absolute entropies. For ions, the entropy of H+(aq) is arbitrarily assigned a standard entropy of zero, and the entropy values for other ions are assigned relative to this value.

A numerical value for entropy can be determined for any substance under a given set of conditions. The greater the dispersal of energy, the greater the entropy and the larger the value of S. The point of reference for entropy values is established by the third law of thermodynamics. Defined by Ludwig Boltzmann, the third law states that  a perfect crystal at 0 K has zero entropy; that is, S = 0.  The entropy of an element or compound under any other set of conditions is the entropy gained by converting the substance from 0 K to those conditions. To determine the value of S, it is necessary to measure the energy transferred as heat under reversible conditions for the conversion from 0 K to the defined conditions and then to use Equation 18.1 (∆S = qrev/T). Because it is necessary to add energy as heat to raise the temperature,  all substances have positive entropy values at temperatures above 0 K.  Negative values of entropy cannot occur. Recognizing that entropy is directly related to energy added as heat allows you to predict several general features of entropy values:

(a)

The entropy of liquid bromine, Br2(ℓ), is 152.2 J/K ∙ mol, and that for bromine vapor is 245.47 J/K ∙ mol.

(b)

The entropy of ice, which has a highly ordered ­ olecular arrangement, is smaller than the m entropy of ­liquid water.

Figure 18.7  Entropy and states of matter.

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Chapter 18 / Principles of Chemical Reactivity: Entropy and Free Energy

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Standard Entropy Values, S°

S° (J/K ∙ mol)

The concept of standard states was introduced into the earlier discussion of enthalpy (Section 5.5), and the entropy of a substance in its standard state can be similarly defined. The standard molar entropy, S°, of a substance is the entropy gained by converting 1 mol of it from a perfect crystal at 0 K to standard state conditions (1 bar, 1 molal for a solution) at the specified temperature. The units for standard molar entropy values are J/K ∙ mol. Generally, values of S° found in tables of data refer to a temperature of 298 K. Appendix L lists many standard molar entropies at 298 K. More extensive lists of S° values can be found in standard reference sources such as the NIST tables (webbook.nist.gov). Scanning a list of standard entropies (such as those in Appendix L) shows that  large molecules generally have larger entropies than small molecules.  With a larger molecule, there are more ways for the molecule to rotate and vibrate, which provides a larger number of energy microstates over which energy can be distributed. For example, the standard entropies for methane (CH 4), ­ethane (C2H6), and propane (C3H8) are 186.3, 229.2, and 270.3 J/K ∙ mol, ­respectively. Also,  molecules with more complex structures have larger entropies than ­m olecules with simpler structures.  The effect of molecular structure can also be seen when comparing atoms or molecules of similar molar mass: Gaseous argon, CO 2 , and C 3 H 8 have entropies of 154.9, 213.7, and 270.3 J/K ∙ mol, respectively. Tables of entropy values also show that  entropies of gases are larger than those for liquids, and entropies of liquids are larger than those for solids.  In a solid, the particles have fixed positions in the solid lattice. When a solid melts, these particles have more freedom to assume different positions, resulting in an increase in the number of microstates available and an increase in entropy. When a liquid evaporates, constraints due to forces between the particles nearly disappear, the volume increases greatly, and a large entropy increase occurs. For example, the standard entropies of I2(s), Br2(ℓ), and Cl2(g) are 116.1, 152.2, and 223.1 J/K ∙ mol, respectively. Finally, as illustrated in Figure 18.7,  for a given substance, a large increase in entropy accompanies changes of state,  reflecting the relatively large energy transfer as heat required to carry out these processes (as well as the dispersion of energy over a larger number of available microstates). For example, the entropies of liquid and gaseous water are 69.95 and 188.84 J/K ∙ mol, respectively.

186.3 methane

229.2

ethane

270.3 propane

Problem Solving Tip 18.2 A List of Common Entropy-Favored Processes The discussion to this point allows the listing of several generalizations involving entropy changes:

• The entropy of a substance increases in going from a solid to a liquid to a gas. • The entropy of any substance

raised. Energy must be transferred to a system to increase its temperature (that is, q > 0), so qrev /T is necessarily positive.

• The entropy of a gas increases with an increase in volume. A larger volume provides a larger number

of energy states over which to ­disperse energy.

• Reactions that increase the number of moles of gases in a system are accompanied by an increase in entropy.

increases as the temperature is



18.3  Entropy Measurement and Values

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E xamp le 18.1

Entropy Comparisons Problem  Which substance has the higher entropy under standard conditions at 25 °C? Explain your reasoning. (a) NO2(g) or N2O4(g)

(b) I2(g) or I2(s)

What Do You Know?  Larger molecules of related substances have greater entropies than smaller molecules, and entropy decreases in the order gas > liquid > solid.

Strategy  For each part, identify the difference between the two substances and relate this to the general rules for entropy given above. Solution (a) Both NO2 and N2O4 are gases.  N2O4 is a larger molecule than NO2 and so is expected to have the higher standard entropy.  (b) For a given substance,  gases have higher entropies than solids, so I2(g) is expected to have the greater standard entropy. 

Think about Your Answer S° values in Appendix L confirm these predictions. At 25 °C, S° for NO2(g) is 240.04 J/K ∙ mol, and S° for N2O4(g) is 304.38 J/K ∙ mol. S° for I2(g) is 260.69 J/K ∙ mol; S° for I2(s) is 116.135 J/K ∙ mol.

Check Your Understanding  Predict which substance in each pair has the higher entropy and explain your reasoning. (a) O2(g) or O3(g)

(b) SnCl4(ℓ) or SnCl4(g)

Determining Entropy Changes in Physical and Chemical Processes It is possible to use standard molar entropy values quantitatively to calculate the change in entropy that occurs in various processes under standard conditions. The standard entropy change for a reaction (∆r S°) is the sum of the standard molar entropies of the products, each multiplied by its stoichiometric coefficient, minus the sum of the standard molar entropies of the reactants, each multiplied by its ­stoichiometric coefficient. © Charles D. Winters/Cengage



∆rS° = ΣnS°(products) − ΣnS°(reactants)

(18.3)

 This equation allows you to calculate entropy changes for a system in which reactants are completely converted to products, under standard conditions.  For example, for the oxidation of NO with O2 2 NO(g) + O2(g) n 2 NO2(g) The reaction of NO with O2. The entropy of the system decreases when two molecules of gas are produced from three molecules of gaseous reactants. (Here NO gas is bubbled into water. When the gas emerges from the water it encounters atmospheric oxygen and forms NO2.)

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∆r S° = (2 mol NO2/mol-rxn) S°[NO2(g)] −  {(2 mol NO(g)/mol-rxn) S°[NO(g)] + (1 mol O2/mol-rxn) S°[O2(g)]} = (2 mol NO2/mol-rxn)(240.0 J/K ∙ mol) −  [(2 mol NO(g)/mol-rxn)(210.8 J/K ∙ mol) + (1 mol O2/mol-rxn)(205.1 J/K ∙ mol)] = −146.7 J/K ∙ mol-rxn

The entropy of the system decreases, as is generally observed when some number of gaseous reactants has been converted to fewer molecules of gaseous products.

Chapter 18 / Principles of Chemical Reactivity: Entropy and Free Energy

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Ex am p le 18.2

Predicting and Calculating 𝚫rS° for a Reaction Problem  Using standard molar entropies, calculate the standard entropy changes for the following processes. (a) Evaporation of 1.00 mol of liquid ethanol to ethanol vapor: C2H5OH(ℓ) n C2H5OH(g) (b) Formation of ammonia from hydrogen and nitrogen based on the following equation: N2(g) + 3 H2(g) n 2 NH3(g)

What Do You Know?   For each part, you are given a balanced chemical equation and asked to determine the standard entropy change for the reaction (∆r S°). Values of standard molar entropies for the substances can be found in Appendix L. Strategy  Entropy changes for each system can be calculated from values of standard entropies (Appendix L) using Equation 18.3.

Solution (a) Evaporation of ethanol

Amount of Substance and Thermodynamic Calculations 

∆r S° = ΣnS°(products) − ΣnS°(reactants) = (1 mol C2H5OH(g)/mol-rxn) S°[C2H5OH(g)] −   (1 mol C2H5OH(ℓ)/mol-rxn) S°[C2H5OH(ℓ)] = (1 mol C2H5OH(g)/mol-rxn)(282.7 J/K ∙ mol) −  (1 mol C2H5OH(ℓ)/mol-rxn)(160.7 J/K ∙ mol) =  +122.0 J/K ∙ mol-rxn 

In the calculation here and in all others in this chapter, when we write, for example, 282.70 J/K ∙ mol

for the standard entropy of ­ethanol gas at 298 K, we mean 282.70 J/[K ∙ mol C2H5OH(g)]

(b) Formation of ammonia

The identifying formula has been left off for the sake of simplicity.

∆r S°  = ΣnS°(products) − ΣnS°(reactants)  = (2 mol NH3(g)/mol-rxn) S°[NH3(g)] −   {(1 mol N2(g)/mol-rxn) S°[N2(g)] + (3 mol H2(g)/mol-rxn) S°[H2(g)]}  = (2 mol NH3(g)/mol-rxn)(192.77 J/K ∙ mol) −  [(1 mol N2(g)/mol-rxn)(191.56 J/K ∙ mol) +  (3 mol H2(g)/mol-rxn)(130.7 J/K ∙ mol)]  =  −198.1 J/K ∙ mol-rxn 

Think about Your Answer   Predictions for the signs of these entropy changes can be made using the guidelines given in the text. In part (a), a large positive value for the entropy change is expected because the process converts ethanol from a liquid to a vapor. In part (b), a decrease in entropy is predicted because the number of moles of gases decreases from four to two.

Check Your Understanding  Calculate the standard entropy changes for the following processes using the entropy values in Appendix L. Are the signs of the calculated values of ∆r S° in accord with predictions? (a) Dissolving 1 mol of NH4Cl(s) in water: NH4Cl(s) n NH4Cl(aq) (b) Oxidation of ethanol: C2H5OH(g) + 3 O2(g) n 2 CO2(g) + 3 H2O(g)



18.3  Entropy Measurement and Values

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18.4 Entropy Changes and Spontaneity Goals for Section 18.4 • Calculate the change in entropy for a system, its surroundings, and the universe to determine whether a process is spontaneous under standard conditions.

• Recognize how reaction conditions influence whether a reaction is spontaneous. As illustrated by Example 18.2, the standard entropy change for the system in a physical or chemical change can be either positive (evaporation of ethanol) or negative (synthesis of ammonia from nitrogen and hydrogen). How does this information contribute to determining the spontaneity of the process? As discussed previously (Section 18.1), spontaneity is determined by the  ­second law of thermodynamics,  which states that  a spontaneous process is one that results in an increase of entropy in the universe.  The universe has two parts: the system and its surroundings (Section 5.1), and so the entropy change for the universe is the sum of the entropy changes for the system and the surroundings. That is, under standard conditions, the entropy change for the universe, ∆S°(universe) is Using 𝚫S°(universe)  For a process that is spontaneous under standard conditions: ΔS°(universe) > 0

For a process at equilibrium under standard conditions: ΔS°(universe) = 0

For a process that is not spontaneous under standard conditions: ΔS°(universe) < 0



∆S°(universe) = ∆S°(system) + ∆S°(surroundings)

(18.4)

The calculation in Example 18.2 solved for the entropy change under standard conditions for a system, only half of the information needed. You will also have to determine how the change affects the entropy of the surroundings. The value of ∆S°(universe) calculated from Equation 18.4 is the entropy change when reactants are converted completely to products, with all species at standard conditions.  A process is spontaneous under standard conditions if ∆S°(universe) is greater than zero.  As an example of the determination of reaction spontaneity, consider the reaction currently used to manufacture methanol, CH3OH CO(g) + 2 H2(g) n CH3OH(ℓ)

If ∆S°(universe) is positive, the conversion of 1 mol of CO(g) and 2 mol of H2(g) to 1 mol of CH3OH(ℓ) will be spontaneous under standard conditions. Calculating ∆S° (system)  To calculate ∆S°(system), start by defining the system to include the reactants and products. This means that ∆S°(system) corresponds to the entropy change for the reaction, ∆r S°. Calculation of this entropy change follows the procedure given in Example 18.2. ∆S°(system) = ∆r S° = ΣnS°(products) − ΣnS°(reactants)  = (1 mol CH3OH(ℓ)/mol-rxn) S°[CH3OH(ℓ)] −   {(1 mol CO(g)/mol-rxn) S°[CO(g)] + (2 mol H2(g)/mol-rxn) S°[H2(g)]}  = (1 mol CH3OH(ℓ)/mol-rxn)(127.2 J/K ∙ mol) − [(1 mol CO(g)/mol-rxn)(197.7 J/K ∙ mol) +  (2 mol H2(g)/mol-rxn)(130.7 J/K ∙ mol)]  = −331.9 J/K ∙ mol-rxn

A decrease in entropy for the system is expected because 3 mol of gaseous reactants are converted to 1 mol of a liquid product. Calculating ∆S° (surroundings)  Recall from Equation 18.1 that for a reversible change, ∆S is equal to qrev/T. Under constant pressure conditions (and when the only type of work possible is P-V work) and assuming a reversible process, the entropy change in the surroundings results from the fact that the enthalpy change for the reaction (qrev = ∆rH) affects the surroundings. For example, the energy associated with an exothermic chemical reaction is dispersed into the surroundings. Recognizing that

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Chapter 18 / Principles of Chemical Reactivity: Entropy and Free Energy

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∆H°(surroundings) = −∆H°(system) = −∆rH°, the entropy change for the surroundings can be calculated by the equation ∆S°(surroundings) = −∆r H°/T

For the synthesis of methanol by the reaction given, the enthalpy change can be ­calculated from enthalpy of formation data using Equation 5.7. ∆r H° = Σn∆f H°(products) − Σn∆f  H°(reactants)  = (1 mol CH3OH(ℓ)/mol-rxn) ∆f  H°[CH3OH(ℓ)] −  {(1 mol CO(g)/mol-rxn) ∆f  H°[CO(g)] + (2 mol H2(g)/mol-rxn) ∆f  H°[H2(g)]}  = (1 mol CH3OH(ℓ)/mol-rxn)(−238.4 kJ/mol) −  [(1 mol CO(g)/mol-rxn)(−110.5 kJ/mol) + (2 mol H2(g)/mol-rxn)(0 kJ/mol)]  = −127.9 kJ/mol-rxn

The entropy change for the surroundings in the methanol synthesis is ­calculated as follows. ∆S°(surroundings)  = −∆r H°/T  = −[(−127.9 kJ/mol-rxn)/298 K)](1000 J/kJ)  = +429.2 J/K ∙ mol-rxn

Calculating ∆S°(universe)  The pieces are now in place to calculate the entropy change in the universe. For the formation of CH3OH(ℓ) from CO(g) and H2(g), ∆S°(universe) is ∆S°(universe) = ∆S°(system) + ∆S°(surroundings)  = −331.9 J/K ∙ mol-rxn + 429.2 J/K ∙ mol-rxn  = +97.3 J/K ∙ mol-rxn

A Closer Look

The positive value indicates an increase in the entropy of the universe. It follows from the second law of thermodynamics that this reaction is spontaneous under standard conditions (see “A Closer Look: Entropy and Spontaneity?”).

Entropy and Spontaneity? In the text, ∆S°(universe) was calculated for the reaction

CO(g) + 2 H2(g) n CH3OH(ℓ)

This calculation was for the complete conversion of reactants in their standard states into products in their standard states. The value of ∆S°(universe) is positive for this transformation, and thus you can predict that this reaction is spontaneous. Does  this mean that the reaction will proceed spontaneously all the way from reactants to products? The answer is no. The  reaction will be spontaneous only to the point at which it is at equilibrium,



where there will be a mixture of reactants and products. When ∆S°(universe) is positive, this equilibrium position will be closer to the products than to the ­reactants—the reaction will be productfavored at equilibrium. Now consider a reaction that has a negative value for ∆S°(universe). The conclusion would be that the reaction completely converting reactants in their standard states into products in their standard states is not spontaneous. Does this mean that the reaction does not proceed at all? Again, the answer is no. Starting with pure reactants, the reaction will proceed spontaneously until it

reaches equilibrium, but the equilibrium position now will occur closer to the reactants than to the products—the reaction will be ­reactant-favored at equilibrium. References For more on the issues of spontaneity and equilibrium, see a series of articles by Lionel Raff: • L. Raff, Journal of Chemical Education, 2014, 91, 386–395. • L. Raff, Journal of Chemical Education, 2014, 91, 839–847. • L. Raff, Journal of Chemical Education, 2014, 91, 2128–2136. • L. Silverberg and L. Raff, Journal of Chemical Education, 2015, 92, 655–659.

18.4  Entropy Changes and Spontaneity

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E xamp le 18.3

Determining Whether a Process Is Spontaneous Problem Calculate ∆S°(universe) for the process of dissolving NaCl in water at 298 K. What Do You Know?  The process occurring is NaCl(s) n NaCl(aq). ∆S°(universe) is equal to the sum of ∆S°(system) and ∆S°(surroundings). Values of S° and ∆f H° for NaCl(s) and NaCl(aq) are given in Appendix L.

Strategy Step 1. Calculate ∆S°(system).  The entropy change for the system, ∆S°(system), can be calculated from values of S° for the two species using Equation 18.3. Step 2. Calculate ∆S°(surroundings).  ∆r H° can be calculated from values of ∆f  H° for the two species using Equation 5.7. ∆S°(surroundings) is determined by dividing −∆r H° for the process by the Kelvin temperature. Step 3. Calculate ∆S°(universe).  The sum of ∆S°(system) and ∆S°(surroundings) is ∆S°(universe). If this value is positive, then the process is spontaneous under standard conditions.

Solution Step 1. Calculate ∆S°(system). ∆r S°  = ΣnS°(products) − ΣnS°(reactants)  = (1 mol NaCl(aq)/mol-rxn) S°[NaCl(aq)] −  (1 mol NaCl(s)/mol-rxn) S°[NaCl(s)]  = (1 mol NaCl(aq)/mol-rxn)(115.5 J/K ∙ mol) −   (1 mol NaCl(s)/mol-rxn)(72.11 J/K ∙ mol)  = +43.39 J/K ∙ mol-rxn Step 2. Calculate ∆S°(surroundings). ∆r H°  = Σn∆f  H°(products) − Σn∆f  H°(reactants)  = (1 mol NaCl(aq)/mol-rxn) ∆f  H°[NaCl(aq)] −  (1 mol NaCl(s)/mol-rxn) ∆f  H°[NaCl(s)]  = (1 mol NaCl(aq)/mol-rxn)(−407.27 kJ/mol) −   (1 mol NaCl(s)/mol-rxn)(−411.12 kJ/mol)  = +3.85 kJ/mol-rxn The entropy change of the surroundings is determined by dividing −∆r H° by the Kelvin temperature. ∆S°(surroundings) = −∆r H°/T  = −[3.85 kJ/mol-rxn/298 K](1000 J/kJ)  = −12.92 J/K ∙ mol-rxn Step 3. Calculate ∆S°(universe). The overall entropy change—the change of entropy in the universe—is the sum of the values for the system and the surroundings. ∆S°(universe) = ∆S°(system) + ∆S°(surroundings)  = (+43.39 J/K ∙ mol-rxn) + (−12.92 J/K ∙ mol-rxn)  = +30.5 J/K ∙ mol-rxn 

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Chapter 18 / Principles of Chemical Reactivity: Entropy and Free Energy

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Think about Your Answer   The sum of the two entropy quantities is positive, indicating that the entropy in the universe increases; thus, the process is spontaneous under standard conditions. Notice that the spontaneity of the process results from ∆S°(system) and not from ∆S°(surroundings).

Check Your Understanding  Based on ∆r H° and ∆r S°, predict the spontaneity of the reaction of hydrogen and chlorine to give hydrogen chloride gas under standard conditions (at 298 K). Calculate ∆S°(universe) to verify your prediction. H2(g) + Cl2(g) n 2 HCl(g)

Spontaneous or Not? In the preceding examples, predictions about the spontaneity of a process under standard conditions were made using values of ∆S°(system) and ∆H°(system) ­c alculated from tables of thermodynamic data. It will be useful to look at all ­possibilities that result from the interplay of these two quantities. There are four possible outcomes when these two quantities are paired (Table 18.1). In two, ∆H°(system) and ∆S°(system) work in concert (Types 1 and 4 in Table 18.1). In the other two, the two quantities are opposed (Types 2 and 3). Processes in which both the standard enthalpy and entropy changes favor energy dispersal (Type 1) are always spontaneous under standard conditions. Processes disfavored by both their standard enthalpy and entropy changes in the system (Type 4) can never be spontaneous under standard conditions. Some examples that illustrate each situation are given below. Combustion reactions are always exothermic and often produce a larger number of gaseous product molecules from a few reactant molecules. They are Type 1 reactions. The combustion of butane is an example.

𝚫S°(universe), Spontaneity, and Standard Conditions  It is

important to reiterate that ΔH° and ΔS° values for a reaction are for the complete conversion of reactants to products under standard conditions. If ΔS°(universe) > 0, the reaction as written is spontaneous under standard conditions. However, you can calculate values for ΔS(universe) (without the superscript zero) for nonstandard conditions. If ΔS(universe) > 0, the reaction is spontaneous under those conditions.

2 C4H10(g) + 13 O2(g) n 8 CO2(g) + 10 H2O(g)

For this reaction, ∆H°(system) = ∆r H° = −5315.1 kJ/mol-rxn, and ∆S°(system) = ​ ∆r S° = ​312.4 J/K ∙ mol-rxn. Both contribute to this reaction being spontaneous under standard conditions.

Table 18.1 Reaction Type



Predicting Whether a Reaction Will Be Spontaneous under Standard Conditions

Spontaneous Process? (Standard Conditions)

𝚫H°(system)

𝚫S°(system)

1

Exothermic, < 0

Positive, > 0

Spontaneous at all temperatures. ∆S °(universe) > 0.

2

Exothermic, < 0

Negative, < 0

Depends on relative magnitudes of ∆H ° and ∆S °. Spontaneous at lower temperatures.

3

Endothermic, > 0

Positive, > 0

Depends on relative magnitudes of ∆H ° and ∆S °. Spontaneous at higher temperatures.

4

Endothermic, > 0

Negative, < 0

Not spontaneous at any ­temperature. ∆S °(universe) < 0. 18.4  Entropy Changes and Spontaneity

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Hydrazine, N2H4, is used as a high-energy rocket fuel. Synthesis of N2H4 from gaseous N2 and H2 would be attractive because these reactants are inexpensive.

© Charles D. Winters/Cengage

N2(g) + 2 H2(g) n N2H4(ℓ)

Figure 18.8 Thermal decomposition of NH4Cl(s).  ​ White, solid ammonium chloride, NH4Cl(s), is heated in a spoon. At high temperatures, decomposition to form NH3(g) and HCl(g) is spontaneous. At lower tem­peratures, the reverse reaction, forming NH4Cl(s), is spontaneous. As gaseous HCl(g) and NH3(g) cool, they recombine to form solid NH4Cl, the white “smoke” seen in this photo.

However, this reaction fits into the Type 4 category. The reaction is endothermic (∆H°(system) = ∆rH° = +50.63 kJ/mol-rxn), and the entropy change is negative (∆S°(system) = ∆r S° = −331.4 J/K ∙ mol-rxn) (1 mol of liquid is produced from 3 mol of gases), so the reaction is not spontaneous under standard conditions, and complete conversion of reactants to products will not occur without outside intervention. In the two other possible outcomes, entropy and enthalpy changes oppose each other. A process could be favored by the enthalpy change but disfavored by the entropy change (Type 2), or vice versa (Type 3). In either instance, whether a process is spontaneous depends on which factor is more important.

How Temperature Affects ∆S°(universe) Temperature also influences the value of ∆S°(universe). Because the enthalpy change for the surroundings is divided by the temperature to obtain ∆S°(surroundings), the numerical value of ∆S°(surroundings) will be smaller (either less positive or less negative) at higher temperatures. In contrast, ∆S°(system) and ∆H°(system) do not vary much with temperature. Thus, the effect of ∆S°(surroundings) relative to ∆S°(system) is diminished at higher temperature. Stated another way, at higher temperature, the enthalpy change becomes a less important factor in determining the overall entropy change. Consider the two cases where ∆H°(system) and ∆S°(system) are in opposition (Table 18.1): •

Type 2: Exothermic processes with ∆S°(system) < 0. Such processes become less favorable with an increase in temperature.

•

Type 3: Endothermic processes with ∆S°(system) > 0. These processes become more favorable as the temperature increases.

The effect of temperature is illustrated by two examples. The first (Type 2) is the reaction of N2 and H2 to form NH3. The reaction is exothermic, and thus it is favored by energy dispersal to the surroundings. The entropy change for the system is unfavorable, however, because the reaction, N2(g) + 3 H2(g) n 2 NH3(g), converts 4 mol of gaseous reactants to 2 mol of gaseous products. The favorable enthalpy effect [∆r S°(surroundings) = −∆H°(system)/T] becomes less important at higher temperatures. It is therefore reasonable to expect that the reaction will not be spontaneous if the temperature is too high. The second example (Type 3) considers the thermal decomposition of NH4Cl (Figure 18.8). At room temperature, NH4Cl is a stable, white, crystalline salt. When heated strongly, it decomposes to NH3(g) and HCl(g). The reaction is endothermic (enthalpy-disfavored) but entropy-favored because of the formation of 2 mol of gas from 1 mol of a solid reactant. The reaction is increasingly favored at higher temperatures.

18.5 Gibbs Free Energy Goals for Section 18.5 • Define the Gibbs free energy change, ∆G, and relate ∆rG° to ∆rH° and ∆rS°. • Relate ∆rG, ∆rG°, Q, and K to reaction spontaneity and product- and reactant-favorability. The method used so far to determine whether a process is spontaneous requires evaluation of two quantities, ∆S°(system) and ∆S°(surroundings). Wouldn’t it be convenient to have a single thermodynamic function that serves the same purpose?

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Chapter 18 / Principles of Chemical Reactivity: Entropy and Free Energy

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G = H − TS

where H is enthalpy, T is the Kelvin temperature, and S is entropy. In this equation, G, H, and S all refer to the system. Because enthalpy and entropy are state functions (Section 5.4), free energy is also a state function. Every substance possesses free energy, but the actual quantity is seldom known. Instead, just as with enthalpy (H) and internal energy (U), it is the changes in free energy, ∆G, that are measured in chemical and physical processes.

INTERFOTO/Alamy Stock Photo

A function associated with the system only—one that does not require assessment of the surroundings—would be even better. Such a function exists. It is called the Gibbs free energy, with the name honoring  J. Willard Gibbs (1839–1903). Gibbs free energy, G, often referred to simply as free energy, is defined mathematically as

J. Willard Gibbs (1839–1903) 

Gibbs received a Ph.D. from Yale University in 1863. His was the first Ph.D. in science awarded from a U.S. university.

The Change in the Gibbs Free Energy, 𝚫G Recall the equation defining the entropy change for the universe: ∆S(universe) = ∆S(surroundings) + ∆S(system)

The entropy change of the surroundings equals the negative of the change in enthalpy of the system divided by T. Thus, ∆S(universe) = −∆H(system)/T + ∆S(system)

Multiplying through this equation by −T, gives the equation −T∆S(universe) = ∆H(system) − T∆S(system)

Gibbs defined the free energy function so that ∆G(system) = −T∆S(universe). Thus, the general expression relating changes in free energy to the enthalpy and entropy changes in the system is the following: ∆G = ∆H − T∆S

Under standard conditions, the Gibbs free energy equation is written as

∆G° = ∆H° − T∆S°

(18.5)

Gibbs Free Energy, Spontaneity, and Chemical Equilibrium Because ∆G is related directly to ∆S(universe), the Gibbs free energy can be used as a criterion of spontaneity for physical and chemical changes. In each part of ­Figure 18.9 the free energy of pure, unmixed reactants is indicated on the left, and the free energy of the pure, unmixed products is indicated on the right. The extent of reaction, plotted on the x-axis, goes from zero to one.  ∆G° is the change in free energy accompanying the complete conversion of reactants into products under standard conditions (∆G° = G°products − G°reactants).  It has units of kJ. A closely related parameter, ∆rG°, is the change of G° as a function of the reaction composition on going from reactants to products. This corresponds to the slope of a line connecting reactants and products on this graph and has units of kJ/mol-rxn. As you shall see below, ∆rG° is related to the equilibrium constant, K. As the reaction proceeds, the free energy of the system (now a mixture of reactants and products) changes. The red lines in Figures 18.9a and 18.9b indicate the value of G for the system at each point during two different reactions. At any one point along the way from reactants to products the difference between the free energy of the system at that point and another is ∆G. In both cases in Figure 18.9, the free energy initially declines as reactants begin to form products; it reaches a minimum at equilibrium and then increases again from the equilibrium position to pure products.  The free energy at equilibrium, where there is a mixture of reactants and products, is always lower than the free

Free Energy, Spontaneity, and Equilibrium  ΔrG° is an intensive

quantity that describes how G° changes per mole of reaction. For more on these relationships, see J. Quilez, Journal of Chemical Education, 2012, 89, 87–93.

18.5  Gibbs Free Energy

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Reaction is reactant-favored at equilibrium ∆rG° is positive, K < 1

Increasing free energy, G

Increasing free energy, G

Reaction is product-favored at equilibrium ∆rG° is negative, K > 1

∆G° < 0 QK Slope > 0 ∆rG > 0

Q=K Slope = 0 ∆rG = 0 Equilibrium mixture

0 Reactants only (a)

Extent of reaction

1 Products only

∆G° > 0

QK Slope > 0 ∆rG > 0

1 Products only

Extent of reaction

© Charles D. Winters/Cengage

Figure 18.9  Free energy changes in the course of a reaction.  ​The difference in free energy between the pure reactants in their standard states and the pure products in their standard states is ΔG°. Here, Q is the reaction quotient, and K is the equilibrium constant. (See Section 15.2 for a discussion of the reaction quotient.)

A reactant-favored process.  If a sample of yellow lead(II) iodide is placed in pure water, a small amount of the compound will dissolve spontaneously (Δr G < 0 and Q < K) until equilibrium is reached. Because PbI2 is quite insoluble (Ksp = 9.8 × 10−9), however, the process of dissolving the compound is reactant-favored at equilibrium. Thus, the value of Δr G° is positive for the dissolving process.

904

energy of the pure reactants and of the pure products. A reaction proceeds spontaneously toward the minimum in free energy.  Now consider what happens to the instantaneous slope of the curve in ­Figure 18.9 as the reaction proceeds spontaneously toward equilibrium. Initially, this slope is negative. That is, the change in G per extent of reaction, ∆rG, has a ­negative value.  For all spontaneous reactions ∆rG < 0.  Eventually the free energy reaches a minimum. At this point, the instantaneous slope of the graph is zero and ∆rG = 0; the reaction has reached equilibrium. Beyond the equilibrium point, the instantaneous slope is positive (∆ rG > 0). Proceeding further toward products is not spontaneous. In fact, the reverse reaction will occur spontaneously (because ∆rG will be negative in the reverse direction), and the reaction will once again proceed toward equilibrium. An important observation in Figure 18.9a is that the equilibrium position occurs closer to the product side than to the reactant side. This is a product-favored reaction at equilibrium, and there is a connection between this observation and the quantity ∆rG°. That is,  a reaction with ∆rG° < 0 is product-favored at equilibrium.  In Figure 18.9b, you find the opposite. The reaction is reactant-favored at equilibrium.  A reaction with ∆rG° > 0 is reactant-favored at equilibrium.  There is a very useful relationship (Equation 18.6) between the change in free energy per extent of reaction under standard conditions (∆rG°) and the change in free energy per extent of reaction under nonstandard conditions (∆rG)

∆rG = ∆rG° + RT lnQ

(18.6)

where R is the universal gas constant (usually expressed in units of J/mol ∙ K or kJ/mol ∙ K), T is the temperature in kelvins, and Q is the reaction quotient (­Section 15.2). For a general reaction, a A + b B n c C + d D: Q

[C]c [D]d [A]a[B]b

Equation 18.6 indicates that, at a given temperature, ∆rG is determined by the values of ∆rG° and Q. When the system reaches equilibrium, no further net change in concentration of reactants and products will occur; at this point ∆rG = 0 and Q = K. Substituting these values into Equation 18.6 gives 0 = ∆r G° + RT lnK (at equilibrium)

Chapter 18 / Principles of Chemical Reactivity: Entropy and Free Energy

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Rearranging this equation leads to an important relationship between the standard free energy change for a reaction and the equilibrium constant, K, Equation 18.7:

∆rG° = −RT lnK

(18.7)

When ∆rG° is negative, K is greater than 1, and the reaction is said to be productfavored at equilibrium. The more negative the value of ∆rG°, the larger the equilibrium constant. This makes sense because, as described in Chapter 15, large equilibrium constants are associated with product-favored reactions. The converse is also true: For reactions that are reactant-favored at equilibrium, ∆rG° is positive, and K is less than 1. Finally, if K = 1 (a special set of conditions), then ∆rG° = 0. Equation 18.6 can also lead to the relationships between Q and K that were introduced in Chapter 15. ∆r G = ∆r G° + RT lnQ

Substituting −RT lnK for ∆rG° (Equation 18.7) gives ∆r G = −RT lnK + RT lnQ

This equation can be rearranged as follows: ∆r G = RT (lnQ − lnK ) ∆r G = RT ln (Q/K )

This means that for a spontaneous reaction where ∆rG is negative, Q must be less than K (Q < K), just as has been stated earlier. A similar analysis shows that if ∆rG is positive, then Q > K.

A Summary: Gibbs Free Energy (𝚫rG and 𝚫rG°), the Reaction Quotient (Q) and Equilibrium Constant (K), and Reaction Favorability Here is a summary of the relationships among ∆rG°, ∆rG, Q, and K. •

Free energy decreases to a minimum as a system approaches equilibrium (see Figure 18.9).

•

When ∆rG < 0, the reaction is proceeding spontaneously toward equilibrium and Q < K.

•

When ∆rG > 0, the reaction is not spontaneous and Q > K. It will be spontaneous in the reverse direction.

•

When ∆rG = 0, the reaction is at equilibrium; Q = K.

•

A reaction for which ∆rG° < 0 will proceed to an equilibrium position at which point the products will dominate in the reaction mixture because K > 1. That is, the reaction is product-favored at equilibrium.

•

A reaction for which ∆rG° > 0 will proceed to an equilibrium position at which point the reactants will dominate in the equilibrium mixture because K < 1. That is, the reaction is reactant-favored at equilibrium.

•

For the special condition where a reaction has ∆rG° = 0, the reaction is at equilibrium at standard conditions, with K = 1.

What Is “Free” Energy? The term free energy was not arbitrarily chosen. In any given process, the free energy represents the maximum energy available to do useful work (mathematically, ∆G calculated starting from the reactants and proceeding to the equilibrium position corresponds to wmax). In this context, the word free means available. To illustrate the reasoning behind this relationship, consider a reaction carried out under standard conditions and in which energy is evolved as heat (∆rH° < 0)

18.5  Gibbs Free Energy

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905

and entropy decreases (∆r S° < 0), such as the combustion of H2 gas. At first glance, it might seem reasonable that all the energy released as heat would be available to do work. This is not the case, however. A negative entropy change in this reaction means that energy is less dispersed in the products than in the reactants. A portion of the energy released from the reaction must be used to reverse energy dispersal in the system; that is, to concentrate energy in the product. The energy left over is free, or available to perform work.

18.6 Calculating and Using Standard Free Energies, 𝚫rG° Goals for Section 18.6 • Use free energies of formation, ∆f G°, to calculate the standard free energy change for a reaction.

• Use the equation ∆rG° = ∆rH° − T∆rS ° to assess the effect of changes in temperature on the favorability of a reaction.

• Relate ∆rG° and the equilibrium constant for a reaction.

Standard Free Energy of Formation The standard free energy of formation of a compound, D f G8, is the free energy change that occurs when forming one mole of the compound from the component elements, with products and reactants in their standard states. By defining ∆f G° in this way,  the free energy of formation of an element in its standard state is zero.  Just as the standard enthalpy or entropy change for a reaction can be calculated using values of ∆f H° (Equation 5.7) or S° (Equation 18.3), the standard free energy change for a reaction can be calculated from values of ∆f G° using a similar equation, where n represents the stoichiometric coefficient of the material in the ­balanced chemical equation under consideration:

∆rG° = Σn∆f G°(products) − Σn∆f  G°(reactants)

(18.8)

Calculating 𝚫rG°, the Free Energy Change for a Reaction Under Standard Conditions The free energy change for a reaction under standard conditions can be calculated from thermodynamic data in two ways, either from standard enthalpy and entropy changes using values of ∆rH° and ∆rS° or directly from values of ∆f G° found in tables. These calculations are illustrated in the following two examples.

E xamp le 18.4

Calculating 𝚫rG° from 𝚫rH° and 𝚫rS° Problem  Calculate the standard free energy change, ∆r G°, for the formation of methane from carbon and hydrogen at 298 K, using tabulated values of ∆f  H° and S°. Is the reaction product-favored or reactant-favored at equilibrium? C(graphite) + 2 H2(g) n CH4(g)

906

Chapter 18 / Principles of Chemical Reactivity: Entropy and Free Energy

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What Do You Know?   You are given a balanced chemical equation. Standard molar enthalpies of formation and standard molar entropies can be found in Appendix L. Strategy  The values of ∆f H° and S° are first combined to find ∆r H° and ∆r S°. With these values known, ∆r G° can be calculated using Equation 18.5. (Remember that S° values are given in units of J/K ∙ mol, whereas ∆f H° values are given in units of kJ/mol.) Solution C(graphite)

+

2 H2(g)

n

CH4(g)

∆f H° (kJ/mol)

 0

  0

  −74.9

S° (J/K ∙ mol)

+5.6

+130.7

+186.3

From these values, you can find both ∆r H° and ∆r S° for the reaction: ∆r H°  = Σn∆f  H°(products) − Σn∆f  H°(reactants)  = (1 mol CH4(g)/mol-rxn) ∆f  H°[CH4(g)] − {(1 mol C(graphite)/mol-rxn) ∆f  H°[C(graphite)] +   (2 mol H2(g)/mol-rxn) ∆f  H°[H2(g)]}  = (1 mol CH4(g)/mol-rxn)(−74.9 kJ/mol) −  [(1 mol C(graphite)/mol-rxn)(0 kJ/mol) + (2 mol H2(g)/mol-rxn)(0 kJ/mol)]  = −74.9 kJ/mol-rxn ∆r S°  = ΣnS°(products) − ΣnS°(reactants)  = (1 mol CH4(g)/mol-rxn) S°[CH4(g)] − {(1 mol C(graphite)/mol-rxn) S°[C(graphite)] +   (2 mol H2(g)/mol-rxn) S°[H2(g)]}  = (1 mol CH4(g)/mol-rxn)(186.3 J/K ∙ mol) − [1 mol C(graphite)/mol-rxn](5.6 J/K ∙ mol) +   (2 mol H2(g)/mol-rxn)(130.7 J/K ∙ mol)]  = −80.7 J/K ∙ mol-rxn Combining the values of ∆r H° and ∆r S° using Equation 18.5 gives ∆r G°. ∆r G°  = ∆r H° − T∆r S°  = −74.9 kJ/mol-rxn − [(298 K)(−80.7 J/K ∙ mol-rxn)](1 kJ/1000 J)  =  −50.9 kJ/mol-rxn  ∆r G° is negative at 298 K, so the reaction is predicted to be  product-favored at equilibrium. 

Think about Your Answer   In this example, the product T∆ rS° is negative (−24.0 kJ/mol-rxn) and disfavors the reaction. However, the entropy change is relatively small, and ∆r H° = −74.9 kJ/mol-rxn is the dominant term. Chemists call this an enthalpydriven reaction.

Check Your Understanding  Using values of ∆f H° and S° to find ∆r H° and ∆r S°, calculate the free energy change, ∆r G°, for the formation of 2 mol of NH3(g) from the elements at standard conditions and 25 °C. N2(g) + 3 H2(g) n 2 NH3(g)



18.6  Calculating and Using Standard Free Energies, ∆r G°

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907

E xamp le 18.5

Calculating 𝚫rG° Using Free Energies of Formation Problem  Calculate the standard free energy change for burning one mole of methane using values for standard free energies of formation of the products and reactants and assuming the products are all gases. Is the reaction product-favored at equilibrium?

What Do You Know?   You are asked to determine the standard free energy change for a reaction (∆r G°). Values of standard molar free energies of formation for the substances involved in the reaction can be found in Appendix L.

Strategy   Write a balanced equation for the reaction. Then, use Equation 18.8 with ­values of ∆f G° obtained from Appendix L. Solution  The balanced equation and values of ∆f G° for each reactant and product are CH4(g) ∆f  G°(kJ/mol)

−50.8

+

2 O2(g)

n

0

2 H2O(g) −228.6

+

CO2(g) −394.4

These values can then be substituted into Equation 18.8. ∆r G°  = Σn∆f  G°(products) − Σn∆f  G°(reactants)  = {(2 mol H2O(g)/mol-rxn) ∆f  G°[H2O(g)] + (1 mol CO2(g)/mol-rxn) ∆f  G°[CO2(g)]} − {(1 mol CH4(g)/mol-rxn) ∆f  G°[CH4(g)] + (2 mol O2(g)/mol-rxn) ∆f  G°[O2(g)]}  = [(2 mol H2O(g)/mol-rxn)(−228.6 kJ/mol) + (1 mol CO2(g)/mol-rxn)(−394.4 kJ/mol)] − [(1 mol CH4(g)/mol-rxn)(−50.8 kJ/mol) + (2 mol O2(g)/mol-rxn)(0 kJ/mol)]  =  −800.8 kJ/mol-rxn  The large negative value of ∆ r G° indicates that the reaction is  product-favored at equilibrium. 

Think about Your Answer  Common errors made by students in this calculation are (1) overlooking the stoichiometric coefficients in the equation and (2) confusing the signs for the terms when using Equation 18.8.

Check Your Understanding  Calculate the standard free energy change for the oxidation of 1.00 mol of SO2(g) to form SO3(g) using values of ∆f G°.

Free Energy and Temperature The definition of free energy, G = H − TS, informs you that free energy is a function of temperature, so ∆rG° will change as the temperature changes (Figure 18.10). A consequence of this dependence on temperature is that, in certain instances, reactions can be product-favored at equilibrium at one temperature and reactantfavored at another. Those instances arise when the ∆r H° and T∆r S° terms work in opposite directions: •

Processes that are entropy-favored (∆ r   S° > 0) but enthalpy-disfavored (∆ r H° > 0)

•

Processes that are enthalpy-favored (∆ r H° < 0) but entropy-disfavored (∆ r  S°  0

Dehydration of CuSO4 ∙ 5 H2O and other hydrates is favorable only at higher temperatures.

© Charles D. Winters/Cengage

© Charles D. Winters/Cengage

Reaction of potassium with water is favorable at all temperatures.

∆rG° > 0

∆rG° > 0

∆rH° < 0 ∆rS° < 0 0

0

Increasing Temperature

Reactant-favored reactions ∆rG° = ∆rH° − T∆rS° > 0

0

∆rH° < 0 ∆rS° > 0 ∆rG° < 0

∆rS° < 0 ∆rH° > 0

Product-favored reactions ∆rG° = ∆rH° − T∆rS° < 0

∆rH° > 0 ∆rS° > 0 ∆rG° < 0

Increasing Temperature

∆rG° < 0

Blue line: ∆rH° < 0 and ∆rS° < 0. Favored at low T. Red line: ∆rH° > 0 and ∆rS° > 0. Favored at high T.

∆rH° < 0 and ∆rS° > 0. Product-favored at all temperatures.

Increasing Temperature ∆rH° > 0 and ∆rS° < 0. Reactant-favored at all temperatures.

Figure 18.10  The variation in 𝚫rG° with temperature.

component of limestone, marble, and seashells. Heating CaCO 3 produces lime, CaO, an important chemical, along with gaseous CO2. The data below from ­Appendix L are at 298 K (25 °C). CaCO3(s)

n

CaO(s)

+

CO2(g)

∆f G° (kJ/mol)

−1129.16

−603.42

−394.36

∆f H° (kJ/mol)

−1207.6

−635.09

−393.51

S° (J/K ∙ mol)

    91.7

    38.2

  213.74

For the conversion of 1 mol of CaCO 3(s) to 1 mol of CaO(s) under standard conditions, ∆ r G° = +131.38  kJ/mol-rxn, ∆ r H° = +179.0  kJ/mol-rxn, and ∆r S° = +160.2 J/K ∙ mol-rxn. Although the reaction is entropy-favored, the large positive and unfavorable enthalpy change dominates at 298 K. Thus, the standard free energy change is positive at 298 K and 1 bar, indicating that the reaction is reactant-favored at equilibrium. The temperature dependence of ∆ rG° provides a means to turn the CaCO 3 decomposition into a product-favored reaction. Notice that the entropy change in the reaction is positive as a result of the formation of CO2 gas in the reaction. Thus, raising the temperature results in the value of T∆r S° becoming increasingly larger. At a high enough temperature, T∆r S° will outweigh the enthalpy effect, and the process (shown by the red line in the middle diagram in Figure 18.10) will become product-favored at equilibrium. How high must the temperature be for this reaction to become product-favored? An estimate of the temperature can be obtained using Equation 18.5, by calculating the temperature at which ∆rG° = 0. Above that temperature, ∆rG° will have a negative value. ∆r G° = ∆r H° − T∆r S° 0 = (179.0 kJ/mol-rxn)(1000 J/kJ) − T(160.2 J/K ∙ mol-rxn) T = 1117 K (or 844 °C)

CaCO3 Decomposition Experiments show that the pressure of CO2 in an equilibrium system [CaCO3(s) uv CaO(s) + ​ CO2(g)] is 1 bar at about 900 °C (K = 1 and Δr G° = 0), close to the estimated temperature.

18.6  Calculating and Using Standard Free Energies, ∆r G°

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909

How accurate is this result? As noted earlier, this answer is only an estimate of the temperature needed. One source of error is the assumption that ∆r H° and ∆r S° do not vary with temperature, an assumption that is not strictly true. There is always a small variation in these values when the temperature changes—not large enough to be important if the temperature range is narrow, but potentially a problem over wider temperature ranges such as seen in this example. As an estimate, however, a temperature in the range of 850 °C for this reaction is reasonable.

Exampl e 1 8 .6

Effect of Temperature on 𝚫rG° Problem   The decomposition of liquid Ni(CO)4 to produce nickel metal and carbon monoxide has a ∆r G° value of 40 kJ/mol-rxn at 25 °C. Ni(CO)4(ℓ) n Ni(s) + 4 CO(g) Use values of ∆f H° and S° for the reactant and products to estimate the temperature at which the reaction becomes product-favored at equilibrium.

What Do You Know?   You are given a balanced chemical equation. Values of standard molar enthalpies of formation and standard molar entropies can be found in the chemical literature. You also know that the objective is to determine the temperature at which ∆r G° is zero.

Strategy Map Problem Determine the temperature above which a chemical ­reaction is product-favored at equilibrium. Data/Information • Balanced chemical equation • Temperature • Values of ∆f H° and S°

Step 1

Strategy  The reaction is reactant-favored at equilibrium at 298 K, as indicated by the positive value for ∆r G° at this temperature. However, if the entropy change is positive for the reaction and the reaction is endothermic (with a positive value of ∆r H°), then a higher temperature may allow the reaction to become product-favored at equilibrium. Step 1. Calculate ∆r H° and ∆r S° to see if their values meet these criteria. Step 2. If the necessary criteria are met in Step 1, then determine the temperature at which the following condition is met: 0 = ∆r H° − T∆r S°.

Solution   Values for ∆r H° and ∆rS° are obtained from the chemical literature for the ­substances involved. Ni(CO)4(ℓ)

n

Ni(s)

∆f H°(kJ/mol)

−632.0

 0

S°(J/K ∙ mol)

  320.1

29.87

+

4 CO(g) −110.525    197.67

Calculate ∆r H° using Equation 5.7 and calculate ∆r S° using Equation 18.3. For a process in which 1 mol of liquid Ni(CO)4 is converted to 1 mol of Ni(s) and 4 mol of CO(g), the standard enthalpy of reaction and standard entropy of reaction are ∆r H° = +189.90 kJ/mol-rxn ∆r S°  = +500.45 J/K mol-rxn

Step 2

Determine the temperature at which ∆r G° 5 0. When both ∆rH° and ∆rS° are positive, increasing the temperature high enough will result in a product-favored reaction at equilibrium. ∆r G° = ∆r H° − T∆r S° 0 = (189.90 kJ/mol-rxn)(1000 J/kJ) − T(500.45 J/K ∙ mol-rxn) T =  379.5 K (or 106.3 °C) 

910

Chapter 18 / Principles of Chemical Reactivity: Entropy and Free Energy

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Think about Your Answer   At 298 K, the reaction is reactant-favored at equilibrium largely because it is quite endothermic. However, the positive entropy change allows the reaction to be product-favored at equilibrium at a slightly higher temperature.

Check Your Understanding  Oxygen was first prepared by Joseph Priestley (1733–1804) by heating HgO. Use data in Appendix L to estimate the temperature required to decompose HgO(s) into Hg(ℓ) and O2(g).

Using the Relationship between 𝚫rG° and K Equation 18.7 provides a direct route to determine the standard free energy change from experimentally determined equilibrium constants.

Ex am p le 18.7

Calculating 𝚫rG° from Ksp for an Insoluble Solid Problem   The value of Ksp for AgCl(s) at 25 °C is 1.8 × 10−10. Determine ∆r G° for the process Ag+(aq) + Cl−(aq) uv AgCl(s) at 298.15 K.

What Do You Know  You are given the value of Ksp. The process of interest is the reverse of the chemical equation for Ksp. You also know the temperature.

Strategy  The chemical equation given is the opposite of the equation used to define Ksp; therefore, the equilibrium constant for this reaction is 1/Ksp. This value is used in Equation 18.7 to calculate ∆r G°. Solution For Ag+(aq) + Cl−(aq) uv AgCl(s), K = 1/Ksp = 1/ 1.8 × 10−10 = 5.56 × 109 ∆r G° = −RT lnK = −(8.3145 J/K ∙ mol-rxn)(298.15 K) ln(5.56 × 109)  = −55,600 J/mol-rxn =  −55.6 kJ/mol-rxn 

Think about Your Answer  The negative value of ∆rG° indicates that the precipitation of AgCl from Ag+(aq) and Cl−(aq) is product-favored at equilibrium.

Check Your Understanding  The value of Ksp for PbI2(s) at 25 °C is 9.8 × 10−9. Determine ∆rG° for the process Pb2+(aq) + 2 l−(aq) uv PbI2(s) at 298.15 K.

You can solve Equation 18.7 for K by the following steps. First, divide both sides of the equation by −RT 

r G   ln K RT

and then take the antilog of both sides of the equation.



K e

rG  RT



(18.9)

This equation can be used to calculate the value of K if the value of ∆rG° is known.



18.6  Calculating and Using Standard Free Energies, ∆r G°

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911

E xamp le 18.8

Calculating Kp from 𝚫rG° Problem   Determine the standard free energy change, ∆r G°, for the environmentally important reaction of nitrogen monoxide and oxygen to form nitrogen dioxide, according to the following chemical equation at 25.0 °C. 2 NO(g) + O2(g) uv 2 NO2(g) Use this value to calculate the equilibrium constant for this reaction at this temperature.

What Do You Know  Values of ∆f G° are given in Appendix L. Strategy   Free energy of formation values are used to calculate ∆r G° for the reaction using Equation 18.8. The equilibrium constant for this reaction is calculated from ∆r G° using Equation 18.9. Because the reactants and products are gases, the calculated value will be Kp.

Solution  2 NO(g) ∆f G° (kJ/mol)

+

86.58

O2(g)

uv

0

2 NO2(g) 51.23

These values can be substituted into Equation 18.8. ∆r G°  = Σn∆f  G°(products) − Σn∆f  G°(reactants)  = (2 mol NO2(g)/mol-rxn) ∆f  G°[NO2(g)] − {(2 mol NO(g)/mol-rxn) ∆f  G°[NO(g)] + (1 mol O2(g)/mol-rxn) ∆f  G°[O2(g)]}  = (2 mol NO2(g)/mol-rxn)(51.23 kJ/mol)  − {(2 mol NO(g)/mol-rxn)(86.58 kJ/mol) + (1 mol O2(g)/mol-rxn)(0 kJ/mol)}  =  −70.70 kJ/mol-rxn  This value is then used in Equation 18.9 to solve for Kp. 

r G  RT



70.70 kJ/mol-rxn (0.0083145 kJ/K  mol)(298.2 K)

Kp  e e

 Kp = 2.4 × 1012 

Think about Your Answer   The value of ∆r G° is less than zero, indicating that this reaction is product-favored at equilibrium. The value of Kp calculated is greater than 1 as it should be for such a process.

Check Your Understanding  Determine the value of the equilibrium constant, Kp, for the decomposition of liquid water according to the following equation at 25.0 °C. 2 H2O(ℓ) uv 2 H2(g) + O2(g)

Calculating 𝚫rG, the Free Energy Change for a Reaction Using 𝚫rG° and the Reaction Quotient Equation 18.6 allows you to calculate the value of ∆rG, the free energy change for a reaction at nonstandard conditions, given the values of ∆rG° and the reaction ­quotient, Q.

912

Chapter 18 / Principles of Chemical Reactivity: Entropy and Free Energy

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Ex am p le 18.9

Calculating 𝚫rG from 𝚫rG° and Q Problem  Gaseous iodine monochloride can be decomposed into iodine and chlorine gases according to the following chemical equation 2 ICl(g) n I2(g) + Cl2(g) (a) Calculate ∆r G° for this reaction at 298 K using values for the standard free energies of formation of the products and reactant. Is this reaction reactant-favored or productfavored at equilibrium? (b) Calculate the value of ∆r G at 298 K for this reaction if the reactant and products are mixed with the following partial pressures: 1.0 atm ICl, 1.0 × 10−3 atm I2, and 1.0 × 10−3 atm Cl2. Is the reaction spontaneous under these conditions?

What Do You Know?  You are given a balanced chemical equation. Standard free energies of formation are listed in Appendix L.

Strategy  For part (a), use Equation 18.8 with values of ∆f G° to obtain the value of ∆r G°. For part (b), use Equation 18.6 to obtain the value of ∆r G under the conditions given.

Solution 2 ICl(g)

(a) ∆f G° (kJ/mol)

n

−5.73

I2(g) 19.33

+

Cl2(g) 0

These values can be substituted into Equation 18.8. ∆r G° = Σn∆f  G°(products) − Σn∆f  G°(reactants)  = {(1 mol I2(g)/mol-rxn) ∆f  G°[I2(g)] + (1 mol Cl2(g)/mol-rxn) ∆f  G°[Cl2(g)]} −   (2 mol ICl(g)/mol-rxn) ∆f  G°[ICl(g)]  = [(1 mol I2(g)/mol-rxn)(19.33 kJ/mol) + (1 mol Cl2(g)/mol-rxn)(0 kJ/mol)] −   (2 mol ICl(g)/mol-rxn)(−5.73 kJ/mol) =  30.79 kJ/mol-rxn  ∆r G° is positive at 298 K, so the reaction is predicted to be  reactant-favored at equilibrium.  (b) Equation 18.6 can be used to calculate ∆ rG for this reaction under the given conditions. ∆r G = ∆r G° + RT lnQ  = ∆r G° + RT ln(PI2PCl2/PICl2)  = 30.79 kJ/mol-rxn +  (0.0083145 kJ/K ∙ mol-rxn)(298 K)ln[(1.0 × 10−3)(1.0 × 10−3)/(1.0)2]  =  −3.44 kJ/mol-rxn  ∆r G is negative at 298 K, so  the reaction will proceed spontaneously until the r­ eactant-favored equilibrium position is reached. 

Think about Your Answer  Even though the equilibrium position for this reaction favors the reactants, the reaction is spontaneous under the conditions presented in part (b) of the question. Even a reactant-favored reaction is spontaneous until equilibrium is attained.



18.6  Calculating and Using Standard Free Energies, ∆r G°

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913

Check Your Understanding  Nitrogen and oxygen can react to form nitrogen monoxide according to the following chemical equation N2(g) + O2(g) n 2 NO(g) (a) Calculate ∆r G° for this reaction at 298 K using values for the standard free energies of formation of the product and reactants. Is this reaction reactant-favored or productfavored at equilibrium? (b) Calculate the value of ∆r G at 298 K for this reaction if the reactants and product are mixed with the following partial pressures: 0.10 atm N2, 0.10 atm O2, and 0.010 atm NO. Is this reaction spontaneous under these conditions?

18.7 The Interplay of Kinetics and Thermodynamics Goal for Section 18.7 • Understand how both thermodynamics and kinetics affect chemical reactions. You have learned about both thermodynamics and kinetics in this text. Thermodynamics tells you whether a chemical reaction can occur on its own. Kinetics tells you how fast a reaction occurs. Just because a reaction is product-favored at equilibrium, that is, thermodynamically favored, does not mean that it will necessarily proceed quickly. For example, at 25 °C and atmospheric pressure, ∆rG° for the conversion of diamond to graphite has a value of −2.9 kJ/mol-rxn. Thermodynamics favors the transformation. Because it has a lower free energy, graphite is more thermodynamically stable than diamond under these conditions, but the activation energy for this process, a kinetics parameter, is very large. This causes the reaction to be so slow that it does not occur on an ordinary timescale. A diamond’s long existence under normal conditions is due to the slow kinetics of the process, rather than to its thermodynamics. While diamond is thermodynamically unstable compared to graphite, diamond is said to be kinetically stable due to the large activation energy for the process. Likewise the transformation of hydrogen gas and oxygen gas to water 2 H2(g) + O2(g) n 2 H2O(g)

is thermodynamically very favored with a ∆rG° value of −457 kJ/mol-rxn at 25 °C. Yet a balloon can be filled with a mixture of hydrogen and oxygen and float in a room for a long time until some external source of energy, such as a lighted candle, is provided (page 886). Hydrogen and oxygen are thermodynamically unstable with respect to water at this temperature but are kinetically stable because the activation energy for the reaction is large enough that ambient conditions do not provide enough energy to overcome this barrier to reaction. Once some energy is provided, however, the reaction of some molecules proceeds; the energy released in the reaction of this small number of molecules then provides the energy needed for more of the reaction to occur, and the reaction proceeds explosively. Yet another example of the interplay of thermodynamics and kinetics is the addition of hydrogen bromide to 1,3-butadiene. Two products are possible: 3-bromo-1-butene and 1-bromo-2-butene. H2CPCHCHPCH2 + HBr n H2CPCHCHBrCH3 + BrCH2CHPCHCH3 3-bromo-1-butene 1-bromo-2-butene

914

Yield at −80 °C

81%

19%

Yield at 45 °C

15%

85%

Chapter 18 / Principles of Chemical Reactivity: Entropy and Free Energy

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Activation energy for formation of 3-bromo-1-butene is lower than that for formation of 1-bromo-2-butene

Br– H H3C

C H

CHCH

C

H

H

CH2 + HBr

1-bromo-2-butene has a lower free energy and so is thermodynamically more stable than 3-bromo-1-butene

Energy

CH2

C +

CH3CHCH

CH2

Br 3-bromo-1-butene CH3CH CHCH2Br 1-bromo-2-butene

Figure 18.11  Reaction coordinate diagram for the addition of HBr to 1,3-butadiene.

Depending on the temperature, different ratios of these two products are obtained. At a low temperature, such as –80 °C, 3-bromo-1-butene predominates, whereas at a higher temperature, such as 45 °C, 1-bromo-2-butene is the major product. Why is this so? The reaction coordinate diagram for this reaction is shown in Figure 18.11. Butadiene first forms an intermediate (called an allylic carbocation) by the addition of a hydrogen ion from the hydrogen bromide. This intermediate can then form either of the two products by addition of the bromide ion. Br– H

Favored at lower temperatures

H3C

C H

C +

Favored at higher temperatures

C

H

H

The point of attack of Br– on the allylic carbocation depends on temperature.

At –80 °C, the product ratio is influenced by the rates of the chemical reactions and is said to be under kinetic control. The activation energy for the formation of 3-bromo-1-butene is smaller, so this product forms faster and is thus present in a larger amount. As the temperature is raised, the number of molecules following the two paths is increased and the ratio of products changes. More molecules have sufficient energy to form both products. In addi­tion, reformation of the intermediate and Br− from the two product molecules (the reverse of the second step of this reaction) also occurs. This provides a mechanism allowing the two products to reach equilibrium. 3-bromo-1-butene uv 1-bromo-2-butene

At 45 °C, the ratio of the two products is established by this equilibrium. A ­reaction run under these conditions, in which the ratio of products is defined by the equilibrium between these two species, is said to be under thermodynamic ­control, that is, thermodynamics is controlling the ratio of products. In studying chemical reactions, therefore, both kinetics and thermodynamics play important roles.

18.7  The Interplay of Kinetics and Thermodynamics

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915

Applying Chemical Principles

The laws of thermodynamics apply to all chemical reactions. It should come as no surprise, therefore, that issues of spontaneity and calculations involving ∆ r G also arise in studies of biochemical reactions. For biochemical processes, however, a different standard state is often used. Most of the usual definition is retained: 1 bar pressure for gases and 1 m concentration for aqueous solutes with the exception of one very important solute. Rather than using a standard state of 1 molal for hydronium ions (corresponding to a pH of about 0), biochemists use a hydronium concentration of 1 × 10−7 M, corresponding to a pH of 7. This pH is more useful for biochemical reactions. When biochemists use this as the standard state, they write the symbol ′ next to the thermodynamic function. For example, they write ∆G°′ (pronounced delta G zero prime). Living things require energy to perform their many functions. One of the main reactions providing this energy is that of adenosine triphosphate (ATP) with water, a reaction for which ∆r G°′ = −30.5 kJ/mol-rxn. NH2 N

N

−

O

O−

O−

O−

+

+

+

P

O

O−

P

O

O−

P

N

N O

CH2

O−

O

H

H

HO

OH

ATP, adenosine triphosphate

One of the key functions of the process of respiration is to produce molecules of ATP for our bodies to use. ATP is produced in the reaction of adenosine diphosphate (ADP) with hydrogen phosphate (HPi = HPO42−), ADP + HPi + H n ATP + H2O

∆r G°′ = +30.5 kJ/mol-rxn

+

a reaction that is reactant-favored at equilibrium. How then do our bodies get this reaction to occur? The answer is to couple the production of ATP with another reaction that is even more product-favored than ATP production is reactant-favored. For example, organisms carry out the oxidation of carbohydrates in a multistep process, producing energy. One of the compounds produced in the process called glycolysis is phosphoenolpyruvate (PEP).

−O

O−

CH2 O

P+ O

C

C

O−

O− PEP, phosphoenolpyruvate

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Its reaction with water is product-favored at equilibrium PEP + H2O n Pyruvate + HPi   ∆r G°′ = −61.9 kJ/mol-rxn This reaction and ATP formation are linked through the HPi that is produced in the PEP reaction. The two reactions can be added together. PEP + H2O n Pyruvate + HPi

  ∆r G°′ = −61.9 kJ/mol-rxn

ADP + HPi + H n ATP + H2O

  ∆r G°′ = +30.5 kJ/mol-rxn

+

H

H

John C. Kotz

18.1  Thermodynamics and Living Things

PEP + ADP + H n Pyruvate + ATP   ∆r G°′ = −31.4 kJ/mol-rxn +

The overall reaction has a negative value for ∆r G°′ and thus is product-favored at equilibrium. ATP is formed in this process. The coupling of reactions to produce a system that is product-favored is used in a multitude of reactions that occur in our bodies.

Questions

1. Consider the hydrolysis reactions of creatine phosphate and adenosine-5′-mono­phosphate. Creatine phosphate + H2O n Creatine + HPi ∆r G°′ = −43.3 kJ/mol-rxn Adenosine-5′-monophosphate + H2O n Adenosine + HPi ∆r G°′ = −9.2 kJ/mol-rxn Which of the following combinations produces a reaction that is product-favored at equilibrium: for creatine phosphate to transfer phosphate to adenosine or for adenosine5′-monophosphate to transfer phosphate to creatine? 2. Assume the reaction A(aq) + B(aq) n C(aq) + H3O+(aq) produces one hydronium ion. What is the mathematical relationship between ∆r G°′ and ∆r G° at 25 °C? (Hint: Use the equation ∆r G = ∆r G° + RT lnQ and substitute ∆r G°′ for ∆r G.)

Chapter 18 / Principles of Chemical Reactivity: Entropy and Free Energy

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Natural diamonds are created over 150 km beneath the Earth’s surface. At these depths, enormous pressures and high temperatures convert graphite, the more stable allotrope of carbon at lower pressure, into diamond. It is believed that volcanic eruptions then push diamonds to the Earth’s surface. If a diamond is pushed to the surface quickly, it will cool before it has a chance to revert to graphite. Although diamonds are thermodynamically unstable at room temperature, the activation barrier for conversion is too high for the process to occur. Diamonds have been synthesized in laboratories since the 1950s by duplicating the conditions that exist deep ­u nderground. To produce a synthetic diamond, a seed ­c rystal of diamond is combined with graphite and nickel (or  ­c hromium), then subjected to pressures greater than 50,000 atm and temperatures around 1500 °C. The nickel, now molten, s­ imultaneously acts as a solvent for carbon and a catalyst for the formation of diamond. If conditions are maintained near the equilibrium line between diamond and ­g raphite (­F igure 1), single crystals of synthetic diamonds may be grown to sizes larger than one carat in approximately three days. Synthetic diamonds may be produced in a variety of ­colors. One of the most common colors is yellow, which results from a small amount of nitrogen replacing carbon in diamond’s ­lattice. Is this a problem? Probably not. Natural yellow diamonds are extremely rare and sell for high prices. Currently, a technique known as chemical vapor deposition (CVD) is being used to grow diamonds in a vacuum and at lower temperatures (800 °C). In CVD, methane and hydrogen gases are atomized above a diamond seed crystal. Hydrogen atoms bond with carbon atoms on the diamond’s surface, preventing the carbon atoms from forming double bonds with their neighbors. Carbon radicals, from the decomposed ­methane, slowly replace the hydrogen atoms on the diamond’s surface, building a larger diamond, one atomic layer at a time. In a­ ddition to gemstones, the CVD method may be used to 1000 Diamond

100

P (GPa)

10

Liquid

1 0.1 Graphite

0.01 0.001

0

1

2

3

Vapor

4

5

6

7

8

9

T/1000 (K)

Figure 1  Phase diagram for carbon. The hatched areas are regions where the two phases can coexist. Pressures are in gigapascals (where 1 GPa is about 9900 atm).



10

TOSHIFUMI KITAMURA/AFP/Getty Images

18.2  Are Diamonds Forever?

Figure 2  Diamond wafers. The chemical vapor deposition (CVD) technique allows the synthesis of diamond in the shape of extended disks or wafers. Under optimized growth conditions the properties of these disks approach those of perfect diamond single crystals.

create diamond windows or thin diamond films over a variety of ­substrates (Figure 2).

Questions

1. The decomposition of diamond to graphite [C(diamond) n C(graphite)] is thermodynamically favored, but occurs slowly at room temperature. (a) Use ∆f G° values from Appendix L to calculate ∆r G° and Keq for the reaction under standard conditions and 298.15 K. (b) Use ∆f H° and S° values from Appendix L to estimate ∆r G° and Keq for the reaction at 1000 K. Assume that the enthalpy and entropy values in Appendix L are valid at these temperatures. Does heating shift the equilibrium toward the formation of diamond or graphite? (c) Why is the formation of diamond favored at high pressures? (d) The phase diagram shows that diamond is thermodynamically favored over graphite at 20,000 atm pressure (about 2 GPa) at room temperature. Why is this conversion actually done at much higher temperatures and pressures? 2. It has been demonstrated that buckminsterfullerene (C60), another allotrope of carbon (Section 2.3), may be converted into diamond at room temperature and 20,000 atm pressure (about 2 GPa). The standard enthalpy of formation, ∆f H°, for buckminsterfullerene is 2320 kJ/mol at 298.2 K. (a) Calculate ∆ r H° for the conversion of C60 to diamond at standard state conditions and 298.2 K. (b) Assuming that the standard entropy per mole of carbon in both C60 and diamond is comparable (both about 2.3 J/K mol), is the conversion of C60 to diamond product-favored at room temperature?

Applying Chemical Principles

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917

1. What does it mean to say that a process is spontaneous? Give an example of a spontaneous process. 2. Give an example of a chemical reaction or a physical change that is reversible and one that is irreversible. Explain why the irreversible process is not reversible. 3. A system consists of solid sodium chloride and a beaker of ­ issolves water at the same temperature. When the NaCl d in the water, the temperature of the resulting solution decreases slightly from that of the initial water temperature. Is the dissolution of the NaCl in water enthalpy or entropy driven? Explain. 4. On the extent-of-reaction diagram, indicate (a) ∆G° and (b) the portion of the graph where ∆rG is negative for the forward reaction. Is the reaction product-favored or reactantfavored at equilibrium? Explain how ∆r G° and ∆r G differ.

Increasing free energy, G

Think–Pair–Share

1 Products only

0 Reactants only Extent of reaction

Chapter Goals Revisited The goals for this chapter are keyed to specific Study Questions to help you organize your review:

18.1  Spontaneity and Dispersal of Energy: Entropy • Understand the concept of spontaneity: spontaneous processes occur without outside intervention and proceed to equilibrium, a result of energy dispersal. 1, 2.

• Know the second law of thermodynamics: a spontaneous process is one that leads to an increase in entropy in the universe. 3, 4.

• Recognize that an entropy change is the energy transferred as heat for a reversible process divided by the kelvin temperature. 7–10, 60, 61.

18.2  Entropy: A Microscopic Understanding • Understand the statistical basis of entropy, that entropy is proportional to the number of ways that energy can be dispersed, that is, to the number of microstates available to a system. 11, 12.

18.3  Entropy Measurement and Values • Recognize that assigning a perfect crystal at 0 K to have zero entropy

(the third law of thermodynamics) establishes a means of evaluating the entropy of a substance. 13, 88.

• Define standard molar entropies, S°, and evaluate and compare factors

affecting S° values (temperature, volume, molecular structure, states of matter). 14–16, 51.

• Use values of standard molar entropies to calculate changes in entropy, ∆rS°, for a chemical reaction. 17–22, 52, 78, 93.

918

Chapter 18 / Principles of Chemical Reactivity: Entropy and Free Energy

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18.4  Entropy Changes and Spontaneity • Calculate the change in entropy for a system, its surroundings, and the

universe to determine whether a process is spontaneous under standard conditions. 23–26, 55, 58, 74.

• Recognize how reaction conditions influence whether a reaction is spontaneous. 27, 28.

18.5  Gibbs Free Energy • Define the Gibbs free energy change, ∆G, and relate ∆rG° to ∆rH° and ∆rS°. 29–32.

• Relate ∆rG, ∆rG°, Q, and K to reaction spontaneity and product- and reactant-favorability. 29–32, 49, 50, 53, 54, 66, 67, 75.

Q

𝚫rG

Spontaneous?

QK

∆r G > 0

Not spontaneous to the right; spontaneous to the left

18.6  Calculating and Using Standard Free Energies, ∆rG° • Use free energies of formation, ∆fG°, to calculate the standard free energy change for a reaction. 33–36, 56, 95, 103, 104.

• Use the equation ∆rG° = ∆rH° − T∆rS° to assess the effect of changes in temperature on the favorability of a reaction. 37–40, 63, 73, 79, 100, 101.

• Relate ∆rG° and the equilibrium constant for a reaction. 41–48, 59, 69, 71, 72, 84, 90.

K

𝚫rG°

Reactant-Favored or Product-Favored at Equilibrium?

K >> 1

∆r G°  H+ > Ni2+ > Zn2+ strong

weak

Reducing agents: Zn > Ni > H2 > Ag > Cl− strong

weak

Finally, notice that the value of E°cell is greater the farther apart the oxidizing and reducing agents are on the potential ladder. For example, Zn(s) + Cl2(g) n Zn2+(aq) + 2 Cl−(aq)  E° = +2.12 V

is more strongly product-favored than the reduction of hydrogen ions with nickel metal. Ni(s) + 2 H+(aq) n Ni2+(aq) + H2(g)  E° = +0.25 V

E xamp le 19.5

Ranking Oxidizing and Reducing Agents Problem   Use the table of standard reduction potentials (Table 19.1) to do the following: (a) Rank the halogens in order of their strength as oxidizing agents. (b) Decide whether hydrogen peroxide (H2O2) in acid solution is a stronger oxidizing agent than Cl2. (c) Decide which of the halogens is capable of oxidizing gold metal to Au3+(aq).

What Do You Know?   A table of electrode potentials, such as Table 19.1 or ­Appendix M, contains the information needed to answer these questions.

Strategy  The ability of a species on the left side of the E° table to function as an oxidizing agent declines on descending the list.

958

Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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Solution (a) Ranking halogens according to oxidizing ability: The halogens (F2, Cl2, Br2, and I2) appear in the upper-left portion of Table 19.1, with F2 being highest, followed in order by the other three species. Their  strengths as oxidizing agents are F2 > Cl2 > Br2 > I2.  (The ability of bromine to oxidize iodide ions to molecular iodine is illustrated in Figure 19.16.) (b) Comparing hydrogen peroxide and chlorine: H2O2 lies just below F2 but well above Cl2 in the potential ladder (Table 19.1). Thus,  H2O2 is a weaker oxidizing agent than F2 but a stronger one than Cl2.  (Note that the E° value for H2O2 refers to an acidic solution and standard conditions.) (c) Which halogen will oxidize gold metal to gold(III) ions? The Au3+ | Au half-reaction is listed below the F2 | F− half-reaction and just above the Cl2 | Cl− half-reaction. This means that, among the halogens,  only F2 is capable of oxidizing Au to Au3+ under standard conditions.  That is, for the reaction of Au and F2,

Oxidation, anode:

2[Au(s) n Au3+(aq) + 3 e−]

Reduction, cathode:

3[F2(g) + 2 e− n 2 F−(aq)]

Net ionic equation:

3 F2(g) + 2 Au(s) n 6 F−(aq) + 2 Au3+(aq)

E°cell = E°cathode − E°anode = +2.87 V − (+1.50 V) = +1.37 V F2 is a stronger oxidizing agent than Au3+, so the reaction proceeds from left to right as written. (This is confirmed by a positive value of E°cell.) For the reaction of Cl2 and Au, Table  19.1 shows that Cl2 is a weaker oxidizing agent than Au3+, so the reaction would be expected to proceed in the opposite direction under standard conditions.

Oxidation, anode:

2[Au(s) n Au3+(aq) + 3 e−]

Reduction, cathode:

3[Cl2(aq) + 2 e− n 2 Cl−(aq)]

Net ionic equation:

3 Cl2(aq) + 2 Au(s) n 6 Cl−(aq) + 2 Au3+(aq)

E°cell = E°cathode − E°anode = +1.36 V − (+1.50 V) = −0.14 V This is confirmed by the negative value for E°cell.

Think about Your Answer   In part (c), the value of E°cell was calculated for two reactions. To achieve a balanced net ionic equation, the half-reactions were added but only after multiplying the gold half-reaction by 2 and the halogen half-reaction by 3. (This means 6 mol of electrons were transferred from 2 mol Au to 3 mol Cl2.) Notice that this multiplication does not change the value of E° for the half-reactions because cell potentials do not depend on the quantity of material.

Photos: © Charles D. Winters/Cengage

The test tube contains an aqueous solution of KI (top layer) and immiscible CCl4 (bottom layer).

Add Br2 to solution of KI and shake.

After adding a few drops of Br2 in water, the I2 produced collects in the bottom CCl4 layer and gives it a purple color. (The top layer contains excess Br2 in water.) The presence of I2 in the bottom layer indicates that the added Br2 was able to oxidize the iodide ions to molecular iodine (I2).

Figure 19.16  The reaction of bromine and iodide ion.  This experiment proves that Br2 is a better oxidizing agent than I2.

19.4  Standard Electrochemical Potentials

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959

Check Your Understanding (a) Rank the following metals in their ability to function as reducing agents: Hg, Sn, and Pb. (b) Which halogens will oxidize mercury to mercury(II)?

19.5 Electrochemical Cells Under Nonstandard Conditions Goals for Section 19.5 • Use the Nernst equation to calculate the cell potential under nonstandard conditions.

• Use cell potential to determine the pH and other ion concentrations. In the real world, electrochemical cells seldom operate under standard conditions. Even if the cell is constructed with all dissolved species at 1 M, reactant concentrations decrease and product concentrations increase in the course of the reaction. Changing concentrations of reactants and products, as well as the temperature, will affect the cell voltage. This section explores what happens to cell potentials under nonstandard conditions.

The Nernst Equation Based on both theory and experimental results, cell potentials are related to concentrations of reactants and products and to temperature, as follows: E = E° − (RT/nF  ) lnQ



(19.2)

In this equation, which is known as the Nernst equation, R is the gas constant (8.314462618 J/K ∙ mol); T is the temperature (K); and n is the number of moles of electrons transferred between oxidizing and reducing agents (as determined by the balanced equation for the reaction). The symbol F represents the Faraday constant (9.648533212 × 104 C/mol).  One Faraday is the quantity of electric charge carried by one mole of electrons. The term Q is the reaction quotient (Equation 15.2, ­Section 15.2). Substituting values for the constants in Equation 19.2, and using 298 K as the temperature, gives E  E 



0.0257 ln Q n

at 298 K

(19.3)

or using base-10 logarithms, E  E 

0.0592 log Q n

In essence, the term (RT/nF )lnQ corrects the standard potential E° for nonstandard conditions or concentrations.

E xamp le 19.6

Using the Nernst Equation Problem   A voltaic cell is set up at 25 °C with the half-cells Al3+(0.0010 M) | Al and Ni2+(0.50 M) | Ni. Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.

960

Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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What Do You Know?  You know the temperature and the identity and concentrations of the reactants and products.

Strategy  Step 1.  D  etermine which substance is reduced (Al3+ or Ni2+) by looking at the appropriate half-reactions in Table 19.1 and deciding which is the better oxidizing agent (Example 19.5). Step 2.  Add the half-reactions to determine the net ionic equation and calculate E °cell. Step 3.  Use the Nernst equation to calculate E, the potential.

Solution  Step 1.  Determine which substance is reduced. Nickel ion (Ni2+) is a stronger oxidizing agent than Al3+. Therefore, Ni2+ is reduced, and the Ni2+ | Ni compartment is the cathode. (Conversely, Al metal is a better reducing agent than Ni metal. Therefore, Al is oxidized, and the Al | Al3+ compartment is the anode.) Step 2.  Add the half-reactions to determine the net ionic equation and calculate E °cell. Cathode, reduction:

3 [Ni2+(aq) + 2 e− n Ni(s)]

Anode, oxidation:

2 [Al(s) n Al3+(aq) + 3 e−]

Net ionic equation:

 2 Al(s)  + 3 Ni2+(aq) n 2 Al3+(aq) + 3 Ni(s)  E°cell = E°cathode − E°anode E°cell = (−0.25 V) − (−1.66 V) = 1.41 V

Step 3.  Use the Nernst equation to calculate E, the potential. The expression for Q is written based on the cell reaction. In the net reaction, Al3+(aq) has a coefficient of 2, so this concentration is squared. Similarly, [Ni2+(aq)] is cubed. Solids are not included in the expression for Q (Section 15.2). Q

[Al 3]2 [Ni2]3

The net equation requires transfer of 6 mol of electrons from 2 mol of Al atoms to 3 mol of Ni2+ ions, so n = 6, and the Nernst equation gives Ecell  E cell 

0.0257 [Al 3]2 ln 2 3 [Ni ] n

 1.41 V 

0.0257 [0.0010]2 ln 6 [0.50]3

= +1.41 V − 0.004283 ln(8.00 × 10−6) = +1.41 V − 0.004283 (−11.736) =  1.46 V 

Think about Your Answer  The concentrations of Al3+ and Ni2+ both affect the cell potential. Analysis of the lnQ term in the Nernst equation shows that if [Ni2+] = 1 M but [Al3+] < 1 M, then Ecell > E°cell. The reaction is more product-favored in this situation. The reverse situation (with [Ni2+] < 1 M and [Al3+] = 1 M) would lead to Ecell < E°cell. In this example, the very low value of [Al3+] has the greater effect, and Ecell is greater than E°cell.

Check Your Understanding  A voltaic cell is set up with an aluminum electrode in a 0.025 M Al(NO3)3(aq) solution and an iron electrode in a 0.50 M Fe(NO3)2(aq) solution. Determine the cell potential, Ecell , at 298 K.



19.5  Electrochemical Cells Under Nonstandard Conditions

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Example 19.6 demonstrated the calculation of a cell potential if concentrations are known. It is also useful to apply the Nernst equation in the opposite sense, using a measured cell potential to determine an unknown concentration.

E xamp le 19.7

Variation of Cell Potential with Concentration Problem  A voltaic cell is set up with copper and hydrogen half-cells. Standard conditions are used in the copper half-cell, Cu2+(aq, 1.00 M) | Cu(s). The hydrogen gas pressure is 1.00 bar. A value of 0.490 V is recorded for Ecell at 298 K. Determine the concentration of H+ and the pH of the solution. What Do You Know?   You know the temperature, the identity of the reactants and products, the concentration of the Cu2+ ion, the partial pressure of H2, and Ecell.

Strategy  Step 1.  Determine which substance is reduced (Cu2+ or H+) by looking at the appropriate half-reactions in Table 19.1 and deciding which is the better oxidizing agent. Step 2. Add the half-reactions to determine the net ionic equation and calculate E °cell (Example 19.5). Step 3. Use the Nernst equation with the given Cu2+ ion concentration to calculate the hydrogen ion concentration. Finally, calculate the pH.

Solution  Step 1.  Determine which substance is reduced. Based on their positions in a table of standard reduction potentials, Cu2+ is a better ­oxidizing agent than H+, so Cu2+ (aq, 1.00 M) | Cu(s) is the cathode, and H2(g, 1.00 bar) | H+(aq, ? M) is the anode. Step 2.  Add the half-reactions to determine the net ionic equation and calculate E °cell. Cathode, reduction:

Cu2+(aq) + 2 e− n Cu(s)

Anode, oxidation:

H2(g) n 2 H+(aq) + 2 e−

Net ionic equation:

H2(g) + Cu2+(aq) n Cu(s) + 2 H+(aq) E°cell = E°cathode − E°anode

E°cell = (+0.337 V) − (0.00 V) = +0.337 V Step 3.  Use the Nernst equation to calculate [H+]. Calculate the pH. The reaction quotient, Q, is derived from the balanced net ionic equation. Q

[H]2 [Cu2]PH2

The net equation requires the transfer of two moles of electrons, so n = 2. The value of [Cu2+] is 1.00 M, and the pressure of H2 is 1.0 bar, but [H+] is unknown. Substitute this information into the Nernst equation (and do not overlook the fact that [H+] is squared in the expression for Q). E  E 

0.0257 [H]2 ln [Cu2]PH2 n

0.490 V  0.337 V 

962

0.0257 [H]2 ln 2 (1.00)(1.00)

Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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−11.91 = ln[H+]2 [H+] = 2.6 × 10−3 M = 3 × 10−3 M  pH  = 2.6

Think about Your Answer   Be sure to write the balanced equation. Without it you may not have the correct exponents in the term for Q in the Nernst equation.

Check Your Understanding   The half-cells Ag+(aq, 1.0 M) | Ag(s) and H2(1.0 bar) | H+(aq, ? M) are linked by a salt bridge to create a voltaic cell. With the silver electrode as the cathode, a value of 0.902 V is recorded for Ecell at 298 K. Determine the concentration of H+ and the pH of the solution.

A pH meter is a device that uses a measured cell potential to determine hydrogen ion concentrations. In an electrochemical cell in which H+(aq) is a reactant or product, the cell voltage will vary predictably with the hydrogen ion concentration. The cell voltage is measured and the value used to calculate pH. Example 19.7 illustrates how Ecell can depend on the hydrogen ion concentration in a simple cell. In the real world, using a hydrogen electrode in a pH meter is not practical. The apparatus is clumsy; it is anything but robust; and platinum (for the electrode) is costly. Common pH meters use a glass electrode, so called because it contains a thin glass membrane separating the cell from the solution whose pH is to be measured (Figure 19.17). Inside the glass electrode is a silver wire coated with AgCl and a solution of HCl; outside is the solution of unknown pH to be evaluated. An Ag/AgCl or calomel electrode—the latter a common reference electrode using a mercury(I)–mercury redox couple (Hg2Cl2 | Hg)—serves as the second electrode of the cell. The potential across the glass membrane depends on [H+]. Common pH meters give a direct readout of pH.

Coaxial cable

Nonconductive glass or plastic electrode body

Photos: © Charles D. Winters/Cengage

Reference electrode Porous ceramic diaphragm Internal solution Internal electrode (Ag/AgCl) pH-sensitive glass membrane 1.0 M H+(aq)

Figure 19.17  Measuring pH.



19.5  Electrochemical Cells Under Nonstandard Conditions

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19.6 Electrochemistry and Thermodynamics Goal for Section 19.6 • Use the relationship between cell potential (E °cell) and free energy (∆rG °) and between E °cell and an equilibrium constant for the cell reaction.

Work and Free Energy The first law of thermodynamics states that the internal energy change in a system (∆U) is related to two quantities, heat (q) and work (w): ∆U = q + w (Section 5.4). This equation also applies to chemical changes that occur in a voltaic cell. As current flows, energy is transferred from the system (the voltaic cell) to the surroundings. In a voltaic cell, the decrease in internal energy in the system will manifest itself ideally as electrical work done on the surroundings by the system. The maximum work done by an electrochemical system (ideally, assuming no heat is generated) is proportional to the potential difference (volts) and the quantity of charge (coulombs):  wmax = nFE 



(19.4)

In this equation, E is the cell potential, and nF is the quantity of electric charge transferred from anode to cathode. The free energy change for a process is, by definition, the maximum amount of work that can be obtained (Section 18.5). Because the maximum work and the cell potential are related, E° and ∆rG° can be related mathematically (taking care to assign signs correctly). The maximum work done on the surroundings when electricity is produced by a voltaic cell is +nFE, with the positive sign denoting an increase in energy in the surroundings. The energy content of the cell decreases by this amount. Thus, ∆rG for the voltaic cell has the opposite sign. Units in Equation 19.6  n has units

of mol e−, and F has units of (C/mol e−). Therefore, nF has units of coulombs (C). Because 1 J = 1 C ∙ V, the product nFE will have units of energy (J).

 ΔrG = −nFE 

(19.5)

Under standard conditions, the appropriate equation is

 ΔrG° = −nFE° 

(19.6)

This expression shows that the more positive the value of E°, the more negative the value of ∆rG° for the reaction. Also, because of the relationship between ∆rG° and the equilibrium constant (K), the farther apart the half-reactions are on the potential ladder, the more strongly product-favored the reaction is at equilibrium.

E xamp le 19.8

Relating E° and 𝚫rG° Problem  The standard cell potential, E°cell, for the reduction of silver ions with copper metal (Figure 19.5) is +0.462 V at 25 °C. Calculate ∆rG° for this reaction. What Do You Know?  You know the cell potential under standard conditions and therefore know to use Equation 19.6 to calculate the standard free energy change. In this equation F is the Faraday constant (96,485 C/mol e−), but n has to be determined from the balanced equation for the cell reaction. Strategy  Use Equation 19.6 where E°cell and F are known. The value of n, the number of moles of electrons transferred between copper metal and silver ions, comes from the balanced equation.

964

Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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Solution   In this cell, copper is the anode, and silver is the cathode. The overall cell reaction is Cu(s) + 2 Ag+(aq) n Cu2+(aq) + 2 Ag(s) which means that each mole of copper transfers 2 mol of electrons to 2 mol of Ag + ions. That is, n = 2. Now use Equation 19.6. ∆rG° = −nFE° = −(2 mol e−)(96,485 C/mol e−)(0.462 V) = −89,200 C ∙ V Because 1 C ∙ V = 1 J,  ΔrG° = −89,200 J or −89.2 kJ 

Think about Your Answer  This example demonstrates an effective method of obtaining thermodynamic values from electrochemical experiments. Keep in mind that a positive E° implies a negative ∆rG°.

Check Your Understanding  The following reaction has an E° value of −0.76 V: H2(g) + Zn2+(aq) n Zn(s) + 2 H+(aq) Calculate ∆rG° for this reaction. Is the reaction product- or reactant-favored at equilibrium?

E° and the Equilibrium Constant When a voltaic cell produces an electric current, the reactant concentrations decrease, and the product concentrations increase. The cell voltage also changes. As reactants are converted to products, the value of Ecell decreases and the cell potential eventually reaches zero; no further net reaction occurs, and equilibrium is achieved. When Ecell = 0, the reactants and products are at equilibrium, and the reaction quotient, Q, is equal to the equilibrium constant, K. Substituting the appropriate symbols and values into the Nernst equation, E  0  E 

0.0257 ln K n

and collecting terms gives an equation that relates the standard cell potential and equilibrium constant:

ln K 

nE    at 25°C (298 K) 0.0257

(19.7)

Equation 19.7 can be used to determine values for equilibrium constants, as illustrated in Example 19.9.

Ex am p le 19.9

E° and Equilibrium Constants Problem  Calculate the equilibrium constant for the reaction at 298 K: Ag+(aq) + Fe2+(aq)  uv Ag(s) + Fe3+(aq)

What Do You Know?   You have the balanced chemical equation and know that Equation 19.7 is required. You need to determine E° (from standard reduction potential values in Table 19.1) and n, the number of electrons transferred.

19.6  Electrochemistry and Thermodynamics

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Strategy Map

Strategy 

Problem Calculate an ­equilibrium ­constant using ­electrochemical data.

Step 1. Determine the moles of electrons transferred (n) by balancing and adding the two half-reactions. Step 2. Determine E °cell by subtracting E ° for the oxidation half-reaction from E ° of the reduction half-reaction. Step 3.  Calculate the equilibrium constant, K, using Equation 19.7.

Data/Information • Balanced equation

Solution  Step 1

Determine the moles of electrons transferred (n) by balancing and adding the two half-reactions. The balanced equation indicates that the oxidation number of silver changes from +1 to 0, so the silver ions gain one electron each and are reduced. Likewise, the oxidation number of the iron increases from +2 to +3, so the iron(II) ions lose one electron each and are oxidized. Add the two half-reactions to determine the overall chemical equation. Determine the number of moles of electrons (n) transferred in the reaction.

Step 2

Cathode, reduction:

Ag+(aq) + e− n Ag(s)

Anode, oxidation:

Fe2+(aq) n Fe3+(aq) + e−

Net ionic equation:

Ag+(aq) + Fe2+(aq) uv Fe3+(aq) + Ag(s)

Determine E °cell by subtracting E ° for the oxidation half-reaction from E ° of the reduction half-reaction. E°cell = E°cathode − E°anode E°cell = (0.799 V) − (0.771 V) = +0.028 V

Step 3

Calculate the equilibrium constant, K, using Equation 19.7. Substitute n = 1 and E °cell into Equation 19.7. Solve for K. ln K 

(1)(0.028 V) nE    1.09 0.0257 0.0257  K = 3 

Think about Your Answer  The relatively small positive voltage (0.028 V) for the cell indicates that the cell reaction is only slightly product-favored at equilibrium. A value of 3 for the equilibrium constant is in accord with this observation.

Check Your Understanding Calculate the equilibrium constant at 25 °C for the reaction 2 Ag+(aq) + Hg(ℓ) uv 2 Ag(s) + Hg2+(aq)

The relationships between E°, K, and ∆rG° are summarized in Table 19.2. ­Values of E° can be used to obtain equilibrium constants for many different c­ hemical systems. For example, the solubility product constant, Ksp, for AgCl can be determined with the aid of a half-cell based on the reduction of AgCl. ­Figure 19.18 ­illustrates how the potential for the reduction of AgCl in the presence of Cl− ion (1.00 M) can be determined. AgCl(s) + e− n Ag(s) + Cl−(aq)  E° = +0.222 V

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Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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Table 19.2

Summary of the Relationship of K, 𝚫rG°, and E°

K

𝚫rG°

E°

Reactant-Favored or Product-Favored at Equilibrium?

K >> 1

ΔrG° < 0

E° > 0

Product-favored

K = 1

ΔrG° = 0

E° = 0

[C]c[D]d = [A]a[B]b at equilibrium

K 0

E° < 0

Reactant-favored

The standard reduction potential for the AgCl | Ag half-cell is +0.222 V. If this halfreaction is paired with a standard silver electrode in a hypothetical voltaic cell, the cell reactions could be written as Cathode, reduction:

AgCl(s) + e− n Ag(s) + Cl−(aq)

Anode, oxidation:

Ag(s) n Ag+(aq) + e−

Net ionic equation:

AgCl(s) n Ag+(aq) + Cl−(aq)

The equation for the net reaction represents the equilibrium of solid AgCl and its ions. The cell potential is negative, E°cell = E°cathode − E°anode = (+0.222 V) − (+0.799 V) = −0.577 V

indicating a reactant-favored process, as would be expected based on the low solubility of AgCl. Using Equation 19.7, the value of Ksp can then be obtained from E°cell. ln K 

nE  (1)( 0.577 V)    22.45 0.0257 0.0257

K sp  e22.45  2  1010

(− )

(+) Salt bridge

(+)

H2(g)

Ag coated with AgCl

KCl solution saturated with AgCl

Chemically inert Pt electrode

H2(g)



rounded to the proper number of significant figures, ln Ksp = −22.5. This number has only one significant figure after the decimal point. Thus, the inverse logarithm has only one significant figure.

Figure 19.18  Measurement of the standard electrode potential for the Ag| AgCl electrode.

Voltmeter (−)

Significant Figures in Calculations Involving Logarithms When

2 H+(aq) + 2 e−

AgCl(s) + e−

Ag(s) + Cl−(aq)

19.6  Electrochemistry and Thermodynamics

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19.7 E lectrolysis: Chemical Change Using Electrical Energy Goal for Section 19.7 • Describe the chemical processes occurring in an electrolysis. Recognize the factors that determine which substances are oxidized and reduced at the electrodes.

The electrochemical cells described thus far use product-favored redox reactions to generate an electric current. Equally important, however, is the opposite process, electrolysis, the use of electrical energy to bring about chemical change.

Electrolysis of Molten Salts All electrolysis experiments are set up in a similar manner. The material to be electrolyzed, either a molten salt or a solution, is contained in an electrolysis cell (Figure 19.19). As was the case with voltaic cells, ions must be present in the liquid or solution for a current to flow. The movement of ions constitutes the electric current within the cell. The cell has two electrodes that are connected to a source of DC (direct-current) voltage. If the applied voltage is high enough, chemical reactions occur at the two electrodes. Reduction occurs at the negatively charged cathode, with electrons transferred from that electrode to a chemical species in the cell. ­Oxidation occurs at the positive anode, with electrons from a chemical species transferred to that electrode. The electrolysis of molten sodium chloride to produce sodium metal and chlorine gas is a process of commercial interest. Sodium chloride melts at about 800  °C, and in the molten state sodium ions (Na+) and chloride ions (Cl−) are freed from their rigid arrangement in the crystalline lattice. If a potential is applied to the electrodes, sodium ions are attracted to the negative electrode, and chloride ions are attracted to the positive electrode (Figure 19.20). If the potential is high Oxygen—gas

Hydrogen—gas Battery (+)

(−)

e−

e−

(−)

Cu2+

Cu2+

© Charles D. Winters/Cengage

Cu2+ Cu2+ Cu2+ Anode, impure copper

Cu2+ Cathode, object to be plated with pure copper

Greenshoots Communications/Alamy Stock Photo

(+) Aqueous solution with free copper ions

Water—liquid (a) Electrolysis of water produces hydrogen (at the cathode) and oxygen gas (at the anode).

(b) (left) Purifying copper by electrolysis. Copper ions are produced at an anode made of impure copper. The ions migrate to the cathode where they are reduced to ­copper metal. (right) A commercial electrolysis unit producing copper.

Figure 19.19 Electrolysis. ​Electrical energy is used to carry out reactions that are otherwise reactant favored.

968

Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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Figure 19.20  The preparation of sodium and chlorine by the electrolysis of molten NaCl.  ​In the molten state, sodium ions migrate to the negative cathode, where they are reduced to sodium metal. Chloride ions migrate to the positive anode, where they are oxidized to elemental chlorine.

Battery (−)

(+) e−

e− Cathode (−)

Anode (+)

e−

e−

Molten NaCl

e− e−

e− + Sodium ion migrates to cathode.

− Reduced to sodium metal

−

Chloride migrates to anode.

Oxidized to chlorine

enough, chemical reactions occur at each electrode. At the negative cathode, Na + ions accept electrons and are reduced to sodium metal (a liquid at this temperature). S­ imultaneously, at the positive anode, chloride ions give up electrons and form elemental chlorine. Cathode (−), reduction:

2 Na+ + 2 e− n 2 Na(ℓ)

Anode (+), oxidation:

2 Cl− n Cl2(g) + 2 e−

Net ionic equation:

2 Na+ + 2 Cl− n 2 Na(ℓ) + Cl2(g)

Electrons move through the external circuit under the force exerted by the applied potential, and the movement of positive and negative ions in the molten salt constitutes the current within the cell. The energy required for this reactant-favored reaction to occur is provided by an external source such as a battery.

Electrolysis of Aqueous Solutions Sodium ions (Na +) and chloride ions (Cl −) are the primary species present in molten NaCl. Only chloride ions can be oxidized, and only sodium ions can be reduced. Electrolysis of a substance in aqueous solution is more complicated than

Problem Solving Tip 19.3 Electrochemical Conventions: Voltaic Cells and Electrolysis Cells

Whether you are describing a voltaic cell or an electrolysis cell, the terms anode and cathode always refer to the electrodes at which oxidation and reduction occur, respectively. However, the electrodes in the two types of electrochemical cells have different polarities.



Type of Cell

Electrode

Function

Polarity

Voltaic

Anode

Oxidation

−

Cathode

Reduction

+

Anode

Oxidation

+

Cathode

Reduction

−

Electrolysis

19.7  Electrolysis: Chemical Change Using Electrical Energy

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969

H2 gas and OH− form at the cathode. A drop of phenolphthalein has been added to the solution in this experiment so that the formation of OH−(aq) can be detected (by the pink color of the indicator in basic solution).

Cathode (−): 2 e− + 2 H2O(ℓ)

H2(g) + 2 OH−(aq)

∙ e− Cathode

e−

∙

Photos: © Charles D. Winters/Cengage

Figure 19.21  Electrolysis of aqueous NaI.

Iodine forms at the anode.

A solution of NaI(aq) is electrolyzed, a potential applied using an external source of electricity.

I2(aq) + 2 e−

Anode (+): 2 I−(aq)

the electrolysis of a molten salt, however, because water is now present. Water is an electroactive substance; that is, it can be oxidized or reduced in an electrochemical process. Consider the electrolysis of aqueous sodium iodide (Figure 19.21). In this experiment, the electrolysis cell contains Na +(aq), I−(aq), and H2O molecules. ­Possible reduction reactions at the negative cathode include Na+(aq) + e− n Na(s)

Electrochemistry and Michael Faraday

The terms anion, cation, electrode, and electrolyte originated with Michael Faraday (1791–1867), one of the most influential people in the history of chemistry. Faraday was apprenticed to a bookbinder in ­London when he was 13. This situation suited him perfectly, as he enjoyed reading the books sent to the shop for binding. By chance, one of these volumes was a small book on chemistry, which whetted his appetite for science, and he began performing experiments on electricity. In 1812, a patron of the shop invited ­Faraday to accompany him to the Royal Institution to attend a lecture by one of the most famous chemists of the day, Sir Humphry Davy. Faraday was intrigued by Davy’s lecture and wrote to ask Davy for a position as an assistant. His request was granted and he began work in 1813. ­Faraday was so talented that his work proved extraordinarily

970

fruitful, and only 12 years later he was made the director of the laboratory of the Royal Institution. It has been said that Faraday’s contributions were so enormous that, had there been Nobel Prizes when he was alive, he would have received at least six. These could have been awarded for discoveries such as the following: • The laws of electrolysis (the effect of electric current on chemicals) • Electromagnetic induction, which led to the first transformer and electric motor • The magnetic properties of matter • Benzene and other organic chemicals (which led to important chemical industries) • The Faraday effect (the rotation of the plane of polarized light by a magnetic field) • The introduction of the concept of electric and magnetic fields

Oesper Collection in the History of Chemistry/University of Cincinnati

A Closer Look

2 H2O(ℓ) + 2 e− n H2(g) + 2 OH−(aq)

Michael Faraday (1791–1867) In addition to making discoveries that had profound effects on science, ­Faraday was an educator. He wrote and spoke about his work in memorable ways, especially in lectures to the general public that helped to popularize science. A transcript of Faraday’s lectures, The Chemical History of a Candle, was published in 1867. This small book, still widely available, is a beautiful and readable account of scientific thinking.

Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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Possible oxidation reactions at the positive anode are 2 I−(aq) n I2(aq) + 2 e− 2 H2O(ℓ) n O2(g) + 4 H+(aq) + 4 e−

In the electrolysis of aqueous NaI, H2(g) and OH−(aq) are formed at the cathode, and iodine is formed at the anode. Thus, the overall cell process can be summarized by the following equations: Cathode (−), reduction:

2 H2O(ℓ) + 2 e− n H2(g) + 2 OH−(aq)

Anode (+), oxidation:

2 I−(aq) n I2(aq) + 2 e−

Net ionic equation:

2 H2O(ℓ) + 2 I−(aq) n H2(g) + 2 OH−(aq) + I2(aq)

where E°cell has a negative value. E°cell = E°cathode − E°anode = (−0.8277 V) − (+0.621 V) = −1.449 V

This process is reactant favored at equilibrium, and a potential of at least 1.45 V must be applied to the cell for these reactions to occur. If the process had involved the oxidation of water instead of iodide ion at the anode, the required potential would be greater, −2.057 V [E°cathode − E°anode = (−0.8277 V) − (+1.229 V)], and if the reaction involving the reduction of Na+ and the oxidation of I− had occurred, the required potential would be −3.335 V [E°cathode − E°anode = (−2.714 V) −  (+0.621 V)]. The reaction occurring is the one requiring the smallest applied potential, so the net cell reaction in the electrolysis of NaI(aq) is the oxidation of iodide ion and reduction of water. While comparisons of required E°cell values often allow proper predictions to be made about which half-reactions will occur in electrolytic cells, there are exceptions. Consider, for example, the electrolysis of SnCl 2. In this case, aqueous Sn2+ ion is much more easily reduced (E° = −0.14 V) than water (E° = −0.83 V) at the cathode, so tin metal is produced. At the anode, two oxidations are possible: Cl −(aq) to Cl 2(g) or H 2O(ℓ) to O 2(g). Experiments show that chloride ion is ­oxidized in preference to water, so the reactions occurring on electrolysis of aqueous tin(II) chloride are (Figure 19.22) Cathode (−), reduction:

Sn2+(aq) + 2 e− n Sn(s)

Anode (+), oxidation:

2 Cl−(aq) n Cl2(g) + 2 e−

Net ionic equation:

Sn2+(aq) + 2 Cl−(aq) n Sn(s) + Cl2(g)

E°cell = E°cathode − E°anode = (−0.14 V) − (+1.36 V) = −1.50 V

SnCl2(aq)

© Charles D. Winters/Cengage

Anode (+)

Cathode (−)

Cl2



Sn

Figure 19.22  Electrolysis of aqueous tin(II) chloride.  ​Tin metal collects at the negative cathode. Chlorine gas is formed at the positive anode. Elemental chlorine is formed in the cell, in spite of the fact that the potential for the oxidation of Cl− is more negative than that for oxidation of water. (That is, chlorine should be less easily oxidized than water.) This is the result of chemical kinetics and illustrates the complexity of some aqueous electrochemistry.

19.7  Electrolysis: Chemical Change Using Electrical Energy

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971

Formation of Cl2 at the anode in the electrolysis of SnCl2(aq) is contrary to a prediction based on E° values. If the electrode reactions were Cathode (−), reduction:

Sn2+(aq) + 2 e− n Sn(s)

Anode (+), oxidation:

2 H2O(ℓ) n O2(g) + 4 H+(aq) + 4 e−

E°cell = (−0.14 V) − (+1.23 V) = −1.37 V

a smaller applied potential is required. To explain the formation of chlorine instead of oxygen, the rates of reaction must be taken into account. The oxidation of Cl−(aq) is much more rapid than the oxidation of H2O. In the electrolysis of aqueous NaCl, a voltage high enough to oxidize both Cl− and H2O is used. Because chloride ion is oxidized much faster than H2O, Cl2 is the major product in this electrolysis. Electrolysis of aqueous NaCl is the predominant means by which chlorine is produced commercially. Another instance in which rates are important concerns electrode materials. Graphite, commonly used to make inert electrodes, can be oxidized. For the halfreaction CO2(g) + 4 H+(aq) + 4 e− n C(s) + 2 H2O(ℓ), E° is +0.20 V, indicating that carbon is slightly easier to oxidize than copper (E° = +0.34 V). Based on this value, oxidation of a graphite electrode might reasonably be expected to occur during an electrolysis. And indeed it does, albeit slowly; graphite electrodes used in electrolysis cells slowly deteriorate and eventually have to be replaced. One other factor—the concentration of electroactive species in solution—must be taken into account when discussing electrolysis. As shown in Section 19.5, the potential at which a species in solution is oxidized or reduced depends on concentration. Unless standard conditions are used, predictions based on E° ­values are merely qualitative. In addition, the rate of a half-reaction depends on the ­concentration of the electroactive substance at the electrode surface. At a very low concentration, the rate of the redox reaction may depend on the rate at which an ion diffuses from the solution to the electrode surface.

E xamp le 19.10

Electrolysis of Aqueous Solutions Problem   Predict how products of the electrolysis of aqueous solutions of NaF, NaCl, NaBr, and NaI are likely to be different and predict E°cell for each electrolysis. (The electrolysis of NaI is illustrated in Figure 19.21.)

What Do You Know?  You know the identity of the compounds to be electrolyzed, but you will need to know the E° values for their half-reactions and for water electrolysis.

Strategy  The main criterion used to predict the chemistry in an electrolytic cell should be the ease of oxidation and reduction, an assessment based on E° values. Solution   Water is reduced to hydroxide ion and H2 gas in preference to reduction of Na+(aq) (as in the electrolysis of aqueous NaI). Thus, the  primary cathode reaction in all cases is   2 H2O(ℓ) + 2 e− n H2(g) + 2 OH−(aq)   E°cathode = −0.83 V  At the anode, you need to assess the ease of oxidation of the halide ions relative to water. Based on E° values, this should be I−(aq) > Br−(aq) > Cl−(aq) >> F−(aq). Fluoride ion is much more difficult to oxidize than water, and electrolysis of an aqueous solution containing this ion results exclusively in O2 formation. That is,  the primary anode reaction for NaF(aq) is 

972

Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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 2 H2O(ℓ) n O2(g) + 4 H+(aq) + 4 e−   E°anode = +1.23 V  Therefore, for NaF,  E°cell = (−0.83 V) − (+1.23 V) = −2.06 V  Recall that  chlorine is the primary product at the anode in the electrolysis of aqueous solutions of chloride salts  (as in Figure 19.22). Therefore, the primary anode reaction for NaCl(aq) is  2 Cl−(aq) n Cl2(g) + 2 e−   E°cell = (−0.83 V) − (+1.36 V) = −2.19 V  Bromide ions are considerably easier to oxidize than chloride ions, so Br2 may be expected as the primary product in the electrolysis of aqueous NaBr.  For NaBr(aq), the primary anode reaction is   2 Br−(aq) n Br2(ℓ) + 2 e−  so E°cell is  E°cell = (−0.83 V) − (+1.08 V) = −1.91 V  Thus,  the electrolysis of NaBr resembles that of NaI (Figure 19.21) in producing the halogen, hydrogen gas, and hydroxide ion.  The half-reactions and the cell potential for aqueous NaI were given on page 971.

Think about Your Answer  As described above, you would predict from E° values the ease of oxidation of halide ions is I−(aq) > Br−(aq) > Cl−(aq) >> F−(aq). This is confirmed by the results.

Check Your Understanding  Predict the chemical reactions that will occur at the two electrodes in the electrolysis of an aqueous sodium hydroxide solution.

19.8 Counting Electrons Goal for Section 19.8 • Relate the quantity of a substance oxidized or reduced to the amount of current and the time the current flows.

In the electrolysis of aqueous AgNO3, one mole of electrons is required to produce one mole of silver. In contrast, two moles of electrons are required to produce one mole of tin (Figure 19.22): Sn2+(aq) + 2 e− n Sn(s)

If the number of moles of electrons flowing through the electrolysis cell can be measured, the amount of silver or tin produced can be calculated. Conversely, if the amount of silver or tin produced is known, then the number of moles of electrons moving through the circuit can be calculated. The number of moles of electrons consumed or produced in an electron transfer reaction is obtained by measuring the current flowing in the external electric circuit in a given time. The current flowing in an electrical circuit is the amount of charge (in units of coulombs, C) per unit time, and the usual unit for current is the ampere (A). One ampere equals the passage of one coulomb of charge per second.

19.8  Counting Electrons

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973

Current (amperes, A) 



Faraday Constant  The Faraday constant is the charge carried by 1 mol of electrons: 9.648533212 × 104 C/mol e−.

electric charge (coulombs, C) time (seconds, s)

(19.8)

The current passing through an electrochemical cell and the time for which the current flows are measurable quantities. If the current is constant, the charge (in coulombs) that passes through a cell can be obtained by multiplying the current (in amperes) by the time (in seconds). Knowing the charge and using the Faraday constant as a conversion factor, you can calculate the number of moles of electrons that passed through an electrochemical cell. In turn, you can use this quantity to calculate the quantities of reactants and products. The following example illustrates this type of calculation.

E xamp le 19.11

Using the Faraday Constant Problem  A constant current of 2.40 A is passed through a solution containing Cu2+(aq) for 30.0 minutes, with copper metal being deposited at the cathode. What mass of copper, in grams, is deposited? Assume that only Cu2+(aq) is reduced at the cathode.

What Do You Know?  You know the current passed through the cell and the time over which it was passed.

Strategy  Step 1. The product of the current (amperes = C/s) and time (s) is the amount of charge (C) passed through the cell. Step 2. Determine the amount (moles) of electrons transferred to Cu2+(aq) using the charge and the Faraday constant. Step 3. Relate the amount (moles) of electrons transferred to the amount (moles) of Cu(s), and ultimately the mass of Cu(s) deposited on the cathode.

Solution Step 1. The product of the current and time is the amount of charge passed through the cell. Charge (C)  current (A)  time (s)  (2.40 A)(30.0 min)(60.0 s/min)  4.320  103 C Step 2. Determine the amount (moles) of electrons transferred to Cu2+(aq) using the charge and the Faraday constant.  1 mol e  (4.320  103 C)   4.477  102 mol e  96, 485 C  Step 3. Relate the amount (moles) of electrons transferred to the amount (moles) of Cu(s), and finally the mass of Cu(s) deposited on the cathode. Reducing one mole of Cu2+(aq) to Cu(s) requires two moles of electrons, as shown by the balanced half-reaction for the process Cu2+(aq) + 2 e– 88n Cu(s)  1 mol Cu   63.55 g Cu   1.42 g  mass of copper  (4.477  102 mol e)    2 mol e   1 mol Cu 

974

Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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Think about Your Answer   The key relationship in this calculation is current =  charge/time. Most situations will involve knowing two of these three quantities from experiment and calculating the third.

Check Your Understanding 1. Calculate the mass of O2 produced in the electrolysis of water, using a current of 0.445 A for a period of 45 minutes. 2. In the commercial production of sodium by electrolysis, the cell operates at 7.0 V and a current of 2.5 × 104 A. What mass of sodium can be produced in 1 hour?

19.9 Corrosion: Redox Reactions in the Environment Goal for Section 19.9 • Understand the oxidation–reduction reactions responsible for the corrosion of metals.

Corrosion is the deterioration of a substance, usually a metal, in the environment by a product-favored oxidation–reduction reaction. Among common examples are the tarnishing of silver and the conversion of copper (on a roof or drain spout) to an attractive green patina. However, perhaps the most important example of corrosion occurs with iron, which, depending on conditions, coats the metal with a black deposit of Fe(OH)2 or Fe3O4 or reddish deposits of hydrated iron(III) oxide, Fe2O3 ⋅ x H2O ­(Figure 19.23). Because iron is used in many products, including automobiles, bridges, and building, corrosion has significant economic and safety implications. The rusting (corrosion) of iron is expected to be a product-favored reaction. Air and water are reactants in this process; both are sufficiently strong oxidizing agents to oxidize iron. Corrosion occurs in what can be described as a voltaic cell with a cathode and anode, with movement of electrons and ions. First consider the metal surface. Although it may appear smooth and homogeneous, there are significant irregularities at the microscopic level. Stressing and bending during manufacturing lead to regions that are more susceptible to oxidation, as do surface irregularities and the presence of impurities in the metal (Figure 19.24). Now, imagine a piece of iron

© Charles D. Winters/Cengage

Fe2+ ions form at tip and react with [Fe(CN)6]3− to give Prussian blue (Fe4[Fe(CN)6]3).

© Charles D. Winters/Cengage

Corrosion: An Electrochemical Process

Figure 19.23  The corrosion of iron results in major economic loss. The Economic Costs of Corrosion ​ It is estimated that corrosionrelated expenses to the United States economy exceed $1 trillion per year. This estimate (of around 5% of the U.S. Gross Domestic Product) includes the costs of corrosion prevention as well as repair and replacement of corroded products.

H2 and OH− are formed. OH− is detected by pink color of indicator.

Anodic region

Cathodic region

Figure 19.24  Anode and cathode reactions in iron corrosion.  ​Two iron nails were placed in an agar gel that contains K3[Fe(CN)6] and the acid–base indicator phenolphthalein. In this electrochemical cell, regions of stress—the ends and the bent region of the nail—act as anodes, and the remainder of the surface serves as the cathode.

19.9  Corrosion: Redox Reactions in the Environment

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975

Figure 19.25  Corrosion of iron in the presence of both oxygen and water.  Initially, Fe is oxidized by O2, pitting the anodic region of the surface. Electrons migrate to the cathodic region where water and/or O2 is reduced, forming OH− ions. Fe(OH)2 precipitates as the Fe2+ and OH− ions interact.

Oxidation 3 Fe(OH)2 + ½ O2 88n Fe3O4 + 3 H2O Precipitation Fe 2+ + 2 OH− 88n Fe(OH)2

Oxidation 2 Fe(OH)2 + ½ O2 88n Fe2O3 · H 2O + H 2O

Protective layer

Anode Fe 88n Fe 2+ + 2 e−

Cathode ½ O2 + 2 e− + H2O 88n 2 OH−

Iron

metal exposed to H2O and O2 (Figure 19.25). Sites that are more susceptible to oxidation (anodic sites) become the anode in the cell and the less reactive surfaces of the metal act as a cathode. Electrons can migrate through the metal to the cathodic sites where reduction of H2O and/or O2 occurs. The circuit is completed by migration of ions within the aqueous solution and formation of the corrosion products on the electrode surface. The reaction that occurs at the cathode determines the products of corrosion and this in turn is controlled by the rate of the cathodic process. Several cathodic reactions are possible, and the one that is fastest will be determined by acidity and the amount of oxygen present. If little or no oxygen is present—as when a piece of iron is buried in soil such as moist clay—hydronium ions or water are reduced, and H2(g) and hydroxide ions are the products. Black iron(II) hydroxide is relatively insoluble and will precipitate on the metal surface, inhibiting further formation of Fe2+(aq). Anode

Fe(s) 88n Fe2+(aq) + 2 e−

Cathode

2 H2O(ℓ) + 2 e− 88n H2(g) + 2 OH−(aq)

Precipitation

Fe2+(aq) + 2 OH−(aq) 88n Fe(OH)2(s)

Net reaction

Fe(s) + 2 H2O(ℓ) 88n H2(g) + Fe(OH)2(s)

If both water and O2 are present, the chemistry of iron corrosion is different, and the corrosion reaction is about 100 times faster than without oxygen. Anode

2 Fe(s) 88n 2 Fe2+(aq) + 4 e−

Cathode

O2(g) + 2 H2O(ℓ) + 4 e− 88n 4 OH−(aq)

Precipitation

2 Fe2+(aq) + 4 OH−(aq) 88n 2 Fe(OH)2(s)

Net reaction

2 Fe(s) + 2 H2O(ℓ) + O2(g) 88n 2 Fe(OH)2(s)

Further oxidation of the Fe(OH)2 leads to the formation of magnetic iron oxide Fe3O4 (which can be thought of as a mixed oxide of Fe2O3 and FeO) if some additional oxygen is present but not in great excess. 3 Fe(OH)2(s) + 1⁄2 O2(g) 88n Fe3O4 ∙ H2O(s) + 2 H2O(ℓ) green hydrated magnetite

Fe3O4 ∙ H2O(s) 88n H2O(ℓ) + Fe3O4(s) black magnetite

976

Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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If the iron object has free access to oxygen and water, as in the open or in flowing water, red-brown iron(III) oxide will form. 2 Fe(OH)2(s) + 1⁄2 O2(g) 88n Fe2O3 ∙ H2O(s) + H2O(ℓ) red-brown

The overall process of converting Fe(s) to Fe2O3(s) is illustrated in Figure 19.25. This is the familiar rust you see on cars and buildings and the substance that colors the water red in some mountain streams or in your home. Other substances in air and water can assist in corrosion. Chlorides, from sea air or from salt spread on roads in winter, are notorious. Because the chloride ion is relatively small, it can diffuse into and through a protective metal oxide coating. Metal chlorides, which are more soluble than metal oxides or hydroxides, can then form. These chloride salts leach back through the oxide coating, and a path is now open for oxygen and water to further attack the underlying metal. (See Figure 19.2 for an example of aluminum foil corrosion in the presence of aqueous Cu(NO3)2 and NaCl.) This is the reason that you often see small pits on the surface of a corroded metal.

Protecting Metal Surfaces from Corrosion There are many methods for stopping a metal object from corroding, some more effective than others, but none totally successful. The general approach is to inhibit either or both the anodic and the cathodic processes. The usual method is anodic inhibition, attempting to directly prevent the oxidation reaction by painting the metal surface or by allowing a thin oxide film to form. One method of protecting iron surfaces is by reaction with chromate ion: 2 Fe(s) + 2 Na2CrO4(aq) + 2 H2O(ℓ) 88n Fe2O3(s) + Cr2O3(s) + 4 NaOH(aq)

(a) Zinc phosphate (gray) and zinc chromate (gold) adhere well to steel screws and protect them from corrosion.

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John C. Kotz

© Charles D. Winters/Cengage

The iron surface is oxidized by the chromate ion to give iron(III) and chromium(III) oxides. These form a coating impervious to oxygen and water, and further atmospheric oxidation is inhibited. Another inhibition method of protecting metal surfaces is phosphating. The process involves treating the metal with an acidic solution containing dissolved metal phosphates, such as zinc phosphate [Zn3(PO4)2] or iron(III) phosphate (FePO4). Phosphating creates a thin metal phosphate coating on the surface of the metal ­(Figure 19.26a). These coatings, which adhere well to surfaces and have low solubilities in water, directly provide protection against corrosion. Further protection is provided by one or more layers of paint, which adhere well to the phosphate coating. (“Lead” in Chapter 25 (page 1260) shows the consequences of not using phosphate salts to protect pipes from corrosion in the municipal water system in Flint, Michigan.)

(b) A galvanized steel bucket can be identified by the mottled appearance of its surface.

(c) This photograph shows zinc metal bars attached to the hull and rudder of a ship. They act as sacrificial anodes that are oxidized instead of the iron of the hull.

Figure 19.26  A corrosion inhibitor can be a coating, such as zinc phosphate, that protects a metal from contact with corrosive chemicals. The inhibitor can also be another metal that is preferentially oxidized.

19.9  Corrosion: Redox Reactions in the Environment

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977

Another way to inhibit metal oxidation is to force the metal to become the cathode, instead of the anode, in an electrochemical cell. This is called cathodic protection (Figure 19.26b and 19.26c). A more readily oxidized metal is attached to the metal in question. An example of this is galvanized iron, iron that has been coated with a film of zinc. The E° value for zinc is more negative than the E° value for iron, so the zinc film is oxidized prior to any of the iron. Thus, the zinc ­coating forms what is called a sacrificial anode. Over time, the zinc reacts with ­oxygen and water to form Zn(OH)2(s), which adheres to the surface and continues to ­protect it from corrosion. Further protection is provided as the Zn(OH)2(s) reacts with ­carbon dioxide to form an even more protective coating, ZnCO3(s). Thus, the iron is protected from corrosion even after the zinc anode has been fully oxidized. ­Alternatively, a block of the sacrificial anode is attached so that it is in electrochemical contact with the metal to be preserved (see “Applying Chemical Principles 19.2: Sacrifice!”).

Applying Chemical Principles

Plug-in electric vehicles are increasingly available, and some manufacturers will no longer produce gasoline-powered vehicles after about 2030. Plug-in electric vehicles have some advantages over those with the more common combustion engine. The cost of electricity per mile driven is one-third to one-quarter that of gasoline. The vehicles are environmentally friendly, producing far fewer greenhouse gases and other pollutants than combustion engine vehicles. In addition, electric engines are quiet and the cars accelerate quickly. However, fully electric vehicles suffer from some shortcomings. Most of these vehicles have ranges of less than 400 miles between charges. They are generally better suited to urban driving than long distance driving. Furthermore, complete battery recharging times can be several hours. Nonetheless, as battery technology has improved, both driving range and recharging times have improved, and many believe that it is simply a matter of time before many gasoline-fueled vehicles are replaced with plug-in ones. Despite this, it is true that fossil fuels such as gasoline contain far more chemical energy (per unit mass or volume) than any chemical battery. Lithium-ion batteries are a popular choice for electric vehicles as they provide high energy content with a relatively low mass. Lithium is a low-density element, but it is only one of many components in a battery. A battery that contains around 4.0 kg of lithium has a total weight of 300 kg. If all the lithium can be oxidized when discharging of the battery, 200 MJ (megajoules) of energy are produced. This amounts to 0.67 MJ per kilogram of battery weight. In

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19.1  Electric Batteries versus Gasoline

The chassis of a Tesla Model S electric car showing the battery packs lining the chassis.

contrast, the combustion of gasoline produces over 46 MJ/kg. Electric motors are more energy efficient than gasoline-­powered motors, but the increase in efficiency is not nearly enough to compensate for the large energy content difference.

Questions

1. A lithium ion battery produces a voltage of 3.6 V. (a) What is the energy released for each mole of lithium oxidized during the discharge of the battery? (b) What is the energy released per kilogram of lithium? 2. Use the energy produced by the combustion of gasoline (46 MJ/kg) to calculate the energy contained in 15 gallons of gasoline. (Assume the density of gasoline is 0.70 kg/L.) 3. What mass of lithium ion batteries would produce the same amount of energy as the combustion of 15 gallons of gasoline?

19.2 Sacrifice! In the latter half of the eighteenth century the British sheathed the hulls of their naval fleet with copper to protect the hulls from deterioration. Unfortunately, the copper corroded in sea water, so the navy called upon Sir Humphry Davy to determine

978

the cause of the corrosion and to find a cure. Although Davy initially assumed that impurities in the copper caused the corrosion, he soon learned that high-purity copper corroded faster. Turning his attention to the sea water, Davy determined

Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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Cu(s) + 1⁄2 O2(g) + H2O(ℓ) n Cu(OH)2(s) produces copper(II) hydroxide, which falls from the hull to the sea floor. Davy, a pioneer in the field of electrochemistry, quickly determined that the corrosion of copper can be prevented by attaching small pieces of a more easily oxidized metal (such as zinc, tin, or iron) to the copper sheathing. The quantity of the metal may be small in comparison to the copper, but to be effective, the metal must be positioned below the water line and in direct electrical contact with the copper. Why is corrosion prevented by attaching a small piece of metal, such as zinc, to the copper? When the two different metals are submerged in seawater they form a galvanic cell. The copper serves as the cathode and the zinc as a sacrificial anode. The zinc anode is oxidized instead of the copper. Although the zinc anodes corrode over time and must be replaced, the cost is cheaper than replacing the copper on an entire ship’s hull. In modern fleets, copper-sheathed hulls have given way to steel hulls. Though surface coatings, such as paint, provide some protection against corrosion, sacrificial anodes are still used to protect the steel from corrosion. In addition, sacrificial anodes are now used to protect the steel in underground pipes as well as in home water heaters, outboard motors, and boilers.

Questions

1. If an electrically insulating material, such as paint, is placed between the zinc and the copper, the zinc will still corrode in seawater, but it will not protect the copper from corrosion. Explain. 2. Use standard reduction potentials to determine which of the following metals could serve as a sacrificial anode on a ­copper-sheathed hull of a ship. Indicate all correct responses. (a) tin (b) silver (c) iron (d) nickel (e) chromium 3. Use standard reduction potentials to determine which of the following metals could serve as a sacrificial anode on a steel hull of a ship. Assume the standard reduction potential

for steel is the same as that of iron. Indicate all correct responses. (a) tin (b) silver (c) iron (d) nickel (e) chromium 4. The overall reaction for the production of Cu(OH)2 from Cu in oxygenated water can be broken into three steps: an oxidation half-reaction, a reduction half-reaction, and a precipitation reaction. (a) Complete and balance the two missing half-reactions to give the overall equation for the oxidation of copper in seawater. Oxidation half-reaction: ? Reduction half-reaction: ? Precipitation: Cu2+(aq) + 2 OH−(aq) n Cu(OH)2(s) Overall: Cu(s) + 1⁄2 O2(g) + H2O(ℓ) n Cu(OH)2(s) (b) Determine the equilibrium constant for the overall reaction at 25 °C using standard reduction potentials and the solubility product constant (Ksp) of Cu(OH)2(s). 5. Assume the following electrochemical cell simulates the ­galvanic cell formed by copper and zinc in seawater at pH 7.90 and 25 °C.

Zn | Zn(OH)2(s) | OH−(aq) || Cu(OH)2(s) | Cu(s) (a) Write a balanced equation for the reaction that occurs at the cathode. (b) Write a balanced equation for the reaction that occurs at the anode. (c) Write a balanced chemical equation for the overall reaction. (d) Determine the potential (in volts) of the cell.

John C. Kotz

that the copper is oxidized by oxygen in the sea water. The overall reaction,

Sacrificial anode attached to a steel-hulled ship. Sacrificial anodes of aluminum or magnesium are also used in home electric hot-water heaters to prevent corrosion in the heater.

Think–Pair–Share 1. Alkaline batteries rely on the following (unbalanced) half-reactions. Cathode:  MnO2(s) n Mn2O3(s) Anode:    Zn(s) n ZnO(s) (a) Balance each half-reaction. (The term alkaline refers to the fact that the reactions occur in a basic electrolyte.)

(b) Write an overall balanced chemical equation for the reaction. (c) For alkaline batteries, the reduction of MnO2(s) produces another solid, Mn2O3(s) that deposits on the cathode. Likewise, the oxidation product, ZnO, is also a solid. It is common in batteries for all electroactive species to be solids. Identify one or more advantages to having solids for both reactants and products.

Think–Pair–Share

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2. An electrochemical cell is constructed using the following half-reactions: AgCl(s) + e− 88n Ag(s) + Cl−(aq)  E° = 0.222 V Zn2+(aq) + 2 e− 88n Zn(s)     E° = −0.763 V (a) Write the reduction half-reaction that will occur in a voltaic cell. (b) Write the oxidation half-reaction that will occur in a voltaic cell. (c) Write the balanced chemical equation. (d) Calculate E°cell. 3. For the following electrochemical cell E °cell = 1.08 V: Co(s) | Co2+(aq, 1.0 M) || Ag+(aq, 1.0 M) | Ag(s) (a) Write a balanced reduction half-reaction. (b) Write a balanced oxidation half-reaction. (c) Will Ecell increase or decrease if the concentration of Ag+ is decreased? Explain. (d) Will Ecell increase or decrease if the concentration of Co2+ is decreased? Explain. 4. For the following electrochemical cell E °cell = −0.728 V: Ag(s) | Ag+(aq) || Br−(aq) | AgBr | Ag(s) (a) Write a balanced chemical equation for the overall reaction. (b) Calculate ∆rG° for the reaction. (c) Calculate the equilibrium constant, K, for the reaction. (d) Is the reaction product- or reactant-favored at equilibrium?

5. The surface of silver tarnishes in the presence of sulfur-­ containing gases, forming Ag 2 S. Silver sulfide can be ­polished off the surface, but polishing slowly wears away the silver. Another method of cleaning tarnished silver is to place it in an aluminum pan filled with dissolved baking soda (sodium hydrogen carbonate, NaHCO3). When in contact with the aluminum surface, the silver sulfide is reduced. (a) Write a balanced half-reaction for the reduction of Ag2S. (b) The aluminum is oxidized. One possible product of the oxidation is Al2S3, an insoluble compound. Write a balanced oxidation half-reaction for the oxidation of Al to Al2S3. (c) The hydrogen carbonate ion can also act as an acid in aqueous solutions. Can you identify some ways in which this ion helps the reaction proceed at a reasonable rate? 6. Assume an electrochemical cell is based on the following overall chemical equation. 2 AgCl(s) + Fe(s) 88n 2 Ag(s) + FeCl2(aq) (a) Sketch the electrochemical cell. Identify the cathode and the anode. (b) Write balanced reduction and oxidation half-reactions. (c) Use cell notation to depict the electrochemical cell. 7. LCD (liquid crystalline display) watches often operate for years before their batteries die. Assume that an LCD watch is powered by a lithium-ion battery, which during operation oxidizes Li to Li+. If the current used to run the watch remains steady at 0.10 milliamperes (mA), how long (in years) will it take to oxidize 1.0 g of Li in the battery?

Chapter Goals Revisited The goals for this chapter are keyed to specific Study Questions to help you organize your review.

19.1  Oxidation–Reduction Reactions • Balance equations for oxidation–reduction reactions in acidic or basic conditions using the half-reaction approach. 1–6, 57, 58, 92.

19.2  Voltaic Cells • In a voltaic cell, identify the half-reactions occurring at the anode and

the cathode, the polarity of the electrodes, the direction of electron flow in the external circuit, and the direction of ion flow in the salt bridge. 7–10, 59.

19.3  Commercial Voltaic Cells • Understand the chemistry and advantages and disadvantages of dry cells, alkaline batteries, lead storage batteries, lithium batteries, and Ni-cad batteries. 15, 16, 101.

• Understand how fuel cells work, and recognize the difference between batteries and fuel cells. 113.

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Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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19.4  Standard Electrochemical Potentials • Use standard reduction potentials to determine cell potentials for cells under standard conditions. 17–20, 60, 61.

• Use a table of standard reduction potentials to rank the strengths of

oxidizing and reducing agents, to predict which substances can reduce or oxidize another species, and to predict whether redox reactions will be product-favored or reactant-favored at equilibrium. 21–28, 63, 64.

19.5  Electrochemical Cells Under Nonstandard Conditions • Use the Nernst equation to calculate the cell potential under nonstandard conditions. 29–32, 83–87.

• Use cell potential to determine the pH and other ion concentrations. 33, 34.

19.6  Electrochemistry and Thermodynamics • Use the relationship between cell potential (E°cell) and free energy (∆rG°) and between E°cell and an equilibrium constant for the cell reaction. 35–40, 69–71, 88, 90.

19.7  Electrolysis: Chemical Change Using Electrical Energy • Describe the chemical processes occurring in an electrolysis. Recognize

the factors that determine which substances are oxidized and reduced at the electrodes. 41–46, 82.

19.8  Counting Electrons • Relate the quantity of a substance oxidized or reduced to the amount of current and the time the current flows. 47–52, 67, 72–74.

19.9  Corrosion: Redox Reactions in the Environment • Understand the oxidation–reduction reactions responsible for the corrosion of metals. 53–56, 103, 104.

Key Equations Equation 19.1 (page 954) Calculating a standard cell potential, E°cell, from standard half-cell reduction potentials.

E°cell = E°cathode − E°anode

Equation 19.2 (page 960) The Nernst equation, the relationship of the cell potential under nonstandard conditions (E) to that under standard conditions (E°). R is the gas constant (8.314462618 J/K · mol); T is the temperature (K); and n is the number of moles of electrons transferred between oxidizing and reducing agents. F is the Faraday constant (9.648533212 × 104 C/mol of e−), and Q is the reaction quotient. E = E° − (RT/nF  ) lnQ

Equation 19.3 (page 960) Nernst equation at 298 K. E  E 



0.0257 ln Q n Key Equations

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Equation 19.4 (page 964) The work done (w) by an electrochemical system. wmax = nFE

Equations 19.5 and 19.6 (page 964) Relationship between the free energy change and the cell potential under nonstandard or standard conditions, respectively. ΔrG = −nFE or ΔrG° = −nFE°

Equation 19.7 (page 965) Relationship between the equilibrium constant and the standard cell potential for a reaction (at 298 K). ln K 

nE  0.0257

Equation 19.8 (page 974) Relationship between current, electric charge, and time. Current (amperes, A) 

electric charge (coulombs, C) time (seconds, s)

Study Questions denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully worked solutions in the Student Solutions Manual.

▲

Practicing Skills Balancing Equations for Oxidation–Reduction Reactions (See Section 19.1 and Examples 19.1–19.2.) When balancing the following redox equations, it may be necessary to add H+(aq) or H+(aq) plus H2O for reactions in acid, and OH−(aq) or OH−(aq) plus H2O for reactions in base. 1. Write balanced equations for the following halfreactions. Specify whether each is an oxidation or reduction. (a) Cr(s) n Cr3+(aq) (in acid) (b) AsH3(g) n As(s) (in acid) − 2+ (c) VO3 (aq) n V (aq) (in acid) (d) Ag(s) n Ag2O(s) (in base) 2. Write balanced equations for the following halfreactions. Specify whether each is an oxidation or reduction. (a) H2O2(aq) n O2(g) (in acid) (b) H2C2O4(aq) n CO2(g) (in acid) (c) NO3−(aq) n NO2(g) (in acid) (d) MnO4−(aq) n MnO2(s) (in base) 3. Balance the following redox equations. All occur in acid solution. (a) Ag(s) + NO3−(aq) n NO2(g) + Ag+(aq) (b) MnO4−(aq) + HSO3−(aq) n Mn2+(aq) + SO42−(aq) (c) Zn(s) + NO3−(aq) n Zn2+(aq) + N2O(g) (d) Cr(s) + NO3−(aq) n Cr3+(aq) + NO(g)

4. Balance the following redox equations. All occur in acid solution. (a) Sn(s) + H+(aq) n Sn2+(aq) + H2(g) (b) Cr2O72−(aq) + Fe2+(aq) n Cr3+(aq) + Fe3+(aq) (c) MnO4−(aq) + Cl−(aq) n Mn2+(aq) + Cl2(g) (d) CH2O(aq) + Ag+(aq) n HCO2H(aq) + Ag(s) 5. Balance the following redox equations. All occur in basic solution. (a) Al(s) + H2O(ℓ) n Al(OH)4−(aq) + H2(g) (b) CrO42−(aq) + SO32−(aq) n Cr(OH)3(s) + SO42−(aq) (c) Zn(s) + Cu(OH)2(s) n [Zn(OH)4]2−(aq) + Cu(s) (d) HS−(aq) + ClO3−(aq) n S(s) + Cl−(aq) 6. Balance the following redox equations. All occur in basic solution. (a) MnO4−(aq) + I−(aq) n MnO2(s) + IO3−(aq) (b) NiO2(s) + Zn(s) n Ni(OH)2(s) + Zn(OH)2(s) (c) Fe(OH)2(s) + CrO42−(aq) n Fe(OH)3(s) + [Cr(OH)4]−(aq) (d) N2H4(aq) + Ag2O(s) n N2(g) + Ag(s)

Constructing Voltaic Cells (See Section 19.2 and Examples 19.3 and 19.4.) 7. A voltaic cell is constructed using the reaction of chromium metal and iron(II) ions. 2 Cr(s) + 3 Fe2+(aq) n 2 Cr3+(aq) + 3 Fe(s)

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Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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Complete the following sentences: Electrons in the external circuit flow from the electrode to the electrode. Negative ions move in the salt bridge from the half-cell to the half-cell. The half-reaction at the anode is  , and that at the cathode is  . 8. A voltaic cell is constructed using the reaction Mg(s) + 2 H+(aq) n Mg2+(aq) + H2(g)

(a) Write equations for the oxidation and reduction half-reactions. (b) Which half-reaction occurs in the anode compartment, and which occurs in the cathode compartment? (c) Complete the following sentences: Electrons in the external circuit flow from the electrode to the electrode. Negative ions move in the salt bridge from the half-cell to the half-cell. The half-reaction at the anode is  , and that at the cathode is  . 9. The half-cells Fe2+(aq) | Fe(s) and O2(g) | H2O (in acid solution) are linked to create a voltaic cell. (a) Write equations for the oxidation and reduction half-reactions and for the overall (cell) reaction. (b) Which half-reaction occurs in the anode compartment, and which occurs in the cathode compartment? (c) Complete the following sentences: Electrons in the external circuit flow from the electrode to the electrode. Negative ions move in the salt bridge from the half-cell to the half-cell. 10. The half-cells Sn2+(aq) | Sn(s) and Cl2(g) | Cl−(aq) are linked to create a voltaic cell. (a) Write equations for the oxidation and reduction half-reactions and for the overall (cell) reaction. (b) Which half-reaction occurs in the anode com­ partment, and which occurs in the cathode compartment? (c) Complete the following sentences: Electrons in the external circuit flow from the electrode to the electrode. Negative ions move in the salt bridge from the half-cell to the half-cell. 11. For each of the following electrochemical cells, write equations for the oxidation and reduction half-reactions and for the overall reaction. (a) Cu(s) | Cu2+(aq) || Fe3+(aq), Fe2+(aq) | Pt(s) (b) Pb(s) | PbSO4(s) | SO42−(aq) || Fe3+(aq), Fe2+(aq) | Pt(s)



12. For each of the following electrochemical cells, write equations for the oxidation and reduction half-reactions and for the overall reaction. (a) Pb(s) | Pb2+(aq) || Fe3+(aq), Fe2+(aq) | C(s) (b) Hg(ℓ) | Hg2Cl2(s) | Cl−(aq) || Ag+(aq) | Ag(s) 13. Use cell notation to depict an electrochemical cell based upon the following reaction that is productfavored at equilibrium. Cu(s) + Cl2(g) n 2 Cl−(aq) + Cu2+(aq)

14. Use cell notation to depict an electrochemical cell based upon the following reaction that is productfavored at equilibrium. Fe(s) + AgCl(s) n Fe2+(aq) + Ag(s) + Cl−(aq)

Commercial Electrochemical Cells (See Section 19.3.) 15. What are the similarities and differences between dry cells, alkaline batteries, and Ni-cad batteries? 16. What reactions occur when a lead storage battery is recharged?

Standard Electrochemical Potentials (See Section 19.4 and Example 19.5.) 17. Calculate the value of E°cell for each of the following reactions. Decide whether each is productfavored at equilibrium in the direction written. (a) 2 I−(aq) + Zn2+(aq) n I2(s) + Zn(s) (b) Zn2+(aq) + Ni(s) n Zn(s) + Ni2+(aq) (c) 2 Cl−(aq) + Cu2+(aq) n Cu(s) + Cl2(g) (d) Fe2+(aq) + Ag+(aq) n Fe3+(aq) + Ag(s) 18. Calculate the value of E°cell for each of the following reactions. Decide whether each is productfavored at equilibrium in the direction written. [Reaction (d) is carried out in basic solution.] (a) I2(s) + Mg(s) n Mg2+(aq) + 2 I−(aq) (b) Zn2+(aq) + Ni(s) n Zn(s) + Ni2+(aq) (c) Sn2+(aq) + 2 Ag+(aq) n Sn4+(aq) + 2 Ag(s) (d) 2 Zn(s) + O2(g) + 2 H2O(ℓ) + 4 OH−(aq) n  2 [Zn(OH)4]2−(aq) 19. Balance each of the following unbalanced ­equations; then calculate the standard potential, E°cell, and decide whether each is product-favored at equilibrium as written. (All reactions are carried out in acid solution.) (a) Sn2+(aq) + Ag(s) n Sn(s) + Ag+(aq) (b) Al(s) + Sn4+(aq) n Sn2+(aq) + Al3+(aq) (c) ClO3−(aq) + Ce3+(aq) n Cl2(g) + Ce4+(aq) (d) Cu(s) + NO3−(aq) n Cu2+(aq) + NO(g) 20. Balance each of the following unbalanced ­equations; then calculate the standard potential, E°cell, and decide whether each is product-favored Study Questions

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at equilibrium as written. (All reactions are carried out in acid solution.) (a) I2(s) + Br−(aq) n I−(aq) + Br2(ℓ) (b) Fe2+(aq) + Cu2+(aq) n Cu(s) + Fe3+(aq) (c) Fe2+(aq) + Cr2O72−(aq) n Fe3+(aq) + Cr3+(aq) (d) MnO4−(aq) + HNO2(aq) n Mn2+(aq) + NO3−(aq)

Ranking Oxidizing and Reducing Agents (See Section 19.4 and Example 19.5. Use a table of standard reduction potentials [Table 19.1 or Appendix M] to answer Study Questions 21–28.) 21. Consider the following half-reactions: Half-Reaction

E° (V)

Cu2+(aq) + 2 e− n Cu(s)

+0.34

Sn2+(aq) + 2 e− n Sn(s)

−0.14

Fe2+(aq) + 2 e− n Fe(s)

−0.44

Zn2+(aq) + 2 e− n Zn(s)

−0.76

Al3+(aq) + 3 e− n Al(s)

−1.66

22. Consider the following half-reactions: E° (V)

MnO4 (aq) + 8 H (aq) + 5 e n Mn2+(aq) + 4 H2O(ℓ)

+1.51

BrO3−(aq) + 6 H+(aq) + 6 e− n Br−(aq) + 3 H2O(ℓ)

+1.47

Cr2O72−(aq) + 14 H+(aq) + 6 e− n  2 Cr3+(aq) + 7 H2O(ℓ)

+1.33

NO3−(aq) + 4 H+(aq) + 3 e− n NO(g) + 2 H2O(ℓ)

+0.96

SO42−(aq) + 4 H+(aq) + 2 e− n SO2(g) + 2 H2O(ℓ)

+0.20

−

+

−

25. Which of the following ions is most easily reduced? (a) Cu2+(aq) (d) Ag+(aq) (b) Zn2+(aq) (e) Al3+(aq) 2+ (c) Fe (aq) 26. From the following list, identify the ions that are more easily reduced than H+(aq). (a) Cu2+(aq) (d) Ag+(aq) (b) Zn2+(aq) (e) Al3+(aq) 2+ (c) Fe (aq) 27. (a) Which halogen is most easily reduced in acidic solution: F2, Cl2, Br2, or I2? (b) Identify the halogens that are better oxidizing agents in acidic solution than MnO2(s). 28. (a) Which ion is most easily oxidized to the elemental halogen in acidic solution: F−, Cl−, Br−, or I−? (b) Identify the halide ions that are more easily oxidized in acidic solution than H2O(ℓ).

Electrochemical Cells Under Nonstandard Conditions

(a) Identify the strongest and weakest oxidizing agents among the reactants in these half-reactions. (b) Which of the oxidizing agents listed is (are) capable of oxidizing Cr3+(aq) to Cr2O72−(aq) (in acid solution)?

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23. Which of the following elements is the best reducing agent under standard conditions? (a) Cu (d) Ag (b) Zn (e) Cr (c) Fe 24. From the following list, identify those elements that are easier to oxidize than H2(g). (a) Cu (d) Ag (b) Zn (e) Cr (c) Fe

(a) Based on E° values, which metal is the most easily oxidized? (b) Which metals on this list are capable of reducing Fe2+(aq) to Fe(s)? (c) Write a balanced chemical equation for the reaction of Fe2+(aq) with Sn(s). Is this reaction product-favored or reactant-favored at equilibrium? (d) Write a balanced chemical equation for the reaction of Zn2+(aq) with Sn(s). Is this reaction ­product-favored or reactant-favored at equilibrium?

Half-Reaction

(c) Write a balanced chemical equation for the reaction of NO3−(aq) with SO2(g) in acid solution. Is this reaction product-favored or reactant-favored at equilibrium? (d) Write a balanced chemical equation for the reaction of Cr2O72−(aq) with Mn2+(aq). Is this reaction product-favored or reactant-favored at equilibrium?

(See Section 19.5 and Examples 19.6 and 19.7.) 29. Calculate the potential delivered by a voltaic cell using the following reaction if all dissolved species are 2.5 × 10−2 M and the pressure of H2 is 1.0 bar. Zn(s) + 2 H2O(ℓ) + 2 OH−(aq) n [Zn(OH)4]2−(aq) + H2(g)

30. Calculate the potential developed by a voltaic cell using the following reaction if the cell contains 0.20 M Fe2+, 0.20 M Fe3+, and is at a pH of 1.00. 2 Fe2+(aq) + H2O2(aq) + 2 H+(aq) n   2 Fe3+(aq) + 2 H2O(ℓ)

Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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31. One half-cell in a voltaic cell is constructed from a silver wire electrode in a 0.25 M solution of AgNO3. The other half-cell consists of a zinc ­electrode in a 0.010 M solution of Zn(NO3)2. ­Calculate the cell potential. 32. One half-cell in a voltaic cell is constructed from a copper wire electrode in a 4.8 × 10−3 M ­solution of Cu(NO3)2. The other half-cell consists of a zinc electrode in a 0.40 M solution of Zn(NO3)2. ­Calculate the cell potential. 33. One half-cell in a voltaic cell is constructed from a silver wire electrode in a AgNO3 solution of unknown concentration. The other half-cell consists of a zinc electrode in a 1.0 M solution of Zn(NO3)2. A potential of 1.48 V is measured for this cell. Use this information to calculate the concentration of Ag+(aq). 34. One half-cell in a voltaic cell is constructed from an iron electrode in an Fe(NO3)2 solution of unknown concentration. The other half-cell is a standard hydrogen electrode. A potential of 0.49 V is measured for this cell. Use this information to calculate the concentration of Fe2+(aq).

Electrochemistry, Thermodynamics, and Equilibrium (See Section 19.6 and Examples 19.8 and 19.9.) 35. Calculate ΔrG° and the equilibrium constant for the following reactions. (a) 2 Fe3+(aq) + 2 I−(aq) uv  2 Fe2+(aq) + I2(aq) − (b) I2(aq) + 2 Br (aq) uv 2 I−(aq) + Br2(ℓ) 36. Calculate ΔrG° and the equilibrium constant for the following reactions. (a) Zn2+(aq) + 2 Ag(s) + 2 Br−(aq) uv Zn(s) + 2 AgBr(s) (b) Ni(s) + 2 Ag+(aq) uv Ni2+(aq) + 2 Ag(s) 37. Use standard reduction potentials (Appendix M) for the half-reactions AgBr(s) + e− n Ag(s) + Br−(aq) and Ag+(aq) + e− n Ag(s) to calculate the value of Ksp for AgBr. 38. Use the standard reduction potentials (­Appendix M) for the half-reactions Hg2Cl2(s) + 2 e− n 2 Hg(ℓ) + 2 Cl−(aq) and Hg22+(aq) + 2 e− n 2 Hg(ℓ) to calculate the value of Ksp for Hg2Cl2. 39. Use the standard reduction potentials (Appendix M) for the half-reactions [AuCl4]−(aq) + 3 e− n Au(s) + 4 Cl−(aq) and Au3+(aq) + 3 e− n Au(s) to calculate the value of Kformation for the complex ion [AuCl4]−(aq). 40. Use the standard reduction potentials (­Appendix M) for the half-reactions [Zn(OH)4]2−(aq) + 2 e− n Zn (s) + 4 OH−(aq) and Zn2+(aq) + 2 e− n Zn(s) to calculate the value of Kformation for the complex ion [Zn(OH)4]2−.

Electrolysis (See Section 19.7 and Example 19.10.) 41. Diagram the apparatus used to electrolyze molten NaCl. Identify the anode and the cathode. Trace the movement of electrons through the external circuit and the movement of ions in the electrolysis cell. 42. Diagram the apparatus used to electrolyze aqueous CuCl2. Identify the reaction products, the anode, and the cathode. Trace the movement of electrons through the external circuit and the movement of ions in the electrolysis cell. 43. Which product, O2 or F2, is more likely to form at the anode in the electrolysis of an aqueous solution of KF? Explain your reasoning. 44. Which product, Ca or H2, is more likely to form at the cathode in the electrolysis of aqueous CaCl2? Explain your reasoning. 45. An aqueous solution of KBr is placed in a beaker with two inert platinum electrodes. When the cell is attached to an external source of electrical energy, electrolysis occurs. (a) Hydrogen gas and hydroxide ion form at the cathode. Write an equation for the halfreaction that occurs at this electrode. (b) Bromine is the primary product at the anode. Write an equation for its formation. 46. An aqueous solution of Na2S is placed in a beaker with two inert platinum electrodes. When the cell is attached to an external battery, electrolysis occurs. (a) Hydrogen gas and hydroxide ion form at the cathode. Write an equation for the halfreaction that occurs at this electrode. (b) Sulfur is the primary product at the anode. Write an equation for its formation.

Counting Electrons (See Section 19.8 and Example 19.11.) 47. In the electrolysis of a solution containing Ni2+(aq), metallic Ni(s) deposits on the cathode. Using a current of 0.150 A for 12.2 minutes, what mass of nickel will form? 48. In the electrolysis of a solution containing Ag+(aq), metallic Ag(s) deposits on the cathode. Using a current of 1.50 A for 3.50 hours, what mass of silver forms? Assume that only Ag+(aq) is reduced at the cathode. 49. Electrolysis of a solution of CuSO4(aq) to give copper metal is carried out using a current of 0.66 A. Assuming only Cu2+(aq) is reduced at the cathode, how long should electrolysis continue to produce 0.50 g of copper? 50. Electrolysis of a solution of Zn(NO3)2(aq) to give zinc metal is carried out using a current of 2.12 A. Study Questions

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Assuming that only Zn2+(aq) is reduced at the cathode, how long should electrolysis continue to prepare 2.5 g of zinc? 51. A voltaic cell can be built using the reaction between Al metal and O2 from the air. If the Al anode of this cell consists of 55 g of aluminum, how much charge (in coulombs) is required to oxidize all the aluminum? 52. Assume the specifications of a Ni-Cd voltaic cell include delivery of 0.25 A of current for 1.00 hour. What is the minimum mass of the cadmium that must be used to make the anode in this cell?

Corrosion: Redox Reactions in the Environment (See Section 19.9.) 53. Use E° values to predict which of the following metals, if coated on iron, will provide cathodic protection against corrosion to iron. (a) Cu (c) Ni (b) Mg (d) Sn 54. Use E° values to predict which of the following metals, if coated on nickel, will provide cathodic protection against corrosion to nickel. (a) Cu (c) Zn (b) Mg (d) Cr

(a) UO2+(aq) n U4+(aq) (b) ClO3−(aq) n Cl−(aq) (c) N2H4(aq) n N2(g) (d) ClO−(aq) n Cl−(aq)

58. Balance the following equations. (a) Zn(s) + VO2+(aq) n Zn2+(aq) + V3+(aq)  (acid solution) (b) Zn(s) + VO3−(aq) n V2+(aq) + Zn2+(aq)  (acid solution) (c) Zn(s) + ClO−(aq) n Zn(OH)2(s) + Cl−(aq)  (basic solution) (d) ClO−(aq) + [Cr(OH)4]−(aq) n Cl−(aq) + CrO42−(aq)  (basic solution) 59. Magnesium metal is oxidized, and silver ions are reduced in a voltaic cell using Mg2+(aq, 1 M) | Mg and Ag+(aq, 1 M) | Ag half-cells. Voltmeter Ag

O2(g) + 2 H2O(ℓ) + 4 e n 4 OH (aq) −

−

Calculate the standard potential, E°cell, and decide whether the reaction is product-favored at equilibrium. Will decreasing the pH make the reaction less thermodynamically product-favored at equilibrium? 56. In the presence of oxygen and acid, two halfreactions responsible for the corrosion of iron are Fe(s) n Fe2+(aq) + 2 e− O2(g) + 4 H+(aq) + 4 e− n 2 H2O(ℓ)

Calculate the standard potential, E°cell, and decide whether the reaction is product-favored at equilibrium. Will decreasing the pH make the reaction less thermodynamically product-favored at equilibrium?

General Questions These questions are not designated as to type or location in the chapter. They may combine several concepts. 57. Write balanced equations for the following half-reactions.

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NO3−

Ag+

55. In the presence of oxygen and water, two halfreactions responsible for the corrosion of iron are Fe(s) n Fe2+(aq) + 2 e−

(acid solution) (acid solution) (basic solution) (basic solution)

NO3−

Mg

Na+

Mg2+ NO3−

(a) Label which electrode is the anode and which is the cathode. (b) Write equations for the half-reactions occurring at the anode and the cathode, and write an equation for the net reaction in the cell. (c) Trace the movement of electrons in the external circuit. Assuming the salt bridge contains NaNO3, trace the movement of the Na+ and NO3− ions in the salt bridge that occurs when a voltaic cell produces current. Why is a salt bridge required in a cell? 60. You want to set up a series of voltaic cells with specific cell potentials. A Zn2+(aq, 1.0 M) | Zn(s) half-cell is in one compartment. Identify several half-cells that you could use so that the cell potential will be close to (a) 1.5 V and (b) 0.5 V. ­Consider cells in which the zinc cell can be either the cathode or the anode. 61. You want to set up a series of voltaic cells with specific cell potentials. The Ag+(aq, 1.0 M) | Ag(s) half-cell is one of the compartments. Identify several half-cells that you could use so that the cell potential will be close to (a) 1.7 V and (b) 0.5 V. Consider cells in which the silver cell can be either the cathode or the anode.

Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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62. Which of the following reactions is (are) productfavored at equilibrium? (a) Cu(s) + I2(s) n Cu2+(aq) + 2 I−(aq) (b) 2 Cl−(aq) + I2(s) n Cl2(g) + 2 I−(aq) (c) 2 K+(aq) + 2 Cl−(aq) n 2 K(s) + Cl2(g) (d) 2 K(s) + 2 H2O(ℓ) n  2 K+(aq) + H2(g) + 2 OH−(aq) 63. In the table of standard reduction potentials, locate the half-reactions for the reductions of the following metal ions to the metal: Sn2+(aq), Au+(aq), Zn2+(aq), Co2+(aq), Ag+(aq), Cu2+(aq). Among the metal ions and metals that make up these half-reactions: (a) Which metal ion is the weakest oxidizing agent? (b) Which metal ion is the strongest oxidizing agent? (c) Which metal is the strongest reducing agent? (d) Which metal is the weakest reducing agent? (e) Will Sn(s) reduce Cu2+(aq) to Cu(s)? (f) Will Ag(s) reduce Co2+(aq) to Co(s)? (g) Which metal ions on the list can be reduced by Sn(s)? (h) What metals can be oxidized by Ag+(aq)? 64.

In the table of standard reduction potentials, locate the half-reactions for the reductions of the following nonmetals: F2, Cl2, Br2, I2 (reduction to halide ions), and O2, S, Se (reduction to H2X in aqueous acid). Among the elements, ions, and compounds that make up these half-reactions: (a) Which element is the weakest oxidizing agent? (b) Which ion or H2X is the weakest reducing agent? (c) Which of the elements listed is (are) capable of oxidizing H2O to O2? (d) Which of these elements listed is (are) capable of oxidizing H2S to S? (e) Is O2 capable of oxidizing I− to I2, in acid solution? (f) Is S capable of oxidizing I− to I2? (g) Is the reaction H2S(aq) + Se(s) n H2Se(aq) + S(s) product-favored at equilibrium? (h) Is the reaction H2S(aq) + I2(s) n 2 H+(aq) + 2 I−(aq) + S(s) product-favored at equilibrium?

Linking any two half-cells makes a voltaic cell. Given four different half-cells, six voltaic cells are possible. These are labeled, for simplicity, Ag-Zn, Ag-Cu, Ag-Ni, Zn-Cu, Zn-Ni, and Cu-Ni. (a) In which of the voltaic cells does the copper electrode serve as the cathode? In which of the voltaic cells does the nickel electrode serve as the anode? (b) Which combination of half-cells generates the highest potential? Which combination generates the lowest potential? 67. The reaction occurring in the cell in which Al2O3 and aluminum salts are electrolyzed is Al3+(aq) + 3 e− n Al(s). If the electrolysis cell operates at 5.0 V and 1.0 × 105 A, what mass of aluminum metal can be produced in a 24-hour day? 68.

▲ A cell is constructed using the following half-reactions:

▲

65. Four voltaic cells are set up. In each, one half-cell contains a standard hydrogen electrode. The second half-cell is one of the following: (i) Cr3+(aq, 1.0 M) | Cr(s) (ii) Fe2+(aq, 1.0 M) | Fe(s) (iii) Cu2+(aq, 1.0 M) | Cu(s) (iv) Mg2+(aq, 1.0 M) | Mg(s) (a) In which of the voltaic cells does the hydrogen electrode serve as the cathode? (b) Which voltaic cell produces the highest potential? Which produces the lowest potential?

66. The following half-cells are available: (i) Ag+(aq, 1.0 M) | Ag(s) (ii) Zn2+(aq, 1.0 M) | Zn(s) (iii) Cu2+(aq, 1.0 M) | Cu(s) (iv) Ni2+(aq, 1.0 M) | Ni(s)

Ag+(aq) + e− n Ag(s) 

Ag2SO4(s) + 2 e− n 2 Ag(s) + SO42−(aq)  E° = 0.653 V

(a) What reactions should be observed at the anode and cathode? (b) Calculate the solubility product constant, Ksp, for Ag2SO4. 69.

▲ A potential of 0.142 V is recorded (under standard conditions) for a voltaic cell constructed using the following half reactions:

Cathode: Pb2+(aq) + 2 e− n Pb(s) Anode: PbCl2(s) + 2 e− n Pb(s) + 2 Cl−(aq) Net: Pb2+(aq) + 2 Cl−(aq) n PbCl2(s) (a) What is the standard reduction potential for the anode reaction? (b) Calculate the solubility product, Ksp, for PbCl2. 70. What is the value of E° for the following half-reaction? Ag2CrO4(s) + 2 e− n 2 Ag(s) + CrO42−(aq)

71. The standard potential, E°cell, for the reaction of Zn(s) and Cl2(g) is +2.12 V. What is the standard free energy change, ΔrG°, for the reaction? 72.

▲ An electrolysis cell for aluminum production operates at 5.0 V and a current of 1.0 × 105 A. Calculate the number of kilowatt-hours of energy required to produce 1 metric ton (1.0 × 103 kg) of aluminum. (1 kWh = 3.6 × 106 J and 1 J = 1 C ∙ V). Study Questions

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73.

74.

75.

▲ Electrolysis of molten NaCl is done in cells operating at 7.0 V and 4.0 × 104 A. What mass of Na(s) and Cl2(g) can be produced in 1 day in such a cell? What is the energy consumption in kilowatthours? (1 kWh = 3.6 × 106 J and 1 J = 1 C ∙ V). ▲

A current of 0.0100 A is passed through a solution of rhodium sulfate, causing reduction of the metal ion to the metal. After 3.00 hours, 0.038 g of Rh has been deposited. What is the charge on the rhodium ion, Rhn+? What is the formula for rhodium sulfate? ▲ A current of 0.44 A is passed through a solution of ruthenium nitrate causing reduction of the metal ion to the metal. After 25.0 minutes, 0.345 g of Ru has been deposited. What is the charge on the ruthenium ion, Run+? What is the formula for ruthenium nitrate?

76. The total charge that can be delivered by a large dry cell battery before its voltage drops too low is usually about 35 amp-hours. (One amp-hour is the charge that passes through a circuit when 1 A flows for 1 hour.) What mass of Zn is consumed when 35 amp-hours are drawn from the cell? 77. Chlorine gas is obtained commercially by electrolysis of brine (a concentrated aqueous solution of NaCl). If the electrolysis cells operate at 4.6 V and 3.0 × 105 A, what mass of chlorine can be produced in a 24-hour day? 78. Write equations for the half-reactions that occur at the anode and cathode in the electrolysis of molten KBr. What are the products formed at the anode and cathode in the electrolysis of aqueous KBr? 79. The products formed in the electrolysis of aqueous CuSO4 are Cu(s) and O2(g). Write equations for the anode and cathode reactions. 80. Predict the products formed in the electrolysis of an aqueous solution of CdSO4. 81. In the electrolysis of HNO3(aq), hydrogen is produced at the cathode. According to a table of reduction potentials, NO3−(aq) is easier to reduce than H+(aq). Suggest a possible reason why H2 formed rather than NO. What is the product formed at the anode? Write an equation for the anode reaction. 82. The metallurgy of aluminum involves electrolysis of Al2O3 dissolved in molten cryolite (Na3AlF6) at about 950 °C. Aluminum metal is produced at the cathode. Predict the anode product and write equations for the reactions occurring at both electrodes. 83. Two half-cells, Pt | Fe3+(aq, 0.50 M), Fe2+(aq, 1.0 × 10−5 M) and Hg2+(aq, 0.020 M) | Hg, are constructed and then linked together to form a

988

voltaic cell. Which electrode is the anode? What will be the potential of the voltaic cell at 298 K? 84. A voltaic cell is set up using the reaction Cu(s) + 2 Ag+(aq) n Cu2+(aq) + 2 Ag(s) Cu(s) | Cu2+(aq, 1.0 M) || Ag+(aq, 0.001 M) | Ag(s)

Under standard conditions, the expected potential is 0.45 V. Predict whether the potential for the voltaic cell will be higher, lower, or the same as the standard potential. Verify your prediction by calculating the new cell potential. 85. Calculate the cell potential for the following cell: Pt | H2(P = 1 bar) | H+(aq, 1.0 M) ||  Fe3+(aq, 1.0M), Fe2+(aq, 1.0M) | Pt

Will this reaction be more or less favorable at higher pH? To determine this, calculate the cell potential for a reaction in which [H+(aq)] is 1.0 × 10−7 M. 86. A voltaic cell set up using the reaction Cu(s) + 2 Ag+(aq) n Cu2+(aq) + 2 Ag(s)

has a cell potential of 0.45 V at 298 K. Describe how the potential of this cell will change as the cell is discharged. At what point does the cell potential reach a constant value? Explain your answer. 87. Two Ag+(aq) | Ag(s) half-cells are constructed. The first has [Ag+] = 1.0 M, the second has [Ag+] = 1.0 × 10−5 M. When linked together with a salt bridge and external circuit, a cell potential is observed. (This kind of voltaic cell is referred to as a concentration cell.) (a) D  raw a picture of this cell, labeling all components. Indicate the cathode and the anode, and indicate in which direction electrons flow in the external circuit. (b) Calculate the cell potential at 298 K. 88. Calculate equilibrium constants for the following reactions at 298 K. Indicate whether the equilibrium as written is reactant- or product-favored at equilibrium. (a) Co(s) + Ni2+(aq) uv Co2+(aq) + Ni(s) (b) Fe3+(aq) + Cr2+(aq) uv Cr3+(aq) + Fe2+(aq) 89. Calculate equilibrium constants for the following reactions at 298 K. Indicate whether the equilibrium as written is reactant- or product-favored at equilibrium. (a) 2 Cl−(aq) + Br2(ℓ) uv Cl2(aq) + 2 Br−(aq) (b) Fe2+(aq) + Ag+(aq) uv Fe3+(aq) + Ag(s) 90. Use the table of standard reduction potentials (Appendix M) to calculate ΔrG° for the following reactions at 298 K.

Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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(a) ClO3−(aq)+ 5 Cl−(aq) + 6 H+(aq) n   3 Cl2(g) + 3 H2O(ℓ) − (b) AgCl(s) + Br (aq) n AgBr(s) + Cl−(aq) 91. Use the table of standard reduction potentials (Appendix M) to calculate ΔrG° for the following reactions at 298 K. (a) 3 Cu(s) + 2 NO3−(aq) + 8 H+(aq) n  3 Cu2+(aq) + 2 NO(g) + 4 H2O(ℓ) (b) H2O2(aq) + 2 Cl−(aq) + 2 H+(aq) n Cl2(g) + 2 H2O(ℓ) 92.

93.

▲ Write balanced equations for the following reduction half-reactions involving organic compounds. (a) HCO2H n CH2O (acid solution) (b) C6H5CO2H n C6H5CH3 (acid solution) (c) CH3CH2CHO n CH3CH2CH2OH  (acid solution) (d) CH3OH n CH4 (acid solution) ▲ Balance the following equations involving organic compounds. (a) Ag+(aq) + C6H5CHO(aq) n  Ag(s) + C6H5CO2H(aq)  (acid solution) (b) CH3CH2OH + Cr2O72−(aq) n  CH3CO2H(aq) + Cr3+(aq)  (acid solution)

94. A voltaic cell is constructed in which one half-cell consists of a silver wire in an aqueous solution of AgNO3. The other half-cell consists of an inert platinum wire in an aqueous solution containing Fe2+(aq) and Fe3+(aq). (a) Calculate the cell potential, assuming standard conditions. (b) Write the net ionic equation for the reaction occurring in the cell. (c) Which electrode is the anode and which is the cathode? (d) If [Ag+] is 0.10 M and [Fe2+] and [Fe3+] are both 1.0 M, what is the cell potential? Is the net cell reaction still that used in part (a)? If not, what is the net reaction under the new conditions? 95. An expensive but lighter alternative to the lead storage battery is the silver-zinc battery. Ag2O(s) + Zn(s) + H2O(ℓ)n Zn(OH)2(s) + 2 Ag(s)

The electrolyte is 40% KOH, and silver–silver oxide electrodes are separated from zinc–zinc hydroxide electrodes by a plastic sheet that is permeable to hydroxide ions. Under normal operating conditions, the battery has a potential of 1.59 V. (a) How much energy can be produced per gram of reactants in the silver-zinc battery? Assume the battery produces a current of 0.10 A.

(b) How much energy can be produced per gram of reactants in the standard lead storage battery? Assume the battery produces a current of 0.10 A at 2.0 V. (c) Which battery (silver-zinc or lead storage) produces the greater energy per gram of reactants? 96. The specifications for a lead storage battery include delivery of a steady 1.5 A of current for 15 hours. (a) What is the minimum mass of lead that will be used in the anode? (b) What mass of PbO2 must be used in the cathode? (c) Assume that the volume of the battery is 0.50 L. What is the minimum molarity of H2SO4 necessary? 97. Manganese may play an important role in chemical cycles in the oceans. Two reactions involving manganese (in acid solution) are the reduction of nitrate ions (to NO) with Mn2+ ions and the oxidation of ammonium ions (to N2) with MnO2. (a) Write balanced chemical equations for these reactions (in acid solution). (b) Calculate E°cell for the reactions. (One halfreaction potential you need is for the reduction of N2 to NH4+, E° = −0.272 V). 98.

▲ You want to use electrolysis to plate a cylindrical object (radius = 2.50 cm and length = 20.00 cm) with a coating of nickel metal, 4.0 mm thick. You place the object in a bath containing a salt (Na2SO4). One electrode is impure nickel, and the other is the object to be plated. The electrolyzing potential is 2.50 V. (a) Which is the anode and which is the cathode in the experiment? What half-reaction occurs at each electrode? (b) Calculate the number of kilowatt-hours (kWh) of energy required to carry out the electrolysis. (1 kWh = 3.6 × 106 J and 1 J = 1 C × 1 V).

99.

▲ Iron(II) ion undergoes a disproportionation reaction to give Fe(s) and the iron(III) ion. That is, iron(II) ion is both oxidized and reduced within the same reaction.

3 Fe2+(aq) uv Fe(s) + 2 Fe3+(aq)

(a) What two half-reactions make up the disproportionation reaction? (b) Use the values of the standard reduction potentials for the two half-reactions in part (a) to determine whether this disproportionation reaction is product-favored at equilibrium. (c) What is the equilibrium constant for this reaction? 100. ▲ Copper(I) ion disproportionates to copper metal and copper(II) ion. (See Study Question 99.) 2 Cu+(aq) uv Cu(s) + Cu2+(aq) Study Questions

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(a) What two half-reactions make up the disproportionation reaction? (b) Use values of the standard reduction potentials for the two half-reactions in part (a) to determine whether this disproportionation reaction is product-favored at equilibrium. (c) What is the equilibrium constant for this reaction? If you have a solution that initially contains 0.10 mol of Cu+ in 1.0 L of water, what are the concentrations of Cu+ and Cu2+ at equilibrium? 101. One way to write the reaction for discharge in a lithium-ion battery is Li(on carbon)(s) + CoO2(s) n 6 C(s) + LiCoO2(s)

(a) What are the oxidation numbers for cobalt in the two substances in the battery? (b) In such a battery, what reaction occurs at the cathode? At the anode? (c) An electrolyte is needed for ion conduction within the battery. From what you know about lithium chemistry, can the electrolyte in the battery be dissolved in water? 102. A lithium-ion camera battery is rated at 7500 mAh. That is, it can deliver 7500 milliamps (mA) or 7.5 amps of steady current for an hour. (a) How many moles of electrons can the battery deliver in one hour? (b) What mass of lithium is oxidized under these conditions in 1.0 hour? 103. Can either sodium or potassium metal be used as a sacrificial anode to protect the iron hull of a ship? Explain your answer. 104. Galvanized steel pipes are used in the plumbing of many older homes. When copper plumbing is added to a system consisting of galvanized steel pipes it is necessary to place an insulator between the copper and the steel to avoid corrosion. Write a balanced oxidation-reduction equation for the reaction that occurs if the pipes are directly connected. What is the standard potential between the metals?

In the Laboratory 105. Consider an electrochemical cell based on the halfreactions Ni2+(aq) + 2 e− n Ni(s) and Cd2+(aq) + 2 e− n Cd(s). (a) Diagram the cell, and label each of the components (including the anode, cathode, and salt bridge). (b) Use the equations for the half-reactions to write a balanced, net ionic equation for the overall cell reaction. (c) What is the polarity of each electrode?

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(d) What is the value of E°cell? (e) In which direction do electrons flow in the external circuit? (f) Assume that a salt bridge containing NaNO3 connects the two half-cells. In which direction do the Na+(aq) ions move? In which direction do the NO3−(aq) ions move? (g) Calculate the equilibrium constant for the reaction. (h) If the concentration of Cd2+ is reduced to 0.010 M and [Ni2+] = 1.0 M, what is the value of Ecell? Is the net reaction still the reaction given in part (b)? (i) If 0.050 A is drawn from the battery, how long can it last if you begin with 1.0 L of each of the solutions and each was initially 1.0 M in dissolved species? Each electrode weighs 50.0 g in the beginning. 106. An old method of measuring the current flowing in a circuit was to use a silver coulometer. The current passed first through a solution of Ag+(aq) and then into another solution containing an electroactive species. The amount of silver metal deposited at the cathode was weighed. From the mass of silver, the number of atoms of silver was calculated. Since the reduction of a silver ion requires one electron, this value equaled the number of electrons passing through the circuit. If the time was noted, the average current could be calculated. If, in such an experiment, 0.052 g of Ag is deposited during 450 s, what was the average current flowing in the circuit? 107. A silver coulometer (Study Question 106) was used in the past to measure the current flowing in an electrochemical cell. Suppose you found that the current flowing through an electrolysis cell deposited 0.089 g of Ag metal at the cathode after exactly 10 min. If this same current then passed through a cell containing gold(III) ion in the form of [AuCl4]−, how much gold was deposited at the cathode in that electrolysis cell? 108. ▲ Four metals, A, B, C, and D, exhibit the following properties: (a) Only A and C react with l.0 M hydrochloric acid to give H2(g). (b) When C is added to solutions of the ions of the other metals, metallic B, D, and A are formed. (c) Metal D reduces Bn+ to give metallic B and Dn+. Based on this information, arrange the four metals in order of increasing ability to act as reducing agents.

Chapter 19 / Principles of Chemical Reactivity: Electron Transfer Reactions

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109. ▲ A solution of KI is added dropwise to a pale blue solution of Cu(NO3)2. The solution changes to a brown color, and a precipitate of CuI forms. In contrast, no change is observed if solutions of KCl and KBr are added to aqueous Cu(NO3)2. Consult the table of standard reduction potentials to explain the dissimilar results seen with the different halides. Write an equation for the reaction that occurs when solutions of KI and Cu(NO3)2 are mixed. 110.

▲

The amount of oxygen, O2, dissolved in a water sample at 25 °C can be determined by titration. The first step is to add solutions of MnSO4 and NaOH to the water to convert the dissolved oxygen to MnO2. A solution of H2SO4 and KI is then added to convert the MnO2 to Mn2+, and the iodide ion is converted to I2. The I2 is then titrated with standardized Na2S2O3. (a) Balance the equation for the reaction of Mn2+ ions with O2 in basic solution. (b) Balance the equation for the reaction of MnO2 with I− in acid solution. (c) Balance the equation for the reaction of S2O32− with I2. (d) Calculate the amount of O2 in 25.0 mL of water if the titration requires 2.45 mL of 0.0112 M Na2S2O3 solution.

Summary and Conceptual Questions The following questions may use concepts from this and ­previous chapters. 111. Fluorinated organic compounds are used as herbicides, flame retardants, and fire-extinguishing agents, among other things. A reaction such as CH3SO2F + 3 HF n CF3SO2F + 3 H2

is carried out electrochemically in liquid HF as the solvent. (a) If you electrolyze 150 g of CH3SO2F, what mass of HF is required, and what mass of each product can be isolated? (b) Is H2 produced at the anode or the cathode of the electrolysis cell? (c) A typical electrolysis cell operates at 8.0 V and 250 A. How many kilowatt-hours of energy does one such cell consume in 24 hours? 112. ▲ The free energy change for a reaction, ΔrG°, is the maximum energy that can be extracted from the process as work, whereas ΔrH° is the total chemical potential energy change. The efficiency of a fuel cell is the ratio of these two quantities. Efficiency 



r G  100% r H 

Consider the hydrogen-oxygen fuel cell, where the net reaction is H2(g) + 1⁄2 O2(g) n H2O(ℓ)

(a) Calculate the efficiency of the fuel cell under standard conditions. (b) Calculate the efficiency of the fuel cell if the product is water vapor instead of liquid water. (c) Does the efficiency depend on the state of the reaction product? Why or why not? 113. A hydrogen-oxygen fuel cell operates on the simple reaction H2(g) + 1⁄2 O2(g) n H2O(ℓ)

If the cell is designed to produce 1.5 A of current and if the hydrogen is contained in a 1.0-L tank at 200 atm pressure at 25 °C, how long can the fuel cell operate before the hydrogen runs out? (Assume there is an unlimited supply of O2.) 114. ▲ Consider the half-reaction 2 H2O(ℓ) + 2 e− n 2 OH−(aq) + H2(g)  E° = −0.83 V

Assuming the pressure of H2 is 1.0 bar, what is the reduction potential for water in solutions at pH = 7 (neutral) and pH = 1 (acid)? Is it easier to reduce water in acid or base? Comment on the value of E° at pH = 1. 115. ▲ Living organisms derive energy from the oxidation of food, typified by glucose. C6H12O6(aq) + 6 O2(g) n 6 CO2(g) + 6 H2O(ℓ)

Electrons in this redox process are transferred from glucose to oxygen in a series of at least 25 steps. It is instructive to calculate the total daily current flow in a typical organism and the rate of energy expenditure (power). (See T. P. Chirpich, Journal of Chemical Education, 1975, 52, 99–100.) (a) The molar enthalpy of combustion of glucose is −2800 kJ/mol-rxn. If you are on a typical daily diet of 2400 Cal (kilocalories), what amount of glucose (in moles) must be consumed in a day if glucose is the only source of energy? What amount of O2 must be consumed in the oxidation process? (b) How many moles of electrons must be supplied to reduce the amount of O2 calculated in part (a)? (c) Based on the answer in part (b), calculate the current flowing, per second, in your body from the combustion of glucose. (d) If the average standard potential in the electron transport chain is 1.0 V, what is the rate of energy expenditure in watts? Study Questions

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NASA

20 Nuclear Chemistry

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C hapt e r O ut li n e 20.1 Natural Radioactivity 20.2 Nuclear Reactions and Radioactive Decay 20.3 Stability of Atomic Nuclei 20.4 The Origin of the Elements: Nucleosynthesis 20.5 Rates of Nuclear Decay 20.6 Artificial Nuclear Reactions 20.7 Nuclear Fission and Nuclear Fusion 20.8 Radiation Health and Safety 20.9 Applications of Nuclear Chemistry

Nuclear chemistry is the study and use of radioactive elements and their compounds. It is a subject that bridges many areas of science and has a significant impact on modern society. Radioactive elements are widely used in medicine, and some of the latest diagnostic techniques, such as PET (positron emission tomography), depend on radioactivity. You may have had a stress test using radioactive technetium, and your home may be protected by a smoke detector, which contains a small amount of the radioactive element americium. Nuclear power plants generate electricity around the world. In the United States, there are over 90 nuclear power reactors, and over 400 are in use worldwide. Some of the most prominent scientists in the twentieth century were involved in nuclear chemistry and physics. Marie Curie (page 82) was a major figure in the field, and others will be introduced in this chapter. The world became more aware of their work around the time of World War II in the 1940s. On August 2, 1939, on the brink of the war, Albert Einstein sent a letter to President Franklin D. Roosevelt that changed history. Einstein called attention to work being done in nuclear p ­ hysics and chemistry. He said that he and others believed this work suggested the possibility that “uranium may be turned into a new and important source of energy … and [that it was] conceivable … that extremely powerful bombs of a new type may thus be constructed …” Einstein’s letter was the beginning of the Manhattan Project, which led to the development of the first atomic bomb. The world has not been the same since.

◀ The James Webb Space Telescope began sending images to Earth in 2022.  This image of the Southern Ring nebula is among the first of many to be sent from the telescope. The telescope can capture images, in the near infrared region of the electromagnetic spectrum, of galaxies formed early in the life of the universe. From these data, scientists can now learn even more about the synthesis of chemical elements in stars.

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Table 20.1

Characteristics of 𝛂, 𝛃, and 𝛄 Radiation

Name

Symbols

Charge

Mass (g/particle)

Alpha

4 4 2He, 2α

2+

6.65 × 10−24

Beta

0 0 − 1e, − 1β

1−

9.11 × 10−28

Gamma

γ

0

0

20.1 Natural Radioactivity Goal for Section 20.1 • Identify α, β, and γ radiation, the three major types of radiation in natural radioactive decay. Discovery of Radioactivity ​The

discovery of radioactivity by Henri Becquerel and the isolation of radium and polonium from pitchblende, a uranium ore, by Marie Curie were described in ”Key Experiments: The Nature of the Atom and Its Components,” page 73 and in “A Closer Look: Marie Curie (1867–1934),” page 82.

Common Symbols: 𝛂 and 𝛃 ​ Symbols used to represent alpha and beta particles do not include a plus or minus superscript on the right for the charge.

In the late nineteenth century, while studying radiation emanating from uranium and thorium, Ernest Rutherford (1871–1937) stated, “There are present at least two distinct types of radiation—one that is readily absorbed, which will be termed for convenience 𝛂 (alpha) radiation, and the other of a more penetrative character, which will be termed 𝛃 (beta) radiation.” Subsequently, charge-to-mass ratio measurements showed that α radiation is composed of helium nuclei (He2+) and β radiation is composed of electrons (e−) (Table 20.1). A third type of radiation was later discovered by the French scientist Paul Villard (1860–1934); he named it 𝛄 (gamma) radiation, using the third letter in the Greek alphabet in keeping with Rutherford’s scheme. Unlike α and β radiation, γ radiation is not affected by electric and magnetic fields. Rather, it is a form of electromagnetic radiation that is more energetic than X-rays. Early studies measured the penetrating power of the three types of radiation (Figure 20.1). Alpha radiation is the least penetrating; it can be stopped by several sheets of ordinary paper or clothing. Beta particles can penetrate several millimeters of living bone or tissue, but about 0.5 cm of lead will stop the particles. Gamma radiation is the most penetrating. This type of radiation can pass completely through the human body, so thick layers of lead or concrete are required to shield the body from this radiation. Alpha and beta particles typically possess high kinetic energies. The energy of γ radiation is always very high. When α and β particles are stopped by a material, or when γ radiation is absorbed, the energy associated with the radiation is transferred to the material. This fact is important because the damage caused by radiation is related to the energy absorbed (Sections 20.8 and 20.9).

Figure 20.1  The relative penetrating ability of 𝛂, 𝛃, and 𝛄 radiation.  Highly charged α particles interact strongly with matter and are stopped by a piece of paper. Beta particles, with less mass and a lower charge, interact to a lesser extent with matter and thus can penetrate farther. Gamma radiation is the most penetrating.

994

10 cm of lead 0.5 cm of lead Paper

Alpha (α)

Beta (β) Gamma (γ)

Chapter 20 / Nuclear Chemistry

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20.2 Nuclear Reactions and Radioactive Decay Goals for Section 20.2 • Write balanced equations for nuclear reactions (alpha, beta, gamma, and positron emission; electron capture; and fission).

• Recognize that radioactive isotopes decay by a series of nuclear reactions, called a radioactive decay series, eventually ending with a stable isotope.

Equations for Nuclear Reactions In 1903, Rutherford and Frederick Soddy (1877–1956) proposed that radioactivity is the result of a natural change of an isotope of one element into an isotope of a different element. Such processes are called nuclear reactions. Consider a reaction in which radium-226 (the isotope of radium with mass number 226) emits an α particle to form radon-222. The equation for this ­reaction is 226 88Ra

n 24α + 22862Rn

In a nuclear reaction, the sum of the mass numbers of reacting particles must equal the sum of the mass numbers of products. Furthermore, to maintain nuclear charge balance, the sum of the atomic numbers of the products must equal the sum of the atomic numbers of the reactants. These principles are illustrated using the preceding nuclear equation: 226 88Ra

4 2α

n

+

α particle

radium-226

Symbols Used in Nuclear Equations ​

The mass number is included on the element symbol as a superscript, and the atomic number is given as a subscript.

222 86Rn

radon-222

Mass number

226

=

4

+

222

Atomic number

 88

=

2

+

 86

Be sure to notice that  α-particle emission leads to decrease of two units in atomic number and four units in the mass number. Similarly, nuclear mass and nuclear charge balance accompany β-particle emission, as illustrated by the radioactive decay of uranium-239: 239 92U

0 − 1β

n

+

β particle

uranium-239

239 93Np

neptunium-239

Mass number

239

=

0

+

239

Atomic number

 92

=

1−

+

 93

The β particle has a charge of 1− and a mass number of 0. For beta radiation, charge balance requires that the atomic number of the product is one unit greater than the atomic number of the reacting nucleus. The mass number does not change in this process. How does a nucleus, composed of protons and neutrons, eject an electron? It is a complex process, but the net result is the conversion within the nucleus of a neutron to a proton and an electron. 1 0n

8777n −10 e

neutron

electron

+

Atomic Number of Beta Particles ​In

Chapter 2, atomic number was defined as the number of protons in the nucleus of an atom. In nuclear equations, the atomic number is the charge of the nuclear particle. Thus, the atomic number of a beta particle (an electron) in nuclear equations is 1–.

1 1p

proton

Notice that the mass and atomic numbers balance in this equation. What is the origin of the γ radiation that accompanies many nuclear reactions? Recall that a photon of visible light is emitted when an atom undergoes a transition from an excited electronic state to a lower energy state (Section 6.3). Gamma radiation similarly originates from transitions between nuclear energy levels. Nuclear reactions

20.2  Nuclear Reactions and Radioactive Decay

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often result in the formation of a product nucleus in an excited nuclear state. The product nucleus can return to the ground state by emitting a photon. The energy of γ radiation is a measure of the large energy difference between the energy levels in the nucleus.

Radioactive Decay Series Several naturally occurring radioactive isotopes decay to form a product that is also radioactive. When this happens, the initial nuclear reaction is followed by a second nuclear reaction; if the situation is repeated, a third and a fourth nuclear reaction occur; and so on. Eventually, a nonradioactive isotope is formed to end the series. Such a sequence of nuclear reactions is called a radioactive decay series. In each step of this sequence, the reactant nucleus is called the parent, and the product is called the daughter. Uranium-238, the most abundant of three naturally occurring uranium isotopes, is the origin of one of four radioactive decay series (Figure 20.2). This series begins with the loss of an α particle from 23982U to form radioactive 23940Th. Thorium-234 then decomposes by β emission to 23941Pa, which emits a β particle to give 23942U. Uranium-234 is an α emitter, forming 23900Th. Further α and β emissions follow, until the series ends with formation of the stable, nonradioactive isotope, 20862Pb. In all, this radioactive decay series converting 23982U to 20862Pb has 14 decays, with eight α and six β particles being emitted. An equation can be written for each step in the sequence. Equations for the first four steps in the uranium-238 radioactive decay series are as follows: Step 1.

238 92U

Step 2.

234 90Th

n 23914Pa + −10β

Step 3.

234 91Pa

n 23924U + −10β

Step 4.

234 92U

n 23900Th + 24α

n 23904Th + 24α

Uranium ore always contains trace quantities of the radioactive elements formed in the radioactive decay series. A significant development in nuclear chemistry was Marie Curie’s discovery in 1898 of radium (226Ra) and polonium (210Po) as trace components

Tl

Pb

Bi

Po

At

Rn

Fr

Ra

Ac

Th

Pa

U 109 y

238  s = m= h = d = y =

234 230 Mass number (A)

U 238

seconds minutes hours days years

226 222

Rn 3.8 d

222

234

Th 24 d

230Th 104 y 226Ra 1600 y



234 Pa 6.7 h



234 U 105 y





 Each -decay step lowers the atomic number by two units and the mass number by four units.

 218  214Pb

214 210 206

27 m

210Tl 1.3 m

210Pb

206Tl

206Pb

22 y

4.2 m

stable

81

82

214Bi

218Po 3.1 m

Beta-particle emission does not change the mass number but raises the atomic number by one unit.

218At

1.4 s

214Po

 27 m  162 μs  210Bi



5d

210Po

 138 d



83

84

85

86 87 88 Atomic number (Z)

89

90

91

92

Figure 20.2  The uranium-238 radioactive decay series.  ​The steps in this radioactive decay series are

shown graphically in this plot of mass number versus atomic number. Half-lives of the isotopes are in­cluded on the chart. Notice that several of the isotopes in this series decompose by two different pathways.

996

Chapter 20 / Nuclear Chemistry

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A Closer Look

Radioactive Decay Series

Figure 20.2 illustrates the u ­ ranium-238 decay series, which is sometimes referred to as the uranium series. It is also called the 4n 1 2 series because each nuclide mass, M, in the series fits this equation where n is an integer. For the U-238 series in Figure 20.2, n = 59 for the first member of this series: 4(59) + 2 = 238. Because the mass number changes in this series only when an α particle is lost, all the mass numbers in a decay series are some multiple of four less than the first isotope in the series. For the series beginning with 238U, the mass numbers are: 238 (n = 59), 234 (n = 58), 230 (n = 57), . . ., and 206 (n = 51).



Series member, 227 89Ac: 4(56) + 3 = 227



Last member of the series, 207 82Pb: 4(51) + 3 = 207

There is another decay series with the equation 4n + 1 = M, called the neptunium series. Many of its members have only a fleeting existence. Neptunium-237 (which has a half-life of over 2 million years) has the equation 4(59) + 1 = 237. The series ends 209 at 205 81Tl (n = 51) or 83Bi (n = 52). Tl

First member of the series, 235 92 U: 4(58) + 3 = 235

Bi

Po

s = m= h = d = y =

228 Mass number (A)

The actinium series is a decay series headed by 235 92 U. The series includes actinium, astatine, and bismuth, among others, and it also ends at lead (207Pb). Intermediate nuclides have the mass numbers 235, 231, 227, . . ., 207. The equation for this series is 4n + 3.

Pb

232

First member of the series, 238 92U: 4(59) + 2 = 238 Last member of the series, 206 82Pb: 4(51) + 2 = 206

Finally, there is the 4n series, or thorium series, which begins with 232 90 Th (see ­Figure). The series includes ­actinium, radium, radon, polonium, bismuth, ­ resent lead, and thallium. All can be p in any naturally occurring sample of ­thorium. This decay series ends at 208 82Pb. The equation for this series is 4n = M where n = 58 for the first member of the series, 232 90Th.

224

At

Rn

Fr

Ra

Ac

Th 232Th

seconds minutes hours days years

1010 y

 228

Ra 5.7 y

224

Ra 3.6 d

220Rn

220



228Ac 6.1 h



228Th

1.9 y





55 s

 216Po

216  212

Pb Bi 10.6 h  61 m

212

 208

212





0.14 s 212 Po 10–7 s

208Tl 208Pb 3.1 m  stable

81

82

83

84

85 86 87 Atomic number (Z)

88

89

90

Figure  The thorium decay series.  The series begin with 232 90 Th, and the mass number of each successive member is given by 4n where n = 58, 57, 56, 55, . . ., 52.

of pitchblende, a uranium ore. The amount of each of these elements is small because their isotopes have short half-lives; for example, 210Po has a half-life of 138 days. Consequently, Curie was able to isolate only a single gram of radium from 7 tons of ore. It is a credit to her skill as a chemist that she extracted sufficient amounts of radium and polonium from uranium ore to identify these elements.

Ex am p le 20.1

Radioactive Decay Series Problem  Another uranium radioactive decay series begins with 235 92U and ends with 207 82Pb.

(a) How many α and β particles are emitted in this series? (b) The first three steps of this series are (in order) α, β, and α emission. Write an equation for each of these steps.

What Do You Know?  You are given the starting (long-lived radioactive) isotope and the final (stable) isotope. A series of α and β reactions is the link between these two s­ pecies. You also know the composition of α and β particles and the rules for writing nuclear equations.

20.2  Nuclear Reactions and Radioactive Decay

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997

Strategy  In part (a), find the total change in atomic number and mass number. A com­ bination of α and β particles is required that will decrease the mass number and atomic number by these amounts. In part (b), write the individual nuclear equations for the indicated steps, keeping in mind that in each equation, the sum of the mass numbers for reactants and products must be equal, and the sum of the atomic numbers for reactants and products must also be equal. Solution (a) The mass number declines by 28 (235 − 207). Because a decrease of four in the mass number occurs with each α emission,  seven α particles must be emitted.  Also, for each α emission, the atomic number decreases by two. Emission of seven α particles would cause the atomic number to decrease by 14, but the actual decrease in atomic number is 10 (92 − 82). This means that  four β particles must also have been emitted  because each β emission increases the atomic number of the product by one. Thus, the radioactive decay sequence involves emission of seven α and four β particles. (b) Step 1.

235 92U

Step 2.

231 90Th

Step 3.

231 91Pa

n 23901Th + 24α n 23911Pa + −10β n 22897Ac + 24α

Think about Your Answer   Because the mass number changes in these series only when an α particle is lost, all of the mass numbers in a given decay series are some multiple of four less than the first isotope in the series. For the series beginning with 238U, the mass numbers are in the series 238, 234, 230, . . . , 206. This series is sometimes called the 4n + 2 series because each mass number (M) fits the equation 4n + 2 = M, where n is an integer (n is 59 for the first member of this series and 51 for the last member). Other decay series are also possible, as outlined in “A Closer Look: ­Radioactive Decay Series.”

Check Your Understanding (a) Six α and four β particles are emitted in the thorium-232 radioactive decay series before a stable isotope is reached. What is the final product in this series? (b) The first three steps in the thorium-232 decay series (in order) are α, β, and β emission. Write an equation for each step.

Other Types of Radioactive Decay Positrons  Positrons were

discovered by Carl Anderson (1905–1991) in 1932. The positron is one of a group of particles known as antimatter. If matter and antimatter particles collide, mutual annihilation occurs, with energy being emitted.

Most naturally occurring radioactive elements decay by emission of α, β, and γ radiation. Other nuclear decay processes became known, however, when new radioactive elements were synthesized by artificial means. These include positron (+10𝛃) emission and electron capture. Positrons (+01β) and electrons have the same mass but opposite charge. The positron is the antimatter analog to an electron. Positron emission by polonium-207, for example, results in the formation of bismuth-207. 207 84Po

n

polonium-207

0 +1β

+

207 83Bi

bismuth-207

positron

Mass number

207

=

0

+

207

Atomic number

 84

=

1

+

 83

To retain charge balance,  positron decay results in a decrease in the atomic number.

998

Chapter 20 / Nuclear Chemistry

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Chemistry in Your Career

Sophia Zahn

Sophia Zahn Although Sophia Zahn loved the chemistry courses she took while obtaining her degrees (B.S. and M.S. from Carnegie Mellon University), she knew she didn’t want a career as a researcher working in a lab. Instead, she went to law school to become a patent attorney with a focus on ­p atent prosecution in the chemical industries. “I like that patent law allows me to engage with science in a way that is a good fit for me.” Zahn explains what her job typically involves: “The patent drafting process begins

with the inventor writing an invention disclosure. It is my job to take this disclosure and draft a patent that protects my client’s invention and advances their goals.” She emphasizes how her chemistry education makes her better at her job. “My chemistry background, especially my research experience, is essential to understanding my client’s research in order to effectively protect their invention.”

In electron capture, an extranuclear electron is captured by the nucleus. The mass number is unchanged, and the atomic number is reduced by one. (In an old nomenclature, the innermost electron shell was called the K shell, and electron capture was called K capture.) 7 4Be

+

0 −1e

beryllium-7

n

7 3Li

lithium-7

electron

Mass number

7

+

0

=

7

Atomic number

4

+

1−

=

3

In summary, most unstable nuclei decay by one of four paths: α or β decay, positron emission, or electron capture. Gamma radiation often accompanies these processes. Section 20.7 introduces a fifth way that nuclei decompose, fission.

Ex am p le 20.2

Neutrinos and Antineutrinos  Beta particles with a wide range of energies are emitted. To balance the energy and momentum associated with β decay, it is necessary to postulate the concurrent emission of another particle, the antineutrino. Similarly, neutrino emission accompanies positron emission. Much study has gone into detecting neutrinos and antineutrinos. These particles do not have a charge. They are believed to have mass, but their mass is so small it has not been possible to measure it. Neutrinos and antineutrinos are not usually included when writing nuclear equations.

Nuclear Reactions Problem   Complete the following equations. Give the symbol, mass number, and atomic number of the product species. (a)

37 18Ar

(b)

11 6C

+ −10e n ?

n 115B + ?

(c)

35 16S

n 1375Cl + ?

(d)

30 15P

n +10β + ?

What Do You Know?   You are given the mass numbers and atomic numbers for two of the three species involved in each reaction. You know that both the masses and charges must balance in a balanced nuclear equation. Strategy   The missing product in each reaction can be determined by recognizing that the sums of mass numbers and atomic numbers for products and reactants must be equal. When you know the nuclear mass and nuclear charge of the product, you can identify it with the appropriate symbol.

Solution (a) This is an electron-capture reaction. The product has a mass number of 37 + 0 = 37 and an atomic number of 18 − 1 = 17. Therefore, the symbol for the product is 1377Cl.

20.2  Nuclear Reactions and Radioactive Decay

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999

(b) This missing particle has a mass number of zero and a charge of 1+; these are the characteristics of a positron, +10β. If this particle is included in the equation, the sums of the atomic numbers (6 = 5 + 1) and the mass numbers (11) on both sides of the equation are equal. (c) A β particle, −10β, is required to balance the mass numbers (35) and atomic numbers (16 = 17 − 1) in the equation. (d) The product nucleus has mass number 30 and atomic number 14. This identifies the unknown as 1340Si.

Think about Your Answer  Be sure to double check that the mass numbers and atomic numbers add to the same values on both sides of the equation.

Check Your Understanding  Indicate the symbol, the mass number, and the atomic number of the missing product in each of the following nuclear reactions. (a)

13 7N

n 136C + ?

(b)

41 20Ca

+ −10e n ?

(c)

90 38Sr

n 3990Y + ?

(d)

22 11Na

n ? + +10β

Radon: An Environmental Problem

John R. Townsend

The uranium-238 radioactive decay series is the source of the environmental hazard radon. Trace quantities of uranium are often present naturally in soil and rocks, and radon-222 is being continuously formed. Because radon is chemically inert, it is not trapped by chemical processes in soil or water but is instead free to seep into mines or homes through pores in cement block walls, cracks in basement floors or walls, or around pipes. Because it is denser than air, radon tends to collect in low spots, and its concentration can build up in the basement of a building if steps are not taken to remove it. The major health hazard from radon, when it is inhaled by humans, arises not from radon itself but from its decomposition product, polonium.

Radon detector.  This monitor detects radon gas in the home. The units reported are picocuries per liter, pCi/L. (See Section 20.8 for a discussion of the units used to report levels of radioactivity.)

222 86Rn

4 n 218 84Po + 2a

218 84Po

4 n 214 82Pb + 2a

Radon does not undergo chemical reactions or form compounds that can be taken up in the body. Polonium, however, is not chemically inert. Polonium-218 can lodge in body tissues, where it undergoes α decay to yield lead-214, another radioactive isotope. The range of an α particle in body tissue is quite small, perhaps 0.7 mm. However, this is approximately the thickness of lung epithelial cells so α-particle radiation can seriously damage lung tissue. Virtually every home in the United States has some level of radon, and kits can be purchased to test for the presence of this gas. If a significant amount of radon gas is detected in your home, you should take corrective actions such as sealing cracks around the foundation and in the basement.

20.3 Stability of Atomic Nuclei Goals for Section 20.3 • Assess nuclear stability based on the numbers of neutrons and protons in a nucleus. • Predict possible modes of decay of unstable nuclei based on n/p ratios. • Calculate the mass defect for an isotope and from this determine the nuclear binding energy Eb and the binding energy per nucleon.

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140

Figure 20.3  Stable and unstable isotopes. ​A graph of the number of neutrons (n) versus the number of protons (p) for stable (black circles) and radioactive (red circles) isotopes from hydrogen through mercury. This graph is used to assess criteria for nuclear stability and to predict modes of decay for unstable nuclei.

130 120 Isotopes in this neutronrich region (with n/p >> 1) decay by β emission to give a lower n/p ratio.

110 100

Isotopes in this region decay by α emission to give a lower atomic number.

α emission

Number of neutrons

90 80 β emission

70

n/p = 1 60 50 Stable Radioactive

40 30 20 10 0

Isotopes in this neutronpoor region decay by positron emission or electron capture to give a higher n/p ratio.

Positron emission or electron capture

0

10

20

30

40 50 60 Number of protons

70

80

90

100

Figure 20.3 is a plot of the composition of the known isotopes of the elements. The horizontal axis represents the number of protons, and the vertical axis gives the number of neutrons. Each circle represents an isotope identified by the n ­ umber of neutrons and protons contained in its nucleus. The black circles indicate stable ­(nonradioactive) isotopes, 254 in number, and the red circles indicate some of the known radioactive isotopes. For example, the three isotopes of hydrogen are 11H and 2 3 1H ­(neither is radioactive) and 1H (tritium, radioactive). For lithium, the third element, isotopes are known with mass numbers 4, 5, 6, and 7. The isotopes with masses of 6 and 7 (black) are stable, whereas the other two isotopes (red) are radioactive. Figure 20.3 illustrates the following information about nuclear stability: •

Stable isotopes fall in a very narrow range called the band of stability. It is remarkable how few isotopes are stable.

•

Only two stable isotopes (11H and 32He) have more protons than neutrons.

•

Up to calcium (Z = 20), stable isotopes often have equal numbers of protons and neutrons or only one or two more neutrons than protons.

•

Beyond calcium, the neutron-to-proton ratio is always greater than 1. As the mass increases, the band of stable isotopes deviates more and more from a 1∶1 neutron-to-proton ratio (the line in Figure 20.3 for which n/p = 1).

•

Beyond bismuth (83 protons), all isotopes are unstable and radioactive. The forces of attraction between nuclear particles are not strong enough to hold heavy nuclei together.

•

Isotopes that fall farther from the band of stability tend to have shorter halflives than do unstable isotopes nearer to the band of stability.

•

Isotope stability is associated with even numbers of protons and even numbers of neutrons. Among the stable isotopes, 148 have an even number of protons and neutrons, 53 have an even number of protons and an odd number of neutrons, and 48 have an odd number of protons and an even number of neutrons. Only five stable isotopes (12H, 36Li, 105B, 147N, and 18730Ta) have odd numbers of both protons and neutrons.



20.3  Stability of Atomic Nuclei

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1001

The Band of Stability and Radioactive Decay Besides being a criterion for stability, the neutron-to-proton ratio can assist in predicting what type of radioactive decay will be observed. Unstable nuclei decay in a manner that brings them toward a stable neutron-to-proton ratio—that is, toward the band of stability. All elements beyond bismuth (Z = 83) are unstable. For an element beyond this atomic number, a process that decreases the atomic number is needed to reach the band of stability. Alpha emission is an effective way to do this because each emission decreases the atomic number by two. For example, americium, the radioactive element used in smoke detectors, decays by α emission:

•

243 95Am

n 24α + 23939Np

Beta emission occurs for isotopes that have a high neutron-to-proton ratio— that is, isotopes above the band of stability. With β decay, the atomic number increases by one, and the mass number remains constant, resulting in a lower neutron-to-proton ratio:

•

60 27Co

n

0 −1β

+ 2680Ni

Isotopes with a low neutron-to-proton ratio, below the band of stability, decay by positron emission or by electron capture. Both processes lead to product nuclei with a lower atomic number and the same mass number and move the product to a greater n/p ratio and closer to the band of stability:

•

13 7N

n +10β + 136C

41 20Ca

+ −10e n 1491K

E xamp le 20.3

Predicting Modes of Radioactive Decay Problem   Identify probable mode(s) of decay for each isotope and write an equation for the decay process. (a) oxygen-15, 158O

(c) fluorine-20, 209F

(b) uranium-234, 23924U

(d) manganese-56, 2556Mn

What Do You Know?  The possible modes of decomposition are α, β, or positron emission and electron capture. The preferred mode will give a more stable isotope and create a nucleus closer to the band of stability.

Strategy   There are two main ideas to consider. First, if Z is greater than 83, then α decay is likely. Second, consider the n/p ratio. If the ratio is much greater than 1, then β decay is probable. If the ratio is less than 1, then positron emission or electron capture is the more likely process. It is not possible to choose between the latter two modes of decay without further information.

Solution (a) Oxygen-15 has 7 neutrons and 8 protons, so the neutron-to-proton (n/p) ratio is less than 1—too low for 15O to be stable. Nuclei with too few neutrons are expected to decay by  either positron emission or electron capture.  In this instance, the process is 0 +1β or positron emission. 15 8O

n +10β + 157N.

(b)  Alpha emission  is a common mode of decay for isotopes of elements with atomic numbers higher than 83. The decay of uranium-234 is one example: 234 92U

n 23900Th + 24α

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(c) Fluorine-20 has 11 neutrons and 9 protons, a high n/p ratio. The ratio is lowered by  β emission:  20 9F

n −10β + 1200Ne

(d) The mass number of 56Mn is higher than the atomic weight of the element (54.85). This suggests that this radioactive isotope has an excess of neutrons (giving a high n/p ratio), in which case it would be expected to decay by  β emission:  56 25Mn

n −10β + 2566Fe

Think about Your Answer   Decomposition can occur by four possible paths, α emission, β emission, positron emission, or electron capture. Whether the decay f­ ollows beta emission or positron emission (or electron capture) is predicted by the ­neutron/proton ratio, relative to the n/p ratio in stable isotopes.

Check Your Understanding  Write an equation for the probable mode of decay for each of the following unstable ­isotopes, and write an equation for that nuclear reaction. (a) silicon-32, 1342Si

(c) plutonium-239, 23949Pu

(b) titanium-45, 2425Ti

(d) potassium-42, 1492K

Nuclear Binding Energy An atomic nucleus can contain as many as 83 protons and be stable. For stability, nuclear binding (attractive) forces must be greater than the electrostatic repulsive forces between the closely packed protons in the nucleus. Nuclear binding energy, Eb, is defined as the energy required to separate the nucleus of an atom into protons and neutrons. For example, the nuclear binding energy for deuterium is the energy required to convert one mole of deuterium (21H) nuclei into one mole of protons and one mole of neutrons. 2 1H

n 11p + 01n  Eb = +2.146 × 108 kJ/mol

The positive sign for Eb indicates that energy is required for this process. A deuterium nucleus is more stable than an isolated proton and an isolated neutron, just as the H2 molecule is more stable than two isolated H atoms. Recall, however, that the H—H bond energy is only 436 kJ/mol. The energy holding a proton and a neutron together in a deuterium nucleus, 2.146 × 108 kJ/mol, is about 500,000 times larger than the typical covalent bond energies. Nuclear binding energy can be explained using a combination of experimental observation and a theory. The experimental observation is that the mass of a nucleus is always less than the sum of the masses of its constituent protons and neutrons. The theory is that the missing mass, called the mass defect, is equated with the energy that holds the nuclear particles together. The mass defect for deuterium is the difference between the mass of a deuterium nucleus and the sum of the masses of a proton and a neutron. Mass spectrometric measurements (Section 2.2) give the masses of these particles to a high level of accuracy, providing the numbers needed to carry out calculations of mass defects. Masses of atomic nuclei are not generally listed in reference tables, but masses of atoms are. Nonetheless, calculation of the mass defect can be carried out using atom masses instead of masses of nuclei. (By using atomic masses, the calculation includes the masses of extranuclear electrons in the reactants and the products. However, because the same number of extranuclear electrons appears in products



20.3  Stability of Atomic Nuclei

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1003

and reactants, this does not affect the result.) Thus, for one mole of deuterium nuclei, the mass defect is found as follows: 2 1H

1 1H

n

2.014102 g/mol

1 0n

+

1.007825 g/mol

1.008665 g/mol

Mass defect = ∆m = mass of products − mass of reactants = [1.007825 g/mol + 1.008665 g/mol] − 2.014102 g/mol = 0.002388 g/mol

The relationship between mass and energy is given by Einstein’s 1905 theory of special relativity, which holds that mass and energy are different manifestations of the same quantity. Einstein showed that energy is equivalent to mass times the square of the speed of light; that is, E = mc2. In the case of atomic nuclei, it is assumed that the missing mass (the mass defect, ∆m) is equal to the binding energy holding the nucleus together. Eb = (∆m)c2



(20.1)

If ∆m is given in kilograms and the speed of light is given in meters per second, Eb will have units of joules (because 1 J = 1 kg ∙ m2/s2). For the decomposition of one mole of deuterium nuclei to one mole of protons and one mole of neutrons, the energy required is Eb = (2.388 × 10−6 kg/mol)(2.9979 × 108 m/s)2 = 2.146 × 1011 J/mol of 12H nuclei (= 2.146 × 108 kJ/mol of 12H nuclei)

The nuclear stabilities of different elements are compared using the binding energy per mole of nucleons. (Nucleon, n, is the general name given to nuclear particles—that is, protons and neutrons.) A deuterium nucleus contains two nucleons, so the binding energy per mole of nucleons, Eb/n, is 2.146 × 108 kJ/mol divided by two, or 1.073 × 108 kJ/mol nucleon.  2.146 3 108 kJ   1 mol 21H nuclei     mol 21H nuclei   2 mol nucleons 

Eb/n = 

Eb/n = 1.073 × 108 kJ/mol nucleons

The binding energy per nucleon can be calculated for any atom whose mass is known. Then, to compare nuclear stabilities, binding energies per nucleon are plotted as a function of mass number (Figure 20.4). The greater the binding energy per nucleon, the greater the stability of the nucleus. Notice in Figure 20.4 that the point of maximum nuclear stability occurs at a mass of 56, that is, at iron in the periodic table. 9.0 × 108

Figure 20.4  Relative stability of nuclei.  ​Binding energy per

Binding energy (kJ/mol of nucleons)

nucleon for the most stable isotope of elements between hydrogen and uranium is plotted as a function of mass number. (Fission and fusion are discussed in Section 20.7.)

4 2 He

8.0 × 108

56 26 Fe

7.0 × 108

238 92 U

6.0 × 108 5.0 × 108 Fusion

4.0 × 108

Region of greatest stability

Fission

100

150

3.0 × 108 2.0 × 108 1.0 × 108 0

50

200

250

Mass number

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Ex am p le 20.4

Nuclear Binding Energy Problem   Calculate the binding energy per mole of nucleons, E b (in kJ/mol), and the binding energy per nucleon, E b /n (in kJ/mol nucleon), for carbon-12. The molar masses needed are: carbon-12 (12.000000 g/mol, hydrogen (11H, 1.007825 g/mol), and neutrons ( 10n, 1.008665 g/mol).

What Do You Know?   The molar masses of carbon-12, hydrogen, and neutrons are given in the problem.

Strategy  Step 1. Calculate the mass defect, ∆m. Step 2. Calculate the binding energy using Equation 20.1. Step 3. Calculate the binding energy per nucleon.

Solution   The molar mass of 11H is 1.007825 g/mol, and the molar mass of 10n is

1.008665 g/mol. Carbon-12, 126C, has a molar mass of 12.000000 g/mol. Step 1. Calculate the mass defect, ∆m.

The mass defect is the difference between the mass of carbon-12 and the masses of six protons, six neutrons, and six electrons. The mass of 1 mol of protons and 1 mol of electrons can be taken into account by using the molar mass of 11H. ∆m = [(6 × mass 11H) + (6 × mass 01n)] − mass 126C = [(6 × 1.007825 g/mol) + (6 × 1.008665 g/mol)] − 12.000000 g/mol = 9.8940 × 10−2 g/mol Step 2. Calculate the binding energy using Equation 20.1 Using the mass in kilograms and the speed of light in meters per second gives the binding energy in joules: Eb = (∆m)c2 = (9.8940 × 10−5 kg/mol)(2.99792 × 108 m/s)2 = 8.89226 × 1012 J/mol (=  8.8923 × 109 kJ/mol ) Step 3. Calculate the binding energy per nucleon. The binding energy per nucleon, Eb/n, is determined by dividing the binding energy by 12 (the number of nucleons). Eb 8.89226 3 109 kJ/mol  n 12 mol nucleons/mol  = 7.4102 × 108 kJ/mol nucleons 

Think about Your Answer   The binding energy is a very large quantity of energy compared to those of ordinary chemical reactions. For comparison, ∆rH° for the very e­ xothermic chemical reaction of hydrogen and oxygen to form water vapor is only −242 kJ per mol of water vapor formed at 25 °C.

Check Your Understanding  Calculate the binding energy per nucleon, in kilojoules per mole, for lithium-6. The molar mass of 36Li is 6.015125 g/mol.



20.3  Stability of Atomic Nuclei

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1005

20.4 Origin of the Elements: Nucleosynthesis Goal for Section 20.4 • Understand the steps that can lead to the synthesis of some elements. Presently, 118 elements are known on planet Earth. Of these, the first 92—except for technetium and promethium—occur naturally. The remainder are synthetized by well-defined methods in laboratories (page 1016). But what is the origin of these 90 naturally occurring elements? Why are some very abundant on Earth while ­others are found in only trace amounts? And why do elements with odd atomic numbers have lower abundance than their close neighbors on the periodic table with even atomic numbers (Figure 20.5)? Only recently have astronomers, p ­ hysicists, and chemists begun to understand the origin of the elements in our universe.

The Big Bang The most widely accepted theory of the origin of the universe, including the elements, is that it began with the Big Bang. It is thought that the universe began about 14 billion years ago from a point of infinite density called a singularity. For some unknown reason, it expanded and stretched. At a time of about 5 × 10−44 seconds, called the Planck time, the density of the universe was about 1093 g/cm3 and the temperature was 1032 K. Within one second, the universe contained matter that we are familiar with: protons, electrons, and neutrons. According to the Big Bang theory, the first nuclei containing more than one nuclear particle were deuterium nuclei, which formed from the combination of a proton with a neutron. 1 1H

Deuterium formation:  

+ 01n n 12H

These could undergo further reactions to form helium-4 (42He) nuclei, which is now the most abundant isotope of helium in the universe. There are several ways this could happen. One way is for the deuterium to combine with another ­proton to form a helium-3 nucleus, which then combined with another neutron to form the helium-4 nucleus. 2 1H

+ 11H n 32He

Helium-4 formation:  

3 2He

+ 10n n 42He

109 Abundance, atoms of element per 106 atoms of Si

Figure 20.5 Relative abundance of the elements.  ​ Elements with odd atomic numbers are less abundant than their neighbors with even atomic numbers. The abundances refer to the number of atoms of each element relative to silicon taken as 106. (Source: G. B. Haxel, J. B. Hendrick, and G. J. Orris. Fact Sheet 087-02, P. H. Stauffer and J. W. Hendley II, eds., U.S. Geological Survey, 2002. https://pubs.usgs.gov/fs/2002 /fs087-02/)

Helium-3 formation:  

Rock-forming elements O Al Na

H

F

Li B

K

Ca

Mg

C

103

Major industrial metals in red Precious metals in cyan Rare-earth elements in blue

Si

106

N

Be

100

Fe

Ti P S

Mn

Cl

Sr Zr Cu Zn Rb V Ga Nb Cr Sc Co Y Ni As Ge Br Mo Ag

Se

Ba Ce

Sb

Cd

Rh

Cs Pr

Sm Dy Eu

I

Pd

Hf Yb Ta W

Tb Ho Tm Lu

Te Re

0

10

20

30

40

Th U

Tl

Hg

Bi

Au Pt

Os Ir

Rarest “metals” 10–6

Pb

Er

In

Ru

10–3

La

Sn

Gd Nd

50

60

70

80

90

Atomic number (Z)

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Another method has the addition of particles in the opposite order: deuterium adds a neutron to form tritium (31H) and then this adds the additional proton to form helium-4. Tritium formation:  

2 1H

+ 10n n 31H

Helium-4 formation:  

3 1H

+ 11H n 42He

Calculations indicate that within about eight minutes after the Big Bang, about one-quarter of the mass of the universe was helium nuclei and about three-quarters was hydrogen.

Nucleosynthesis in Stars Matter was not evenly distributed in the universe, and this allowed gravity to coalesce the matter into galaxies and stars, where nucleosynthesis occurs by different means. There are two principal mechanisms for the fusion of hydrogen in stars: the proton-proton chain process and the carbon-nitrogen-oxygen process. The mass of the star determines which process predominates. In smaller stars, including ones like our sun, the proton-proton chain ­process ­predominates. In this process, protons rapidly combine to form d ­ euterium atoms, Deuterium formation:  

1 1H

+ 11H n 21H + +10β

The deuterium combines with another proton to form helium-3. Two helium-3 nuclei then combine to form a helium-4 nucleus, with the release of two protons. Helium-3 formation:  

2 1H

+ 11H n 32He

Helium-4 formation:  2 23He n 24He + 2 11H

The net result of this sequence is the transformation of four hydrogen atoms into one helium atom. But, importantly, the one helium atom has less mass than the four hydrogen atoms combined. The missing mass is converted into energy. This process accounts for most of the energy emitted by stars like the sun. For stars more than 1.3 times as massive as the sun, the carbon-nitrogen-oxygen process (or C-N-O process) predominates. Carbon-12 to nitrogen-13:  

12 6C

+ 11H n 137N

Nitrogen-13 to carbon-13:  

13 7N

n 136C + +1   0β

Carbon-13 to nitrogen-14:  

13 6C

+ 11H n 147N

Nitrogen-14 to oxygen-15:  

14 7N

+ 11H n 158O

Oxygen-15 to nitrogen-15:  

15 8O

n 157N + +  01β

Nitrogen-15 to helium-4 and carbon-12:  

15 7N

+ 11H n 42He + 126C

The end product of this process is again helium-4. The process began with a ­carbon-12 nucleus and four hydrogen atoms. In the end, a carbon-12 is left to begin the process again. Once again four protons led to a helium atom and an enormous amount of energy. Both the proton-proton chain reaction and the C-N-O process result in hydrogen nuclei forming helium nuclei. Now, 14 billion years after the Big Bang, helium accounts for about 11.3% of the atoms in the universe and hydrogen for 88.6%. That is, they account for 99.9% of the atoms in the universe and nearly 99% of the mass. How do other elements form? Regardless of which process is used to convert 11H to 42He in a star, once the hydrogen runs out, the star contracts and the temperature increases. If the temperature at the core reaches about 108 K, the helium nuclei begin to react in the ­triple-­alpha process (where three alpha particles [42He] are involved). Beryllium-4 formation:  2 24He n 48Be Carbon-12 formation:  

4 2He

+ 48Be n 126C 20.4  Origin of the Elements: Nucleosynthesis

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1007

The process yields carbon-12. Carbon is the fourth most abundant element in the universe (after H, He, and O). At even higher temperatures in larger stars, heavier elements are formed. Oxygen-16 formation:  

12 6C

+ 24He n 168O

4 Oxygen-16 to silicon-28:   2168O n 28 14Si + 2He

Silicon-28 to magnesium-24:   Silicon-28 to sulfur-32:   Sulfur-32 to argon-36:  

28 14Si

28 14Si 32 16S

4 n 24 12Mg + 2He

+ 24He n 32 16S

+ 24He n 36 18Ar

Notice that in most of these processes, the end product is an element with an even atomic number, a result of the build-up of elements from helium nuclei (alpha particles). Some observations about the lighter elements and their abundances follow from an understanding of the origin of the elements. •

Among lighter elements, those having a mass number divisible by 4 are more abundant than their periodic neighbors: 16O, 20Ne, 24Mg, 28Si, 32S, 36Ar, and 40Ca.

•

Because the lighter elements are largely formed by combinations of helium nuclei, elements of even atomic number are more abundant than their neighbors of odd atomic number (Figure 20.5).

Among the mysteries of element formation is why lithium, beryllium, and boron should exist at all. They are by-passed in the processes described so far (the 8 4Be in the triple-alpha process has a very fleeting existence), but one theory is that they are formed by spallation or fragmentation of heavier nuclei after collisions with high energy protons or alpha particles. Figure 20.4 shows that iron (atomic number 26) is the most stable atomic nucleus. Beyond this, the stability and abundance of elements declines. How are these elements formed? It is believed that they are formed by accretion of neutrons in supernova. One example is the formation of cobalt-59. 56 26Fe

59 0 + 3 10n n 59 26Fe n 27Co + −1β

There is still much to learn about nucleosynthesis. But it is astonishing that out of a bewildering number of possible reactions, during the Big Bang and within stars and supernovae, that there are only 270 known stable isotopes. The most ­massive is 209 83Bi. What if this had not been the case? What if there were many more i­ sotopes of even the known elements? It may have been that precise quantitative chemistry would not be possible and the universe would have been very different.

20.5 Rates of Nuclear Decay Goals for Section 20.5 • Carry out calculations involving rates of radioactive decay using equations defining first-order kinetics for these processes.

• Understand the process by which carbon-14 is used to date artifacts.

Half-Life When a new radioactive isotope is identified, its half-life is usually determined. ­Half-life (t1/2) is used in nuclear chemistry in the same way it is used when d ­ iscussing the kinetics of first-order chemical reactions (Section 14.4): It is the time required for half of a sample to decay (Figure 20.6). Recall that for first-order kinetics, the half-life is independent of the amount of sample.

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Figure 20.6  Decay of 20.0 𝛍g of oxygen-15.  ​After each half-life period of 2.0 minutes, the mass of oxygen-15 decreases by one half. (Oxygen-15 decays by positron emission.)

Mass 15 8 O (μg)

20 15

10 5 0

0

2 First half-life

4 Second half-life

6 Third half-life

etc.

Time (minutes)

Half-lives for radioactive isotopes cover a wide range of values. Uranium-238 has one of the longer half-lives, 4.47 × 109 years, a length of time close to the age of the Earth (estimated at 4.5–4.6 × 109 years). Thus, roughly half of the uranium-238 present when the planet was formed is still around. At the other end of the range of half-lives are isotopes such as copernicium-227 (element 112), which has a half-life of 240 μs (240 × 10−6 s). Half-life provides one way to estimate the time required before a radioactive element is no longer a health hazard. Strontium-90, for example, is a β-emitter with a half-life of 29.1 years. Significant quantities of strontium-90 were dispersed into the environment during atmospheric nuclear bomb tests in the 1950s and 1960s. Based on its half-life, it is known that about one-fourth still remains. The health problems associated with strontium-90 arise because calcium and strontium have similar chemical properties. Strontium-90 is taken into the body and deposited in bone, taking the place of calcium. Radiation damage by strontium-90 in bone is directly linked to bone-related cancers.

Ex am p le 20.5

Using Half-Life Problem   Radioactive iodine-131, used to treat hyperthyroidism, has a half-life of 8.04 days. (a) If you have 8.8 μg (micrograms) of this isotope, what mass remains after 32.2 days? (b) How long will it take for a sample of iodine-131 to decay to one-eighth of its activity? (c) Estimate the length of time necessary for the sample to decay to 10% of its original activity.

What Do You Know?  The half-life of 131I, 8.04 days, is given. Strategy  In part (a), determine how many half-lives have elapsed and use this information to determine how much iodine remains. In part (b), use the fraction or percentage of iodine remaining to determine how many half-lives have elapsed. In part (c), estimate the number of half-lives that have elapsed based on the percentage of iodine remaining. Solution (a) The time elapsed, 32.2 days, is four half-lives (32.2/8.04 = 4). The amount of iodine-131 has decreased to one-sixteenth of the original amount [1/2 × 1/2 × 1/2 × 1/2 = (1/2)4 = 1/16]. The amount of iodine remaining is 8.8 μg × (1/2)4 or  0.55 μg.  (b) One-eighth equals one-half cubed. Thus, it will take  three half-lives  for the sample to decay to one-eighth of its activity. For iodine-131, this is 3(8.04 days) =  24.12 days. 

20.5  Rates of Nuclear Decay

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1009

(c) After three half-lives, one-eighth (12.5%) of the sample remains; after four half-lives, one-sixteenth (6.25%) remains. It will take between three and four half-lives,  between 24.12 and 32.2 days,  to decrease the amount of sample to 10% of its original value.

Think about Your Answer   You will often find it useful to make approximations as was done in (c). An exact time can be calculated from the first-order rate law (Equation 20.5).

Check Your Understanding  Tritium (13H), a radioactive isotope of hydrogen, has a half-life of 12.3 years. (a) Starting with 1.5 mg of this isotope, what mass (mg) remains after 49.2 years? (b) How long will it take for a sample of tritium to decay to one-eighth of its activity? (c) Estimate the length of time necessary for the sample to decay to 1% of its original activity.

Kinetics of Nuclear Decay The rate of nuclear decay is determined from measurements of the activity (A) of a sample. Activity refers to the number of disintegrations observed per unit of time, a quantity that can be measured readily with devices such as a Geiger–­Müller counter (Figure 20.7).  Activity is proportional to the number of ­radioactive atoms present (N). A∝N



(20.2)

If the number of radioactive nuclei N is reduced by half, the activity of the sample will be half as large. Doubling N will double the activity. This means that the rate of decomposition is first order with respect to N. Consequently, the equations describing rates of radioactive decay are the same as those used to describe the kinetics of first-order chemical reactions; the change in the number of radioactive atoms N per unit of time is proportional to N (Section 14.4) DN 5 2kN Dt



(20.3)

−

Thin window through which radiation enters

© Charles D. Winters/Cengage

+

Figure 20.7  A Geiger–Müller counter.  ​A charged particle (an α or β particle) or a gamma ray enters the gas-filled tube (left) and ionizes the gas. The gaseous ions migrate to electrically charged electrodes and are recorded as a pulse of electric current. The current is amplified and used to operate a counter (right).

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The integrated rate equation can be written in two ways depending on the data used:



N  ln   5 2kt  N0 

(20.4)

 A  ln   5 2kt  A0 

(20.5)

or



Here, N0 and A0 are the number of atoms and the initial activity of the sample, respectively, and N and A are the number of atoms and the activity of the sample after time t, respectively. Thus, N/N0 is the fraction of atoms remaining after a given time (t), and A/A0 is the fraction of the activity remaining after the same period. In these equations, k is the rate constant (decay constant) for the isotope in question. The relationship between half-life and the first-order rate constant is the same as seen with chemical kinetics (Equation 14.4): t1/2 5



0.693 k

(20.6)

Equations 20.3–20.6 are useful in several ways: •

If the activity (A) [which is proportional to the number of radioactive nuclei (N)] is measured in the laboratory over some period t, k can be calculated. The decay constant k can then be used to determine the half-life of the sample.

•

If k is known, the fraction of a radioactive sample (N/N0) still present after some time t has elapsed can be calculated.

•

If k is known, the time required for that isotope to decay to a fraction of the original activity (A/A0) can be calculated.

Ex am p le 20.6

Kinetics of Radioactive Decay Problem   A sample of radon-222 has an initial α-particle activity (A0) of 7.0 × 104 dps (disintegrations per second). After 6.6 days, its activity (A) is 2.1 × 104 dps. What is the halflife of radon-222?

What Do You Know?  You are given the initial and final activities of the sample of Rn and the time elapsed.

222

Strategy  Values for A, A0, and t are given. The problem can be solved using Equation 20.5 with k as the unknown. Once k is found, the half-life can be calculated using Equation 20.6. Solution ln(2.1 × 104 dps/7.0 × 104 dps) = −k (6.6 days) ln(0.300) = −k(6.6 days)

k = 0.182 days−1

From k the half-life is t1/2 = 0.693/0.182 days−1 =  3.8 days

Think about Your Answer  Notice that the activity decreased to between onehalf and one-fourth of its original value. The 6.6 days of elapsed time represents one full half-life and part of another half-life.

20.5  Rates of Nuclear Decay

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1011

Check Your Understanding (a) A sample of Ca3(PO4)2 containing phosphorus-32 has an activity of 3.35 × 103 dpm. Exactly 2 days later, the activity is 3.18 × 103 dpm. Calculate the half-life of phosphorus-32. (b) A highly radioactive sample of nuclear waste products with a half-life of 200. years is stored in an underground tank. How long will it take for the activity to diminish from an initial activity of 6.50 × 1012 dpm to a fairly harmless activity of 3.00 × 103 dpm?

Radiocarbon Dating In certain situations, the age of a material can be determined based on the rate of decay of a radioactive isotope. The best-known example of this procedure is the use of carbon-14 to date historical artifacts. Naturally occurring carbon is primarily carbon-12 and carbon-13 with isotopic abundances of 98.9% and 1.1%, respectively. In addition, traces of a third isotope, carbon-14, are present to the extent of about 1 in 1012 atoms in atmospheric CO2 and in living materials. Carbon-14 is a β emitter with a half-life of 5730 years. A 1-gram sample of carbon from living material will show about 14 disintegrations per minute, not a lot of radioactivity but nevertheless detectable. Carbon-14 is formed in the upper atmosphere by nuclear reactions initiated by neutrons in cosmic radiation: 14 7N

+ 01n n 146C + 11H

Once formed, carbon-14 is oxidized to 14CO2, and this circulates through the atmosphere, oceans, and biosphere. Plants absorb CO2 and convert it to organic compounds, thereby incorporating carbon-14 into living tissue. As long as a plant remains alive, this process will continue, and the percentage of carbon that is carbon-14 in the plant will equal the percentage in the atmosphere. When the plant dies, carbon-14 is no longer taken up. Radioactive decay continues, however, with the carbon-14 activity decreasing over time. After 5730 years, the activity will be 7 dpm/g; after 11,460 years, it will be 3.5 dpm/g; and so on. By measuring the activity of a sample, and knowing the h ­ alf-life of carbon‑14, it is possible to calculate when a plant (or a plant-eating a­ nimal) died. Using carbon-14 data to determine the age of an object is more complicated than simply measuring the rate of C-14 decay. Over thousands of years, the amounts of C-14 in the atmosphere have varied by as much as 10% due to changes in cosmic ray activity from the sun. Fortunately, these atmospheric changes have been measured by measuring C-14 concentrations in artifacts of known ages ­(Figure 20.8) BC AD

10 Percent change in 14C from nineteenth-century value

Figure 20.8  Variation of atmospheric carbon-14 activity.  ​ The amount of carbon-14 has changed with the variation in cosmic ray activity. The pre-1990 data for this graph were obtained using carbon-14 dating of tree rings.

5

0

−5

7000

6000

5000

4000

3000

2000

1000

0

1000

2000

Year of tree ring growth Source: Hans E. Suess, La Jolla Radiocarbon Laboratory

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Ex am p le 20.7

Radiochemical Dating Problem   To test the concept of carbon-14 dating, J. R. Arnold and W. F. Libby applied this technique to analyze samples of acacia and cyprus wood whose ages were already known. (The acacia wood, which was supplied by the Metropolitan Museum of Art in New York, came from the tomb of Zoser, the first Egyptian pharaoh to be entombed in a pyramid. The cyprus wood was from the tomb of Sneferu.) The average activity based on five determinations on one of these wood samples was 7.04 dpm per gram of carbon. Assume (as Arnold and Libby did) that the original activity of carbon-14, A0, was 12.6 dpm per gram of carbon. Calculate the approximate age of the sample.

Jean Louis Pradels/MaxPPP/ RODEZ AVEYRON France/Newscom

and corrections can be made. Another limitation is that it is not possible to use carbon-14 to date an object less than about 100 years old; the radiation level from carbon-14 will not change enough in this short time period to permit accurate detection of a difference from the initial value. In most instances, the accuracy of the measurement is only about ±100 years. Finally, it is not possible to determine ages of objects much older than about 60,000 years. By then, after more than 10 half-lives, the 14C radioactivity will have decreased virtually to zero. But for the span of time between 100 and 60,000 years, this technique has provided important information (Figure 20.9).

Figure 20.9  The Iceman.  ​ Europe’s oldest naturally preserved mummy was discovered in the ice of a glacier high in the Alps. Carbon-14 dating techniques allowed scientists to determine that he lived about 5300 years ago. See page 117 for more information on the Iceman.

What Do You Know?  You are given the initial and final 14C activity. The first-order rate constant can be calculated from the half-life.

Strategy   First, determine the rate constant for the decay of carbon-14 from its half-life (t1/2 for 14C is 5.73 × 103 years). Then, use Equation 20.5 to determine the age of the wood. Solution k = 0.693/t1/2 = 0.693/5730 yr = 1.209 × 10−4 yr−1 ln(A/A0) = −kt  7.04 dpm/g  ln  = (−1.209 × 10−4 yr−1)t  12.6 dpm/g   t = 4.81 × 103 yr  The wood is about 4800 years old.

Think about Your Answer  This problem uses real data from an early research paper in which the carbon-14 dating method was being tested. The age of the wood was known to be 4750 ± 250 years. (See J. R. Arnold and W. F. Libby, Science, 1949, 110, 678–680.)

Check Your Understanding  A sample of the inner part of a redwood tree felled in 1874 was shown to have 14C activity of 9.32 dpm/g. Calculate the approximate age of the tree when it was cut down. Compare this age with that obtained from tree ring data, which estimated that the tree began to grow in 979 ± 52 b.c. Use 13.4 dpm/g for the value of A0.



20.5  Rates of Nuclear Decay

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1013

© Lawrence Berkeley Laboratory

Glenn T. Seaborg (1912–1999)  Seaborg figured out that thorium and the elements that followed it fit under the lanthanides in the periodic table. For this insight, he and Edwin McMillan shared the 1951 Nobel Prize in Chemistry. Over a 21-year period, Seaborg and his colleagues synthesized 10 new transuranium elements. To honor Seaborg’s scientific contributions, the name “seaborgium” was assigned to element 106 in 1997. It marked the first time an element was named for a living person.

20.6 Artificial Nuclear Reactions Goals for Section 20.6 • Describe the experimental procedures used to carry out nuclear reactions. • Describe procedures used to prepare new transuranic elements. How many different isotopes are found on Earth? All of the stable isotopes occur naturally as well as a few unstable (radioactive) isotopes that have long half-lives; the best-known examples of the latter type are uranium-235, uranium-238, and thorium-232. Trace quantities of other radioactive isotopes with short half-lives are also present because they are being formed continuously by nuclear reactions. These include isotopes of radium, polonium, and radon, along with other elements produced in various radioactive decay series, and carbon-14, formed in a nuclear reaction initiated by cosmic radiation. Naturally occurring radioactive isotopes account for only a very small fraction of the currently known radioactive isotopes. The rest—several thousand— have been synthesized via artificial nuclear reactions, sometimes referred to as transmutations. The first artificial nuclear reaction was identified by Patrick Blackett (1897– 1974) in 1925 while a colleague of Ernest Rutherford at Cambridge University in England. Recall that Rutherford explained the experiment in which gold foil was bombarded with α particles by proposing the nuclear model of the atom, page 74. In the years following the gold foil experiment, Rutherford and colleagues bombarded many other elements with α particles. In 1919, Rutherford reported that one of these experiments led to an unexpected result: When nitrogen atoms were bombarded with α particles, protons were detected among the products. After much work, Blackett correctly concluded that a nuclear reaction had occurred and identified the products of the reaction in 1925. Nitrogen had undergone a transmutation to oxygen. 4 2He

+ 147N n 178O + 11H

Other nuclear reactions were discovered by bombarding other elements with α particles. Progress was slow, however, because in most cases α particles are simply scattered by target nuclei. The bombarding particles cannot get close enough to the nucleus to react because of the strong repulsive forces between the positively charged α particle and the positively charged atomic nucleus. Two advances were made in 1932 that greatly extended nuclear reaction chemistry. The first involved the use of particle accelerators to create high-energy particles as projectiles. The second was the use of neutrons as the bombarding particles. The α particles used in the early studies on nuclear reactions came from naturally radioactive materials such as uranium and had relatively low energies. Particles with higher energy were needed, so J. D. Cockcroft (1897–1967) and E. T. S. Walton (1903–1995), working in Rutherford’s laboratory in Cambridge, England, turned to protons. Protons are formed when hydrogen atoms ionize in a cathode-ray tube, and it was known that they could be accelerated to higher energy by applying a high voltage. Cockcroft and Walton found that when energetic protons struck a lithium target, helium-4 was the result. 7 3Li

+ 11p n 2 24He

This was the first example of a reaction initiated by a particle that had been artificially accelerated to high energy. Since this experiment was done, the technique has been developed much further, and particle accelerators in nuclear chemistry are now commonplace. Particle accelerators operate on the principle that a charged particle placed between charged plates will be accelerated to a high speed and high energy. A modern example of this process is the synthesis of the

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transuranium elements, as is described in more detail in “The Search for New ­Elements,” page 1016. Experiments using neutrons as bombarding particles were first carried out in the United States and Great Britain in 1932. Nitrogen, oxygen, fluorine, and neon were bombarded with energetic neutrons, and α particles were detected among the products. Using neutrons made sense: Because neutrons have no charge, it was reasoned that these particles would not be repelled by the positively charged nuclear particles and so did not need high energies to react. In 1934, Enrico Fermi (1901–1954) and his coworkers showed that nuclear reactions using neutrons are more favorable if the neutrons have low energy. A lowenergy neutron is simply captured by the nucleus, giving a product in which the mass number is increased by one unit. Because of the low energy of the bombarding particle, the product nucleus does not have sufficient energy to fragment in these reactions. The new nucleus is produced in an excited state, however, and when the nucleus returns to the ground state, a γ ray is emitted. Reactions in which a neutron is captured and a γ ray is emitted are called (n, 𝛄) reactions. The (n, γ) reactions are the source of many radioisotopes used in medicine and chemistry. An example is radioactive phosphorus, 3125P, which is used in chemical studies such as tracing the uptake of phosphorus in the body. 31 15P

Discovery of Neutrons Neutrons had been predicted to exist for more than a decade before they were identified in 1932 by James Chadwick (1891–1974) by the interaction of alpha particles with beryllium-9. See Example 20.8.

+ 01n n 1352P + γ

The products of (n, γ) reactions provide further access to new isotopes. Addition of a neutron to a nucleus raises the neutron-to-proton ratio (and often results in a change in the number of neutrons from even to odd) so that the resulting product is often unstable. With a higher n/p ratio the new isotopes are beta emitters, which form an isotope of the element with the next atomic number. An example of this process is the preparation of technetium-99m (“Applying Chemical Principles 20.2: Technetium-99m and Medical Imaging,” page 1028). Transuranium elements, elements with an atomic number greater than 92, were first made in a nuclear reaction sequence beginning with an (n, γ) reaction followed by beta decay. Scientists at the University of California at ­Berkeley bombarded uranium-238 with neutrons. Among the products identified were neptunium-239 and plutonium-239, which were formed when 239U decayed by β emission. 238 92U 239 92U

+ 01n n 23929U

n

239 93Np

239 93Np

+

Transuranium Elements in Nature ​

0 −1β

n 23949Pu + −10β

A similar reaction sequence was used to make americium-241. Plutonium-239 was found to add two neutrons to form plutonium-241, which decays by β emission to give americium-241.

Neptunium, plutonium, and americium were unknown prior to their preparation via nuclear reactions in the laboratory. Later, these elements were found in trace quantities within uranium ores.

Ex am p le 20.8

Nuclear Reactions Problem  Write equations for the nuclear reactions described below. (a) Fluorine-19 undergoes an (n, γ) reaction to give a radioactive product that decays by β emission. (Write equations for both nuclear reactions.) (b) When an atom of beryllium-9 (the only stable isotope of beryllium) reacts with an α particle emitted by a plutonium-239 atom, a neutron is ejected.

What Do You Know?   Reactants and one of two products are given for each reaction.

Strategy  The equations are written so that both mass and charge are balanced.

20.6  Artificial Nuclear Reactions

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1015

Solution (a)

19 9F

+ 01n n 209F + γ

(b)

239 94Pu



20 9F

n 1200Ne + −10β



4 9 2α + 4Be

n 23925U + 24α n 126C + 01n

Think about Your Answer   Both answers involve neutrons. The equation given for part (a), an (n, γ) reaction, illustrates a process that is easy to carry out. In a lab, the neutrons might be produced in a small device called a Pu-Be neutron source. The answer to part (b) describes the nuclear reactions in such a device.

Check Your Understanding  Technetium is one of the two elements with atomic numbers less than 83 for which there is no stable isotope. (Promethium, element 61, is the other.) Nevertheless, technetium is a very important element because of its extensive use in medical imaging (page 1028). It is produced in a two-step process. First, 98Mo undergoes an (n, γ ) ­reaction; then, the resulting unstable isotope decomposes to 99Tc. Write equations for these two reactions.

The Search for New Elements By 1936, guided first by Mendeleev’s predictions and later by atomic theory, chemists had identified all but two of the elements with atomic numbers between 1 and 92. These two gaps in the periodic table were filled when radioactive technetium and promethium were identified in 1937 and 1942, respectively. From this point onward, all new elements to be discovered came from artificial nuclear reactions. The first success in the search for elements with atomic numbers higher than 92 came with the 1940 discovery of neptunium and plutonium. Since 1950, laboratories in the United States (Lawrence Berkeley National ­Laboratory), Russia (Joint Institute for Nuclear Research at Dubna, near Moscow), and Europe (Institute for Heavy Ion Research at Darmstadt, Germany) have competed to make new elements. Syntheses of new transuranium elements use a standard methodology. An element of fairly high atomic number is bombarded with a beam of high-energy particles. Initially, neutrons were used. Later, helium nuclei, and then larger nuclei, such as 11B and 12C, were used. More recently, highly charged ions of elements such as calcium, chromium, cobalt, and zinc have been chosen. The bombarding particle fuses with the nucleus of the target atom, forming a new nucleus that lasts for a short time before decomposing. New elements are detected by their decomposition products, a signature of particles with specific masses and energies. By using bigger particles and higher energies, the list of known elements reached 106 by the end of the 1970s. To further extend the search, Russian scientists used a new idea: Precisely matching the energy of the bombarding particle with the energy required to fuse the nuclei. This technique enabled the synthesis of elements 107, 108, and 109 in Darmstadt in the early 1980s, and the synthesis of elements 110, 111, and 112 in the following decade. Lifetimes of these elements were in the millisecond range; copernicium-277, 217172Cn, for example, has a half-life of 240 μs. Yet another breakthrough was needed to extend the list further. Scientists have long known that isotopes with specific magic numbers of neutrons and protons are more stable. Elements with 2, 8, 20, 50, and 82 protons are members of this category, as are elements with 126 neutrons. The magic numbers correspond to filled shells in the nucleus. Their significance is analogous to the significance of filled shells for electronic structure. Theory had predicted that the next magic numbers would be 114 protons and 184 neutrons. Using this information, researchers discovered element 114 in early 1999. The Dubna group reporting this discovery found that the mass 289 isotope had an exceptionally long half-life, about 20 seconds. Element 114 was named flerovium, Fl, after the name of the laboratory in Dubna, Russia.

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Element 116 was also discovered in 1999. About 35 atoms of this element were obtained; the longest-lived isotope, with mass number 293, had a half-life of about 60 ms. As research progressed, element 113 was discovered in 2003 by workers in Russia and Japan, and several atoms of element 118 were discovered in 2006 by bombarding atoms of 249Cf with nuclei of 48Ca. Finally, the discovery of element 117 was announced in April 2010 by an international team in Dubna, Russia. They obtained six atoms of the element by bombarding 249Bk with 48Ca nuclei. The right to name new elements goes to the labs that discovered them. In December, 2016, the International Union of Pure and Applied Chemistry (IUPAC) accepted the names proposed for the remaining four unnamed elements. They are: Element 113, Nihonium, Nh, after the Japanese word for Japan Element 115, Moscovium, Mc, after Moscow Element 117, Tennessine, Ts, after Tennessee Element 118, Oganesson, Og, after Yuri Oganessian, a Russian scientist Element 117 was the second element to be named after a state (the first was Cf, Californium), and element 118 was the second element to be named after a living scientist. (The first was seaborgium, Sb, element 106, named for Glenn Seaborg.) Notice the endings for elements 117 and 118 match the endings on other elements in their representative groups in the periodic table (fluorine, chlorine, bromine, iodine, astatine, tennessine; and neon, argon, krypton, xenon, radon, oganesson). With this action all the elements from 1 to 118 have been assigned official names. An important question is how scientists can be sure of the identity of a new element. Element 113 (nihonium, Nh), for example, is produced by the collision of atoms of 209Bi and 70Zn. The process gives an atom of element 113, 278Nh, and a neutron (Figure 20.10). 209 83Bi

+ 3700Zn n

278 113Nh

+ 01n

The atom then decays to give six alpha particles and atoms of six well-known elements, the last of which is mendelevium. The scientists observed, for example, that lawrencium-258 decayed in the final step by emitting an alpha particle to give mendelevium-254. The observed half-life of this decay was 3.9 s, which corresponds to the known value for 254Md. Figure 20.10 Identifying Element 113, 278Nh.  Scientists at the Riken Nishina Center for Accelerator-Based Science in Japan first produced atoms of 278Nh in an accelerator. The atom decayed by emitting six alpha particles to finally give 254Md. Element 113 was confirmed when the measured half-lives in each of the last five steps agreed with known half-lives for these isotopes.

274Rg Roentgenium

Element 113 278Nh Nihonium 270Mt

Meitnerium #2

Alpha particle #1

254Md Mendelevium

#3 #6 266Bh Bohrium

#4 #5 258Lr

Lawrencium 262Db Dubnium



20.6  Artificial Nuclear Reactions

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1017

20.7 Nuclear Fission and Nuclear Fusion Goal for Section 20.7 • Describe nuclear chain reactions, nuclear fission, and nuclear fusion. In 1938, two chemists, Otto Hahn (1879–1968) and Fritz Strassman (1902–1980), isolated and identified barium in a sample of uranium that had been bombarded with neutrons. How was barium formed? The answer, the physicist Lise Meitner (1878–1968) realized, was that that the uranium nucleus had split into smaller pieces in a process now called nuclear fission. This was one of the most significant scientific discoveries of the twentieth century.

Nuclear Fission The details of nuclear fission were unraveled through the work of a number of scientists. They determined that a uranium-235 nucleus initially captured a neutron to form uranium-236. This isotope then broke apart, or fissioned, to produce two new nuclei, one with a mass around 140 and the other with a mass around 90, along with several neutrons (Figure 20.11). The nuclear reactions that led to formation of barium when a sample of 235U was bombarded with neutrons are 235 92U 236 92U

Chain Reactions  You might want to review chain reactions in the kinetics chapter (Section 14.7, page 715).

+ 01n n 23926U

n 14561Ba + 3962Kr + 3 01n

An important aspect of fission reactions is that they produce more neutrons than are used to initiate the process. Under the right circumstances, these neutrons then serve to continue the reaction. If one or more of these neutrons is captured by another 235U nucleus, a further reaction can occur, releasing still more neutrons. This sequence repeats over and over. Such a mechanism, in which each step generates a reactant to continue the reaction, is called a chain reaction. A nuclear fission chain reaction has three general steps: 1. Initiation. The reaction of a single atom is needed to start the chain. Following absorption of a neutron to form uranium-236, the reaction is initiated by decomposition of this atom. 2. Propagation. Absorption of neutrons and splitting of uranium atoms repeats over and over, with each step yielding more product. The fission of 236U releases neutrons that initiate the fission of other uranium atoms. 3. Termination. Eventually, the chain will end. Termination could occur if the reactant (235U) is consumed or if the neutrons that continue the chain escape from the sample without being captured by 235U. A nuclear reactor uses the energy generated by fission as heat (Figure 20.12). To harness the energy, it is necessary to control the rate at which a fission reaction occurs. This is managed by balancing the propagation and termination steps

Figure 20.11 Nuclear fission.  ​Neutron capture by 23 5 23 6 9 2U produces 9 2U. This isotope undergoes fission, which yields several fragments along with sev­eral neutrons. These neutrons initiate further nuclear reactions by adding to other 23 5 9 2U nuclei. The process is highly exothermic, producing about 2 × 1010 kJ/mol.

92 36 Kr

Neutron 2 × 1010 235 92 U

kJ mol

236 92 U

(Unstable nucleus) 141 56 Ba

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Figure 20.12  Schematic of a nuclear power plant.  ​The heat from the fission reaction is used to generate steam, which in turn powers a turbine.

Containment shell Steam generator

Electric generator

Steam turbine

Condenser Steam from turbine condenses on cooling coil

Fuel rods

Pump

Reactor Pump Pump

River or lake Hot water (350°C) under pressure

Cold water

Shipman, James T, Jerry D. Wilson, Charles A. Higgins, and Omar J. Torres. An Introduction to Physical Science, 14ed, 2016, Cengage Learning.

Steam Control rods

Warm water

to limit the number of neutrons available. This is accomplished by using cadmium rods to absorb neutrons in a nuclear reactor. By withdrawing or inserting the rods, the number of neutrons available to propagate the chain can be changed, and the rate of the fission reaction (and the rate of energy production) can be increased or decreased. Uranium-235 and plutonium-239 are the fissionable isotopes most commonly used in power reactors. Natural uranium contains only 0.72% of uranium-235; more than 99% of the natural element is uranium-238. The percentage of uranium-235 in natural uranium is too small to sustain a chain reaction, however, so the uranium used for nuclear fuel must be enriched in this isotope. One way to do so is by gaseous centrifugation (Figure 20.13).

Enriched UF6

Depleted UF6

UF6 feed Depleted UF6

Figure 20.13  Isotope separation by gas centrifuge.

UF6 gas is injected into the centrifuge from a tube passing down through the center of a tall, spinning cylinder. The heavier 238UF6 molecules experience more centrifugal force and move to the outer wall of the cylinder; the lighter 235UF6 molecules stay closer to the center. A temperature difference inside the rotor causes the 235UF molecules to move 6 to the top of the cylinder and the 238UF6 molecules to move to the bottom.

20.7  Nuclear Fission and Nuclear Fusion

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1019

ullstein bild/Getty Images

A Closer Look

Lise Meitner (1878–1968)

Lise Meitner made one of the fundamental discoveries of the twentieth century, nuclear fission. This led to a way to generate energy for nations around the world as well as weapons of unimaginable destruction. Element 109, meitnerium, was named in her honor. Meitner was born in 1878, the third child of eight in a gifted and liberal family in Vienna, Austria. Early in life, she had an interest in the natural world, but the education of girls ended at 14 in Austria. As her biographer said, the Viennese educational system allowed girls to learn “enough to run a household, raise children, and converse charmingly with a husband.” Women were not permitted to attend academic high schools in Austria until 1899, the year Meitner turned 21. During the years she should have been in high school, Meitner earned a certificate to teach French in a finishing school. Her father told her that, upon completion, he would hire a tutor to prepare her for university. She completed eight years of high school work in two years and was admitted to the University of Vienna in 1901 at the age of 23. One of Meitner’s professors at the university was Ludwig Boltzmann, an

enthusiastic lecturer who inspired her to study physics for the remainder of her life. In 1905, she was the second woman ever to receive a doctorate in physics in Vienna and began to study radioactivity. In 1907, Meitner decided to pursue further studies in physics, and she found a place with Max Planck in Berlin, where Einstein was also in residence. Planck did not approve of women students, but they later became friends, and she admired him greatly. Meitner wanted to do experimental work and met Otto Hahn, a young German chemist who was looking for a collaborator. However, the chemistry institute in Berlin where Hahn worked did not allow women in the building. Only when Ernest Rutherford, who was Hahn’s previous mentor, recommended the project did the institute allow Meitner to work there. Her workspace was an abandoned carpenter’s shop in the basement, and she had to use the toilet at a nearby hotel. Finally, in 1908, Berlin universities opened to women, and she moved upstairs. Her career truly began, and she stayed in Berlin until just before World War II. She and Hahn discovered element 91, protactinium, and she was head of the radiophysics department, which she ran with a firm hand. “She was a highly esteemed woman, and she wasn’t shy.” At age 48, she became Germany’s first female physics professor. The 1930s brought two events that changed her career and world history. In Italy, Enrico Fermi found interesting results when he bombarded uranium with neutrons. In Germany, Adolph Hitler came to power. Meitner was fascinated by Fermi’s results and asked Hahn to again collaborate with her on similar experiments. She also brought in a talented analytical chemist, Fritz Strassmann. In 1936, Strassmann found that repeating Fermi’s experiments also yielded barium. Irène Curie also found smaller elements when

repeating Fermi’s experiments. While Meitner and Curie both assumed the results were wrong, they continued the experiments. By July 1938, Meitner decided she could no longer remain in Berlin. Her grandparents were Jewish, which made her ineligible to retain a position at the university. She left behind her belongings and scientific papers, going first to Copenhagen to work with Niels Bohr and finally to Sweden. Though Meitner had left Berlin, Hahn and Strassmann continued their experiments and kept in touch with Meitner. She finally realized that barium must indeed have been a product, and in January 1939, Hahn and Strassmann published the results of their experiments. That same month, Otto Robert Frisch, Meitner’s nephew, visited her in Sweden. He was also a physicist, then working with Niels Bohr in Copenhagen. Meitner and Frisch discussed the barium result and on December 30 realized what it meant: that the uranium nucleus had come apart. They gave it the name “nuclear fission.” The world would never be the same. Meitner and Hahn were nominated many times for the Nobel Prize. In 1944, the Nobel committee voted to award the chemistry prize to Otto Hahn. Few chemists would argue with Hahn’s chemistry award, but physicists are nearly unanimous that Meitner should have been awarded the physics prize. Despite this significant insult, she and Hahn remained good friends. Lise Meitner died in England just before her ninetieth birthday. The physicist Peter Armbruster, a co-discoverer of elements 107 to 112, said that “She should be honored as the most significant woman scientist of [the twentieth] century.” For an excellent account of the life of Lise Meitner and other women ­scientists, see Sharon B. McGrayne, Nobel Prize Women in Science, New York: Carol Publishing Group, 1993.

Plutonium, which occurs naturally in trace quantities, must be made by a nuclear reaction. The raw material for this nuclear synthesis is the more abundant uranium isotope, 238U. Addition of a neutron to 238U gives 239U, which, as noted earlier, undergoes two β emissions to form 239Pu. In 2021, there were 55 commercial nuclear power plants with a total of 93 reactors operating in the United States with locations in 28 states. There are about 440

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nuclear reactors worldwide. About 20% of the electricity in the United States (and 10% of the world’s electricity) comes from nuclear power. Although one might imagine that nuclear energy would be called upon to meet the ever-increasing needs of society, there is reluctance to move in this direction. Among other things, serious reactor accidents (at Fukushima in Japan in 2011, Chernobyl in the Ukraine in 1986, and Three Mile Island in the United States in 1979) have sensitized the public to potential safety issues. The cost to construct a nuclear power plant (measured in terms of dollars per kilowatt-hour of power) is considerably more than the cost for a natural gas– powered facility, and there are severe regulatory restrictions. Disposal of highly radioactive nuclear waste is yet another thorny problem, with 20 metric tons of waste being generated per year at each reactor. In addition to technical problems, nuclear energy production brings with it significant geopolitical security concerns. The process for enriching uranium for use in a reactor is the same process used to generate weapons-grade uranium. Also, some nuclear reactors are designed to yield the isotope plutonium-239 as a by-product, which can be removed and used in a nuclear weapon. Despite these problems, nuclear fission is an important part of the energy profile in a number of countries. For example, about 70% of electrical production in France and 25% in South Korea is nuclear generated.

Nuclear Fusion In a nuclear fusion reaction, several small nuclei react to form a larger nucleus and generate tremendous amounts of energy. An example is the fusion of deuterium and tritium nuclei to form 42He and a neutron: 2 1H

+ 13H n 24He + 01n  ∆E = −1.7 × 109 kJ/mol

Other fusion reactions provide the energy of the sun and other stars (Section 20.4). Scientists have long dreamed of being able to harness fusion to provide power. To do so, a temperature of 106 to 107 K, like that in the interior of the sun, would be required to bring the positively charged nuclei together with enough energy to overcome nuclear repulsions. At the very high temperatures needed for a fusion reaction, matter does not exist as atoms or molecules; instead, matter is a plasma made up of unbound nuclei and electrons. Three critical requirements must be met before nuclear fusion could become a viable energy source. First, the temperature must be high enough for fusion to occur. Second, the plasma must be confined long enough to release a net output of energy. Third, the energy must be recovered in some usable form. Harnessing a nuclear fusion reaction for a peaceful use has been difficult to achieve, but in 2022 it was announced that the Joint European Torus tokomak reactor had fused tritium and deuterium and produced 59 megajoules of energy in a 5 s pulse. This came after several decades of research at this facility. Newer facilities have been built in Europe and the United States so improved results are anticipated.

20.8 Radiation Health and Safety Goal for Section 20.8 • Describe the units used to measure levels of radiation (curie, becquerel), the amount of energy absorbed by human tissue (rad, gray), and the damage to human tissue (rem, sievert).

Units for Measuring Radiation Scientists measure radiation in different ways. Sometimes they need to measure the amount of radiation a person absorbs, but at other times, they may be interested in the total amount of radioactivity in a sample of soil or water. There are four

20.8  Radiation Health and Safety

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1021

Names for Units of Radiation The roentgen (R) was named for Wilhelm Roentgen (1845–1923), who first produced and detected X radiation. Element 111 is named roentgenium in his honor. The curie is named for Marie (1867–1934) and Pierre (1859–1906) Curie (page 82), who are also honored by the name given element 96. The becquerel is named for Henri Becquerel (1852–1908), the gray is named for the English physicist Louis H. Gray (1905– 1965), and the sievert is named for a Swedish physicist, Rolf Sievert (1896–1966).

radioactivity measurement types, each with its own units of measurement, and different units are used internationally and in the United States. Radioactivity, or activity, is the amount of ionizing radiation released by a source. This can be alpha or beta particles, gamma rays or X-rays, or neutrons. The activity is related to the number of atoms in the measured substance that decay during a set time period. The international unit, or SI unit, of activity is the ­becquerel (Bq), which is equal to 1 disintegration per second (dps). Radiochemists in the United States use the curie (Ci) where 1 Ci is 3.7 × 1010 dps. Many types of radiation monitors measure exposure, the amount of radiation in the air. The international unit of exposure is coulombs/kg (C/kg). The roentgen (R) is the unit used in the United States, with 1 R equaling 0.000258 C/kg. The amount of radiation an object or person absorbs is the absorbed dose. The international unit is the gray (Gy), which denotes the absorption of 1 J per kilogram of tissue. The unit in the United States is the rad (radiation absorbed dose). One gray is equivalent to 100 rads. Different forms of radiation cause different amounts of biological damage. The amount of damage depends on how strongly a form of radiation interacts with matter. Alpha particles cannot penetrate the body any farther than the outer layer of skin. If α particles are emitted within the body, however, they will cause between 10 and 20 times the amount of damage done by γ rays, which can pass through a human body without being stopped. In determining the amount of biological damage to living tissue, differences in damaging power are accounted for with a quality factor. This quality factor has been set at 1 for β and γ radiation, 5 for low-energy protons and neutrons, and 20 for α particles or high-energy protons and neutrons. Biological damage is quantified in the United States by the rem (roentgen equivalent man). A dose of radiation in rem is determined by multiplying the energy absorbed in rads by the quality factor for that kind of radiation. Commonly, the effective dose of radiation is small, so biological damage is often given in terms of millirems where 1 millirem is one-thousandth of a rem. The SI equivalent of the rem is the sievert (Sv), determined by multiplying the dose in grays by the quality factor. One sievert is equal to 100 rems.

Radiation: Doses and Effects

Atomic Bomb Tests  The first atomic

bomb test (Trinity) occurred in New Mexico on July 16, 1945. Since then, well over 1300 nuclear test explosions were carried out in the atmosphere, underground, and underwater. Radioactive remnants from those explosions are with us today. North Korea continues to test nuclear weapons.

Exposure to a small amount of radiation is unavoidable. Earth is constantly bombarded with radioactive particles from outer space. There is also some exposure to radioactive elements that occur naturally on Earth, including 14C, 40K (a radioactive isotope that occurs naturally in 0.0117% abundance), 238U, and 232Th. Radioactive elements in the environment that were created artificially (in the fallout from nuclear bomb tests, for example) also contribute to this exposure. For some people, medical procedures using radioisotopes are a significant contributor. The average dose of background radioactivity to which a person in the United States is exposed is about 200 mrem per year (Table 20.2). Well over half of that amount comes from natural sources over which you have no control. Of the 60–70 mrem per year exposure that comes from artificial sources, nearly 90% is delivered in medical procedures such as X-ray examinations and radiation therapy. Less than 0.5% of the total annual background dose of radiation that the average person receives can be attributed to the nuclear power industry. Describing the biological effects of a dose of radiation precisely is not simple. The amount of damage done depends on the kind of radiation, the amount of energy absorbed, the tissues exposed, and the rate at which the dose builds up. Information is more accurate when dealing with single, large doses than it is for the effects of chronic, smaller doses of radiation. A great deal has been learned about the effects of radiation on the human body by studying the survivors of the bombs dropped over Japan in World War II and the workers exposed to radiation from the reactor disaster at Chernobyl. From studies of the health of these survivors, radiation scientists have learned that the effects of radiation are not generally observable

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Table 20.2

Radiation Exposure of an Individual for One Year from Natural and Artificial Sources

Millirem/Year

Percentage

Cosmic radiation

 50.0

25.8

The Earth

 47.0

24.2

Building materials

  3.0

 1.5

Inhaled from the air

  5.0

 2.6

Elements found naturally in human tissue

 21.0

10.8

Subtotal

126.0

64.9

Diagnostic X-rays

 50.0

25.8

Radiotherapy

 10.0

 5.2

Internal diagnosis

  1.0

 0.5

Subtotal

 61.0

31.5

   0.85

 0.4

Luminous watch dials, TV tubes

  2.0

 1.0

Fallout from nuclear tests

  4.0

 2.1

Subtotal

  6.9

 3.5

Total

193.9

99.9

Natural Sources

Medical Sources

Other Artificial Sources Nuclear power industry

below a single dose of 25 rem. At the other extreme, a single dose of >200 rem is fatal to about half the population exposed (Table 20.3).

20.9 Applications of Nuclear Chemistry Goal for Section 20.9 • Recognize some uses of radioactive isotopes in science and medicine. People tend to think about nuclear chemistry in terms of power plants and bombs. In truth, radioactive elements are now used in many areas of science and medicine, and they are of ever-increasing importance to our lives. Because describing all of Table 20.3 Dose (rem)

Effect

  0–25

No effect observed

 26–50

Small decrease in white blood cell count

 51–100

Significant decrease in white blood cell count, lesions

101–200

Loss of hair, nausea

201–500

Hemorrhaging, ulcers, death in 50% of population

500

Effects of a Single Dose of Radiation

Death 20.9  Applications of Nuclear Chemistry

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their uses would take several books, a few examples have been selected to illustrate the diversity of applications of radioactivity.

Nuclear Medicine: Medical Imaging

Courtesy of Brookhaven National Laboratory

Diagnostic procedures using nuclear chemistry are essential in medical imaging, the creation of images of specific parts of the body. There are three principal components to constructing a radioisotope-based image:

Dr. Joanna Fowler is a pioneer in the area of PET technology. She

was a recipient of the National Medal of Science in 2008 and spent her career at Brookhaven National Laboratory in New York. She was instrumental in developing radioactive tracers (based on 11C and 18F) to study cancer using positron emission tomography (PET). She also used these radioactive isotopes to examine the brain, particularly when affected by addictive substances such as cocaine.

•

A radioactive isotope, administered as the element or incorporated into a compound that concentrates the radioactive isotope in the tissue to be imaged

•

A method of detecting the type of radiation involved

•

A computer to assemble the information from the detector into a meaningful image

The choice of a radioisotope and the manner in which it is administered are determined by the tissue to be studied. A compound containing the isotope must be absorbed more by the target tissue than by the rest of the body. Table 20.4 lists radioisotopes that are commonly used in nuclear imaging processes, their half-lives, and the tissues they are used to image. All the isotopes in the table are γ emitters because γ radiation is preferred for imaging; it is less damaging to the body in small doses than either α or β radiation. Technetium-99m is used in more than 85% of the diagnostic scans done in hospitals each year (“Applying Chemical Principles 20.2: Technetium-99m and M ­ edical Imaging,” page 1028). The “m” stands for metastable, a term used to identify an excited state of the nucleus that exists for a finite period of time. Recall that atoms in excited electronic states emit visible, infrared, and ultraviolet radiation (Chapter 6). Similarly, a nucleus in an excited state gives up its excess energy, but in this case a much higher energy is involved, and the emission occurs as γ radiation. Another medical imaging technique based on nuclear chemistry is positron ­emission tomography (PET). In PET, an isotope that decays by positron emission is incorporated into a carrier compound and given to the patient. The emitted positron travels no more than a few millimeters before undergoing matter–antimatter annihilation. 0 +1β

+ −10e n 2γ

The two emitted γ rays travel in opposite directions. By determining where high numbers of γ rays are being emitted, one can construct a map showing where the positron emitter is located in the body. An isotope often used in PET is 15O. A patient is given gaseous O2 that contains 15 O. This isotope travels throughout the body in the bloodstream, allowing images of the brain and bloodstream to be obtained (Figure 20.14). Because positron emitters are typically very short-lived, PET facilities must be located near a cyclotron where the radioactive nuclei are prepared and then immediately incorporated into a carrier compound. Table 20.4

Radioisotopes Used in Medical Diagnostic Procedures

Radioisotope

Half-Life (h)

Imaging

Tc

 6.0

Thyroid, brain, kidneys

Tl

73.0

Heart

I

13.2

Thyroid

Ga

78.2

Various tumors and abscesses

F

 1.8

Brain, sites of metabolic activity

99m 201 123 67 18

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WDCN/Univ. College London/Science Source

Figure 20.14  PET scans of the brain.  These scans show the left side of the brain; red

indicates an area of highest activity. (upper left) Sight activates the visual area in the occipital cortex at the back of the brain. (upper right) Hearing activates the auditory area in the superior temporal cortex of the brain. (lower left) Speaking activates the speech centers in the insula and motor cortex. (lower right) Thinking about verbs, and speaking them, generates high activity, including in the hearing, speaking, temporal, and parietal areas.

Nuclear Medicine: Radiation Therapy

59 27Co

60 0 + 10n n 60 27Co n 28Ni + −1β + γ

A beam of gamma rays from a 60Co source external to the body is focused on the cancerous growth. While successful, 60Co has largely been replaced by linear accelerators that produce a beam of high energy radiation. Instead of focusing a beam of gamma rays on a cancerous growth, another technique places the radioactive source in your body, within or near the tumor. This technique is called brachytherapy, and the source is usually contained in seeds, a ribbon, or capsules. Iodine-131 is commonly used in brachytherapy to treat ­thyroid cancer as well as non-malignant thyroid disorders, and iridium-192 implants are used to treat head and breast cancers. Two other isotopes—iodine-125 and ­palladium-103—are used in permanently implanted seeds.

Carbon-14 as a Tracer Carbon-14, a beta emitter, can be incorporated into an organic compound by standard synthetic methods. The fate of a drug or environmental contaminant in a natural system can then be analyzed by searching for levels of 14C in the system that are higher than naturally occurring 14C. While the presence of this carbon isotope can be assessed by its beta emission, its presence and amount can alternatively be determined with a mass spectrometric analysis of a metabolic product. One example of using 14C as a radioactive tracer involves triclocarban (TCC, C13H9Cl3N2O), an antimicrobial found in personal care products such as soap and shampoo and one of the most commonly detected contaminants in wastewater. The suspicion that it could affect a developing embryo and fetus was investigated by administering 14C-labeled TCC to pregnant mice through their drinking water. The investigators found that the TCC readily transferred to the offspring through breast milk and that offspring were heavier in weight than those not exposed to TCC. In general, they found that TCC may interfere with lipid metabolism early in life, with implications for human health. The concentration of 14C in the atmosphere has also been investigated. Figure 20.8 shows the change of 14C levels in the atmosphere over many centuries. For the last 1000 years or so it was relatively steady, but it recently increased significantly. This spike in 14C is generally referred to as the radiocarbon bomb pulse. The amount of 14C in the atmosphere nearly doubled between 1955 and 1963 because of atmospheric atomic bomb tests by the United States, the Soviet Union, and the

© Thomson G (2020) Food Irradiation: A Way of Food Preservation an Image Article . J Food Process Technol. 11:850.

Cancerous growths are relatively sensitive to irradiation and so can be controlled or eliminated by irradiating the area of the growth. Physicians realized this should be possible decades ago and used the best form of ionizing radiation then available, X-rays. They also realized that radium could be used, but only a few grams of this element were available worldwide and at an enormous price. When nuclear reactors were built during World War II, other radioactive isotopes, especially cobalt-60, became available. Cobalt-60 has since been used widely in medical radiotherapy as well as for sterilizing medical equipment and food irradiation, among other applications. The isotope does not occur naturally but is made when cobalt-59 undergoes neutron capture. The resulting cobalt-60 has a half-life of 5.27 years and decays by beta and gamma emission to nickel-60.

Irradiation of food. Strawberries were harvested and some were irradiated (left) while others were not (right). Irradiation damages the DNA of pathogenic organisms and slows deterioration of the food. It has the added advantage of being done without heating the food. Making a 14C Tracer. Carbon-14 to be used as a tracer is usually produced in a nuclear reactor by bombarding a nitrogen-14containing compound such as aluminum nitride, AlN. 14 7N

+ 10n n

14 6C

+ 11H

The carbon-14 is then typically incorporated into, and sold as, barium carbonate, Ba14CO3. This can be decomposed to 14 CO2, which is then used to synthesize an organic molecule.

20.9  Applications of Nuclear Chemistry

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O H HO OCH3 Vanillin.  The compound is obtained from natural sources or synthesized. The amount of 14C in the molecule depends on the source.

United Kingdom. However, 14C levels declined after 1963 when the tests were halted by the Limited Test Ban Treaty. Knowledge of the bomb pulse has been useful. Chemicals produced from petroleum contain extremely low levels of 14C because the carbon in petroleum comes from plants that died millennia ago. In contrast, chemicals, vitamins, or food additives from contemporary sources reflect the elevated amount of 14C in the atmosphere in which they grew. One example is vanilla. The taste of vanilla comes from the compound vanillin, C8H8O3. A 2011 study found that vanilla extracted from fresh vanilla beans had a 14C content that reflected contemporary atmospheric levels of CO2. In contrast, an imitation vanilla, made from petroleum products, had a much lower 14C content.

Other Radioactive Tracers Other radioactive isotopes can help determine the fate of compounds in the body or environment. In biology, for example, scientists can use radioactive isotopes to measure the uptake of nutrients. Plants take up phosphorus-containing compounds from the soil through their roots. By adding a small amount of radioactive 32P, a β emitter with a half-life of 14.3 days, to fertilizer then measuring the rate at which the radioactivity appears in the leaves, plant biologists can determine the rate at which phosphorus is taken up. The outcome can assist scientists in identifying hybrid plant strains that can absorb phosphorus quickly, resulting in faster-­ maturing crops, better yields per acre, and more food or fiber at less expense. To measure pesticide levels, a pesticide can be tagged using a radioisotope with an appropriate half-life and then applied to a test field. By counting the disintegrations of the radioactive tracer, information can be obtained about how much pesticide accumulates in the soil, how much is taken up by the plant, and how much is carried off in runoff surface water. After these tests are completed, the radioactive isotope decays to harmless levels in a few days or weeks because of the short halflives of the isotopes used.

Analytical Methods: Isotope Dilution Imagine, for the moment, that you want to estimate the volume of blood in a live animal subject. How might you do this? Draining the blood and measuring its volume in volumetric glassware is not a desirable option. One technique uses a method called isotope dilution. In this process, a small amount of radioactive isotope is injected into the bloodstream. After allowing the isotope time to become distributed throughout the body, a blood sample is taken and its radioactivity measured. The calculation used to determine the total blood volume is illustrated in the next example.

E xamp le 20.9

Analysis Using Radioisotopes Problem   A 1.00-mL solution containing 0.240 μCi of tritium is injected into a dog’s bloodstream. After allowing the isotope to disperse, a 1.00-mL sample of blood is drawn. The radioactivity of the sample is found to be 4.3 × 10−4 μCi/mL. What is the total volume of blood in the dog? What Do You Know?  You know the concentration (activity) and volume of a concentrated solution and you then measure the concentration of the dilute solution. The unknown is the volume of the dilute solution. Strategy  In this problem, the activity of the sample (in Ci) is related to the amount of the radioisotope present. The total amount of solute is 0.240 μCi, and the concentration (measured on the small sample of blood) is 4.3 × 10−4 μCi/mL. The unknown is the total volume of blood, V.

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Solution   The blood contains a total of 0.240 μCi of radioactive material. Exactly 1.00  mL containing this amount was injected. After dilution in the bloodstream, 1.00 mL of blood, representative of the total volume V, is found to have an activity of 4.3 × 10−4 μCi/mL. The activity in the original sample is the same as that in the total diluted volume. In both cases, the amounts are calculated by multiplying the concentration by the volume. (0.240 μCi/mL)(1.00 mL) = (4.3 × 10−4 μCi/mL)(V)  V = 560 mL 

Think about Your Answer   This is solved as a classic dilution problem, where Cconc × Vconc = Cdil × Vdil.

Check Your Understanding  To measure the solubility for PbCrO4 you mix a small amount of a lead(II) salt having radioactive 212Pb with a lead salt containing 0.0100 g of lead. To this mixture you add enough K2CrO4 to completely precipitate the lead(II) ions as PbCrO4. The supernatant solution still contains a trace of lead, and when you evaporate 10.00 mL of this solution to solid PbCrO4, you find a radioactivity that is 4.17 × 10−5 of what it is for the pure 212Pb salt. Calculate the solubility of PbCrO4 in mol/L. (Adapted from C. E. Housecraft and A. G. Sharpe, Inorganic Chemistry, 3rd ed., p. 84, Pearson, 2008.)

Applying Chemical Principles

In 1972, a French scientist noticed that the uranium taken from a mine in Oklo, Gabon, West Africa, was strangely deficient in 235U. Uranium exists in nature as two principal isotopes, 238 U (99.274% abundant) and 235U (0.720% abundant). 235U most readily undergoes nuclear fission and is used to fuel nuclear power plants around the world. But the 235U found in the Oklo mines was less abundant than expected, only 0.717% abundant. Based on this tiny discrepancy, and other evidence, scientists concluded that a natural fission process occurred in the bed of uranium ore nearly 2 billion years ago. But intriguing questions remained: Why could fission occur to a significant extent in this natural deposit of uranium and why didn’t the impromptu reactor explode? Apparently, there must have been a moderator of neutron energy and a regulation mechanism. In a modern nuclear power reactor, control rods slow down the ­neutrons from nuclear fission so that they can induce fission in other 235U nuclei. Without a moderator, the neutrons just fly off. The reason the Oklo reactor did not explode is that water could have been the moderator. As the fission process heated the water, it boiled off as steam. This caused the fission to stop, but it began again when more water seeped in. Scientists now believe that this natural reactor would activate for about 30 minutes and then stop for several hours before activating again. There is evidence this natural reactor functioned intermittently for about 1 million years, until the concentration of uranium isotopes was too low to keep the reaction going.

Courtesy of Francois Gauthier-Lafaye

20.1  A Primordial Nuclear Reactor

The natural nuclear reactor in Oklo, Gabon (West Africa).  Nearly 2 billion years ago, a natural formation containing uranium oxide (the yellow material) underwent fission that started and stopped over a period of a million years.

Questions

1. There are two major isotopes of uranium, 235U (half-life 7.038 × 10 8 yr, 0.720%) and 238U (half-life 4.468 × 10 9  yr, 99.274%). What was the relative abundance of 235 U two billion years (2.0 × 109 years) ago? Applying Chemical Principles

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1027

2. Rates of decomposition can be measured based on decompositions per minute. What is the ratio of the rates of decomposition of the two major uranium isotopes (rate for 238U / rate for 235U)? 3. Uranium-235 is an alpha emitter. Write an equation for the nuclear decay process.

4. Calculate the molar mass (atomic weight) of naturally occurring uranium from the masses of the two major isotopes of uranium (235U = 235.0439; 238U = 238.0508) and their natural abundances.

20.2  Technetium-99m and Medical Imaging alumina, Al 2O 3. The MoO 42− ion is continually converted into the pertechnetate ion, 99mTcO4− by β emission. When it is needed, the 99mTcO4− is washed from the column using a saline solution. Technetium-99m may be used directly as the pertechnetate ion (as NaTcO 4) or converted into other compounds. The pertechnetate ion or radiopharmaceuticals made from it are administered intravenously to the patient. Such small quantities are needed that 1 μg (microgram) of technetium-99m is sufficient for the average hospital’s daily imaging needs. One use of 99mTc is for imaging the thyroid gland. Because − I (aq) and TcO4−(aq) ions have very similar sizes, the thyroid will (mistakenly) take up TcO4−(aq) along with iodide ion. This uptake concentrates 99mTc in the thyroid and allows a physician to obtain images such as the one shown here.

Questions

1. Write an equation for the β decay of 99Mo to 99Tc. 2. What is the oxidation number of Tc in the pertechnetate ion? What is the electron configuration of a Tcn+ ion with a charge equal to this oxidation number? Should the TcO4− ion be paramagnetic or diamagnetic? 3. What amount of Na99mTcO4 is there in 1.0 μg of the salt? What mass of 99mTc? 4. If you have 1.0 μg of 99mTc, what mass remains at the end of 24 hours? 5. 99Tc decays to 99Ru. What particle is produced in this decay? 6. Speculate on the reason the TcO4− ion is held less strongly to a column of Al2O3 than in the MoO42− ion.

Photos: CNRI/Science Photo Library/ Science Source

Technetium was the first element to be made artificially. One might think that this would make the element a chemical rarity, but it is not. Technetium is the product of nuclear reactions that can be done in a laboratory. As a consequence, it is readily available and even inexpensive (about $60 per gram). This has also led to widespread use in medical diagnosis, and it is now used all over the world in evaluations of the thyroid gland, the heart, kidneys, and lungs. Technetium-99m is formed when molybdenum-99 decays by β emission. Technetium-99m (99mTc) then decays to its ground state (forming 99Tc) with a half-life of 6.01 hours, giving off a 140-KeV γ ray in the process. (Technetium-99 is also radioactive, decaying to stable 99Ru with a half-life of 2.1 × 105 years.) Molybdenum-99 is not a component of naturally occurring molybdenum. It is made in an (n, γ ) reaction from molybdenum-98 (23.8% abundant in naturally occurring samples of the element). Nuclear power reactors utilizing U-235 provide the source of neutrons for this synthesis. But there was a problem with this route beginning about 2010 when the main reactors producing the molybdenum (in Canada and The Netherlands) were closed for repairs. Because technetium is so important worldwide for medical procedures, scientists immediately began looking for new sources. Technetium-99m is produced in hospitals using a molybdenum–technetium generator. Shielded in lead, the generator contains the isotope 99 Mo in the form of the molybdate ion, MoO42−, adsorbed on a column of

(a) Healthy human thyroid gland.

(b) Thyroid gland showing effect of hyperthyroidism.

Thyroid imaging with technetium-99m.  The radioactive isotope 99mTc concentrates in sites of high activity. Images of this gland, which is located at the base of the neck, were obtained by recording γ-ray emission after the patient was given radioactive technetium-99m. Current ­technology creates a computer color-enhanced scan.

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20.3  The Age of Meteorites Meteorites have a significant value for collectors, but they are also of scientific interest. Meteorites are generally regarded as some of the oldest materials in the solar system. But how is the age of a meteorite determined? The age refers to the time since the meteor condensed to a solid, locking the components into a solid matrix. There are a number of methods to measure the age of a meteor, all involving the decay of longlived radioactive elements. One dating procedure involves vaporizing samples of the material and measuring the amounts of 86Sr, 87Sr, and 87Rb by mass spectrometry. Both 86Sr and 87Sr are stable isotopes.

The rubidium isotope, which decays to 87Sr, has a half-life of 4.965 × 1010 years. The amount of 87Sr provides a measure of the amount of 87Sr initially present (87Sr0) when the meteorite was formed plus the amount formed by decay of 87 Rb. The 86Sr remains constant and is used as a check on the original amounts of the other isotopes. The three measured amounts are related by an equation (which can be derived from the first-order rate law): [87Sr/86Sr] = (ekt − 1) [87Rb/86Sr] + [87Sr0/86Sr] The equation has the form of the equation of a straight line, y = mx + b. Multiple samples are analyzed from each meteorite. The amounts of each isotope may vary between samples, but a graph of [87Sr/86Sr] versus [87Rb/86Sr] will give a straight line with a slope m = (ekt − 1) and a y-intercept of [87Sr0/86Sr]. The symbol k refers to the rate constant for the radioactive decomposition. This method is called isochron dating.

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Questions

Meteorites.  Meteorites are materials that originate in outer space and survive the fall to the Earth’s surface. They can be large or small, but as they enter Earth’s atmosphere they are warmed to a high temperature by friction in the atmosphere and emit light, forming a fireball. There are three general categories: stony meteorites, which are silicate minerals; iron meteorites, largely composed of iron and nickel; and stony-iron meteorites. (See the photo of an iron meteorite, Figure 7.7.)



1. Write a balanced equation for the radioactive decomposition of 87Rb. 2. The decomposition of 87Rb occurs by which of the following processes? (a) alpha emission (b) beta emission (c) gamma emission (d) electron capture (e) positron emission 3. Determine the rate constant for the radioactive decay of 87Rb. 4. The oldest dated meteorites have ages of about 4.5 billion years. What fraction of the initial 87Rb has decayed? 5. The relative abundances of 86Sr, 87Sr, and 87Rb were measured for four samples of a meteorite; the results are tabulated below. Make a strontium-rubidium isochron plot of these data, then use the slope of the line to determine the age of the meteorite.

Relative Abundance Sample #

86

Sr

87

Sr

87

Rb

1

1.000

0.819

0.839

2

1.063

0.855

0.506

3

0.950

0.824

1.929

4

1.011

0.809

0.379

6. Derive the equation given in the text for [87Sr/86Sr]. (Hint: Start with the rate equation in the form [87Rb0] = [87Rb] ekt.)

Applying Chemical Principles

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Think–Pair–Share 1. Lead-210 is part of the uranium decay series. In the final three steps of the series 210Pb decays by two beta emissions, followed by one alpha emission. (a) Write a nuclear equation for each of these steps. (b) What is the final product of this decay series? 2. Uranium-234 is an isotope of uranium found in the u ­ ranium-238 decay chain. As part of that series, 234U decays by a series of five alpha particle emissions. (a) What element is the result of the first step in this series? (b) What is the final product of the five steps? 3. Samarium-153 is used to treat severe pain associated with bone cancer. It is a beta-emitter with a half-life of 1.9 days. (a) What is the decay product of 153Sm? (b) If you begin with 3 mg of the isotope, approximately how much remains after 1 week? 4. Place the following isotopes in order of their increasing binding energy per nucleon: beryllium-9; gold-197; cobalt-59. Explain your reasoning. 5. The carbon-14 dating method only works for objects up to about 70,000 years old, so other methods have been developed for older objects. For example, the decay of 40K to 40Ar has been used to date objects from 1000 years old to a billion years old. The half-life of 40K is 1.25 × 10 9 years, and its natural abundance is 0.0117%. [The other isotopes of potassium are stable: 39K (93.26%) and 41K (6.730%).]

(a) Write a nuclear equation for the decay of 40K to 40Ar. (b) The K-Ar dating method is valuable for measuring the age of igneous rocks, which are formed in a volcanic eruption (like the one on the cover of this text). Explain how the age of a volcanic rock (and thus, perhaps, a fossil buried in that rock) could be estimated by measuring the relative amounts of 40K and 40Ar in the rock. 6. Carbon-14 activity is widely used to estimate the age of objects. Charcoal from an ancient campfire was found to have an activity of 6.22 dpm/g (where dpm = disintegrations per minute). In contrast, modern day carbon has an activity of 13.56 dpm/g. Calculate the approximate age of the wood (k = 1.209 × 10−4 yr−1). 7. Thorium is found in the environment as a product of radioactive decay: 230Th occurs in the 238U decay chain and decays to 226Ra. Its half-life is 75,380 yr. In contrast, 232Th is in the 232 Th-206Pb decay chain. It decays to 228Ra with a half-life of 1.40 × 1010 yr. Geochemists assume both thorium isotopes behave chemically the same way in the environment and that the initial ratio of the isotopes was constant in the environment for the last several hundred-thousand years. However, once trapped in marine sediments, this is no longer the case. (a) Write a nuclear equation for the decay of both thorium ­isotopes to the appropriate radium isotope. (b) Explain how the thorium isotopes can be used to estimate the age of a sediment layer.

Chapter Goals Revisited The goals for this chapter are keyed to specific Study Questions to help you organize your review.

20.1  Natural Radioactivity • Identify α, β, and γ radiation, the three major types of radiation in natural radioactive decay. 1, 2.

20.2  Nuclear Reactions and Radioactive Decay • Write balanced equations for nuclear reactions (alpha, beta, gamma, and positron emission; electron capture; and fission). 11,14, 19, 21.

• Recognize that radioactive isotopes decay by a series of nuclear

reactions, called a radioactive decay series, eventually ending with a stable isotope. 9, 10, 23, 24.

20.3  Stability of Atomic Nuclei • Assess nuclear stability based on the numbers of neutrons and protons in a nucleus. 3, 4.

• Predict possible modes of decay of unstable nuclei based on n/p ratios. 25, 27. • Calculate the mass defect for an isotope and from this determine the

nuclear binding energy Eb and the binding energy per nucleon. 31, 33.

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20.4  The Origin of the Elements: Nucleosynthesis • Understand the steps that can lead to the synthesis of some elements. 37, 38.

20.5  Rates of Nuclear Decay • Carry out calculations involving rates of radioactive decay using equations defining first order kinetics for these processes. 39, 41, 43.

• Understand the process by which carbon-14 is used to date artifacts. 7, 8, 65.

20.6  Artificial Nuclear Reactions • Describe the experimental procedures used to carry out nuclear reactions. 5, 6.

• Describe procedures used to prepare new transuranic elements. 49, 50.

20.7  Nuclear Fission and Nuclear Fusion • Describe nuclear chain reactions, nuclear fission, and nuclear fusion. 11–14.

20.8  Radiation Health and Safety • Describe the units used to measure levels of radiation (curie, becquerel), the amount of energy absorbed by human tissue (rad, gray), and the damage to human tissue (rem, sievert). 15.

20.9  Applications of Nuclear Chemistry • Recognize some uses of radioactive isotopes in science and medicine. 69, 70, 73.

Key Equations Equation 20.1 (page 1004) The equation relating interconversion of mass (m) and energy (E). This equation is applied in the calculation of binding energy (Eb) for nuclei. Eb = (∆m)c2

Equation 20.2 (page 1010) The activity of a radioactive sample (A) is proportional to the number of radioactive atoms (N). A~N

Equation 20.3 (page 1010) The change in the number of radioactive elements with time is equal to the product of the rate constant (k, decay constant) and number of atoms present (N). ∆N/∆t = −kN

Equation 20.4 (page 1011) The rate law for nuclear decay based on number of radioactive atoms initially present (N0) and the number N after time t. ln(N/N0) = −kt

Key Equations

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1031

Equation 20.5 (page 1011) The rate law for nuclear decay based on the measured activity of a sample (A). ln(A/A0) = −kt

Equation 20.6 (page 1011) The relationship between the half-life and the rate constant for a nuclear decay process. t1/2 = 0.693/k

Study Questions denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully worked solutions in the Student Solutions Manual. ▲

Practicing Skills Important Concepts 1. Rank the three types of natural radiation (α, β, γ): (a) In order of increasing mass (b) In order of increasing penetrating power 2. What information was used to identify α and β particles? 3. A graph of binding energy per nucleon is shown in Figure 20.4. Explain how the data used to construct this graph were obtained. 4. How is Figure 20.3 used to predict the type of decomposition for unstable (radioactive) isotopes? 5. Outline how the synthesis of new isotopes is carried out in the laboratory. Why are neutrons so effective when used as a bombarding particle? 6. Cobalt-60, used as a source of high-energy gamma radiation in medical procedures, is made in a nuclear reactor by neutron irradiation of cobalt-59. Write an equation for this reaction. 7. Describe the process that uses carbon-14 for ­estimating the ages of archeological artifacts. 8. Explain why carbon-14 dating is limited to the range of 100 to about 60,000 years. 9. How many alpha and beta emissions occur in the uranium-238 radioactive decay series ending with lead-206? See Figure 20.2. 10. The uranium-235 radioactive decay series includes 7 alpha emissions and 4 beta emissions. Identify the stable isotope at the end of this series. 11. Chain reactions occur in a series of steps described as initiation, propagation, and termination. Describe each of these steps for the fission of uranium-235.

12. The fission of uranium-235 releases 2 × 1010 kJ/mol. Calculate the quantity of mass converted to energy in this process. 13. In a nuclear reactor, what is a moderator and what is its function? 14. Identify the other element generated in the reaction 235 U + 1n n 141Ba + 2 1n + ? 15. Define each of the following units:  (a) curie  (b) rad 16. The interaction of radiation with matter has both positive and negative consequences. Discuss briefly the hazards of radiation and the way that radiation can be used in medicine. 17. Oxygen-15 is used in the medical procedure called positron emission tomography. This isotope is prepared on a cyclotron from irradiation of nitrogen-14 with deuterium nuclei. It decays to nitrogen-15 by positron emission, and the positrons are annihilated when they collide with electrons, forming two gamma rays. Write equations for the three reactions described here. 18. Write equations that describe the preparation of technetium-99m from molybdenum-98.

Nuclear Reactions (See Section 20.2 and Examples 20.1 and 20.2.) 19. Complete the following nuclear equations. Write the mass number and atomic number for the remaining particle, as well as its symbol. 4 (a) 54 → 2 10n 1 ? 26 Fe 1 2 He   27 4 30 (b) 13 Al 1 2He  → 15 P1? 32 1 1 (c) 16 S 1 0 n  → 1H 1 ? 96 2 (d) 42 Mo 1 1 H  → 10 n 1 ? 1 (e) 98 → 99 42 Mo 1 0 n   43 Tc 1 ? 18 18 (f) 9 F  → 8O 1 ?

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20. Complete the following nuclear equations. Write the mass number, atomic number, and symbol for the remaining particle. (a) 94 Be 1 ?  → 63Li 1 42 He 24 (b) ? 1 10n  → 11 Na 1 42 He 40 (c) 40 → 19 K 1 11H 20 Ca 1 ?   241 4 (d) 95 Am 1 2He  → 243 97 Bk 1 ? 246 12 (e) 96 Cm 1 6 C  → 4 10 n 1 ? 249 (f) 238 → 100 Fm 1 5 10n 92 U 1 ?   21. Complete the following nuclear equations. Write the mass number, atomic number, and symbol for the remaining particle. (a) 111 → 111 47 Ag   48 Cd 1 ? 87 0 (b) 36 Kr  → 21b + ? (c) 231 Pa   → 227 91 89 Ac 1 ? 230 4 (d) 90 Th  → 2He 1 ? 82 (e) 35 Br  → 82 36Kr 1 ? 24 (f) ?  → 12 Mg 1 201b 22. Complete the following nuclear equations. Write the mass number, atomic number, and symbol for the remaining particle. (a) 19 → 101b + ? 10 Ne   (b) 59 → 201b + ? 26 Fe   40 (c) 19 K  → 201b + ? 37 (d) 18 Ar 1 201e (electron capture)  →? 55 0 (e) 26 Fe 1 21e (electron capture)  →? 26 25 (f) 13 Al  → 12Mg 1 ? 23. The uranium-235 radioactive decay series, beginning with 23952U and ending with 20872Pb, occurs in the following sequence: α, β, α, β, α, α, α, α, β, β, α. Write an equation for each step in this series. 24. The thorium-232 radioactive decay series, beginning with 23920Th and ending with 20882Pb, occurs in the following sequence: α, β, β, α, α, α, α, β, β, α. Write an equation for each step in this series.

Nuclear Stability and Nuclear Decay (See Section 20.3 and Examples 20.3 and 20.4.) 25. What particle is emitted in the following nuclear reactions? Write an equation for each reaction. (a) Gold-198 decays to mercury-198. (b) Radon-222 decays to polonium-218. (c) Cesium-137 decays to barium-137. (d) Indium-110 decays to cadmium-110. 26. What is the product of the following nuclear decay processes? Write an equation for each process. (a) Gallium-67 decays by electron capture. (b) Potassium-38 decays with positron emission.

(c) Technetium-99m decays with γ emission. (d) Manganese-56 decays by β emission. 27. Predict the probable mode of decay for each of the following radioactive isotopes, and write an equation to show the products of decay. (a) bromine-80 (c) cobalt-61 (b) californium-240 (d) carbon-11 28. Predict the probable mode of decay for each of the following radioactive isotopes, and write an equation to show the products of decay. (a) manganese-54 (c) silver-110 (b) americium-241 (d) mercury-197m 29. (a) Which of the following nuclei decay by 0 −1β decay? 3

H 16O 20F 13N (b) Which of the following nuclei decay by 0 +1β decay? 238

U

19

F

22

24

Na

Na

30. (a) Which of the following nuclei decay by 0 −1β decay? 1

H 23Mg 32P 20Ne (b) Which of the following nuclei decay by 0 +1β decay? 235

U

35

Cl

38

24

K

Na

31. Boron has two stable isotopes, B and 11B. Calculate the binding energies per mole of nucleons of these two nuclei. The required masses (in g/mol) are 11H = 1.00783, 10n = 1.008665, 105B = 10.01294, and 115B = 11.00931. 10

32. Calculate the binding energy in kilojoules per mole of nucleons of P for the formation of 30P and 31P. The required masses (in g/mol) are 1 1 30 1H = 1.00783, 0n = 1.008665, 1 5P = 29.97832, 31 and 15P = 30.97376. 33. Calculate the binding energy per mole of nucleons for calcium-40, and compare your result with the value in Figure 20.4. Masses needed for this calculation are (in g/mol) 11H = 1.00783, 10n = 1.008665, and 4200Ca = 39.96259. 34. Calculate the binding energy per mole of nucleons for iron-56. Masses needed for this calculation (in g/mol) are 11H = 1.00783, 10n = 1.008665, and 5266Fe = 55.9349. Compare the result of your calculation to the value for iron-56 in the graph in Figure 20.4. 35. Calculate the binding energy per mole of nucleons for 168O. Masses needed for this calculation are 11H = 1.00783, 10n = 1.008665, and 168O = 15.99491.

Study Questions

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1033

36. Calculate the binding energy per mole of nucleons for nitrogen-14. The mass of nitrogen-14 is 14.003074.

Nucleosynthesis of the Elements (See Section 20.4) 37. It is not completely understood how some of the lightest elements formed, specifically Li, Be, and B, but these two reactions are plausible. Write ­balanced nuclear equations for the reactions. (a) The reaction of helium-3 and helium-4 to give beryllium-7 and a gamma ray. (b) Electron capture by beryllium-7 to give lithium-7. 38. The triple-alpha process is a nuclear fusion process in which three helium-4 nuclei form carbon-12. Write balanced equations for each process described below. (a) The fusion of two helium-4 nuclei to form beryllium-8. (b) The fusion of beryllium-8 with a helium-4 nucleus to give carbon-12. (c) The fusion of carbon-12 with another helium-4 nucleus to give oxygen-16 and a gamma ray.

Rates of Radioactive Decay (See Section 20.5 and Examples 20.5–20.7.)

44. Iodine-131 (t½ = 8.02 days), a β emitter, is used to treat thyroid cancer. (a) Write an equation for the decomposition of 131I. (b) If you ingest a sample of NaI containing 131I, how much time is required for the activity to decrease to 35.0% of its original value? 45. Radon has been the focus of much attention because it is often found in homes. Radon-222 emits α particles and has a half-life of 3.82 days. (a) Write a balanced equation to show this process. (b) How long does it take for a sample of 222Rn to decrease to 20.0% of its original activity? 46. Strontium-90 is a hazardous radioactive isotope that resulted from atmospheric testing of nuclear weapons. A sample of strontium carbonate containing 90Sr is found to have an activity of 1.0 × 103 dpm. One year later, the activity of this sample is 975 dpm. (a) Calculate the half-life of strontium-90 from this information. (b) How long will it take for the activity of this sample to drop to 1.0% of the initial value?

40. Gold-198 is used in the diagnosis of liver problems. The half-life of 198Au is 2.69 days. If you begin with 2.8 μg of this gold isotope, what mass remains after 10.8 days?

47. Radioactive cobalt-60 is used extensively in nuclear medicine as a γ-ray source. It is made by a neutron capture reaction from cobalt-59 and is a β emitter; β emission is accompanied by strong γ radiation. The half-life of cobalt-60 is 5.27 years. (a) How long will it take for a cobalt-60 source to decrease to one eighth of its original activity? (b) What fraction of the activity of a cobalt-60 source remains after 1.0 year?

41. Iodine-131 is used to treat thyroid cancer. (a) The isotope decays by β-particle emission. Write a balanced equation for this process. (b) Iodine-131 has a half-life of 8.02 days. If you begin with 2.4 μg of radioactive 131I, what mass remains after 40.2 days?

48. Scandium occurs in nature as a single isotope, scandium-45. Neutron irradiation produces scandium-46, a β emitter with a half-life of 83.8 days. If the initial activity is 7.0 × 104 dpm, draw a graph showing disintegrations per minute as a function of time during a period of 1 year.

39. Copper(II) acetate containing 64Cu is used to study brain tumors. This isotope has a half-life of 12.7 hours. If you begin with 25.0 μg of 64Cu, what mass remains after 63.5 hours?

42. Phosphorus-32 is used in the form of Na2HPO4 in the treatment of chronic myeloid leukemia, among other things. (a) The isotope decays by β-particle emission. Write a balanced equation for this process. (b) The half-life of 32P is 14.3 days. If you begin with 4.8 μg of radioactive 32P in the form of Na2HPO4, what mass remains after 28.6 days (about 1 month)? 43. Gallium-67 (t½ = 78.27 hours) is used in the medical diagnosis of certain kinds of tumors. If you ingest a compound containing 0.015 mg of this isotope, what mass (in milligrams) remains in your body after 13 days? (Assume none is excreted.)

Nuclear Reactions (See Section 20.6 and Examples 20.8 and 20.9.) 49. Americium-240 is made by bombarding plutonium-239 with α particles. In addition to 240 Am, the products are a proton and two neutrons. Write a balanced equation for this process. 50. There are two isotopes of americium, both with half-lives sufficiently long to allow the handling of large quantities. Americium-241, with a half-life of 432 years, is an α emitter used in smoke detectors. The isotope is formed from 239Pu by absorption of two neutrons followed by emission of a β particle. Write a balanced equation for this process.

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51. The superheavy element 287Fl (element 114) was made by firing a beam of 48Ca ions at 242Pu. Three neutrons were ejected in the reaction. Write a balanced nuclear equation for the synthesis of 287Fl. 52. To synthesize the heavier transuranium elements, a nucleus must be bombarded with a relatively large particle. If you know the products are californium-246 and four neutrons, with what particle would you bombard uranium-238 atoms? 53. Deuterium nuclei (21H) are particularly effective as bombarding particles to carry out nuclear reactions. Complete the following equations: 2 (a) 114 → ? 1 11H 48 Cd 1 1H   (b) 63Li 1 21 H  → ? 1 01n 40 2 (c) 20 Ca 1 1H  → 38 19K 1 ? 2 65 (d) ? 1 1H  → 30 Zn 1  54. Some important discoveries in scientific history that contributed to the development of nuclear chemistry are listed below. Briefly, describe each discovery, identify prominent scientists who contributed to it, and comment on the significance of the discovery to the development of this field. (a) 1896, the discovery of radioactivity (b) 1898, the identification of radium and polonium (c) the first artificial nuclear reaction 55. Boron is an effective absorber of neutrons. When boron-10 is bombarded by neutrons, an α particle is emitted. Write an equation for this nuclear reaction. 56. Identify the unknown species in each reaction below. (a) 147 N 1 42He  → 178O 1 ? (b) 94 Be 1 24He  → ? 1 01n 30 (c) ? 1 42He  → 15 P 1 01n 4 (d) 239 → ? 1 01n 94 Pu 1 2He  

General Questions These questions are not designated as to type or location in the chapter. They may combine several concepts. 57. ▲ A technique to date geological samples uses rubidium-87, a long-lived radioactive isotope of rubidium (t½ = 4.965 × 1010 years). Rubidium-87 decays by β emission to strontium-87. If rubidium-87 is part of a rock or mineral, then strontium-87 will remain trapped within the crystalline structure of the rock. The age of the rock dates back to the time when the rock solidified. Chemical analysis of the rock gives the amounts of 87 Rb and 87Sr. From these data, the fraction of 87Rb that remains can be calculated. Suppose analysis of a stony meteorite determined that 1.8 mmol of 87Rb and 1.6 mmol of 87Sr (the portion of 87Sr formed by decomposition of 87Rb) were present. Estimate the

age of the meteorite. (Hint: The amount of 87Rb at t0 is moles 87Rb + moles 87Sr.) 58. Tritium, 31H, is one of the nuclei used in fusion reactions. This isotope is radioactive, with a half-life of 12.3 years. Like carbon-14, tritium is formed in the upper atmosphere from cosmic radiation, and it is found in trace amounts on Earth. To obtain the amounts required for a fusion reaction, however, it must be made via a nuclear reaction. The reaction of 63Li with a neutron produces tritium and an α particle. Write an equation for this nuclear reaction. 59. Phosphorus occurs in nature as a single isotope, phosphorus-31. Neutron irradiation of phosphorus-31 produces phosphorus-32, a β emitter with a half-life of 14.28 days. Assume you have a sample containing phosphorus-32 that has a rate of decay of 3.2 × 106 dpm. Draw a graph showing disintegrations per minute as a function of time over a period of 1 year. 60. In June 1972, natural fission reactors, which operated billions of years ago, were discovered in Oklo, Gabon (page 1027). At present, natural uranium contains 0.72% 235U. How many years ago did natural uranium contain 3.0% 235U, the amount needed to sustain a natural reactor? (t½ for 235U is 7.04 × 108 years.) 61. If a shortage in worldwide supplies of fissionable uranium arose, it would be possible to use other fissionable nuclei. Plutonium, one such fuel, can be made in breeder reactors that manufacture more fuel than they consume. The sequence of reactions by which plutonium is made is as follows: (a) A 238U nucleus undergoes an (n, γ) reaction to produce 239U. (b) 239U decays by β emission (t½ = 23.5 min) to give an isotope of neptunium. (c) This neptunium isotope decays by β emission to give a plutonium isotope. (d) The plutonium isotope is fissionable. On collision of one of these plutonium isotopes with a neutron, fission occurs, with at least two neutrons and two other nuclei as products. Write an equation for each of the nuclear reactions. 62. When a neutron is captured by an atomic nucleus, energy is released as γ radiation. This energy can be calculated based on the change in mass in converting reactants to products. For the nuclear reaction 63 Li 1 01 n  → 73 Li 1  : (a) Calculate the energy evolved in this reaction (per atom). Masses needed (in g/mol) are 63Li = 6.01512, 10n = 1.008665, and 73Li = 7.01600. (b) Use the answer in part (a) to calculate the wavelength of the γ rays emitted in the reaction. Study Questions

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1035

63. The synthesis of livermorium: (a) Lv-296 is made from the collision of a relatively light atom with curium-248. What is the lighter particle? (b) Lv-296 is not stable, decaying to Lv-293. What particles are emitted in this process? 64. Calculate the energy involved in the fusion of a deuterium nucleus and a tritium nucleus to form helium-4 and a neutron. The masses needed for this calculation are: 21H = 2.01410, 31H = 3.01605, 4 1 2He = 4.00260, and 0n = 1.008665.

In the Laboratory 65. A piece of charred bone found in the ruins of a Native American village has a 14C/12C ratio that is 72% of the ratio found in living organisms. ­Calculate the age of the bone fragment. (t½ for 14 C is 5.73 × 103 years.) 66. A sample of wood from a Thracian chariot found in an excavation in Bulgaria has a 14C activity of 11.2 dpm/g. Estimate the age of the chariot and the year it was made. (t½ for 14C is 5.73 × 103 years, and the activity of 14C in living material is 14.0 dpm/g.) 67. The isotope of polonium that was most likely present in the samples used by Marie Curie in her pioneering studies is polonium-210. A sample of this element was prepared in a nuclear reaction. Initially, its activity (α emission) was 7840 dpm. Measuring radioactivity over time produced the data below. Determine the half-life of polonium-210. Activity (dpm)

Time (days)

7840

 0

7570

 7

7300

14

5920

56

5470

72

68. Sodium-23 (in a sample of NaCl) is subjected to neutron bombardment in a nuclear reactor to produce 24Na. When removed from the reactor, the sample is radioactive, with β activity of 2.54 × 104 dpm. The decrease in radioactivity over time was studied, producing the following data: Activity (dpm)

Time (hours)

2.54 × 10

 0

2.42 × 10

 1

2.31 × 10

 2

2.00 × 10

 5

1.60 × 10

10

1.01 × 10

20

4 4 4 4 4 4

1036 Chapter 20 / Nuclear Chemistry

(a) Write equations for the neutron capture reaction and for the reaction in which the product of this reaction decays by β emission. (b) Determine the half-life of sodium-24. 69. The age of minerals can sometimes be determined by measuring the amounts of 206Pb and 238U in a sample. This determination assumes that all of the 206 Pb in the sample comes from the decay of 238U. The date obtained identifies when the rock solidified. Assume that the ratio of 206Pb to 238U in an igneous rock sample is 0.33. Calculate the age of the rock. (t½ for 238U is 4.5 × 109 years.) 70. To measure the volume of the blood system of an animal, the following experiment was done. A 1.0-mL sample of an aqueous solution containing tritium, with an activity of 2.0 × 106 dps, was injected into the animal’s bloodstream. After time was allowed for complete circulatory mixing, a 1.0-mL blood sample was withdrawn and found to have an activity of 1.5 × 104 dps. What was the volume of the circulatory system? (The half-life of tritium is 12.3 years, so this experiment assumes that only a negligible amount of tritium has decayed in the time of the experiment.)

Summary and Conceptual Questions The following questions may use concepts from this and ­previous chapters. 71. The average energy output of a good grade of coal is 2.6 × 107 kJ/ton. Fission of 1 mol of 235U releases 2.1 × 1010 kJ. Find the number of tons of coal needed to produce the same energy as 1 pound of 235 U. (See Appendix C for conversion factors.) 72. Collision of an electron and a positron results in formation of two γ rays. In the process, their masses are converted completely into energy. (a) Calculate the energy evolved from the annihilation of an electron and a positron, in kilojoules per mole. (b) Using Planck’s equation (Equation 6.2), determine the frequency of the γ rays emitted in this process. 73. The principle underlying the isotope dilution method of analysis can be applied to many kinds of problems. Suppose that you, a marine biologist, want to estimate the number of fish in a lake. You release 1000 tagged fish, and after allowing an ­adequate amount of time for the fish to disperse evenly in the lake, you catch 5250 fish and find that 27 of them have tags. How many fish are in the lake? 74. ▲ Radioactive isotopes are often used as tracers to follow an atom through a chemical reaction. The following is an example of this process:

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Acetic acid reacts with methanol, CH3OH, by eliminating a molecule of H2O to form methyl acetate, CH3CO2CH3. Explain how you would use the radioactive isotope 15O to show whether the oxygen atom in the water product comes from the OOH of the acid or the OOH of the alcohol.

Natasja Weitsz/Getty Images News/ Getty Images

75. On November 1, 2006, Alexander Litvinenko suddenly fell ill and went to a London hospital. He died three weeks later of radiation poisoning induced by polonium-210. Litvinenko had been an officer in the Russian Federal Service, formerly called the KGB. However, he had left Russia to avoid prosecution and sought political asylum in the United Kingdom. On October 31, 2006, Litvinenko met with two other former KGB officers, and it is thought that these two agents put a tiny amount of polonium-210 in Litvinenko’s tea. It was later determined that Litvinenko had been given about 10 µg of 210Po. The radiation from this sample would have been about 2 GBq or 50 mCi, about 200 times the lethal dose of the radionuclide. Polonium-210 is an alpha emitter made by bombarding 209Bi with neutrons. This produces 210Bi, which decays by beta emission to 210Po. This isotope decays by alpha emission to stable 206Pb with a half-life of 138 days.

(a) Write nuclear equations for the synthesis of 210 Po and its decay to 206Pb. (b) Litvinenko died in three weeks. If he had consumed 10 micrograms of 210Po, how much remained after three weeks? 76. ▲ The thorium decay series includes the isotope 228 90Th. Determine the sequence of nuclei going from 232 228 90Th to 90Th by alpha and beta emission. 77. ▲ The last unknown element between bismuth and uranium was discovered by Lise Meitner (1878–1968) and Otto Hahn (1879–1968) in 1918. They obtained 231Pa by chemical extraction of pitchblende, in which its concentration is about



1 ppm (part per million). This isotope, an α emitter, has a half-life of 3.27 × 104 years. (a) Which radioactive decay series (the uranium-235, uranium-238, or thorium-232 series) contains 231Pa as a member? (b) Suggest a possible sequence of nuclear reactions starting with 235U and ending with 231Pa. (c) What quantity of ore would be required to isolate 1.0 g of 231Pa, assuming 100% yield? (d) Write an equation for the radioactive decay process for 231Pa. 78. ▲ You might wonder how it is possible to determine the half-life of long-lived radioactive isotopes such as 238U. With a half-life of more than 109 years, the radio­activity of a sample of uranium will not measurably change in your lifetime. In fact, you can calculate the half-life using the mathematics governing first-order reactions. A 1.0-mg sample of 238U decays at the rate of 12 α emissions per second. Set up a mathematical equation for the rate of decay, ∆N/∆t = −kN, where N is the number of nuclei in the 1.0-mg sample and ∆N/∆t is 12 dps. Solve this equation for the rate constant for this process, and then relate the rate constant to the half-life of the reaction. Carry out this calculation, and compare your result with the literature value, 4.5 × 109 years. 79. ▲ Marie and Pierre Curie identified radium and polonium in a uranium ore (pitchblende, which contains 238U and 235U). Which of the following isotopes of radium and polonium can be found in the uranium ore? (Hint: Consider both the isotope half-lives and the decay series starting with 238U and 235U.) Isotope

Half-Life

Ra

1600 y

Ra

  14.8 d

Ra

   5.75 y

Po

   0.15 s

Po

  138.4 d

226 225 228

216 210

80. The very heavy element tennessine (element 117) is made by a collision of a berkelium-249 nucleus and a calcium-48 atom. It produces tennessine-293. Write the nuclear equation for this process.

Study Questions

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1037

John C. Kotz

21

The Chemistry of the Main Group Elements

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C hapt e r O ut li n e 21.1

Abundance of the Elements

21.2

The Periodic Table: A Guide to the Elements

21.3 Hydrogen 21.4

The Alkali Metals, Group 1A (1)

21.5

The Alkaline Earth Elements, Group 2A (2)

21.6

Boron, Aluminum, and the Group 3A (13) Elements

21.7

Silicon and the Group 4A (14) Elements

21.8

Nitrogen, Phosphorus, and the Group 5A (15) Elements

21.9

Oxygen, Sulfur, and the Group 6A (16) Elements

21.10 The Halogens, Group 7A (17) 21.11 The Noble Gases, Group 8A (18)

As you read this chapter, think about the chemicals within your reach. For example, the air you breathe contains the elements oxygen and nitrogen, and you exhale a carbon compound, CO2. Your body is largely composed of carbon, hydrogen, nitrogen, and oxygen. The pen in your hand may contain a plastic material, largely carbonbased. Your cell phone is powered by a silicon chip and has a battery that uses lithium chemistry. All these elements fall into a category chemists call main group elements, elements in the A groups in the periodic table (or those in Groups 1–2 and 13–18 using the 1–18 numbering system). These elements are the subject of this chapter.

21.1 Abundance of the Elements Goal for Section 21.1 • Recognize patterns in element abundance. In Chapter 7 you learned that the main group elements have valence electrons in the s and p atomic orbitals of the highest energy level, as opposed to transition elements whose valence electrons include not only those in the highest energy level s orbital but also in partially-filled d and f subshells of lower energy levels. As the main group elements include metals, metalloids, and nonmetals, it is not surprising that their chemical and physical properties vary greatly. Many main group elements, especially the lighter ones, are among the most abundant elements in the universe. Figure 21.1 shows the abundances in the ◀ Aragonite, CaCO3.  This form of calcium carbonate is less stable than the calcite form and is slightly more water-soluble (page 1276). It is a component of sea shells and pearls.  John C. Kotz

1039

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The vertical axis uses a logarithmic scale. This means, for example, there are 1012 H atoms for every 100 B atoms.

Number of atoms per 1012 atoms of H

1014 1012 1010 108 106 104 102 0

H

He

Li

Be

B

Li, Be, and B have relatively low abundances because they are circumvented when elements are made in stars.

C

N

O

F

Ne

Na Mg

Al

Si

P

S

Cl

Ar

The common elements such as C, O, and Ne are made in stars by the accretion of alpha particles (helium nuclei). Helium has an atomic number of 2. If three He atoms combine, they produce an atom with atomic number 6 (carbon). Adding yet another He atom gives an atom with atomic number 8 (oxygen), and so on.

Figure 21.1  Abundance of elements 1–18 in the solar system.  See Section 20.4 for a more complete discussion of the formation of the elements and reasons for the variation in abundance.

Table 21.1 The 10 Most Abundant Elements in Earth’s Crust

Rank Element

Abundance (ppm)*

 1

Oxygen

474,000

 2

Silicon

277,000

 3

Aluminum

 82,000

 4

Iron

 56,300

 5

Calcium

 41,000

 6

Sodium

 23,600

 7

Magnesium

 23,300

 8

Potassium

 21,000

 9

Titanium

  5,600

10

Hydrogen

  1,520

*ppm = g per 1000 kg. Data taken from J. Emsley, The Elements, 3rd ed., New York: Oxford University Press, 1998.

solar system of the elements of the first three periods; this figure is a portion of ­Figure 20.5, which showed more elements. Hydrogen and helium are the most abundant elements in the solar system by a wide margin because most of this mass resides in the sun, and these elements are the sun’s primary components. Beyond helium, the light elements lithium, ­beryllium, and boron are low in abundance, but carbon and oxygen have high abundances (over 1 million times more abundant than boron). From this point on, there is ­considerable variation in element abundance in the solar system. Ten ­elements account for 99% of the aggregate mass of the Earth’s crust (Table 21.1), and oxygen, silicon, and aluminum represent more than 80% of this mass. Oxygen and n ­ itrogen are the primary components of the atmosphere, and oxygen-­containing water is highly abundant on Earth’s surface, underground, and as a vapor in the atmosphere. Calcium, aluminum, and silicon are found in some of the most c­ ommon m ­ inerals on the Earth. For example, the photograph opening this chapter is the mineral ­aragonite, a form of calcium carbonate found in seashells and pearls. Finally, as Figure 21.1 shows, elements with even atomic numbers are more abundant than their neighbors with odd numbers (compare Si and Al, for ­example). As described in Section 20.4, the reason is that elements are formed in the stars by the successive addition of helium nuclei (alpha particles, with an atomic number of 2). E ­ lements of odd atomic number are formed from those of even number by another process.

21.2 The Periodic Table: A Guide to the Elements Goal for Section 21.2 • Know the general features of the chemistry of main group elements. The similarities in the chemical and physical properties of certain elements guided Mendeleev when he created the first periodic table (described in Section 2.3). He placed elements in groups based partly on the composition of their common

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Table 21.2

Similarities Within Periodic Groups*

Group

1A (1)

2A (2)

3A (13)

4A (14)

5A (15)

6A (16)

7A (17)

Common oxide

M2O

MO

M2O3

EO2

E4O10

EO3

E2O7

Common hydride

MH

MH2

MH3

EH4

EH3

EH2

EH

Highest oxidation state

+1

+2

+3

+4

+5

+6

+7

compounds with oxygen and hydrogen (Table 21.2). With some discrepancies, this worked well at the time, but the elements are now grouped according to the arrangements of their valence electrons. Recall that the metallic character of the elements decreases moving from left to right in the periodic table. Thus, elements in Group 1A (1), the alkali metals, are the most metallic elements in the periodic table. Elements on the far right are nonmetals, and in between are the metalloids. Metallic character increases from the top of a group to the bottom, a feature well illustrated by Group 4A (14). Carbon, at the top of the group, is a nonmetal; silicon and germanium are metalloids; and tin and lead are metals (Figure 21.2). It is useful to be aware of the metallic character of an element or group of elements because metals typically react with nonmetals to form ionic compounds, whereas compounds composed only of nonmetals are covalent. Recall from ­Chapter 12 that ionic compounds are typically crystalline solids that have high melting points and conduct electricity in the molten state. Covalent compounds, on the other hand, can be gases, liquids, or solids and often have low melting and boiling points.

Valence Electrons for Main Group Elements

© Charles D. Winters/Cengage

*M denotes a metal and E denotes a nonmetal or metalloid.

Figure 21.2  Group 4A (14) elements from nonmetals to metals.  ​A nonmetal, carbon (graphite crucible); a metalloid, silicon (round, lustrous bar); and metals tin (chips of metal) and lead (bullets, a toy, and a sphere).

The chemical behavior of an element is determined by the valence electrons, and you know from Chapter 7 that the valence electrons for main group elements are the ns and np electrons (where n is the period in which the element is found). When considering electronic structure, the noble gases (Group 8A [18]) are useful reference points. Helium has an electron configuration of 1s 2; the other noble gases have ns2np6 valence electron configurations. The dominant characteristic of the noble gases is their lack of reactivity. Indeed, the first two elements in the group do not form any compounds that can be isolated. The other Group 8A (18) elements have limited chemistry, however, and the discovery of xenon compounds in the 1960s ranks as one of the most interesting developments in modern chemistry.

Ionic Compounds of Main Group Elements Ions of main group elements with filled s and p subshells are common—­justifying the often-seen statement that elements react in ways that achieve a noble gas ­configuration. The elements in Groups 1A (1) and 2A (2) form 1+ and 2+ ions with electron configurations that are the same as those for the previous noble gases. All common compounds of these elements (such as NaCl or CaCO3) are ionic. The metallic elements in Group 3A (13) (aluminum, gallium, indium, and thallium) form compounds containing 3+ ions. Elements of Groups 6A (16) and 7A (17) can achieve a noble gas configuration by adding electrons. The Group 7A (17) elements (halogens) form anions with a 1− charge (the halide ions, F−, Cl−, Br−, I−), and the Group 6A (16) elements form anions with a 2− charge (O2−, S2−, Se2−, Te2−). In Group 5A (15) chemistry, 3− ions with a noble gas configuration (such as the nitride ion, N3−) are also known.

21.2  The Periodic Table: A Guide to the Elements

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1041

E xamp le 21.1

Reactions of Main Group Elements Problem  Give the formula and name for the product in each of the following ­reactions. Write a balanced chemical equation for the reaction. (a) Ca(s) + S8(s)

(c) lithium and chlorine

(b) Rb(s) + I2(s)

(d) aluminum and oxygen

What Do You Know?  The reactions listed are between metals in periodic groups 1A (1), 2A (2), and 3A (13) and nonmetals in periodic groups 6A (16) and 7A (17). The products formed will be ionic compounds made up of cations formed from metals by loss of electrons and anions formed from nonmetals by the gain of electrons.

Strategy   Predictions are based on the assumption that ions are formed with the electron configuration of the nearest noble gas. Group 1A (1) elements form 1+ ions; Group 2A (2) elements form 2+ ions; and metals in Group 3A (13) form 3+ ions. In their reactions with metals, halogen atoms add a single electron to give anions with a 1− charge; Group 6A (16) elements add two electrons to form anions with a 2− charge. For names of products, refer to the nomenclature discussion in Section 2.6.

Solution Balanced Equation

Product Name

(a) 8 Ca(s) + S8(s) n 8 CaS(s)

calcium sulfide

(b) 2 Rb(s) + I2(s) n 2 RbI(s)

rubidium iodide

(c) 2 Li(s) + Cl2(g) n 2 LiCl(s)

lithium chloride

(d) 4 Al(s) + 3 O2(g) n 2 Al2O3(s)

aluminum oxide

Think about Your Answer  It will be useful to review the formulas and names of common ions in Section 2.5.

Check Your Understanding 

© Charles D. Winters/Cengage

Write balanced chemical equations for the reactions forming the following compounds from the elements.

Figure 21.3  Boron halides— molecular compounds.  Liquid BBr3 (left) and solid BI3 (right). Formed from a metalloid and a nonmetal, both are molecular compounds. The two compounds are sealed in glass ampules to prevent them from reacting with water in the air.

(a) NaBr

(c) PbO

(b) CaSe

(d) AlCl3

Molecular Compounds of Main Group Elements Many avenues of reactivity are open to nonmetallic main group elements. When they react with metals, the usual result is the formation of an ionic compound, whereas compounds containing only metalloids and nonmetallic elements are generally molecular in nature. Molecular compounds are encountered with the Group 3A (13) element boron (Figure 21.3), and the chemistry of carbon in Group 4A (14) is dominated by molecular compounds with covalent bonds (Chapter 23). Similarly, nitrogen chemistry is dominated by molecular compounds. Consider ammonia, NH3; the various nitrogen oxides (such as laughing gas, N2O); and nitric acid, HNO3. In each of these compounds, nitrogen bonds covalently to another nonmetallic element. Also in Group 5A (15), phosphorus reacts with chlorine to produce the molecular compounds PCl3 (Figure 3.1) and PCl5.

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Chemistry in Your Career

Charles Ichoku

Charles Ichoku

Source: Charles Ichoku

I conduct research about aerosols at NASA. My goal is to compare, understand and integrate ­aerosol retrievals from a lot of different satellite instruments. I also work on remote sensing of wildfires and other types of global, open biomass burning. I analyze the satellite fire measurements from the entire world together with the satellite aerosol measurements to estimate the smoke emissions from these wildfires. Understanding wildfires is important because they affect our lives and environment in a variety of ways. Their emitted smoke is composed of many constituents, which can adversely

affect air quality, human health, the environment and climate. I generally work on several different projects at the same time. Each project is supported by a team of scientists and programmers who process and analyze the data. On a daily basis, they work with approximately one terabyte of data which can fill up about 200 DVDs or 20 times the full storage capacity of a 64 GB iPhone 6 or an equivalent digital camera. I apply the satellite measurements of aerosols and fires to study their impacts on the environment and climate on a global level and sometimes on a regional level.

The valence electron configuration of an element determines the composition of its molecular compounds. Involving all the valence electrons is a frequent occurrence in main group element chemistry, and many compounds exist in which the central element has the highest possible oxidation number (such as P in PF 5). The highest oxidation number equals the group number (when using the Group A-B nomenclature). Thus, the highest (and only) oxidation number of sodium in its compounds is +1; the highest oxidation number of C is +4; and the highest oxidation number of phosphorus is +5 (Tables 21.2 and 21.3).

Ex am p le 21.2

Predicting Formulas for Molecular Compounds of Main Group Elements

Table 21.3 Fluorine Compounds Formed by Main Group Elements

Group Compound Bonding 1A (1)

NaF

Ionic

2A (2)

MgF2

Ionic

3A (13)

AlF3

Ionic

4A (14)

SiF4

Covalent

5A (15)

PF5

Covalent

6A (16)

SF6

Covalent

Problem  Predict the formula for each of the following:

7A (17)

IF7

Covalent

(a) The product of the reaction between germanium and excess oxygen

8A (18)

XeF4

Covalent

(b) The product of the reaction of arsenic and excess fluorine (c) A compound formed from phosphorus and excess chlorine (d) The fully deprotonated anion of selenic acid

What Do You Know?  You know the symbol and location of each of the elements in the periodic table.

Strategy  You can predict that in each reaction the element bonded to the halogen or oxygen in the product will achieve its most positive oxidation number, a value equal to the number of its periodic group in the A-B system of numbering. Solution (a) The Group 4A (14) element germanium should have a maximum oxidation number of +4. Thus, its oxide has the formula  GeO2.  (b) Arsenic, in Group 5A (15), reacts vigorously with fluorine to form  AsF5,  in which arsenic has an oxidation number of +5. (c)  PCl5  is formed when the Group 5A (15) element phosphorus reacts with excess chlorine. (d) The chemistries of the Group 6A (16) elements S and Se are similar. Sulfur has a maximum oxidation number of +6, so it forms SO3 and sulfuric acid, H2SO4. Selenium has analogous chemistry, forming SeO3 and selenic acid, H2SeO4. The fully deprotonated anion of this acid is the selenate ion,  SeO42−. 

21.2  The Periodic Table: A Guide to the Elements

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1043

Think about Your Answer  Notice that in several parts of this example the use of excess oxidizing agent was specified. This is important because, for some of the n ­ onmetals, several different products are possible. For example, the oxidation of carbon can produce either CO2 or CO depending on whether there is an excess or a deficiency of oxygen. You will encounter other examples in later sections of this chapter.

Check Your Understanding  Write the formula for each of the following: (a) hydrogen telluride (b) sodium arsenate (c) selenium hexachloride (d) perbromic acid

Using Group Similarities There are many similarities among elements in the same periodic table group. This means you can use compounds of more common elements as examples when you encounter compounds of elements with which you are less familiar. For example, water, H2O, is the simplest hydrogen compound of oxygen. Thus, you can reasonably expect the hydrogen compounds of other Group 6A (16) elements to be H 2S, H2Se, and H2Te; all are well known.

E xamp le 21.3

Predicting Formulas Problem  Predict the formula for each of the following: (a) A compound of hydrogen and phosphorus (b) The hypobromite ion (c) Germane (the simplest hydrogen compound of germanium) (d) Two oxides of tellurium

What Do You Know?  Elements in a periodic group are expected to behave similarly in their chemistry. Thus, you can use known examples from the more common elements found early in the periodic table and extrapolate to predict the behavior of the less common elements found in the same group.

Strategy  Recall as examples some of the compounds of lighter elements in a group, and then assume other elements in that group will form analogous compounds. Solution (a) Phosphine,  PH3,  has a composition analogous to ammonia, NH3. (b) Hypobromite ion,  BrO−,  is similar to the hypochlorite ion, ClO−, the anion of hypochlorous acid (HClO). (c)  GeH4  (germane) is analogous to CH4 (methane) and SiH4 (silane), two other Group 4A (14) hydrogen compounds. (d) Te and S are in Group 6A (16).  TeO2 and TeO3  are analogs of the oxides of sulfur, SO2 and SO3.

1044 Chapter 21 / The Chemistry of the Main Group Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Think about Your Answer  Keep in mind there can be differences between elements within a periodic group. One difference is that second-period elements generally obey the octet rule, whereas in the third period and below, compounds with expanded octets can form (Section 8.6). For example, in reactions with fluorine, nitrogen reacts with F2 to form only NF3 whereas phosphorus forms both PF3 and PF5.

Check Your Understanding  Identify a compound or ion of a second-period element that has a formula and Lewis structure analogous to each of the following: (a) PH4+

(b) S22−

(c) P2H4

(d) PF3

Ex am p le 21.4

Recognizing Incorrect Formulas Problem  One formula is incorrect in each of the following groups. Find the incorrect formula, and indicate why it is incorrect. (a) CsSO4, KCl, NaNO3, Li2O

(c) CO, CO2, CO3

(b) MgO, CaI2, BaPO4, CaCO3

(d) PF5, PF4+, PF7, PF6−

What Do You Know?  You know the highest oxidation state for a main group element is equal to its group number (using the A-B group numbers).

Strategy   Look for errors such as incorrect charges on ions or an oxidation number exceeding the maximum possible for the periodic group.

Solution (a)  CsSO4.  Sulfate ion has a 2− charge, so this formula would require a Cs2+ ion. Cesium, in Group 1A (1), forms only 1+ ions. The formula of cesium sulfate is Cs2SO4. (b)  BaPO4.  This formula implies a Ba3+ ion (because the phosphate anion is PO43−). The cation charge does not equal the group number. The formula of barium phosphate is Ba3(PO4)2. (c)  CO3.  Given that O has an oxidation number of −2, carbon would have to have an oxidation number of +6. Carbon is in Group 4A (14), however, and can have a maximum oxidation number of +4. (d)  PF7.  The formula implies P has an oxidation state of +7, whereas +5 is expected for the maximum possible P oxidation state.

Think about Your Answer   There are a lot of facts to learn in chemistry. Fortunately, many can be organized in ways that make learning easier. Using the periodic table to predict the chemistry of each element, and recognizing similarities among elements in groups, are essential aspects.

Check Your Understanding  Explain why compounds with the following formulas would not be expected to exist: Na2Cl, CaCH3CO2, Mg2O.



21.2  The Periodic Table: A Guide to the Elements

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1045

21.3 Hydrogen Goal for Section 21.3 • Know the basic properties of hydrogen and its preparation.

Chemical and Physical Properties of Hydrogen Hydrogen has three isotopes, two of them stable (protium and deuterium) and one of them radioactive (tritium). Isotopes of Hydrogen

John C. Kotz

Isotope Mass (u)

Figure 21.4  Heavy water.  This

exhibit in the Nobel Museum in Stockholm, Sweden shows that the sample of D2O (d = 1.107 g/mL) on the left is heavier than a sample of H2O with the same volume on the right.

Symbol

1.0078

1

2.0141

2

3.0160

3

Name

H (H)

Hydrogen (protium)

H (D)

Deuterium

H (T)

Tritium

Of the three isotopes, only H and D are found in nature in significant quantities. Radioactive tritium (half-life, 12.26 years), which is produced by cosmic ray bombardment of nitrogen in the atmosphere, occurs to the extent of only about 1 atom per 1018 atoms of ordinary hydrogen. Under standard conditions, hydrogen (H2) is a colorless gas. Its very low boiling point, 20.7 K, reflects its nonpolar character and low molar mass. As the least dense gas known, it is ideal for filling lighter-than-air craft. Deuterium compounds are widely used in research. One important observation is that, because D has twice the mass of H, reactions involving D atom transfer are slightly slower than those involving H atoms. This knowledge led to a way to obtain D2O, which is sometimes called heavy water (Figure 21.4). Hydrogen can be produced, albeit expensively, by electrolysis of water (Figure 19.19). 2 H2O(ℓ) + electrical energy n 2 H2(g) + O2(g)

Any sample of natural water always contains a tiny concentration of D2O. When electrolyzed, H2O is electrolyzed more rapidly than D2O. Thus, as the electrolysis proceeds, the liquid remaining is enriched in D 2O, and repeating the process many times will eventually give pure D2O. Large amounts of D2O are now produced because this compound is used as a moderator in some nuclear reactors. Hydrogen combines chemically with virtually every other element, except the noble gases. There are three different types of binary hydrogen-containing compounds. Ionic metal hydrides are formed in the reaction of H2 with a Group 1A (1) or 2A (2) metal. 2 Na(s) + H2(g) n 2 NaH(s)

© Charles D. Winters/Cengage

Ca(s) + H2(g) n CaH2(s)

Figure 21.5  The reaction of H2 and Br2.  Hydrogen gas burns in an atmosphere of bromine vapor to give hydrogen bromide, HBr.

These compounds contain the hydride ion, H−, in which hydrogen has a −1 oxidation number. Molecular compounds (such as H2O, HF, and NH3) are generally formed by direct combination of hydrogen with nonmetallic elements (Figure 21.5). The oxidation number of the hydrogen atom in these compounds is +1, and the nonmetal and hydrogen are linked by covalent bonds. N2(g) + 3 H2(g) n 2 NH3(g) F2(g) + H2(g) n 2 HF(g)

Hydrogen is absorbed by many metals to form interstitial hydrides, the third general class of hydrogen compounds. This name refers to the structures of these species,

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in which the hydrogen atoms reside in the spaces between the metal atoms (called interstices) in the crystal lattice (Figure 12.21b). Palladium metal, for example, can soak up 1000 times its volume of hydrogen (at STP). Most interstitial hydrides are non-stoichiometric; that is, the ratio of metal and hydrogen is not a whole number. When interstitial metal hydrides are heated, H2 is driven out. This behavior means these materials can be used to store H2, just as a sponge can store water, and they have been suggested as one way to store hydrogen for use in automobiles.

Preparation of Hydrogen Over 100 million metric tons of hydrogen gas are produced annually in the United States, and virtually all is used immediately in the manufacture of ammonia (­Section 21.8), methanol (CH3OH), or other chemicals. Some hydrogen is made from coal and steam, a reaction that has been used for more than 100 years. C(s) + H2O(g) 88n H2(g) + CO(g)

   ∆rH° = +131 kJ/mol-rxn

water gas or synthesis gas

The reaction is carried out by injecting water into a bed of red-hot coke (a carbonrich, coal-based fuel). The mixture of gases produced, called water gas or synthesis gas, was used until about 1950 as a fuel for cooking, heating, and lighting. However, it has serious drawbacks. It produces only about half as much heat as an equal amount of methane, and the flame is nearly invisible. Moreover, because it contains carbon monoxide, water gas is toxic. The largest quantity of hydrogen gas is now produced by the catalytic steam reformation of methane in natural gas (Figure 21.6). Methane reacts with steam at high temperature to give H2 and CO. CH4(g) + H2O(g) n 3 H2(g) + CO(g)   ∆rH° = +206 kJ/mol-rxn

The reaction is rapid at 900–1000 °C and goes nearly to completion. More hydrogen can be obtained in a second step in which the CO formed in the first step reacts with more water. This so-called water gas shift reaction is run at 400–500 °C and is slightly exothermic. H2O(g) + CO(g) n H2(g) + CO2(g)  ∆rH° = −41 kJ/mol-rxn

The CO2 formed in the process is removed by reaction with CaO (to give solid CaCO3), leaving fairly pure hydrogen. Perhaps the cleanest way to make hydrogen on a relatively large scale is the electrolysis of water (Figure 21.7). According to the U.S. Department of Energy, “­Electrolysis is a leading hydrogen production pathway to achieve the …. goal of reducing the cost of clean hydrogen by 80% to $1 per 1 kilogram in 1 decade. Hydrogen produced via electrolysis can result in zero greenhouse gas emissions, depending on the source of the electricity used.”

Gas

Oil

Coal

Bio mass

Steam reforming

Synthesis gas

Partial oxidation

CO + H2

Gasification

(syngas)

Water-gas-shift reaction

H2

Fischer-Tropsch synthesis

Synthetic fuels

Methanol synthesis

CH3OH methanol

Figure 21.6  Production of water gas.  Water gas, also called synthesis gas, is a mixture of CO and H2. It is produced by treating coal, coke, or a hydrocarbon like methane with steam at high temperatures. Methane has the advantage that it gives more total H2 per gram than other hydrocarbons, and the ratio of the by-product CO2 to H2 is lower. The syngas mixture is used to make synthetic fuels or methanol. Or, the mixture is treated further to isolate H2, which is then used to make NH3. 21.3 Hydrogen

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1047

4 H+ + 4 e–

2 H2

2 H2O

O2 + 4 H+ + 4 e– O2

2 H2

Cathode

H+

Anode

H2O

Table 21.4

Source: Illustration is from: https://www.energy.gov/eere/fuelcells/hydrogen-production-electrolysis

Figure 21.7 Producing hydrogen by electrolysis.  The hydrogen electrolysis cells can be powered by wind, solar, geothermal, or hydroelectric energy sources. The goal is to eventually reduce the cost of producing hydrogen to $1/kg.

Methods for Preparing H2 in the Laboratory

1. Metal + Acid n metal salt + H2 Mg(s) + 2 HCl(aq) n MgCl2(aq) + H2(g) 2. Metal + H2O n metal hydroxide or oxide + H2

2 Na(s) + 2 H2O(ℓ) n 2 NaOH(aq) + H2(g)



2 Fe(s) + 3 H2O(ℓ) n Fe2O3(s) + 3 H2(g)

3. Metal hydride + H2O n metal hydroxide + H2 CaH2(s) + 2 H2O(ℓ) n Ca(OH)2(s) + 2 H2(g)

Table 21.4 and Figure 21.8 give examples of reactions used to produce H2 gas in the laboratory. The most often used method is the reaction of a metal with an acid. Alternatively, the reaction of aluminum with aqueous NaOH (Figure 21.8b) can be used.

© Charles D. Winters/Cengage

2 Al(s) + 2 NaOH(aq) + 6 H2O(ℓ) n 2 Na[Al(OH)4](aq) + 3 H2(g)

(a) The reaction of Mg and an acid to produce H2 gas and a magnesium salt is significantly exothermic. This reaction is used in meals-ready-to-eat (MREs) (page 299) because it is significantly exothermic.

(b) The reaction of aluminum and aqueous NaOH. The products of this reaction are hydrogen gas and a solution of Na[Al(OH)4].

(c) The reaction of CaH2 and water. The products are hydrogen gas and Ca(OH)2.

Figure 21.8 ​Producing hydrogen gas.

1048 Chapter 21 / The Chemistry of the Main Group Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

In 1783, Jacques Charles used hydrogen to fill a balloon large enough to float him over the French countryside (see page 508). In the 1920s and 30s, the Graf Zeppelin, a passenger-carrying, hydrogen-filled dirigible built in Germany carried more than 13,000 people between Germany and the U.S. The Graf Zeppelin was replaced by another lighter-than-air craft, the Hindenburg, in 1937. The Hindenburg was designed to be filled with nonflammable helium. However, World War II was approaching, and the United States, which has much of the world’s supply of helium, would not sell the gas to Germany. Instead, the Hindenburg had to use hydrogen. Unfortunately, the Hindenburg exploded and burned

when landing in Lakehurst, New Jersey in May 1937, killing 35 of the 97 people on board. As a result of the Hindenburg disaster, hydrogen acquired a reputation for being dangerous. Actually, it is as safe to handle as many other fuels, and it is again being used. Thus far, the most visible use of hydrogen is as a fuel in the rockets used to launch space vehicles. However, with the success of electric vehicles, there is now more interest in autos, trucks, and buses  ​A production Toyota with a hydrogen fuel cell. that have fuel cells (Figure 19.13) fueled by hydrogen. Like batteries, fuel cells effi- though, is that there are very few places to ciently convert chemical potential energy “gas up” with hydrogen. directly into electricity, with little energy An alternative to storing hydrogen gas itself lost as heat. And, unlike the rechargeable for use in fuel cells is described in more batteries used in vehicles, fuel cells can detail in “Applying Chemical ­Principles be refueled rapidly. Their disadvantage, 21.2: Hydrogen Storage” (page 1090).

During World War II, this reaction was used to obtain hydrogen to inflate small balloons for weather observation and to raise radio antennas. Metallic aluminum was plentiful at the time because it came from damaged aircraft. The combination of a metal hydride and water (Figure 21.8c) is an efficient but expensive way to synthesize H2 in the laboratory. The reaction is also commonly used in laboratories to dry organic solvents because the metal hydride reacts with traces of water present in the solvent.

Group 1A (1) Alkali metals Lithium 3

Li

20 ppm

21.4 The Alkali Metals, Group 1A (1)

Sodium 11

Na

Goal for Section 21.4

23,600 ppm

• Know the general chemical properties of the alkali metals. Sodium and potassium are, respectively, the sixth and eighth most abundant elements in the Earth’s crust. In contrast, lithium is relatively rare, as are rubidium and cesium. Only traces of radioactive francium occur in nature. Its longest-lived isotope (223Fr) has a half-life of only 22 minutes. The Group 1A (1) elements are metals, and all are highly reactive with oxygen, water, and other oxidizing agents (described on page 1054). In all cases, these reactions yield compounds containing the Group 1A (1) metals as a 1+ ion. Because the elements are so reactive, the free metals are never found in nature. Most sodium and potassium compounds are water-soluble (see the solubility guidelines in Figure 3.10), so it is not surprising that sodium and potassium compounds are found either in the oceans or in underground deposits that are the residue of ancient seas. To a much smaller extent, these elements are also found in minerals, such as Chile saltpeter (NaNO3). Despite the fact that sodium is only slightly more abundant than potassium on the Earth, seawater contains significantly more sodium than potassium (2.8% NaCl versus 0.8% KCl). Why the great difference? Most compounds of both elements are water-soluble, so why doesn’t rain dissolve Na- and K-containing minerals and carry them to the sea to appear in equivalent proportions as on land? The answer

Potassium 19

K

21,000 ppm Rubidium 37

Rb

90 ppm Cesium 55

Cs

0.0003 ppm Francium 87

Fr

trace Element abundances are in parts per million in the Earth’s crust.

21.4  The Alkali Metals, Group 1A (1)

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A Closer Look

Hydrogen in Transportation

© Charles D. Winters/Cengage

 The importance of salt.  All animals, including humans, need a certain amount of salt in their diet. Sodium ions are important in maintaining electrolyte balance and in regulating osmotic pressure. For an account of the importance of salt in society, culture, history, and economy, see M. Kurlansky, Salt, A World History, New York: Penguin Books, 2003.

is that potassium is an important factor in plant growth. Most plants contain four to six times as much combined potassium as sodium. Thus, most of the potassium ions in groundwater from dissolved minerals are taken up preferentially by plants, whereas sodium ions continue on to the oceans. Some NaCl is essential in the diet of humans and other animals because many biological functions are controlled by the concentrations of Na+ and Cl− ions. The fact that salt has long been recognized as important is evident in surprising ways. For example, workers are paid a salary. This word is derived from the Latin salarium, which meant “salt money” because Roman soldiers were paid in salt.

Preparation of Sodium and Potassium Sodium is produced by reducing sodium ions in sodium salts. This is usually done by electrolysis, because few chemical reducing agents are powerful enough to convert sodium ions to sodium metal. The English chemist Sir Humphry Davy first isolated sodium in 1807 by the electrolysis of molten sodium carbonate. However, the element remained a laboratory curiosity until 1824 when chemists found sodium could be used to reduce aluminum chloride to aluminum metal. At that time, metallic aluminum was rare and very valuable, so this discovery inspired considerable interest in manufacturing sodium. By 1886, a practical method of sodium production had been devised (the reduction of NaOH with carbon). Unfortunately for sodium producers, in this same year Charles Hall and Paul Heroult invented a better way to produce aluminum (by electrolysis, page 1060), thereby eliminating this market for sodium. Sodium is currently produced by the electrolysis of molten NaCl (Section 19.7). The Downs cell for the electrolysis of molten NaCl operates at 7 to 8 V with currents of 25,000 to 40,000 amps (Figure 21.9). The cell is filled with a mixture of dry NaCl, CaCl2, and BaCl2. [Adding other salts to NaCl lowers the melting point from that of pure NaCl (800.7 °C) to about 600 °C. Recall that solutions have lower melting points than pure solvents (Section 13.4).] Sodium is produced at a copper or iron cathode that surrounds a circular graphite anode. Directly over the cathode is an inverted trough in which the low-density, molten sodium (melting point, 97.8 °C) collects. Chlorine, a valuable by-product, collects at the anode.

Figure 21.9  A Downs cell for preparing sodium.

Cl2 gas Inlet for NaCl

Liquid Na metal

Na outlet

Cl2 output

Chlorine gas is produced at the anode and collected inside the inverted cone in the center of the cell.

At the temperature of the electrolysis, about 600 °C, sodium is a liquid. It rises to the top and is drawn off continuously.

Iron screen

Cathode (−)

A circular iron cathode is separated from the graphite anode by an iron screen.

Anode (+)

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Potassium can also be made by electrolysis, but separation of the metal is difficult because molten potassium is soluble in molten KCl. The preferred method for potassium preparation is the reaction of sodium vapor with molten KCl, with potassium being continually removed from the equilibrium mixture. Na(g) + KCl(ℓ) uv K(g) + NaCl(ℓ)

Properties of Sodium and Potassium Sodium and potassium are silvery metals that are soft and easily cut with a knife (Figure  2.6). They are just a bit less dense than water. Their melting points, 97.8 °C for sodium and 63.7 °C for potassium, are quite low. When exposed to moist air, the surface of an alkali metal is quickly coated with a film of oxide or hydroxide. Consequently, the metals must be stored in a way that avoids contact with air, typically by placing them in kerosene or mineral oil. The great reactivity of Group 1A (1) metals is exemplified by their reaction with water, which generates an aqueous solution of the metal hydroxide and hydrogen gas (see page 1054 and “A Closer Look: The Reactivity of the Alkali Metals”), 2 Na(s) + 2 H2O(ℓ) n 2 Na+(aq) + 2 OH−(aq) + H2(g)

and their reaction with any of the halogens to yield a metal halide (Figure 1.2), 2 Na(s) + Cl2(g) n 2 NaCl(s) 2 K(s) + Br2(ℓ) n 2 KBr(s)

Chemistry often produces surprises. Group 1A (1) metal oxides, M 2O, are known, but they are not the principal products of the reaction of Group 1A (1) elements and oxygen. Instead, the primary product of the reaction of sodium and oxygen is sodium peroxide, Na2O2, whereas the principal product from the reaction of potassium and oxygen is KO2, potassium superoxide. 2 Na(s) + O2(g) n Na2O2(s)

Both Na2O2 and KO2 are ionic compounds where the Group 1A (1) cation is paired with either the peroxide ion (O22−) or the superoxide ion (O2−). The peroxide and superoxide are not merely laboratory curiosities. Potassium superoxide in particular is used in oxygen-generation devices in places where ­people are confined, such as submarines, aircraft, and spacecraft, or when an emergency supply is needed. The KO2 closed-circuit breathing equipment in Figure 21.10 is self-contained, that is, it is able to both consume exhaled CO2 and provide a continuous supply of oxygen. On average, a person exhales 0.82 L of CO2 and some water vapor for every 1.0 L of O2 inhaled. The reaction of KO2 and exhaled water vapor produces O2 and KOH, and the latter absorbs exhaled CO2. 4 KO2(s) + 2 H2O(g) n 4 KOH(aq) + 3 O2(g) 4 KOH(aq) + 2 CO2(g) n 2 K2CO3(s) + 2 H2O(g)

Notice that, for every 4 mol of KO2, 3 mol O 2 is produced and 2 mol CO2 is absorbed.

Important Lithium, Sodium, and Potassium Compounds Lithium compounds have a wide range of significant uses: in ceramics and glass, in reagents for the production of pharmaceuticals, in lubricating greases, and most recently in batteries (Chapter 19) for electric and hybrid cars as well as aircraft. Indeed, lithium has become so important in recent years that the supply of lithium is a major factor in the worldwide economy. A significant amount of lithium salts come from Australia, northern Chile, southern Bolivia, and China. Lithium in Australia comes from mining the ore, but in Chile and Argentina, groundwater

Courtesy Neutronics, Inc.

K(s) + O2(g) n KO2(s)

Figure 21.10  A closed-circuit breathing apparatus.  ​Both carbon dioxide and moisture are exhaled by the wearer into a breathing tube. Three moles of oxygen are produced from the reaction of 2 mol of water with 4 mol of KO2, replacing the oxygen consumed in respiration. In addition, the 4 mol of KOH formed by this reaction can react with 2 mol of CO2, removing the CO2 produced by respiration.

21.4  The Alkali Metals, Group 1A (1)

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Lithium and lithium salts. The high plateau in Bolivia where ground water contains high concentrations of Group 1A (1) salts, among them lithium carbonate. The water is left in the sun to deposit the salts by evaporation. Lithium Batteries The New York

Times (A. Rubin, “Lithium-Ion Batteries in E-Bikes and Other Devices Pose Fire Risks,” New York Times, Late Edition, November 14, 2022.) reported that almost 200 fires occurred in New York City in 2022 because of lithium batteries. These are often larger batteries used in e-bikes and similar vehicles. It is recommended that you use only chargers and cables recommended by a safety testing lab, and replace batteries that expand.

Figure 21.11 Producing chlorine and sodium hydroxide in a membrane cell.  ​This is the basis of the chlor-alkali industry. Brine (a concentrated aqueous solution of NaCl) is fed into the anode compartment and dilute sodium hydroxide or water into the cathode compartment. Pipes carry the evolved gases and NaOH away from the chambers of the electrolysis cell.

with high concentrations of Group 1A (1) salts is pumped into ponds. The solution slowly evaporates, and after about a year, the highly concentrated solution is taken to a chemical plant where it is evaporated to give white, powdered lithium carbonate. One estimate is that Bolivia alone has a reserve of about 73 million metric tons of Li2CO3. To produce lithium metal, the carbonate is converted to LiCl, which is then electrolyzed to produce the metal. Electrolysis of aqueous sodium chloride (brine) is the basis of one of the largest chemical industries in the United States. 2 NaCl(aq) + 2 H2O(ℓ) n Cl2(g) + 2 NaOH(aq) + H2(g)

Two of the products from this process—chlorine and sodium hydroxide—give the industry its name: the chlor-alkali industry. More than 10 billion kg of Cl2 is ­produced annually in the United States (as well as NaOH). Oxidation of chloride ion to Cl2 gas in the electrolysis process occurs at the anode, and reduction of water to hydroxide ion and H2 occurs at the cathode (Figure 21.11). Anode reaction (oxidation):

2 Cl−(aq) n Cl2(g) + 2 e−

Cathode reaction (reduction):

2 H2O(ℓ) + 2 e− n H2(g) + 2 OH−(aq)

Activated titanium is used for the anode, and stainless steel or nickel is preferred for the cathode. The anode and cathode compartments are separated by a membrane that allows Na+ ions to pass to maintain the charge balance. Thus, the membrane performs a similar function in the cell as a salt bridge (Section 19.2). The energy consumption of these cells is in the range of 2000–2500 kWh per ton of NaOH produced. Another important compound of sodium is sodium carbonate, Na2CO3. Known by two common names, soda ash and washing soda, it is used in making glass, soap, and paper. In the past, it was largely manufactured by combining NaCl, ammonia, and CO2 in the Solvay process (which remains the method of choice in many countries). In the United States, however, sodium carbonate is obtained from naturally occurring deposits of the mineral trona, Na2CO3 ∙ NaHCO3 ∙ 2 H2O (Figure 21.12). Sodium bicarbonate, NaHCO3, also known as baking soda, is another common sodium compound. Not only is NaHCO3 used in cooking, but it is also added in small amounts to table salt. NaCl is often contaminated with small amounts of MgCl2. The magnesium salt is hygroscopic; that is, it picks water up from the air and, Anode (+)

Cathode (−) Ion-permeable membrane

Depleted brine

Cl2

Cl−

H2

Water

Na+ OH−

Brine

Cl− ions are oxidized to Cl2 gas at the anode.

H2O

H2O

The anode and cathode compartments are separated by an ion-permeable membrane that allows Na+ ions to pass but not Cl– or OH–.

NaOH(aq)

H2O is reduced to H2 gas at the cathode.

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Figure 21.12  Soda ash.  ​

AP Images/Matt Joyce

Trona mined in Wyoming and California is processed into soda ash (Na2CO3) and other sodiumbased chemicals. Soda ash is the ninth most widely used chemical in the United States. Domestically, about half of all soda ash produced is used in making glass. Most of the remainder goes to make chemicals such as sodium silicate, sodium phosphate, and sodium cyanide. Some is also used to make detergents, in the pulp and paper industry, and in water treatment.

in doing so, causes the NaCl to clump. Adding NaHCO3 converts MgCl2 to magnesium carbonate, a nonhygroscopic salt. MgCl2(s) + 2 NaHCO3(s) n MgCO3(s) + 2 NaCl(s) + H2O(ℓ) + CO2(g)

Sodium nitrate, NaNO3, is another common sodium compound, and large deposits are found in Chile, which explains its common name of Chile saltpeter. These deposits are thought to have formed eons ago by bacterial action on organisms in shallow seas. The initial product was ammonia, which was subsequently oxidized to nitrate ion; its combination with sea salt led to sodium nitrate. Because nitrates in general, and alkali metal nitrates in particular, are highly water-soluble, deposits of NaNO3 are found only in areas with very little rainfall. Sodium nitrate can be converted to potassium nitrate by an exchange reaction. NaNO3(aq) + KCl(aq) uv KNO3(aq) + NaCl(s)

Equilibrium favors the products here because, of the four salts involved in this reaction, NaCl is least soluble in hot water. Sodium chloride precipitates, and the KNO3 that remains in solution can be recovered by evaporating the water. Potassium nitrate has been used for centuries as the oxidizing agent in gunpowder. A mixture of KNO3, charcoal, and sulfur will react when ignited. 2 KNO3(s) + 4 C(s) n K2CO3(s) + 3 CO(g) + N2(g)

© Charles D. Winters/Cengage

2 KNO3(s) + 2 S(s) n K2SO4(s) + SO2(g) + N2(g)



Gunpowder.  Gunpowder was developed well over 1000 years ago by the Chinese. One of the three ingredients is KNO3, commonly called saltpeter. Although there are places in the world where it is prevalent and can be mined, for centuries it was obtained by mixing manure or human waste with wood ashes. Animal and human waste contain ammonia, which is converted by bacterial oxidation to nitrates. Wood ashes contain potash, potassium carbonate, the source of potassium in KNO3. 21.4  The Alkali Metals, Group 1A (1)

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1053

Mitch Jacob

A favorite demonstration in chemistry classes is to put a small piece of one of the alkali metals in water. Lithium reacts sluggishly, but as you can see on page  344 sodium reacts much more vigorously. Potassium (Figure 1), rubidium, and cesium are even more reactive, exploding on contact with water. One question is why the reaction is so rapid, and this was answered by clever experiments and high speed filming when liquid Na/K alloy was dropped into water (Figure  2). A fraction of a millisecond after a tiny drop of the metal contacts water, “spikes” appear from the drop. The researchers showed that there is nearly instantaneous transfer of electrons from the metal through these “spikes” to the water, thus creating metal ions. The ions

© Charles D. Winters/Cengage

A Closer Look

The Reactivity of the Alkali Metals

Figure 1  Potassium reacts vigorously with water. strongly repel each other and create what is called a “Coulomb explosion,” blowing the metal drop apart and creating more, even tinier drops, which react further.

Figure 2  The effect of dropping Na/K alloy into water.  ​Spikes spread explosively from the drop of Na/K alloy. Times are noted in milliseconds.

Notice that both reactions (which are doubtless more complex than described in these equations) produce gases. These gases propel the bullet from a gun or cause a firecracker to explode. Group 2A (2) Beryllium 4

Be

2.6 ppm Magnesium 12

Mg

23,300 ppm Calcium 20

Ca

41,000 ppm Strontium 38

Sr

370 ppm Barium 56

Ba

500 ppm Radium 88

Ra

6 × 10−7 ppm Element abundances are in parts per million in the Earth’s crust.

21.5 The Alkaline Earth Elements, Group 2A (2) Goals for Section 21.5 • Know some of the common, commercially important compounds of the Group 2A (2) elements.

• Know the general chemical properties of the alkaline earth elements. The “earth” part of the name alkaline earth dates back to the days of medieval alchemy. To alchemists, any solid that did not melt and was not changed by fire into another substance was called an earth. The word “alkaline” applies because compounds of Group 2A (2) elements, such as CaO, are alkaline according to experimental tests conducted by the alchemists: The compounds had a bitter taste and neutralized acids. Finally, with very high melting points, these compounds were unaffected by fire. Calcium and magnesium rank fifth and seventh, respectively, in abundance in Earth’s crust. Both elements are the basis of many commercially important compounds. Like the Group 1A (1) elements, Group 2A (2) elements are very reactive, and the free elements are not found in nature. Unlike most of the compounds of the Group 1A (1) metals, however, many compounds of the Group 2A (2) elements are poorly soluble in water, which explains their occurrence in various minerals ­(Figure 21.13). Calcium minerals include limestone (CaCO 3), gypsum (CaSO 4 ∙ 2 H 2O), and fluorite (CaF 2). Magnesite (MgCO 3), talc or soapstone (3 MgO ∙ 4 SiO 2 ∙ H 2O), and asbestos (3 MgO ∙ 4 SiO 2 ∙ 2 H 2O) are common magnesium-containing minerals. The mineral dolomite, MgCa(CO3)2, contains both magnesium and calcium.

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© Charles D. Winters/Cengage

© A. N. Palmer

Icelandic spar. This mineral, one of a number of crystalline forms of CaCO3, displays birefringence, a property in which a double image is formed when light passes through the crystal.

Evaporative deposit of gypsum (CaSO4 ∙ 2H2O) in a cave within ­Carlsbad Caverns National Park, New Mexico.

Figure 21.13  Various minerals containing calcium.

Limestone, a sedimentary rock, is found widely on the Earth’s surface. Many of these deposits contain the fossilized remains of marine life. Other forms of calcium carbonate include marble, aragonite (page 1038), and Iceland spar (a form of calcite). Iceland spar occurs as large, clear crystals with the interesting optical property of birefringence (Figure 21.13).

Beryllium–The Lightest Group 2A (2) Element The James Webb Space Telescope was launched on December 25, 2021 and arrived at its station to observe the deep universe in January 2022. It is expected to last at least 10 years, but the results of its observations have already been startling. To collect light in the near infrared region of the spectrum, the telescope consists of 18 hexagonal mirror segments. Combined, these create a 6.5-m diameter mirror. Each mirror segment is made of beryllium metal that is plated with a thin, reflective film of gold. The Group 2A (2) metal beryllium, although toxic, was chosen for the mirror supports for several reasons. It is the second lightest of all the metals (d = 1.848 g/cm3), and it is six times stiffer than steel. Most importantly, it holds its shape much better than other metals over a broad range of temperatures. Beryllium metal does not change its shape below 100 K. This is especially important because the mirror must be kept very cold (at about 50 K).

BEST-BACKGROUNDS/Shutterstock.com

James Webb Space Telescope.  The mirror of the telescope consists of 18 goldplated beryllium segments.



21.5  The Alkaline Earth Elements, Group 2A (2)

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© Charles D. Winters/Cengage

Properties of Magnesium and Calcium

Figure 21.14  The reaction of calcium and warm water.  ​ Hydrogen gas bubbles are seen rising from the metal surface. The other reaction product is Ca(OH)2. The inset is a model of hexagonal close-packed calcium metal (Section 12.1). Hard Water  Cal­cium and magnesium ions are present in some common minerals and so often occur in natural waters. There they can interact with atmospheric CO2 to form insoluble carbonates, and these can deposit in unwanted places, like the water pipes in your home. For more on this problem, see “Applying Chemical Principles 25.2: Hard Water,” page 1283.

Magnesium and calcium are fairly high-melting, silvery metals, and share chemical properties with other elements of Group 2A (2). They are oxidized by a wide range of oxidizing agents to form ionic compounds that contain the M2+ ion. For example, these elements combine with halogens to form MX2, with oxygen or sulfur to form MO or MS, and with water to form hydrogen and the metal hydroxide, M(OH)2 (Figure 21.14). With acids, hydrogen is evolved (Figure 21.8a), and a salt of the metal cation and the anion of the acid results.

The Production of Magnesium Almost one million tons of magnesium are produced worldwide annually. It is used largely with other elements such as aluminum, zinc, manganese, and silicon to ­produce alloys that have low density, high strength, can be machined, and resist corrosion. Almost half is combined with aluminum to make containers such as b ­ everage cans. In the United States, most beverage cans are aluminum with 5% magnesium. Other alloys with much higher magnesium content are used in car and aircraft parts. Magnesium is obtained from two primary sources: from the minerals magnesite (MgCO3) and dolomite (MgCO3 ∙ CaCO3) or from seawater, in which Mg2+ ion has a concentration of about 0.05 M. To obtain magnesium from seawater, the first step is to precipitate its hydroxide [Ksp for Mg(OH)2 = 5.6 × 10−12] by mixing with dolomite that has been heated to form the mixed oxides (MgO ∙ CaO). Mg2+(aq) + MgO ∙ CaO(s) + 2 H2O(ℓ) n 2 Mg(OH)2(s) + Ca2+(aq)

Magnesium hydroxide is isolated by filtration and then converted to magnesium chloride by first heating to form MgO, which is then treated with carbon and chlorine at a high temperature in an electric furnace. 2 MgO(s) + C(s) + 2 Cl2(g) n 2 MgCl2(s) + CO2(g)

Finally, molten MgCl2 is electrolyzed to give the metal and chlorine. MgCl2(ℓ) n Mg(s) + Cl2(g)

China now manufactures about 90% of the magnesium produced in the world, primarily by another method, the direct chemical reduction of magnesium minerals at high temperatures (1500–1800 K). The ore is treated with ferrosilicon, an alloy of iron and silicon where the silicon functions as the reducing agent. The product is magnesium vapor, which is condensed to the solid metal. 2 MgO(s) + Si(s) uv SiO2(s) + 2 Mg(g)

© Charles D. Winters/Cengage

Interestingly, this endothermic reaction is not thermodynamically favored. ­However, in the manufacturing process, the magnesium vapor is removed continuously, ­shifting the equilibrium to the right. The thermal process used in China for producing magnesium requires ­enormous amounts of energy and so is not used in many countries where energy costs are high. In fact, the metal is no longer produced in western Europe.

The mineral fluorite, CaF2. ​Cal­ cium fluoride is found in a wide variety of colors, and the crystals are frequently octahedral in shape.

Calcium Minerals and Their Applications The most common calcium-based minerals are its fluoride, phosphate, and ­carbonate salts. Fluorite, CaF2, and fluoroapatite, Ca5F(PO4)3, are important as ­commercial sources of fluorine. Almost half of the CaF2 mined is used in the steel industry, where it is added to the mixture of materials that is melted to make crude iron. The CaF2 acts to remove some impurities and improves the separation of ­molten iron from silicates and other by-products resulting from the reduction of iron ore to the metal (Chapter 22). A second major use of fluorite is in the manufacture of ­hydrofluoric acid by a reaction of the mineral with concentrated sulfuric acid. CaF2(s) + H2SO4(ℓ) n 2 HF(g) + CaSO4(s)

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Alkaline Earth Metals and Biology Plants and animals derive energy from the oxidation of a sugar, glucose, with oxygen. Plants, however, can synthesize glucose from CO 2 and H2O using sunlight as an energy source. This process is initiated by chlorophyll, a very large, magnesiumbased molecule. Magnesium

A molecule of chlorophyll. ​Magnesium is its central element. In your body, the alkaline earth metal ions Mg2+ and Ca2+ serve regulatory functions. Although these two metal ions are required by living systems, the other Group 2A (2)



Blue apatite. ​Apatite can be found in gem-quality deposits.

elements are toxic. Beryllium compounds are carcinogenic, and soluble barium salts are poisons. For this reason, you may be concerned if your physician asks you to drink a “cocktail” containing BaSO 4 to check the condition of your digestive tract. Don’t be afraid, because the cocktail contains very insoluble BaSO4 (Ksp = 1.1 × 10−10), so it passes through your digestive tract without a significant amount being absorbed. Barium sulfate is opaque to X-rays, so its path through your organs appears on the developed X-ray image. The calcium-containing compound hydroxyapatite is the main component of tooth enamel. Cavities in your teeth form when acids (such as those in soft drinks) decompose the weakly basic hydroxyapatite coating.

Ca5(OH)(PO4)3(s) + 4 H3O+(aq) n   5 Ca2+(aq) + 3 HPO42−(aq) + 5 H2O(ℓ) This reaction can be prevented by converting hydroxyapatite to the much more acid-resistant coating of fluoroapatite.

Ca5(OH)(PO4)3(s) + F−(aq) n  Ca5F(PO4)3(s) + OH−(aq)

CNRI/Science Source

A Closer Look

Hydrofluoric acid is used to make cryolite, Na3AlF6, a material needed in aluminum production (Section 21.6) and in the manufacture of fluorine-containing ­compounds such as tetrafluoroethylene (F2CPCF2), the precursor to Teflon™. Apatites have the general formula Ca 5X(PO4)3 (X = F, Cl, OH). More than 255 million tons of phosphate rock, the source of apatite, are mined annually, with ­Florida alone accounting for about one-third of the world’s output. Most of this material is converted to phosphoric acid by reaction with sulfuric acid. Phosphoric acid is needed in the manufacture of a multitude of products, including fertilizers and detergents, baking powder, and various food products (Section 21.8). Limestone, which consists mostly of calcium carbonate, has been used in ­agriculture for centuries. It is spread on fields to neutralize acidic compounds in the soil and to supply Ca2+, an essential nutrient. Because magnesium carbonate is often present in limestone, liming a field also supplies Mg2+, another important nutrient for plants. Calcium carbonate and calcium oxide (lime) are of special interest. The t­ hermal decomposition of CaCO3 to give lime (and CO2) is one of the oldest chemical ­reactions known. Lime is one of the top 10 industrial chemicals produced today, with about 420 million metric tons produced globally each year. For several thousand years, lime has been used in mortar (a lime, sand, and water paste) to secure stones to one another in building houses, walls, and roads. The Chinese used it to set stones in the Great Wall. The Romans perfected its use, and the fact that many of their constructions still stand today is testament both to their skill and to the usefulness of lime. The famous Roman road, the Appian Way, used lime mortar between several layers of its stones.

X-ray of a gastrointestinal tract using BaSO4 to make the organs visible. The source of the fluoride ion can be sodium fluor­ide or sodium monofluorophosphate (Na2FPO3, commonly known as MFP) in your toothpaste. Alternatively, a fluoride-containing material can be ­applied by your dentist.

21.5  The Alkaline Earth Elements, Group 2A (2)

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The utility of mortar depends on some simple chemistry. Mortar consists of one part lime to three parts sand, with water added to make a thick paste. The first reaction, referred to as slaking, occurs after the solids are mixed with water. This produces a slurry containing calcium hydroxide, which is known as slaked lime. CaO(s) + H2O(ℓ) uv Ca(OH)2(s)

When the wet mortar mix is placed between bricks or stone blocks, it slowly reacts with CO2 from the air, and the slaked lime is converted to calcium carbonate. Ca(OH)2(s) + CO2(g) uv CaCO3(s) + H2O(ℓ)

A Closer Look

The sand grains are bound together by the particles of calcium carbonate.

Cement—The Second Most Used Substance

Cement is such a commonplace material that it is easy to take it for granted. The preparation of cement has origins more than 2000 years ago and was used in Roman roads that still exist today. The current recipe for cement was patented in 1824 by a British bricklayer, Joseph Aspdin, and has not been changed much since that time. Aspdin called his product Portland cement because it was similar in appearance to Portland stone, from the Isle of Portland in Dorset, England. In 2020, the world production of Portland cement was 4.1 billion tons; this amount is second only to water among substances people use.

According to Chemical & Engineering News, the world is using more concrete per capita than ever. Global per capita consumption has about tripled in the last 40 years. The preparation of Portland cement involves heating a mixture of CaCO3, SiO2, and aluminosilicate (clay) at 1450 °C. This produces a solid mass called a clinker. Grinding the material with a little CaSO4 gives cement as a fine gray powder; its main components are CaO (60– 67%), SiO 2 (17–25%), Al 2O 3 (3–8%), and Fe 2O3 (0–6%). Mixing cement with sand and gravel produces concrete. The usefulness of cement in these products comes from the fact that when it is mixed

Calcium carbonate/ limestone (CaCO3)

About half of the CO2 emissions in this process comes from decomposition of CaCO3.

Sand (SiO2)

Clay/alumina silicate

High temperature required to prepare clinker releases about 900 kg of CO2 for each 1000 kg of cement.

Kiln (1450 °C)

Clinker

with water a series of not-well-understood reactions occur to produce a hard, strong, solid product that is ideal for use in construction. The high temperature required to prepare the clinker makes this process energy intensive. Furthermore, it releases about 800 kg of CO2 for each 1000 kg of cement, which is estimated to be about 8% of the anthropogenic emissions of CO2. (About half of the CO2 comes from decomposition of CaCO 3 to form CO 2 ; the other half is from fuel combustion.) Improving these figures would help address two current environmental concerns, energy consumption and greenhouse gas evolution.

Gypsum (CaSO4)

Grinder Cement

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21.6 Boron, Aluminum, and the Group 3A (13) Elements

Group 3A (13) Boron 5

B

10 ppm

Goal for Section 21.6

Aluminum 13

• Know the general chemical properties of the Group 3A (13) elements.

Al

With Group 3A (13), you see the first evidence of a change from metallic behavior of the elements at the left side of the periodic table to nonmetal behavior on the right side of the table. Boron is a metalloid, whereas all the other elements of Group 3A (13) are metals. The elements of Group 3A (13) vary widely in their relative abundances on Earth. Aluminum is the third most abundant element in Earth’s crust (82,000 ppm), whereas the other elements of the group are relatively rare, and except for boron, their compounds have limited commercial uses.

82,000 ppm Gallium 31

Ga

18 ppm Indium 49

In

0.05 ppm

Chemistry of the Group 3A (13) Elements

Thallium 81

One way to remember some of the chemistry of the elements is to recognize that there are chemical similarities between some elements diagonally situated in the periodic table. This diagonal relationship means that lithium and magnesium share some chemical properties, as do Be and Al, and B and Si. For example: •

Boric oxide, B2O3, and boric acid, B(OH)3, are weakly acidic, as are SiO2 and its acid, orthosilicic acid (H4SiO4). Boron–oxygen compounds, borates, are often chemically similar to silicon–oxygen compounds, silicates.

•

Both Be(OH)2 and Al(OH)3 are amphoteric, dissolving in both strong bases and strong acids (Figure 16.11).

•

Chlorides, bromides, and iodides of boron and silicon (such as BCl3 and SiCl4) react vigorously with water.

•

The hydrides of boron and silicon are simple, molecular species; are volatile and flammable; and react readily with water.

•

Beryllium hydride and aluminum hydride are colorless, nonvolatile solids that are extensively polymerized through BeOHOBe and AlOHOAl three-center, two-electron bonds (page 484).

Tl

0.6 ppm Element abundances are in parts per million in the Earth’s crust.

1A Li

Diagonal Relationship 2A 3A Be B Mg

Al

4A

Si

Diagonal Relationship The

chemistries of elements diagonally situated in the periodic table are often quite similar.

Finally, the Group 3A (13) elements are characterized by electron configurations of the type ns2np1. This means that each atom may lose three electrons to form compounds in which the element has a +3 oxidation number, although the heavier elements, especially thallium, also form compounds with an oxidation number of +1.

Boron Minerals and Production of the Element Although boron has a low abundance on Earth, its minerals are found in ­concentrated deposits. For example, large deposits of borax, Na2B4O7 ∙ 10 H2O, are currently mined in the Mojave Desert near the town of Boron, California. Isolation of pure, elemental boron from boron-containing minerals is difficult and is done only in small quantities. Like most metals and metalloids, boron can be obtained by chemically or electrolytically reducing an oxide or halide. Magnesium has often been used for chemical reductions, but the product of this reaction is noncrystalline boron of low purity. B2O3(s) + 3 Mg(s) n 2 B(s) + 3 MgO(s)

Another way to prepare elemental boron leads to an important industrial product: boron fibers. A boron halide, usually BCl3 in the gas phase, is reduced by hydrogen gas on tungsten fiber. 2 BCl3(g) + 3 H2(g) n 2 B(s) + 6 HCl(g)

21.6  Boron, Aluminum, and the Group 3A (13) Elements

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The resulting tungsten boride fiber is used in composites in aerospace structures and even in golf clubs and fishing poles. Boron has several allotropes, some characterized by having an icosahedron of boron atoms as a structural element (Figure 21.15). Elemental boron is a very hard and refractory (resistant to heat) semiconductor. In this regard, it differs from the other Group 3A (13) elements; Al, Ga, In, and Tl are all relatively low-melting, rather soft metals with high electrical conductivity.

Figure 21.15 Boron Icosahedron.  Several allotropes

Oesper Collection in The History of Chemistry/University of Cincinnati

of elemental boron have an icosahedron (a 20-sided polyhedron) of 12 covalently linked boron atoms as a structural element.

Charles Martin Hall (1863–1914)  Hall was only 22 years old when, in a woodshed behind the family home in Oberlin, Ohio, he worked out the electrolytic process for extracting aluminum from Al2O3. He went on to found a company that eventually became ALCOA, the Aluminum Corporation of America.

Metallic Aluminum and Its Production The low cost of aluminum and the excellent characteristics of its alloys with other metals (low density, strength, ease of handling in fabrication, and inertness toward corrosion, among others) have led to its widespread use. You probably know it best in the form of aluminum foil, aluminum cans, and parts of aircraft. Pure aluminum is soft and weak; moreover, it loses strength rapidly at temperatures higher than 300 °C. What you might refer to as aluminum is actually aluminum alloyed with small amounts of other elements to strengthen the metal and improve its properties. A typical alloy may contain about 4% copper with smaller amounts of silicon, magnesium, and manganese. Softer, more corrosionresistant alloys for window frames, furniture, highway signs, and cooking utensils may include only manganese. The standard reduction potential of aluminum [Al 3+(aq) + 3 e − n Al(s); E° = −1.66 V] tells you that aluminum is easily oxidized and so might be expected to be highly susceptible to corrosion. In fact, it is quite resistant to corrosion due to the formation of a thin, tough, and transparent skin of Al2O3 that adheres to the metal surface. However, if you penetrate the surface coating by scratching it or using a chemical agent, aluminum on the exposed surface can react immediately with oxygen or other oxidizing agent. In Figure 21.16, the surface is breached by chloride ions, and the exposed aluminum reacts rapidly with aqueous copper(II) ions. Aluminum was first prepared by reducing AlCl3 with sodium or potassium. This was a costly process, so aluminum was considered a precious metal in the nineteenth century. At the 1855 Paris Exposition, a sample of aluminum was exhibited along with the crown jewels of France. In 1886, in an interesting coincidence, Frenchman Paul Heroult (1863–1914) and American Charles Martin Hall (1863–1914) simultaneously and independently conceived of the less expensive electrochemical method used today. The Hall–Heroult electrolysis method for the industrial production of aluminum bears the names of the two discoverers.

Copper metal collects on the surface of the aluminum.

Photos: © Charles D. Winters/Cengage

The reaction is rapid and so exothermic that the water can boil on the surface of the foil. The blue color of aqueous Cu2+ ions fades as they are consumed in the reaction.

(a) A ball of aluminum foil is added to a solution of copper(II) nitrate and sodium chloride. Normally, the coating of chemically inert Al2O3 on the surface of aluminum protects the metal from further oxidation.

(b) In the presence of the Cl− ion, the coating of Al2O3 is breached, and aluminum reduces copper(II) ions to copper metal in an exothermic reaction.

Figure 21.16  Corrosion of aluminum.

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Aluminum is found in nature as aluminosilicates, minerals such as clay that are based on aluminum, silicon, and oxygen. As these minerals weather, they break down to various forms of hydrated aluminum oxide, Al2O3 ∙ n H2O, called bauxite. Mined in huge quantities, bauxite is the raw material from which aluminum is obtained. The first step is to purify the ore, separating Al2O3 from iron and silicon oxides. This is done by the Bayer process, which relies on the amphoteric, basic, or acidic nature of the various oxides. (Silica, SiO2, is an acidic oxide, whereas Al2O3 is amphoteric, and Fe2O3 is a basic oxide.) Silica and Al2O3 dissolve in a hot concentrated solution of caustic soda (NaOH), leaving insoluble Fe2O3 to be filtered out.

Bauxite  This mineral is named for the village of Les Baux-deProvence in southern France. The picturesque village is on a rocky outcrop topped by a ruined castle. The mineral was first discovered near the village in 1821.

Al2O3(s) + 2 NaOH(aq) + 3 H2O(ℓ) n 2 Na[Al(OH)4](aq) SiO2(s) + 2 NaOH(aq) + 2 H2O(ℓ) n Na2[Si(OH)6](aq)

If a solution containing aluminate and silicate anions is treated with CO2, Al2O3 precipitates, and the silicate ion remains in solution. Recall that CO2 is an acidic oxide that forms the weak acid H2CO3 in water, so the Al2O3 precipitation in this step is an acid–base reaction. H2CO3(aq) + 2 Na[Al(OH)4](aq) n Na2CO3(aq) + Al2O3(s) + 5 H2O(ℓ)

Metallic aluminum is obtained from purified bauxite by electrolysis (­Figure 21.17). Bauxite is first mixed with cryolite, Na3AlF6, to give a lower-melting mixture (melting temperature = 980 °C) that is electrolyzed in a cell with graphite electrodes. The cell operates at a relatively low voltage (4.0–5.5 V) but with an extremely high current (50,000–150,000 A). Aluminum is produced at the cathode and oxygen at the anode. To produce 1 kg of aluminum requires 13 to 16 kilowatthours of energy plus the energy required to maintain the high temperature.

Boron Compounds Borax, Na2B4O7 ∙ 10 H2O, is the most important boron–oxygen compound and is the form of the element most often found in nature (Figure 21.18). It has been used for centuries in metallurgy because of the ability of molten borax to dissolve other metal oxides. Because of this, borax is used as a flux that cleans the surfaces of metals to be joined and thereby permits a good metal-to-metal contact.

Frozen electrolyte crust

+ Graphite anode

Carbon lining

Oxygen is produced at a graphite anode, and the gas reacts slowly with the carbon to give CO2, leading to eventual destruction of the electrode.

Al2O3 in Na3AlF6(ℓ) Molten Al

Cathode(−)

The aluminum-containing substances are reduced at a graphite cathode to give molten aluminum.

Figure 21.17  Industrial production of aluminum.  ​Purified aluminum-containing ore (bauxite) is essentially Al2O3. In the Hall–Heroult process this is mixed with cryolite (Na3AlF6) and other fluorides such as AlF3. (The additives serve to lower the melting point of the mixture and increase the conductivity.)

21.6  Boron, Aluminum, and the Group 3A (13) Elements

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Figure 21.18  Borax, the mineral source of elemental boron.

© Charles D. Winters/Cengage

B atom surrounded by 4 electron pairs

B atom surrounded by 3 electron pairs

(a) The structure of the ion [B4O5(OH)4]2− found in borax.

(b) The mineral named ulexite, a form of crystalline borax, Na2B4O7 ∙ 10 H2O.

The formula for borax gives little information about its structure. The anion is better described by the formula [B4O5(OH)4]2−, the structure of which illustrates two commonly observed structural features in inorganic chemistry. First, many minerals consist of MOn groups where two atoms share O atoms. Second, this sharing of O atoms between two metals or metalloids often leads to MO rings. Borax can be treated with sulfuric acid to produce boric acid, B(OH)3, Na2B4O7 ∙ 10 H2O(s) + H2SO4(aq) n 4 B(OH)3(aq) + Na2SO4(aq) + 5 H2O(ℓ)

an acid that exhibits both Lewis and Brønsted acid behavior. Hydronium ions are produced by a Lewis acid–base interaction between boric acid and water.

NICK FIELDING/Alamy Stock Photo

H

Boron carbide in fireworks Barium

salts and chlorinated organic compounds are frequently used in fireworks to produce a green color. The problem is that these ingredients are toxic. Fortunately, a solution was recently found: boron carbide, B4C, gives off an intense green color when combined with an oxidizing agent such as KNO3 at high temperatures.

O H

OH

OH B (aq) HO OH

HO

B

OH OH

−

(aq) + H3O+(aq)

Ka = 7.3 × 10−10 Because of its weak acid properties and slight biological activity, boric acid has been used for many years as an antiseptic. Furthermore, because boric acid is a weak acid, salts of borate ions, such as the [B4O5(OH)4]2− ion in borax, are weak bases. Boric acid is dehydrated to boric oxide when strongly heated. 2 B(OH)3(s) n B2O3(s) + 3 H2O(ℓ)

By far the largest use for the oxide is in the manufacture of borosilicate glass (the type of glass used in most laboratory glassware). This type of glass is composed of 76% SiO2, 13% B2O3, and much smaller amounts of Al2O3 and Na2O. The presence of boric oxide gives the glass a higher softening temperature, imparts a better resistance to attack by acids, and limits the expansion of the glass upon heating. This last property makes borosilicate glass less prone to breakage when heated or cooled. Like its metalloid neighbor silicon, boron forms a series of molecular compounds with hydrogen. Because boron is slightly less electronegative than hydrogen, these compounds are described as hydrides, in which the H atoms bear a partial negative charge. More than 20 neutral boron hydrides, or boranes, with the general formula BxHy are known. The simplest of these is diborane, B2H6, a colorless, gaseous compound with a boiling point of −92.6 °C. This molecule is described as electron deficient because there are not enough electrons to attach all of the atoms using two-electron bonds. Instead, the description of bonding uses three-center two-electron bonds in the BOHOB bridges (page 484).

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133 pm

Diborane has an endothermic enthalpy of formation (∆fH° = +41.0 kJ/mol), so its combustion in air (to give boric oxide and water) is extremely exothermic.

119 pm

B2H6(g) + 3 O2(g) n B2O3(s) + 3 H2O(ℓ)  ∆rH° = −2170.4 kJ/mol-rxn

Because of this, diborane and other boron hydrides were once considered as possible rocket fuels. However, this idea was abandoned because they are difficult to handle. Furthermore, one combustion product is solid B 2O 3, which means a rocket engine could spew massive amounts of white, powdered B 2O3 into the environment. Diborane can be synthesized from sodium borohydride, NaBH4, the only BOH compound produced in ton quantities.

H B H

H

H 97°

B

H

122°

H

2 NaBH4(s) + I2(s) n B2H6(g) + 2 NaI(s) + H2(g)

Sodium borohydride, NaBH4, a white, crystalline, water-soluble solid, is made from NaH and borate esters such as B(OCH3)3.

Diborane, B2H6, the simplest member of a large family of boron hydrides.

4 NaH(s) + B(OCH3)3(g) n NaBH4(s) + 3 NaOCH3(s)

The main use of NaBH4 is as a reducing agent in organic synthesis (Chapter 23).

Aluminum Compounds Aluminum is an excellent reducing agent, so it reacts readily with hydrochloric acid (forming Al3+ ions and H2). In contrast, it does not react with nitric acid because the acid oxidizes the surface of aluminum and produces a film of Al2O3 that protects the metal from further attack. In fact, this means nitric acid can be shipped in aluminum tanks. Various salts of aluminum dissolve in water, giving the hydrated Al 3+(aq) ion, which is a weak Brønsted acid (Table 16.2). Adding acid shifts the equilibrium to the left, whereas adding base causes the equilibrium to shift to the right. Addition of sufficient hydroxide ion results ultimately in precipitation of the hydrated oxide Al2O3 ∙ 3 H2O. Aluminum oxide, Al2O3, formed by dehydrating the hydrated oxide, is quite insoluble in water and generally resistant to chemical attack. In the crystalline form, aluminum oxide is known as corundum (Figure 21.19). This material is extraordinarily hard, a property that leads to its use as an abrasive in grinding wheels, “sandpaper,” and toothpaste. Some gems are impure aluminum oxide. Rubies, beautiful red crystals (page 94) prized for jewelry and used in some lasers, are composed of Al2O3 contaminated with a small amount of Cr3+. The Cr3+ ions replacing some of the Al3+ ions in the crystal lattice are the source of the red color. Synthetic rubies were first made in 1902, and the worldwide capacity is now several hundred thousand kg/ year; much of this production is used for jewel bearings in watches and instruments. Blue sapphires consist of Al2O3 with Fe2+ and Ti4+ impurities in place of Al3+ ions. Boron forms halides such as gaseous BF3 and BCl3 that have the expected trigonal planar molecular geometry of halogen atoms surrounding an sp2 hybridized boron atom. In contrast, the aluminum halides are all solids and have different structures. Aluminum bromide, which is made by the very exothermic reaction of aluminum metal and bromine (Figure 21.20), 2 Al(s) + 3 Br2(ℓ) n Al2Br6(s)

has the formula Al2Br6. The structure resembles that of diborane in that bridging atoms appear between the two Al atoms. However, unlike B2H6, Al2Br6 is not electron deficient; each bridge is formed when a Br atom on one AlBr3 uses a lone pair



© Charles D. Winters/Cengage

[Al(H2O)6]3+(aq) + H2O(ℓ) uv [Al(H2O)5(OH)]2+(aq) + H3O+(aq)

Figure 21.19 Corundum, Al2O3.  ​Corundum is a crystalline form of aluminum oxide. Both rubies and sapphires are a form of corundum in which a few Al3+ ions have been replaced by ions such as Cr3+, Fe2+, or Ti4+.

21.6  Boron, Aluminum, and the Group 3A (13) Elements

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© Charles D. Winters/Cengage

Aluminum metal foil and liquid bromine, Br2.

The reaction produces white solid Al2Br6. The heat of the reaction is so high that any aluminum in excess melts.

When the foil is added to the bromine a vigorous reaction occurs.

Figure 21.20  Reaction of aluminum metal and bromine to give Al2Br6.

to form a coordinate covalent bond to a neighboring tetrahedral, sp3-­hybridized aluminum atom.

Br Br

93°

Al 221 pm

233 pm

Br

Al Br

Br Br

115°

87°

A Closer Look

The structure of Al2Br6.  The bonding in Al2Br6 is not unique to the aluminum halides. Metal–halogen–metal bridges are found in many other metal–halogen compounds.

Complexity in Boron Chemistry

In addition to commercially important boron compounds like boric acid and borax, chemists who study boron compounds have synthesized a wide range of materials. A chemist working in this field said “no other p-block elements shows a similar complexity of chemical bonding, which is well displayed within the cluster chemistry of boron hydrides and carboranes.” The reaction of boric acid and ammonia in a nitrogen atmosphere at high temperature gives h-BN, hexagonal boron nitride, with a structure very similar to that of graphite. The material, which is stable up to 900 ˚C, even in air, is a lubricant like graphite, but it is even more useful. Graphite requires water or gas molecules trapped between the layers to be an effective lubricant, but h-BN does not. It can be used, therefore, in space applications. Boron nitride also has commercial uses: h-BN is found in nearly all cosmetics (foundations, eye shadows, lipsticks, and so on) as well as in pencil leads and dental cements. It has also been used as a lubricant for bullets and the barrels of precision target rifles.

Hexagonal boron nitride, h-BN.  There are layers of interlaced BN rings. In h-BN the B atom of one layer lies above the N of a lower layer. In contrast, in graphite, the C atom of one layer is over the middle of the C6 ring in the lower layer (Figure 12.18).

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Both aluminum bromide and aluminum iodide have this structure. Aluminum chloride has a different solid state structure, but it exists as dimeric molecules in the vapor state. Aluminum fluoride, in contrast, has an ionic lattice structure, built from Al3+ and F− ions. Aluminum chloride can react with a chloride ion to form the tetrahedral anion [AlCl4]−. Aluminum fluoride, in contrast, can accommodate three additional F− ions to form an octahedral [AlF6]3− ion. This is the anion found in cryolite, Na3AlF6, the compound added to aluminum oxide in the electrolytic production of aluminum metal. Apparently, the Al3+ ion can bind to six of the smaller F− ions, whereas only four of the larger Cl−, Br−, or I− ions can surround an Al3+ ion.

21.7 Silicon and the Group 4A (14) Elements

The octahedral [AlF6]3− anion.  This is the anion in the mineral cryolite.

Goals for Section 21.7 • Know how elemental silicon is produced. • Know the general chemical properties of the Group 4A (14) elements. The elements of Group 4A (14) have a broad range of chemical behavior: carbon is a nonmetal; silicon and germanium are metalloids; and tin and lead are metals. The Group 4A (14) elements are characterized by half-filled valence shells with two electrons in the ns orbital and two electrons in np orbitals. The bonding in carbon and silicon compounds is largely covalent and involves sharing four electron pairs with neighboring atoms. In germanium compounds, the +4 oxidation state is common (GeO2 and GeCl4), but some +2 oxidation state compounds exist (GeI2). Oxidation numbers of both +2 and +4 are common in compounds of tin and  lead (such as SnCl 2 , SnCl 4 , PbO, and PbO 2 ). Oxidation numbers two units less than the group number are often encountered for heavier elements in Groups 3A–7A (13–17).

Silicon

Group 4A (14)

Silicon is second after oxygen in abundance in the Earth’s crust, so it is not surprising we are surrounded by silicon-containing materials: bricks, pottery, porcelain, lubricants, sealants, computer chips, and solar cells. The modern technological revolution is based on the semiconducting properties of silicon (Section 12.4). Reasonably pure silicon can be made in large quantities by heating pure silica sand with purified coke to approximately 3000 °C in an electric furnace.

480 ppm

SiO2(s) + 2 C(s) n Si(ℓ) + 2 CO(g)

277,000 ppm

The molten silicon is drawn off the bottom of the furnace and allowed to cool to a shiny blue-gray solid. Because extremely high-purity silicon is needed for the electronics industry, purifying raw silicon requires additional steps. First, the silicon in the impure sample is treated with chlorine to convert the silicon to liquid silicon tetrachloride.

Germanium 32

Si(s) + 2 Cl2(g) n SiCl4(ℓ)

C

Silicon 14

Si

Ge

1.8 ppm Tin 50

Sn

Silicon tetrachloride (boiling point of 57.6 °C) is carefully purified by distillation and then reduced to silicon with magnesium. SiCl4(g) + 2 Mg(s) n 2 MgCl2(s) + Si(s)

2.2 ppm Lead 82

Pb

The by-product, magnesium chloride, is washed out with water, and the silicon is remelted and cast into bars. A final purification is carried out by zone refining, a process in which a special heating device is used to melt a narrow segment of the silicon rod. As the heater is moved slowly along the rod, impurities contained in

Carbon 6

14 ppm Element abundances are in parts per million in the Earth’s crust.

21.7  Silicon and the Group 4A (14) Elements

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1065

Impure material

Molten zone © Charles D. Winters/Cengage

Heater moves slowly upward Purified silicon

(a) To purify silicon, a rod of impure material is heated in an inert atmosphere by a circular heater that travels slowly along the rod. The molten zone is a solution of silicon and impurities. This solution has a lower melting point than the pure element, so the pure element crystallizes. The impurities stay in the molten zone and so travel along the rod with the heater.

(b) A rod of silicon after purification. After cutting off the ends where impurities have accumulated, thin wafers of silicon are cut from the bars and are the basis for the semiconducting chips in computers and other devices. (Colors on the surface of the bar are from light refraction by a very thin surface film of silicon dioxide.)

Figure 21.21  Pure silicon by zone refining.

the silicon tend to remain in the liquid phase. The silicon that crystallizes from the heated zone is therefore of a higher purity (Figure 21.21).

Silicon Dioxide

Synthetic quartz.  These crystals were grown from silica in sodium hydroxide. The colors come from added Co2+ ions (blue) or Fe2+ ions (brown).

John C. Kotz

© Charles D. Winters/Cengage

The simplest oxide of silicon is SiO2, commonly called silica, a constituent of many rocks such as granite and sandstone. Quartz is a pure crystalline form of silica, but impurities in quartz produce gemstones such as amethyst (Figure 21.22). Silica and CO2 are oxides of two elements in the same chemical group, so similarities between them might be expected. However, SiO2 is a high-melting solid (quartz melts at 1610 °C), whereas CO2 is a gas at room temperature and 1 atm. This great disparity arises from the different structures of the two oxides. Carbon dioxide is a molecular compound, with the carbon atom linked to each oxygen

Natural quartz.  Pure quartz is colorless, but impurities such as iron add color to produce, for example, purple amethyst.

Quartz, a network solid.  Each Si atom is bound tetrahedrally to four O atoms, each linked to another Si atom.

Figure 21.22  Various forms of quartz.

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atom by a double bond. In contrast, SiO2 is a network solid, which is the preferred structure because the bond energy of two SiPO double bonds is much less than the bond energy of four Si—O single bonds. The contrast between SiO2 and CO2 is an example of a more general phenomenon. Multiple bonds, often encountered between second period elements, are rare among elements in the third period and beyond. There are many compounds with multiple bonds to carbon but very few compounds featuring multiple bonds to silicon. Quartz crystals are used to control the frequency of radio and television transmissions. Because these and related applications use so much quartz, there is not enough natural quartz to fulfill demand, and quartz must be synthesized. Noncrystalline, or vitreous, quartz, made by melting pure silica sand, is placed in a steel bomb, and dilute aqueous NaOH is added. A seed crystal is placed in the mixture, just as you might use a seed crystal in a hot sugar solution to grow rock candy. When the mixture is heated to a temperature above 400 °C at a pressure of about 1700 atm (the critical point of water occurs at 374 °C and 218 atm) over a period of days, pure quartz crystallizes. Silicon dioxide also dissolves slowly in hot, molten NaOH or Na2CO3 to give Na4SiO4, sodium silicate.

Glass, a silicon-oxygen network  Window glass, drinking glasses, bottles, and laboratory glassware are all networks of silicon and oxygen. “A Closer Look: Glass,” page 605.

Forms of sodium silicate  Sodium silicate is the name given to several variations of sodium salts of silicate anions: SiO32− is the metasilicate anion; SiO44− is orthosilicate; Si2O76− is the pyrosilicate anion. The metaand pyrosilicate anions are polymers of Si and O.

SiO2(s) + 2 Na2CO3(ℓ) n Na4SiO4(s) + 2 CO2(g)

Silicate Minerals with Chain and Ribbon Structures The structure and chemistry of silicate minerals is an enormous topic in geology and chemistry. Although all silicates are built from tetrahedral SiO4 units, they have different properties and a wide variety of structures because of the way these tetrahedral SiO4 units link together. The simplest silicates, orthosilicates, contain SiO44− anions. The 4− charge of the anion is balanced by four M+ ions, two M2+ ions, or a combination of ions. Olivine, an important mineral in the Earth’s mantle, contains Mg2+ and Fe2+, with the Fe2+ ion giving the mineral its characteristic olive color, and gem-like zircons are ZrSiO4. Calcium orthosilicate, Ca2SiO4, is a component of Portland cement (“A Closer Look: Cement—The Second Most Used Substance,” page 1058). Pyroxenes are minerals with the empirical formula (MSiO3)x, where M is a divalent metal cation such as Mg21. They are characterized by having a chain of SiO 4 tetrahedra as the basic structural unit.

A chain of linked SiO4 tetrahedra.

© Charles D. Winters/Cengage

After the molten mixture has cooled, hot water is added under pressure. This partially dissolves the material to give a solution of sodium silicate. After filtering off insoluble sand or glass, the solvent is evaporated to leave sodium silicate, called water glass. The biggest single use of this material is in household and industrial detergents because a sodium silicate solution can maintain pH by its buffering ability. Additionally, sodium silicate is used in various adhesives and binders, especially for gluing corrugated cardboard boxes. If sodium silicate is treated with acid, a gelatinous precipitate of SiO2 called silica gel is obtained. Washed and dried, silica gel is a highly porous material with many uses. It is a drying agent, readily absorbing up to 40% of its own weight of water. Small packets of silica gel are often placed in packing boxes of merchandise. The material is frequently stained with (NH4)2CoCl4, a humidity detector that is pink when hydrated and blue when dry.

Asbestos, a Silicate.  The asbestos minerals are fibrous silicates. The family includes the minerals actinolite, amosite, anthophyllite, chrysotile (photo above), crocidolite, and tremolite. About 90% of the asbestos used in the United States is chrysotile. More than 50 countries have banned the use of asbestos because of its link to mesothelioma, a lung cancer. There have been significant efforts in the U.S. to clean up places where it has been used in insulation materials, floor tiles, brake linings, and cement sheeting.

21.7  Silicon and the Group 4A (14) Elements

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1067

“Books” of mica are stacks of silicate sheets.

John C. Kotz

Six-member rings of Si atoms with O atoms in edges are a common structural element in silicates.

Figure 21.23  Mica, a sheet silicate.  ​The sheet-like molecular structure of mica explains its physical appearance. As in the pyroxenes, each silicon is bonded to four oxygen atoms, but the Si and O atoms form a sheet of six-member rings of Si atoms with O atoms in each edge. The ratio of Si to O in this structure is 1 to 2.5. A formula of SiO2.5 requires a positive ion, such as Na+, to counterbalance the charge. Thus, mica and other sheet silicates, and aluminosilicates such as talc and many clays, have positive ions between the sheets.

If two such chains are linked together by sharing oxygen atoms, the result is an amphibole, of which some asbestos minerals are one example. The molecular chain makes asbestos a fibrous material.

Silicates with Sheet Structures and Aluminosilicates

Remedy for stomach upset. One of the ingredients in Kaopectate is kaolin, one form of clay. The large white disc is clay purchased in a market in Ghana, West Africa. It is eaten for an upset stomach, a practice widespread round the world.

Figure 21.24  Clay, an aluminosilicate.

Kaolinite clay. The basic structural feature of many clays, and kaolinite in particular, is a sheet of SiO4 tetrahedra (black and red spheres) bonded to a sheet of AlO6 octahedra (gray and green spheres).

© Charles D. Winters/Cengage

© Charles D. Winters/Cengage

Linking many silicate chains together produces a sheet of SiO 4 tetrahedra (­Figure 21.23). This sheet is the basic structural feature of some of Earth’s most important minerals, particularly the clay minerals (such as china clay), mica, talc, and the chrysotile form of asbestos. However, these minerals do not contain just silicon and oxygen. Rather, they are often referred to as aluminosilicates because they frequently have Al3+ ions in place of some Si4+ ions (which means that other positive ions such as Na+, K+, and Mg2+ must also be present in the lattice to balance the net negative and positive charges). In kaolinite clay, for example, the sheet of SiO4 tetrahedra is bonded to a sheet of AlO6 octahedra (Figure 21.24). Another example is muscovite, a form of mica. Aluminum ions have replaced some Si 4+ ions, and there are charge-balancing K+ ions, so it is best represented by the formula KAl2(OH)2(Si3AlO10). There are some interesting uses of clays, one being in medicine (Figure 21.24). In certain cultures, clay is eaten for medicinal purposes. Several remedies for the

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© Charles D. Winters/Cengage

Igor Petrushenko/Shutterstock.com

Figure 21.25  Zeolites are aluminosilicates.

Zeolite structure. Each edge consists of Si—O—Si, Al—O—Si, or Al—O—Al bonds. The channels in the framework can selectively capture ions or small molecules.

Apophyllite, a crystalline zeolite.

relief of upset stomach contain highly purified clays that absorb excess stomach acid as well as potentially harmful bacteria and their toxins by exchanging the intersheet cations in the clays for the toxins, which are often organic cations. Other aluminosilicates include the feldspars, common minerals that make up about 60% of the Earth’s crust, and zeolites (Figure 21.25). Both materials are composed of SiO4 tetrahedra in which some of the Si atoms have been replaced by Al atoms, along with alkali and alkaline earth metal ions for charge balance. The main feature of zeolite structures is their regularly shaped tunnels and cavities. Hole diameters are between 300 and 1000 pm, and small molecules such as water can fit into the cavities of the zeolite structure. As a result, zeolites can be used as drying agents to selectively absorb water from air or a solvent. Small amounts of zeolites are often sealed into multipane windows to keep the air dry between the panes. Zeolites are also used as catalysts. For example, methanol, CH3OH, is converted to gasoline in the presence of specially tailored zeolites. In addition, zeolites are added to detergents, where they function as water-softening agents because the sodium ions of the zeolite are exchanged for Ca2+ ions in hard water, effectively removing Ca2+ ions from the water. Finally, you will find them in consumer products that remove odor-causing molecules from air.

Silicone Polymers Silicon and chloromethane (CH3Cl) react at 300 °C in the presence of a catalyst, copper powder. The primary product of this reaction is (CH3)2SiCl2. Si(s) + 2 CH3Cl(g) n (CH3)2SiCl2(ℓ)

Halides of Group 4A (14) elements other than carbon hydrolyze readily, so the reaction of (CH3)2SiCl2 with water initially produces (CH3)2Si(OH)2. On standing, these molecules combine to form a polymer by eliminating water. The polymer is called polydimethylsiloxane, a member of the silicone family of polymers (Figure 21.26). (CH3)2SiCl2 + 2 H2O n (CH3)2Si(OH)2 + 2 HCl n (CH3)2Si(OH)2 n [O(CH3)2SiOO]n + n H2O

21.7  Silicon and the Group 4A (14) Elements

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1069

CH3

Figure 21.26  A silicone polymer.  ​This silicone polymer has the formula (R2SiO)x, where R

represents an organic group such as CH3 (shown here as a black sphere). In this structure, the silicon atoms have tetrahedral geometry, bonding to two O atoms and two organic groups. Groups other than CH3 can be used, and chains can be cross-linked.

Silicone polymers are nontoxic and have good stability to heat, light, and oxygen; they are chemically inert and have valuable antistick and antifoam properties. They can take the form of oils, greases, and resins. Some have rubber-like properties (Silly Putty, for example, is a silicone polymer). Over 2 million tons of silicone polymers are made worldwide annually. These materials are used in a wide variety of products: lubricants, peel-off labels, lipstick, suntan lotion, car polish, and building caulk.

The Heavier Elements of Group 4A (14): Ge, Sn, and Pb

Lead in Drinking Water  See page 1260 for a description of the problem with lead in drinking water in Flint, Michigan.

Just as silicon chemistry is dominated by oxides, so too is the chemistry of the heavier elements in Group 4A (14). For example, the source of tin is an oxide, the mineral cassiterite, SnO2. Thousands of tons are used in the ceramics industry in glazes, enamels, and pigments. A yellow glaze, for example, is a mixture of oxides, SnO2 and V2O5. Lead(II) oxide, PbO, is a red solid at room temperature. The red form is commonly called litharge, but above 488 ˚C a yellow form (called massicot) with a different unit cell is stable. PbO is the most important lead compound, one large use being in the manufacture of lead glass, a glass with a high index of refraction that gives it a special brilliance. The other major use for PbO is in lead storage batteries (Figure 19.11). The oxide is applied as a paste in sulfuric acid to lead battery plates. The positive plate ­(cathode) is prepared by oxidizing PbO to lead(IV) oxide, PbO2. The negative plate (anode) is elemental lead. If lead(II) oxide is heated in air, it forms red lead, Pb3O4, which is best formulated as a mixture of lead(II) and lead(IV) oxides. It was once widely used in rustinhibiting paint and primers for iron and steel. Other lead compounds that were used in paints include brilliant yellow PbCrO4, which was used to paint the yellow lines on roadways. Finally, a mixed carbonate, 2 PbCO3·Pb(OH)2, is a white pigment. Paints with lead-containing pigments were useful because they dried faster, resisted moisture, and were durable. But lead is known to be toxic, and lead-based paints were banned in the United States in 1978. Nonetheless, they are used in other countries and remain in our environment (“Applying Chemical Principles 21.1: Lead in the Environment”).

21.8 Nitrogen, Phosphorus, and the Group 5A (15) Elements Goal for Section 21.8 • Know common compounds of nitrogen and phosphorus, particularly the oxides. Group 5A (15) elements are characterized by the ns2np3 configuration with its halffilled np subshell. Although nitrogen in its compounds displays a range of oxidation

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Group 5A (15)

numbers from −3 to +5, the primary oxidation numbers for the heavier Group 5A (15) elements are +3 and +5. Once again, as in Groups 3A (13) and 4A (14), the lower oxidation state is more common in the heavier elements than in the lighter ones. In many arsenic, antimony, and bismuth compounds, the element has an oxidation number of +3. Compounds of these heavier elements with oxidation numbers of +5 are powerful oxidizing agents. This section concentrates on the chemistries of nitrogen and phosphorus. Nitrogen is found primarily as N2 in the atmosphere, where it constitutes 78.1% by volume (75.5% by weight). In contrast, phosphorus occurs in Earth’s crust in a wide variety of solids. More than 200 different phosphorus-containing minerals are known, and all contain the tetrahedral phosphate ion, PO43−, or a derivative of this ion. By far, the most abundant phosphorus-containing minerals are apatites such as Ca5X(PO4)3 (where X can be OH–, F–, Cl–, or some other anion). Nitrogen and its compounds play a key role in modern economies, with ammonia being particularly important. Phosphoric acid is an important commodity chemical, and it finds its greatest use in fertilizers. Both phosphorus and nitrogen are part of every living organism. Phosphorus is contained in nucleic acids and phospholipids, and nitrogen occurs in proteins and nucleic acids (see Chapter 24).

Nitrogen 7

N

25 ppm Phosphorus 15

P

1000 ppm Arsenic 33

As

1.5 ppm Antimony 51

Sb

0.2 ppm Bismuth 83

Bi

0.048 ppm

Properties of Elemental Nitrogen and Phosphorus

Element abundances are in parts per million in the Earth’s crust.

Nitrogen (N2) is a colorless gas that liquefies at 77 K (−196 °C) (Figure 11.1). Its most notable feature is its reluctance to react with other elements or compounds because the NqN triple bond has a large bond-dissociation enthalpy (945 kJ/ mol) and because the molecule is nonpolar. Nitrogen does, however, react with hydrogen to give ammonia in the presence of a catalyst (“Applying Chemical Principles 15.1: ​Applying Equilibrium Concepts—The Haber–Bosch Ammonia Process,” page 762) and with a few metals (notably lithium and magnesium) to give metal nitrides, compounds containing the N3− ion. 3 Mg(s) + N2(g) 87n Mg3N2(s) Elemental nitrogen is very useful. Because of its lack of reactivity, it can provide a nonoxidizing atmosphere for packaged foods and wine and is used to pressurize electric cables and telephone wires. Liquid nitrogen (bp, –196 ˚C) is valuable as a coolant in freezing biological samples such as blood and semen and in freezedrying food. Dermatologists use it to remove pre-cancerous growths from skin by freezing a tiny patch of skin. Elemental phosphorus was first derived from human waste (“A Closer Look: Alchemists Making Phosphorus,” page 1072), but it is now produced by the reduction of phosphate ­minerals in an electric furnace. 2 Ca3(PO4)2(s) + 10 C(s) + 6 SiO2(s) n P4(g) + 6 CaSiO3(s) + 10 CO(g)

The phosphorus vapor is cooled under water to prevent its spontaneous combustion and eventually yields the solid white waxy allotrope. Although white ­phosphorus is defined as the standard state of the element, it is the least stable form thermodynamically. Rather than occurring as a diatomic molecule with a triple bond, like its ­second-period neighbor nitrogen (N2), white phosphorus is composed of tetrahedral P4 molecules in which each P atom is joined to three others by single bonds.

white phosphorus, P4

© Charles D. Winters/Cengage

magnesium nitride

The red and white allotropes of phosphorus. ​White phosphorus is stored under water; this form of phosphorus burns when exposed to air. In contrast, red phosphorus is stable in air.

polymeric red phosphorus 21.8  Nitrogen, Phosphorus, and the Group 5A (15) Elements

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Compound and Oxidation Number of N Ammonia, −3

Hydrazine, −2

The red allotrope of phosphorus, a polymer of P4 units, is made by heating the white form out of contact with air. It is much less reactive than the white form and is easier to handle. If this allotrope is heated under pressure, it eventually produces the thermodynamically most stable allotrope, black phosphorus, which is a complex polymer of phosphorus atoms.

Nitrogen Compounds

Dinitrogen, 0

A notable feature of the chemistry of nitrogen is the wide diversity of its compounds. Compounds are known with nitrogen in all oxidation numbers between −3 and +5 (Figure 21.27).

Dinitrogen monoxide, +1

Hydrogen Compounds of Nitrogen: Ammonia and Hydrazine

Nitrogen monoxide, +2

Ammonia is a gas at room temperature and pressure. It has a very penetrating odor and condenses to a liquid at −33 °C under 1 atm of pressure. Aqueous solutions of ammonia, often referred to as ammonium hydroxide, are basic due to the reaction of ammonia with water (Figure 3.13). NH3(aq) + H2O(ℓ) uv NH4+(aq) + OH−(aq)  Kb = 1.8 × 10−5 at 25 °C

Nitrogen dioxide, +4

Nitric acid, +5

Figure 21.27 Compounds and oxidation numbers for nitrogen.  ​In its compounds, the N atom can have oxidation states ranging from −3 to +5.

Ammonia is a major industrial chemical and is prepared by the Haber-Bosch process (page 762), largely for use as a fertilizer. Hydrazine, N2H4, is a colorless, fuming liquid with an ammonia-like odor (mp, 2.0 °C; bp, 113.5 °C). Almost 1 million kg of hydrazine are produced annually by several processes, one of the important avenues being the Raschig process. In one variation of this method, ammonia reacts with alkaline sodium hypochlorite in the presence of a catalyst. 2 NH3(aq) + NaClO(aq) uv N2H4(aq) + NaCl(aq) + H2O(ℓ)

Hydrazine, like ammonia, is a base,

Alchemists Making Phosphorus He stoked his small furnace with more charcoal and pumped the bellows until his retort glowed red hot. S ­ uddenly something strange began to happen. Glowing fumes filled the vessel and from the end of the retort dripped a shining liquid that burst into flames. J. Emsley, The 13th Element, p. 5, New York: John Wiley, 2000.

John Emsley begins his story of phosphorus, its discovery, and its uses, by imagining what the German alchemist Hennig Brandt must have seen in his laboratory that day in 1669. He was in search of the philosopher’s stone, the magic elixir that would turn the crudest substance into gold. (Some may recall that the first Harry Potter novel was titled Harry Potter and the Philosopher’s Stone when it was published in Great Britain.) Brandt was experimenting with urine, which had served as the source of useful

chemicals since Roman times. It is not surprising that phosphorus could be extracted from this source. Humans consume much more phosphorus, in the form of phosphate, than they require, and the excess phosphorus (about 1.4  g∕day) is excreted in the urine. It is nonetheless extraordinary that Brandt was able to isolate the element. According to an eighteenth-century chemistry book, about 30 g of phosphorus could be obtained from 60 gallons of urine. And the process was not simple. Another eighteenthcentury recipe states that “50 or 60 pails full” of urine was to be used. “Let it lie steeping . . . till it putrefy and breed worms.” The chemist was then to reduce the volume until a paste was obtained and finally to heat the paste very strongly in a retort, a piece of chemical glassware used in some distillations. After some days, phosphorus distilled from the mixture and was collected in water. (Modern

chemists know now that carbon from the organic compounds in the urine reduces phosphate ions to elemental phosphorus.) Phosphorus was made in this manner for more than 100 years.

© Charles D. Winters/Cengage

A Closer Look

N2H4(aq) + H2O(ℓ) uv N2H5+(aq) + OH−(aq)  Kb = 8.5 × 10−7

The glow of phosphorus burning in air.

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but it is also a strong reducing agent, as reflected in the reduction potential for the following half-reaction in basic solution: N2(g) + 4 H2O(ℓ) + 4 e− n N2H4(aq) + 4 OH−(aq)  E° = −1.15 V

Hydrazine’s reducing ability is exploited in its use in treating wastewater from chemical plants. It removes oxidizing ions such as CrO 42− by reducing them, thus preventing them from entering the environment. A related use is the treatment of water boilers in large electric-generating plants. Oxygen dissolved in the water is a serious problem in these plants because the dissolved gas can corrode the metal of the boiler and pipes. Hydrazine reduces the amount of dissolved oxygen in water. N2H4(aq) + O2(g) n N2(g) + 2 H2O(ℓ)

Most hydrazine is used as a precursor to what is known in industry as a blowing or foaming agent. These are compounds that decompose to a variety of gases, which are then blown through a liquid to create a foam. The main example is in the manufacture of polystyrene foams. Hydrazine is used to make blowing agents such as azodicarbonamide (C2H 4N 4O 2), which decomposes to N2, CO, CO2, and NH3.

Oxides and Oxoacids of Nitrogen Nitrogen is unique among all elements in the number of binary oxides it forms (Table 21.5). All are thermodynamically unstable with respect to decomposition to N2 and O2; that is, all have positive ∆fG° values. Most are slow to decompose, however, and so are described as kinetically stable.

Table 21.5

Some Oxides of Nitrogen

Formula Name N2O

Dinitrogen monoxide (nitrous oxide)

NO

Nitrogen monoxide (nitric oxide)

N2O3

Dinitrogen trioxide

Nitrogen Oxidation Number Description

Structure† N

N

O

+1

Colorless gas (laughing gas)

+2

Colorless gas; odd-electron molecule (paramagnetic)

+3

Blue solid (mp, −100.7 °C); reversibly dissociates to NO and NO2 above its mp

+4

Brown, paramagnetic gas; odd-electron molecule

+4

Colorless liquid/gas; dissociates to NO2 (Figure 15.7)

+5

Colorless solid

linear

*

O

O N

N

O planar

NO2

Nitrogen dioxide

N O

O O

O N2O4

Dinitrogen tetraoxide

N

N O

O planar

O N2O5

Dinitrogen pentaoxide

N O

O

O N O

Only one resonance structure is shown for each structure. *It is not possible to draw a Lewis structure that accurately represents the electronic structure of NO (Chapter 8). †



21.8  Nitrogen, Phosphorus, and the Group 5A (15) Elements

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Wellcome Images CC/DIOMEDIA

Dinitrogen monoxide, N2O, commonly called nitrous oxide, is a nontoxic, odorless, and tasteless gas in which nitrogen has the lowest oxidation number (+1) among the nitrogen oxides. It can be made by the careful decomposition of ammonium nitrate at 250 °C.

Nitrous oxide, N2O, laughing gas ​The famous English chemist Humphry Davy (1778–1829) was researching nitrous oxide, which he produced by heating NH4NO3. As chemists did in those days, he wanted to see what its effect might be on him, so he took a deep breath. He later recorded that ”The objects around me became dazzling and my hearing more acute.” In others, it produced continual laughter. It was soon apparent it actually dulled the senses, and it was used early on in dentistry and is still used today as a paindeadening substance. (For a history of pain-killing chemicals see https://www.sciencehistory. org/distillations/painless-dreams.)

NH4NO3(s) n N2O(g) + 2 H2O(g)

It is used as an anesthetic in minor surgery and has been called laughing gas because of its euphoriant effects. Because it is soluble in vegetable fats, the largest commercial use of N2O is as a propellant and aerating agent in cans of whipped cream. Nitrogen monoxide, NO, is an odd-electron molecule. It has 11 valence electrons, giving it one unpaired electron and making it a free radical. The compound has been the subject of intense research because it is important in a number of biochemical processes. Nitrogen dioxide, NO2, is the brown gas you see when a bottle of nitric acid is allowed to stand in the sunlight, 2 HNO3(aq) n 2 NO2(g) + H2O(ℓ) + 1⁄2 O2(g)

and as a product in many reactions involving nitric acid (Figure 21.28). Nitrogen dioxide is also a culprit in air pollution (Section 25.1). The primary source of atmospheric NO2 is the reaction of atmospheric nitrogen and oxygen in internal combustion engines, a reaction that produces NO. When released into the atmosphere, NO rapidly reacts with O2 to form NO2. 2 NO(g) + O2(g) n 2 NO2(g)

Nitrogen dioxide has 17 valence electrons, so, like NO, it is an odd-electron molecule. Because the odd electron largely resides on the N atom, two NO2 molecules can combine, forming an N—N bond and producing colorless N2O4, dinitrogen tetraoxide. 2 NO2(g) 878n N2O4(g) deep brown gas

colorless (mp, −11.2 °C)

Solid N2O4 consists entirely of N2O4 molecules. However, as the solid melts and the temperature increases to the boiling point, the color darkens as some N2O4 dissociates to form brown NO2. At the normal boiling point (21.5 °C), the distinctly brown gas consists of 15.9% NO2 and 84.1% N2O4 (Figure 15.7). When NO2 is bubbled into water, nitric acid and nitrous acid form. 2 NO2(g) + H2O(ℓ) n HNO3(aq) + HNO2(aq)

nitric acid

nitrous acid

Nitric acid has been known for centuries and is an important compound in modern economies. The oldest way to make the acid is to treat NaNO3 with sulfuric acid (Figure 21.28a). 2 NaNO3(s) + H2SO4(ℓ) n 2 HNO3(ℓ) + Na2SO4(s)

However, enormous quantities of nitric acid are now produced industrially by the oxidation of ammonia in the multistep Ostwald process. The acid has many applications, but by far the greatest amount is turned into ammonium nitrate (for use as a fertilizer) by the reaction of nitric acid and ammonia. Nitric acid is a powerful oxidizing agent, as the large, positive E° values for the following half-reactions illustrate: NO3−(aq) + 4 H3O+(aq) + 3 e− n NO(g) + 6 H2O(ℓ)

E° = + 0.96 V

NO3−(aq) + 2 H3O+(aq) + e− n NO2(g) + 3 H2O(ℓ)

E° = + 0.80 V

Concentrated nitric acid attacks and oxidizes most metals. (Aluminum is one of the exceptions; page 1060.) In this process, the nitrate ion is reduced to one of the nitrogen oxides. Which oxide is formed depends on the metal and on reaction conditions.

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Figure 21.28  The preparation and properties of nitric acid.

HNO3 was distilled and collected in this flask cooled with ice.

© Charles D. Winters/Cengage

H2SO4 and NaNO3 were heated in this flask.

Brown NO2 gas fills the apparatus and colors the liquid in the distillation flask.

(a) Nitric acid can be made from the reaction of concentrated sulfuric acid and sodium nitrate.

(b) When concentrated nitric acid reacts with copper, the metal is oxidized to copper(II) ions, and NO2 gas is a reaction product.

In the case of copper, for example, either NO or NO2 is produced, depending on the concentration of the acid (Figure 21.28b). In dilute acid:

Ammonium Nitrate—A Mixed Blessing

NH4NO3 is a widely used fertilizer as it is an excellent source of nitrogen. However, if mixed with organic material, it is also a powerful explosive, and many fatal accidents have happened over the past hundred years. Among the best known occurred on April 16, 1947 when almost 600 people died when a ship blew up in the harbor in Texas City, Texas. In April 2015, a plant making the material in West, Texas blew up, killing a number of people. And in 2020, a storage facility at the docks in Beirut, Lebanon exploded, causing at least 218 deaths, many injuries, and costing US$15 billion in property damage. Ammonium nitrate has also been used in weapons. On April 19, 1995 a bomb made from ammonium nitrate and fuel oil destroyed the Federal Building in Oklahoma City. In recent years it has been used in IEDs (improvised explosive devices).



There have been attempts to ban ammonium nitrate because of its ­w eaponization, but this is detrimental to people dependent on its use as a fertilizer. A possible solution is to modify the chemical by adding a substance that prevents it from acting as a

strong oxidant. One suggestion is to add a small amount of FeSO4, which reacts with NH4NO3 to produce Fe(NO3)2 and (NH4)2SO4, substances still useful in a fertilizer but that will not function in an explosive device.

Ali Chehade/Shutterstock.com

A Closer Look

3 Cu(s) + 8 H3O+(aq) + 2 NO3−(aq) n 3 Cu2+(aq) + 12 H2O(ℓ) + 2 NO(g)

Damage to the port of Beirut, Lebanon after the explosion of a facility storing 2750 tons of ammonium nitrate on August 4, 2020.

21.8  Nitrogen, Phosphorus, and the Group 5A (15) Elements

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In concentrated acid: Cu(s) + 4 H3O+(aq) + 2 NO3−(aq) n Cu2+(aq) + 6 H2O(ℓ) + 2 NO2(g)

Four metals (Au, Pt, Rh, and Ir) that are not attacked by nitric acid are often described as the noble metals. The alchemists of the fourteenth century, however, knew that if they mixed HNO3 with HCl in a ratio of about 1:3, this aqua regia, or “kingly water,” would attack even gold, the noblest of metals. Au(s) + NO3−(aq) + 4 Cl−(aq) + 4 H3O+(aq) n [AuCl4]−(aq) + NO(g) + 6 H2O(ℓ)

Hydrogen Compounds of Phosphorus and Other Group 5A (15) Elements The phosphorus-containing analog of ammonia, phosphine (PH3), is a poisonous, highly reactive gas with a faint garlic-like odor. Industrially, it is made by the reaction of white phosphorus and aqueous KOH. P4(s) + 3 KOH(aq) + 3 H2O(ℓ) n PH3(g) + 3 KH2PO2(aq)

The hydrides of the heavier Group 5A (15) elements are also toxic and become more unstable as the atomic number of the element increases. Nonetheless, arsine (AsH3) is important in the semiconductor industry as a starting material in the preparation of gallium arsenide (GaAs) semiconductors.

Phosphorus Oxides and Sulfides The most important compounds of phosphorus are those with oxygen, and there are at least six binary compounds containing just phosphorus and oxygen. All of them can be thought of as being derived structurally from the P4 tetrahedron of white phosphorus. For example, if P4 is carefully oxidized, P4O6 is formed; an O atom has been placed into each P—P bond in the tetrahedron (Figure 21.29).

165.6 pm

143 pm

221 pm 127.0°

123.0°

O2

160 pm

O2 99.5°

P4

102.0°

P4O6 P4O10 H2O

H2O

H3PO3

H3PO4

phosphorous acid

phosphoric acid

Figure 21.29  Phosphorus oxides.  ​Other binary POO compounds have formulas between P4O6 and P4O10. They are formed by starting with P4O6 and adding O atoms successively to the P atom vertices.

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90%

95%

Phosphate rock

Phosphoric acid (impure)

10%

5% 80%

Elemental phosphorus

Food phosphates

80% Phosphoric acid (pure)

Industrial phosphates Detergent phosphates

20%

20% Phosphorus sulfides Phosphorus chlorides Organic phosphorus compounds

Fertilizers

Used as such in metal treatment, etc.

The most common and important phosphorus oxide is tetraphosphorus decaoxide (P4O10). In P4O10, each phosphorus atom is surrounded tetrahedrally by O atoms. Phosphorus also forms a similar series of compounds with sulfur. Of these, the most important is P4S3, a molecule in which S atoms are inserted in three POP bonds. The principal use of P4S3 is in strike-anywhere matches, the kind that light when you rub the head against a rough object. The active ingredients are P4S3 and the powerful oxidizing agent potassium chlorate, KClO3. The safety match is now more common than the strike-anywhere match. In safety matches, the head is predominantly KClO3, and the material on the match book is red phosphorus (about 50%), Sb2S3, Fe2O3, and glue.

100° 103° 103°

4 P(s, red allotrope) + 3/8 S8(s)

209 pm

60°

233 pm

P4S3

© Charles D. Winters/Cengage

Figure 21.30  Uses of phosphate rock, phosphorus, and phosphoric acid.

Matches.  The head of a strikeanywhere match contains P4S3 and the oxidizing agent KClO3. (Other components are ground glass, Fe2O3, ZnO, and glue.) Safety matches have sulfur (3–5%) and KClO3 (45–55%) in the match head and red phosphorus in the striking strip.

Enormous quantities of phosphorus compounds are used around the world, and most of this begins with phosphate rock, which is largely Ca3(PO4)2 or apatite. As outlined in Figure 21.30, phosphate rock is converted to impure phosphoric acid (and then to other products) or to elemental phosphorus (from which pure phosphoric acid and other products are made).

Phosphorus Oxoacids and Their Salts A few of the many known phosphorus oxoacids are illustrated in Table 21.6. Indeed, there are so many acids and their salts in this category that structural principles have been developed to organize and understand them. •

All P atoms in the oxoacids and their anions (conjugate bases) are fourcoordinate and tetrahedral.

•

All the P atoms in the acids have at least one P—OH group. (This is the acidic hydrogen atom in these compounds.)

•

Some oxoacids have one or more P—H bonds. These H atoms are not ionizable as H+.



21.8  Nitrogen, Phosphorus, and the Group 5A (15) Elements

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1077

Table 21.6 Formula

Phosphorus Oxoacids

Name

pKa

Structure O

H3PO4

Orthophosphoric acid

P

HO OH O

H4P2O7

Pyrophosphoric acid (diphosphoric acid)

P

HO OH O

(HPO3)3

Metaphosphoric acid

HO

P O O

2.21, 7.21, 12.67

OH O

O O

P

P

P

OH OH

0.85, 1.49, 5.77, 8.22

O OH

O OH

O

H3PO3

Phosphorous acid (phosphonic acid)

P

H OH

OH

2.00, 6.59

O

H3PO2

Hypophosphorous acid (phosphinic acid)

P

H H

OH

1.24

•

Polymerization can occur by P—O—P bond formation to give both linear and cyclic species. Two P atoms are never joined by more than one P—O—P bridge.

•

When a P atom is surrounded only by O atoms (as in H 3PO4), its oxidation number is +5. For each P—OH that is replaced by P—H, the oxidation number drops by 2. For example, the oxidation number of P in H3PO2 is +1.

Orthophosphoric acid, H 3PO 4, and its salts are far more important commercially than other P—O acids. Millions of tons of phosphoric acid are made annually, some using white phosphorus as the starting material. The element is burned in oxygen to give P4O10, and the oxide reacts with water to produce the acid (Figure 21.31).

© Charles D. Winters/Cengage

P4O10(s) + 6 H2O(ℓ) n 4 H3PO4(aq)

Figure 21.31  Reaction of P4O10 and water.  ​The white solid oxide reacts vigorously with water to give orthophosphoric acid, H3PO4. (The heat generated vaporizes the water, so steam is visible.)

This approach gives a pure product, so it is used to make the acid found in food products in particular. The acid is nontoxic, and it gives the tart or sour taste to carbonated soft drinks, such as various colas (about 0.05% H3PO4) or root beer (about 0.01% H3PO4). A major use for phosphoric acid is to impart corrosion resistance to metal objects such as nuts and bolts, tools, and car-engine parts by plunging the object into a hot acid bath. Car bodies are similarly treated with metal phosphates (such as Zn3(PO4)2, page 977) and aluminum trim is polished by treating it with phosphoric acid. The reaction of H3PO4 with strong bases produces salts such as NaH 2PO4, Na2HPO4, and Na3PO4. In industry, the monosodium and disodium salts are produced using Na2CO3 as the base, but an excess of the stronger (and more expensive) base NaOH is required to remove the third proton to give Na3PO4. Sodium phosphate (Na3PO4) is used in scouring powders and paint strippers because the anion PO43− is a relatively strong base in water (Kb = 0.028). Sodium monohydrogen phosphate, Na2HPO4, which has a less basic anion than PO43−, is widely used in food products. Kraft has patented a process using this compound in the manufacture of pasteurized cheese, for example. Thousands of tons of Na2HPO4

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are still used for this purpose, even though the function of the salt in this process is not completely understood. In addition, a small amount of Na2HPO4 in pudding mixes enables the mix to gel in cold water, and the basic anion raises the pH of cereals to provide quick-cooking breakfast cereal. (The OH− ion from HPO42− hydro­ lysis accelerates the breakdown of the cellulose material in the cereal.) Calcium phosphates are used in a broad spectrum of products. For example, the weak acid Ca(H2PO4)2 ∙ H2O is used as the acid leavening agent in baking powder. A typical baking powder contains (along with inert ingredients) 28% NaHCO3, 10.7% Ca(H2PO4)2 ∙ H2O, and 21.4% NaAl(SO4)2 (also a weak acid). The weak acids react with sodium bicarbonate to produce CO2 gas. For example, Ca(H2PO4)2 ∙ H2O(s) + 2 NaHCO3(aq) n 2 CO2(g) + 3 H2O(ℓ) + Na2HPO4(aq) + CaHPO4(aq)

Finally, calcium monohydrogen phosphate, CaHPO 4, is used as an abrasive and polishing agent in toothpaste.

21.9 Oxygen, Sulfur, and the Group 6A (16) Elements Goal for Section 21.9 • Know common compounds of oxygen and sulfur, particularly the oxides of sulfur. Oxygen is by far the most abundant element in Earth’s crust, representing slightly less than 50% of its weight. It is present as elemental oxygen (O2) in the atmosphere and is combined with other elements in water and in many minerals. Scientists believe that elemental oxygen did not appear on this planet until about 3.5 billion years ago, when it was formed by plants through the process of photosynthesis. Sulfur, seventeenth in abundance in the Earth’s crust, is also found in its ele­ mental form in nature, but only in certain concentrated deposits. Sulfur-containing compounds occur in natural gas, coal, and oil. In minerals, sulfur occurs as the sulfide ion (Figure 21.32) and as sulfate ion (for example, in gypsum, CaSO4 ∙ 2 H2O). Sulfur oxides (SO2 and SO3) also occur in nature, primarily as products of volcanic activity. In the United States, several million tons per year of sulfur are obtained from deposits of the element found along the Gulf of Mexico. These deposits occur typically at a depth of 150 to 750 m below the surface in layers about 30 m thick. They are thought to have been formed by anaerobic (without elemental oxygen) bacteria acting on sedimentary sulfate deposits such as gypsum.

Group 6A (16) Oxygen 8

O

474,000 ppm Sulfur 16

S

260 ppm Selenium 34

Se

0.5 ppm Tellurium 52

Te

0.005 ppm Polonium 84

Po

trace

© Steven Hyatt

Element abundances are in parts per million in the Earth’s crust.

Figure 21.32  Sulfide-containing minerals.  ​Many minerals contain the sulfide or disulfide ion, and examples include (left) orpiment, As2S3; (center) iron pyrite, FeS2; and (right) stibnite, Sb2S3.

21.9  Oxygen, Sulfur, and the Group 6A (16) Elements

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Preparation and Properties of the Elements Pure oxygen is obtained by the fractional distillation of air and is among the top five industrial chemicals produced in the United States. Oxygen can be made in the laboratory by electrolysis of water and by the catalyzed decomposition of metal chlorates such as KClO3. catalyst

© Charles D. Winters/Cengage

© Charles D. Winters/Cengage

Selenium. ​Glass takes on a brilliant red color when a mixture of cadmium sulfide/selenide (CdS, CdSe) is added to it.

At room temperature and pressure, oxygen is a colorless gas, but it is pale blue when condensed to a liquid at −183 °C (Figure 21.33a). As described in Section 9.2, diatomic oxygen is paramagnetic because it has two unpaired electrons. An allotrope of oxygen, ozone (O3), is a blue, diamagnetic gas with an odor so strong that it can be detected in concentrations as low as 0.05 ppm. Ozone is synthesized by passing O2 through an electric discharge or by irradiating O2 with ultraviolet light. It is often in the news because of the realization that the Earth’s protective layer of ozone in the stratosphere is disrupted by chlorofluorocarbons and other chemicals (Section 25.1). Sulfur has numerous allotropes. The most common and most stable allotrope is the yellow, orthorhombic form, which consists of S8 molecules with the sulfur atoms arranged in a crown-shaped ring (Figure 21.33b). Less stable allotropes are known that have rings of 6 to 20 sulfur atoms. Another form of sulfur, called plastic sulfur, has a molecular structure with chains of sulfur atoms (Figure 21.33c). Selenium and tellurium are comparatively rare on Earth, having abundances about the same as those of silver and gold, respectively. Because their chemistry is similar to that of sulfur, they are often found in minerals associated with the sulfides of copper, silver, iron, and arsenic, and they are recovered as by-products of the industries devoted to those elements. Selenium has a range of uses, including glass making. A cadmium sulfide/ selenide mixture is added to glass to give it a brilliant red color. The most familiar use of selenium is in xerography, meaning “dry printing,” a process at the heart of the modern copy machine. Most photocopy machines use an aluminum plate or roller coated with selenium. Light coming from the imaging lens selectively discharges a static electric charge on the selenium surface, and the black toner sticks only on the areas that remain charged. A copy is made when the toner is transferred to a sheet of plain paper.

(a) Liquid oxygen (bp, –183 ˚C) is pale blue.

© Charles D. Winters/Cengage

© Charles D. Winters/Cengage

2 KClO3(s) 88888n 2 KCl(s) + 3 O2(g)

(b) At room temperature, sulfur exists as a bright yellow solid composed of S8 rings.

Figure 21.33  Oxygen and sulfur.

(c) When heated, the S8 rings break open and eventually form chains of S atoms in a material described as plastic sulfur.

1080 Chapter 21 / The Chemistry of the Main Group Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

The heaviest element of Group 6A (16), polonium, is radioactive and found only in trace amounts on Earth. It was discovered in Paris, France in 1898 by Marie Skłodowska Curie (1867–1934), a native of Poland, and her husband Pierre Curie (1859–1906; see page 82).

Sulfur Compounds Hydrogen sulfide, H2S, has a bent molecular geometry, like water. Unlike water, however, H2S is a gas under standard conditions (mp, −85.6 °C; bp, −60.3 °C) because its intermolecular forces are weak compared with the strong hydrogen bonding in water (Figure 11.4). Hydrogen sulfide is poisonous, comparable in toxicity to hydrogen cyanide, but fortunately it has a strong odor and is detectable in concentrations as low as 0.02 ppm. You must be careful with H2S, though. Because it has an anesthetic effect, your nose rapidly loses its ability to detect it. Death occurs at H2S concentrations of 100 ppm. Sulfur is often found as the sulfide ion in conjunction with metals, and most metal sulfides (except those based on Group 1A (1) metals) are not water soluble. The recovery of metals from their sulfide ores usually begins by heating the ore in air. For example, lead(II) sulfide (the mineral galena) is converted to lead(II) oxide. 2 PbS(s) + 3 O2(g) n 2 PbO(s) + 2 SO2(g)

Bad Breath  Halitosis or bad

breath is due to three sulfurcontaining compounds: H2S, CH3SH (methyl mercaptan), and (CH3)2S (dimethyl sulfide). All three can be detected in very tiny concentrations. For example, your nose knows if as little as 0.2 microgram of CH3SH is present per liter of air. The compounds result from bacteria’s attack on the sulfurcontaining amino acids cysteine and methionine in food particles in the mouth.

The lead(II) oxide is then reduced to lead using carbon or carbon monoxide in a blast furnace. PbO(s) + CO(g) n Pb(ℓ) + CO2(g)

Alternatively, the oxide can be reduced to elemental lead by combining it with fresh lead sulfide. 2 PbO(s) + PbS(s) n 3 Pb(s) + SO2(g)

H2S

SO2

SO3

H2SO4

Models of some common sulfur-containing molecules.

Sulfur dioxide (SO2), a colorless, toxic gas with a sharp odor, is produced on an enormous scale by the combustion of sulfur and by roasting sulfide ores in air. The combustion of sulfur in sulfur-containing coal and fuel oil creates particularly large environmental problems. It has been estimated that about 2.0 × 108 tons of sulfur oxides (primarily SO2) are released into the atmosphere each year by human activities; this is more than half of the total emitted by all other natural sources of sulfur in the environment. The most important reaction of sulfur dioxide is its oxidation to SO3. SO2(g) + 1⁄2 O2(g) n SO3(g)   ∆rH° = −98.9 kJ/mol-rxn

Sulfur trioxide is almost never isolated but is converted directly to sulfuric acid, H2SO4, by reaction with water. This acid is the compound manufactured in ­largest quantity by the chemical industry (“A Closer Look: Sulfuric Acid,” page 162). Approximately 37 million tons of sulfuric acid is produced annually in the United States. In the United States, roughly 70% of the sulfuric acid is used to manufacture superphosphate fertilizer from phosphate rock. Plants need a soluble form of phosphorus for growth, but calcium phosphate and apatite [Ca5X(PO4)3, X = F, OH, Cl]

Environmental Problems with SO2  Atmospheric SO2 can affect respiration, particularly in adults and children with asthma. It can also harm trees and plants and decrease growth, react with other airborne compounds forming a visible haze, and damage buildings. As recently as 1980, air in the United States contained 50–250 parts per billion SO2 depending on location. However, owing to more stringent environmental standards over the past few decades there has been a 94% decrease in atmospheric SO2.

21.9  Oxygen, Sulfur, and the Group 6A (16) Elements

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are insoluble. Treating phosphate-containing minerals with sulfuric acid produces a mixture of soluble phosphates. The balanced equation for the reaction of excess sulfuric acid and calcium phosphate, for example, is Ca3(PO4)2(s) + 3 H2SO4(ℓ) n 2 H3PO4(ℓ) + 3 CaSO4(s)

(This reaction is explained by Brønsted acid–base theory: sulfuric acid is a stronger acid than H3PO4 (Table 16.2), so the equilibrium lies to the right and the PO43− ion is protonated by sulfuric acid.) This reaction, however, does not tell the whole story. Concentrated superphosphate fertilizer is actually mostly CaHPO4 or Ca(H2PO4)2 plus some H3PO4 and CaSO4. Smaller amounts of sulfuric acid are used in the conversion of ilmenite, a titanium-bearing ore, to TiO2, which is then used as a white pigment in paint, plastics, and paper. The acid is also used to manufacture iron and steel as well as petroleum products, synthetic polymers, and paper. Group 7A (17)

21.10 The Halogens, Group 7A (17)

Halogens Fluorine 9

F

Goals for Section 21.10

950 ppm

• Know the methods for the preparation of fluorine and chlorine. • Understand the general chemical properties of the halogens and their compounds.

Chlorine 17

Cl

Fluorine and chlorine are the most abundant halogens in Earth’s crust, with fluorine somewhat more abundant than chlorine. If their abundance in seawater is measured, however, the situation is quite different. Chlorine has an abundance in seawater of 18,000 ppm, whereas the abundance of fluorine is only 1.3 ppm. This variation is a result of the differences in the solubility of their salts and plays a role in the methods used to obtain the elements themselves.

130 ppm Bromine 35

Br

0.37 ppm Iodine 53

Preparation of the Elements

I

0.14 ppm

Fluorine

Astatine 85

The water-insoluble mineral fluorspar or fluorite (calcium fluoride, CaF2, page 580) is one of the many sources of fluorine. Because the mineral was originally used as a flux in metalworking, its name comes from the Latin word meaning “to flow.” In the seventeenth century, it was discovered that solid fluorspar would emit light when heated, and the phenomenon was called fluorescence. In the early 1800s, when it was recognized that a new element was contained in fluorspar, A. M. Ampère (1775–1836) suggested that the element be called fluorine. Although fluorine was recognized as an element by 1812, it was not until 1886 that it was isolated by the French chemist Henri Moissan (1852–1907) in elemental form as a very pale yellow gas by the electrolysis of KF dissolved in anhydrous HF. Electrolysis is still the only practical way to obtain gaseous F2 (Figure 21.34) because F2 is such a powerful oxidizing agent that chemical oxidation of F− to F2 is not feasible. Even so, because F2 is so reactive, it corrodes the electrolysis equipment and reacts violently with traces of grease or other contaminants. In addition, the electrolysis products, F2 and H2, can recombine explosively, so they must not be allowed to come into contact. (Compare with the reaction of another halogen, Br2, with H2 in Figure 21.5.) Current U.S. production of fluorine is approximately 5000 metric tons per year.

At

trace Element abundances are in parts per million in the Earth’s crust.

F2 –

+

H2 +

Anode Cathode Skirt Cooling tube

Figure 21.34  Schematic of an electrolysis cell for producing fluorine.

Chlorine Chlorine is a strong oxidizing agent, so to prepare this element from chloride ion by a chemical reaction requires an even stronger oxidizing agent. Elemental chlorine was first made by the Swedish chemist Karl Wilhelm Scheele (1742–1786) in 1774,

1082 Chapter 21 / The Chemistry of the Main Group Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

by the reaction of sodium chloride with an oxidizing agent (MnO2) in an acidic solution. Now, permanganate ions or dichromate ions in acid solution are used in laboratory preparations (Figure 21.35). Industrially, chlorine is made by electrolysis of brine (concentrated aqueous NaCl), a process described in Figure 21.11.

Bromine The standard reduction potentials of the halogens indicate that their strength as oxidizing agents decreases going from F2 to I2.

−

Reduction Potential (E°, V) −

F2(g) + 2 e n 2 F (aq)

2.87

−

Cl2(g) + 2 e n 2 Cl (aq)

1.36

Br2(ℓ) +2 e n 2 Br (aq)

1.08

−

−

−

−

−

I2(s) + 2 e n 2 I (aq)

© Charles D. Winters/Cengage

Half-Reaction

0.535

This means that Cl2 will oxidize Br− ions to Br2 in aqueous solution, for example. Cl2(g) + 2 Br−(aq) n 2 Cl−(aq) + Br2(aq) E°net = E°cathode − E°anode = 1.36 V − 1.08 V = +0.28 V

In fact, this is the commercial method of preparing bromine from NaBr obtained from natural brine wells in Arkansas and Michigan.

Figure 21.35 Chlorine preparation.  ​Chlorine is prepared by oxidation of chloride ion using a strong oxidizing agent. Here, oxidation of NaCl is accomplished using K2Cr2O7 in H2SO4. (The Cl2 gas is bubbled into water in a receiving flask.)

Iodine Iodine is a lustrous, purple-black solid, easily sublimed at room temperature and atmospheric pressure (Figure 12.25). The element was first isolated in 1811 from seaweed and kelp, extracts of which had long been used for treatment of goiter, the enlargement of the thyroid gland. It is now known that the thyroid gland produces a growth-regulating hormone (thyroxine) that contains iodine. Consequently, most table salt in the United States has 0.01% NaI added to provide the necessary iodine in the diet (“A Closer Look: Iodine and Your Thyroid Gland”). A laboratory method for preparing I2 is illustrated in Figure 21.36. The commercial preparation depends on the source of I− and its concentration. In one approach, iodide ions are first precipitated with silver ions to give insoluble AgI. I−(aq) + Ag+(aq) n AgI(s)

Figure 21.36  The preparation of iodine.  ​A mixture of sodium iodide and manganese(IV) oxide was placed in the flask (left). On adding concentrated sulfuric acid (right), brown iodine vapor is evolved.

© Charles D. Winters/Cengage

2 NaI(s) + 2 H2SO4(aq) + MnO2(s) n  Na2SO4(aq) + MnSO4(aq) + 2 H2O(ℓ) + I2(g)



21.10  The Halogens, Group 7A (17)

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A Closer Look

Iodine and Your Thyroid Gland

The primary function of the thyroid gland, located in your neck, is the production of thyroxine (3,5,3′,5′-​ tetraiodothyronine) and 3,5,3′-​ triiodothyronine. These chemical compounds are hormones that help to regulate the rate of metabolism, a term that refers to all of the chemical reactions that take place in the body. In particular, the thyroid hormones play an important role in the processes that release energy from food. Abnormally low levels of thyroxine result in a condition known as hypothyroidism. Its symptoms include lethargy and feeling cold much of the time. The remedy for this condition is medication, consisting of thyroxine pills. The opposite condition, hyperthyroidism, also occurs in some people. In this condition, the body produces too much of the hormone. Hyperthyroidism is diagnosed by symptoms such as nervousness, heat intolerance, increased appetite, and muscle weakness and fatigue when blood sugar is too rapidly depleted. The standard remedy for hyperthyroidism is to destroy part of the thyroid gland, and one way to do so is to use a compound containing radioactive iodine-123 or iodine-131. To understand this procedure, you need to know something about iodine in the body. Iodine is an essential element. Some diets provide iodine naturally

I

H C

HO

C

C

O

C C

H

C

C

C

I

H C

C C

I

I

C H

H H

N

H

C

C

C

H

H

O

O

H

3,5,3′,5′-tetraiodothyronine (thyroxine)

Thyroxine.  The hormone thyroxine exerts a stimulating effect on metabolism.

(seaweed, for example, is a good source of iodine), but in the Western world most iodine taken up by the body comes from iodized salt, NaCl containing about 0.01% of NaI. An adult of average size should take in about 150 μg (micrograms, 1 μg = ​10−6 g) of iodine in the daily diet. In the body, iodide ion is transported to the thyroid, where it serves as one of the raw materials in making thyroxine.

The fact that iodine concentrates in the thyroid tissue is essential to the procedure for using radioiodine therapy as a treatment for hyperthyroidism. Typically, an aqueous NaI solution is used in which a small fraction of iodide is the radioactive isotope iodine-131 or iodine-123, and the rest is nonradioactive iodine-127. The radioactivity destroys thyroid tissue, resulting in a decrease in thyroid activity.

This is reduced by clean scrap iron to give iron(II) iodide and metallic silver. 2 AgI(s) + Fe(s) n FeI2(aq) + 2 Ag(s)

The silver is recycled by oxidizing it with nitric acid (forming silver nitrate which is then reused). Finally, iodide ion from water-soluble FeI2 is oxidized to iodine with chlorine [with iron(III) chloride as a by-product]. 2 FeI2(aq) + 3 Cl2(aq) n 2 I2(s) + 2 FeCl3(aq)

Fluorine Compounds Bond-Dissociation Enthalpies of Some Halogen Compounds (kJ/mol)

X

XOX

HOX

COX (in CX4)

F

155

565

485

Cl

242

432

339

Br

193

366

285

I

151

299

213

Fluorine is the most reactive of all of the elements, forming compounds with every element except He and Ne. In most cases, the elements combine directly, and some reactions can be so vigorous as to be explosive. This can be explained by at least two features of fluorine chemistry: the relatively weak F—F bond compared with chlorine and bromine, and in particular, the relatively strong bonds formed by fluorine to other elements. This is illustrated by the table of bond-dissociation enthalpies in the margin. In addition to its oxidizing ability, another notable characteristic of fluorine is its small size. These properties lead to the formation of compounds where a number of F atoms can be bonded to a central element in a high oxidation state. Examples include PtF6, UF6, IF7, and XeF6.

1084 Chapter 21 / The Chemistry of the Main Group Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A Closer Look

The Many Uses of Fluorine-Containing Compounds The American Chemical Society reported in 2020 that about 20% of pharmaceuticals contain at least one fluorine atom in the molecule. In addition, three common anesthetic agents, sevoflurane (C4H3F7O), isoflurane (C3H2ClF5O), and desflurane (C3H2F6O) are more than 50% F by mass.

Sevoflurane, C4H3F7O, an anesthetic

But the incorporation of fluorine in pharmaceuticals is just the beginning. You may have a non-stick pan coated with Teflon™, a polymer with fluorine atoms along the carbon chain. You also may have access to air conditioning, either in your home or car. If your unit is an older one, it may contain CHF2Cl (also known as R-22

or HCFC-22) as the working fluid. This is now banned, however, because of its effect on Earth’s ozone layer. It has been replaced to some extent by HFC-134a, the hydrofluorocarbon CH2FCF3. This is not as harmful to the ozone layer, but it is still a planet-warming gas. To avert a planet-wide warming trend, an agreement was reached in 2016 to freeze or decrease production of HFCs. The extent of decrease depends on the country. For example, the United States and ­European Union agreed to freeze production and consumption by 2018 and to reduce HFCs to 15% of 2012 levels by 2036. Now what? What can be used as a refrigerant? One substitute is CH2CFCF3 (HFO-1234yf), which is now used in most new car air conditioners.

Among the many other c­ ommercial fluorine-containing compounds is a family of compounds called polyfluoroalkyl and perfluoroalkyl substances (collectively referred to as PFAS) that contain a large number of fluorine atoms in their chemical structures. The most prevalent compounds are based on perfluorooctanoic acid (PFOA, CF3(CF2)6CO2H) and perfluorooctane sulfonic acid (PFOS, CF 3 (CF 2 ) 7 SO 3 H). PFAS have been widely used as fire retardants, on stain-resistant carpets, and coatings on fast food containers and non-stick cookware, among other things. In the past few years, however, environmental concern about these compounds has increased as well. One of the authors of this book has been working with biologists studying alligators and crocodiles around the world, and measurable amounts of PFAS are found in these creatures, even in remote locations. For more on environmental issues surrounding these compounds, see Chapter 25.

CH2CFCF3 (HFO-1234yf) a refrigerant in air conditioners.

Hydrogen fluoride is an important industrial chemical. More than 1 million tons of hydrogen fluoride is produced annually worldwide, almost all by the action of concentrated sulfuric acid on fluorspar. CaF2(s) + H2SO4(ℓ) n CaSO4(s) + 2 HF(g)

The U.S. capacity for HF production is approximately 210,000 metric tons, but demand often exceeds supply. Anhydrous HF is used in a broad range of industries: in the production of refrigerants, herbicides, pharmaceuticals, high-octane gasoline, aluminum, plastics, electrical components, and frosted light bulbs. The fluorspar used to produce HF must be very pure and free of SiO2 because HF reacts readily with silicon dioxide. SiO2(s) + 4 HF(aq) n SiF4(g) + 2 H2O(ℓ) SiF4(g) + 2 HF(aq) n H2SiF6(aq)

This series of reactions also explains why HF can be used to etch or frost glass (such as the inside of light bulbs). And it explains why HF is not shipped in glass containers. The aluminum industry consumes about 10–40 kg of cryolite, Na3AlF6, per metric ton of aluminum produced. The reason is that cryolite is added to aluminum oxide to produce a lower-melting mixture that can be electrolyzed. Cryolite is found

21.10  The Halogens, Group 7A (17)

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in only small quantities in nature, so it is also manufactured in various ways, among them the following reaction: 6 HF(aq) + Al(OH)3(s) + 3 NaOH(aq) n Na3AlF6(s) + 6 H2O(ℓ)

About 3% of the hydrofluoric acid produced is used in uranium fuel production. To separate uranium isotopes in a gas centrifuge (Section 20.7), the uranium must be in the form of a volatile compound. Naturally occurring uranium ore is processed to give UO2. This oxide is treated with hydrogen fluoride to give UF4, which is then reacted with F2 to produce the volatile solid UF6. UO2(s) + 4 HF(aq) n UF4(s) + 2 H2O(ℓ) UF4(s) + F2(g) n UF6(s)

This last step consumes 70%–80% of the fluorine produced annually.

Chlorine Compounds Hydrogen Chloride Muriatic Acid  Hydrochloric acid

is sold to consumers under the common name “muriatic acid.”

Hydrochloric acid, an aqueous solution of hydrogen chloride, is a valuable industrial chemical. Hydrogen chloride gas can be prepared by the reaction of hydrogen and chlorine, but the rapid, exothermic reaction is difficult to control. The classical method of making HCl in the laboratory is the reaction of NaCl and sulfuric acid, a procedure that takes advantage of the facts that HCl is a gas and that H2SO4 will not oxidize the chloride ion. 2 NaCl(s) + H2SO4(ℓ) n Na2SO4(s) + 2 HCl(g)

Hydrogen chloride gas has a sharp, irritating odor. Both gaseous and aqueous HCl react with metals and metal oxides to give metal chlorides and, depending on the reactant, hydrogen or water. Mg(s) + 2 HCl(aq) n MgCl2(aq) + H2(g) ZnO(s) + 2 HCl(aq) n ZnCl2(aq) + H2O(ℓ)

Oxoacids of Chlorine Oxoacids of chlorine range from HClO, in which chlorine has an oxidation number of +1, to HClO4, in which the oxidation number is equal to the group number (using the A-B numbering system), +7. All are strong oxidizing agents. Oxoacids of Chlorine Acid HClO HClO2 HClO3 HClO4

Disproportionation  A reaction in which an element or compound is simultaneously oxidized and reduced is called a disproportionation reaction. Here, Cl2 is oxidized to ClO− and reduced to Cl−.

Name Hypochlorous Chlorous Chloric Perchloric

Anion ClO

−

Name Hypochlorite ion

−

Chlorite ion

−

Chlorate ion

−

Perchlorate ion

ClO2 ClO3 ClO4

Hypochlorous acid, HClO, forms when chlorine dissolves in water. In this reaction, some of the chlorine is oxidized to hypochlorite ion and an equal amount is reduced to chloride ion in a disproportionation reaction. Cl2(g) + 2 H2O(ℓ) uv H3O+(aq) + HClO(aq) + Cl−(aq)

If Cl2 is dissolved in cold aqueous NaOH instead of in pure water the products are hypochlorite ion and chloride ion. Cl2(g) + 2 OH−(aq) uv ClO−(aq) + Cl−(aq) + H2O(ℓ)

1086 Chapter 21 / The Chemistry of the Main Group Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Under basic conditions, the equilibrium lies far to the right. The resulting alkaline solution is the liquid bleach used in home laundries. The bleaching action of this solution is a result of the oxidizing ability of ClO−. Most dyes are colored organic compounds, and the hypochlorite ion oxidizes many dyes to colorless products. Solid Ca(ClO)2 is produced when calcium hydroxide is combined with Cl2. This compound is easily handled and is the “chlorine” that is sold for swimming pool disinfection. When a basic solution of hypochlorite ion is heated, another disproportionation occurs, forming chlorate ion and chloride ion: Sodium and potassium chlorates are made in large quantities this way. Sodium chlorate can be reduced to ClO2, a compound used for bleaching paper pulp. Some NaClO3 is also converted to potassium chlorate, KClO3, an oxidizing agent in fireworks and a component of safety matches. Perchlorates, salts containing ClO4−, are powerful oxidants. Pure perchloric acid, HClO4, is a colorless liquid that explodes if shocked. It explosively oxidizes organic materials and rapidly oxidizes silver and gold. Dilute aqueous solutions of the acid are safe to handle, however. More than half of the sodium perchlorate currently manufactured is converted to the ammonium salt. The process for making this conversion is an exchange reaction that takes advantage of the fact that ammonium perchlorate is less soluble in water than sodium perchlorate: NaClO4(aq) + NH4Cl(aq) uv NaCl(aq) + NH4ClO4(s)

Perchlorate salts of most metals are usually relatively stable, albeit unpredictable. Great care should be used when handling any perchlorate salt. Ammonium perchlorate, for example, bursts into flame if heated above 200 °C.

John C. Kotz

3 ClO−(aq) n ClO3−(aq) + 2 Cl−(aq)

Fireworks and oxoacids of chlorine. ​Potassium chlorate, KClO3, was commonly used as the oxidizer in fireworks. However, this was the source of numerous accidents because it formed friction-sensitive compounds in contact with sulfur and powdered metals. Now potassium perchlorate, KClO4, is used, even though it is more difficult to ignite.

2 NH4ClO4(s) n N2(g) + Cl2(g) + 2 O2(g) + 4 H2O(g)

The strong oxidizing ability of the ammonium salt is the reason it has been used as the oxidizer in the solid booster rockets for space vehicles. The solid propellant in these rockets is largely NH4ClO4, the remainder being the reducing agent, powdered aluminum.

21.11 The Noble Gases, Group 8A (18) Goal for Section 21.11 • Apply the principles of structure and bonding to noble gas compounds. The noble or rare gases have already been mentioned in various places in this text, and most notably their lack of chemical reactivity. At one time these elements were called the inert gases but that name was dismissed with the 1962 discovery of compounds of xenon. Since that time, an elaborate chemistry of this element has been developed. Helium was first detected spectroscopically in the sun’s spectrum in 1868, but it was not isolated until 1895. (The element’s name is derived from the Greek word for sun, helios.) It is now obtained mostly from natural-gas wells where it may be present in amounts up to 7%. Helium is the most commercially important of these elements, with its main use as a coolant. Part of this use is in chemical and medical instrumentation such as MRIs, which have superconducting magnets that require cooling to liquid helium temperatures (4.2 K) to function. Because of this widespread use, though, there are serious concerns about the supply of helium gas, and labs are encouraged to have helium recovery facilities in place.



Xe2F3+

XeO2F2

21.11  The Noble Gases, Group 8A (18)

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1087

Group 8A (18) Rare Gases Helium 2

He

0.008 ppm © Charles D. Winters/Cengage

Neon 10

Ne

7 × 10−5 ppm Argon 18

Ar

1.2 ppm Krypton 36

Kr

The noble or rare gases. ​Neon and other gases are used in advertising signs, and xenon gas is found in automobile headlights. Radioactive radon gas occurs naturally in minerals in many parts of the United States. You can test your home for radon with a simple kit.

1 × 10−5 ppm

Xenon 54

Xe

2 × 10−6 ppm Radon 86

Rn

trace

The Noble Gases—Not So Inert

Before 1962, most chemists thought that compounds of noble gases could not exist. In fact, few chemists gave this issue much thought, but the data were there to predict that these compounds just might exist, had scientists looked for them. An assessment of bond energy data is the clue to predicting that compounds of xenon should exist. Consider the formation of XeF2 from Xe and F2. The enthalpy of this reaction can be related to bond energies (Section 8.10).

Xe(g) + F2(g) n XeF2(g) ∆rH = D(bonds broken) − D(bonds formed) = D(FOF) − 2 D(XeOF) = 155 kJ/mol − 2 D(XeOF) From these data, one can conclude that if the XeOF bond energy is 78 kJ/mol or greater this reaction will be exothermic (∆rH will be negative).

A similar extrapolation predicted that How might you predict the bond energy of the Xe—F bond? The logical way is to XeF4 should exist, and it is in fact now well look at trends within the periodic table. In known. On the other hand, compounds this case, the trend in bond energies in such as XeCl 2 , KrF 2 , and KrF 4 are all the known compounds TeF4 and IF3 allows predicted to have positive enthalpies you to extrapolate a value for XeOF. The of formation, and they have not been bond-dissociation enthalpies of the TeOF found. bond in TeF 4 (D = 335 kJ/mol) and the IOF bond in IF 3 (D = 272  kJ/mol) are known. Using these numbers, a chemist might predict a bond energy for XeOF to be substantially more than 78 kJ/ mol. This would, in turn, predict the reaction to be exothermic and likely product-favored. The enthalpy of formation of XeF2(g) was found by e­ xperiment to be −108 kJ/mol. From this the XeOF bond-dissociation enthalpy is calculated to be 132 kJ/mol, Preparing XeF2.  If a flask containing Xe and significantly more than is required F2 gas is placed in the sunlight for a few days, to make this an exothermic crystals of solid XeF2 are found in the flask. reaction. © Gary J. Schrobilgen

A Closer Look

Element abundances are in parts per million in the Earth’s atmosphere.

Argon is the most abundant of the elements of this group, present in air to the extent of 0.93% by volume. The discovery of argon was an interesting experimental exercise (“Applying Chemical Principles 2.3: Argon—An Amazing Discovery,” page 118), requiring highly accurate density measurements. Argon is mainly used to provide an inert atmosphere when required. Soon after the discovery and characterization of argon in 1895, the other nonradioactive noble gases (neon, krypton, and xenon) were found as trace components of air. These gases are best known from their use in lamps: neon for the red color in neon lamps, krypton and xenon for the intense light in lasers, and in xenon headlights for cars. Radioactive radon is found in trace quantities as a product of the decay of uranium minerals (Chapter 20).

1088 Chapter 21 / The Chemistry of the Main Group Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

An unexpected property of xenon is that it is an inhalation anesthetic. It is more effective than nitrous oxide but less so than commonly used compounds such as sevoflurane (C4H3F7O, page 1085). Xenon is nonpolluting, has analgesic properties and cardiovascular stability, and induces anesthesia rapidly.

Xenon Compounds The first evidence that xenon compounds could exist was reported in 1962. Neil Bartlett, then at the University of British Columbia, found that PtF 6, an exceedingly strong oxidizing agent, reacted with molecular oxygen to give O2+PtF6−. Noticing the similarity of the ionization energies of Xe and O2, Bartlett next carried out a reaction between PtF6 and Xe, and obtained a mixture of several compounds including XeF+PtF6− and XeF+Pt2F11−. Bartlett’s discovery was followed by rapid experimentation that resulted in the identification of a number of new xenon compounds. After chemists had for years predicted that the noble gases would not form compounds, it was particularly notable that XeF2, XeF4, and XeF6 could be prepared readily by direct reaction of the elements. The chemistry of these compounds has been further extended, and a number of xenon oxides (such as XeOF4, XeO2F2, and XeO3) as well as ionic species (for example, Xe2F3+ and XeF5+) are now known.

Applying Chemical Principles 21.1  Lead in the Environment The two heaviest elements in Group 4A (14) are lead (Pb) and flerovium (Fl). Flerovium was made in the laboratory in 1998 whereas lead is one of a handful of elements known since ancient times. It is the most abundant nonessential (not required for life processes) element in Earth’s crust and ranks fifth among metals in usage behind iron, copper, aluminum, and zinc. The major uses of lead and its compounds are in storage batteries (Section 19.3), pigments, ammunition, ­solders, plumbing, and bearings. Unfortunately, lead and its compounds are c­ umulative ­p oisons, particularly in children. At a blood level as low as 50 ppb (parts per billion), blood pressure is elevated; ­intelligence is affected at 100 ppb; and blood levels higher than 800 ppb can lead to coma and possible death. The symptoms of lead poisoning include nausea, abdominal pain, irritability, ­headaches, and excess lethargy or hyperactivity. Health experts believe more than 200,000 children become ill from lead poisoning annually, a problem caused chiefly by children eating paint containing lead-based pigments. Paint containing white lead [2 PbCO3 ∙ Pb(OH)2] was used until about 50 years ago, when white lead was replaced by TiO2. Lead salts have a sweet taste, which may contribute to the tendency of children to chew on painted objects. Until 50 years ago one of the most important uses of lead was as a gasoline additive, tetraethyllead (TEL), (C2H5)4Pb. It was introduced in the 1920s as an octane booster for gasoline. However, after it became clear that enormous amounts of lead were entering the environment from this source, TEL was phased out in the U.S. beginning in the 1970s. Other countries around the world have also banned TEL, but some did so only recently.

20

Lead Enrichment, Pb:Th ratio

15 10 5 0 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000 2010

Figure  A plot of the ratio of 210Pb to Th for the last 100 years in the environment in northern Alberta, Canada.  Lead in the environment was elevated from 1960 to 1990 until the ban on TEL fully took effect. (The amount of lead is compared with thorium, an indicator of general mineral abundance.)

Fortunately, recent research has found that the ban on leadedgasoline has allowed the environmental lead level to return to levels comparable to those observed before the use of TEL. Unfortunately, lead is a persistent problem affecting the environment as well as human and wildlife health. In 2015, people living in Flint, Michigan discovered elevated lead levels in their drinking water. The problem arose from a change in water supply and insufficient water treatment that leached lead from an old and poorly maintained water system using lead pipes (see page 1260). The water switch is now also linked to deaths caused by the bacterial pathogen Legionella pneumophila. Wildlife biologists recently uncovered another issue involving lead contamination in the environment when they examined a Applying Chemical Principles

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large number of living and dead bald eagles and golden eagles for signs of lead poisoning. Almost half had signs of chronic poisoning, and a large number had acute poisoning. The source of the lead is fragments from lead ammunition used to bring down animals on which eagles prey. While eagles hunt for living prey, they are also scavengers. Especially in the winter, eagles consume what is left after hunters dress out the animal.

The bald eagle population in the United States has increased from about 72,000 in 2009 to almost 320,000 in 2022, but that increase is showing signs of slowing due to lead poisoning. If hunters would change from lead bullets to steel or copper ammunition, this could be averted.

Questions

John C. Kotz

1. If blood contains 50. ppb lead, how many atoms of lead are there in 1.0 L of blood? (Assume d(blood) = 1.0 g/mL.) 2. Research has found that port wine stored for a year in leadglass decanters contains 2000 ppm lead. If the decanter contains 750 mL of wine (d = 1.0 g/mL), what mass of lead has been extracted into the wine? 3. Suppose you consume a 0.15-g chip of paint with white lead pigment (12% by weight). What is the concentration of lead in your blood system if your body contains 4.7 L of blood?

American bald eagle.  Almost half are threatened by chronic and acute lead poisoning.

The high energy content of hydrogen makes it an attractive energy source. However, there are obstacles to overcome before wide use of hydrogen as a fuel is a reality. One obstacle is storage. Storage as a gas at 1 atm requires a large volume, a problem that could be addressed by storing the gas under high pressure. However, then specialized containment is needed, and there are potential hazards involved. There are several criteria necessary for a successful hydrogen storage device. Foremost is hydrogen density. A useful storage device would contain a large amount of hydrogen in a small volume. Also, the hydrogen must be released easily, and, once hydrogen has been removed from the device, it should be possible to refill it. For a benchmark to use to consider this subject, consider the density of gaseous hydrogen: at 1.0 atm and 20 °C, d = 0.084 g/L. The value improves under high pressure: at 20 °C and 700 bar (691 atm) the value is 40 g/L. Ammonia borane, H3NBH3, is being studied as a possible answer to the storage problem. Ammonia borane is a colorless crystalline solid, stable in air and water. The compound is isoelectronic and isostructural with ethane, H3CCH3. However, its properties are in stark contrast to ethane. Most notably, ammonia borane is solid (mp, 104 °C) whereas ethane is a gas. Its chemical properties are also significantly different. When mildly warmed, ammonia borane loses hydrogen, forming first an aggregate species (H 2NBH2)x and then B3N3H6 (borazine, a cyclic compound with a structure like benzene). Further heating at 500 °C results in the formation of boron nitride, (BN)x. In contrast, ethane doesn’t dissociate hydrogen when heated. Recent studies on ammonia borane have used a variety of metal compounds as catalysts to speed up the decomposition to form gaseous hydrogen. An important question about

khaffizzul hakim/Shutterstock.com

21.2  Hydrogen Storage

Figure  Hydrogen-powered Toyota Mirai. ammonia borane is the proportion of hydrogen gas that can be released.

Questions

1. Draw the Lewis structure of ammonia borane. What are the formal charges on each of the atoms? 2. Calculate the mass of hydrogen in 1.00 kg of ammonia borane. 3. What is the hydrogen density (mass H2 per liter) in ammonia borane? The density of ammonia borane is 0.780 g/cm3. 4. The best catalysts used to accelerate the decomposition of ammonia borane result in release of 140 g H2 per kg. What proportion of hydrogen is released? 5. Draw a Lewis structure for borazine. Note that the molecule is a six-member ring of alternating B and N atoms, each with an attached H atom. Is the compound isoelectronic with benzene?

1090 Chapter 21 / The Chemistry of the Main Group Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Think–Pair–Share 1. You are given a white solid (A) and are told it is a salt of a Group 2A (2) element. When a crystal of A is placed in a Bunsen burner flame, you observe a bright red flame. When a 1.00-g sample of A is strongly heated, a colorless gas is evolved and a white solid (B) remains. Solid B also produces a bright red flame color, and the gas that was evolved has a pressure of 168 mm Hg in a 750 mL container at 25 °C. When the gas is bubbled into a solution of Ca(OH)2, a white solid precipitates. The unknown solid B is tested further. All of your sample of B dissolves in hydrochloric acid. Upon evaporating this solution, 1.07 g of another white solid (C) is isolated. Solid C can be redissolved in water, but when ­sulfuric acid is added, another white solid (D) precipitates (1.24 g). Identify the unknown compounds. 2. Silicon-oxygen and phosphorus-oxygen rings are a common structural feature in silicates and phosphates. (a) Draw the structure of the anion [Si3O9]6−, which is found in the mineral benitoite. Is the ring expected to be planar? (b) What are the expected O–Si–O bond angles? (c) What hybridization is expected for the Si atoms? (d) Draw the structure for the anion [P 4O 12] 4−. Is the ring expected to be planar? 3. Over Antarctica, polar vortex clouds form in the middle of winter, and some chemical reactions occur on the clouds. Among these are reactions of HCl and ClONO2: HCl + ClONO2 n Molecule A + Cl2 H2O + ClONO2 n Molecule A + HOCl HCl + HOCl n H2O + Molecule B

(a) What are molecules A and B? (b) Draw an electron dot structure for the major resonance structure of ClONO2, and give the formal charge for each atom in the structure. 4. “The global nitrogen cycle is complex. It includes chemical changes and nitrogen transport between oceans, land, and atmosphere, and involves natural and man-made sources

of nitrogen.” [C. E. Housecraft and A. G. Sharpe, Inorganic Chemistry, 3rd ed., Harlow, England, Pearson, 2008.] Part of the complex web of reactions is the removal of nitrates from the environment by enzymatic denitrification, ending with nitrogen gas. NO3− n NO2− n NO n N2O n N2

(a) Are any of these ions or molecules odd-electron species? (b) Draw an electron dot structure for each of these molecules or ions that is not an odd-electron molecule. (c) What is the oxidation state of the nitrogen atom in each molecule or ion? (d) Sketch a molecular orbital diagram for NO. What is the highest occupied molecular orbital (HOMO)? (Base your answer on Figure 9.16.) 5. Phosphorus-based oxoacids have an extensive chemistry and are important in industry. Consider just three of these acids: H3PO4 phosphoric acid H3PO3 phosphonic acid H3PO2 phosphinic acid (a) Specify the oxidation number of phosphorus in each acid. (b) In phosphorus oxoacids, there is always one P−O bond with no H atom attached to the O atom. There is always at least one P−OH group, and there may be one or more P−H bonds. With this knowledge, draw a dot structure for each of the acids. (c) Two of the acids are potential reducing agents. H3PO3 + 2 H+ + 2 e− n H3PO2 + H2O Eo = −0.499 V H3PO4 + 2 H+ + 2 e− n H3PO3 + H2O Eo = −0.276 V Phosphinic acid (H3PO2) can be used to reduce metal ions in solution. Which of the following ions can be reduced to the metal by phosphinic acid? Sn2+(aq) + 2 e− n Sn(s) Eo = −0.14 V 2+ − Ni (aq) + 2 e n Ni(s) Eo = −0.25 V 2+ − Zn (aq) + 2 e n Zn(s) Eo = −0.763 V Al3+(aq) + 3 e− n Al(s) Eo = −1.66 V

Chapter Goals Revisited The goals for this chapter are keyed to specific Study Questions to help you organize your review.

21.1  Abundance of the Elements • Recognize patterns in element abundance. 14, 41.

21.2  The Periodic Table: A Guide to the Elements • Know the general features of the chemistry of main group elements. 1, 3, 10, 12, 89.

21.3 Hydrogen • Know the basic properties of hydrogen and its preparation. 19, 21.

Chapter Goals Revisited

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21.4  The Alkali Metals, Group 1A (1) • Know the general chemical properties of the alkali metals. 27, 28, 119.

21.5  The Alkaline Earth Elements, Group 2A (2) • Know some of the common, commercially important compounds of the Group 2A (2) elements. 33, 34, 37.

• Know the general chemical properties of the alkaline earth elements. 33, 35, 92, 113.

21.6  Boron, Aluminum, and the Group 3A (13) Elements • Know the general chemical properties of the Group 3A (13) elements. 41–43, 47, 105, 123.

21.7  Silicon and the Group 4A (14) Elements • Know how elemental silicon is produced. 56. • Know the general chemical properties of the Group 4A (14) elements. 53, 54, 57, 125.

21.8  Nitrogen, Phosphorus, and the Group 5A (15) Elements • Know common compounds of nitrogen and phosphorus, particularly the oxides. 59–61, 67, 104.

21.9  Oxygen, Sulfur, and the Group 6A (16) Elements • Know common compounds of oxygen and sulfur, particularly the oxides of sulfur. 69, 70, 73.

21.10  The Halogens, Group 7A (17) • Know the methods for the preparation of fluorine and chlorine. 80, 81. • Understand the general chemical properties of the halogens and their compounds. 75–79, 82.

21.11  The Noble Gases, Group 8A (18) • Apply the principles of structure and bonding to noble gas compounds. 83, 84.

Study Questions denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully worked solutions in the Student Solutions Manual. ▲

Practicing Skills Properties of the Elements (See Sections 21.1 and 21.2.) 1. Which of the following formulas is incorrect? (a) CaH2 (c) CaS (b) CaI2 (d) Ca2O3

2. The reaction of elemental phosphorus and excess oxygen produces P4O10. Which of the following is the name the compound? (a) phosphorus oxide (b) phosphoric acid (c) phosphorus decaoxide (d) tetraphosphorus decaoxide

1092 Chapter 21 / The Chemistry of the Main Group Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

3. Like sulfur, selenium forms compounds in several different oxidation states. Which of the following is NOT likely to be an oxidation state of selenium in its compounds? (a) −2 (b) +3 (c) +6 (d) +4 4. What is the highest oxidation state that antimony can have in its compounds? (a) 0 (b) +1 (c) +3 (d) +5 5. Give examples of two basic oxides. Write chemical equations illustrating the formation of each oxide from its component elements. Write other chemical equations that illustrate the basic character of each oxide.

17. Complete and balance the equations for the following reactions. [Assume an excess of oxygen for (d).] (a) Na(s) + Br2(ℓ) n (c) Al(s) + F2(g) n (b) Mg(s) + O2(g) n (d) C(s) + O2(g) n 18. Complete and balance the equations for the following reactions: (a) K(s) + I2(g) n (c) Al(s) + S8(s) n (b) Ba(s) + O2(g) n (d) Si(s) + Cl2(g) n

Hydrogen (See Section 21.3.)

6. Give examples of two acidic oxides. Write chemical equations illustrating the formation of each oxide from its component elements. Write other chemical equations that illustrate the acidic character of each oxide.

19. Which of the following elements does not react with hydrogen? (a) neon (c) potassium (b) nitrogen (d) fluorine

7. Give the name and symbol of each element having the valence configuration [noble gas] ns2np1.

20. Which of the methods below is the most suitable for the preparation of large quantities of hydrogen (such as the amounts needed for the industrial synthesis of compounds such as ammonia)? (a) Electrolysis of water (b) The reaction of metal hydrides with water (c) The high-temperature reaction of methane and water (d) The reaction of zinc and hydrochloric acid

8. Give symbols and names for four monatomic ions that have the same electron configuration as argon. 9. Select one of the alkali metals, and write a balanced chemical equation for its reaction with chlorine. Is the reaction likely to be exothermic or endothermic? Is the product ionic or molecular? 10. Select one of the alkaline earth metals and write a balanced chemical equation for its reaction with oxygen. Is the reaction likely to be exothermic or endothermic? Is the product ionic or molecular? 11. For the product of the reaction you selected in Study Question 9, predict the following physical properties: color, state of matter (s, ℓ, or g), solubility in water. 12. For the product of the reaction you selected in Study Question 10, predict the following physical properties: color, state of matter (s, ℓ, or g), solubility in water. 13. Would you expect to find calcium occurring naturally in the Earth’s crust as a free element? Why or why not? 14. Which of the first 10 elements in the periodic table are found as free elements in the Earth’s crust? Which elements in this group occur in the Earth’s crust only as part of a chemical compound? 15. Place the following oxides in order of increasing basicity: CO2, SiO2, SnO2.



16. Place the following oxides in order of increasing basicity: Na2O, Al2O3, SiO2, SO3.

21. Write balanced chemical equations for the reaction of hydrogen gas with oxygen, chlorine, and nitrogen. 22. Write a balanced chemical equation for the reaction of potassium and hydrogen. Name the product. Is it ionic or covalent? Predict one physical property and one chemical property of this compound. 23. Write a balanced chemical equation for the preparation of H2 (and CO) by the reaction of CH4 and water. Using data in Appendix L, calculate ΔrH°, ΔrG°, and ΔrS° for this reaction at 298 K. 24. Using data in Appendix L, calculate ΔrH°, ΔrG°, and ΔrS° for the reaction of carbon and water to give CO and H2 at 298 K. 25. A method suggested for the preparation of hydrogen (and oxygen) from water proceeds as follows: (a) Sulfuric acid and hydrogen iodide are formed from sulfur dioxide, water, and iodine. (b) The sulfuric acid from the first step is decomposed by heat to water, sulfur dioxide, and oxygen.

Study Questions

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(c) The hydrogen iodide from the first step is decomposed with heat to hydrogen and iodine. Write a balanced equation for each of these steps, and show that their sum is the decomposition of water to form hydrogen and oxygen. 26. Compare the mass of H2 expected from the reaction of steam (H2O) per mole of ­methane, petroleum, and coal. (Assume complete reaction in each case. Use CH2 and CH as representative formulas for petroleum and coal, respectively.)

Alkali Metals (See Section 21.4.) 27. Which of the following is not a property of sodium? (a) Reacts with Cl2 to form NaCl (b) Has a high melting point (>400 °C) (c) Has a silvery color (d) Conducts an electric current 28. The compound Na2O2 consists of (a) two Na+ ions and two O2− ions (b) molecules of Na2O2 (c) two Na+ ions and one O22− ion (d) Na2+ and O2− ions 29. Write balanced chemical equations for the reaction of sodium with each of the halogens. Predict at least two physical properties that are common to all of the alkali metal halides. 30. Write balanced chemical equations for the reaction of lithium, sodium, and potassium with O2. Specify which metal forms an oxide, which forms a peroxide, and which forms a superoxide. 31. The electrolysis of aqueous NaCl gives NaOH, Cl2, and H2. (a) Write a balanced chemical equation for the process. (b) In the United States, 1.19 × 1010 kg of NaOH and 1.14 × 1010 kg of Cl2 were produced in a recent year. Does the ratio of masses of NaOH and Cl2 produced agree with the ratio of masses expected from the balanced equation? If not, what does this tell you about the way in which NaOH and Cl2 are actually produced? Is the electrolysis of aqueous NaCl the only source of these chemicals?

32. (a) Write balanced chemical equations for the half-reactions that occur at the cathode and the anode when an aqueous solution of KCl is electrolyzed. Which chemical species is ­oxidized, and which chemical species is reduced in this reaction? (b) Predict the products formed when an aqueous solution of CsI is electrolyzed.

Alkaline Earth Elements (See Section 21.5.) 33. Which of the following insoluble calcium ­compounds does not dissolve in hydro­chloric acid? (a) limestone, CaCO3 (b) slaked lime, Ca(OH)2 (c) gypsum, CaSO4 ∙ 2 H2O (d) hydroxyapatite, Ca5(OH)(PO4)3 34. Calcium minerals are the raw materials for a ­variety of large-scale industrial processes. Which of the following is not an industrial process? (a) Converting limestone, CaCO3, to lime (b) Converting fluorite, CaF2, to HF (c) Converting slaked lime, Ca(OH)2, to lime (d) Converting apatite minerals to phosphate fertilizers 35. When magnesium burns in air, it forms both an oxide and a nitride. Write balanced equations for the formation of both compounds. 36. Calcium reacts with hydrogen gas at 300−400 °C to form a hydride. This compound reacts readily with water, so it is an excellent drying agent for organic solvents. (a) Write a balanced chemical equation showing the formation of calcium hydride from Ca and H2. (b) Write a balanced chemical equation for the reaction of calcium hydride with water (Figure 21.8). 37. Name three uses of limestone. Write a balanced chemical equation for the reaction of limestone with CO2 in water. 38. Explain what is meant by hard water. What causes hard water, and what problems are associated with it? 39. Calcium oxide, CaO, is used to remove SO2 from power plant exhaust. These two compounds react to give solid CaSO3. What mass of SO2 can be removed using 1.2 × 103 kg of CaO?

1094 Chapter 21 / The Chemistry of the Main Group Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

40. Ca(OH)2 has a Ksp of 5.5 × 10−6, whereas Ksp for Mg(OH)2 is 5.6 × 10−12. Calculate the equilibrium constant for the reaction

46. Diborane can be prepared by the reaction of NaBH4 and I2. Which substance is oxidized, and which is reduced?

Ca(OH)2(s) + Mg2+(aq) uv Ca2+(aq) + Mg(OH)2(s)

47. Write balanced chemical equations for the reactions of aluminum with HCl(aq), Cl2, and O2.

Explain why this reaction can be used in the commercial isolation of magnesium from seawater.

Boron and Aluminum (See Section 21.6.) 41. In terms of abundance of the elements in the Earth’s crust, aluminum ranks (a) first (c) third (b) second (d) fourth 42. The element below aluminum in Group 3A (13) is gallium, and there are numerous ­similarities in the chemistry of these two elements. For ­example, the hydroxides of both elements are amphoteric. A consequence of this is that both gallium hydroxide and aluminum hydroxide (choose one) (a) are soluble in water (b) dissolve only in acid (c) dissolve only in base (d) dissolve in acid and in base 43. Draw a possible structure for the cyclic anion in the salt K3B3O6 and the anion in Ca2B2O5. 44. The boron trihalides (except BF3) hydrolyze completely to boric acid and the acid HX. (a) Write a balanced equation for the reaction of BCl3 with water. (b) Calculate ΔrH° for the hydrolysis of BCl3 using data in Appendix L and the following information: ΔfH° [BCl3(g)] =  −403 kJ/mol; ΔfH° [B(OH)3(s)] = −1094 kJ/mol.

45. When boron hydrides burn in air, the reactions are very exothermic. (a) Write a balanced chemical equation for the combustion of B5H9(g) in air to give B2O3(s) and H2O(ℓ). (b) Calculate the enthalpy of combustion for B5H9(g) (ΔfH° = 73.2 kJ/mol), and compare it with the enthalpy of combustion of B2H6 (−2170.4 kJ/mol). (The enthalpy of formation of B2O3(s) is −1271.9 kJ/mol.) (c) Compare the enthalpy of combustion of C2H6(g) with that of B2H6(g). Which transfers more energy as heat per gram?



48. (a) Write a balanced chemical equation for the reaction of Al and H2O(ℓ) to produce H2 and Al2O3. (b) Using thermodynamic data in Appendix L, calculate ΔrH°, ΔrS°, and ΔrG° for this reaction. Do these data indicate that the reaction should favor the products at equilibrium? (c) Why is aluminum metal unaffected by water? 49. Aluminum dissolves readily in hot aqueous NaOH to give the aluminate ion, [Al(OH)4]−, and H2. Write a balanced chemical equation for this reaction. If you begin with 13.2 g of Al, what volume (in liters) of H2 gas is produced when the gas is measured at 22.5 °C and a pressure of 735 mm Hg? 50. Alumina, Al2O3, is amphoteric. Among examples of its amphoteric character are the reactions that occur when Al2O3 is heated strongly or fused with acidic oxides and basic oxides. (a) Write a balanced equation for the reaction of alumina with silica, an acidic oxide, to give aluminum metasilicate, Al2(SiO3)3. (b) Write a balanced equation for the reaction of alumina with the basic oxide CaO to give calcium aluminate, Ca(AlO2)2. 51. Aluminum sulfate is the most commercially important aluminum compound, after aluminum oxide and aluminum hydroxide. It is produced from the reaction of aluminum oxide and sulfuric acid. What masses (in kilograms) of aluminum oxide and sulfuric acid must be used to manufacture 1.00 kg of aluminum sulfate? 52. Aerated concrete bricks are widely used building materials. They are obtained by mixing gas-forming additives with a moist mixture of lime, cement, and possibly sand. Industrially, the following reaction is important: 2 Al(s) + 3 Ca(OH)2(s) + 6 H2O(ℓ) n   3 CaO ∙ Al2O3 ∙ 6 H2O(s) + 3 H2(g)

Assume the mixture of reactants contains 0.56 g of Al (as well as excess calcium hydroxide and water) for each brick. What volume of hydrogen gas do you expect at 26 °C and a pressure of 745 mm Hg?

Study Questions

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1095

Silicon (See Section 21.7.) 53. Elemental silicon is oxidized by O2 to give unknown A. Compound A is dissolved in molten Na2CO3 giving B. When B was treated with aqueous hydrochloric acid, C is produced. Identify compound C. (a) SiH4 (c) SiO2 (b) H4SiO4 (d) SiCl4 54. Silicon and oxygen form a six-member ring in the silicate anion [Si3O9]6−. The rare blue mineral benitoite, the California state gemstone, has the formula BaTiSi3O9. (a) What is the oxidation state of silicon in this anion? (b) Draw the structure of the anion. Is it expected to be planar? 55. Describe the structure of pyroxenes (see page 1067). What is the ratio of silicon to oxygen in this type of silicate? 56. Describe how ultrapure silicon can be produced from sand. 57. Silicate structures: Draw a structure, and give the charge for a cyclic silicate anion with the formula [Si6O18]n−. 58. Silicates often have chain, ribbon, cyclic, or sheet structures. One of the simpler ribbon structures is [Si2O52−]n. Draw a structure for this anionic material.

Nitrogen and Phosphorus (See Section 21.8.) 59. Construct Lewis structures for the several resonance forms of N2O. What is the predicted N—N bond order? (a) 1 (c) between 2 and 3 (b) 2 (d) 3 60. Which statement about ammonia is not correct? (a) Ammonia can be made by a direct reaction of the elements. (b) Aqueous solutions of ammonia are acidic. (c) Ammonia is a gas at room temperature and atmospheric pressure. (d) Ammonia is used as a reactant in the synthesis of nitric acid. 61. Consult the data in Appendix L. Are any of the nitrogen oxides listed there stable with respect to decomposition to N2 and O2?

62. Use data in Appendix L to calculate the enthalpy and free energy changes at 25 °C for the reaction 2 NO2(g) n N2O4(g)

Is this reaction exothermic or endothermic? Is the reaction product- or reactant-favored at equilibrium? 63. Use data in Appendix L to calculate the enthalpy and free energy changes at 25 °C for the reaction 2 NO(g) + O2(g) n 2 NO2(g)

Is this reaction exothermic or endothermic? Is the reaction product- or reactant-favored at equilibrium? 64. The overall reaction involved in the industrial ­synthesis of nitric acid is NH3(g) + 2 O2(g) n HNO3(aq) + H2O(ℓ)

Calculate ΔrG° for this reaction and its equilibrium constant at 25 °C. 65. A major use of hydrazine, N2H4, is in steam boilers in power plants. (a) The reaction of hydrazine with O2 dissolved in water gives N2 and water. Write a balanced equation for this reaction. (b) O2 dissolves in water to the extent of 0.0044 g in 100. mL of water at 20 °C. What mass of N2H4 is needed to consume all of the dissolved O2 in 3.00 × 104 L of water (enough to fill a small swimming pool)? 66. Before hydrazine came into use to remove ­dissolved oxygen in the water in steam boilers, Na2SO3 was commonly used for this purpose: 2 Na2SO3(aq) + O2(aq) n 2 Na2SO4(aq)

What mass of Na2SO3 is required to remove O2 from 3.00 × 104 L of water as outlined in Study Question 65? 67. The structure of phosphorous acid is given in Table 21.6. Draw the structure of diphosphorous acid, H4P2O5. What is the maximum number of ionizable protons in a molecule of this acid? 68. Unlike carbon, which can form extended chains of atoms, nitrogen can form chains of very limited length. Draw the Lewis electron dot structure of the azide ion, N3−. Is the ion linear or bent?

1096 Chapter 21 / The Chemistry of the Main Group Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Oxygen and Sulfur (See Section 21.9.) 69. Which of the following is not a common oxidation number for sulfur in its compounds? (a) −2 (b) +6 (c) +3 (d) +4 70. Which statement about oxygen is not true? (a) Liquid oxygen is attracted to a magnet. (b) The allotropes of oxygen are O2 and O3. (c) Oxygen is the most abundant element in the Earth’s crust. (d) All electrons in O2 are paired. 71. In the contact process for making sulfuric acid, sulfur is first burned to SO2. Environmental restrictions allow no more than 0.30% of this SO2 to be vented to the atmosphere. (a) If enough sulfur is burned in a plant to produce 1.80 × 106 kg of pure, anhydrous H2SO4 per day, what is the maximum amount of SO2 allowed to be exhausted to the atmosphere? (b) One way to prevent any SO2 from reaching the atmosphere is to scrub the exhaust gases with slaked lime, Ca(OH)2: Ca(OH)2(s) + SO2(g) n CaSO3(s) + H2O(ℓ) 2 CaSO3(s) + O2(g) n 2 CaSO4(s)

What mass of Ca(OH)2 (in kilograms) is needed to remove the SO2 calculated in part (a)? 72. A sulfuric acid plant produces an enormous amount of heat. To keep costs as low as possible, much of this heat is used to make steam to generate electricity. Some of the electricity is used to run the plant, and the excess is sold to the local electrical utility. Three reactions are important in sulfuric acid production: (1) burning S to SO2; (2) oxidation of SO2 to SO3; and (3) reaction of SO3 with H2O: SO3(g) + H2O (in 98% H2SO4) n H2SO4(ℓ)

The enthalpy change of the third reaction is −130 kJ/mol. Estimate the enthalpy change when 1.00 mol of S is used to produce 1.00 mol of H2SO4. How much energy is produced per metric ton (1.00 × 103 kg) of H2SO4? 73. Sulfur forms anionic chains of S atoms called polysulfides. Draw a Lewis electron dot structure for the S22− ion, an analogue of the peroxide ion. It occurs in iron pyrites, such as FeS2.



74. Sulfur forms a range of compounds with fluorine. Draw Lewis electron dot structures for S2F2 (connectivity is FSSF), SF2, SF4, SF6, and S2F10. What is the oxidation number of sulfur in each of these compounds?

The Halogens (See Section 21.10.) 75. Which halogen has the highest bond-dissociation enthalpy? (a) F2 (b) Cl2 (c) Br2 (d) I2 76. Which of the following statements is not correct? (a) The ease of oxidation of the halide ions is F− < Cl− < Br− < I−. (b) Fluorine is the most abundant halogen in the Earth’s crust. (c) F2 is prepared industrially by electrolysis of aqueous NaF. (d) HF is used to etch glass. 77. The halogen oxides and oxoanions are good oxidizing agents. For example, the reduction of bromate ion has an E° value of 1.44 V in acid solution: 2 BrO3−(aq) + 12 H+(aq) + 10 e− n Br2(aq) + 6 H2O(ℓ)

Is it possible to oxidize aqueous 1.0 M Mn2+ to aqueous MnO4− with 1.0 M bromate ion? 78. The hypohalite ions, XO−, are the anions of weak acids. Calculate the pH of a 0.10 M solution of NaClO. What is the concentration of HClO in this solution? 79. Bromine is obtained from brine wells. The process involves treating water containing bromide ion with Cl2 and extracting the Br2 from the solution using an organic solvent. Write a balanced ­chemical equation for the reaction of Cl2 and Br−. ­What are the oxidizing and reducing agents in this reaction? Using the table of standard reduction potentials (Appendix M), verify that this is a product-favored reaction at equilibrium. 80. To prepare chlorine from chloride ion a strong oxidizing agent is required. The dichromate ion, Cr2O72−, is one example (Figure 21.35). Consult the table of standard reduction potentials (Appendix M), and identify several other oxidizing agents that may be suitable. Write balanced chemical equations for the reactions of these substances with chloride ion.

Study Questions

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1097

81. If an electrolytic cell for producing F2 (Figure 21.34) operates at 5.00 × 103 A (at 10.0 V), what mass of F2 can be produced per 24-hour day? Assume the conversion of F− to F2 is 100%. 82. Halogens combine with one another to produce inter­halogens such as BrF3. Sketch a possible molecular structure for this molecule, and decide if the FOBrOF bond angles will be less than or greater than ideal.

The Noble Gases (See Section 21.11.) 83. In which of the following species is Xe in the +4 oxidation state? (a) XeOF4 (c) XeF3+ + (b) Xe2F3 (d) XeO64− 84. Predict the electron-pair geometry for the xenon atom in XeOF4. (a) linear (b) trigonal bipyramid (c) trigonal planar (d) octahedral (e) tetrahedral 85. The standard enthalpy of formation of XeF4 is −218 kJ/mol. Use this value and the enthalpy of dissociation of the F—F bond to calculate the Xe—F bond-dissociation enthalpy. 86. Draw the Lewis electron dot structure for XeO3F2. What is its electron-pair geometry and its molecular geometry? 87. Argon is present in dry air to the extent of 0.93% by volume. What quantity of argon is present in 1.00 L of air? If you wanted to isolate 1.00 mol of argon, what volume of air would you need at 1.00 atm pressure and 25 °C? 88. The reaction of XeF6 with water gives a yellow solution from which solid xenon trioxide, XeO3, can be isolated (using extreme caution). (a) Write a balanced equation for this reaction. (b) What is the molecular geometry of XeO3? (c) Xenon trioxide is described as treacherous, exploding with little provocation. Even so, it has been possible to determine the enthalpy of formation, +402 kJ/mol, for this compound. Use this value, and the value of the O=O bond-dissociation enthalpy, to determine the bond-dissociation enthalpy of the Xe−O bond.

General Questions The questions are not designated as to type or location in the chapter. They may combine several concepts. 89. For each of the third-period elements (Na through Ar), identify the following: (a) whether the element is a metal, nonmetal, or metalloid (b) the color and appearance of the element (c) the state of the element (s, ℓ, or g) under standard conditions 90. When BCl3 gas is passed through an electric discharge, small amounts of the reactive molecule B2Cl4 are produced. (The molecule has a BOB covalent bond.) (a) Draw a Lewis electron dot structure for B2Cl4. (b) Describe the hybridization of the B atoms in the molecule and the geometry around each B atom. 91. Complete and balance the following chemical equations. (a) KClO3 + heat n (b) H2S(g) + O2(g) n (c) Na(s) + O2(g) n (d) P4(s) + KOH(aq) + H2O(ℓ) n (e) NH4NO3(s) + heat n (f) In(s) + Br2(ℓ) n (g) SnCl4(ℓ) + H2O(ℓ) n 92. (a) Heating barium oxide in pure oxygen gives barium peroxide. Write a balanced chemical equation for this reaction. (b) Barium peroxide is an excellent oxidizing agent. Write a balanced chemical equation for the reaction of iron with barium peroxide to give iron(III) oxide and barium oxide. 93. Worldwide production of silicon carbide, SiC, is several hundred thousand tons annually. If you want to produce 1.0 × 105 metric tons of SiC, what mass (metric tons) of silicon sand (SiO2) will you use if 70% of the sand is converted to SiC? (A metric ton is exactly 1000 kg.) 94. To store 2.88 kg of gasoline with an energy ­equivalence of 1.43 × 108 J requires a volume of 4.1 L. In comparison, 1.0 kg of H2 has the same energy equivalence. What volume is required if this quantity of H2 is to be stored at 25 °C and 1.0 atm of pressure?

1098 Chapter 21 / The Chemistry of the Main Group Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

95. Using data in Appendix L, calculate ΔrG° values for the decomposition of MCO3 to MO and CO2 where M = Mg, Ca, Ba. What is the relative tendency of these carbonates to decompose? 96. Ammonium perchlorate is used as the oxidizer in solid-fuel rockets. Assume that one launch requires 700. tons (6.35 × 105 kg) of the salt. 2 NH4ClO4(s) n N2(g) + Cl2(g) + 2 O2(g) + 4 H2O(g)

(a) What mass of water is produced? What mass of O2 is produced? (b) If all the O2 produced is assumed to react with the powdered aluminum present in the rocket engine, what mass of aluminum is required to use up all of the O2? (c) What mass of Al2O3 is produced? 97.

▲ Metals react with hydrogen halides (such as HCl) to give the metal halide and hydrogen:

M(s) + n HX(g) n MXn(s) + 1⁄2n H2(g)

The free energy change for the reaction is ΔrG° = ΔfG°(MXn) − nΔfG°[HX(g)]. (a) ΔfG° for HCl(g) is −95.1 kJ/mol. What must be the value for ΔfG° for MXn for the reaction to be product-favored at equilibrium? (b) Which of the following metals is (are) predicted to have product-favored reactions with HCl(g): Ba, Pb, Hg, Ti? 98. Halogens form polyhalide ions. Sketch Lewis electron dot structures and molecular structures for the following ions: (a) I3− (b) BrCl2− (c) ClF2+ (d) An iodide ion and two iodine molecules form the I5− ion. Here, the ion has five I atoms in a row, but the ion is not linear. Draw the Lewis dot structure for the ion, and propose a structure for the ion. 99. The standard enthalpy of formation of OF2 gas is +24.5 kJ/mol. Calculate the average OOF bonddissociation enthalpy. 100. Calcium fluoride can be used in the fluoridation of municipal water supplies. If you want to achieve a fluoride ion concentration of 2.0 × 10−5 M, what mass of CaF2 must you use for 1.0 × 106 L of water? (Ksp for CaF2 is 5.3 × 10−11.)



101. The steering rockets in space vehicles use N2O4 and a derivative of hydrazine, 1,1-dimethylhydrazine. This mixture is called a hypergolic fuel because it ignites when the reactants come into contact: H2NN(CH3)2(ℓ) + 2 N2O4(ℓ) n   3 N2(g) + 4 H2O(g) + 2 CO2(g)

(a) Identify the oxidizing agent and the reducing agent in this reaction. (b) The same propulsion system was used by the Lunar Lander on moon missions in the 1970s. If the Lander used 4100 kg of H2NN(CH3)2, what mass (in kilograms) of N2O4 was required to react with it? What mass (in kilograms) of each of the reaction products was generated? 102. ▲ Liquid HCN is dangerously unstable with respect to trimer formation—that is, formation of (HCN)3 with a cyclic structure. (a) Propose a structure for this cyclic trimer. (b) Estimate the energy of the trimerization reaction using bond-dissociation enthalpies (Table 8.8). 103. Use ΔfH° data in Appendix L to calculate the enthalpy change of the reaction 2 N2(g) + 5 O2(g) + 2 H2O(ℓ) n 4 HNO3(aq)

Speculate on whether such a reaction could be used to fix nitrogen. Would research to find ways to accomplish this reaction be a useful endeavor? 104. ▲ Phosphorus forms an extensive series of oxoanions. (a) Draw a structure, and give the charge for an oxophosphate anion with the formula [P4O13]n−. How many ionizable H atoms should the completely protonated acid have? (b) Draw a structure, and give the charge for a cyclic oxophosphate anion with the formula [P4O12]n−. How many ionizable H atoms should the completely protonated acid have? 105. ▲ Boron and hydrogen form an extensive family of compounds, and the following diagram shows how five of them (labeled A–E) are related by reaction.

Study Questions

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1099

+ H2 at 60 °C

B

C

100 °C 120 °C

100 °C 200 °C 180 °C

A

109. Metal sulfides roasted in air produce metal oxides. 2 ZnS(s) + 3 O2(g) n 2 ZnO(s) + 2 SO2(g)

25 °C

120 °C

D

E 95 °C

The following table gives the weight percent of boron in each of the compounds. Derive the empirical and molecular formulas of compounds A−E.

Substance

State (at STP)

(b) Gallium ion in water, Ga3+(aq), has a Ka value of 1.2 × 10−3. Is this ion a stronger or a weaker acid than Al3+(aq)?

Use thermodynamics to decide if the reaction is product- or reactant-favored at equilibrium at 298 K. Will the reaction be more or less productfavored at a high temperature? 110. Metals generally react with hydrogen halides to give the metal halide and hydrogen. Determine whether this is true for silver by calculating ΔrG° for the reaction with each of the hydrogen halides. Ag(s) + HX(g) n AgX(s) + 1⁄2 H2(g)

The required free energies of formation are:

Mass Percent B

Molar Mass (g/mol)

HX

−𝚫f G°(kJ/mol)

AgX

−𝚫f G°(kJ/mol)

HF

273.2

AgF

193.8

A

Gas

78.3

 27.7

HCl

95.09

AgCl

109.76

B

Gas

81.2

 53.3

HBr

53.45

AgBr

 96.90

C

Liquid

83.1

 65.1

HI

 1.56

AgI

 66.19

D

Liquid

85.7

 63.1

E

Solid

88.5

122.2

106. In 1774, both Carl Scheele and Joseph Priestley isolated oxygen. (a) Oxygen reacts with cesium to produce several compounds, among them one with an oxygen content of 19.39%. What is the formula of this compound? (b) The compound found in part (a) reacts with hydrogen peroxide to give oxygen and a cesium compound. Write a balanced equation for the reaction. 107. When palladium metal is exposed to H2 gas, the metal becomes brittle because H2 molecules dissociate and H atoms fill some of the octahedral holes in the face-centered cubic lattice. To find the value of x in the formula PdHx, you perform the following experiment: H2 gas in a 2.25-L flask has a pressure of 113 mm at 23.0 ˚C. After exposing the gas to 0.192 g of Pd, the pressure is now 108 mm at 23 ˚C. What is the value of x in PdHx? 108. The chemistry of gallium: (a) Gallium hydroxide, like aluminum hydroxide, is amphoteric. Write a balanced chemical equation to show how this hydroxide can ­dissolve in both HCl(aq) and NaOH(aq).

In the Laboratory 111. One material needed to make silicones is dichloro­dimethylsilane, (CH3)2SiCl2. It is made by treating silicon powder at about 300 °C with CH3Cl in the presence of a copper-containing catalyst. (a) Write a balanced chemical equation for the reaction. (b) Assume you carry out the reaction on a small scale with 2.65 g of silicon. To measure the CH3Cl gas, you fill a 5.60-L flask at 24.5 °C. What pressure of CH3Cl gas must you have in the flask to have the stoichiometrically correct amount of the compound? (c) What mass of (CH3)2SiCl2 can be produced from 2.65 g of Si and excess CH3Cl? 112. Sodium borohydride, NaBH4, reduces many metal ions to the metal. (a) Write a balanced chemical equation for the reaction of NaBH4 with AgNO3 in water to give silver metal, H2 gas, boric acid, and sodium nitrate. (The chemistry of NaBH4 is described in Section 21.6.) (b) What mass of silver can be produced from 575 mL of 0.011 M AgNO3 and 13.0 g of NaBH4?

1100 Chapter 21 / The Chemistry of the Main Group Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

© Charles D. Winters/Cengage

113. When 1.00 g of a white solid A is strongly heated, you obtain another white solid, B, and a gas. An experiment is carried out on the gas, showing that it exerts a pressure of 209 mm Hg in a 450-mL flask at 25 °C. Bubbling the gas into a solution of Ca(OH)2 gives another white solid, C. If the white solid B is added to water, the resulting solution turns red litmus paper blue. Addition of aqueous HCl to the solution of B and evaporation of the resulting solution to dryness yields 1.055 g of a white solid D. When D is placed in a Bunsen burner flame, it colors the flame green. Finally, if the aqueous solution of B is treated with sulfuric acid, a white precipitate, E, forms. Identify the lettered compounds in the reaction scheme.

The salts of CaCl2, SrCl2, and BaCl2 were suspended in methanol.  When the methanol was set ablaze, the heat of combustion causes the salts to emit light of characteristic wavelengths: Calcium salts produce yellow flames (center); strontium salts produce red flames (right); and barium salts produce green-yellow flames (left). (CAUTION: This experiment should be done only with appropriate safety precautions and only by professional chemists.)

114. A common analytical method for hydrazine involves its oxidation with iodate ion, IO3−, in acid solution. In the process, hydrazine acts as a four-electron reducing agent. N2(g) + 5 H3O+(aq) + 4 e− n N2H5+(aq) + 5 H2O(ℓ)  E° = −0.23 V

Write the balanced chemical equation for the reaction of hydrazine in acid solution (N2H5+) with IO3−(aq) to give N2 and I2. Calculate E° for this reaction. 115. ▲ A yellow-orange bromine oxide (BrO2) can be prepared by treating Br2 with ozone in a fluorocarbon solvent. Upon heating, this oxide decomposes to two other oxides, a less volatile golden yellow oxide (A) and a more volatile deep brown



oxide (B). Oxide B was later identified as Br2O. To determine the formula for oxide A, a sample was treated with sodium iodide. The reaction liberated iodine, which was titrated to an equivalence point with 17.7 mL of 0.065 M sodium thiosulfate. I2(aq) + 2 S2O32−(aq) n 2 I−(aq) + S4O62−(aq)

Compound A was also treated with AgNO3, and 14.4 mL of 0.020 M AgNO3 was required to completely precipitate the bromine from the sample. (a) What is the formula of the unknown bromine oxide A? (b) Draw Lewis structures for A and Br2O. Speculate on their molecular geometry. 116. A mixture of PCl5 (12.41 g) and excess NH4Cl was heated at 145 °C for 6 hours. The two reacted in equimolar amounts and evolved 5.14 L of HCl (at STP). Three substances (A, B, and C) were isolated from the reaction mixture. The three substances had the same elemental composition but differed in their molar mass. Substance A had a molar mass of 347.7 g/mol and B had a molar mass of 463.5 g/mol. Give the empirical and molecular formulas for A and B and draw a reasonable Lewis structure for A.

Summary and Conceptual Questions The following questions may use concepts from this and ­previous chapters. 117. Dinitrogen trioxide, N2O3, has the structure shown here. O 114.2 pm

O N N

121 pm

130°

O

The oxide is unstable, decomposing to NO and NO2 in the gas phase at 25 °C. N2O3(g) n NO(g) + NO2(g)

(a) Draw resonance structures for N2O3 and explain why one NOO bond distance is 114.2 pm, whereas the other two bonds are longer (121 pm) and nearly equal to each other. (b) For the decomposition reaction, ΔrH° = +40.5 kJ/mol and ΔrG° = −1.59 kJ/mol. Calculate ΔrS° and K for the reaction at 298 K. (c) Calculate ΔfH° for N2O3(g).

Study Questions

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1101

118. ▲ The density of lead is 11.350 g/cm3, and the metal crystallizes in a face-centered cubic unit cell. Estimate the radius of a lead atom. 119. How would you extinguish a sodium fire in the laboratory? What normal fire-fighting technique should you not use? 120. Tin(IV) oxide, cassiterite, is the main ore of tin. It crystallizes in a rutile-like unit cell with tin(IV) ions taking the place of Ti4+ ions (Study Question 12.6). (The O2− ions marked x are wholly within the unit cell.) Sn4+

O2−

124. (a) Magnesium is obtained from seawater. If the concentration of Mg2+ in seawater is 0.050 M, what volume of seawater (in liters) must be treated to obtain 1.00 kg of magnesium metal? What mass of lime (CaO; in kilograms) must be used to precipitate the magnesium in this volume of seawater? (b) When 1.2 × 103 kg of molten MgCl2 is electrolyzed to produce magnesium, what mass (in kilograms) of metal is produced at the cathode? What is produced at the anode? What is the mass of this product? What is the total number of faradays of electricity used in the process? (c) One industrial process has an energy consumption of 18.5 kWh/kg of Mg. How many joules are required per mole (1 kWh = 1 kilowatt-hour = 3.6 × 106 J)? How does this energy compare with the energy of the following process? MgCl2(s) n Mg(s) + Cl2(g)

(a) How many tin(IV) ions and oxide ions are there per unit cell of this oxide? (b) Is it thermodynamically feasible to transform solid SnO2 into liquid SnCl4 by reaction of the oxide with gaseous HCl? What is the equilibrium constant for this reaction at 25 °C? 121. You are given a stoppered flask that could contain hydrogen, nitrogen, or oxygen. Suggest an experiment you could do to identify the gas. 122. The structure of nitric acid is illustrated on pages 809 and 1072. (a) Why are the NOO bonds the same length, and why are both shorter than the NOOH bond length? (b) Rationalize the bond angles in the molecule. (c) What is the hybridization of the central N atom? Which orbitals overlap to form the NOO π bond? 123. The reduction potentials for the Group 3A (13) metals, E°, are given below. What trend or trends do you observe in these data? What can you ­conclude about the strengths of Group 3A (13) elements as reducing agents from these data? Half-Reaction

−1.66

Ga (aq) + 3 e 88n Ga(s)

−0.53

In (aq) + 3 e 88n In(s)

−0.338

Tl (aq) + 3 e 88n Tl(s)

+0.72

3+

3+

−

−

−

126. ▲ Boron nitride, BN, has the same solid state structure as ZnS (Figure 12.9).

Reduction Potential (E°, V)

Al3+(aq) + 3 e− 88n Al(s) 3+

125. Comparing the chemistry of carbon and silicon. (a) Write balanced chemical equations for the reactions of H2O(ℓ) with CH4 (forming CO2 and H2) and SiH4 (forming SiO2 and H2). (b) Using thermodynamic data, calculate the standard free energy change for the reactions in (a). Is either reaction product-favored at equilibrium? (c) Look up the electronegativities of carbon, silicon, and hydrogen. What conclusion can you draw concerning the polarity of C—H and Si—H bonds? (d) Carbon and silicon compounds with the formulas (CH3)2CO (acetone) and [(CH3)2SiO]n (a silicone polymer) also have quite different structures. Draw Lewis structures for these species. This difference, along with the difference between structures of CO2 and SiO2, suggests a general observation about silicon compounds. Based on that observation, do you expect that a silicon compound with a structure similar to ethene (C2H4) exists?

N atoms define face B atom in lattice hole centered cubic lattice

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You can consider it a face-centered cube of nitrogen atoms with boron atoms in one half of the tetrahedral holes of the lattice. If the density of this form of BN is 3.45 g/cm3, what is the B−N bond length? 127. ▲ Xenon trioxide, XeO3, reacts with aqueous base to form the xenate anion, HXeO4−. This ion reacts further with OH− to form the perxenate anion, XeO64−, in the following reaction:

Identify the elements that are oxidized and reduced in this reaction. The equation is balanced with respect to the number of atoms on either side. Verify that the redox part of this equation is also balanced, that is, that the extents of oxidation and reduction are also equal.

2 HXeO4−(aq) + 2 OH−(aq) n  XeO64−(aq) + Xe(g) + O2(g) + 2 H2O(ℓ)



Study Questions

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1103

John C. Kotz

22

The Chemistry of the Transition Elements

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C hapt e r O ut li n e 22.1 Overview of the Transition Elements 22.2 Periodic Properties of the Transition Elements 22.3 Metallurgy 22.4 Coordination Compounds 22.5 Structures of Coordination Compounds 22.6 Bonding in Coordination Compounds

From earliest times, people have been fascinated by the colors of gems and ­minerals. Indeed, it is part of the reason your book authors became chemists. Many ­gemstones and minerals owe their color to transition metal ions in their structures. It is copper that gives the blue color to azurite, the mineral on the opposing page, or to malachite and turquoise. The red color of rubies or the green of emeralds is due to the presence of chromium. The objective of this chapter is to introduce the basic chemistry of the transition metals and to provide some insight into how so many of their compounds can exhibit a wide range of colors.

22.1 Overview of the Transition Elements Goals for Section 22.1 • Identify the general classes of transition metals based on their location in the periodic table. • Recognize general features of transition metals and their compounds.

© Charles D. Winters/Cengage

22.7 Colors of Coordination Compounds

Colors in gemstones.  The red color of rubies is due to traces of chromium ions in an aluminum oxide crystal lattice.

The transition elements are the large block of elements in the central portion of the periodic table, bridging the s-block elements at the left and the p-block elements on the right (Figure 22.1). The transition elements are divided into two groups: the d‑block elements, whose occurrence in the periodic table coincides with the filling of the d orbitals, and the f-block elements, characterized by filling of the f orbitals. Contained within this latter group of elements are two subgroups: the ­lanthanides, elements that occur between La and Hf, and the actinides, elements that occur between Ac and Rf. The d-block metals include elements with a wide range of properties. They encompass the most common metal used in construction and manufacturing (iron), metals valued for their beauty (gold, silver, and platinum), and metals used in coins (nickel, copper, and zinc). There are metals used in modern technology (titanium) and metals known and used in early civilizations (copper, silver, gold, and iron). The d-block contains the densest elements (osmium, d = 22.59 g/cm3, ◀ Azurite.  A mineral containing copper(II) ions with hydroxide and carbonate anions: Cu3(CO3)2(OH)2.

It was mined in ancient Egypt and used as a source of copper and probably as a pottery glaze. It was also used as a pigment in European art from the fifteenth to the eighteenth centuries.

1105

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d-block

Lanthanides Actinides f-block

Group 8B(8): Iron (Fe)

Group 4B(4): Titanium (Ti)

Group 5B(5): Vanadium (V)

Group 6B(6): Chromium (Cr)

Group 7B(7): Manganese (Mn)

Group 8B(9): Cobalt (Co)

Group 8B(10): Nickel (Ni)

Group 1B(11): Copper (Cu)

Group 2B(12): Zinc (Zn)

© Charles D. Winters/Cengage

Ti V Cr Mn Fe Co Ni Cu Zn

Figure 22.1  The transition metals.  ​The d-block elements and f-block elements (the transition elements) are highlighted in a darker shade of blue.

© Charles D. Winters/Cengage

and iridium, d = 22.56 g/cm3), the metals with the highest and lowest melting points (tungsten, mp = 3422 °C, and mercury, mp = −38.9 °C), and one element that has only radioactive isotopes, technetium (43Tc). With the exception of mercury, the transition elements are solids under ambient conditions. They have a metallic sheen and conduct electricity and heat. They react with various oxidizing agents to give ionic compounds, although there is considerable variation in such reactions among the elements. In contrast to most of these elements, gold and platinum resist oxidation; consequently, they are used for jewelry and decorative items. Certain d-block elements are important in living organisms. Cobalt is a c­ rucial element in vitamin B12, which is part of a catalyst essential for several biochemical reactions. Hemoglobin and myoglobin, oxygen-carrying and storage proteins, contain iron. Molybdenum and iron, together with sulfur, form the reactive portion of nitrogenase, a biological catalyst used by nitrogen-fixing organisms to convert atmospheric nitrogen into ammonia. Many transition metal compounds are highly colored, which makes them useful as pigments in paints and dyes (Figure 22.2). Prussian blue, Fe4[Fe(CN)6]3 ∙ ​14 H2O,

(a) Paint pigments: (left to right) yellow, CdS; green, Cr2O3; white, TiO2 and ZnO; purple, Mn3(PO4)2; blue, Co2O3; ochre, Fe2O3.

(b) Small amounts of transition metal com­ pounds are used to color glass: blue, Co2O3; green, copper or chromium oxides; purple, nickel or cobalt oxides; red, copper oxide; iridescent green, uranium oxide.

(c) Traces of transition metal ions are responsible for the colors of (counterclockwise from top) green jade (iron), red corundum ­(chromium), blue azurite (copper), blue-green turquoise (copper), and purple amethyst (iron).

Figure 22.2  Colorful chemistry.  ​Transition metal compounds have a wide variety of colors.

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Chemistry in Your Career

Shubha Banavar

Shubha Banavar Shubha Banavar is an artist and small-business owner who uses traditional Indian forms of painting to create artwork, jewelry, and home décor. Although she loves science and her educational background is in biology and biotechnology, her true passion is creating art. Banavar explains the importance of understanding the chemical reactions involved in different types of paint. For example, in traditional Indian painting natural pigments derived from sources such as plants and minerals are mixed with water, gum Arabic, and copper(II) sulfate

crystals (CuSO4). “The gum acts as the binding agent between the water and the pigments and the copper sulfate provides antifungal and antimicrobial properties to the final product. This makes the paint last for years without the necessity of a protective top layer such as varnish.” Different chemical reactions produce acrylic and oil paints, which produce artwork with distinct properties. “Studying chemistry in depth helped me get a much better understanding of the very complex processes involved in my artwork and the different materials I use.”

is a “bluing agent” used in engineering blueprints and in the laundry. A common pigment used by artists (cadmium yellow) contains cadmium sulfide, CdS, and the white pigment in most white paints is titanium(IV) oxide, TiO2. The presence of transition metal ions in crystalline silicates or alumina transforms these common materials into gemstones. Iron(III) ions produce the yellow color in citrine and the purple color of amethyst, and chromium(III) ions are responsible for the red color of a ruby. Transition metal compounds in small quantities add color to glass. Blue glass contains a small amount of a cobalt(III) oxide, and addition of chromium(III) oxide or uranium oxide gives glass a green color.

22.2 Periodic Properties of the Transition Elements Goal for Section 22.2 • Assess periodic trends of the transition metals based on electronic structure.

Electron Configurations Because chemical behavior is related to electronic structure, it is important to know the electron configurations of the d-block elements (Table 22.1 and Section 7.3) and their common ions (Sections 7.4). Recall that the configuration of these metals has the general form [noble gas core](n − 1)dxnsy; that is, valence electrons for the transition elements reside in the ns and (n − 1)d subshells.

Common Oxidation States A characteristic chemical property of all metals is that they undergo oxidation by a wide range of oxidizing agents such as oxygen, the halogens, and some aqueous acids. Standard reduction potentials for the elements of the first transition series can be used to predict which elements will be oxidized by a given oxidizing agent in aqueous solution. For example, all of these metals except vanadium and copper are oxidized by aqueous HCl (Table 22.2). When a transition metal is oxidized, the outermost s electrons are transferred to the oxidizing agent, followed by one or more d electrons. With a few exceptions, transition metal ions have the electron configuration [noble gas core](n − 1)dx

22.2  Periodic Properties of the Transition Elements

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1107

Table 22.1

Table 22.2

Electron Configurations of the Fourth-Period Transition Elements

spdf Configuration Box Notation 3d Sc Ti V Cr Mn Fe Co Ni Cu Zn

[Ar]3d14s2 [Ar]3d24s2 [Ar]3d34s2 [Ar]3d54s1 [Ar]3d54s2 [Ar]3d64s2 [Ar]3d74s2 [Ar]3d84s2 [Ar]3d104s1 [Ar]3d104s2

4s

Products from Reactions of the Elements in the First Transition Series with O2, Cl2, or Aqueous HCl

Element

Reaction with O2*

Reaction with Cl2

Reaction with Aqueous HCl

Scandium

Sc2O3

ScCl3

Sc3+(aq)

Titanium

TiO2

TiCl4

Ti3+(aq)

Vanadium

V2O5

VCl4

NR†

Chromium

Cr2O3

CrCl3

Cr2+(aq)

Manganese

MnO2

MnCl2

Mn2+(aq)

Iron

Fe2O3

FeCl3

Fe2+(aq)

Cobalt

Co2O3

CoCl2

Co2+(aq)

Nickel

NiO

NiCl2

Ni2+(aq)

Copper

CuO

CuCl2

NR†

Zinc

ZnO

ZnCl2

Zn2+(aq)

*Product obtained with excess oxygen. † NR = no reaction.

© Charles D. Winters/Cengage

where x can vary from 0 (Ti4+) to 10 (Cu+). In contrast to ions formed by main group elements, transition metal cations often possess unpaired electrons, resulting in paramagnetism (Section 7.4). Their compounds are frequently colored as well, due to the absorption of light in the visible region of the electromagnetic spectrum. Color and magnetism figure prominently in a discussion of the properties and bonding of these elements, as you shall see shortly. In the first transition series, the most commonly encountered compounds have metal ions with oxidation numbers of +2 and +3 (Table 22.2). With iron, for example, oxidation converts Fe([Ar]3d64s2) to either Fe2+([Ar]3d6) or Fe3+([Ar]3d5). Iron reacts with chlorine to give FeCl3, and it reacts with aqueous acids to produce Fe2+(aq) and H2 (Figure 22.3). Despite the preponderance of 2+ and 3+ ions in compounds of the first transition metal series, the range of possible oxidation states for these compounds is broad (Figure 22.4). For example, chromium has an oxidation number of +6 in CrO42− and Cr2O72−, manganese has an oxidation number of

(a) Steel wool reacts with O2.

(b) Steel wool reacts with chlorine gas, Cl2.

(c) Iron chips react with aqueous HCl.

Figure 22.3  Typical reactions of transition metals.  ​Iron reacts with oxygen, with halogens, and with acids under appropriate conditions.

1108 Chapter 22 / The Chemistry of the Transition Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Cr(NO3)3

CrCl3

K2CrO4

K2Cr2O7

7

5 4 3 © Charles D. Winters/Cengage

Oxidation states

6

2 1 Sc 3B (3)

Ti 4B (4)

V 5B (5)

Cr 6B (6)

Mn 7B (7)

Fe

Co

Ni

(8)

8B (9)

(10)

Cu 1B (11)

(a) The most common oxidation states are indicated with red squares; less common oxidation states are indicated with blue dots.

Figure 22.4  Oxidation states of the transition elements in the first transition series.  ​

(b) Aqueous solutions of chromium compounds with two different oxidation numbers: +3 in Cr(NO3)3 (violet) and CrCl3 (green), and +6 in K2CrO4 (yellow) and K2Cr2O7 (orange). The two Cr(III) species have different colors in solution because there are different complex ions based on Cr(III) in solution. The complex ion in the purple solution is [Cr(H2O)6]3+; the complex ion in the green solution is [Cr(H2O)4Cl2]+.

+7 in MnO4−, silver and copper often form 1+ ions, and vanadium has oxidation numbers that can range from +5 to +1 (see Figure 19.3). Higher oxidation numbers are common in compounds of the elements in the second and third transition series. For example, the naturally occurring sources of molybdenum and tungsten are the ores molybdenite [MoS 2, Mo(IV)] and wolframite [WO 3, W(VI)]. This general trend to form compounds in high oxidation states is carried over in the f-block. Most lanthanide compounds have the metal ion in the +3 oxidation state. In contrast, actinide elements usually have higher oxidation numbers in their compounds; +4 and even +6 are typical. For example, UO3 [(U(VI)]is a common oxide of uranium, and UF6 [(U(VI)] is a compound important in processing uranium fuel for nuclear reactors (Section 20.7).

The Highest Oxidation Number  Iridium, element 77,

is found in compounds where the element has an oxidation number ranging from –3 to +8. Recently the cation IrO4+ was characterized, the first time an oxidation number of +9 has been observed for any element.

Periodic Trends in the d-Block: Size, Density, Melting Point Atomic radii, density, and melting point are three physical properties of the transition elements that vary in a periodic manner.

Metal Atom Radii The radii of the transition elements vary over a fairly narrow range, with a small decrease to a minimum being observed around the middle of the d-block of elements (Figure 7.9). Atom size is determined by the electrons in the outermost orbital, which for these elements is the ns orbital (n = 4, 5, or 6). Progressing from left to right in the periodic table, the size decline expected from increasing the number of protons in the nucleus is mostly canceled out by an opposing effect, repulsion from additional electrons in the (n − 1)d orbitals. Another important observation is that the radii of the d-block elements in the fifth and sixth periods in each group are almost identical, an effect that is known as the lanthanide contraction. Lanthanum (57La) and hafnium (72Hf) are bridged by the lanthanide elements [cerium (58Ce) through lutetium (71Lu)]. The filling of

22.2  Periodic Properties of the Transition Elements

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1109

4f orbitals in the lanthanide elements is accompanied by a steady contraction in size, consistent with the general trend of decreasing size from left to right in the periodic table. At the point where the 5d orbitals begin to fill again in the sixth period (at hafnium), the radii have decreased to a size similar to that of elements in the fifth period. Thus, taking Group 4B (4) as an example, the radius of the fifth period element zirconium (40Zr) is 160 pm and that of its sixth period group ­member, hafnium (72Hf), is 156 pm. The similar sizes of the second and third series d-block elements have significant consequences for their chemistry. For example, the platinum group metals (Ru, Os, Rh, Ir, Pd, and Pt) all have radii in the range 134–138 pm. Furthermore, their 2+ ions are nearly identical in size (85–89 pm), and their chemical properties are very similar. Thus, it is not surprising that minerals containing these metals are found in the same geological zones on Earth. Nor is it surprising that it is difficult to separate these elements from one another.

Density The variation in metal radii causes the densities of the transition elements to first increase and then decrease across a period (Figure 22.5, left). Although the overall change in radii among these elements is small, the effect is magnified because the volume is actually changing with the cube of the radius [V = (4/3)πr3]. You can see this effect by comparing the fourth period elements titanium (d = 4.51 g/cm3; r = 145 pm) in Group 4B (4) and nickel (d = 8.91 g/cm3; r = 125 pm) in Group 8B (10). The lanthanide contraction explains why elements in the sixth period have the highest densities. The relatively small radii of sixth-period transition metals, combined with the fact that their atomic masses are considerably larger than their counterparts in the fifth period, causes sixth-period metal densities to be larger.

Melting Point Melting points reflect the forces of attraction between the atoms, molecules, or ions that compose the solid. With transition elements, the melting points rise to a ­maximum around the middle of the series (Figure 22.5, right), then descend, indicating that the strongest metallic bonds occur when the d subshell is about half filled. This is also the point at which the largest number of electrons occupy the bonding molecular orbitals in the metal. (See the discussion of bonding in metals in Section 12.4.) 25

Fourth period

Fifth period

4000

Sixth period

Fourth period

Fifth period

Sixth period

3500 3000 Melting point (K)

Density (g/mL)

20

15

10

2500 2000 1500 1000

5

500 0

3B (3)

4B (4)

5B (5)

6B (6)

7B 8B (7) (8, 9, 10) Group

1B 2B (11) (12)

0

3B (3)

4B (4)

5B (5)

6B (6)

7B 8B (7) (8, 9, 10) Group

1B 2B (11) (12)

Figure 22.5  Periodic properties in the transition series.  ​Density (left) and melting point (right) of the d-block elements.

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22.3 Metallurgy Goal for Section 22.3 • Describe the metallurgy of iron and copper. Most transition metals are found in nature as oxides, sulfides, halides, carbonates, or other ionic compounds (Figure 22.6), but metals such as copper, silver, and gold can occur uncombined. Some metal-containing mineral deposits have little ­economic value, either because the concentration of the metal is too low or because the metal is difficult to separate from impurities. The relatively few minerals from which elements can be obtained profitably are called ores. Metallurgy is the general name given to the process of obtaining metals from their ores. Very few ores are chemically pure substances. Instead, the desired mineral is ­usually mixed with large quantities of impurities such as sand and clay, called gangue (­pronounced “gang”). Generally, the first step in a metallurgical process is to s­ eparate the mineral from the gangue. Then the ore is converted to the metal, a reduction process. Pyrometallurgy and hydrometallurgy are two methods of recovering metals from their ores. Pyrometallurgy involves high temperatures and hydrometallurgy uses aqueous solutions (and thus is limited to the relatively low temperatures at which water is a liquid). Iron and copper metallurgy illustrate these two methods of metal production.

Pyrometallurgy: Iron Production Iron is produced from its ores in a blast furnace (Figure 22.7). The furnace is loaded, or charged, with a mixture of ore (usually hematite, Fe2O3), coke (which is primarily carbon), and limestone (CaCO3). A blast of hot air forced in at the bottom of the furnace causes the coke to burn with such an intense heat that the temperature at the bottom is almost 1500 °C. The quantity of air forced into the furnace is controlled so that carbon monoxide is the primary product. Both carbon and carbon monoxide participate in the reduction of iron(III) oxide to give impure metal.

Coke: A Reducing Agent  Coke is

made by heating coal in a tall, narrow oven that is sealed to keep out oxygen. Heating drives off volatile chemicals, including benzene and ammonia. What remains is nearly pure carbon.

Fe2O3(s) + 3 C(s) 88n 2 Fe(ℓ) + 3 CO(g) Fe2O3(s) + 3 CO(g) 88n 2 Fe(ℓ) + 3 CO2(g)

Much of the carbon dioxide formed in the reduction process (and from heating limestone) is reduced on contact with unburned coke and produces carbon monoxide, which can then function as a reducing agent. CO2(g) + C(s) 88n 2 CO(g)

© Charles D. Winters/Cengage

Li

Copper occurs in nature as the metal (native copper) and as minerals such as blue azurite [2 CuCO3 ∙ Cu(OH)2] and green malachite [CuCO3 ∙ Cu(OH)2].

H

He

Be

B

C

N

O

F

Ne

Na Mg

Al

Si

P

S

Cl

Ar

K

Ca

Sc

Ti

V

Rb

Sr

Y

Zr

Nb Mo

Cs

Ba

La

Hf

Ta

W

Lanthanides

Ce

Cr

Fe

Co

Ni

Cu

Zn

Ga

Ge

As

Se

Br

Kr

Ru

Rh

Pd

Ag

Cd

In

Sn

Sb

Te

I

Xe

Re

Os

Ir

Pt

Au Hg

Tl

Pb

Bi

Pr

Nd

Sm

Eu

Tb

Dy Ho

Mn

Gd

Rn Er

Tm

Yb

Lu

Key

Sulfides

Oxides

Can occur uncombined

Halide salts

Phosphates Silicates

C from coal, B from borax

Carbonates

Figure 22.6  Sources of the elements.  ​A few transition metals, such as copper and gold, occur in nature as the metal. Most other elements are found naturally as oxides, sulfides, or other salts. 22.3 Metallurgy

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1111

Figure 22.7  A blast furnace.  ​ The largest modern furnaces have hearths 14 meters in diameter. They can produce as much as 10,000 tons of iron per day.

Charge of ore, coke, and limestone

Flue gas

230 °C

525 °C Hot gases used to preheat air

Reducing zone 945 °C

Heated air 1510 °C Slag Molten iron

The molten iron flows down through the furnace and collects at the bottom, where it is tapped off through an opening in the side. This impure iron is called cast iron or pig iron. Usually, the impure metal is either brittle or soft (undesirable properties for most uses) due to the presence of impurities such as elemental carbon, phosphorus, and sulfur. Iron ores generally contain silicate minerals and silicon dioxide. Lime (CaO), formed when limestone is heated, reacts with these materials to give calcium silicate.

Ammit Jack/Shutterstock.com

SiO2(s) + CaO(s) 88n CaSiO3(ℓ)

Figure 22.8  Molten iron being poured from a basic oxygen furnace.  ​

(This is an acid–base reaction because CaO is a basic oxide and SiO2 is an acidic oxide.) The calcium silicate, molten at the temperature of the blast furnace and less dense than molten iron, floats on the iron. Other nonmetal oxides dissolve in this layer and the mixture, called slag, is easily removed. Pig iron from the blast furnace may contain as much as 4.5% carbon, 0.3% phosphorus, 0.04% sulfur, 1.5% silicon, and some other elements as well. The impure iron must be purified to remove these nonmetal components, and several purification processes are available. The most important of these uses the basic ­oxygen furnace (Figure 22.8). Pure oxygen is blown into the molten pig iron and oxidizes phosphorus to P4O10, sulfur to SO2, and carbon to CO2. These n ­ onmetal oxides either escape as gases or react with basic oxides such as CaO that are added or are used to line the furnace. For example, P4O10(g) + 6 CaO(s) 88n 2 Ca3(PO4)2(ℓ)

The result is ordinary carbon steel. Almost any degree of flexibility, hardness, strength, and malleability can be achieved in carbon steel by reheating and cooling in a process called tempering. The resulting material can then be used in a wide variety of applications. The major disadvantages of carbon steel are that it corrodes easily and that it loses its properties when heated strongly.

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Other transition metals, such as chromium, manganese, and nickel, can be added during the steel-making process, giving alloys (solids made up of two or more metals, page 606) that have specific physical, chemical, and mechanical properties. One well-known alloy is stainless steel, which contains 18% to 20% Cr and 8% to 12% Ni. Stainless steel is much more resistant to corrosion than carbon steel. Another alloy of iron is the permanent magnet Alnico 5, which contains five elements: Al (8%), Ni (14%), Co (24%), Cu (3%), and Fe (51%). Alnico magnets are used in consumer products such as motors, guitar pickups, loud speakers, and microphones.

Hydrometallurgy: Copper Production In contrast to iron ores, which are usually oxides, most copper minerals are ­sulfides. Copper-bearing minerals include chalcopyrite (CuFeS2), chalcocite (Cu2S), and co­vel­lite (CuS). Because ores containing these minerals generally have a very low percentage of copper, enrichment is necessary. This step is carried out by a process known as flotation. First, the ore is finely powdered. Next, oil is added and the mixture is agitated with soapy water in a large tank (Figure 22.9a). At the same time, compressed air is forced through the mixture, so that the lightweight, oil-covered copper sulfide particles are carried to the top as a frothy mixture. The heavier gangue settles to the bottom of the tank, and the copper-laden froth is skimmed off. After evaporation what remains is an enriched ore. Hydrometallurgy can be used to obtain copper from an enriched ore. In one method, enriched chalcopyrite ore (CuFeS2) is treated with a solution of copper(II) chloride. These react to leave copper in the form of solid, insoluble copper(I) ­chloride, CuCl, which is easily separated from the iron that remains in solution as aqueous FeCl2. CuFeS2(s) + 3 CuCl2(aq) 88n 4 CuCl(s) + FeCl2(aq) + 2 S(s)

Aqueous NaCl is then added, and CuCl dissolves upon formation of the soluble complex ion [CuCl2]−. CuCl(s) + Cl−(aq) 88n [CuCl2]−(aq)

Copper(I) compounds in solution are unstable with respect to Cu(0) and Cu(II). Thus, [CuCl2]− disproportionates to the metal and copper(II) chloride, CuCl2, and the latter is used to treat further ore. 2 [CuCl2]−(aq) 88n Cu(s) + Cu2+(aq) + 4 Cl−(aq)

(a) Flotation.  The less dense ­particles of Cu2S are trapped in the soap bubbles and float. The denser gangue settles to the bottom.

Anode +

Tom Hollyman/Science Source

John C. Kotz

− Cathode

Thin sheets of pure copper Solution of CuSO4 and H2SO4

(b) Electrolysis.

Slabs of impure copper

(c) Producing pure copper by electrolysis.

Figure 22.9  Copper metallurgy.  ​

22.3 Metallurgy

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1113

Approximately 10% of the copper produced in the United States is obtained with the aid of bacteria. Acidified water is sprayed onto copper-mining wastes that contain low levels of copper. As the water trickles down through the crushed rock, the bacterium Thiobacillus ferrooxidans breaks down the iron sulfides in the rock and converts iron(II) to iron(III). Iron(III) ions oxidize the sulfide ion of copper ­sulfide to sulfate ions, leaving copper(II) ions in solution. Then the copper(II) ion is reduced to metallic copper by reaction with iron. Cu2+(aq) + Fe(s) 88n Cu(s) + Fe2+(aq)

The purity of the copper obtained by these metallurgical processes is about 99%, but even this is not acceptable because traces of impurities greatly diminish the electrical conductivity of the metal. Consequently, a further purification step involving electrolysis is needed (Figures 22.9b and 22.9c). Thin sheets of pure copper metal and slabs of impure copper are immersed in a solution containing CuSO4 and H2SO4. The pure copper sheets serve as the cathode of an electrolysis cell, and the impure slabs are the anode. Copper in the impure sample is oxidized to copper(II) ions at the anode, and copper(II) ions in solution are reduced to pure copper at the cathode.

22.4 Coordination Compounds Goals for Section 22.4 • Know the names and formulas of common monodentate and bidentate ligands. • Given the formula for a coordination complex, identify the metal and its oxidation state, the ligands, the coordination number and coordination geometry, and the overall charge on the complex.

• Given a formula, name a complex; given a name of a complex, write its formula.

Complexes and Ligands A green solution formed by dissolving nickel(II) chloride in water contains Ni2+(aq) and Cl−(aq) ions (Figure 22.10). If the solvent is removed, a green crystalline solid remains. The formula of this solid is often written as NiCl2 ∙ 6 H2O, and the classical name for this compound is nickel(II) chloride hexahydrate. Addition of ammonia to the aqueous nickel(II) chloride solution gives a lilac-colored solution from which another compound, NiCl2 ∙ 6 NH3, can be isolated. This formula looks very similar to the formula for the hydrate, with ammonia substituted for water. What are these two nickel species? The formulas identify the compositions of the compounds but give no information about their structures. Because properties of compounds derive from their structures, the structures should be evaluated in more detail. Typically, metal compounds are ionic, and solid ionic compounds have structures with cations and anions arranged in a regular array. The structure of hydrated nickel(II) chloride contains cations with the formula [Ni(H 2O) 6] 2+ and chloride anions. The structure of the ammonia-containing compound is similar to the hydrate; it is made up of [Ni(NH3)6]2+ cations and chloride anions. Ions such as [Ni(H 2O) 6] 2+ and [Ni(NH 3) 6] 2+, in which a metal ion and ­molecules (or ions) such as water or ammonia compose a single structural unit, are examples of coordination complexes, also known as complex ions (Figure 22.11). Compounds containing a coordination complex as part of the structure are called coordination compounds, and their chemistry is known as coordination ­chemistry.

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Figure 22.10  Coordination compounds of Ni2+ ion.  ​The transition metals and their

[Ni(H2O)6]2+

ions form a wide range of compounds, often with beautiful colors and interesting structures. One pur­pose of this chapter is to explore some commonly observed structures and explain how these compounds can be so colorful.

Add NH3 Add NaOH

[Ni(NH3)6]2+

Insoluble Ni(OH)2 Add dimethylglyoxime (dmg)

Photos: © Charles D. Winters/Cengage

Add ethylenediamine NH2CH2CH2NH2

[Ni(NH2CH2CH2NH2)3]2+

Ni(dmg)2

Although the older hydrate formulas are still used, the preferred method of writing the formula for coordination compounds places the metal atom or ion and the molecules or anions directly bonded to it within brackets to show that it is a single structural entity. Thus, the formula for the nickel(II)–ammonia compound is better written as [Ni(NH3)6]Cl2. All coordination complexes contain a metal atom or ion as the central part of the structure. Bonded to the metal are molecules or ions called ligands (from the Latin verb ligare, meaning “to bind”). In the preceding examples, water and



22.4  Coordination Compounds

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1115

ammonia are the ligands. The number of ligand atoms attached to the metal defines the coordination number of the metal. The geometry around the metal described by the attached ligands is called the coordination geometry. In [Ni(NH 3) 6] 2+ (­F igure 22.11), nickel has a coordination number of six and the six ligands are arranged in a regular octahedral geometry around the central metal ion. Ligands can be either neutral molecules or anions (or, in rare instances, cations). The characteristic feature of a ligand is that it contains a lone pair of electrons. In the classic description of bonding in a coordination complex, the lone pair of electrons on a ligand is shared with the metal cation. That is, this is an interaction of a Lewis base (the ligand) with a Lewis acid (the metal ion or atom), and the bond formed between the acid and the base is called a coordinate covalent bond (Section 8.6). The name “coordination complex” derives from the name given to this kind of bonding. The net charge on a coordination complex is the sum of the charges on the metal and its attached groups. Complexes can be cations (as in the two nickel complexes used as examples here), anions, or neutral species. Ligands such as H 2O and NH 3, which coordinate to the metal via a single Lewis base atom, are termed monodentate. The word “dentate” comes from the Latin word dentis, meaning “tooth,” so NH3 is a one-toothed ligand. Some ligands—called polydentate ligands—attach to the metal with more than one donor atom. Ethylenediamine (1,2-diaminoethane), H 2NCH 2CH 2NH 2, often abbreviated as en; oxalate ion, C 2O 42− (ox 2−); and phenanthroline, C 12H 8N 2 (phen) are examples of bidentate ligands (Figure 22.12). Structures and examples of some complex ions with bidentate ligands are shown in Figure 22.13. Be sure to notice that  all common bidentate ligands bind to adjacent sites on the metal.  Polydentate ligands are also called chelating ligands, or just chelates (pronounced “key-lates”). The name derives from the Greek chele, meaning “claw.” ­Because two or more bonds need to be broken to separate the ligand from the metal, complexes with chelated ligands often have greater stability than those with monodentate ligands. Complexes with chelating ligands are important in many applications. For example, one way to clean the rust out of water-cooled automobile engines and steam boilers is to add a solution of oxalic acid. Iron(III) oxide reacts with oxalic acid to give a water-soluble iron oxalate complex ion (the green compound on the left in Figure 22.13):

Sum of metal ion and ligand charges Coordination complex

2+

H N

Ni2+

Coordinated metal ion

Ligand

[Ni(NH3)6]2+

Figure 22.11  A coordination complex.  ​In the [Ni(NH3)6]2+ ion, the ligands are NH3 molecules. Because the metal has a 2+ charge and the ligands have no charge, the charge on the complex ion is 2+.

3 H2O(ℓ) + Fe2O3(s) + 6 H2C2O4(aq) 88n 2 [Fe(C2O4)3]3−(aq) + 6 H3O+(aq)

Ethylenediaminetetraacetate ion (EDTA4−), a hexadentate ligand­(­Figure 22.14), is an excellent chelating ligand because it can wrap around a metal ion,

− −

(a) H2NCH2CH2NH2, en ethylenediamine

−

(b) C2O42−, ox oxalate ion

(c) CH3COCHCOCH3−, acac− acetylacetonate ion

(d) C12H8N2, phen phenanthroline

Figure 22.12  Common bidentate ligands.  ​Coordination of these bidentate ligands to a transition metal ion results in five- or six-member metal-containing rings and no ring strain. (Only one resonance structure for the acetylacetonate and oxalate ions and the phenanthroline molecule are shown.)

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© Charles D. Winters/Cengage

[Fe(C2O4)3]3−

[Co(en)3]3+

Cr(acac)3

Figure 22.13  Complex ions with bidentate ligands.  ​The three ligands shown are the oxalate anion (C2O42−), ethylenediamine (en), and the acetylacetonate anion (acac) (Figure 22.12).

encapsulating it. Salts of this anion are often added to commercial salad dressings to remove traces of free metal ions from solution; otherwise, these metal ions could act as catalysts for the oxidation of the oils in the dressing. Without EDTA 4−, the dressing would quickly become rancid. Another use is in bathroom cleansers. The EDTA4− ion removes deposits of CaCO3 and MgCO3 left by hard water by coordinating to Ca2+ or Mg2+ ions to create soluble complex ions. Complexes with polydentate ligands play important roles in biochemistry; one example is described in “A Closer Look: Hemoglobin: A Molecule with a Tetradentate Ligand.”

Formulas of Coordination Compounds It is useful to be able to predict the formula of a coordination complex, given the metal ion and ligands, and to derive the oxidation number of the coordinated metal ion, given the formula of a coordination compound. The following examples explore these issues.

−

−

− −

(a) ethylenediaminetetraacetate, EDTA42

(b) [Co(EDTA)]2

Figure 22.14 EDTA , a hexadentate ligand.  ​(a) Ethyl­ene­diamine­tetraacetate, EDTA4−. 4−

(b) The complex ion [Co(EDTA)]−, in which EDTA4− binds to Co3+. Notice the five-member rings created when this ligand bonds to the metal.



22.4  Coordination Compounds

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1117

A Closer Look

Hemoglobin: A Molecule with a Tetradentate Ligand

Metal-containing coordination compounds are important in many biochemical reactions. Perhaps the best-known example is hemoglobin, the molecule responsible for O2 transport in the blood. It is also one of the most thoroughly studied bio­ inorganic compounds. Hemoglobin (Hb) is a large ironcontaining protein (see Figures 24.4 and 24.5). It includes four polypeptide segments, each containing an iron(II) ion locked inside a porphyrin ring system and coordinated to a nitrogen atom from another part of the protein. A sixth site is available to attach to oxygen. Hemoglobin functions by reversibly adding oxygen as a ligand to the sixth coordination position of each iron, giving a complex called oxyhemoglobin. Because hemoglobin has four iron centers, a maximum of four molecules of  oxygen can bind to the molecule.

O

O

120°

Fe

N N

Base from protein

Figure  Oxygen binding. Oxygen

binds to the iron of the heme group in oxyhemoglobin (and in myoglobin). Interestingly, the FeOOOO angle is bent. The binding to oxygen is cooperative; that is, binding one O 2  molecule enhances the tendency to bind the second, third, and fourth O 2 molecules. Formation of the oxygenated complex is favored, but not too highly, because oxygen must also be released by the molecule to body tissues. Interestingly, an increase in acidity

leads to a decrease in the stability of the oxygenated complex. This phenomenon is known as the Bohr effect, named for Christian Bohr, the father of Niels Bohr (Section 6.3). Release of oxygen in tissues is facilitated by an increase in acidity that results from the presence of CO2 formed by metabolism. Among the properties of hemoglobin is its ability to form a complex with carbon monoxide. This complex is very stable, with the equilibrium constant for the following reaction being about 200 (where Hb is hemoglobin):

HbO2(aq) + CO(g) uv HbCO(aq) + O2(g) When CO complexes with iron, the oxygen-carrying capacity of hemoglobin is lost. Consequently, CO is highly toxic to humans. Exposure to even small amounts greatly reduces the capacity of the blood to transport oxygen.

Figure  Porphyrin ring of the heme group. The tetradentate N N

H

N H

N

porphyrin

+

−2 H

N

N

N

N

porphyrin2−

ligand surrounding the iron(II) ion in hemoglobin is a dianion of a molecule called a porphyrin. Because of the double bonds in this structure, all of the carbon and nitrogen atoms in the dianion of the porphyrin lie in a plane. In addition, the nitrogen lone pairs are directed toward the center of the ion, and the molecular dimensions are such that a metal ion may fit nicely into the cavity.

E xamp le 22.1

Formulas of Coordination Complexes Problem  Give the formula for each of the following coordination complexes: (a) A Ni2+ ion is bound to two water molecules and two bidentate oxalate ions. (b) A Co3+ ion is bound to one Cl − ion, one ammonia molecule, and two bidentate ethylenediamine (en) molecules.

What Do You Know?  The composition of each coordination complex is given. Strategy  The problem requires determining the net charge, which equals the sum of the charges of the various parts of the complex ion. The metal ion and the formulas of the ligands are placed in square brackets, and the net charge indicated.

Solution (a) This complex ion is constructed from two neutral H2O molecules, two C2O42− ions, and one Ni2+ ion, so the net charge on the complex is 2−. The formula for the complex ion is [Ni(C2O4)2(H2O)2]2−

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(b) This cobalt(III) complex ion combines two ethylenediamine molecules and one NH3 molecule, none having a charge, as well as one Cl− ion and a Co3+ ion. Thus, the net charge is 2+. The formula for this complex (writing out the formula for ethylenediamine) is [Co(H2NCH2CH2NH2)2(NH3)Cl]2+

Think about Your Answer   Enclosing the formula of a complex in brackets is done to designate that this is a single unit. Species outside the brackets are counterions (when shown). In general, you can calculate the charge on a complex ion as follows: Net charge = charge on central cation + (charge on anionic ligands) × (number of anionic ligands).

Check Your Understanding (a) What is the formula of a complex ion composed of one Co3+ ion, three ammonia ­molecules, and three Cl− ions? (b) What is the formula for the coordination complex of an Fe2+ ion having two ethylenediamine ligands and two bromide ion ligands?

Ex am p le 22.2

Coordination Compounds Problem  In each of the following coordination compounds, determine the charge on the central metal ion and coordination number. (a) [Co(en)2(NO2)2]Cl (b) Pt(NH3)2(C2O4) (c) Pt(NH3)2Cl4 (d) [Co(NH3)5Cl]SO4

What Do You Know?   The central metal in a coordination complex is a neutral atom or a cation. The coordination number is the number of ligand attachments to the metal. Remember that bidentate ligands coordinate to two adjacent positions.

Strategy   Each formula consists of a complex ion or molecule made up of the metal atom or ion with neutral or anionic ligands. Counterions (shown outside the square ­ rackets) may also be present. The oxidation number of the central metal is the charge b necessary to balance the sum of the negative charges associated with any anionic ligands and counterions. The coordination number is the number of donor atoms in the ligands that are bonded to the metal. Remember that bidentate ligands in these examples (en, oxalate ion) attach to the metal at two sites and any counterions present are not part of the complex ion—that is, they are not ligands.

Solution (a)

Two neutral bidentate en ligands

H2N H2N



NH2 Co NH2

Two NO2− ligands +

NO2

Charge on complex ion because there is one Cl− counterion.

NO2   

The central cobalt ion has a 3+ charge and the coordination number is 6.

22.4  Coordination Compounds

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1119

(b) Two neutral

One C2O42− bidentate ligand

NH3 ligands

O H3N H3N

Pt

O

C

O

C

The central platinum ion has a 2+ charge and the    coordination number is 4.

O (c) Two neutral

Four Cl− ligands

NH3 ligands

H3N H3N

Cl Pt Cl

(d) Five neutral

H3N

The central platinum ion has a 4+ charge and the    coordination number is 6.

Cl

One Cl− ligand

NH3 ligands

H3N

Cl

NH3 Co NH3

2+ Cl

Charge on complex ion because there is one SO42− counterion.

NH3   

The central cobalt ion has a 3+ charge and the coordination number is 6.

Think about Your Answer  Knowing the charge on the central metal in a coordination compound is important. From this, you can determine the electron configuration of the metal. As described in Sections 22.6 and 22.7, this information is needed to determine important properties of a complex such as magnetism and color. Finally, as noted in Example 22.1, you can always determine the central metal atom charge because the net charge on a coordination complex = charge on central cation + (charge on anionic ligands) × (number of anionic ligands).

Check Your Understanding (a) Determine the oxidation number and coordination number for the central metal (Co or Mn) in (i) K3[Co(NO2)6] and (ii) Mn(NH3)4Cl2. (b) Determine the oxidation number and coordination number of cobalt in the complex NH4[Co(EDTA)].

Naming Coordination Compounds Coordination compounds are named according to an established system. The three compounds below are named according to the rules that follow. Compound

Systematic Name

[Ni(H2O)6]SO4

Hexaaquanickel(II) sulfate

[Cr(en)2(CN)2]Cl

Dicyanobis(ethylenediamine)chromium(III) chloride

K[Pt(NH3)Cl3]

Potassium amminetrichloroplatinate(II)

1. Name the cation first and then the anion. (This is how all salts are commonly named.) 2. When giving the name of the complex ion or molecule, name the ligands first, in alphabetical order, followed by the name of the metal. (When determining the alphabetical order of the ligands, ignore any prefix.)

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3. Ligands and their names: (a) If a ligand is an anion whose name ends in -ite or -ate, the final e is changed to o (sulfate 88n sulfato or nitrite 88n nitrito). (b) If the ligand is an anion whose name ends in -ide, the ending is changed to o (chloride 88n chloro, cyanide 88n cyano). (c) If the ligand is a neutral molecule, its common name is usually used with several important exceptions: Water as a ligand is referred to as aqua; ­ammonia is called ammine; and CO is called carbonyl. (d) When there is more than one of a particular monodentate ligand with a simple name, the number of identical ligands is designated by the appropriate prefix: di, tri, tetra, penta, or hexa. If the ligand name is complicated (such as ­ethylenediamine), the prefix changes to bis, tris, tetrakis, pentakis, or hexakis, followed by the ligand name in parentheses. 4. If the coordination complex is an anion, the suffix -ate is added to the metal name. 5. Following the name of the metal, the oxidation number of the metal is given in Roman numerals.

Common Monodentate Ligands and Their Names Ligand

Name

F−

fluoro

Cl−

chloro

Br−

bromo

I−

iodo

H 2O

aqua

NH3

ammine

CO

carbonyl

CN

−

OH

cyano

−

ONO2

hydroxo −

nitro

Ex am p le 22.3

Naming Coordination Compounds Problem  Name the following compounds: (a) [Cu(NH3)4]SO4

(c) Co(phen)2(CN)2

(b) K2[CoCl4]

(d) [Co(en)2(H2O)Cl]Cl2

What Do You Know?  The formula of each compound is given. Strategy  Apply the rules for nomenclature given above. Solution (a) The complex ion (in square brackets) is composed of four NH 3 molecules (named ammine in a complex) and the copper ion. To balance the 2− charge on the sulfate counterion and account for four NH3 ligands with no charge, copper must have a 2+ charge. The compound’s name is tetraamminecopper(II) sulfate (b) The complex ion [CoCl4]2− has a 2− charge to balance two K+ counterions. With four Cl− ligands, the cobalt ion must have a 2+ charge, so the sum of charges is 2−. The name of the compound is potassium tetrachlorocobaltate(II) (c) This is a neutral coordination compound. The ligands include two CN− ions and two neutral bidentate phen (phenanthroline) ligands (page 1116). The metal ion must therefore have a 2+ charge (Co2+). The name, listing ligands in alphabetical order, is dicyanobis(phenanthroline)cobalt(II) (d) The complex ion has a 2+ charge because it is paired with two uncoordinated Cl− ions. The cobalt ion is Co3+ because it is bonded to two neutral ethylenediamine bidentate ligands, one neutral water, and one Cl−. The name is aquachlorobis(ethylenediamine)cobalt(III) chloride



22.4  Coordination Compounds

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1121

Think about Your Answer   The earliest studies of coordination compounds c­ reated a colorful array of names, and a few of these are still used (Prussian Blue, page 1137, and Zeise’s salt, K[Pt(C2H 4)Cl 3], are two examples). However, as the depth and breadth of coordination chemistry expands, a systematic nomenclature is essential to communication among scientists working in the area. The rules of nomenclature make it possible for any scientist to derive a formula from the name, or to name a compound given its formula.

Check Your Understanding  Name the following coordination compounds. (a) [Ni(H2O)6]SO4

(b) [Cr(en)2(CN)2]Cl

(c) K[Pt(NH3)Cl3]

(d) K[CuCl2]

22.5 Structures of Coordination Compounds Goals for Section 22.5 • Identify the different types of isomers that are encountered with coordination compounds.

• Given the molecular formula of a coordination compound, determine whether isomers are possible, and draw their structures.

Common Coordination Geometries linear [Ag(NH3)2]+

The geometry of a coordination complex is defined by the arrangement of donor atoms of the ligands around the central metal ion. Metal ions in coordination compounds can have coordination numbers ranging from 2 to 12. Only complexes with coordination numbers of 2, 4, and 6 are common, so many complex ions have formulas such as [ML2]n±, [ML4]n±, and [ML6]n±, where M is the metal ion and L is a monodentate ligand. For complexes having these stoichiometries, the following geometries are encountered: •

All [ML2]n± complexes are linear. The two ligands are on opposite sides of the metal, and the LOMOL bond angle is 180°. Common examples are [Ag(NH3)2]+ and [CuCl2]−.

•

Many [ML 4] n± complexes have square planar geometry. This geometry is most often found for metal ions with eight d electrons. Examples include Pt(NH3)2Cl2, [Ni(CN)4]2−, and the nickel complex with the dimethylglyoximate (dmg−) ligand in Figure 22.10.

•

Tetrahedral geometry occurs in some [ML 4]n± complexes. Examples include TiCl4, [CoCl4]2−, [NiCl4]2−, and [Zn(NH3)4]2+.

•

Octahedral geometry is found in complexes with the stoichiometry [ML6]n± (Figure 22.10).

square planar Pt(NH3)2Cl2

tetrahedral [NiCl4]2−

Isomerism Isomers are compounds with the same molecular formulas but different ­arrangements of atoms. There are several types of isomers. octahedral [Ni(H2O)6]2+

Common Coordination Geometries

•

Structural isomers have the same molecular formula but different bonding arrangements of atoms.

•

Stereoisomers have the same atom-to-atom bonding sequence, but the atoms differ in their arrangement in space. There are two types of stereoisomerism: geometric isomerism and optical isomerism.

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All three types of isomerism—structural, geometric, and optical—are encountered in coordination chemistry.

Structural Isomerism The two most important types of structural isomerism in coordination chemistry are coordination isomerism and linkage isomerism. Coordination isomerism occurs by exchanging a coordinated ligand and the uncoordinated counterion. For example, dark violet [Co(NH3)5Br]SO4 and red [Co(NH3)5SO4]Br are coordination isomers. In the first compound, bromide ion is a ligand and sulfate is a counterion; in the second, sulfate is a ligand and bromide is the counterion. A diagnostic test for this kind of isomer is based on precipitation reactions. Addition of Ba2+(aq) to a solution of [Co(NH3)5Br]SO4 gives a precipitate of BaSO4, indicating the presence of sulfate ion in solution. In contrast, no reaction occurs if Ba2+(aq) is added to a solution of [Co(NH3)5SO4]Br. In this complex, sulfate ion is attached to Co3+ and is not a free ion in solution. [Co(NH3)5Br]SO4 + Ba2+(aq) 88n BaSO4(s) + [Co(NH3)5Br]2+(aq) [Co(NH3)5SO4]Br + Ba2+(aq) 88n no reaction

Linkage isomerism occurs when it is possible to attach a ligand to the metal through different atoms. The two most common ligands with which linkage isom­ er­ism arises are thiocyanate, SCN−, and nitrite, NO2−. The Lewis structure of the thiocyanate ion shows that there are lone pairs of electrons on sulfur and nitrogen. Therefore, the ligand can attach to a metal either through a sulfur lone pair (called S-bonded thiocyanate) or through a nitrogen lone pair (called N-bonded thiocyanate). Nitrite ion can attach either at oxygen or at nitrogen. If the ligand is O-bonded, it is called a nitrito ligand; if it is N-bonded, it is called a nitro ligand (Figure 22.15).

[Ag(NH3)2]F in Dentistry  The cation [Ag(NH3)2]+ is one of the few linear coordination complexes, and the fluoride-containing salt is used in dental care. Brushing [Ag(NH3)2]F onto a cavity is found to stop further decay. Particularly with children, this quick, easy, painless treatment of a dental cavity is often preferable to traditional drilling and filling with a mercury amalgam. In addition, silver ions released by the complex ion kill cavity-forming bacteria in the mouth, and the F− ion promotes fluoroapatite formation.

Ligands forming linkage isomers −

S

C

N

O

N

O

−

Can bond to metal ion using either lone pair.

Can bond to metal ion using either lone pair.

Geometric Isomerism Geometric isomerism results when the atoms bonded directly to the metal have a different spatial arrangement. The simplest example of geometric isomerism in coordination chemistry is cis–trans isomerism, which occurs in both square-planar and octahedral complexes. The square-planar complex Pt(NH3)2Cl2 is an example of

H3N H3N

NH3 Co NH3

2+

NH3

O

N

H3N O

H3N

red-orange, O-bonded NO2−

Cis–Trans Isomerism  Cis–trans isomerism is not possible for tetrahedral complexes. All LOMOL angles are 109.5°, and all positions are equivalent in this three-dimensional structure.

NH3

2+

NH3 O N NH3 O

Co

yellow-orange, N-bonded NO2− © Charles D. Winters/Cengage

Figure 22.15  Linkage isomers, [Co(NH3)5ONO]

2+

and [Co(NH3)5NO2]2+.  ​These complex ions, whose systematic names are pentaamminenitritocobalt(III) and pentaamminenitrocobalt(III), respectively, were the first known examples of this type of isomerism. (The compounds shown in this figure have two chloride ions as counterions.)

22.5  Structures of Coordination Compounds

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1123

cis isomer

cis isomer

cis isomer, purple

fac isomer

trans isomer

trans isomer

trans isomer, green

mer isomer

(a) The square-planar complex Pt(NH3)2Cl2 exists in two geometries, cis and trans.

(b) Cis and trans octahedral isomers for [Co(NH3)4Cl2]+.

Figure 22.16  Geometric isomers. ​

(c) Cis and trans octahedral isomers for [Co(en)2Cl2]+.

(d) Fac and mer isomers of Cr(NH3)3Cl3: fac isomer, the three identical ligands are arranged at the corners of a triangular face. In the mer isomer, the three similar ligands follow a meridian.

cis–trans isomerism (Figure 22.16a). In this complex, the two Cl− ions can be either adjacent to each other (cis) or on opposite sides of the metal (trans). The cis isomer is effective in the treatment of testicular, ovarian, bladder, and osteogenic sarcoma cancers, but the trans isomer has no effect on these diseases (“Applying Chemical Principles 22.2: Cisplatin: Accidental Discovery of a Chemotherapy Agent”). Cis–trans isomerism also occurs in octahedral complexes with the formula MA4B2. In the complex [Co(NH3)4Cl2]+ the two Cl− ions can occupy positions that are either adjacent (ClOCoOCl angle = 90°) or on opposite sides of the metal (ClOCoOCl angle = 180°) (Figure 22.16b). Cis–trans isomerism in an octahedral complex with two bidentate ethylenediamine ligands is illustrated by [Co(H2NCH2CH2NH2)2Cl2]+. The two Cl− ions occupy positions that are either adjacent (the purple cis isomer) or opposite (the green trans isomer) (Figure 22.16c). Another common type of geometric isomerism occurs for octahedral complexes with the general formula MA3B3. A fac isomer has three identical ligands lying at the corners of a triangular face of an octahedron defined by the ligands ( fac = facial), whereas the ligands follow a meridian in the mer isomer (mer = meridional). Fac and mer isomers of Cr(NH3)3Cl3 are shown in Figure 22.16d.

Optical Isomerism Optical isomerism is a second type of stereoisomerism, where the molecules or ions have nonsuperimposable mirror images. Molecules that have nonsuperimposable mirror images are termed chiral. Pairs of nonsuperimposable, mirror-image ­molecules are called enantiomers. In chemistry a common model for chirality is a carbon-based compound with four different groups attached to a tetrahedral C atom (Figure 22.17). One ­enantiomer is a mirror image of the other, and the two enantiomers cannot be superimposed. (Chirality occurs widely in organic chemistry and biochemistry and will be discussed in Chapters 23 and 24.) There are many examples of chirality in non-chemical contexts. S­ eashells, for example, may spiral in different directions, giving left-handed and right-handed

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Figure 22.17 Stereoisomerism in chemistry and nature.  ​

Isomer II

© Steven Hyatt

Isomer I

Lactic acid enantiomers are nonsuperimposable, just as your right hand is not superimposable on your left.

Left- and right-handed seashells (of different species)

shells (Figure  22.17). Bolts and screws can also have left-handed or righthanded threads. Pure samples of enantiomers have the same physical properties, such as melting point, boiling point, density, and solubility in common solvents, but they differ in one significant way: When a beam of plane-polarized light passes through a solution of a pure enantiomer, the plane of polarization rotates (Figure 22.18). The two enantiomers rotate polarized light to an equal extent, but do so in opposite directions. The terms optical isomerism and optical activity are used because this effect involves light. Chirality occurs for octahedral complexes when the metal ion coordinates to three bidentate ligands or when the metal ion coordinates to two bidentate ligands and two monodentate ligands (the same or different) in a cis position. The complexes [Co(en)3]3+ and cis-[Co(en)2Cl2]+, illustrated in Figure 22.19, are examples of chiral complexes. The diagnostic test for chirality is met with both species: Mirror images of these molecules are not superimposable on one another, and solutions of the optical isomers rotate plane-polarized light in opposite directions.

Plane of polarized light

Monochromatic light (sodium lamp)

Chiral complexes.  The examples

cited in this paragraph are commonly observed formulations, but others are possible.

Tube filled with a solution of an optically Analyzer active compound (Polarizer rotated to pass light)

Vertically oriented Polarizer screen

Figure 22.18  Rotation of plane polarized light by an optical isomer.  ​Monochromatic light (light of only one wavelength) is produced by a sodium lamp. After it passes through a polarizing filter, the light vibrates in only one direction—it is polarized. A solution of an optical isomer placed between the first and second polarizing filters causes rotation of the plane of polarized light. The second filter is rotated to a point where a maximum of light is transmitted and the angle of rotation is calculated. The magnitude and direction of rotation are unique physical properties of the optical isomer being tested.

22.5  Structures of Coordination Compounds

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1125

[Co(en)3]3+

cis-[Co(en)2Cl2]+

[Co(en)3]3+ mirror image

cis-[Co(en)2Cl2]+ mirror image

Figure 22.19  Chiral metal complexes.  ​Both [Co(en)3]3+ and cis-[Co(en)2Cl2]+ are chiral. Notice that the mirror images are not superimposable.

E xamp le 22.4

Isomerism in Coordination Chemistry Problem  For which of the following compounds or complex ions do isomers exist? If isomers are possible, identify the type of isomerism (structural, geometric, or optical). (a) [Co(NH3)4Cl2]+

(d) K3[Fe(C2O4)3]

(b) Pt(NH3)2(CN)2 (square-planar)

(e) Zn(NH3)2Cl2 (tetrahedral)

(c) Co(NH3)3Cl3

(f ) [Co(NH3)5SCN]2+

What Do You Know?  You can infer coordination number and geometry from the formulas. Structural isomers are those for which there are different attachments of atoms; geometric isomers are those in which the spatial arrangement of atoms is different. Complexes that are optically active have nonsuperimposable mirror images.

Strategy   The following checklist is useful when deciding what type of isomerism a compound may exhibit. (1) Drawing pictures of the molecules will help you visualize the isomers. (2) Decide how many ligands are attached to the metal ion and whether the ligands are monodentate, bidentate, tridentate, and so on. (3) The only stereoisomerism possible for square-planar complexes is geometric (cis and trans). (4) Tetrahedral coordination complexes do not have geometric isomers. Optical isomers are possible but exceedingly rare. (5) Six-coordinate complexes of the formula MA4B2 can be either cis or trans. (6) Mer and fac isomers are possible for octahedral complexes with a stoichiometry of MA3B3. (7) Optical activity is possible for octahedral metal complexes such as cis-M(bidentate)2X2, cis-M(bidentate)2XY, and M(bidentate)3 (where X and Y are monodentate ligands), though some other octahedral complexes can also have optical isomers. (8) Linkage isomerism, a type of structural isomerism, occurs for only a few different ligands (for example, SCN– and NO2–).

Solution (a) Two geometric isomers can be drawn for octahedral complexes with a formula of MA4B2. One isomer has two Cl− ions in cis positions (adjacent positions, at a 90° angle), and the other isomer has the Cl− ligands in trans positions (with a 180° angle between the ligands). Optical isomers are not possible. H3N H3N

NH3 Co Cl

+

NH3

H3N

Cl

H3N

cis isomer

Cl Co Cl

+

NH3 NH3

trans isomer

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(b) In this square-planar complex, the two NH3 ligands (and the two CN− ligands) can be either cis or trans. These are geometric isomers. Optical isomers are not possible. NC

Pt

NC

NH3

H3N

NH3

NC

CN

Pt

NH3

trans isomer

cis isomer

(c) Two geometric isomers of this octahedral complex, with chloride ligands either fac or mer, are possible. In the fac isomer, the three Cl− ligands are all at 90° to each other; in the mer isomer, two Cl− ligands are at 180°, and the third is 90° from the other two. Optical isomers are not possible. H3N H3N

NH3 Co Cl

Cl

H3N

Cl

H3N

fac isomer

Cl

Cl

Co

NH3

Cl

mer isomer

(d) Ignore the counterions, K+. The anion is an octahedral complex. Each bidentate oxalate ion occupies two adjacent coordination sites of the metal. As there are three oxalate ions, the metal must have a coordination number of 6. Mirror images of complexes of the stoichiometry M(bidentate)3 (Figure 22.19) are not superimposable; therefore, two optical isomers are possible. (Here the ligands, C2O42−, are drawn abbreviated as OOO.) O O

O Fe O

3−

O

O

O

O

O Fe O

3−

O O

nonsuperimposable mirror images of [Fe(ox)3]3−

(e) Only a single structure is possible for tetrahedral complexes such as Zn(NH3)2Cl2. Cl Zn Cl

NH3 NH3

(f ) Only linkage isomerism (structural isomerism) is possible for this octahedral cobalt complex. Either the sulfur or the nitrogen of the SCN− anion can be attached to the cobalt(III) ion in this complex. H3N H3N

NH3 Co NH3

2+

NH3

H3N

SCN

H3N

S-bonded SCN−

NH3 Co NH3

2+

NH3 NCS

N-bonded SCN−

Think about Your Answer  More complex situations occur when a complex has three or more different ligands. In those instances it is important to be able to visualize the three-dimensional molecule by drawing its structure and to organize your approach.

Check Your Understanding  What types of isomers are possible for the following compounds or complex ions? (a) K[Co(NH3)2Cl4]

(d) [Ru(phen)3]Cl3

(b) Pt(en)Cl2 (square-planar)

(e) Na2[MnCl4] (tetrahedral)

(c) [Co(NH3)5Cl]

(f ) [Co(NH3)5NO2]2+

2+



22.5  Structures of Coordination Compounds

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1127

22.6 Bonding in Coordination Compounds Goals for Section 22.6 • Use ligand field theory to describe metal-ligand bonding in coordination compounds.

• Draw an energy-level diagram representing splitting of the metal d orbitals by the ligand field.

• Predict and rationalize magnetism in coordination complexes. Metal-ligand bonding in a coordination complex was described earlier in this chapter as being covalent, resulting from the sharing of an electron pair between the metal and the ligand donor atom. Although frequently used, this description is not capable of explaining the color and magnetic behavior of these complexes. As a consequence, the covalent bonding picture has now largely been superseded by two other bonding models: molecular orbital theory and ligand field theory. The bonding model based on molecular orbital theory assumes that the metal and the ligand bond through the molecular orbitals formed by atomic orbital o ­ verlap between metal and ligand. The ligand field model, in contrast, focuses on repulsion (and destabilization) of electrons in the metal coordination sphere. The ligand field model also assumes that the positive metal ion and the ­negative ligand lone pair are attracted electrostatically; that is, the bond results from the attraction between a positively charged metal ion and a negative ion or the negative end of a polar molecule. For the most part, the molecular orbital and ligand field models predict similar, qualitative results regarding color and magnetic behavior. Here, the focus will be on the ligand field approach and using it to explain the color and magnetic behavior of transition metal complexes.

The d Orbitals: Ligand Field Theory To understand ligand field theory, it is necessary to look at the d orbitals (­Figure 22.20), specifically with regard to their orientation relative to the positions of ligands in a metal complex. In an isolated atom or ion, the five d orbitals have the same energy. For a metal atom or ion in a coordination complex, however, the d orbitals have different energies. According to the ligand field model, repulsion between d electrons on the metal and electron pairs of the ligands destabilizes electrons that reside in the d orbitals; that is, it causes the orbital energy to increase. Electrons in the various d orbitals are not affected equally, however, because of their different orientations in space relative to the position of the ligand lone pairs. For octahedral complexes, assume the ligands lie along the x-, y-, and z-axes. Next, subdivide the five d orbitals into two sets: the dx2−y2 and dz2 orbitals in one set and the dxy, dxz, and dyz orbitals in the second (Figure 22.20). The dx2−y2 and dz2 ­orbitals are directed along the x-, y-, and z-axes, whereas the orbitals of the second

Ligand z

Ligand

z

Ligand

Ligand

x

y Ligand

Ligand

x

y dx2−y2

dz2

Orbitals are along x, y, z.

dxy

dyz

dxz

Orbitals are between x, y, z.

Figure 22.20 The d orbitals.  ​The five d orbitals and their spatial relation to the ligands on the x-, y-, and z-axes.

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dx2−y2, dz2 ∆o

ENERGY

dxy, dxz, dyz

The five d orbitals in a free transition metal ion have the same energy.

d orbitals of metal ion in octahedral ligand field

Figure 22.21  Ligand field splitting for an octahedral complex.  ​The d-orbital energies

increase as the ligands approach the metal along the x−, y−, and z-axes. The dxy , dxz , and dyz orbitals, not pointed toward the ligands, are less destabilized than the dx 2−y 2 and dz 2 orbitals. Thus, the dxy , dxz , and dyz orbitals are at lower energy. (∆o stands for the difference in d-orbital energies in an octahedral ligand field.)

group are aligned between these axes. Electrons in the dx2−y2 and dz2 orbitals experience a larger repulsion because these orbitals point directly at the ligand electron pairs. A smaller repulsive effect is experienced by electrons in the dxy, dxz, and dyz orbitals. The difference in degree of repulsion means that an energy difference exists between the two sets of orbitals in an octahedral complex (Figure 22.21). This difference is called the ligand field splitting and is denoted by the symbol ∆ o in an octahedral ligand field. The magnitude of ∆o is a function of the metal and the ligands and varies predictably from one complex to another. A different splitting pattern is encountered with square-planar complexes (Figure 22.22). Assume that the four ligands are along the x- and y-axes. The dx2−y2 orbital also points along these axes, so it is most affected and has the highest energy. The dxy orbital (which also lies in the xy-plane, but does not point at the ligands) is next highest in energy, followed by the dz2 orbital. The dxz and dyz orbitals, both of which partially point in the z-direction, have the lowest energy. The d-orbital splitting pattern for a tetrahedral complex is the reverse of the pattern observed for octahedral complexes. Three orbitals (dxz, dxy, dyz) are higher in energy, whereas the dx2−y2 and dz2 orbitals are below them in energy (Figure 22.22).

dx2−y2 ∆sp dxy

ENERGY

dz2 ∆t

dxy dx2−y2

dxz

dyz

dxz

dyz

dz2

d orbitals of metal ion in tetrahedral ligand field

The five d orbitals in a free transition metal ion have the same energy.

d orbitals of metal ion in square-planar ligand field

Figure 22.22  Splitting of the d orbitals in (left) tetrahedral and (right) square-planar geometries.  ​(∆t and ∆sp are, respectively, the splitting in tetrahedral and square-planar ligand fields.)

22.6  Bonding in Coordination Compounds

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1129

Electron Configurations and Magnetic Properties The d-orbital splitting in coordination complexes provides an explanation of both the magnetic behavior and the color of these complexes. To understand this explanation, however, you will need to know how to assign electrons to the various orbitals in each geometry. A gaseous Cr2+ ion has the electron configuration [Ar]3d4. The term gaseous in this context is used to denote a single, isolated atom or ion with all other particles located an infinite distance away. In this situation, the five 3d orbitals have the same energy. The four electrons reside singly in different d orbitals, according to Hund’s rule, and the Cr2+ ion has four unpaired electrons. Cr(II) electron configuration

Cr2+

[Ar]3d4 3d

Paramagnetism Figure 7.5

(page 362) shows the paramagnetic behavior of a compound of Fe3+ (a high spin d 5 ion).

4s

When the Cr2+ ion is part of an octahedral complex, the five d orbitals do not have identical energies. As illustrated in Figure 22.21, these orbitals divide into two sets, with the dxy, dxz, and dyz orbitals having a lower energy than the dx2−y2 and dz2 orbitals. Having two sets of orbitals means that two different electron configurations are possible for Cr2+ (Figure 22.23). Three of the four d electrons in Cr2+ are assigned to the lower-energy dxy, dxz, and dyz orbitals. The fourth electron can either be assigned to an orbital in the higher-energy dx2−y2 and dz2 set or can pair up with an electron already in the lower-energy set. The first arrangement is called high spin, because it has the maximum number of unpaired electrons, four in the case of Cr2+. The second arrangement is called low spin, because it has the minimum number of unpaired electrons possible. At first glance, a high-spin configuration appears to contradict conventional thinking. It seems logical that the most stable situation would occur when electrons occupy the lowest-energy orbitals. A second factor intervenes, however. Because electrons are negatively charged, repulsion increases when they are assigned to the same orbital. This destabilizing effect is due to the pairing energy (P). (There is an energy cost to pairing electrons.) The preference for an electron to be in the lowest-energy orbital and the pairing energy have opposing effects (Figure 22.23). Low-spin complexes arise when the splitting of the d orbitals by the ligand field is large—that is, when ∆o has a large value. The energy gained by putting all of the electrons in the lowest-energy level is the dominant effect when ∆o > P. In contrast, high-spin complexes occur if the value of ∆o is smaller than the energy required to pair electrons (∆o < P).

Low spin ENERGY

High spin

∆o

∆o

∆o < P

∆o > P

Figure 22.23  High- and low-spin cases for an octahedral chromium(II) complex.  ​(left, high spin) If the ligand field splitting (Δo) is smaller than the pairing energy (P), the electrons are placed in different orbitals, and the complex has four unpaired electrons. (right, low spin) If the splitting is larger than the pairing energy, all four electrons will be in the lower-energy orbital set. This requires pairing two electrons in one of the orbitals, so the complex will have two unpaired electrons.

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Ti3+

Ti2+

V2+

Cr2+

Mn2+, Fe3+

Fe2+, Co3+

Co2+

Ni2+

Cu2+

Zn2+

d1

d2

d3

d4

d5

d6

d7

d8

d9

d 10

Small ∆o

HIGH SPIN

d4

d5

d6

d7

Large ∆o

LOW SPIN

Figure 22.24  High- and low-spin octahedral complexes.  ​d-Orbital occupancy for ­octahedral complexes of metal ions. Only the d 4 through d 7 cases have both high- and low-spin configurations.

For octahedral complexes, high- and low-spin complexes are possible only for configurations d4 through d7 (Figure 22.24). Complexes of the d 6 metal ion, Fe2+, for example, can have either high spin or low spin. The complex formed when the Fe2+ ion is placed in water, [Fe(H2O)6]2+, is high spin, whereas the [Fe(CN)6]4− complex ion is low spin. Electron configuration for Fe2+ in an octahedral complex

dx2−y2,dz2

dx2−y2,dz2 ∆o(CN−)

∆o(H2O)

dxy,dxz,dyz

dxy,dxz,dyz

high spin [Fe(H2O)6]2+

low spin [Fe(CN)6]4−

dx2−y2

It is possible to tell whether a complex is high or low spin by examining its magnetic behavior. The high-spin complex [Fe(H 2O)6]2+ has four unpaired electrons and is paramagnetic (attracted by a magnet), whereas the low-spin [Fe(CN)6]4− complex has no unpaired electrons and is diamagnetic (repelled by a magnet) (Section 7.4). Most complexes of Pd2+ and Pt2+ ions are square-planar, the electron configuration of these metals being [noble gas](n − 1)d 8. In a square-planar complex, there are four sets of orbitals (Figure 22.22). For square-planar d 8 complexes, all except the highest-energy orbital are filled, and all electrons are paired, resulting in diamagnetic (low-spin) complexes. Nickel, which is found above palladium in the periodic table, forms both square-planar and tetrahedral complexes (as well as octahedral complexes). For example, the complex ion [Ni(CN)4]2− is square-planar, whereas the [NiCl4]2− ion is tetrahedral. The magnetic behavior of these complexes allows chemists to ­differentiate between these two geometries. The square-planar cyanide complex is diamagnetic, whereas the tetrahedral chloride complex is paramagnetic with two unpaired electrons.



dxy

dxz

dxy

dyz

dz2 dx2−y2

dz2 dxz 2−

Cl Cl

Ni

dyz

Cl Cl

2−

NC NC

Ni

CN CN

Nickel(II) complexes and magnetism. ​The anion [NiCl4]2− is a paramagnetic tetrahedral complex. In contrast, [Ni(CN)4]2− is a diamagnetic square-planar complex.

22.6  Bonding in Coordination Compounds

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1131

E xamp le 22.5

High- and Low-Spin Complexes and Magnetism Problem   Give the electron configuration for the metal ion in each of the following complexes. How many unpaired electrons are present in each? Are the complexes paramagnetic or diamagnetic? (a) Low-spin [Co(NH3)6]3+ (b) High-spin [CoF6]3−

What Do You Know?   From their formulas, you can presume that these complex ions have six ligands and that they have octahedral geometry. Both are complexes of cobalt(III), a metal ion with a d 6 configuration. One complex is low spin, the other high spin.

Strategy   Set up an energy-level diagram for an octahedral complex. In low-spin complexes, the electrons are added preferentially to the lower-energy set of orbitals. In high-spin complexes, the first five electrons are added singly to each of the five ­o rbitals, then additional electrons are paired with electrons in orbitals in the lower-energy set. Solution

dx2−y2

dz2 dx2−y2

∆o dxy

dxz

dyz

dxy

Electron configuration of lowspin, octahedral [Co(NH3)6]3+

(a)

dz2

dxz

∆o dyz

Electron configuration of highspin, octahedral [CoF6]3−

(b)

(a) In the low-spin [Co(NH3)6]3+ complex ion, the six electrons of the Co3+ ion fill the lower-energy set of orbitals entirely. This d 6 complex ion has no unpaired electrons and is diamagnetic. (b) To obtain the electron configuration in high-spin [CoF6]3−, place one electron in each of the five d orbitals, and then place the sixth electron in one of the lower-energy orbitals. The complex has four unpaired electrons and is paramagnetic.

Think about Your Answer   One objective of research in chemistry is to use experimental results to help predict what will be observed in new experiments. Here you see that the ligand makes a difference in magnetic behavior, with ammonia leading to a low-spin complex whereas the halide ion gives a high spin complex. Many more experiments show this is often the case, a result explored more in the section on The Spectrochemical Series.

Check Your Understanding  For each of the following complex ions, give the oxidation number of the metal, depict possible low- and high-spin configurations, give the number of unpaired electrons in each configuration, and tell whether each is paramagnetic or diamagnetic. (a) [Ru(H2O)6]2+ (b) [Ni(NH3)6]2+

1132 Chapter 22 / The Chemistry of the Transition Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Problem Solving Tip 22.1 Ligand Field Theory This summary of the concepts of ligand field theory may help you to keep the broader picture in mind.

• Ligand-metal bonding results from the electrostatic ­attraction between a metal cation and ligands that are either anions or polar molecules.

• The ligands define a coordination geometry. Common geometries are linear (coordination number = 2), tetrahedral and square-planar (coordination number = 4), and octahedral (coordination number = 6). • The placement of the ligands around the metal causes the

d orbitals on the metal to have ­different energies. In an ­octahedral complex, for example, the d orbitals divide into two groups: a higherenergy group (dx 2−y 2 and dz 2) and a lower-energy group (dxy , dxz , and dyz).

• Electrons are assigned to metal d orbitals in a manner that leads to the lowest total energy. Two competing features determine the placement: the relative energy of the sets of orbitals and the electron pairing energy.

• For the electron configurations of d 4, d 5, d 6, and d 7 in octahedral complexes, two electron

configurations are possible: high spin, which occurs when the orbital splitting is small, and low spin, which occurs with a large orbital splitting. To determine whether a complex is high spin or low spin, its magnetism can be measured to determine the number of unpaired electrons.

• The d-orbital splitting (the energy difference between the metal d-orbital energies) often corresponds to the energy associated with visible light. As a consequence, many metal complexes absorb visible light and thus are colored.

22.7 Colors of Coordination Compounds Goal for Section 22.7 • Know how color and magnetism are related to the electronic structure of a complex.

© Charles D. Winters/Cengage

The range of colors observed for compounds of the transition elements is one of their most interesting features (Figures 22.2 and 22.25), and the underlying reason for this is d-orbital splitting due to the ligands surrounding the metal atom or ion. Before you can understand how d-orbital splitting is involved in determining color, you will need to first explore more closely what the observed color of something means.

Fe3+(aq)

Co2+(aq)

Ni2+(aq)

Cu2+(aq)

Zn2+(aq)

Figure 22.25  Aqueous solutions of some transition metal ions.  ​Com­pounds of transition metal elements are often colored, whereas those of main group metals are usually colorless. Pictured here, from left to right, are solutions of the nitrate salts of Fe3+, Co2+, Ni2+, Cu2+, and Zn2+.

22.7  Colors of Coordination Compounds

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1133

 (nm)

500

Energy increases

600

Wavelength increases

700

400

The ROY G BIV spectrum of colors of visible light.  The colors used in printing this book are cyan, magenta, yellow, and black. The blue in ROY G BIV is actually cyan, according to color industry standards. Magenta doesn’t have its own wavelength region. Rather, it is a mixture of blue and red.

Color Visible light consists of radiation with wavelengths from 400 to 700 nm ­(Section 6.1). Within this region are all the colors you see when white light passes through a prism: red, orange, yellow, green, blue, indigo, and violet (ROY G BIV). Each color is identified with a portion of the wavelength range. Isaac Newton did experiments with light and established that the mind’s perception of color requires only three colors! When you see white light, you are seeing a mixture of all of the colors—in other words, the superposition of red, green, and blue. If one or more of these colors is absent, the light of the other colors that reaches your eyes is interpreted by your mind as a color. Figure 22.26 will help you in analyzing perceived colors. The three primary colors—red, green, and blue—are shown as overlapping disks arranged in a triangle. The secondary colors—cyan, magenta, and yellow—appear where two disks overlap. The overlap of all three disks in the center produces white light. The colors you perceive are determined as follows: •

Light of a single primary color is perceived as that color: Red light is perceived as red, green light as green, blue light as blue.

•

Light made up of two primary colors is perceived as the color shown where the disks in Figure 22.26 overlap: Red and green light together appear yellow, green and blue light together are perceived as cyan; and red and blue light are perceived as magenta.

•

Light made up of the three primary colors is white (colorless).

In discussing the color of a substance such as a coordination complex in ­solution, it is the light that is not absorbed by the solution that is important. •

Red color results when green and blue light are absorbed by the solution.

•

Green color results if red and blue light are absorbed.

•

Blue color results if red and green light are absorbed.

The secondary colors are rationalized similarly. Absorption of blue light gives ­yellow; absorption of red light results in cyan; and absorption of green light results in magenta. These ideas can be applied to explain colors in transition metal complexes. Focus on what kind of light is absorbed. A solution of [Ni(H2O)6]2+ is green. Green light is the result of removing red and blue light from white light. As white light passes through an aqueous solution of Ni2+, red and blue light are absorbed, and

Green Yellow = Red + Green

Cyan = Green + Blue C

Y W

Red

M

Blue

Magenta = Red + Blue

Figure 22.26  Using color disks to analyze colors.  ​The three primary colors are red, green, and blue. Adding light of two primary colors gives the secondary colors yellow (= red + green), cyan (= green + blue), and magenta (= red + blue). Adding all three primary colors results in white light.

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green light is allowed to pass (Figure 22.27). Similarly, the [Co(NH3)6]3+ ion is ­yellow because blue light has been absorbed and red and green light pass through.

Electrons can be excited from lower to higher energy levels by the absorption of energy in atoms, ions, and molecules (Section 6.3). For transition metal complexes, the absorbed energy is often in the visible region of the spectrum, which means the complexes can display a range of colors (see Figures 22.10, 22.13, and 22.25). In coordination complexes, the energy gap in the split d orbitals often corresponds to the energy of visible light, so light in the visible region of the spectrum is absorbed when electrons move from a lower-energy d orbital to a higher-energy d orbital. This change, as an electron moves between two d orbitals having different energies in a complex, is called a d-to-d transition. Qualitatively, such a transition for [Co(NH3)6]3+ might be represented using an energy-level diagram such as that shown here.

dx2−y2

dz2 ∆o

dxy

dxz

dx2−y2

+ energy (∆o)

dz2

(light absorbed)

dyz

dxy

Ground state of low-spin, octahedral Co3+ complex

dxz

dyz

Excited state

© Charles D. Winters/Cengage

The Spectrochemical Series

Figure 22.27  Light absorption and color.  The color of a solution is due to the color of the light not absorbed by the solution. In this photo, a solution of [Ni(H2O)6]2+ is green. This means that green light is not absorbed. Red and blue light from white light are absorbed while green light is transmitted. (Figures 4.13–4.14).

Experiments with coordination complexes reveal that, for a given metal ion, some ligands cause a small energy separation of the d orbitals, whereas others cause a large separation. In other words, some ligands create a small ligand field (small ∆o), and others create a large one. An example is seen in the spectroscopic data for several cobalt(III) complexes presented in Table 22.3. •

Both [Co(NH3)6]3+ and [Co(en)3]3+ are yellow-orange, because they absorb light in the blue portion of the visible spectrum. These compounds have very similar spectra, to be expected because in both the metal is coordinated to six N donor atoms.

Table 22.3

The Colors of Some Co3+ Complexes*

Complex Ion

Wavelength of Light Absorbed (nm)

Color of Light Absorbed

Color of Complex

[CoF6]3−

700

Red

Blue-green

[Co(C2O4)3]

600, 420

Yellow, violet

Dark green

3+

[Co(H2O)6]

600, 400

Yellow, violet

Blue-green

[Co(NH3)6]3+

475, 340

Blue, ultraviolet

Yellow-orange

[Co(en)3]3+

470, 340

Blue, ultraviolet

Yellow-orange

[Co(CN)6]3−

310

Ultraviolet

Very pale yellow, almost colorless

3−

*The complex with fluoride ion, [CoF6]3−, is high spin and has one absorption band. The other c­ omplexes are low spin and have two absorption bands. Although the top of the absorption band for [Co(CN)6]3− is in the UV region, the band is broad and trails into the visible region. The very pale ­yellow color is the result of the small absorption of blue light.

22.7  Colors of Coordination Compounds

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Colors of Coordination Complexes ​

Owing to the fact that complexes often have more than one absorption band, and that the bands are broad, it can be difficult to predict their color.

•

Although [Co(CN)6]3− does not have an absorption band whose maximum falls in the visible region, it is very pale (almost colorless) yellow. The peak of the absorption band occurs in the ultraviolet region, but the absorption band is broad and extends minimally into the visible (blue) region.

•

[Co(C2O4)3]3− and [Co(H2O)6]3+ have similar absorptions, in the yellow and violet regions. Their colors are shades of green with a small difference due to the relative amount of light of each color being absorbed.

•

Of the complex ions in Table 22.3, [CoF6]3− absorbs the lowest energy light (red) and so transmits the highest energy light. (The complex is blue-green.)

The wavelength of maximum absorption among the listed complexes ranges from 700 nm for [CoF6]3− to 310 nm for [Co(CN)6]3−. The ligands change from member to member of this series, and it is reasonable to conclude that the energy of the light absorbed by the complex is related to the different ligand field splittings, ∆o, caused by the different ligands. Fluoride ion causes the smallest splitting of the d orbitals among the complexes listed in Table 22.3 (the lowest energy is required), whereas cyanide causes the largest splitting (the greatest energy required). Spectra of complexes of other metals provide similar results. Based on this information, ligands can be listed in order of their ability to affect the splitting of the d orbitals in a broad range of complexes. This list is called the s­ pectrochemical series because it was determined by spectroscopy. A short list, with some of the more common ligands, follows: F−, Cl−, Br−, I− < C2O42− < H2O < NH3 = en < phen < CN− small d-orbital splitting large d-orbital splitting small ∆o large ∆o

The spectrochemical series is applicable to a wide range of metal complexes. Indeed, the ability of ligand field theory to explain the differences in the colors of the transition metal complexes is one of the strengths of this theory. Based on the relative position of a ligand in the series, predictions can be made about a compound’s magnetic behavior. Recall that d4, d5, d6, and d7 octahedral complexes can be high or low spin, depending on the ligand field splitting, ∆ o. Complexes formed with ligands near the left end of the spectrochemical series are expected to have small ∆o values and, therefore, are likely to be high spin. In contrast, complexes with ligands near the right end are expected to have large ∆o values and low-spin configurations. For example, the complex ion [CoF6]3− is high spin, whereas [Co(NH3)6]3+ and the other complexes in Table 22.3 are low spin.

E xamp le 22.6

© Steven Hyatt

Spectrochemical Series

Green jade.  The rich green color of this New Zealand jade comes from iron(II) and iron(III) ions in the mineral. The iron ions within the mineral are surrounded by oxide ions at the corners of an octahedron, and the difference in energy of the d orbitals means light is absorbed in the visible region.

Problem   An aqueous solution of [Fe(H2O)6]2+ is light blue-green. Do you expect the d6 Fe2+ ion in this complex to have a high- or low-spin configuration? How could you test your prediction experimentally?

What Do You Know?  This is an octahedral complex of Fe2+, an ion with a d6 configuration. The blue-green color is the result of the absorption of visible light due to a d-to-d transition. The complex could be either low spin (diamagnetic with no unpaired electrons) or high spin (paramagnetic with four unpaired electrons).

Strategy  Use the color wheel in Figure 22.26 to determine what color light is transmitted and therefore what color of light has been absorbed. Based on the energy of the light absorbed, determine if ∆o is high or low. If it is low, then the complex is high spin. If it is high, then the complex is low spin.

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Solution  In this case, the blue-green color of the solution indicates that blue and green light are transmitted. This implies that red light was absorbed. The low energy of the light absorbed suggests that ∆o is small and so [Fe(H2O)6]2+ is likely to be a  high-spin complex.  If the complex is high spin, it will have four unpaired electrons and be paramagnetic; if, contrary to the prediction, it is low spin, it will have no unpaired electrons and be diamagnetic. Thus, identifying the presence of four unpaired electrons  by measuring the compound’s magnetism  can be used to verify the high-spin configuration experimentally. Think about Your Answer  The spectrochemical series is useful because it allows you to predict whether a complex is likely to be high or low spin. The more definitive evidence for high or low spin comes from information on the magnetism of the complex.

Check Your Understanding 1. The complex ion [Co(NH3)5OH]2+ has an absorption maximum at 500 nm. What color of light is absorbed and what color is the complex? 2. Figure 22.15 shows a sample of yellow [Co(NH3)5NO2]2+. What color of light is absorbed by this complex? Predict whether this is a high-spin or low-spin complex.

Applying Chemical Principles 22.1  Blue!

Prussian blue.  The deep blue compound called Prussian blue has the formula Fe4[Fe(CN)6]3 ∙ 14 H2O. The color arises from electron transfer between Fe(II) and Fe(III) ions in the compound.

Unfortunately, all the compounds used as blue pigments over the centuries have problems such as durability or expense. Some have environmental problems [such as the release of HCN (hydrogen cyanide) by Prussian blue]. So, imagine the surprise when Professor M. Subramanian and his students at Oregon State University synthesized a stable, durable blue pigment while searching for new solid-state materials in 2009. They combined yttrium(III) oxide (Y2O3), indium(III) oxide (In2O3), and manganese(IV) oxide (MnO2). If just the yttrium and manganese oxides are combined, the product is black YMnO3. If just yttrium and indium oxides are combined, the product is white YInO3. But combining all three in varying amounts produces a vibrant blue pigment YIn 1−xMnxO3 (where x varies from 0.02 to 0.9). At low levels of manganese the color is a startling blue, but it becomes more black as the manganese content increases. The pigment, which is often called YInMn blue, is chemically stable, non-toxic, and the color does not fade.

© M. A. Subramanian

© Charles D. Winters/Cengage

For centuries, craftsmen and artists have collaborated on the development of paints and pigments with colors ranging from subtle to vibrant. One important color is blue, and one of the earliest blue pigments was Egyptian blue (CaCuSi4O10). Later in history, there was Han blue used by Chinese artists ­( BaCuSi 4O 10) and Mayan blue, which was indigo on clay. (Indigo, an organic compound, is described on page 486.) More recently, cobalt blue (CoAl2O4) has been popular, as have azurite [Cu3(CO3)2(OH)2] and Prussian blue (Fe4[Fe(CN)6]3). You can see the blue color of copper(II) ions on page 1133, azurite on page 1104, and Prussian blue in a photo here. The feature that stands out for all these compounds is that they are based on transition-metal ions.

The blue pigment. YIn1−xMnxO3 where x varies from about 0.02 to 0.9. Here x is approximately 0.1. Applying Chemical Principles

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1137

M. A. Subramanian, Oregon State University

M. A. Subramanian, Oregon State University

2. In YInMn blue, the Mn3+ ion is surrounded most directly by five oxide ions at the corners of a trigonal bipyramid.

Structure of YMnO3.  The small cyan spheres are oxide ions, the yellow spheres are yttrium(III) ions, and the dark blue spheres are Mn3+ or In3+ ions. The latter ions are surrounded by a trigonal bipyramid of oxide ions.

Questions

1. In his book Blue, K. Kupferschmidt describes the synthesis of YIn1−xMnxO3. The author combined 4.5392 g of Y2O3, 5.302 g of In2O3, and 0.1585 g of MnO2. After thoroughly grinding the solids to completely mix them, the powder was pressed into a large tablet, which was then heated in an oven at 1300 °C for some hours. What is the value of x in the ­formula of the resulting compound?

The Mn31 ion (dark blue) surrounded by five oxide ions (cyan) in YInMn blue.

The oxide ions split the d orbitals in the pattern shown in the following figure. (a) Knowing that Mn3+ is high spin, place electrons in appropriate d orbitals in the diagram. (b) Explain the order of energy of the orbitals (the dxz, d yz ­orbitals are lowest in energy and the dz2 orbital is highest in energy). dz2 dxy, dx2−y2 dxz, dyz

3. Prussian blue has iron in two different oxidation states bridged by a cyanide ion. What are those different states? References 1.­ K. Kupferschmidt, Blue—In Search of Nature’s Rarest Color, New York: The Experiment, 2021. 2. A. E. Smith, H. Mizoguchi, K. Delaney, N. A. Spaldin, A. W. Sleight, and M. A. Subramanian, J. Am. Chem. Soc., 2009, 131, 17084–17086.

22.2  Cisplatin: Accidental Discovery of a Chemotherapy Agent Chemists have synthesized various metal-based ­compounds for medical purposes. One of these, cisplatin [PtCl2(NH3)2], was known for many years, but it was discovered serendipitously to be effective in treatment of certain kinds of cancers. In 1965, Barnett Rosenberg, a biophysicist at Michigan State University, set out to study the effect of electric fields on living cells, but the results of his experiments were very different from his expectations. He and his students had placed an aqueous suspension of live Escherichia coli bacteria in an electric field between supposedly inert platinum electrodes. Much to their surprise, they found that cell growth was significantly affected. After careful experimentation, the effect on cell division was found to be due to a trace of a complex of platinum, ammonia, and chloride ion formed by an electrolytic process involving the platinum electrode in the presence of ammonia in the growth medium.

To follow up on this interesting discovery, Rosenberg and his students tested the effect of cis- and transPtCl2(NH3)2 on cell growth and found that only the cis isomer was effective. This led Rosenberg and others to study the effect of so-called cisplatin on cancer cell growth, and the result is that cisplatin and similar compounds are now used to treat genitourinary tumors. In fact, testicular cancer is now considered largely curable because of cisplatin chemotherapy. The chemistry of cisplatin has now been thoroughly studied. It has been found that cisplatin has a half-life of 2.5 hours for the replacement of a Cl− ligand by water at 310 K (in a first-order reaction) and that the replacement of a ­second Cl− ligand by water is slightly faster. The aqua species are acidic and damaging to the kidneys, so cisplatin is generally used in a saline solution to prevent the hydrolysis reactions. It has been found that, in blood

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plasma at pH 7.4 and with a Cl− ion concentration of about 1.04 × 10 −5 M, Pt(NH 3) 2Cl 2 and Pt(NH 3) 2(OH)Cl are the dominant species. In the cell nucleus, however, the Cl − ion concentration is lower, and the aqua species are present in higher concentration.

Pt(NH3)2Cl2(aq)

+

H2O(𝓵)

Questions

1.­ If a patient is given 10.0 mg of cisplatin, what quantity remains as cisplatin at 24 hours? 2.­ What is the systematic name for cisplatin? 3.­ How are the d electrons of Pt distributed in a d-orbital splitting diagram?

[Pt(NH3)2(H2O)Cl]+(aq)

+ Cl−(aq)

+

+

22.3  The Rare Earth Elements



China presently produces about 97% of these metals and oxides, and companies in the United States and elsewhere are opening old mines or seeking new sources.

SandiMako/Shutterstock.com

The lanthanide elements are often referred to as rare earth elements. In this book, we include the 14 elements from Ce (cerium) through Lu (lutetium) in the list of rare earths. The ground state electron configurations for these elements utilize the 4f subshell. However, although it does not have a 4f electron, it is common to also include La, lanthanum, with the rare earths. Why are they called rare earth elements and are they in fact rare? In 1803, German and Swedish chemists isolated a new oxide (an “earth”) from a mineral found in Sweden. At first this was thought to be an oxide of a new element, cerium, but several decades later it was discovered that it was a mixture of cerium oxide and oxides of other lanthanides. The term “rare” has been attached to the elements because it is difficult to find them in concentrations great enough for commercial purposes and because they are difficult to separate one from another. But they are in fact rather abundant. Cerium, for example, is the twenty-sixth most abundant in Earth’s crust, about half as abundant as chlorine and five times as abundant as lead. The rare earths have another interesting distinction: there are three rare earth elements named for a single town, Ytterby in Sweden. The elements are terbium (65), erbium (68), and ytterbium (70). There is also a fourth element named for Ytterby: yttrium (39), but it is a d-block element. The rare earth elements were slow to find commercial uses, but some are now crucially important in modern economies. Cerium(IV) oxide is used in catalysts, as an abrasive for polishing glass, and in the coatings of self-cleaning ovens. Neodymium (Nd) is an important component of lasers, and a combination of neodymium, iron, and boron is a powerful magnet. Holmium is used to intensify the magnetic field in MRI machines. One consequence of the growing importance of the rare earths is that they are becoming more expensive.

Neodymium-doped YAG laser.  These lasers are based on yttrium aluminum garnet, Y3Al5O12, in which a tiny fraction of the Y3+ ions are replaced by the rare earth ion Nd3+. Such lasers are widely used, for example in ­ophthalmology for treating vitreous “eye floaters,” among many other things.

Questions

1. What are the electron configurations for Nd and Eu? 2. The lanthanide elements form ionic compounds, with the +3 oxidation state dominating. What are the electron configurations for Ce3+ and Nd3+? 3. The compound (NH4)2Ce(NO3)6 is a widely used oxidizing agent in redox titrations. A 0.181 g sample of pure iron is dissolved in acid, reduced to Fe 2+, and then titrated with 31.33 mL of (NH 4 ) 2 Ce(NO 3 ) 6 (to yield Ce 3+ and Fe 3+). Calculate the concentration of the cerium salt in the titrant. 4. Cerium(IV) oxide is used in catalytic converters for the ­oxidation of CO (to yield cerium(III) oxide and CO2). The cerium(III) oxide is converted back to cerium(IV) oxide by atmospheric oxygen. Write balanced chemical equations for these reactions.

Applying Chemical Principles

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1139

5. Perovskites are a common solid-state structural type. They have the general formula ABO3 in which A and B are metal cations. Recently a rare earth perovskite was discovered, LaWN3, which has interesting magnetic properties. Here, there are nitride ions (N3−) instead of oxide ions, and the A

and B cations are, respectively, La and W. Draw a unit cell of this compound if La cations are at the corners of a cube, a W cation is in the center of the cube, and nitride ions are in the cube faces. What are the charges on the La and W ions?

John C. Kotz

Think–Pair–Share

 Alfred Werner carried out much of the research for which he received the 1913 Nobel Prize in Chemistry in this building at the Swiss Academy of Sciences in Zürich, Switzerland.

1. Alfred Werner (1866–1919) received the 1913 Nobel Prize for his studies of coordination compounds of t­ransition ­m etals, work that opened the way to understanding ­molecular geometry. In one scientific paper (with coauthor A. Miolati), he described an experiment in which they ­dissolved a ­c ompound they called dichrocobalt chloride (with the ­overall formula Co(NH3)3(H2O)Cl3) in water. “A solution is obtained which is colored green for a moment, but only for a moment; the color changes very quickly to blue and then to violet.” They also reported that the electrical conductivity of the solution increased along with the color change. (a) Briefly explain Werner and Miolati’s observations. (Note that chemists now know that dichrocobalt chloride is [Co(NH3)3(H2O)Cl2]Cl.) (b) Do any of the cobalt compounds involved in this series of reactions have cis-trans isomers? Are any chiral? (c) Are any of these cobalt compounds paramagnetic? (d) Is the cobalt ion in the compounds expected to be high spin or low spin?

2. There is an old expression “he is so poor he does not have two nickels to rub together.” Nickels in this case are U.S. coins, which are only 25% nickel, the rest being copper. Nickel is important is the economy, largely because it is used in many important alloys. One is stainless steel, which is mostly iron with about 18% chromium and 8% nickel. Another is nichrome (nickel and chromium) used in the heating wires in ovens. There is also nitinol, the nickel-titanium alloy described on page 1148. This q ­ uestion explores an important nickel-containing alloy, nickel a­ luminide, which is used in high temperature applications. The unit cell of nickel aluminide, NixAly, consists of Al atoms at the corners of a cube with Ni atoms in the cube faces. (a) What are the values of x and y in NixAly. (b) Sketch the solid-state structure. (c) The density of the alloy is 7.16 g/cm3. What is the length of each side of the unit cell (in picometers)? 3. In 1888, chemist Ludwig Mond found that nickel valves in his chemical plant leaked after carbon monoxide, CO, passed through. Upon investigation, he found that the nickel and CO were reacting to form Ni(CO)4, a liquid that boils at 43 °C. This was the first compound of this type discovered. It is a member of what is now a very large class of chemical compounds called metal carbonyls. (a) Ni(CO)4 readily dissociates into the metal and CO so it must be handled with care. If inhaled, one not only inhales CO but also nickel vapor. Despite this, the formation and dissociation of the compound is used to purify nickel. If 1.00 kg of liquid Ni(CO) 4 dissociates, what amount of nickel metal is deposited? What would be the pressure of the CO formed if it is contained in a 25.0 L tank at 19 °C? (b) Over 20 compounds of the type Mx(CO)y are known, and the ones of heavier metals are often polyhedral clusters. One example is a rhodium carbonyl, the red solid Rhx(CO)y. It is 64.75% rhodium and has a molar mass of 954 g/mol. What is the formula of the compound? 4. When CrCl3 dissolves in water, three different compounds can be identified. (1) [Cr(H2O)6]Cl3, violet (2) [Cr(H2O)5Cl]Cl2, pale green (3) [Cr(H2O)4Cl2]Cl, dark green If diethyl ether is added, a fourth complex can be obtained: (4) Cr(H2O)3Cl3 (brown).

(a) Which d orbitals are occupied in these chromium complexes? (b) If complexes 1, 2, and 3 are dissolved in water, what experiment could be done that would differentiate them (aside from their color)? (c) Do any of the four complexes exhibit cis-trans isomerism? Are any expected to be chiral?

1140 Chapter 22 / The Chemistry of the Transition Elements Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Goals Revisited The goals for this chapter are keyed to specific Study Questions to help you organize your review.

22.1  Overview of the Transition Elements • Identify the general classes of transition metals based on their location in the periodic table. 1–4.

• Recognize general features of transition metals and their compounds. 5–8.

22.2  Periodic Properties of the Transition Elements • Assess periodic trends of the transition metals based on electronic structure. 9–14.

22.3 Metallurgy • Describe the metallurgy of iron and copper. 15–18.

22.4  Coordination Compounds • Know the names and formulas of common monodentate and bidentate ligands. 19–22.

• Given the formula for a coordination complex, identify the metal and its oxidation state, the ligands, the coordination number and coordination geometry, and the overall charge on the complex. 23–26.

• Given a formula, name a complex; given a name of a complex, write its formula. 27–32.

22.5  Structures of Coordination Compounds • Identify the different types of isomers that are encountered with coordination compounds. 33, 34.

• Given the molecular formula of a coordination compound, determine whether isomers are possible, and draw their structures. 35–38.

22.6  Bonding in Coordination Compounds • Use ligand field theory to describe metal-ligand bonding in coordination compounds. 39–42.

• Draw an energy-level diagram representing splitting of the metal d orbitals by the ligand field. 43–48.

• Predict and rationalize magnetism in coordination complexes. 43–48.

22.7  Colors of Coordination Compounds • Know how color and magnetism are related to the electronic structure of a complex. 49–52.



Chapter Goals Revisited

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1141

Study Questions challenging questions. Blue-numbered questions have answers in Appendix N and fully worked solutions in the Student Solutions Manual. ▲ denotes

Practicing Skills Properties of Transition Elements and Metallurgy (See Sections 22.1–22.3.) 1. Based on the position in the periodic table, identify the actinide elements among those in the following list: Co, Cm, Cd, Ce, Cf. 2. Based on the position in the periodic table, identify the actinide elements among those in the following list: Ta, Tc, Ti, Th, Tm. 3. Use the periodic table to identify the elements in the following list with unfilled 4d orbitals: ­rhodium, rhenium, ruthenium, rutherfordium, radium. 4. Use the periodic table to identify the elements in the following list with unfilled 3d orbitals: cadmium, cerium, cobalt, chromium, copper. 5. Identify the following as either chemical or physical properties of most transition metals: (a) can be oxidized (b) have unpaired electrons (paramagnetism) (c) solids at 25 °C (d) metallic luster (e) Compounds of the elements are often colored. 6. Iron is the most abundant transition element in nature. Identify common chemical and physical properties of this element. 7. Identity among the transition metals: (a) the most dense element. (b) the metal with the lowest melting point. (c) a radioactive d-block element. (d) an essential element in the human body. 8. Identify among the transition metals: (a) the metal with the highest melting point. (b) a commonly used metal in construction. (c) a radioactive f-block element. (d) one of the metals in nitrogenase, a nitrogenfixing catalyst. 9. Give the electron configuration for each of the ­following ions, and tell whether each is paramagnetic or diamagnetic. (a) Cr3+ (c) Ni2+ 2+ (b) V (d) Cu+

10. Identify two transition metal cations with each of the following electron configurations. (a) [Ar]3d6 (c) [Ar]3d5 (b) [Ar]3d10 (d) [Ar]3d8 11. Identify a cation of a first series transition metal that is isoelectronic with each of the following. (a) Fe3+ (c) Fe2+ (b) Zn2+ (d) Cr3+ 12. Match up the isoelectronic ions on the following list. Cu+ Mn2+ Fe2+ Co3+ Fe3+ Zn2+ Ti2+ V3+

13. The lanthanide contraction is given as an explanation for the fact that the sixth period transition ­metals have (a) lower densities than the fifth period transition elements. (b) atomic radii similar to the fifth period transition elements. (c) lower melting points than the fifth period transition elements. 14. Describe how the atomic radii of the transition metals change across a period and rationalize this change based on electronic structure. 15. What is the most common form in which iron is found in the Earth’s crust? (a) free metal (b) iron oxide (c) iron sulfide (d) iron silicate 16. In the reaction CuFeS2(s) + 3 CuCl2(aq) 88n   4 CuCl(s) + FeCl2(aq) + 2 S(s)

what element is oxidized and what element is reduced? (a) Sulfur is reduced, iron is oxidized. (b) Copper is reduced, sulfur is oxidized. (c) Sulfur is reduced, copper is oxidized. (d) This is not a redox reaction. 17. In the pyrometallurgy of iron, what two species serve as reducing agents? 18. Lime (CaO) is usually added to the blast furnace in the metallurgy of iron where it reacts with impurities

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present in iron ore. What of the following best describes its function? (a) CaO acts as an oxidizing agent. (b) CaO acts as a reducing agent. (c) CaO acts as an acid. (d) CaO acts as a base.

Formulas of Coordination Compounds (See Examples 22.1 and 22.2.) 19. Which of the following ligands are expected to be monodentate? Which might be polydentate? (a) CH3NH2 (d) en (b) CH3CN (e) Br− (c) N3− (f) phen 20. One of the following nitrogen compounds or ions is not capable of serving as a ligand: NH4+, NH3, NH2−. Identify this species, and explain your answer. 21. Give the name or formula for each of the following ligands. Name

Formula

(a) ethylenediamine (b) oxalate ion (c)

NH3

(d)

SCN−

22. Give the name or formula for each of the following ligands. Name

Formula

(a) acetylacetonate (b) nitrite ion (c)

H2O

(d)

NO2−

23. Give the oxidation number of the metal ion in each of the following compounds. (c) [Co(NH3)4Cl2]Cl (a) [Mn(NH3)6]SO4 (b) K3[Co(CN)6] (d) Cr(en)2Cl2 24. Give the oxidation number of the metal ion in each of the following complexes. (a) [Fe(NH3)6]2+ (c) [Co(NH3)5(NO2)]+ (b) [Zn(CN)4]2− (d) [Cu(en)2]2+ 25. Give the formula of a complex constructed from one Ni2+ ion, one ethylenediamine ligand, three ammonia molecules, and one water molecule. Is the complex electrically neutral or is it charged? If charged, give the charge.



26. Give the formula of a complex constructed from one Cr3+ ion, two ethylenediamine ligands, and two ammonia molecules. Is the complex electrically neutral or is it charged? If charged, give the charge.

Naming and Writing Formulas for Coordination Compounds (See Example 22.3.) 27. Write formulas for the following ions or compounds. (a) dichlorobis(ethylenediamine)nickel(II) (b) potassium tetrachloroplatinate(II) (c) potassium dicyanocuprate(I) (d) tetraamminediaquairon(II) 28. Write formulas for the following ions or compounds. (a) diamminetriaquahydroxochromium(II) nitrate (b) hexaammineiron(III) nitrate (c) pentacarbonyliron(0) (where the ligand is CO) (d) ammonium tetrachlorocuprate(II) 29. Name the following ions or compounds. (a) [Ni(C2O4)2(H2O)2]2− (b) [Co(en)2Br2]+ (c) [Co(en)2(NH3)Cl]2+ (d) Pt(NH3)2(C2O4) 30. Name the following ions or compounds. (a) [Co(H2O)4Cl2]+ (c) [Pt(NH3)Br3]− (b) Co(H2O)3F3 (d) [Co(en)(NH3)3Cl]2+ 31. Give the name or formula for each ion or compound, as appropriate. (a) pentaaquahydroxoiron(III) ion (b) K2[Ni(CN)4] (c) K[Cr(C2O4)2(H2O)2] (d) ammonium tetrachloroplatinate(II) 32. Give the name or formula for each ion or compound, as appropriate. (a) tetraaquadichlorochromium(III) chloride (b) [Cr(NH3)5SO4]Cl (c) sodium tetrachlorocobaltate(II) (d) [Fe(C2O4)3]3−

Isomerism (See Example 22.4.) 33. A coordination compound has the formula [Co(en)2Cl2]Cl (en = ethylenediamine, H2NCH2CH2NH2). What types of isomerism

Study Questions

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(­ geometric isomerism, optical isomerism, structural isomerism) are possible with this formula? Identify all possible answers. 34. What types of isomerism (geometric isomerism, optical isomerism, coordination isomerism, linkage isomerism) are possible in the compound diamminechlorothio­cyanato­platinum(II)? Identify all possible answers. 35. Draw all possible geometric isomers for each of the following ions or molecules. (a) Fe(NH3)4Cl2 (b) Pt(NH3)2(SCN)(Br) (SCN− is bonded to Pt2+ through S) (c) Co(NH3)3(NO2)3 (NO2− is bonded to Co3+ through N) (d) [Co(en)Cl4]− 36. In which of the following complexes are geometric isomers possible? If isomers are possible, draw their structures and label them as cis or trans or as fac or mer. (c) [Pt(NH3)Br3]− (a) [Co(H2O)4Cl2]+ (b) Co(NH3)3F3 (d) [Co(en)2(NH3)Cl]2+ 37. Determine whether the following complexes have a chiral metal center. (a) [Fe(en)3]2+ (b) trans-[Co(en)2Br2]+ (c) fac-[Co(en)(H2O)Cl3] (d) square-planar Pt(NH3)(H2O)(Cl)(NO2) 38. Four stereoisomers are possible for [Co(en)(NH3)2(H2O)Cl]+. (a) Draw the structures of all four. (b) Two of the isomers are chiral, meaning that each has a non-superimposable mirror image. Identify those isomers.

Magnetic Properties of Complexes (See Example 22.5.) 39. According to ligand field theory, what is the force of attraction between a metal and a ligand in a coordination complex? 40. Which of the following statements about ligand field theory are correct? (a) Ligands donate an electron pair to a metal ion. (b) Each ligand donates an electron pair to the metal. (c) Coordination of the ligands to a metal causes splitting of the d electron energy levels. (d) Bonding of ligands to a metal results in paramagnetism.

41. In an octahedral complex, the d orbitals split into two groups. Which d orbitals are in the lower energy group? (Assume the ligands lie along the x-, y-, and z-axes.) 42. In a square planar complex, which d orbital is at lowest energy? Which d orbital is at highest energy? (Assume the ligands lie along the x- and y-axes.) 43. The following are low-spin complexes. Use the ligand field model to find the electron configuration of the central metal ion in each ion. Determine which are diamagnetic. Give the number of unpaired electrons for the paramagnetic complexes. (a) [Mn(CN)6]4− (c) [Fe(H2O)6]3+ (b) [Co(NH3)6]Cl3 (d) [Cr(en)3]SO4 44. The following are high-spin complexes. Use the ligand field model to find the electron configuration of the central metal ion in each ion. Determine the number of unpaired electrons, if any, in each. (a) K4[FeF6] (c) [Cr(H2O)6]2+ (d) (NH4)3[FeF6] (b) [MnF6]4− 45. Determine the number of unpaired electrons in the following tetrahedral complexes. All tetrahedral complexes are high spin. (a) [FeCl4]2− (c) [MnCl4]2− (b) Na2[CoCl4] (d) (NH4)2[ZnCl4] 46. Determine the number of unpaired electrons in the following tetrahedral complexes. All tetrahedral complexes are high spin. (a) [Zn(H2O)4]2+ (c) Mn(NH3)2Cl2 (b) VOCl3 (d) [Cu(en)2]2+ 47. For the high-spin complex [Fe(H2O)6]SO4, identify the following: (a) the coordination number of iron (b) the coordination geometry for iron (c) the oxidation number of iron (d) the number of unpaired electrons (e) whether the complex is diamagnetic or paramagnetic 48. For the low-spin complex [Co(en)(NH3)2Cl2]ClO4, identify the following: (a) the coordination number of cobalt (b) the coordination geometry for cobalt (c) the oxidation number of cobalt (d) the number of unpaired electrons (e) whether the complex is diamagnetic or paramagnetic

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49. The anion [NiCl4]2− is paramagnetic, but when CN− ions are added, the product, [Ni(CN)4]2−, is diamagnetic. Explain this observation. [NiCl4]2−(aq) + 4 CN−(aq) 88n paramagnetic[Ni(CN)4]2−(aq) + 4 Cl−(aq) diamagnetic

50. An aqueous solution of iron(II) sulfate is paramagnetic. If NH3 is added, the solution becomes diamagnetic. Why does the magnetism change?

Spectroscopy of Complexes (See Example 22.6.) 51. In water, the titanium(III) ion, [Ti(H2O)6]3+, has a broad absorption band centered at about 500 nm. What color light is absorbed by the ion? 52. In water, the chromium(II) ion, [Cr(H2O)6]2+, absorbs light with a wavelength of about 700 nm. What color is the solution?

General Questions These questions are not designated as to type or location in the chapter. They may contain several concepts. 53. Describe an experiment that would determine whether nickel in K2[NiCl4] is square-planar or tetrahedral. 54. Which of the following high-spin complexes has the greatest number of unpaired electrons? (a) [Cr(H2O)6]3+ (c) [Fe(H2O)6]2+ (b) [Mn(H2O)6]2+ (d) [Ni(H2O)6]2+ 55. How many unpaired electrons are expected for high-spin and low-spin complexes of Fe2+? 56. Excess silver nitrate is added to a solution containing 1.0 mol of [Co(NH3)4Cl2]Cl. What amount of AgCl (in moles) will precipitate? 57. Which of the following complex ions is (are) square-planar? (a) [Ti(CN)4]2− (c) [Zn(CN)4]2− 2− (b) [Ni(CN)4] (d) [Pt(CN)4]2− 58. Which of the following complex ions containing the oxalate ion is (are) chiral? (a) [Fe(C2O4)Cl4]2− (b) cis-[Fe(C2O4)2Cl2]2− (c) trans-[Fe(C2O4)2Cl2]2− 59. How many geometric isomers are possible for the square-planar complex ion [Pt(NH3)(CN)Cl2]−?



60. For the low-spin coordination compound [Fe(en)2Cl2]Cl, identify the following. (a) the oxidation number of iron (b) the coordination number for iron (c) the coordination geometry for iron (d) the number of unpaired electrons per metal atom (e) whether the complex is diamagnetic or paramagnetic (f) the number of geometric isomers 61. For the high-spin coordination compound Mn(NH3)4Cl2, identify the following. (a) the oxidation number of manganese (b) the coordination number for manganese (c) the coordination geometry for manganese (d) the number of unpaired electrons per metal atom (e) whether the complex is diamagnetic or paramagnetic (f) the number of geometric isomers 62. A platinum-containing compound, known as Magnus’s green salt, has the formula [Pt(NH3)4] [PtCl4] (in which both platinum ions are Pt2+). Name the cation and the anion. 63. Early in the twentieth century, coordination compounds sometimes were given names based on their colors. Two compounds with the formula CoCl3 ∙ 4 NH3 were named praseo-cobalt chloride (praseo = green) and violio-cobalt chloride (violet color). Chemists now know that these compounds are octahedral cobalt complexes and that they are cis and trans ­isomers. Draw the structures of these two compounds, and name them using systematic nomenclature. 64. Give the formula and name of a square-planar complex of Pt2+ with one nitrite ion (NO2−, which binds to Pt2+ through N), one chloride ion, and two ammonia molecules as ligands. Are isomers possible? If so, draw the structure of each isomer, and tell what type of isomerism is observed. 65. Give the formula of the coordination complex formed from one Co3+ ion, two ethylenediamine molecules, one water molecule, and one chloride ion. Is the complex neutral or charged? If charged, give the net charge on the ion.

Study Questions

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1145

66. ▲ How many geometric isomers of the complex ion [Cr(dmen)3]3+ can exist? (dmen is the bidentate ligand 1,1-dimethylethylenediamine.) (CH3)2NCH2CH2NH2 1,1-Dimethylethylenediamine, dmen

67. ▲ Diethylenetriamine (dien) is capable of serving as a tridentate ligand. H2NCH2CH2

N CH2CH2NH2 H

Diethylenetriamine, dien

(a) Draw the structures of fac-Cr(dien)Cl3 and mer-Cr(dien)Cl3. (b) Two different geometric isomers of mer-Cr(dien)Cl2Br are possible. Draw the structure for each. (c) Three different geometric isomers are possible for [Cr(dien)2]3+. Two have the dien ligand in a fac configuration, and one has the ligand in a mer orientation. Draw the structure of each isomer. 68. From experiment, it is known that [CoF6]3− is paramagnetic and [Co(NH3)6]3+ is diamagnetic. Using the ligand field model, depict the electron configuration for each ion, and use this model to explain the magnetic property. What can you conclude about the effect of these ligands on the magnitude of Δo? 69. Three geometric isomers are possible for [Co(en)(NH3)2(H2O)2]3+. One of the three is chiral; that is, it has a nonsuperimposable mirror image. Draw the structures of the three isomers. Which one is chiral? 70. The square-planar complex Pt(en)Cl2 has chloride ligands in a cis configuration. No trans isomer is known. Explain why the trans compound is not possible. 71. The complex [Mn(H2O)6]2+ has five unpaired ­electrons, whereas [Mn(CN)6]4− has only one. Using the ligand field model, depict the electron configuration for each ion. What can you conclude about the effects of the different ligands on the magnitude of Δo? 72. Experiments show that K4[Cr(CN)6] is paramagnetic and has two unpaired electrons. The related complex K4[Cr(SCN)6] is paramagnetic and has four unpaired electrons. Account for the magnetism of each compound using the ligand field model. Predict where the SCN− ion occurs in the spectrochemical series relative to CN−.

73. Give a systematic name or the formula for each of the following: (a) (NH4)2[CuCl4] (b) tetraaquadichlorochromium(III) chloride (c) aquabis(ethylenediamine) thiocyanatocobalt(III) nitrate 74. A transition metal coordination compound absorbs 425-nm light. What is its color? (a) red (b) green (c) yellow (d) blue 75. ▲ The complex ion [Co(CO3)3]3−, an octahedral complex with bidentate carbonate ions as ligands, has one absorption in the visible region of the ­spectrum at 640 nm. From this information, (a) Predict the color of this complex and explain your reasoning. (b) Is the carbonate ion a weak- or strong-field ligand? (c) Predict whether [Co(CO3)3]3− is paramagnetic or diamagnetic. 76. The glycinate ion, H2NCH2CO2−, formed by deprotonation of the amino acid glycine, can function as a bidentate ligand, coordinating to a metal through the nitrogen of the amino group and one of the oxygen atoms. Glycinate ion, a bidentate ligand

H

H

O

C

C

O

−

H 2N Site of bonding to transition metal ion

A copper complex of this ligand has the formula Cu(H2NCH2CO2)2(H2O)2. For this complex, determine the following: (a) the oxidation state of copper (b) the coordination number of copper (c) the number of unpaired electrons (d) whether the complex is diamagnetic or paramagnetic 77. ▲ Draw structures for the five possible geometric isomers of Cu(H2NCH2CO2)2(H2O)2. Are any of these species chiral? (See the structure of the ligand in Study Question 76.) 78. A manganese compound has the formula Mn(CO)x(CH3)y. To find the empirical formula of

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the compound, you burn 0.225 g of the solid in oxygen and isolate 0.283 g of CO2 and 0.0290 g of H2O. What is the empirical formula for the compound? That is, what are the values of x and y? 79. Nickel and palladium both form complexes of the general formula M(PR3)2Cl2. (The ligand PR3 is a phosphine such as P(C6H5)3, triphenylphosphine. It is a Lewis base.) The nickel(II) compound is paramagnetic whereas the palladium(II) compound is diamagnetic. (a) Explain the magnetic properties of these compounds. (b) How many isomers of each compound are expected? 80. ▲ The transition metals form a class of compounds called metal carbonyls, an example of which is the tetrahedral complex Ni(CO)4. Given the following thermodynamic data (at 298 K): 𝚫f H° (kJ/mol)

S° (J/K ∙ mol)

Ni(s)

0

29.87

CO(g)

−110.525

+197.674

Ni(CO)4(g)

−602.9

+410.6

Ce4+ O2−

CeO2

82. Neodymium is a component of powerful magnets. (a) Neodymium has three common oxidation states: +2, +3, and +4. Give the electron configurations for Nd and its three positive oxidation states. (b) Are any of the common Nd ions paramagnetic? (c) Neodymium magnets, which are composed of Nd, Fe, and B, were developed because samarium-based magnets (SmCo) were very costly. What is the empirical formula for the neodymium-based magnet material if it is 26.68% Nd, 72.32% Fe, and the remainder B?

81. Cerium, as noted in “Applying Chemical Principles 22.3: The Rare Earth Elements,” is a relatively ­abundant lanthanide element that has some important uses. Cerium(IV) oxide, CeO2, is widely used as a polishing agent for glass. Cerium(III) sulfide, Ce2S3, is becoming more widely used as a red pigment to replace cadmium pigments, which are environmentally less desirable. (a) Give the electron configurations (using the noble gas notation) for Ce, Ce3+, and Ce4+. (b) Is either Ce3+ or Ce4+ paramagnetic? If so, how many unpaired electrons does each have? (c) The solid-state structure for CeO2 is shown below. Describe the unit cell of the compound. How is this structure related to the formula?



© Charles D. Winters/Cengage

(a) Calculate the equilibrium constant for the formation of Ni(CO)4(g) from nickel metal and CO gas. (b) Is the reaction of Ni(s) and CO(g) product- or reactant-favored at equilibrium? (c) Is the reaction more or less product-favored at higher temperatures? How could this reaction be used in the purification of nickel metal? Neodymium magnets

In the Laboratory 83. Two different coordination compounds containing one cobalt(III) ion, five ammonia molecules, one bromide ion, and one sulfate ion exist. The dark violet form (A) gives a precipitate upon addition of aqueous BaCl2. No reaction is seen upon addition of aqueous BaCl2 to the violet-red form (B). ­Suggest structures for these two compounds, and write a chemical equation for the reaction of (A) with aqueous BaCl2. 84. Three different compounds of chromium(III) with water and chloride ion have the same composition:

Study Questions

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1147

85. ▲ A 0.213-g sample of uranyl nitrate, UO2(NO3)2, is dissolved in 20.0 mL of 1.0 M H2SO4 and shaken with Zn. The zinc reduces the uranyl ion, UO22+, to a uranium ion, Un+. To determine the value of n, this solution is titrated with KMnO4. Permanganate is reduced to Mn2+ and Un+ is oxidized back to UO22+. (a) In the titration, 12.47 mL of 0.0173 M KMnO4 was required to reach the equivalence point. Use this information to determine the charge on the ion Un+. (b) With the identity of Un+ now established, write a balanced net ionic equation for the reduction of UO22+ by zinc (assume acidic conditions). (c) Write a balanced net ionic equation for the oxidation of Un+ to UO22+ by MnO4− in acid. 86. Fireworks contain KClO3. To analyze a sample for the amount of KClO3 a chemist first reacts the sample with excess iron(II), ClO3−(aq) + 6 Fe2+(aq) + 6 H3O+(aq) 88n  Cl−(aq) + 9 H2O(ℓ) + 6 Fe3+(aq)

and then titrates the resulting solution with Ce4+ [in the form of (NH4)2Ce(NO3)6] Fe (aq) + Ce (aq) 88n Fe (aq) + Ce (aq) 2+

4+

3+

3+

to determine the quantity of iron(II) that did not react with ClO3−. (This is referred to as a back titration.) Suppose a 0.1342-g sample of a firework was treated with 50.00 mL of 0.0960 M Fe2+. The unreacted Fe2+ ions then required 12.99 mL of 0.08362 M Ce4+. What is the weight percent of KClO3 in the original sample?

Summary and Conceptual Questions The following questions may use concepts from this and ­previous chapters. 87. Nitinol, a nickel-titanium alloy, is used in eyeglass frames or orthodontics. If you bend a pair of eyeglass frames out of shape, they snap back to the proper fit. That is the reason nitinol is often called memory metal.

NASA/Science Source

19.51% Cr, 39.92% Cl, and 40.57% H2O. One of the compounds is violet and dissolves in water to give a complex ion with a 3+ charge and three chloride ions. All three chloride ions precipitate immediately as AgCl upon adding AgNO3. Draw the structure of the complex ion and name the compound. Write a net ionic equation for the reaction of this compound with silver nitrate.

Memory metal is an alloy with roughly the same number of Ni and Ti atoms. When the atoms are arranged in the highly symmetrical austenite phase, the alloy is relatively rigid. In this phase a specific shape is established that will be “remembered.” If the alloy is cooled below its phase transition temperature, it enters a less symmetrical but flexible phase (martensite). Below this transition temperature, the metal is fairly soft and may be bent and twisted out of shape. When warmed above the phase transition temperature, nitinol returns to its original shape. The temperature at which the change in shape occurs varies with small differences in the nickel-to-titanium ratio. Martensite

Austenite Ni

y



x Ti

z

x, y, and z are not equal,  about 96°

CsCl structure x=y=z  =  =  = 90°

Two phases of nitinol.  The austenite form has a structure like CsCl.

(a) What are the dimensions of the austenite unit cell? Assume the Ti and Ni atoms are just touching along the unit cell diagonal. (Atom radii: Ti = 145 pm; Ni = 125 pm.) (b) Calculate the density of nitinol based on the austenite unit cell parameters. Does the calculated density of the austenite unit cell agree with the reported density of 6.5 g/cm3? (c) Are Ti and Ni atoms paramagnetic or diamagnetic? 88. ▲ This question explores the differences between metal coordination by monodentate and bidentate ligands. Formation constants, Kf,

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for [Ni(NH3)6]2+(aq) and [Ni(en)3]2+(aq) are as follows: Ni2+(aq) + 6 NH3(aq) 88n [Ni(NH3)6]2+(aq) 

Kf = 108

Ni2+(aq) + 3 en(aq) 88n [Ni(en)3]2+(aq) 

Kf = 1018

The difference in Kf between these complexes indicates a higher thermodynamic stability for the



chelated complex, caused by the chelate effect. Recall that K is related to the standard free energy of the reaction by ΔrG° = −RT ln K and ΔrG° = ΔrH° − TΔrS°. It is known from experiment that ΔrH° for the NH3 reaction is −109 kJ/mol-rxn, and ΔrH° for the ethylenediamine reaction is −117 kJ/mol-rxn. Is the difference in ΔrH° sufficient to account for the 1010 difference in Kf? Comment on the role of entropy in the second reaction.

Study Questions

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1149

John C. Kotz; © Charles D. Winters/Cengage

23 Carbon: Not Just Another Element

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C hapt e r O ut li n e 23.1 Why Carbon? 23.2 Hydrocarbons 23.3 Alcohols, Ethers, and Amines 23.4 Compounds with a Carbonyl Group 23.5 Polymers

The vast majority of the millions of chemical compounds currently known are organic compounds. It was once thought that organic compounds could only be made from living things, an idea known as vitalism. This is why they were initially called organic. The demise of this idea is usually traced to an experiment performed by Friedrich Wöhler in 1828 in which he was able to react the inorganic compounds lead(II) cyanate, water, and ammonia to form urea, an organic compound that mammals produce as part of protein metabolism and excrete in their urine. Pb(OCN)2 + 2 H2O + 2 NH3 n 2 H2NCONH2 + Pb(OH)2 urea

The term organic chemistry has since come to refer instead to the chemistry of compounds with a carbon framework. Organic compounds vary greatly in size and complexity, from the simplest hydrocarbon, methane (CH4), to molecules made up of many thousands of atoms. You encounter organic chemicals in your everyday life, including the methane or propane you might use as natural gas to heat your home or cook your food, the ethanol in alcoholic beverages, the acetone in nail polish remover, the many plastics that are a part of modern life in an industrialized society, and most of the compounds that make up your body (see Chapter 24: Biochemistry).

23.1 Why Carbon? Goals for Section 23.1 • Understand the factors that contribute to the large numbers of organic compounds and the wide array of structures.

• Recognize the types of isomerism that are seen in organic chemistry. What features of carbon lead to both the abundance and the complexity of organic compounds? The answer revolves around two main characteristics: structural diversity and stability.

◀ Cacao pods on a tree in an island in the Caribbean and the ultimate product.  The model is of theobromine, C7H8N4O2, one of many molecules in chocolate and a smooth muscle stimulator.

1151

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Common and Systematic Names ​ In organic chemistry many compounds have common names that came about when the field was developing. However, chemists have tried to standardize names, so systematic names are also used. For example, C2H4 is commonly called ethylene, but its systematic name is ethene. Similarly, C2H2 is commonly called acetylene, but its systematic name is ethyne.

ethene (ethylene), H2C

Structural Diversity With four electrons in its outer shell, carbon forms four bonds to reach an octet of electrons. In contrast, the elements boron and nitrogen generally form three bonds in molecular compounds; oxygen forms two bonds; and hydrogen and the halogens form one bond. With a larger number of bonds comes the opportunity to create more complex structures. A carbon atom can reach an octet of electrons in various ways (Figure 23.1): •

By forming four single bonds. A carbon atom can bond to four other atoms, which can be either atoms of other elements (often H, N, or O) or other carbon atoms.

•

By forming a double bond and two single bonds. The carbon atoms in ethene (also called ethylene), H2CPCH2, are linked in this way.

•

By forming two double bonds, as in carbon dioxide (OPCPO).

•

By forming a triple bond and a single bond, an arrangement seen in ethyne (also called acetylene), HCqCH.

Recognize, with each of these arrangements, the various possible geometries around carbon: tetrahedral, trigonal-planar, and linear. Carbon’s tetrahedral geometry is of special significance because it leads to three-dimensional chains and rings of carbon atoms, as in propane and cyclopentane.

CH2

ethyne (acetylene), HC CH

Ethene and ethyne. These two-carbon hydrocarbons can be the building blocks of more complex molecules.

propane, C3H8

cyclopentane, C5H10

Similarly, molecules with chain and ring structures are known in which there are double and triple bonds between carbon atoms. H H H O H

C

C

O

H

H

C C

C C

C C

C

N

H

H

H

H

O CH3COH (a) Acetic acid. One carbon atom in this compound is attached to four other atoms by single bonds and has tetrahedral geometry. The other carbon atom, connected by a double bond to one ­oxygen and by single bonds to the other oxygen and to the first carbon, has trigonal-planar geometry.

C6H5C

H

C

C

CH2

C

C

H H

CH2

N

(b) Benzonitrile. Six trigonal-planar carbon atoms make up the benzene ring. The seventh C atom, bonded by a single bond to carbon and a triple bond to nitrogen, has a linear geometry.

(c) Carbon is linked by double bonds to two other carbon atoms in C3H4, a linear molecule commonly called allene.

Figure 23.1  Ways that carbon atoms bond.

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Isomers Isomers are molecules that have the same molecular formula but different atom-to-atom connections or fill space in different ways. A hallmark of carbon chemistry is a remarkable array of isomers, two major groups being structural ­isomers and stereoisomers. Structural isomers are species with the same elemental composition but the atoms are linked in different ways. For example, ethanol and dimethyl ether are structural isomers, as are 1-butene and 2-methylpropene.

ethanol

dimethyl ether

1-butene

2-methylpropene

C2H6O

C2H6O

C4H8

C4H8 CH2

CH3CH2OH

CH3OCH3

CH3CH2CH CH2

CH3CCH3

Isomers  Isomers were discussed

in the context of transition metal chemistry coordination compounds (page 1122).

A Closer Look

Stereoisomers are compounds with the same attachment of atoms but different orientation in space. There are two types of stereoisomers: geometric isomers and optical isomers. Cis- and trans-2-butene are geometric isomers. Geometric isomerism in these compounds occurs as a result of the CPC double bond. Recall that the carbon atom and the attached groups cannot rotate around a double bond (page 471). The geometry around the CPC double bond is fixed in space. Cis–trans isomerism occurs if each carbon atom involved in the double bond has two different groups attached. If, on the adjacent carbons, it turns out that identical groups are on the same side of the double bond, then it is a cis isomer. If those groups appear on opposite sides, a trans isomer is produced.

Writing Formulas and Drawing Structures

In Chapter 2, you learned that there are various ways of presenting structures (pages 85–86). It is appropriate to return to this topic as you begin your study of organic compounds. Consider methane and ethane, for example. These compounds are represented in several ways: (1) Molecular formula: CH4 or C2H6. This type of formula gives information on composition only. (2) Condensed formula: The formula for ethane is often written as CH 3CH 3. This method of writing the formula gives some information on the way atoms are connected. (3) Structural formula: Like a Lewis structure, a structural formula shows the bonds between atoms (but the lone pairs are often omitted). While it does give more information than the condensed formula (2), it does not describe the shapes of molecules.



H H

C

H

H

H methane, CH4

H

H

C

C

H

H

H

ethane, C2H6

(4) Perspective drawings: These drawings are used to convey the threedimensional nature of molecules. Bonds extending out of the plane of the paper are drawn with wedges, and bonds behind the plane of the paper are represented as dashed wedges. Using these guidelines, the structures of methane and ethane can be drawn as follows:

H H

C

H

H H H

H

C H

C

atoms are represented by spheres, and the bonds by sticks. In space-­ filling models partial spheres that have diameters proportional to those of the atoms are used and are joined directly to one another. Space-filling models are a better representation of the relative sizes of atoms and their proximity to each other, but when viewed from certain vantage points, some atoms can be hidden from view.

ball-and-stick

H

H

(5) Ball-and-stick and space-filling ­models: In ball-and-stick models, the

space-filling

23.1  Why Carbon?

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1153

CH3

H3C C H

C H

all naturally occurring amino acids, the basis of proteins, are left-handed (Chapter 24), but right-handed amino acids exist. One such right-handed amino acid is the toxin in a strain of mushrooms in southern China, and the venom in a bite by a duck-billed platypus is based on a right-handed amino acid. See S. Everts, Scientific American, 2013, 308(5), 78–81.

C

H3C

cis-2-butene, C4H8

Chirality and Death Virtually

CH3

H

C

H

trans-2-butene, C4H8

Optical isomers are molecules that have nonsuperimposable mirror images (Figure 23.2). Molecules (and other objects) that have nonsuperimposable mirror images are chiral. These pairs of nonsuperimposable, mirror-image molecules are called enantiomers. Pure samples of enantiomers have the same physical properties, such as melting point, boiling point, density, and solubility in common solvents. However, they differ in one significant way: When a beam of plane-polarized light passes through a solution of a pure enantiomer, the plane of polarization rotates. The two enantiomers rotate polarized light to an equal extent, but in opposite directions (Figure 22.18). The most common examples of chiral organic compounds are those in which four different atoms (or groups of atoms) are attached to a tetrahedral carbon atom. Lactic acid, found in milk and a product of normal human metabolism, is an example of a chiral compound (Figure 23.2). Optical isomerism is particularly important in amino acids and other biologically important molecules. Another example of an optically active compound is frontalin, produced naturally by male elephants (Figure 23.3).

Stability of Carbon Compounds Carbon compounds are notable for their resistance to chemical change. This resistance is a result of two things: strong bonds and slow reactions. Strong bonds are needed for molecules to survive in their environment. Energetic collisions between molecules in gases, liquids, and solutions can provide enough energy to break some bonds, and bonds can be broken if the energy

Central carbon atom surrounded by four different groups

H

Isomer I

Isomer II

C

CH3

CO2H OH

H CO2H

CH3

OH

(a) Lactic acid, CH3CH(OH)CO2H. The ­central atom in lactic acid is surrounded by four ­different groups, which leads to the molecule being chiral.

(b) Lactic acid enantiomers are nonsuperimposable, just as your right hand is not superimposable on your left hand.

Figure 23.2  Optical isomers.  ​Lactic acid, CH3CH(OH)CO2H, is produced when milk is fermented to make cheese. It is also found in other sour foods such as sauerkraut and is a preservative in pickled foods such as onions and olives. In our bodies, it is produced by muscle activity and normal metabolism.

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John C. Kotz

frontalin

One enantiomer of frontalin.

An African elephant. Fluid containing the enantiomers of frontalin flows from a gland between the elephant’s eye and ear.

Figure 23.3  Chirality and elephants.  ​During musth, a time of heightened sexual activity, male elephants produce a secretion containing the enantiomers of frontalin, C8H14O2. Young males produce more of one enantiomer than the other, whereas older elephants produce a more balanced and more concentrated mixture. The secretion from older elephants repels other males while attracting ovulating female elephants.

associated with photons of visible and ultraviolet light exceeds the bond energy. Carbon–carbon bonds are relatively strong, however, as are the bonds between carbon and most other atoms. The average COC bond energy is 346 kJ/mol; the COH bond energy is 413 kJ/mol; and carbon–carbon double and triple bond energies are even higher (Section 8.10). Contrast these values with bond energies for the SiOSi bond (222 kJ/mol) and the SiOH bond (328 kJ/mol). The consequence of high bond energies for bonds to carbon is that, for the most part, organic compounds do not decompose thermally under normal conditions. Oxidation of most organic compounds is strongly product-favored, but most organic compounds survive continual and prolonged contact with O2. The reason is that reactions of most organic compounds with oxygen are very slow. Typically, organic compounds burn only if their combustion is initiated by heat or by a spark. As a consequence, oxidative degradation is not a barrier to the existence of organic compounds.

23.2 Hydrocarbons Goals for Section 23.2 • Draw structural formulas and name hydrocarbons. • Recognize and draw structures of geometric isomers and optical isomers. • Describe chemical and physical properties of hydrocarbons. Hydrocarbons, compounds made only of carbon and hydrogen, are divided into several subgroups: alkanes, cycloalkanes, alkenes, alkynes, and aromatic compounds (Table 23.1).

Alkanes Alkanes have the general formula CnH2n+2, with n having integer values (Table 23.2). Formulas of specific compounds can be generated from this formula, the first four of which are CH4 (methane), C2H6 (ethane), C3H8 (propane), and C4H10 (butane) (Figure 23.4). Methane has four hydrogen atoms arranged tetrahedrally around a single carbon atom. Replacing a hydrogen atom in methane by a OCH3 group gives ethane. If an H atom of ethane is replaced by yet another OCH3 group, propane results. The structure of butane can be obtained by replacing an H atom of one of the chain-ending carbon atoms of propane with a OCH3 group. In all of these compounds, each C atom is attached to four other atoms, either C or H, so alkanes are often called saturated compounds.

Petroleum  Petroleum is a mixture of alkanes and other hydrocarbons. The petroleum industry is described briefly in Section 25.4.

23.2 Hydrocarbons

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1155

Table 23.1

Some Types of Hydrocarbons

Type of Characteristic General Hydrocarbon Features Formula Examples Alkanes

Cycloalkanes

Alkenes

COC single bonds and all C atoms have four single bonds

CnH2n+2

COC single bonds, C atoms arranged in a ring

CnH2n

CPC double bond

Alkynes Aromatics

CH4, methane C2H6, ethane

CnH2n

Selected Hydrocarbons of the Alkane Family, CnH2n12*

Table 23.2

C6H12, cyclohexane

H2CPCH2, ethene

CqC triple bond

CnH2n−2

HCqCH, ethyne

Rings with π bonding extending over several C atoms

—

Benzene, C6H6

Name

Molecular Formula

State at Room Temperature

Methane

CH4

Gas

Ethane

C2H6

Propane

C3H8

Butane

C4H10

Pentane

C5H12 (pent- = 5)

Hexane

C6H14 (hex- = 6)

Heptane

C7H16 (hept- = 7)

Octane

C8H18 (oct- = 8)

Nonane

C9H20 (non- = 9)

Decane

C10H22 (dec- = 10)

Octadecane

C18H38 (octadec- = 18)

Eicosane

C20H42 (eicos- = 20)

Liquid

Solid

*This table lists only selected straight-chain alkanes. At 25 °C straight-chain alkanes with 11–17 C atoms are liquid. Those with more than 17 C atoms are solid.

Structural Isomers

CH3 CH3CH2CH2CH3

butane

CH3CHCH3

2-methylpropane

Structural isomers of butane, C4H10

Structural isomers are possible for all alkanes larger than propane. For example, there are two structural isomers for C4H10 and three for C5H12. As the number of carbon atoms in an alkane increases, the number of possible structural isomers greatly increases; there are 5 isomers possible for C6H14 (Example 23.1), 9 isomers for C7H16, 18 for C8H18, 75 for C10H22, and 1858 for C14H30. To recognize the isomers corresponding to a given formula, keep in mind the following points: •

Each alkane has a framework of tetrahedral carbon atoms, and each carbon has four single bonds.

•

An effective approach to deriving isomer structures is to create a framework of carbon atoms and then fill the remaining positions around carbon with H atoms so that each C atom has four bonds.

H H

C

H

H

methane

H

H

H

C

C

H

H

ethane

H

H

H

H

H

C

C

C

H

H

H

propane

H

H

H

H

H

H

C

C

C

C

H

H

H

H

H

butane

Figure 23.4 Alkanes. ​The lowest–molar-mass alkanes, all gases under normal conditions, are methane, ethane, propane, and butane.

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•

Nearly free rotation occurs around carbon–carbon single bonds. Therefore, when atoms are assembled to form the skeleton of an alkane, the emphasis is on how carbon atoms are attached to one another and not on how they might lie relative to one another in the plane of the paper.

CH3CH2CH2CH2CH3 pentane

CH3 CH3CHCH2CH3 2-methylbutane

CH3

Ex am p le 23.1

H3CCCH3 CH3

Drawing Structural Isomers of Alkanes

2,2-dimethylpropane

Problem   Draw structures of the five isomers of C 6H14. Are any of these isomers chiral?

Structural isomers of pentane, C5H12

What Do You Know?   Each structure must have six carbon and 14 hydrogen atoms. There must be four single bonds to each carbon atom, and each hydrogen will form one bond. Most chiral compounds contain one or more carbon atoms with four different groups attached. Strategy   Focus first on the different frameworks that can be built from six carbon atoms. Having created a carbon framework, fill hydrogen atoms into the structure so that each carbon has four bonds.

Solution Step 1. Placing six carbon atoms in a chain gives the framework for the first isomer. Now fill in hydrogen atoms: three on the carbons on the ends of the chain and two on each of the carbons in the middle. This is the first isomer, hexane.

C

C

C

C

C

C

carbon framework of hexane

H

H

H

H

H

H

H

C

C

C

C

C

C

H

H

H

H

H

H

hexane

Step 2. Draw a chain of five carbon atoms, then add the sixth carbon atom to one of C the carbon atoms in the middle 1 2 3 4 5 C C C C C of this chain. (Adding it to a carbon at the end of the chain gives a six-carbon chain, the same framecarbon framework work drawn in Step 1.) Two difof methylpentane isomers ferent carbon frameworks can be built from the five-carbon chain, depending on whether the sixth C carbon is linked to the 2 or 3 posiC C C C C tion. For each of these frameworks, fill in the hydrogens.

H

H

H

H H

C

C H

H H

H

H

C

C

C

C

H

H

H

H

H

2-methylpentane

H

H

H H

C

H C

C H

H

H H

H

C

C

C

H

H

H

H

3-methylpentane



23.2 Hydrocarbons

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1157

Step 3. Draw a chain of four carbon atoms. Add in the two remaining carbons, again C being careful not to extend 1 2 3 4 C C C C the chain length. Two different structures are possible: one with C the remaining carbon atoms in the 2 and 3 positions, and another with both extra carbon atoms attached carbon atom frameworks at the 2 position. Fill in the 14 for dimethylbutane isomers hydrogens.

H

H

C

C

H H

C

C

C

C

H H

C

H H H

H

H 2,3-dimethylbutane

H

C C

H H

C

H

H

C

C

H H

C

H H

H

C

C

C

C

H H

H H

H

C

H

H 2,2-dimethylbutane

None of the isomers of C6H14 is chiral  because none of them has a carbon atom with four ­ ifferent groups attached. d

Think about Your Answer   Should you look for structures in which the longest Chirality in Alkanes  To be chiral,

a compound must have at least one C atom attached to four different groups. Thus, the C7H16 isomer (3-methylhexane) is chiral. CH3 H

C * CH2CH3 CH2CH2CH3

The center of chirality is often indicated with an asterisk.

chain is three carbon atoms? Try it, but you will see that it is not possible to add the three remaining carbons to a three-carbon chain without creating one of the carbon chains already drawn in a previous step. Thus, only five isomers of this compound are possible. Names have been given to each of these compounds. See the text that follows this Example and Appendix E for guidelines on nomenclature.

Check Your Understanding (a) Draw the nine isomers having the formula C7H16. (Hint: There is one structure with a seven-carbon chain, two structures with six-carbon chains [one is illustrated in the margin], five structures with a five-carbon chain, and one structure with a four-carbon chain.) (b) Identify the isomers of C7H16 that are chiral.

Problem Solving Tip 23.1 Drawing Structural Formulas An error students sometimes make is to suggest that the three carbon skeletons drawn here are different. They are, in fact, the same. All are five-carbon chains with another C atom in the 2 position.

C C

1

C

2

C C

3

C

4

C

5

C

C C

2 3

C

4

C

5

C

5

C

4

C

3

C

2

C

1

C

1

Remember that Lewis structures do not indicate the geometry of molecules.

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Chemistry in Your Career

Dr. Donna Wrublewski

Dr. Donna Wrublewski Dr. Donna Wrublewski worked in industry for several years before she returned to academia as the Head of Research Services and the Chemistry and Chemical Engineering Librarian at The California Institute of Technology. Dr. Wrublewski explains that “being a librarian is the perfect fit for someone who loves chemistry (and science in general) but may not be inclined toward a laboratory career,” especially if you love working with people and helping them find the answers to tough questions.

Dr. Wrublewski describes how her scientific education (B.S. in Chemical Engineering and Ph.D. in Polymer Engineering and Science) informs her work as a librarian. “Understanding chemistry is integral to my work because it allows me to have a scientific frame of reference to better understand the questions my patrons ask. As a graduate student and industrial researcher, I actually used the tools that I now teach to others, so it gives me a deeper insight to explaining how they work and also what tool to recommend to a patron depending on their question.”

Naming Alkanes With so many possible isomers for a given alkane, chemists need a systematic way of naming them. The guidelines for naming alkanes and their derivatives are: •

The names of alkanes end in “-ane.”

•

When naming a specific alkane, the root of the name corresponds to the long­ est carbon chain in the compound. (The names of alkanes with chains of 1 to 10 carbon atoms are given in Table 23.2. After the first four compounds, the names are derived from Greek and Latin numbers—pentane, hexane, heptane, octane, nonane, decane—and this regular naming continues for higher alkanes.)

•

Substituent groups on a hydrocarbon chain are identified by a name and the position of substitution in the carbon chain; this information precedes the root of the name. The position is indicated by a number that refers to the carbon atom to which the substituent is attached. Numbering of the carbon atoms in a chain should begin at the end of the carbon chain that allows the first substituent encountered to have the lowest possible number. (If there is no distinction at this point, then number the carbon atoms so that the second substituent group has the lowest number.)

•

Names of hydrocarbon substituents, called alkyl groups, are derived from the name of the hydrocarbon. The group OCH3, derived by taking a hydrogen from methane, is called the methyl group; the OC2H5 group is the ethyl group.

•

If two or more of the same substituent groups are present in the molecule, the prefixes di-, tri-, and tetra- are added.

•

When different substituent groups are present, they are generally listed in ­alphabetical order (ignoring any prefix).

Systematic and Common Names ​The IUPAC (International Union of Pure and Applied Chemistry) has formulated rules for systematic names, which are generally used in this book (Appendix E). However, many organic compounds are known by common names. For example, 2,2-dimethylpropane is also called neopentane.

CH3 H3C

C

CH3

CH3 2,2-dimethylpropane

This isomer of C5H12 has a three-carbon chain with two OCH3 groups on the second C atom of the chain. Thus, its systematic name is based on ­propane, and both OCH3 groups are located at the 2 position.

Ex am p le 23.2

Naming Alkanes Problem  Give the systematic name for CH3

CH2CH3

CH3CHCH2CH2CHCH2CH3



23.2 Hydrocarbons

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1159

What Do You Know?   You know the condensed formula and can recognize that the compound is an alkane. You also know the rules for naming alkanes.

Strategy  Step 1. Identify the longest carbon chain and base the name of the compound on that alkane. Step 2. I dentify the substituent groups on the chain and their locations. When there are two or more substituents (the groups attached to the chain), number the parent chain from the end that gives the lower number to the substituent encountered first. Step 3. If the substituents are different, list them in alphabetical order.

Solution   Here, the longest chain has seven C atoms, so the root of the name is heptane. There is a methyl group (OCH3) on C-2 and an ethyl group (OC2H5) on C-5. Naming the substituents in alphabetical order and numbering the chain from the end having the methyl group, the systematic name is  5-ethyl-2-methylheptane. 

Think about Your Answer  Notice that the carbon atoms in the longest chain are numbered so that the lower number is given to the substituent encountered first.

Check Your Understanding  Name the nine isomers of C7H16 in “Check Your Understanding” in Example 23.1.

© Charles D. Winters/Cengage

Properties of Alkanes

Figure 23.5  Paraffin wax and mineral oil.  ​These common consumer products are mixtures of alkanes.

Methane, ethane, propane, and butane are gases at room temperature and pressure, whereas the higher-molar-mass compounds are liquids or solids (Table 23.2). This is in agreement with the rule that there is generally an increase in melting point and boiling point with molar mass in a series of similar compounds (Sections 11.4 and 11.6). Several alkanes are common fuels. Natural gas, gasoline, kerosene, fuel oils, and lubricating oils are all mixtures of various alkanes. White mineral oil is also a mixture of alkanes, as is paraffin wax (Figure 23.5). Pure alkanes are colorless and odorless. All are insoluble in water, a property typical of nonpolar compounds. Low polarity is expected for alkanes because the electronegativities of carbon (χ = 2.5) and hydrogen (χ = 2.2) are not greatly different (Section 8.2). All alkanes burn readily in air to give CO2 and H2O in very exothermic reactions. This is the reason they are widely used as fuels. CH4(g) + 2 O2(g) n CO2(g) + 2 H2O(ℓ)  ΔrH° = −890.3 kJ/mol-rxn

Other than in combustion reactions, alkanes exhibit relatively low chemical reactivity. One reaction that does occur, however, is the replacement of the hydrogen atoms of an alkane by chlorine atoms upon reaction with Cl2. It is formally an oxidation because Cl2, like O2, is a strong oxidizing agent. These reactions, which can be initiated by ultraviolet radiation, are free radical reactions (Section 14.7). Highly reactive Cl atoms are formed from Cl2 exposed to ultraviolet (UV) radiation. Reaction of methane with Cl2 under these conditions proceeds in a series of steps, eventually yielding CCl4, commonly known as carbon tetrachloride. (HCl is the other product of these reactions.) Cl2, UV

Cl2, UV

Cl2, UV

Cl2, UV

CH4

CH3Cl

CH2Cl2

CHCl3

CCl4

Systematic name:

chloromethane

dichloromethane

trichloromethane

tetrachloromethane

Common name:

methyl chloride

methylene chloride

chloroform

carbon tetrachloride

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The last three compounds are used as solvents, albeit infrequently anymore because of their toxicity.

Cycloalkanes, CnH2n Cycloalkanes are constructed with tetrahedral carbon atoms joined together to form a ring. Cyclopropane and cyclobutane are the simplest cycloalkanes, although the bond angles in these species are much less than 109.5°. Chemists refer to these compounds as strained hydrocarbons because an unfavorable geometry is imposed around carbon. One of the features of strained hydrocarbons is that the COC bonds are weaker and the molecules readily undergo ring-opening reactions that relieve the bond angle strain. The most common cycloalkane is cyclohexane, C6H12, which has a nonplanar ring with six OCH2 groups. If the carbon atoms in this compound were in the form of a regular hexagon with all carbon atoms in one plane, the COCOC bond angles would be 120°. To have tetrahedral bond angles of 109.5° around each C atom, the ring has to pucker. The C6 ring is flexible and exists in two interconverting forms (“A Closer Look: Flexible Molecules”).

cyclopropane, C3H6

cyclobutane, C4H8

Cyclopropane and cyclobutane. ​ Cyclo­propane was at one time used as a general anesthetic in surgery. However, its explosive nature when mixed with oxygen soon eliminated this application.

Alkenes and Alkynes

A Closer Look

The diversity seen for alkanes is repeated with alkenes, hydrocarbons with one or more CPC double bonds. The presence of a double bond adds two features missing in alkanes: the possibility of geometric isomerism and a higher degree of reactivity. The general formula for alkenes with one double bond is CnH 2n. The first two members of the series are ethene, C2H4 (common name, ethylene), and propene, C3H6 (common name, propylene). Only a single structure can be drawn for these compounds. As with alkanes, the occurrence of isomers begins with species containing four carbon atoms. Four alkene isomers have the formula C4H8, and each has distinct chemical and physical properties (Table 23.3). There are three ­structural isomers, one of which (2-butene, CH 3CHPCHCH3) exists as two ­geometric isomers.

Flexible Molecules Most organic molecules are flexible; that is, they can twist and bend in various ways. Few molecules better illustrate this behavior than cyclohexane. Two structures are possible: chair and boat forms, which can interconvert by partial rotation of several bonds.

The more stable structure is the chair form, which allows the hydrogen atoms to remain as far apart as possible. A side view of this form of cyclohexane reveals two sets of hydrogen atoms in this molecule. Six hydrogen atoms, called the equatorial hydrogens, lie in a plane around the

carbon ring. The other six hydrogens are positioned above and below the plane and are called axial hydrogens. Flexing the ring (a rotation around the COC single bonds) moves the hydrogen atoms between axial and equatorial environments.

axial H atom equatorial H atom

H

H H

H

4

H

6 5

3H

H

H

chair form

H

2

H

1

H

H



H

H

H 4

H

H

6

5

H

2

3

H

H

H 1

H

boat form

H H H

H

H

5

H 3

H

4

H

1

6

H

H

2

H

H

H

H chair form

23.2 Hydrocarbons

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1161

Table 23.3

Properties of Butene Isomers

Boiling Point

Melting Point

Dipole Moment (D)

∆f H° (gas) (kJ/mol)

1-Butene

−6.26 °C

−185.4 °C

—

  −0.63

2-Methylpropene

−6.95 °C

−140.4 °C

0.503

−17.9

cis-2-Butene

3.71 °C

−138.9 °C

0.253

  −7.7

trans-2-Butene

0.88 °C

−105.5 °C

0

−10.8

Name

3

H

1

C

2

4

CH2CH3

C

H

C3H6 Systematic name: propene Common name: propylene

3

1

C H

1-butene

C2H4 Systematic name: ethene Common name: ethylene

H H

2

CH3

1

H3C

C

4

2

C CH3

2-methylpropene

H

3

CH3

C H

cis-2-butene

H 1

H3C

4

2

C

3

CH3

C H

trans-2-butene

Alkene names end in “-ene.” As with alkanes, the root for the name of an alkene is determined by the longest carbon chain that contains the double bond. The position of the double bond is indicated with a number, and when appropriate, the prefix cis or trans is added. Three of the C 4H 8 isomers have four-carbon chains and so are named as butenes. One has a three-carbon chain so the root of the name is propene. Notice that the carbon chain is numbered from the end that gives the double bond the lowest number. In the first isomer at the left, if the chain is numbered with the leftmost carbon as carbon-1, the double bond is between C atoms 1 and 2, so the name is 1-butene. If the rightmost carbon is labeled as carbon-1, then the double bond would be between carbons 3 and 4, so it would be named as 3-butene. Because 1 is less than 3, the correct name is 1-butene.

E xamp le 23.3

Determining Isomers of Alkenes from a Formula Problem   Draw structures for the six possible alkene isomers with the formula C5H10. Give the systematic name of each.

What Do You Know?   When linking the five carbon atoms, two will be joined with a double bond. Each carbon must have four bonds, and hydrogen atoms will fill the remaining positions. Strategy   A procedure that involved drawing the carbon skeleton and then adding hydrogen atoms worked well when drawing structures of alkanes (Example 23.1), and a similar approach can be used here. It will be necessary to put one double bond into the framework and to be alert for cis–trans isomerism.

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Solution Step 1 Draw the possible five-carbon chains containing a double bond. There are two possible five-carbon chains with a double bond, one where the double bond is between carbons 1 and 2 (1-­pentene) and one where it is between carbons 2 and 3 (2-­pentene). Cis–trans isomers are possible for 2-pentene. Fill in the hydrogen atoms to give each carbon atom four bonds.

H C

C

C

C

C

H C

C CH2CH2CH3

H

1-pentene

H

H C

C

H3C C

C

C

C

CH2CH3 cis-2-pentene

C

CH2CH3

H C

C

H3C

H

trans-2-pentene

Step 2   Draw the possible fourcarbon chains containing a double bond. Add the fifth carbon atom to either the 2 or 3 position. When all the possible combinations are found, fill in the hydrogen atoms. This results in three more structures:

C C 1

C 2

H C 3

C

4

CH3 C

C CH2CH3

H

2-methyl-1-butene

C C 1

C 2

C 3

H C

4

H C

C CHCH3

H

CH3 3-methyl-1-butene

C C 4

C 3

C 2

H C

1

CH3 C

H3C

C CH3

2-methyl-2-butene

Think about Your Answer  It is important to be very organized in your approach when drawing isomers. You should look carefully to see that each structure is unique, that is, that no two are the same.

Check Your Understanding  There are 17 possible alkene isomers with the formula C6H12. Draw structures of the five isomers in which the longest chain has six carbon atoms, and give the name of each. Are any of these isomers chiral? (There are also eight isomers in which the longest chain has five carbon atoms, and four isomers in which the longest chain has four carbon atoms. How many can you find?)



23.2 Hydrocarbons

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© Charles D. Winters/Cengage

H2C H2C

H2 C C H

Figure 23.6  Carotene, a naturally occurring compound with 11 CPC bonds.  ​The π electrons can be excited by visible light in the blue-violet region of the spectrum. As a result, carotene appears orange-yellow to the observer. Carotene or carotene-like molecules are partnered with chlorophyll in nature in the role of assisting in harvesting sunlight. Green leaves have a high concentration of carotene. In autumn, green chlorophyll molecules are destroyed, and the yellows and reds of carotene and related molecules are seen. The red color of tomatoes, for example, comes from a molecule very closely related to carotene. As a tomato ripens, its chlorophyll disintegrates, and the green color is replaced by the red of the carotene-like molecule.

Hydrocarbons exist that have two or more double bonds. Butadiene, for example, has two double bonds and is known as a diene. Many natural products have numerous double bonds (Figure 23.6 and theobromine, page 1150). There are also cyclic hydrocarbons (cycloalkenes), such as cyclohexene, with double bonds. Alkynes, compounds with a carbon–carbon triple bond, have the general formula (CnH2n−2). Table 23.4 lists alkynes that have four or fewer carbon atoms. The first member of this family is ethyne (common name, acetylene), a gas used as a fuel in metal cutting torches.

CH2 CH

cyclohexene, C6H10

H2C=CHCH=CH2

Properties of Alkenes and Alkynes Like alkanes, alkenes and alkynes are colorless. Low–molar-mass compounds are gases, whereas compounds with higher molar masses are liquids or solids. Alkenes and alkynes are also oxidized by O2 to give CO2 and H2O. Alkenes and alkynes have an elaborate chemistry because they are unsaturated compounds. Carbon atoms are capable of bonding to a maximum of four other atoms, and they do so in alkanes and cycloalkanes. In alkenes, however, each carbon atom linked by a double bond is bonded to only three atoms. In alkynes, each carbon atom linked by a triple bond is bonded to only two atoms. It is possible to increase the number of groups attached to a carbon atom in an alkene or alkyne by addition reactions, in which molecules with the general formula XOY (such as hydrogen, halogens, hydrogen halides, and water) add across the carbon–carbon double or triple bond (Figure 23.7). For an alkene the result is a compound with four groups bonded to each carbon.

1,3-butadiene, C4H6

An oxy-acetylene torch. ​The reaction of ethyne (acetylene) with oxygen produces a very high temperature. Oxy-acetylene torches, used in welding, take advantage of this fact.

H

H

© Charles D. Winters/Cengage

Cycloalkenes and dienes.  Cyclohexene, C6H10 (top) and 1,3-butadiene (C4H6) (bottom).

C H X

Table 23.4

+X

C

Y

H

H

X

Y

C

C

H

H H

Y = H2, Cl2, Br2; H Cl, H Br, H

OH, HO

Cl

Some Simple Alkynes CnH2n−2

Structure

Systematic Name

Common Name

BP (°C)

HCqCH

ethyne

acetylene

−85

CH3CqCH

propyne

methylacetylene

−23

CH3CH2CqCH

1-butyne

ethylacetylene

CH3CqCCH3

2-butyne

dimethylacetylene

9 27

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Figure 23.7  Bacon fat and addition reactions.  ​The fat in

bacon is partially unsaturated. Like other unsaturated compounds, bacon fat reacts with Br2 in an addition reaction. Here, you see the color of Br2 vapor fade when a strip of bacon is put in the flask.

© Charles D. Winters/Cengage

a few minutes

The products of some addition reactions are substituted alkanes. For example, the addition of bromine to ethene (ethylene) forms 1,2-dibromoethane.

Br Br

H

H C H

+ Br2

C

H

H

C

C

H

H H 1,2-dibromoethane

The addition of 2 mol of chlorine to ethyne (acetylene) gives 1,1,2,2-tetrachloroethane.

Cl Cl HC

CH + 2 Cl2

Cl

C

C

Cl

Nomenclature of Substituted Alkanes ​The substituent groups in

substituted alkanes are identified by the name and position of the substituent on the alkane chain.

H H 1,1,2,2-tetrachloroethane

If the reagent added to a double bond is hydrogen (XOY = H2), the reaction is called hydrogenation. Hydrogenation is usually a very slow reaction, but it can be sped up by adding a catalyst, often a specially prepared form of a metal, such as platinum, palladium, or rhodium. You may have heard the term hydrogenation because certain foods contain hydrogenated or partially hydrogenated ingredients. One brand of crackers has a label that says, “Made with 100% pure vegetable shortening . . . (partially hydrogenated soybean oil with hydrogenated cottonseed oil).” One reason for hydrogenating an oil is to make it less susceptible to spoilage; another is to convert it from a liquid to a solid. During the 1860s, the Russian chemist Vladimir Markovnikov examined a large number of alkene addition reactions. In cases in which two isomeric products were possible, he found that one was more likely to predominate. Based on these results, Markovnikov formulated Markovnikov’s rule, which states that  when a reagent HX adds to an unsymmetrical alkene, the hydrogen atom in the reagent becomes attached to the carbon that already has the larger number of hydrogens.  An example of Markovnikov’s rule is the reaction of 2-methylpropene with HCl that results in formation of 2-chloro-2-methylpropane rather than 1-chloro-2-methylpropane.

Cl

H3C C

CH2 + HCl

H3C 2-methylpropene



H3C

C

Catalysts ​A catalyst is a

substance that causes a reaction to occur at a faster rate without itself being permanently changed in the reaction (Section 14.6).

Remembering Markovnikov’s Rule ​One of the authors of this

book was taught the sentence “Hydrogen goes where hydrogen is.” For example, in the reaction of 2-methylpropene, H attaches to the carbon of the PCH2 group.

H CH3

CH3 2-chloro-2-methylpropane major product

+

H3C

C

CH2Cl

CH3 1-chloro-2-methylpropane minor product 23.2 Hydrocarbons

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1165

E xamp le 23.4

Addition Reactions Problem  Draw the structure(s) and name the product(s) obtained from the following reactions. If more than one product is possible, predict which will be the major product and which will be the minor product. (a) The reaction of Br2 with propene. (b) The reaction of HBr with 2-methyl-2-pentene.

What Do You Know?   Propene is the three-carbon alkene. 2-Methyl-2-pentene has a five-carbon chain with a double bond between carbons 2 and 3 and a methyl group at carbon 2. Addition reactions are among the most common reactions of alkenes. ­Bromine and HBr are common reagents that add to double bonds. When a reagent like HBr adds to an unsymmetrical alkene, Markovnikov’s rule can be used to predict the ­identity of the major product.

Strategy  Add the reagent (Br2 or HBr) across the CPC double bond. The names of the products are based on the name of the carbon chain and indicate the positions of any Br atoms. If more than one product is possible, then use Markovnikov’s rule to predict which will be formed in the greater amount. Solution  (a) Br2 consists of two identical atoms. One will attach to one carbon of the double bond, and the other will attach to the other carbon. Only one product is formed. H

Br Br

H C

+ Br2

C

H

CH3

H

propene

C

C

H

H

CH3

1,2-dibromopropane

(b) When HBr adds across the double bond of 2-methyl-2-pentene, two products are possible. In one, the H attaches to carbon 2 and Br attaches to carbon 3. In the other, the attachments of H and Br are reversed. Markovnikov’s rule predicts that the major product will be the one where the H attaches to the carbon atom that has the greater number of hydrogens already attached. In 2-methyl-2-pentene, carbon 2 has no hydrogens, and carbon 3 has one hydrogen. The  major product  will have the hydrogen attached to carbon 3 and the bromine attached to carbon 2,  2-bromo-2-methylpentane.  H3C C H3C

CH3 Br

CH2CH3 + HBr

C H

H3C

C

C

H

H

CH3 CH2CH3 + H3C

3-bromo-2-methylpentane minor product

C

CH2

CH2CH3

Br 2-bromo-2-methylpentane major product

Think about Your Answer   You might note that the minor product formed in part (b) has four different groups attached to carbon 3, and so this product is chiral, consisting of two enantiomers.

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Check Your Understanding  (a) Draw the structure of the compound obtained from the reaction of HBr with ethylene, and name the compound. (b) Draw the structure of the product of the reaction of Br2 with cis-2-butene, and name this compound.

Aromatic Compounds Benzene, C6H6, is a key molecule in chemistry. It is the simplest aromatic ­compound, a class of compounds originally so named because they often have significant, and usually not unpleasant, odors. Other members of this class, which are based on benzene, include toluene and naphthalene. Coal is a source of many aromatic compounds. These compounds, along with other volatile substances, are released when coal is heated to a high temperature in the absence of air (Table 23.5).

H

O

N

C

C

H H

C

H

H

O

C

H

H

H

C

C C

CH3 C C

C

H

H

H

H

C

C C

C

H

H

benzene

toluene

C C

H

H

H

H

C C

H

H

C

C

C H

C C

C



Formula

Benzene

C

C

H

CH2

C

C

C

C C

H O

C

H

H

C C

H

H

H

H

H

H

naphthalene

Aspartame (C14H18N2O5).  Aspartame, an artificial sweetener, contains an aromatic ring.

H H H

C C

C C H

Some Aromatic Compounds from Coal Tar

Common Name

N

H

Benzene occupies a pivotal place in the history and practice of chemistry. Michael Faraday discovered this compound in 1825 as a by-product in the formation of illuminating gas, a fuel produced by heating coal. Today, benzene is an important industrial chemical, usually ranking among the top 25 chemicals in production annually in the United States. It is used as a solvent and is also the starting point for making thousands of different compounds by replacing the H atoms of the ring. Toluene was originally obtained from tolu balsam, the pleasant-smelling gum of a South American tree, Toluifera balsamum. This balsam has been used in cough syrups and perfumes. Naphthalene has been used as an ingredient in moth balls, although 1,4‑dichlorobenzene is now more commonly used. Aspartame and another artificial sweetener, saccharin, are also benzene derivatives. Table 23.5

O

C

O H

H

H

Boiling Point (°C)

Melting Point (°C)

C6H6

 80

 +6

Toluene

C6H5CH3

111

−95

o-Xylene

1,2-C6H4(CH3)2

144

−25

m-Xylene

1,3-C6H4(CH3)2

139

−48

p-Xylene

1,4-C6H4(CH3)2

138

+13

Naphthalene

C10H8

218

+80

O C C

C NH S O2

Saccharin (C7H5NO3S). ​ Saccharin, an artificial sweetener, contains an aromatic ring.

23.2 Hydrocarbons

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INTERFOTO/Alamy Stock Photo

The Structure of Benzene The formula of benzene suggested to nineteenth-century chemists that this compound should be unsaturated, but if viewed this way, its chemistry was perplexing. Whereas alkenes readily undergo addition reactions, benzene does not do so under similar conditions. Benzene’s different reactivity relates to its structure and bonding, both of which are quite different from the structure and bonding in alkenes. Benzene has six equivalent carbon–carbon bonds, 139 pm in length, intermediate between a COC single bond (154 pm) and a CPC double bond (134 pm). The π bonds are formed by the continuous overlap of the p orbitals on the six carbon atoms (Figure 9.11 and Section 9.2). Using valence bond terminology, the structure is represented by two resonance structures.

or simply

August Kekulé and the Structure of Benzene ​August Kekulé

(1829–1896), one of the most prominent organic chemists in Europe in the late nineteenth century, argued for the ring structure with alternating double bonds based on the number of isomers possible for the structure. The legend in chemistry is that Kekulé proposed the ring structure after dreaming of a snake biting its tail.

representations of benzene, C6H6

Benzene Derivatives Toluene, chlorobenzene, benzoic acid, aniline, styrene, and phenol are common benzene derivatives.

Cl

CO2H

NH2

CH

CH2

OH

Drawing Aromatic Rings ​When

drawing benzene rings chemists often allow the vertices of the hexagon to represent the carbon atoms and do not show the H atoms attached to those carbon atoms. The circle inside the ring is sometimes used to indicate the two resonance structures.

chlorobenzene

benzoic acid

aniline

styrene

phenol

The systematic nomenclature for benzene derivatives with two or more substituent groups involves naming these groups and identifying their positions on the ring by numbering the six carbon atoms (Appendix E). Some common names, which are based on an older naming scheme, are also used. This scheme identified isomers of disubstituted benzenes with the prefixes ortho (o-, substituent groups on adjacent carbons in the benzene ring), meta (m-, substituents separated by one carbon atom), and para (p-, substituent groups on carbons on opposite sides of the ring).

X 1

ortho to X

2

6

3

5 4

meta to X

para to X Cl

CH3

NO2

Cl CH3 NO2 Systematic name: 1,2-dichlorobenzene Common name: o-dichlorobenzene

1,3-dimethylbenzene m-xylene

1,4-dinitrobenzene p-dinitrobenzene

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Ex am p le 23.5

Isomers of Substituted Benzenes Problem  Draw and name the isomers of C6H3Cl3. What Do You Know?  From the formula you can infer that C6H3Cl3 is a substituted benzene with three hydrogen atoms replaced by chlorine atoms. Strategy  Draw the carbon framework of benzene, and attach a chlorine atom to one of the carbon atoms. Place a second Cl atom on the ring in the ortho, meta, and para positions. Add the third Cl in one of the remaining positions, being careful not to repeat a structure already drawn.

Solution  The three isomers of C6H3Cl3 are shown here. They are named as derivatives of benzene by specifying the number of substituent groups with the prefix “tri-,” the name of the substituent, and the positions of the three groups around the six-member ring. Cl

Cl 1 6

2

5 4

3

Cl Cl

Cl

Cl

Cl

Cl

Cl

1,2,3-trichlorobenzene

1,2,4-trichlorobenzene

1,3,5-trichlorobenzene

Think about Your Answer  Are there other possibilities? Try moving the chlorine atoms around in each isomer. In every case, you will find that moving one Cl atom to a ­different position generates one of these three isomers.

Check Your Understanding  Aniline, C 6 H 5 NH 2 , is the common name for aminobenzene. Draw a structure for p‑diaminobenzene, a compound used in dye manufacture. What is the systematic name for p‑diaminobenzene?

Properties of Aromatic Compounds Benzene is a colorless liquid, and simple substituted benzenes are liquids or solids under normal conditions. The properties of aromatic hydrocarbons are typical of hydrocarbons in general: They are insoluble in water, soluble in nonpolar solvents, and are oxidized by O2 to form CO2 and H2O. One of the most important properties of benzene and other aromatic compounds is an unusual stability that is associated with the unique π bonding in this molecule (Figure 9.11 and Section 9.2). Because the π bonding in benzene is typically described using resonance structures, the extra stability is termed resonance stabilization. The extent of resonance stabilization in benzene is evaluated by comparing the energy evolved in the hydrogenation of benzene to form cyclohexane catalyst

C6H6(ℓ) + 3 H2(g) 888888n C6H12(ℓ)  ΔrH° = −206.7 kJ/mol-rxn

with the energy evolved in hydrogenation of three isolated double bonds. 3 H2CPCH2(g) + 3 H2(g) n 3 C2H6(g)  ΔrH° = −410.8 kJ/mol-rxn

The hydrogenation of benzene is about 200 kJ/mol less exothermic than the hydrogenation of three moles of ethene, a difference attributable to the added stability associated with π bonding in benzene.

23.2 Hydrocarbons

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1169

Although aromatic compounds are unsaturated hydrocarbons, they do not undergo the addition reactions typical of alkenes and alkynes. Instead, substitution reactions occur in which one or more hydrogen atoms are replaced by other groups. Such reactions require higher temperatures and a strong Brønsted acid such as H2SO4 or a Lewis acid such as AlCl3 or FeBr3.

Nitration: Alkylation:

C6H6(ℓ) + HNO3(ℓ) C6H6(ℓ) + CH3Cl(ℓ)

Halogenation: C6H6(ℓ) + Br2(ℓ)

H2SO4 AlCl3 FeBr3

C6H5NO2(ℓ) + H2O(ℓ) C6H5CH3(ℓ) + HCl(g) C6H5Br(ℓ) + HBr(g)

23.3 Alcohols, Ethers, and Amines Goals for Section 23.3 • Identify functional groups from organic formulas. • Name and draw structures of alcohols, ethers, and amines. • Describe chemical properties of alcohols and amines. Organic compounds often contain other elements in addition to carbon and hydrogen. Two elements in particular, oxygen and nitrogen, add a rich dimension to ­carbon chemistry. Organic chemistry organizes compounds containing elements other than carbon and hydrogen as derivatives of hydrocarbons. Formulas (and structures) are represented by substituting one or more hydrogens in a hydrocarbon molecule by a functional group. A functional group is an atom or group of atoms attached to a carbon atom in the hydrocarbon. Formulas of hydrocarbon derivatives are then written as ROX, in which R is a hydrocarbon lacking a hydrogen atom and X is the functional group (such as OOH, ONH2, a halogen atom, or OCO2H) that has replaced the hydrogen. The chemical and physical properties of the hydrocarbon derivatives are a blend of the properties associated with hydrocarbons and the group that has been substituted for hydrogen. Table 23.6 identifies some common functional groups and the families of organic compounds resulting from their attachment to a hydrocarbon.

Alcohols and Ethers O H

C

H H H

Methanol, CH3OH, the simplest alcohol. ​Methanol is often called wood alcohol because it was originally produced by heating wood in the absence of air.

If one of the hydrogen atoms of an alkane is replaced by a hydroxyl group (OOH), the result is an alcohol, ROH. To name alcohols, substitute “-ol” for the final “e” in the name of the hydrocarbon and designate the position of the OOH group by the number of the carbon atom. For example, CH3CH2CHOHCH3 is named as a derivative of butane. The OOH group is attached to the second carbon, so the name is 2-butanol. Methanol, CH3OH, and ethanol, CH3CH2OH, are the most important alcohols. In these cases, no number is needed for the position of the OOH group because the OOH must be on the carbon designated as carbon 1. Other alcohols are also commercially important (Table 23.7). Notice that several have more than one OH functional group. More than 5.7 × 108 kg of methanol is produced in the United States annually. Most of this is used to make formaldehyde (CH2O) and acetic acid (CH3CO2H), both important industrial chemicals. Methanol is also used as a solvent, as a de-icer in gasoline, and as a fuel in high-powered racing cars. It is found in low concentration in new wine, where it contributes to the odor, or bouquet. Like ethanol, methanol causes intoxication, but methanol differs in being more poisonous, largely because the human body converts it to formic acid (HCO2H) and formaldehyde

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Table 23.6

Common Functional Groups and Derivatives of Alkanes

Functional Group*

General Formula*

Class of Compound

Examples

F, Cl, Br, I OH OR′ NH2†

RF, RCl, RBr, RI ROH ROR′ RNH2

Haloalkane Alcohol Ether (Primary) Amine

CH3CH2Cl, chloroethane CH3CH2OH, ethanol (CH3CH2)2O, diethyl ether CH3CH2NH2, ethylamine

RCHO

Aldehyde

CH3CHO, ethanal (acetaldehyde)

R′

RCOR′

Ketone

CH3COCH3, propanone (acetone)

OH

RCO2H

Carboxylic acid

CH3CO2H, ethanoic acid (acetic acid)

OR′

RCO2R′

Ester

CH3CO2CH3, methyl acetate

NH2

RCONH2

Amide

CH3CONH2, acetamide

O CH O C O C O C O C

*R and R′ can be the same or different hydrocarbon groups. † Secondary amines (R2NH) and tertiary amines (R3N) are also possible; see discussion in the text.

(CH2O). These compounds attack the cells of the retina in the eye, leading to permanent blindness. Ethanol is the alcohol of alcoholic beverages, in which it is formed by the anaerobic (without air) fermentation of sugar. On a much larger scale, ethanol for use as a fuel is made by fermentation of corn and other plant materials. Some ethanol (about 5%) is made from petroleum by the reaction of ethylene and water.

Table 23.7



Some Important Alcohols

Condensed Formula

BP (°C)

Systematic Name

Common Name

CH3OH

 65.0

Methanol

Methyl alcohol

Fuel, gasoline additive, making formaldehyde

CH3CH2OH

 78.5

Ethanol

Ethyl alcohol

Beverages, gasoline additive, solvent

CH3CH2CH2OH

 97.4

1-Propanol

Propyl alcohol

Industrial solvent

CH3CH(OH)CH3

 82.4

2-Propanol

Isopropyl alcohol

Rubbing alcohol

HOCH2CH2OH

198

1,2-Ethanediol

Ethylene glycol

Antifreeze

HOCH2CH(OH)CH2OH

290

1,2,3-Propanetriol

Glycerol (glycerin)

Moisturizer in consumer products

John C. Kotz

Use

Rubbing alcohol. ​Common rubbing alcohol is 2-propanol, also called isopropyl alcohol.

23.3  Alcohols, Ethers, and Amines

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1171

H H

H

H C

(g) + H2O(g)

C

H

catalyst

H

H

C

OH(ℓ)

C

H H

ethylene

ethanol

Beginning with three-carbon alcohols, structural isomers are possible. For example, 1-propanol and 2-propanol (common names, propyl alcohol and isopropyl alcohol) are different compounds (Table 23.7). Ethylene glycol and glycerol are common alcohols having two and three OOH groups, respectively. Ethylene glycol is used as antifreeze in automobiles. Glycerol’s most common use is as a softener in soaps and lotions. It is also a raw material for the preparation of nitroglycerin (Figure 23.8).

H H H

C

C H

OH OH Systematic name: Common name:

1,2-ethanediol ethylene glycol

H H H H

C

C C H

OH OH OH 1,2,3-propanetriol glycerol or glycerin

Ethers have the general formula ROR′. The best-known ether is diethyl ether, CH3CH2OCH2CH3. Lacking an OOH group, the properties of ethers are in sharp contrast to those of alcohols. Diethyl ether, for example, has a lower boiling point (34.5 °C) than ethanol, CH3CH2OH (78.3 °C), and is only slightly soluble in water. Diethyl ether, commonly referred to as ether, is a colorless liquid that is very flammable and volatile. In the early nineteenth century, diethyl ether was used as a recreational drug. Some people attended “ether frolics,” where they would cover their mouths with cloth soaked in diethyl ether and inhale the vapor. The diethyl ether produced euphoric effects similar to those of dinitrogen ­monoxide, also called nitrous oxide or laughing gas. In 1842, Dr. Crawford Williamson Long, a ­physician in the U.S. state of Georgia was the first to use diethyl ether as an ­inhalant ­anesthetic, but he did not publish his results until 1849. In the m ­ eantime, Dr. William T. G. Morton, a dentist in Boston, Massachusetts, developed a t­ echnique for administering diethyl ether and served as the anesthesiologist in an operation in 1846. Because of this, Morton has often been credited as the first to use diethyl ether as an anesthetic in surgery. Diethyl ether became widely used as an anesthetic, but it has some safety issues such as its flammability. Its use declined beginning in

© Charles D. Winters/Cengage

Courtesy of the Nobel Foundation

Figure 23.8 Nitroglycerin, dynamite, and Nobel.  ​

(a) Nitroglycerin.

(b) Dynamite.

Concentrated nitric acid and glycerin react to form an oily, highly unstable compound called nitroglycerin, C3H5(ONO2)3. It is more stable if absorbed onto an inert solid, a combination called dynamite.

(c) Alfred Nobel. The fortune of Alfred Nobel (1833–1896), built on the manufacture of dynamite, now funds the Nobel Prizes.

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the 1950s when safer compounds were developed. Today, intravenous anesthetics are used more frequently than inhalant anesthetics, but some are still used, including desflurane, isoflurane, and sevoflurane whose chemical structures are based on ethers but in which the hydrogens are replaced by other substituents, leading to safer compounds. F

H

H

H

C

C

H

H

O

H

H

C

C

H

H

H

F

diethyl ether

F

F

C

C

F

H

F O

C

F

F

H

desflurane

F

Cl

C

C

F

H

F O

C

F

F

F F

C

C

C

F

H

H

isoflurane

F H O

C

F

H

sevoflurane

Ex am p le 23.6

Structural Isomers of Alcohols Problem   How many different alcohols with one OOH group are derivatives of pentane? Draw their structures, and name each alcohol.

What Do You Know?   The formula for pentane is C 5H 12. In an alcohol an OOH group will replace one H atom. Strategy  Pentane, C5H12, has a five-carbon chain. An OOH group can replace a hydrogen atom on one of the carbon atoms. Alcohols are named as derivatives of the alkane (pentane) by replacing the “-e” at the end with “-ol” and indicating the position of the OOH group by a numerical prefix (Appendix E). Solution  Three different alcohols are possible, depending on whether the OOH group is placed on the first, second, or third carbon atom in the chain. (The fourth and fifth positions are identical to the second and first positions in the chain, respectively.) H HO

C

1

H

H C

2

H

H C

3

H

H C

4

H

1-pentanol

H C

5

H

H

H

H

OH H

H

H

C

C

C

C

C

H

H

H

H

H

  

2-pentanol

H

H

  

H

H

OH H

H

C

C

C

C

C

H

H

H

H

H

H

3-pentanol

Think about Your Answer   Additional structural isomers with the formula C5H11OH are possible in which the longest carbon chain has three C atoms (one isomer) or four C atoms (four isomers).

Check Your Understanding  Draw the structure of 1-butanol and alcohols that are structural isomers of the compound.



23.3  Alcohols, Ethers, and Amines

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1173

Properties of Alcohols Methane, CH4, is a gas (bp, −161 °C) with low solubility in water. Methanol, CH3OH, by contrast, is a liquid that is miscible with water in all proportions. The boiling point of methanol, 65 °C, is 226 °C higher than the boiling point of methane. What a difference the addition of a single atom into the structure can make in the properties of simple molecules! Alcohols are related to water, with one of the H atoms of H2O being replaced by an organic group. If a methyl group is substituted for one of the hydrogens of water, methanol results. Ethanol has a OC2H5 (ethyl) group, and propanol has a OC3H7 (propyl) group in place of one of the hydrogens of water. Viewing alcohols as related to water helps in understanding their properties. The two parts of methanol, the OCH3 group and the OOH group, contribute to its properties. For example, methanol will burn, a property associated with hydrocarbons. On the other hand, its boiling point is more like that of water. The temperature at which a substance boils is related to intermolecular forces, the forces of attraction between molecules. The stronger the attractive, intermolecular forces in a sample, the higher the boiling point (Section 11.3). These forces are particularly strong in water, a result of the polarity of the OOH group in this molecule (­Section 8.2). Methanol is also a polar molecule, and it is the polar OOH group that leads to a high boiling point. In contrast, methane is nonpolar, and its low boiling point is the result of weak intermolecular forces. It is also possible to explain the differences in the solubility of methane, methanol, and other alcohols in water (Figure 23.9). The solubility of methanol and ethylene glycol is conferred by the polar OOH portion of the molecule. Methane, which is nonpolar, has low water solubility. As the size of the alkyl group in an alcohol increases, the alcohol boiling point rises, a general trend seen in families of similar compounds and related to molar mass (Table 23.7). The solubility in water in this series decreases. Methanol and ethanol are completely miscible in water, whereas 1-propanol is moderately water soluble; 1-butanol is less soluble than 1-propanol. With an increase in the size of the hydrocarbon group, the organic group (the nonpolar part of the molecule) has become a larger fraction of the molecule, and properties associated with nonpolarity begin to dominate. Space-filling models show that in methanol, the polar and nonpolar parts of the molecule are approximately similar in size, but in 1-butanol

polar portion

Methanol is often added to automobile gasoline tanks in the winter to prevent water in the fuel lines from freezing. It is soluble in water and lowers the water’s freezing point.

Ethylene glycol is used in automobile radiators. It is soluble in water, and lowers the freezing point and raises the boiling point of the water in the cooling system. (Section 13.4.)

© Charles D. Winters/Cengage

© Charles D. Winters/Cengage

polar nonpolar hydrocarbon portion portion

Ethylene glycol, a major component of automobile antifreeze, is completely miscible with water.

Figure 23.9  Properties and uses of two alcohols, methanol and ethylene glycol. ​

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the OOH group is less than 20% of the molecule. The molecule is less like water and more “organic.” Electrostatic potential surfaces amplify this point. nonpolar hydrocarbon polar portion portion

nonpolar hydrocarbon portion

methanol

polar portion

1-butanol

Amines It is often convenient to think about water and ammonia as being similar molecules: They are the simplest hydrogen compounds of adjacent second-period elements. Both are polar and exhibit some similar chemistry, such as protonation (to give H3O1 and NH41) and deprotonation (to give OH2 and NH22). This comparison of water and ammonia can be extended to alcohols and amines. Alcohols have formulas related to water in which one hydrogen in H2O is replaced with an organic group (ROOH). In organic amines, one or more hydrogen atoms of NH3 are replaced with an organic group. Amine structures are similar to ammonia’s structure; that is, the geometry about the N atom is trigonal pyramidal. Amines are categorized based on the number of organic substituents as primary (one organic group), secondary (two organic groups), or tertiary (three organic groups). As examples, consider the three amines with methyl groups: CH3NH2, (CH3)2NH, and (CH3)3N.

CH3NH2

(CH3)2NH

(CH3)3N

primary amine methylamine

secondary amine dimethylamine

tertiary amine trimethylamine

Electrostatic potential surface for methylamine. ​The surface for methylamine shows that this water-soluble amine is polar with the partial negative charge on the N atom.

Properties of Amines Amines usually have offensive odors. You know what the odor is if you have ever smelled decaying fish. Two appropriately named amines, putrescine and cadaverine, add to the odor of urine, rotten meat, and bad breath. H2NCH2CH2CH2CH2NH2

H2NCH2CH2CH2CH2CH2NH2

putrescine 1,4-butanediamine

cadaverine 1,5-pentanediamine

The smallest amines are water soluble, but most amines are not. All amines are bases, however, and they react with acids to give salts, many of which are water

23.3  Alcohols, Ethers, and Amines

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1175

soluble. As with ammonia, the reactions involve adding H1 to the lone pair of electrons on the N atom. This is illustrated by the reaction of aniline (aminobenzene) with H2SO4 to give the anilinium cation.

HC HC

H2C

H C

CH

C CH

N

CH2 N

C6H5NH2(aq) + H2SO4(aq)

C6H5NH3+(aq) + HSO4−(aq)

CH2

CH3

nicotine

aniline H+

H+

Nicotine, an amine. ​Two nitrogen atoms in the nicotine molecule can be protonated, which is the form in which nicotine is normally found. The protons can be removed, by treating it with a base. This free-base form is much more poisonous and addictive.

anilinium ion

The facts that an amine can be protonated and that the proton can be removed again by treating the compound with a base have practical and physiological importance. Nicotine in cigarettes is normally found in the protonated form. (This watersoluble form is also used often in insecticides.) Adding a base such as ammonia removes the H1 ion to leave nicotine in its free-base form. NicH221(aq) + 2 NH3(aq) n Nic(aq) + 2 NH41(aq)

In this form, nicotine is much more readily absorbed by the skin and mucous membranes, so the compound is a much more potent poison.

23.4 Compounds with a Carbonyl Group Goals for Section 23.4 • Name and draw structures of carbonyl compounds—aldehydes, ketones, acids, esters, and amides.

• Describe reactions of carbonyl compounds. Formaldehyde, acetic acid, and acetone are important organic compounds, and they share a common structural feature: Each contains a trigonal-planar carbon atom doubly bonded to an oxygen. The CPO group is called the carbonyl group, and these compounds are members of a large class of compounds called carbonyl compounds. O C carbonyl group

formaldehyde

acetic acid

acetone

CH2O aldehyde

CH3CO2H carboxylic acid

CH3COCH3 ketone

Five types of carbonyl compounds (Table 23.6) will be considered in this section: •

Aldehydes (RCHO) have an organic group (OR) and an H atom attached to a carbonyl group.

•

Ketones (RCOR′) have two OR groups attached to the carbonyl carbon; they may be the same organic groups, as in acetone, or different groups.

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•

Carboxylic acids (RCO2H) have an OR group and an OOH group attached to the carbonyl carbon.

•

Esters (RCO2R′) have OR and OOR′ groups attached to the carbonyl carbon.

•

Amides (RCONR2′, RCONHR′, and RCONH2) have an OR group and an amino group (ONH2, ONHR, ONR2) bonded to the carbonyl carbon.

Primary alcohol: ethanol

CH3 H

R

CH2

O

OH

R

primary alcohol

C

H

Secondary alcohol: 2-propanol

CH3 H

R

C

OH

Tertiary alcohol: 2-methyl-2-propanol

CH3 H3C

OH

C

OH

CH3

carboxylic acid

aldehyde

C CH3

O

oxidizing agent

OH

H

Aldehydes, ketones, and carboxylic acids are oxidation products of alcohols and are commonly made by this route. The product obtained through oxidation of an alcohol depends on the alcohol’s structure, which is classified according to the number of carbon atoms bonded to the C atom bearing the OOH group. ­Primary alcohols have one carbon and two hydrogen atoms attached, whereas secondary ­alcohols have two carbon atoms and one hydrogen atom attached. Tertiary alcohols have three carbon atoms attached to the C atom bearing the OOH group. A primary alcohol is oxidized in two steps, first to an aldehyde and then to a ­carboxylic acid: oxidizing agent

C

For example, the air oxidation of ethanol in wine ultimately produces wine vinegar, the most important ingredient of which is acetic acid.

H H H

C

C

oxidizing agent

OH(ℓ)

H

H H

H

O

C

C

OH(ℓ)

H

ethanol

acetic acid

Acids have a sour taste. The word “vinegar” (from the French vin aigre) means sour wine. A device to test a person’s breath for alcohol relies on the oxidation of ethanol (Figure 23.10). Oxidation of a secondary alcohol produces a ketone:

R

C

oxidizing agent

R′

O R

C

R′

H ( R and

secondary alcohol ketone R′ are organic groups. They may be the same or different.)

Common oxidizing agents used for these reactions are reagents such as KMnO4 and K2Cr2O7 (Table 3.4). Finally, tertiary alcohols do not react with the usual oxidizing agents. oxidizing agent

(CH3)3COH 8888888888n no reaction

Aldehydes and Ketones Aldehydes and ketones can have pleasant odors and are often used in fragrances. B ­ enzaldehyde is responsible for the odor of almonds and cherries; cinnamaldehyde is found in the bark of the cinnamon tree; and the ketone 4-(p‑hydroxyphenyl)-2-butanone is responsible for the odor of ripe raspberries. Table 23.8 lists several simple aldehydes and ketones.

© Charles D. Winters/Cengage

OH

Figure 23.10 Alcohol tester.  ​This device for testing a person’s breath for the presence of ethanol relies on the oxidation of the alcohol. If present, ethanol is oxidized by potassium dichromate, K2Cr2O7, to acetaldehyde and then to acetic acid. The yellow-orange dichromate ion is reduced to green Cr31(aq), with the color change indicating that ethanol was present.

23.4  Compounds with a Carbonyl Group

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1177

Table 23.8

Simple Aldehydes and Ketones

Structure

Common Name

Systematic Name

BP (°C)

Formaldehyde

Methanal

Acetaldehyde

Ethanal

20

Acetone

Propanone

56

Methyl ethyl ketone

Butanone

80

Diethyl ketone

3-Pentanone

O HCH

−19

O CH3CH O CH3CCH3 O CH3CCH2CH3 O CH3CH2CCH2CH3

benzaldehyde, C6H5CHO

102

trans-cinnamaldehyde, C6H5CH CHCHO

Aldehydes and ketones are the oxidation products of primary and secondary alcohols, respectively. The reverse reactions—reduction of aldehydes to primary alcohols and reduction of ketones to secondary alcohols—are also known. Commonly used reagents for such reductions are NaBH4 and LiAlH4, although H2 is used on an industrial scale.

OH

O © Charles D. Winters/Cengage

R

Aldehydes, ketones, and odors. ​ The odors of almonds and cinnamon are due to aldehydes, whereas the odor of fresh raspberries comes from a ketone.

C

H

NaBH4 or LiAlH4

R

C

H

H aldehyde

primary alcohol

OH

O R

C

R

NaBH4 or LiAlH4

R

C

R

H ketone

secondary alcohol

Carboxylic Acids Acetic acid is the most common and most important carboxylic acid. For many years, acetic acid was made by oxidizing ethanol produced by fermentation. Now, however, most of the acetic acid for industrial use is made by combining carbon monoxide and methanol in the presence of a catalyst: CH3OH(ℓ) + CO(g) methanol

catalyst

CH3CO2H(ℓ) acetic acid

In 2019, over 3 billion kilograms of acetic acid were produced in the United States for use in plastics, synthetic fibers, and fungicides.

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Problem Solving Tip 23.2 Naming Aldehydes, Ketones, and Carboxylic Acids Ketones:  Substitute “-one” for the final “-e” in the name of the hydrocarbon. The position of the ketone functional group (the carbonyl group) is indicated by the number of the carbon atom. For example, the compound CH3COCH2CH(C2H5)CH2CH3 has the carbonyl group at the ­2 position and an ethyl group at the 4 ­position of a 6-carbon chain; its name is 4-ethyl-2-hexanone.

Aldehydes:  Substitute “-al” for the final “-e” in the name of the hydrocarbon. The carbon atom of an aldehyde is, by definition, carbon-1 in the hydrocarbon chain. For example, CH3CH(CH3)CH2CH2CHO contains a 5-carbon chain with the aldehyde functional group being carbon-1 and the OCH3 group at position 4; thus, the name is 4-methyl­pentanal.

Carboxylic acids (organic acids): ​ Substitute “-oic” for the final “-e” in the name of the hydrocarbon. The carbon atoms in the longest chain are counted beginning with the ­carboxylic carbon atom. For example, trans-CH3CHPCHCH2CO2H is named as a derivative of trans-3-pentene, that is, trans-3-pentenoic acid.

Acids have a sour taste (Figure 23.11), and many occur naturally in common foods: citric acid in fruits, acetic acid in vinegar, and tartaric acid in grapes are just three examples (Table 23.9). Some carboxylic acids have common names derived from the source of the acid (Tables 23.9 and 23.10). Because formic acid is found in ants, its name comes from the Latin word for ant ( formica). Butanoic acid (commonly called butyric acid) gives rancid butter its unpleasant odor, and the name is related to the Latin word for butter (butyrum). Because of the substantial electronegativity of oxygen, the two O atoms of the carboxylic acid group are slightly negatively charged, and the H atom of the OOH group is partially positively charged. This charge distribution has several important implications: •

The polar acetic acid molecule dissolves readily in water, which you already know because vinegar is an aqueous solution of acetic acid. (Acids with larger organic groups are less soluble, however.)

•

The hydrogen of the OOH group is the acidic hydrogen. Recall that acetic acid is a weak acid in water, as are most other organic acids.

Name

Structure CO2H

O

H

acidic H atom

−

H −

+

carboxylic acid group

Acetic acid. ​The H atom of the carboxylic acid group (OCO2H) is the acidic proton of this and other carboxylic acids.

Berries

HO2C

CH2

C

CH2

CO2H

Citrus fruits

CO2H H3C

CH

CO2H

Sour milk

OH Malic acid

HO2C

CH2

CH

CO2H

Oleic acid

CH3(CH2)7

Oxalic acid

HO2C

Stearic acid

CH3(CH2)16

CO2H

Tartaric acid

HO2C

CH

CH

OH

OH

Apples

OH



C

Natural Source

OH

Lactic acid

C

Some Naturally Occurring Carboxylic Acids

Benzoic acid

Citric acid

H

© Charles D. Winters/Cengage

Table 23.9

H O

CH

CH

(CH2)7

CO2H

CO2H

Vegetable oils Rhubarb, spinach, cabbage, tomatoes Animal fats

CO2H

Grape juice, wine

Figure 23.11  Acetic acid in bread.  ​Acetic acid is produced in bread leavened with the yeast Saccharomyces exiguus. Another group of bacteria, Lactobacillus sanfrancisco, contributes to the flavor of sourdough bread. These bacteria metabolize the sugar maltose, excreting acetic acid and lactic acid, CH3CH(OH)CO2H, thereby giving the bread its unique sour taste.

23.4  Compounds with a Carbonyl Group

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1179

Table 23.10 Structure

Some Simple Carboxylic Acids

Common Name

Systematic Name

BP (°C)

Formic acid

Methanoic acid

101

Acetic acid

Ethanoic acid

118

Propionic acid

Propanoic acid

141

Butyric acid

Butanoic acid

163

Valeric acid

Pentanoic acid

187

O HCOH O CH3COH

© Charles D. Winters/Cengage

O CH3CH2COH O CH3(CH2)2COH O CH3(CH2)3COH

Formic acid, HCO2H. ​This acid puts the sting in ant bites.

Carboxylic acids undergo a number of reactions. Among these is the reduction of the acid (with reagents such as LiAlH4 or NaBH4) first to an aldehyde and then to an alcohol. For example, acetic acid is reduced first to acetaldehyde and then to ethanol. CH3CO2H acetic acid

LiAlH4

CH3CHO

LiAlH4

acetaldehyde

CH3CH2OH ethanol

Yet another important aspect of carboxylic acid chemistry is the reaction with bases to give carboxylate anions. For example, acetic acid reacts with hydroxide ions to give acetate ions and water.

Omega-3 Fatty Acids

There is considerable interest in incorporating omega-3 fatty acids in the diet. Although potential health benefits are not fully defined, proponents say there is some evidence that they lower blood pressure and reduce the chances of heart attacks and strokes. Sources of omega-3 fatty acids include fish and plant oils. These compounds are polyunsaturated, long-chain carboxylic acids, and there are a number of different, naturally occurring compounds in this class. The word “omega” in the name refers to the last carbon atom in the carbon chain of the acid, and the number 3 indicates that there is a double bond between the third and fourth carbon atoms when counted from that end of the molecule. An example of an omega-3 fatty acid is all-cis-7, 10, 13-hexadecatrienoic acid,

First C C bond is at 3rd C from omega end.

O OH

H3C Omega end of the acid

C13

C10

whose structure is shown here. Naming large complicated molecules is challenging, but it is possible to decipher the structure from the name. Hexadeca identifies this as a 16-carbon atom compound, triene tells you there are three double bonds (all with cis geometry), and the oic ending defines it as a carboxylic acid. The numbers 7, 10, and 13 indicate the positions of the double bonds relative to the carboxylic carbon atom.

C7

C1

© Steven Hyatt

A Closer Look

CH3CO2H(aq) + OH2(aq) n CH3CO22(aq) + H2O(ℓ)

Omega-3 tablets.

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Esters Esters are the product of the reaction of an acid with an alcohol. Such reactions, called esterification reactions, are usually carried out with an acid catalyst (such as HCl). These reactions do not go to completion, but instead produce an equilibrium mixture of the reactants and products. To obtain the ester in good yield, the reaction can be forced to completion. One way to do this is to remove water from the system (for example, by distilling it from the reaction mixture); this causes the equilibrium to shift to the right, thus forming more of the ester.

O

O

RC

O

H + R′

carboxylic acid

O

H3O+

H

RC

alcohol

R′ + H2O

ester

O

portion from acetic acid

O H3O+

CH3COH + CH3CH2OH acetic acid

O

ethyl acetate, an ester CH3CO2CH2CH3

CH3COCH2CH3 + H2O

ethanol

portion from ethanol

ethyl acetate

In these reactions the O atom of the alcohol ends up as part of the ester. This is known from isotope labeling experiments. If the reaction is run using an alcohol in which the alcohol oxygen is 18O, all of the 18O ends up in the ester molecule. Table 23.11 lists a few common esters and the acid and alcohol from which they are formed. The two-part name of an ester is given by (1) the name of the hydrocarbon group from the alcohol and (2) the name of the carboxylate group derived from the acid name by replacing “-ic” with “-ate.” For example, ethanol (commonly called ethyl alcohol) and acetic acid combine to give the ester ethyl acetate. An important reaction of esters is hydrolysis (literally, reaction with water), a reaction that is the reverse of the reaction forming the ester. In a neutral solution, this reaction also gives an equilibrium mixture of alcohol, acid, and ester. In order for this reaction to reach completion, it is carried out in a basic solution. The base, NaOH for example, reacts with the acid, thus shifting the equilibrium toward the product.

O RCOR′ + NaOH ester

O heat in water

O

O CH3COCH2CH3 + NaOH ethyl acetate

RCO − Na+ + R′OH carboxylate salt alcohol

heat in water

CH3CO − Na+ + CH3CH2OH sodium acetate

ethanol

H

The carboxylic acid can be recovered if the sodium salt is treated with a strong acid such as HCl:

O CH3CO −Na+(aq) + HCl(aq) sodium acetate

C

O O

O CH3COH(aq) + NaCl(aq)

C

CH3

O

acetic acid

Unlike the acids from which they are derived, esters often have pleasant odors (Table 23.11). Typical examples are methyl salicylate, or oil of wintergreen, and ­benzyl acetate. Methyl salicylate is derived from salicylic acid, the parent compound of aspirin.

O

Aspirin, a commonly used analgesic. ​It is based on benzoic acid with an acetate group, OO2CCH3, in the ortho position. Aspirin has both carboxylic acid and ester functional groups.

23.4  Compounds with a Carbonyl Group

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1181

Table 23.11

Some Acids, Alcohols, and Their Esters

Acid

Alcohol

Ester

CH3

Odor of Ester

O

CH3

CH3CO2H

CH3CHCH2CH2OH

CH3COCH2CH2CHCH3

acetic acid

3-methyl-1-butanol

3-methylbutyl acetate

Banana

O CH3CH2CH2CO2H

CH3CH2CH2CH2OH

CH3CH2CH2COCH2CH2CH2CH3

butanoic acid

1-butanol

butyl butanoate

Pineapple

O

© Charles D. Winters/Cengage

butanoic acid

benzyl alcohol

Esters. ​Many fruits such as bananas and strawberries as well as consumer products (here, perfume and oil of wintergreen) contain esters. amide linkage

CH3CH2CH2COCH2

CH2OH

CH3CH2CH2CO2H

benzyl butanoate

O

O

COH + CH3OH

COCH3 + H2O

OH

OH

salicylic acid

methanol

methyl salicylate, oil of wintergreen

Benzyl acetate, the active component of oil of jasmine, is formed from benzyl ­alcohol (C6H5CH2OH) and acetic acid. The chemicals are inexpensive, so synthetic ­jasmine is often used as a fragrance in less-expensive perfumes and toiletries.

O

O

CH3COH + acetic acid this portion from acetic acid

this portion from methylamine

Rose

CH2OH

+ H2O

CH3COCH2

benzyl alcohol

benzyl acetate oil of jasmine

Amides A carboxylic acid and an alcohol react by loss of water to form an ester. In a similar manner, another class of organic compounds—amides—form when a carboxylic acid reacts with an amine, again with loss of water.

O R An amide, N-methylacetamide. ​ The N-methyl portion of the name derives from the amine portion of the molecule, where the N indicates that the methyl group is attached to the nitrogen atom. The “-acet” portion of the name indicates the acid on which the amide is based. The electrostatic potential surface and model show the polarity and planarity of the amide linkage.

C

R′ OH + H

carboxylic acid

N amine

R′

R

O

R′

C

N

R′ + H2O

amide

Amides have an organic group and an amino group (ONH2, ONHR′, or ONR′R) attached to the carbonyl group. The C atom involved in the amide bond has three bonded groups and no lone pairs. You would predict it should be sp2 hybridized with trigonal-planar geometry and bond angles of approximately 120°—and this is what is found. However, the structure of the amide group offers a surprise. The N atom is also observed to have trigonal-planar geometry with bonds to three attached atoms at 120°. Because the amide nitrogen is surrounded by four pairs of electrons, you would probably have predicted the N atom would have sp3 hybridization and bond angles of about 109°.

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Based on the observed geometry of the amide N atom, the atom is assigned sp2 hybridization. To rationalize the observed angle and sp2 hybridization, consider a second resonance form of the amide. O

O R

C

N

H

R

C

−

+

N

R

R

(A)

(B)

H

Form B contains a CPN double bond, and the O and N atoms have negative and positive charges, respectively. The N atom can be assigned sp 2 hybridization, and the π bond in B arises from overlap of p orbitals on C and N. The second resonance structure for an amide link also explains why the carbon– nitrogen bond is relatively short, about 132 pm, a value between that of a CON single bond (149 pm) and a CPN double bond (127 pm). In addition, restricted rotation occurs around the CPN bond, making it possible for isomeric species to exist if the two groups bonded to N are different. The amide grouping is particularly important in some synthetic polymers (Section 23.5) and in proteins, where it is referred to as a peptide link. The compound N-acetyl-p-aminophenol, an analgesic known by the generic name acetaminophen, is another amide. Use of this compound as an analgesic was apparently discovered by accident when a common organic compound called acetanilide (like acetaminophen but without the OOH group) was mistakenly put into a prescription for a patient. Acetanilide acts as an analgesic, but it can be toxic. An OOH group para to the amide group makes the compound nontoxic, an interesting example of how a seemingly small structural difference affects chemical function.

H O H3C

C

H N H

C C

C C

H C C

O H

H

Acetaminophen, N-acetylp-aminophenol. ​This analgesic is an amide. It is used in over-the-counter painkillers such as Tylenol.

Ex am p le 23.7

Functional Group Chemistry Problem (a) Draw the structure of the product of the reaction between propanoic acid and 1-propanol. What is the systematic name of the reaction product, and what functional group does it contain? (b) What is the result of reacting 2-butanol with an oxidizing agent? Give the name, and draw the structure of the reaction product.

What Do You Know?   From the material covered in this chapter, you know the names, structures, and common chemical reactions of organic compounds mentioned in this question. Propanoic acid is a carboxylic acid with three carbon atoms. 1-propanol is an alcohol with three carbon atoms and the OOH group on carbon 1, and 2-butanol is an alcohol with four carbon atoms and the OOH group on carbon 2.

Strategy   Determine the products of these reactions, based on the discussion in the text. The functional groups of organic compounds are usually where chemical changes occur in organic compounds and determine the reactions that are possible. Solution  (a) Carboxylic acids such as propanoic acid react with alcohols to give esters. O



O

CH3CH2COH + CH3CH2CH2OH

CH3CH2COCH2CH2CH3 + H2O

propanoic acid

propyl propanoate, an ester

1-propanol

23.4  Compounds with a Carbonyl Group

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1183

(b) 2-Butanol is a secondary alcohol. Such alcohols are oxidized to ketones. OH CH3CHCH2CH3

O oxidizing agent

CH3CCH2CH3 butanone, a ketone

2-butanol

Think about Your Answer   Students sometimes find themselves overwhelmed by the large amount of information presented in organic chemistry. Your study of this material will be more successful if you carefully organize information based on the type of compound. Functional groups usually react in predictable ways. For example, a secondary alcohol is oxidized to a ketone. This is true whether the starting alcohol is 2-butanol, 2-propanol, 2-pentanol, 3-pentanol, or other secondary alcohol.

Check Your Understanding (a) Name each of the following compounds and its functional group.

O (1) CH3CH2CH2OH

(2) CH3COH

(3) CH3CH2NH2

(b) Name the product from the reaction of compounds 1 and 2 above. (c) What is the name and structure of the product from the oxidation of 1 with an excess of oxidizing agent? (d) Give the name and structure of the compound that results from combining 2 and 3. (e) What is the result of adding an acid (say HCl) to compound 3?

23.5 Polymers Goals for Section 23.5 • Write equations for the formation of addition polymers and condensation polymers and describe their structures.

• Relate properties of polymers to their structure. The very large molecules known as polymers can be either synthetic materials or ­naturally occurring substances such as proteins or nucleic acids. Although many different types of polymers are known with widely different compositions and structures, their properties are understandable, based on the principles already developed for small molecules.

Classifying Polymers The word polymer means “many parts” (from the Greek, poly and meros). Thus, ­ olymers are giant molecules made by chemically joining many small molecules p called monomers. Polymer molar masses typically range from thousands to millions. Extensive use of synthetic polymers is a fairly recent development. A few synthetic polymers (Bakelite, rayon, and celluloid) were made early in the twentieth century, but most of the products with which you are likely to be familiar originated since the middle of the twentieth century. By 1976, synthetic polymers outstripped steel as the most widely used materials in the United States. The production of synthetic polymers in the United States is now about 150 kg or more per person annually.

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The polymer industry classifies polymers in several different ways. One is their response to heating. Thermoplastics (such as polyethylene) soften and flow when they are heated and harden when they are cooled. Thermosetting plastics (such as Formica) are initially soft but set to a solid when heated and cannot be resoftened. Another classification scheme depends on the end use of the polymer—for example, plastics, fibers, elastomers, coatings, and adhesives. A more chemically-oriented approach to polymer classification is based on the method of synthesis. Addition polymers are made by directly adding monomer units together. Condensation polymers are made by combining monomer units and splitting out a small molecule, often water.

Addition Polymers Polyethylene, polystyrene, and polyvinyl chloride (PVC) are common addition polymers (Figure 23.12). They are built by repeated addition reactions of simple alkenes such as ethylene (CH2PCH2), styrene (C6H5CHPCH2), and vinyl chloride (CH2PCHCl). These and other addition polymers (Table 23.12), all derived from alkenes, have widely varying properties and uses.

Polyethylene and Other Polyolefins Polyethylene is by far the leader in amount of polymer produced. Ethylene (C2H4), the monomer from which polyethylene is made, is a product of petroleum refining and one of the top five chemicals produced in the United States. When ethylene is heated to between 100 and 250 °C at a pressure of 1000 to 3000 atm in the presence of a catalyst, polymers with molar masses up to several million are formed. The reaction can be expressed as the chemical equation:

n H2C

CH2

ethylene

H

H

C

C

H

H n

polyethylene

Photos: John C. Kotz

The abbreviated formula of the reaction product, (OCH2CH2O)n, shows that polyethylene is a chain of carbon atoms, each bearing two hydrogens. The chain length for polyethylene can be very long. A polymer with a molar mass of 1 million would contain almost 36,000 linked ethylene molecules. Samples of polyethylene formed under various pressures and catalytic conditions have different properties, as a result of different molecular structures. For example, when chromium(III) oxide is used as a catalyst, the product is almost

(a) High-density polyethylene.

(b) Polystyrene.

(c) Polyvinyl chloride.

Figure 23.12  Common polymer-based consumer products.  ​Recycling information is provided on most plastics (often molded into the bottom of bottles). High-density polyethylene is designated with a “2” inside a triangular symbol and the letters “HDPE.” Polystyrene is designated by “6” with the symbol PS, and polyvinyl chloride, PVC, is designated with a “3” inside a triangular symbol with the symbol “V” or “PVC” below.



23.5 Polymers

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1185

Photos: John C. Kotz

(a) The linear form, high-density polyethylene (HDPE).

(b) Branched chains occur in lowdensity polyethylene (LDPE).

(c) Cross-linked polyethylene (CLPE).

© Charles D. Winters/Cengage

Figure 23.13 Polyethylene. ​

exclusively a linear chain (Figure 23.13a). If ethylene is heated to 230 °C at high pressure, however, irregular branching occurs (Figure 23.13b). Still other conditions lead to cross-linked polyethylene, in which different chains are linked together (Figure 23.13c). The high molar mass chains of linear polyethylene pack closely together and result in a material with a density of 0.97 g/cm3. This material, referred to as highdensity polyethylene (HDPE), is hard and tough, which makes it suitable for items such as milk bottles. If the polyethylene chain contains branches, however, the chains cannot pack as closely, and a lower-density material (0.92 g/cm3) known as low-density polyethylene (LDPE) results. This material is softer and more flexible than HDPE. It is used in plastic wrap and sandwich bags, among other things. Linking up the polymer chains in cross-linked polyethylene (CLPE) causes the material to be more rigid and inflexible. CLPE is often used as insulation for high voltage electrical cables and as a replacement for copper tubing in some home water systems. Polymers formed from substituted ethylenes (CH2PCHX) have a range of properties and uses (Table 23.12). Sometimes, the properties are predictable based on the molecule’s structure. Polymers without polar substituent groups, such as polystyrene, often dissolve in organic solvents, a property useful for some types of fabrication. polymers based on substituted ethylenes, H2C CHX

CH2CH

OH n

Figure 23.14  Slime from polyvinyl alcohol.  ​When boric acid, B(OH)3, is added to an aqueous suspension of polyvinyl alcohol, (CH2CHOH)n, the mixture becomes very viscous because boric acid reacts with the OOH groups on the polymer chain, causing cross-linking to occur. (The model shows an idealized structure of a portion of the polymer.)

CH2CH

CH2CH

OCCH3 n

n

O polyvinyl alcohol

polyvinyl acetate

polystyrene

Polyvinyl alcohol is a polymer with little affinity for nonpolar solvents but an affinity for water, which is not surprising, based on the large number of polar OH groups (Figure 23.14). Vinyl alcohol (CH2PCHOH) is not a stable ­compound (it isomerizes to acetaldehyde CH3CHO), so polyvinyl alcohol cannot be made from this compound. Instead, it is made by hydrolyzing the ester groups in ­polyvinyl acetate.

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Table 23.12

Ethylene Derivatives That Undergo Addition Monomer Common Name

Formula H

Polymer Name (Trade Names)

Uses

H C

C

H

Ethylene

Polyethylene (polythene)

Squeeze bottles, bags, films, toys and molded objects, electric insulation

Propylene

Polypropylene (Vectra, Herculon)

Bottles, films, indooroutdoor carpets

Vinyl chloride

Polyvinyl chloride (PVC)

Floor tile, raincoats, pipe

Acrylonitrile

Polyacrylonitrile (Orlon, Acrilan)

Rugs, fabrics

Styrene

Polystyrene (Styrofoam, Styron)

Food and drink coolers, building material, insulation

Vinyl acetate

Polyvinyl acetate (PVA)

Latex paint, adhesives, textile coatings

Methyl methacrylate

Polymethyl methacrylate (Plexiglas™, Lucite™)

High-quality transparent objects, latex paints, contact lenses

Tetrafluoroethylene

Polytetrafluoroethylene (Teflon™)

Gaskets, insulation, bearings, pan coatings

H H

H C

C CH3

H

H

H C

C Cl

H

H

H C

C CN

H

H

H C

C

H H

H C

C O

H

C

CH3

O H

CH3 C

C C

H

O

CH3

O F

F C

F

C F

H

H

H

H

C

C n + n H2O

C

C n + n CH3CO2H

H

OCCH3

H

OH

O Solubility in water or organic solvents can be a liability for polymers. The many uses of polytetrafluoroethylene [TeflonTM, (OCF2CF2O)n] stem from the fact that it does not interact with water or organic solvents. Polystyrene, with n = 5700, is a clear, hard, colorless solid that can be molded easily at 250 °C. However, you are probably more familiar with the very light, foam-like material known as Styrofoam that is used widely for food and beverage containers and for home insulation (Figure 23.12b). Styrofoam is produced by a process called expansion molding. Polystyrene beads containing 4% to 7% of a low-boiling liquid like pentane are placed in a mold and heated with steam or hot air. Heat causes the solvent to vaporize, creating a foam in the molten polymer that expands to fill the shape of the mold.

23.5 Polymers

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1187

Natural and Synthetic Rubber

CH3 H

C H

C

H C

C

H

H

Gavin Hellier/Alamy Stock Photo

isoprene, 2-methyl-1,3-butadiene

Figure 23.15  Natural rubber.  ​ The sap that comes from the rubber tree is a natural polymer of isoprene. All the linkages in the carbon chain are cis. When natural rubber is heated strongly in the absence of air, it smells of isoprene. This observation provided a clue that rubber is composed of this building block.

Natural rubber was first introduced in Europe in 1740, but it remained a curiosity until 1823, when Charles Macintosh invented a way of using it to waterproof cotton cloth. The macintosh, as rain coats are still sometimes called, became popular despite major problems: Natural rubber is notably weak, soft, and tacky when warm and brittle at low temperatures. In 1839, after 5 years of research on natural rubber, the American inventor Charles Goodyear (1800–1860) discovered that heating gum rubber with sulfur produces a material that is elastic, water-repellent, resilient, and no longer sticky. Rubber is a naturally occurring polymer, the monomers of which are molecules of 2-methyl-1,3-butadiene, commonly called isoprene. In natural rubber, isoprene monomers are linked together through carbon atoms 1 and 4—that is, through the end carbon atoms of the C4 chain (Figure 23.15). This leaves a double bond between carbon atoms 2 and 3. In natural rubber, these double bonds have a cis configuration. In vulcanized rubber, the material that Goodyear discovered, the polymer chains of natural rubber are cross-linked by short chains of sulfur atoms. Cross-linking helps to align the polymer chains, so the material does not undergo a permanent change when stretched and it springs back when the stress is removed. Substances that behave this way are called elastomers. With a knowledge of the composition and structure of natural rubber, chemists began searching for ways to make synthetic rubber. When they first tried to make the polymer by linking isoprene monomers together, however, what they made was sticky and useless. The problem was that synthesis procedures gave a mixture of cis- and transpolyisoprene. In 1955, however, chemists at the Goodyear and Fire­stone companies discovered special catalysts to prepare the all-cis polymer. This synthetic material, which was structurally identical to natural rubber, is now manufactured cheaply. In fact, worldwide production of polyisoprene in 2020 was 1.44 × 1010 kg. Other kinds of polymers have further expanded the repertoire of elastomeric materials now available. Polybutadiene, for example, is currently used in the production of tires, hoses, and belts. Some elastomers, called copolymers, are formed by polymerization of two (or more) different monomers. A copolymer of butadiene and styrene, made with a 3∶1 ratio of these raw materials, is the most important synthetic rubber now made; 8.2 × 109 kilograms of styrene-butadiene rubber (SBR) were produced worldwide in 2020. A major use for this polymer is for making tires, but a little is left over each year to make bubble gum. The stretchiness of bubble gum once came from natural rubber, but SBR is now used to help you blow bubbles. H 3n HC H2C

CH

+ n H2C

C

CH2

1,3-butadiene

styrene

H HC CH2

CH H2C

HC CH2

CH H2C

C

CH2

CH2

HC

CH2 CH

n

styrene-butadiene rubber (SBR)

Condensation Polymers A chemical reaction in which two molecules react by splitting out, or eliminating, a small molecule is called a condensation reaction. The reaction of an alcohol with

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a carboxylic acid to give an ester is an example of a condensation reaction. One way to form a condensation polymer uses two different reactant molecules, each containing two functional groups. Another route uses a single molecule with two different functional groups. Commercial polyesters are made using both types of reactions.

Terephthalic acid contains two carboxylic acid groups, and ethylene glycol contains two alcohol groups. When mixed, the acid and alcohol functional groups at both ends of these molecules can react to form ester linkages, splitting out water. The result is a condensation polymer called polyethylene terephthalate (PET). The multiple ester linkages make this substance a polyester (Figure 23.16). O n HOC

O

O

O

COH + n HOCH2CH2OH

C

COCH2CH2O

terephthalic acid

ethylene glycol

n

© Charles D. Winters/Cengage

Polyesters

+ 2n H2O

Figure 23.16 Polyesters. ​ Polyethylene terephthalate is used to make clothing, soda bottles, car parts, and many other consumer products.

polyethylene terephthalate (PET), a polyester

PET is also made by a reaction between methyl terephthalate and ethylene glycol, a reaction called transesterification. In this reaction one ester is converted to another, and removal of methanol by distillation allows the reaction to reach completion.

Polyester textile fibers made from PET are marketed as Dacron and Terylene. The inert, nontoxic, nonflammable, and non-blood-clotting properties of Dacron polymers make Dacron tubing an excellent substitute for human blood vessels in heart bypass operations, and Dacron sheets are sometimes used as temporary skin for burn victims. A polyethylene terephthalate film, Mylar, has unusual strength and can be rolled into sheets one-thirtieth the thickness of a human hair. You may be familiar with shiny balloons made from Mylar and coated with a thin layer of aluminum foil (Figure 23.17). An advantage of Mylar balloons is that they stay inflated for a longer time than ordinary latex balloon. This is because Mylar balloons are less porous than ordinary latex balloons, and so helium escapes from them by the process of effusion (see Section 10.7) more slowly. A drawback to Mylar balloons, however, is that they are not biodegradable, and there are no practical recycling methods. The best case scenario is that they are disposed of in the trash. In many cases, however, they contribute to plastic pollution in the environment. There is considerable interest in another polyester, polylactic acid (PLA). Lactic acid contains both carboxylic acid and alcohol functional groups, so condensation between molecules of this monomer gives a polymer.

n HO

H

O

C

C

CH3

OH

H

O

C

C

CH3

jesmo5/Shutterstock.com

n CH3O2CC6H4CO2CH3 + n HOCH2CH2OH uv O(O2CC6H4CO2CH2CH2)nO + 2n CH3OH

Figure 23.17  Mylar balloons. 

+ n H2O

O n

polylactic acid (PLA), a polyester

The interest in polylactic acid arises because it is considered a more environmentally conscious choice for several reasons. First, the monomer used to make this polymer is obtained by biological fermentation of plant materials. As a result, formation of this polymer is carbon-neutral. All of the carbon in this polymer came from CO2 in the atmosphere, and degradation at some future time will return the same quantity of CO2 to the environment. In addition, this polymer, which is currently being used in packaging material, is biodegradable, which has the potential to alleviate landfill disposal problems.

23.5 Polymers

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In 2021, the world production of polyethylene terephthalate amounted to 80.9 billion kg. Most of this was used to produce fibers and packaging

O O

C

O H3CO

materials; the remainder was used for ­automobile parts, luggage, filters, and much more. Currently, much of the PET used to make beverage containers is recycled. The recycling process uses some familiar chemistry. Scrap PET is disO solved at over 220  °C in dimethyl C O CH2CH2 phthalate (DMT) and then treated with n methanol at 260−300 °C and 340– 650 kPa. In this process, the methanol DMT and CH3OH at reacts with the polymer to break down high temperature the chains, forming more DMT and ethand pressure ylene glycol. DMT and ethylene ­glycol are separated, purified, and used to make more PET. O

C

C

OCH3

+ HOCH2CH2OH

© Steven Hyatt

A Closer Look

Green Chemistry: Recycling PET

Clothing made of recycled polyester.

Polyamides In 1928, the DuPont Company embarked on a basic research program in polymer chemistry headed by Wallace Carothers (1896–1937). Carothers was interested in high molar mass compounds, such as rubbers, proteins, and resins. In 1935, his research yielded nylon-6,6 (Figure 23.18), a polyamide prepared from ­adipoyl chloride, a derivative of adipic acid (a diacid) and hexamethylenediamine (a diamine):

O

O

O

n ClC(CH2)4CCl + n H2N(CH2)6NH2

O

C(CH2)4C

N(CH2)6N H

© Charles D. Winters/Cengage

adipoyl chloride

Figure 23.18 Nylon-6,6. ​

Hexamethylenediamine is dissolved in water (bottom layer), and adipoyl chloride (a derivative of adipic acid) is dissolved in hexane (top layer). The two compounds react at the interface between the layers to form nylon-6,6, which is being wound onto a glass stirring rod.

hexamethylenediamine

+ 2n HCl

H n

amide link in nylon-6,6, a polyamide

Nylon can be extruded easily into fibers that are stronger than natural fibers and chemically more inert, and this fact jolted the U.S. textile industry at a critical time. Natural fibers were not meeting twentieth-century needs. Silk was expensive and not durable, wool was scratchy, linen crushed easily, and cotton did not have a highfashion image. Perhaps the most identifiable use for the new fiber was in nylon stockings. The first public sale of nylon hosiery took place on October 24, 1939, in Wilmington, Delaware (the site of DuPont’s main office). This use of nylon in commercial products ended shortly thereafter, however, with the start of World War II. All nylon was diverted to making parachutes and other military gear. It was not until about 1952 that nylon reappeared in the consumer marketplace. Figure 23.19 illustrates why nylon makes such a good fiber. To have good tensile strength (the ability to resist tearing), the polymer chains should be able to attract one another, albeit not so strongly that the plastic cannot be drawn into fibers. Ordinary covalent bonds between the chains (cross-linking) would be too strong. Instead, cross-linking occurs by the somewhat weaker intermolecular force called hydrogen bonding (Section 11.3) between the hydrogens of NOH groups on one chain and the carbonyl oxygens on another chain. The polarities of the Nδ2OHδ1 group and the Cδ1POδ2 group lead to attractive forces between the polymer chains of the desired magnitude.

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Figure 23.19 Hydrogen bonding between polyamide chains.  ​Carbonyl oxygen atoms with a partial negative charge on one chain interact with an amine hydrogen with a partial positive charge on a neighboring chain. (Hydrogen bonding is described in more detail in Section 11.3.)

Ex am p le 23.8

Condensation Polymers Problem   What is the repeating unit of the condensation polymer obtained by ­combining HO2CCH2CH2CO2H (succinic acid) and H2NCH2CH2NH2 (1,2-ethylenediamine)?

What Do You Know?  Carboxylic acids and amines react to form amides, splitting out water. Here you have a diacid and diamine that will react. The repeating unit will be the shortest sequence that when repeated gives a long polymer chain.

Strategy   Recognize that the polymer will link the two monomer units through the amide linkage. The smallest repeating unit of the chain will contain two parts, one from the diacid and the other from the diamine. Solution   For each amide linkage, an OH from one of the carboxylic acid groups of s­ uccinic acid and a H from one of the amino groups of 1,2-ethylenediamine are removed as water. The repeating unit of this polyamide is amide linkage

O

O

CCH2CH2C

NCH2CH2N H

H n

Think about Your Answer   Alternating fragments of the diacid and diamine appear in the polymer chain. The fragments are linked by amide bonds, making this a polyamide.

Check Your Understanding  Kevlar is a polymer that is used to make sports equipment and bulletproof vests. The basic structure of the polymer is shown below. Is this a condensation polymer or an addition polymer? What chemicals could be used to make this polymer? Write a balanced equation for the formation of Kevlar. amide group



O

O

C

C

N

N

H

H

n

23.5 Polymers

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Applying Chemical Principles 23.1  An Awakening with l-DOPA From about 1917 to 1928, millions of people worldwide were affected by a condition known as encephalitis lethargica, a form of sleeping sickness. Those who suffered from the condition were in a state of semi-consciousness that lasted for decades. In his book Awakenings, Oliver Sacks wrote about treating a patient with the compound l-DOPA, which “was started in early March 1969 and raised by degrees to 5.0 g a day. Little effect was seen for two weeks, and then a sudden ‘conversion’ took place. . . . Mr. L enjoyed a mobility, a health, and a happiness which he had not known in thirty years. Everything about him filled with delight: he was like a man who had awoken from a nightmare or a serious illness. . . .”

When l-DOPA is ingested, it is metabolized to dopamine in a process that removes the carboxylic acid group, OCO2H, and it is dopamine that is physiologically active. Dopamine is a neurotransmitter that occurs in a wide variety of animals.

AF archive/Alamy Stock Photo

dopamine, C8H11NO2, a neurotransmitter

Interestingly, both l-DOPA and dopamine are closely related to another amine, epinephrine. This is sometimes referred to as adrenaline, the hormone that is released from the adrenal glands when there is an emergency or danger threatens.

Robert DeNiro as Leonard Lowe and Robin Williams as Malcolm Sayer, a fictionalized portrayal of Oliver Sacks, in the movie version of Awakenings.

If you have read the book or have seen the movie of the same name, you know that Mr. L eventually could not t­olerate the treatment, but that Sacks treated many others who ­benefitted from it. l-DOPA is still widely used in the treatment of another condition, Parkinson’s disease, a degenerative disorder of the central nervous system. l -DOPA or levodopa ( l -3,4-dihydroxy–phenylalanine) is ­chiral. The symbol l stands for “levo,” which means a solution of the compound rotates polarized light to the left. The ­compound is also a derivative of phenylalanine, one of many naturally occurring alpha-amino acids that play such an important role in protein formation and other natural processes. l-DOPA also illustrates why chiral molecules are so interesting to chemists: Only the levo enantiomer is physiologically active. The enantiomer that rotates polarized light in the opposite direction has no biological function.

epinephrine or adrenaline, C9H13NO3

Questions

1. l -DOPA is chiral. What is the center of chirality in the molecule? 2. Is either dopamine or epinephrine chiral? If so, what is the center of chirality? 3. If you are treated with 5.0  g of l -DOPA, what amount (in moles) is this? References 1.   Oliver Sacks, Awakenings, New York: Vintage Books, 1999. 2.  N. Angier, “A Molecule of Motivation, Dopamine Excels at Its Task,” New York Times, October 27, 2009.

L-DOPA, C9H11NO4, a treatment for Parkinson’s disease

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Chemist Kaichang Li does research in wood chemistry at ­Oregon State University. Oregon has a beautiful and rugged coast, and Li went there in search of mussels to make a special dish. As the waves pounded onshore, he was struck by the fact that the mussels could cling stubbornly to the rocks in spite of the force of the waves and tides. What glue-like substance enabled them to do this? Back in his lab, Li found that the strands of glue were largely protein-based. Proteins are simply polymers of amino acids with an amide link between units (Chapter 24). Li realized that such polymers could have enormous application in the wood industry. Adhesives, or glues in common terminology, have been known and used for thousands of years. Early glues were based on animal or plant products. Now, however, adhesives are largely synthetic, among them are condensation polymers based on the combination of phenol or urea with formaldehyde. These have been used for well over a half-century in the manufacture of plywood and particle board, and your home or dormitory may contain a significant amount of these building materials. Unfortunately, they have a disadvantage. In their manufacture and use, formaldehyde, a suspected carcinogen, can be released into the air.

Courtesy of Oregon State University

23.2  Green Adhesives

Professor K. Li, a discoverer of “green” adhesives.

Li’s work with mussels eventually led to a new, safer adhesive that could be used in these same wood products. His first problem was how to make a protein-based adhesive in the laboratory. The idea came to him one day at lunch when he was eating tofu, a soy-based food very high in protein. Why not modify soy protein to make a new adhesive? Using mussels as his model, Li did exactly that, and, as he said, “We turned soy proteins into mussel adhesive proteins.” Scientists at Hercules Chemical Company provided expertise to cure (or harden) the new “green” adhesive, and the Columbia Forest Products Company adopted the environmentally friendly adhesive for use in plywood and particle board in 2006. Since then, around 60% of the plywood and veneer industry has also adopted these safer adhesives. In 2007, Li and his coworkers, as well as Columbia Forest Products and Hercules, shared a Presidential Green Chemistry Challenge Award. In 2017, Li received a Golden Goose Award from the American Association for the Advancement of Science, an award for advancements with tremendous human and economic benefits that result from federally funded research of seemingly obscure phenomena.

Questions

1. Draw structures of phenol, urea, and formaldehyde. 2. Describe the bonding in formaldehyde. 3. It has been said that nylon is similar to a protein. Compare and contrast the structures of nylon-6,6 and a protein. (For more information on the structure of proteins, see Chapter 24.) A portion of a protein chain made of repeating glycine molecules (H2NCH2CO2H).

23.3  Bisphenol A (BPA) Bisphenol A (BPA) is an organic compound consisting of two phenol groups linked through a carbon atom. It was first synthesized around 1900, but only recently has it become commercially important. And even more recently it has been very much in the news because of its much-debated possible toxicity.



Bisphenol A is a white solid, slightly water soluble (300 mg/L), and a weak acid (pKa of 9.9). In 2015, the total world demand for BPA was about 7.7 × 109 kg.

Applying Chemical Principles

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Because of this debate the use of BPA polymers has been virtually eliminated in making bottles (especially baby bottles) and for coating food cans. Substitutes are also being found for BPA in cash register receipts. The U.S. Food and Drug Administration (FDA) currently states that it is safe to ingest up to 50 μg per kilogram of body weight per day. However, many scientists believe the safe limit should be much lower, or that BPA should be ­eliminated entirely. To put this in context, some studies have indicated that infants fed from polycarbonate bottles may ingest up to 13 μg/kg/day. A normal adult may take in about 1.5 μg/kg/day. In 2015, the European Food Safety Authority reduced its limit from 50 μg/kg/day to 4 μg/kg/day and in 2021 proposed a further reduction to 0.04 μg/kg/day. In 2022, the U.S. FDA agreed to reassess its limit.

bisphenol A

If BPA is treated with phosgene (Cl2CO), the product is the polymer polycarbonate, and approximately 74% of the BPA is used to make this polymer. Because polycarbonate is lightweight, durable, clear, and nearly shatterproof, it is used to make, among other things, bottles, eyeglass lenses, and football helmets. Another 20% of BPA is converted into epoxy resins by reacting BPA with epichlorohydrin. Until recently epoxy resins were used to coat the inside of some food cans, but they still find use in paints and coatings, adhesives, and composite materials. Some of the remaining 6% of BPA is used in thermal paper. When you get a cash register receipt in a restaurant or store, the coating on the paper may contain BPA. BPA is such a useful material. What could go wrong? The problem is that BPA has estrogenic activity, the ability to simulate the activity of primary female sex hormones. But this is also true for some compounds found naturally in beans, olive oil, and some fruits and vegetables. Nonetheless, there has been a vigorous debate on the biological effects of BPA and its polymers.

C

1. What is the atom economy for the reaction of acetone with phenol to make BPA? (“Applying Chemical Principles 4.1: Atom Economy,” page 232.) 2. Are polycarbonate and epoxy resins addition polymers or condensation polymers? 3. If drinking from a polycarbonate bottle, does a 15 lb infant ingest over 50 μg of BPA per day? (1.00 lb = 454 g) 4. Assume you weigh 156 lb. How much BPA do you likely ingest per day? 5. What quantity of 0.050 M NaOH would be required to react with 300 mg of BPA in 1.00 L of water? (Assume both OOH groups react with NaOH.)

OH

O H3C

Questions

H3C

CH3 C

CH3

+2

+ H 2O HO

Acetone

OH

Phenol

Bisphenol A

O

O

H2C

C

Cl

Cl

Epichlorohydrin

H3C

CH3 C

O

C

CH3 C

O O

Cl

C H2

Phosgene

H3C

CH

O

O

H2 H2 C H C C OH

Polycarbonate

Epoxy resin

Synthesis of BPA and of polycarbonate and epoxy resin from BPA

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Think–Pair–Share 1. Draw the structural formula and give the name of a hydrocarbon containing three carbon atoms in which (a) each carbon is bonded to four other atoms. (b) two carbons are joined by a carbon–carbon double bond. (c) two carbons are joined by a carbon–carbon triple bond. (d) there are two carbon–carbon double bonds. 2. Consider the following two alkanes: CH3 H3C

CH

CH3 CH2

CH H2C

CH2

CH2

CH3

CH3

H 3C

C CH3

CH3 CH2

C

CH2

CH3

4. Draw the structural formulas for the following compounds: 1-butanol, 2-butanol, butanal, and butanone. Predict the structural formula and name of the product, if any, formed when an excess of an oxidizing agent is added to each. 5. Propose a way to prepare ethyl acetate starting with ethene as the only source of carbon atoms. 6. Polymers (a) What is the structural formula of the monomer from which the addition polymer called TeflonTM is formed? What is its name? The formula for TeflonTM can be represented as F

F

C

C

F

F n

CH3

(a) Name each compound. (b) Are these compounds structural isomers? Explain. (c) Is either of these compounds chiral? Explain. 3. You have three test tubes (Tubes 1, 2, and 3), each containing a different colorless liquid, either hexane, 1-hexene, or benzene. (a) Write the molecular and structural formulas for each compound. (b) You add a small amount of bromine to each tube and observe the following: Tube 1: The bromine color fades and eventually disappears, even when the tube is left in the dark. Tube 2: The bromine color does not fade when the tube is in the dark, but does so once you shine a light on it. Tube 3: The bromine color does not fade when the tube is left in the dark or is in the light. Which tube contains which liquid? Explain how you reached your conclusions.

(b) Predict the structure of the condensation polymer that results from the reaction of terephthalic acid and hexamethylenediamine. Which functional group is present in this polymer?

HO

O

O

C

C

OH +

terephthalic acid H2N

CH2

CH2

CH2

CH2

CH2

CH2

NH2

hexamethylenediamine

Chapter Goals Revisited The goals for this chapter are keyed to specific Study Questions to help you organize your review.

23.1  Why Carbon? • Understand the factors that contribute to the large numbers of organic compounds and the wide array of structures. 1–4.

• Recognize the types of isomerism that are seen in organic chemistry. 5–8.

23.2 Hydrocarbons • Draw structural formulas and name hydrocarbons. 11, 13, 15, 17, 31, 32, 37, 38, 43, 46.

• Recognize and draw structures of geometric isomers and optical isomers. 19, 20, 23, 29.

• Describe chemical and physical properties of hydrocarbons. 25, 26, 33–36, 47–50.



Chapter Goals Revisited

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23.3  Alcohols, Ethers, and Amines • Identify functional groups from organic formulas. 77, 102. • Name and draw structures of alcohols, ethers, and amines. 51–58, 108. • Describe chemical properties of alcohols and amines. 59–62, 117.

23.4  Compounds with a Carbonyl Group • Name and draw structures of carbonyl compounds—aldehydes, ketones, acids, esters, and amides. 63–66.

• Describe reactions of carbonyl compounds. 67–72.

23.5 Polymers • Write equations for the formation of addition polymers and condensation polymers and describe their structures. 79–82.

• Relate properties of polymers to their structure. 104, 121, 125.

Study Questions denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully worked ­solutions in the Student Solutions Manual.

▲

Practicing Skills Why Carbon? (See Section 23.1.) 1. In carbon compounds, the carbon atom can achieve an octet configuration in several ways. Draw the Lewis structure of a hydrocarbon compound in which: (a) Carbon is bonded via four single bonds to adjacent atoms; (b) Carbon is bonded via two single bonds and one pi bond; (c) Carbon is bonded via one single bond and one triple bond; (d) Carbon is bonded via two double bonds. 2. Identify the hybridization and the bond angles for the following: (a) Carbon bonded via four single bonds to adjacent atoms; (b) Carbon bonded to adjacent atoms via two single bonds and one pi bond; (c) Carbon bonded to adjacent atoms via one single bond and one triple bond; (d) Carbon bonded to adjacent atoms via two double bonds. 3. Is violet light (with a wavelength of 400 nm) sufficiently energetic to break a carbon–carbon single bond (the average COC bond energy is 346 kJ/mol)? 4. Ethene, C2H4, is unstable with respect to decom­ position to the elements (ΔfH° = 52.47 kJ/mol). It is also unstable with respect to polymerization to polyethylene. Nevertheless, samples of ethene can be stored without decomposition essentially forever. Why is this so?

5. Compounds with the same elemental composition but different atom-to-atom connections are called: (a) stereoisomers, (b) structural isomers, (c) geometric isomers, (d) optical isomers. 6. Compounds with the same attachment of atoms but different orientation in space are called: ­(a) ­stereoisomers, (b) structural isomers, (c) ­cis–trans isomers, (d) optical isomers. 7. Determine how many structural isomers and ­geometric isomers are possible for compounds having the formula C4H10. Are any of the possible isomers chiral? 8. Determine how many noncyclic structural ­isomers and geometric isomers are possible for ­compounds having the formula C4H8. Are any of the possible isomers chiral?

Alkanes and Cycloalkanes (See Section 23.2 and Examples 23.1 and 23.2.) 9. What is the name of the straight (unbranched) chain alkane with the formula C7H16? 10. What is the molecular formula for an alkane with 12 carbon atoms? 11. Which of the following compounds can be an alkane? (a) C2H4 (c) C14H30 (b) C5H12 (d) C7H8

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12. Which of the following compounds can be a cycloalkane? (a) C3H5 (c) C14H30 (b) C5H10 (d) C8H8 13. One of the structural isomers with the formula C9H20 has the name 3-ethyl-2-methylhexane. Draw its structure. Draw and name another s­ tructural isomer of C9H20 in which there is a five-carbon chain. 14. Isooctane, 2,2,4-trimethylpentane, is one of the possible structural isomers with the formula C8H18. Draw the structure of this isomer, and draw and name structures of two other isomers of C8H18 in which the longest carbon chain is five atoms. 15. Give the systematic name for the following alkane: CH3 CH3CHCHCH3 CH3

16. Give the systematic name for the following alkane. Draw a structural isomer of the compound, and give its name. CH3 CH3CHCH2CH2CHCH3 CH2CH3

17. Draw the structure of each of the following compounds: (a) 2,3-dimethylhexane (b) 2,3-dimethyloctane (c) 3-ethylheptane (d) 3-ethyl-2-methylhexane 18. Draw structures for the following compounds. (a) 3-ethylpentane (b) 2,2-dimethylpentane (c) 2,3-dimethylpentane (d) 2,4-dimethylpentane 19. Draw Lewis structures and name all possible alkanes that have a seven-carbon chain with one methyl substituent group. Which of these isomers, if any, has a chiral carbon center? 20. Four (of six possible) dimethylhexanes are named below. Draw the structures of each, and determine which of these isomers, if any, has a chiral carbon center. (a) 2,2-dimethylhexane (b) 2,3-dimethylhexane (c) 2,4-dimethylhexane (d) 2,5-dimethylhexane

21. Draw the structure of the chair form of cyclohexane. Identify the axial and equatorial hydrogen atoms in this drawing. 22. Draw a structure for cycloheptane. Is the sevenmember ring planar? Explain your answer. 23. There are two ethylheptanes (compounds with a seven-carbon chain and one ethyl substituent). Draw the structures, and name these compounds. Is either isomer chiral? 24. Among the 18 structural isomers with the formula C8H18 are two with a five-carbon chain having one ethyl and one methyl substituent group. Draw their structures, and name these two isomers. 25. List several typical physical properties of C4H10. Predict the following physical properties of dodecane, C12H26: color, state (s, ℓ, g), solubility in water, solubility in a nonpolar solvent. 26. Write balanced equations for the following reactions of alkanes. (a) The reaction of methane with excess chlorine. (b) Complete combustion of cyclohexane, C6H12, with excess oxygen.

Alkenes and Alkynes (See Section 23.2 and Examples 23.3 and 23.4.) 27. What is the molecular formula for an alkane that has five carbon atoms? An alkene with five carbon atoms? An alkyne with five carbon atoms? 28. What is the molecular formula for a cycloalkane with six carbon atoms? An alkene with six carbon atoms? 29. Draw structures for the cis and trans isomers of 4-methyl-2-hexene. 30. What structural requirement is necessary for an alkene to have cis and trans isomers? Can cis and trans isomers exist for an alkane? For an alkyne? 31. A hydrocarbon with the formula C5H10 can be either an alkene or a cycloalkane. (a) Draw a structure for each of the six isomers possible for C5H10, assuming it is an alkene. Give the systematic name of each isomer. (b) Draw a structure for a cycloalkane having the formula C5H10. 32. Five alkenes have the formula C7H14 and a sevencarbon chain. Draw their structures and name them. 33. Draw the structure and give the systematic name for the products of the following reactions: (a) CH3CHPCH2 + Br2 n (b) CH3CH2CHPCHCH3 + H2 n Study Questions

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34. Draw the structure and give the systematic name for the products of the following reactions: (a) H3C CH2CH3 C H3C

(b) CH3C

+ H2

C H

CCH2CH3 + 2 Br2

35. The compound 2-bromobutane is a product of the addition of HBr to three different alkenes. Identify the alkenes and write an equation for the reaction of HBr with one of the alkenes. 36. The compound 2,3-dibromo-2-methylhexane is formed by addition of Br2 to an alkene. Identify the alkene, and write an equation for this reaction. 37. Draw structures for the four alkenes that have the formula C3H5Cl, and name each compound. (These are derivatives of propene in which a ­chlorine atom replaces one hydrogen atom.) 38. There are seven possible dichloropropene isomers (molecular formula C3H4Cl2). Draw their structures and name each isomer. (Hint: Don’t overlook cis–trans isomers.) 39. Hydrogenation is an important chemical reaction of compounds that contain double bonds. Write a chemical equation for the hydrogenation of 1-hexene. Hydrogenation reactions are used extensively in the food industry. Describe this reaction and explain its use and importance. 40. Elemental analysis of a colorless liquid has given its formula as C5H10. You recognize that this could be either a cycloalkane or an alkene. A chemical test to determine the class to which this compound belongs involves adding bromine. Explain how this would allow you to distinguish between the two classes. 41. What are the structures and names of the two ­possible products from the addition of HBr to 1-pentene? Which compound is likely the major product? 42. What are the structures and names of the two possible products from the addition of HCl to 2-methyl-2-hexene? Which compound is likely the major product?

Aromatic Compounds (See Section 23.2 and Example 23.5.) 43. Draw structural formulas for the following compounds: (a) 1,3-dichlorobenzene (alternatively called m-dichlorobenzene) (b) 1-bromo-4-methylbenzene (alternatively called p-bromotoluene)

44. Draw structural formulas for the following compounds: (a) 1,2-dinitrobenzene (alternatively called o-dinitrobenzene) (b) 4-chlorophenol (alternatively called p-chlorophenol) 45. Give the systematic name for each of the ­following compounds: (a) Cl Cl

NO2

(b)

O2N (c)

NO2

Br Br

Br 46. Give the systematic name for each of the following compounds: (a) Cl   (b)  NO2  (c)  Cl NO2 C2H5 NO2

47. Write the equation for the reaction of 1,4-dimethyl­benzene with CH3Cl and AlCl3. What is the structure and name of the single organic compound produced? 48. Write an equation for the preparation of hexylbenzene from benzene and other appropriate reagents. 49. Aromatic compounds react with a mixture of nitric and sulfuric acid to form aromatic compounds containing a nitro (−NO2) group. Two isomeric compounds are formed by nitration of 1,2-dimethylbenzene. Draw the structures, and name these compounds. 50. Nitration of toluene gives a mixture of two products, one with the nitro group (−NO2) in the ortho position and one with the nitro group in the para position. Draw structures of the two products.

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Alcohols, Ethers, and Amines (See Section 23.3 and Example 23.6.) 51. Give the systematic name for each of the following alcohols, and tell if each is a primary, secondary, or tertiary alcohol: (a) CH3CH2CH2OH (b) CH3CH2CH2CH2OH (c) CH3 H3C

C

OH

CH3

(d)

CH3 H3C

C

CH2CH3

OH

52. Draw structural formulas for the following ­alcohols, and tell if each is primary, secondary, or tertiary: (a) 1-pentanol (b) 2-pentanol (c) 3,3-dimethyl-2-pentanol (d) 3,3-dimethyl-1-pentanol 53. Write the formula, and draw the structure, for each of the following amines: (a) ethylamine (b) dipropylamine (c) butyldimethylamine (d) triethylamine 54. Write the formula, and draw the structure for each of the following amines: (a) methylamine (b) ethylpropylamine (c) aniline (d) ethylenediamine 55. Name the following amines: (a) (C2H5)2NH (b) (CH3)(CH3CH2CH2)2N (c) CH3CH2CH2CH2NH2 (d) (CH3CH2CH2)3N 56. Name the following amines: (a) CH3CH2CH2NH2 (b) (CH3)3N (c) (CH3)(C2H5)NH (d) C6H13NH2

58. Draw structural formulas for all primary amines with the formula C4H9NH2. 59. Complete and balance the following equations: (a) C6H5NH2(ℓ) + HCl(aq) n (b) (CH3)3N(aq) + H2SO4(aq) n 60. The structure of dopamine, a neurotransmitter, is given on page 1192. Predict its reaction with ­aqueous hydrochloric acid. 61. Draw structures of the product formed by oxidation of the following alcohols. Assume an excess of oxidizing agent is used in each case. (a) 2-methyl-1-pentanol (b) 3-methyl-2-pentanol (c) HOCH2CH2CH2CH2OH (d) H2NCH2CH2CH2OH 62. Aldehydes and carboxylic acids are formed by oxidation of primary alcohols, and ketones are formed when secondary alcohols are oxidized. Give the name and formula for the alcohol that, when oxidized, gives the following products: (a) CH3CH2CH2CHO (b) 2-hexanone

Compounds with a Carbonyl Group (See Section 23.4 and Example 23.7.) 63. Draw structural formulas for (a) 2-pentanone, (b) hexanal, and (c) pentanoic acid. 64. Draw structural formulas for the following acids and esters: (a) 2-methylhexanoic acid (b) pentyl butanoate (which has the odor of apricots) (c) octyl acetate (which has the odor of oranges) 65. Identify the class of each of the following compounds, and give the systematic name for each: (a) CH3 CH3CH2CHCH2CO2H

(b)

O CH3CH2COCH3

(c)

O CH3COCH2CH2CH2CH3

(d)

O Br

COH

57. Draw structural formulas for the four possible alcohols with the formula C4H10O. Give the ­systematic name of each.



Study Questions

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1199

66. Identify the class of each of the following compounds, and give the systematic name for each: (a) O CH3CCH3

(b)

H

O H

CH3CH2CH2CH

(c)

(c) Is this molecule chiral? If so, which carbon atom is chiral? (d) Which hydrogen atom in this compound is acidic?

C

O

1

H

CH3CCH2CH2CH3

67. Give the structural formula and systematic name for the organic product, if any, from each of the following reactions: (a) pentanal and KMnO4 (b) pentanal and LiAlH4 (c) 2-octanone and LiAlH4 (d) 2-octanone and KMnO4 68. Give the structural formula and name for the organic product from the following reactions. (a) CH3CH2CH2CHO + LiAlH4 (b) CH3CH2CH2CH2OH + excess KMnO4 (c) CH3CH2CH2CHO + KMnO4

N C 3C

H 2

H O H   

O

74. The structure of vitamin C, whose chemical name is ascorbic acid, is shown in the figure (without lone pairs of electrons).

HO

H

OH

C

C

H

O

O

C

H H

HO

C O C

C OH   

69. Describe how to prepare propyl propanoate ­beginning with 1-propanol as the only carboncontaining reagent. 70. Give the name and structure of the product of the reaction of benzoic acid and 2-propanol. 71. Draw structural formulas and give the names for the products of the following reaction: O CH3COCH2CH2CH2CH3 + NaOH

72. Draw structural formulas, and give the names for the products of the following reaction: O C

CH3 O

CH + NaOH

?

CH3

73. The structure of phenylalanine, one of the 20 amino acids that make up proteins, is shown in the figure (without lone pairs of electrons). The carbon atoms are numbered for the purpose of this question. (a) What is the geometry of C3? (b) What is the OOCOO bond angle?

(a) What is the approximate value for the OOCOO bond angle? (b) There are four OH groups in this structure. Estimate the COOOH bond angles for these groups. Will they be the same value (more or less), or should there be significant differences in these bond angles? (c) Is the molecule chiral? How many chiral carbon atoms can be identified in this structure? (d) Identify the shortest bond in this molecule. (e) What are the functional groups of the molecule? 75. What is the structure of the product from the reaction of butanoic acid and methylamine? To what class of compounds does this belong? Write a balanced chemical equation for the reaction. 76. The structure of acetaminophen is shown on page 1183. Using structural formulas, write an equation for the reaction of an acid and an amine to form this compound.

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Functional Groups (See Section 23.4 and Example 23.7.) 77. Identify the functional groups in the following molecules. (a) CH3CH2CH2OH (b) O H3CCNHCH3

(c)

O CH3CH2COH

(d)

O CH3CH2COCH3

78. Consider the following molecules: (1) O CH3CH2CCH3

(2)

O CH3CH2COH

(3) H2C (4)

CHCH2OH OH

CH3CH2CHCH3

(a) What is the result of treating compound 1 with NaBH4? What is the functional group in the product? Name the product. (b) Draw the structure of the reaction product from compounds 2 and 4. What is the functional group in the product? (c) What compound results from adding H2 to compound 3? Name the reaction product. (d) What compound results from adding NaOH to compound 2?

Polymers (See Section 23.5 and Example 23.8.) 79. Polyvinyl acetate is the binder in water-based paints. (a) Write an equation for its formation from vinyl acetate. (b) Show a portion of this polymer with three monomer units. (c) Describe how to make polyvinyl alcohol from polyvinyl acetate. 80. Neoprene (polychloroprene, a kind of rubber) is a polymer formed from the chlorinated butadiene H2CPCHCClPCH2. (a) Write an equation showing the formation of polychloroprene from the monomer. (b) Show a portion of this polymer with three monomer units.

81. Polyacrylonitrile is a polymer of acrylonitrile, CH2PCHCN. Write an equation for the synthesis of this polymer. 82. The structure of methyl methacrylate is given in Table 23.12. Draw the structure of a polymethyl methacrylate (PMMA) polymer that has four monomer units. (PMMA has excellent ­optical properties and is used to make hard contact lenses.)

General Questions These questions are not designated as to type or location in the chapter. They may combine several concepts. 83. Three different compounds with the formula C2H2Cl2 are known. (a) Two of these compounds are geometric isomers. Draw their structures. (b) The third compound is a structural isomer of the other two. Draw its structure. 84. Draw the structure of 2-butanol. Identify the chiral carbon atom in this compound. Draw the mirror image of the structure you first drew. Are the two molecules superimposable? 85. An alkene having the formula C6H12 is known to have a six-carbon chain. How many isomeric species can be drawn based on this description? List the names of these species. 86. Draw structures and name the four alkenes that have the formula C4H8. There are also two compounds containing rings of carbon atoms that have the formula C4H8. What are their structures and names? 87. Write equations for the reactions of cis-2-butene with the following reagents, representing the reactants and products using structural formulas. (a) H2O (b) HBr (c) Cl2 88. Draw the structure and name the product formed if the following alcohols are oxidized. Assume an excess of the oxidizing agent is used. If the alcohol is not expected to react with a chemical oxidizing agent, write NR (no reaction). (a) 1-butanol (b) 2-butanol (c) 2-methyl-2-propanol (d) 2-methyl-1-propanol 89. Write equations for the following reactions, representing the reactants and products using structural formulas. (a) The reaction of acetic acid and sodium hydroxide (b) The reaction of methylamine with HCl Study Questions

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1201

90. Write equations for the following reactions, representing the reactants and products using structural formulas. (a) The formation of ethyl acetate from acetic acid and ethanol (b) The hydrolysis of glyceryl tristearate (the triester of glycerol with stearic acid, a fatty acid whose formula is shown in Table 23.9)

99. Voodoo lilies depend on carrion beetles for pollination. Carrion beetles are attracted to dead animals, and because dead and putrefying animals give off the horrible-smelling amine cadaverine, the lily likewise releases cadaverine (and the closely related compound putrescine, page 1175). A biological catalyst, an enzyme, converts the naturally occurring amino acid lysine to cadaverine. H

91. Write an equation for the formation of the following polymers. (a) Polystyrene, from styrene (C6H5CHPCH2) (b) PET (polyethylene terephthalate), from ethylene glycol and terephthalic acid 92. Write equations for the following reactions, representing the reactants and products using structural formulas. (a) The hydrolysis of the amide C6H5CONHCH3 to form benzoic acid and methylamine. (b) The hydrolysis of (OCO(CH2)4CONH(CH2)6NHO)n, (nylon-6,6, a polyamide) to give a di­carboxylic acid and a diamine. 93. Draw the structure of each of the following compounds: (a) 2,2-dimethylpentane (b) 3,3-diethylpentane (c) 3-ethyl-2-methylpentane (d) 3-ethylhexane 94.

95.

▲

Structural isomers. (a) Draw all of the isomers possible for C3H8O. Give the systematic name of each, and tell into which class of compound it fits. (b) Draw the structural formulas for an aldehyde and a ketone with the molecular formula C4H8O. Give the systematic name of each. ▲ Draw structural formulas for possible isomers of the dichlorinated propane, C3H6Cl2. Name each compound.

96. Draw structural formulas for possible isomers with the formula C3H6ClBr, and name each isomer. 97. Give structural formulas and systematic names for the three structural isomers of trimethylbenzene, C6H3(CH3)3. 98. Give structural formulas and systematic names for possible isomers of dichlorobenzene, C6H4Cl2.

H2NCH2CH2CH2CH2

C

NH2

C

OH

O Lysine

What group of atoms must be replaced in lysine to make cadaverine? (Lysine is essential to human nutrition but is not synthesized in the human body.) 100. Benzoic acid occurs in many berries. When humans eat berries, benzoic acid is converted to hippuric acid in the body by reaction with the amino acid glycine H2NCH2CO2H. Draw the structure of hippuric acid, knowing it is an amide formed by reaction of the carboxylic acid group of benzoic acid and the amino group of glycine. Why is hippuric acid referred to as an acid? 101. Consider the reaction of cis-2-butene with H2 (in the presence of a catalyst). (a) Draw the structure, and give the name of the reaction product. Is this reaction product chiral? (b) Draw an isomer of the reaction product. 102. Give the name of each compound below, and name the functional group involved. OH (a) H3C

C

CH2CH2CH3

H O (b) H3C

CCH2CH2CH3 H O

(c) H3C

C

C

H

CH3 O (d) H3CCH2CH2

C

OH

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103. Draw the structure of glyceryl trilaurate, a fat. Lauric acid has the formula CH3(CH2)10CO2H. (a) Write an equation for the base hydrolysis of glyceryl trilaurate. (b) Biodiesel (see Section 25.6), an alternative fuel to diesel fuel from petroleum, often consists of methyl esters of fatty acids obtained from plant and animal fats and oils. The reaction used involves transesterification reactions in which an ester reacts with an alcohol to form a new ester and a new alcohol. Write an equation for the reaction that could be used to prepare biodiesel fuel from glyceryl trilaurate. 104. A well-known company sells outdoor clothing made of recycled polyethylene terephthalate (PET), the principal material in many soft drink bottles. Another company makes PET fibers by treating recycled bottles with methanol to give the diester dimethyl terephthalate and ethylene glycol and then repolymerizes these compounds to give new PET. Write a chemical equation to show how the reaction of PET with methanol can give dimethyl tereph­thalate and ethylene glycol. 105. Identify the reaction products, and write an ­equation for the following reactions of CH2PCHCH2OH. (a) H2 (hydrogenation, in the presence of a catalyst) (b) Oxidation (excess oxidizing agent) (c) Addition polymerization (d) Ester formation, using acetic acid 106. Write a chemical equation describing the reaction between glycerol and stearic acid (Table 23.9) to give glyceryl tristearate. 107. The product of an addition reaction of an alkene is often predicted by Markovnikov’s rule. (a) Draw the structure of the product of adding HBr to propene, and give the name of the product. (b) Draw the structure and give the name of the compound that results from adding H2O to 2-methyl-1-butene. (c) If you add H2O to 2-methyl-2-butene, is the product the same or different than the product from the reaction in part (b)? 108. There are three ethers with the formula C4H10O. Draw their structures. 109. Review the opening photograph about chocolate (which shows the structure of an active ingredient, theobromine) and then answer the following questions.



(a) How do theobromine and caffeine differ structurally?

Caffeine, C8H10N4O2

(b) A 5.00-g sample of Hershey’s cocoa contains 2.16% theobromine. What is the mass of the compound in the sample? 110. Nylon-6 is a polyamide formed by polymerizing H2NCH2CH2CH2CH2CH2CO2H. Write an equation for this reaction.

In the Laboratory 111. Which of the following compounds produces acetic acid when treated with an oxidizing agent such as KMnO4? OH (a) H3C

(c) H3C

CH3

C

H

H O

O (b) H3C

(d) H3C

H

C

C

CH3

112. Consider the reactions of C3H7OH. H H3CCH2

C

O

H

Rxn A H2SO4

H

H

C

C + H2O H

+ CH3CO2H

Rxn B

H H3CCH2

H3C

H

C

O O

CCH3

H

(a) Name the reactant C3H7OH. (b) Draw a structural isomer of the reactant, and give its name. (c) Name the product of reaction A. (d) Name the product of reaction B.

Study Questions

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1203

113. You have a liquid that is either cyclohexene or benzene. When the liquid is exposed to dark-red bromine vapor, the vapor is immediately decolorized. What is the identity of the liquid? Write an equation for the chemical reaction that has occurred. 114. ▲ Hydrolysis of an unknown ester of butanoic acid, CH3CH2CH2CO2R, produces an alcohol A and butanoic acid. Oxidation of alcohol A forms an acid B that is a structural isomer of ­butanoic acid. Give the names and structures for alcohol A and acid B. 115. ▲ You are asked to identify an unknown colorless, liquid carbonyl compound. Analysis has determined that the formula for this unknown is C3H6O. Only two compounds match this formula. (a) Draw structures for the two possible compounds. (b) To decide which of the two structures is correct, you react the compound with an oxidizing agent and isolate from that reaction a compound that is found to give an acidic solution in water. Use this result to identify the structure of the unknown. (c) Name the compound formed by oxidation of the unknown. 116. Describe a simple chemical test to tell the ­difference between CH3CH2CH2CHPCH2 and its ­isomer cyclopentane. 117. Describe a simple chemical test to tell the difference between 2-propanol and its isomer methyl ethyl ether. 118. ▲ An unknown ester has the formula C4H8O2. Hydrolysis gives methanol as one product. Identify the ester, and write an equation for the hydrolysis reaction. 119. ▲ Addition of water to alkene X gives an alcohol Y. Oxidation of Y produces 3,3-dimethyl-​ 2-pentanone. Identify X and Y, and write equations for the two reactions. 120. 2-Iodobenzoic acid, a tan, crystalline solid, can be prepared from 2-aminobenzoic acid. Other required reagents are NaNO2 and KI (as well as HCl).

CO2H

CO2H NH2

2-aminobenzoic acid

NaNO2 HCl, KI

I

(a) If you use 4.0 g of 2-aminobenzoic acid, 2.2 g of NaNO2, and 5.3 g of KI, what is the theoretical yield of 2-iodobenzoic acid? (b) Are other isomers of 2-iodobenzoic acid possible? (c) You titrate the product in a mixture of water and ethanol. If you use 15.62 mL of 0.101 M NaOH to titrate 0.399 g of the product, what is its molar mass? Is it in reasonable agreement with the theoretical molar mass? 121. The transesterification reaction (see Study ­Question 103) between PET and CH3OH forms dimethyl terephthalate and ethylene glycol. If the methanol used in this reaction is labeled with oxygen-18 (18O), in which of the products will the label be found? 122. Vitamin B-5, pantothenic acid, has the structure shown below. Base hydrolysis of this compound followed by acidification gives two compounds. Draw their structures.

H

O

H

CH3 H

O

C

C

C

H

CH3 OH

C

H

H

O

N

C

C

C

H

H

H

O

H

Vitamin B-5, pantothenic acid

Summary and Conceptual Questions The following questions may use concepts from this and ­previous chapters. 123. Carbon atoms appear in organic compounds in several different ways with single, double, and triple bonds combining to give an octet configuration. Describe the various ways that carbon can bond to reach an octet, and give the name and draw the structure of a compound that illustrates that mode of bonding. 124. There is a high barrier to rotation around a ­carbon–carbon double bond, whereas the barrier to rotation around a carbon–carbon single bond is considerably smaller. Use the orbital overlap model of bonding (Chapter 9) to explain why there is restricted rotation around a double bond. 125. What important properties do the following ­characteristics impart to a polymer? (a) Cross-linking in polyethylene (b) The OH groups in polyvinyl alcohol (c) Hydrogen bonding in a polyamide like nylon

2-iodobenzoic acid

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126. One of the resonance structures for pyridine (C5H5N) is illustrated here. Draw another resonance structure for the molecule. Comment on the similarity between this compound and benzene. N

Pyridine

127. Write balanced equations for the combustion of ethane gas and liquid ethanol (to give carbon dioxide gas and liquid water). (a) Calculate the enthalpy of combustion of each compound. Which has the more negative enthalpy change for combustion per gram? (b) If ethanol is assumed to be partially oxidized ethane, what effect does this have on the enthalpy of combustion? 128. Plastics make up about 20% of the volume of landfills. There is, therefore, considerable interest in reusing or recycling these materials. To ­identify common plastics, a set of universal symbols is now used, illustrated here. They symbolize low- and high-density polyethylene, polyvinyl ­chloride, polypropylene, polystyrene, polyethylene ­terephthalate, and other plastic material.



1

2

3

PETE

HDPE

V

4

5

LDPE

PP

6

7

PS

OTHER

(a) Tell which symbol belongs to which type of plastic. (b) Find an item in the grocery store, hardware store, or drug store made from each of these plastics. List the type of product used for each plastic. (c) Properties of three plastics are listed in the table. Based on this information, describe how to separate samples of these three plastics from one another. Density (g/cm3)

Melting Point (°C)

Polypropylene

0.92

170

High-density polyethylene

0.97

135

1.34–1.39

245

Plastic

Polyethylene terephthalate

129. ▲ Maleic acid is prepared by the catalytic oxidation of benzene. It is a dicarboxylic acid; that is, it has two carboxylic acid groups. (a) Combustion of 0.125 g of the acid gives 0.190 g of CO2 and 0.0388 g of H2O. Calculate the empirical formula of the acid. (b) A 0.261-g sample of the acid requires 34.60 mL of 0.130 M NaOH for complete titration (so that the H ions from both ­carboxylic acid groups are used). What is the molecular formula of the acid? (c) Draw a Lewis structure for the acid. (d) Describe the hybridization used by the C atoms. (e) What are the bond angles around each C atom?

Study Questions

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1205

Science Photo Library/Alamy Stock Photo

24 Biochemistry

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C hapt e r O ut li n e 24.1 Proteins 24.2 Carbohydrates 24.3 Nucleic Acids 24.4 Lipids and Cell Membranes 24.5 Metabolism

You are a marvelously complex biological organism, and so is every other living thing on Earth. What molecules are present in you, and what are their properties? How is genetic information passed from generation to generation? How does your body carry out the numerous reactions that are needed for life? These questions and many others fall into the realm of biochemistry, a rapidly expanding area of science. As the name implies, biochemistry exists at the interface of two scientific disciplines: biology and chemistry. Biochemists tend to concentrate on the specific molecules involved in biological processes and on how chemical reactions occur in an organism (Figure 24.1). They use the strategies of chemistry to understand processes in living things. ORGANISM :

ORGAN :

BIOLOGY

BIOCHEMISTRY

CELL :

Pancreas

Pancreatic cell

ORGANELLE :

MOLECULE :

TRADITIONAL CHEMISTRY

Figure 24.1  The human body with areas of interest to biologists, biochemists, and chemists.

Human

Nucleus

DNA

ATOMS SUBATOMIC PARTICLES

© Charles D. Winters/Cengage

Scientific Disciplines and Perspectives

◀ CRISPR-Cas9 Complex.  The CRISPR-Cas9 complex (“A Closer Look: Genetic Engineering

with CRISPR-Cas9,” page 1222) consists of a protein (white) and an RNA guide sequence (pink). This complex is able to cut DNA (green) at a particular sequence of nucleotides complementary to a portion of the guide RNA. In the drawing of the protein, the curly ribbons represent helices, and the arrows represent sheets formed by amino acids (Section 24.1).

  1207

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The main goal of this chapter is to consider how chemistry is involved in answering important biological questions. In particular, four major classes of biological compounds will be examined: proteins, carbohydrates, nucleic acids, and lipids.

24.1 Proteins Goals for Section 24.1 • Draw the Lewis structures for the most common naturally occurring amino acids. • Understand that under physiological conditions, amino acids exist as zwitterions in aqueous solution.

• Draw the Lewis structures for short polypeptides. • Identify the categories of protein structure (primary, secondary, tertiary, and quaternary).

Amino Acids Are the Building Blocks of Proteins Your body contains thousands of different proteins, and about 50% of the dry weight of your body consists of proteins. Proteins provide structural support (­muscle, collagen), help organisms move (muscle), store and transport chemicals from one area to another (hemoglobin), regulate when certain chemical reactions will occur (some hormones), and catalyze a host of chemical reactions (enzymes). Proteins are condensation polymers (Section 23.5) formed from amino acids. Amino acids are organic compounds that contain an amino group (ONH2) and a carboxylic acid group (OCO2H). Almost all amino acids that make up proteins are α-amino acids, in which the amino group is at one end of the molecule, the acid group is at the other end, and in between, a single carbon atom (the α-carbon) has attached to it a hydrogen atom and either another hydrogen atom or an organic group, denoted R. Naturally occurring proteins are predominantly built using 20 amino acids, which differ only in the identity of the organic group, R. These organic groups can be nonpolar (groups derived from alkanes or aromatic hydrocarbons) or polar (with alcohol, acidic, basic, or other polar functional groups) (Figure 24.2). Depending on which amino acids are present, a region in a protein may be nonpolar, very polar, or anything in between.

Chemistry in Your Career

Dr. Tao Li

Dr. Tao Li Using his M.S. in microbiology and a Ph.D. in medicinal chemistry, Dr. Tao Li enjoys the experimental aspect of chemistry research, which he sees as an opportunity for creativity and imagination. Dr. Li has been guided in his career by his sense that the mission of science is to serve humanity. “We live in a world with finite resources. The best thing we can pass on to the future generations is a world of expanded potential.” Dr. Li’s work as a chemist with the U.S. Environmental Protection Agency (EPA) focuses on

sustainability, including green chemistry, c­ hemical process assessment, and the beneficial use of renewable wastes from agriculture and ­forestry. He is currently developing a biosensor for field ­s urveys of arsenic pollution in groundwater. Industrial activities such as mining, manufacturing, and fossil fuels pollute groundwater with arsenic, ­causing serious environmental and health problems. A sensor designed to detect arsenic ­contamination in the field will help researchers quantify and ­ameliorate the problem.

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Polar R

H H3N+

Electrically charged R

CH2

O−

H H3N+

CH HO

−O

O

C C

O−

H3N+

H

O−

H3N+

C C CH2

C C

O

CH3 Methionine (Met)

O−

H +

H3N

CH3

CH2

Alanine (Ala)

H3

N+

H H3

N+

O

C C CH2

O−

H3N+

O

C C CH

CH2

O−

H

CH3

CH2

Phenylalanine (Phe)

O−

CH3

H H3

N+

Valine (Val)

OH

Lysine (Lys)

Tyrosine (Tyr)

H H3N+

H3N

O

C C CH2

CH2

O−

CH3

CH3

+

+

NH2

NH2 H H3N+

O

C C CH2

O

CH2 C H2N

O

Glutamine (Gln)

−

H H3N

H

O

C C CH2

O−

NH N Histidine (His)

H2N

C C

H2C

CH2

O O−

CH2

Arginine (Arg)

+

H

Leucine (Leu)

C

Asparagine (Asn)

Tryptophan (Trp)

CH

NH

O

O

C C CH2

CH2

C H2N

O−

H H3N+

CH2

O−

O−

CH2

NH

O

C C

O

C C

NH3+

H

O−

CH2

CH2

+

O

C C

Basic

O

C C

O

H

O−

O−

Cysteine (Cys)

H3N+

O

C C

Glutamic acid (Glu)

SH

H

CH2 S

CH2

O

O−

Figure 24.2  The 20 most common amino acids.  ​The amino acids are shown in the form most common at a pH of about 7, physiological pH. (Histidine is shown in the electrically charged column because the unprotonated N in the organic group can easily be protonated.)

O

CH2

−

O

C C CH2

Glycine (Gly)

C H3N

O−

H3

N+

O H

Threonine (Thr)

+

C C

Aspartic acid (Asp)

CH3

H

H

O

C

Serine (Ser)

H3N+

O

C C CH2

OH

H

H

Acidic

O

C C

Nonpolar R

H3N+

C C

H3C CH

O

Proline (Pro)

O−

CH2 CH3 Isoleucine (Ile) 24.1 Proteins

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1209

Amino group

H2N

H

O

C

C

OH

R

Chiral carbon

Carboxylic acid group

All α-amino acids, except glycine, have four different groups attached to the α-carbon. The α-carbon is thus a chiral center (Section 23.1), and two enantiomers exist. Interestingly, all these amino acids occur in nature in a single enantiomeric form—the form with the l-configuration (“Applying Chemical Principles 23.1: An Awakening with l-DOPA,” page 1192). Both the amino group and the carboxylic acid group can exist in two different states: an ionized form (ONH3+ and OCO2−) and an un-ionized form (ONH2 and OCO2H). At low pH, both groups will be in their protonated forms, that is, the amino group will be present as ONH3+ and the carboxylic acid group will be present as OCO2H. As the pH is raised, the carboxylic acid group loses its proton and becomes the ionized form (OCO2−). The resulting species, which is the form prevalent in an aqueous environment at physiological pH (about 7.4), contains both a positive and a negative charge and is called a zwitterion. At higher pH values, the protonated amino group loses its ionizable proton and becomes the un-ionized form (ONH2).

H3N+

H

O

C

C

−H+

OH

H3N+

+H+

R low pH

H

O

C

C

−H+

O−

H2N

+H+

R zwitterion

H

O

C

C

O−

R high pH

The pH behavior of amino acids that have acidic or basic R groups is more complicated. A condensation reaction between two amino acids results in the elimination of water and the formation of an amide linkage (Figure 24.3). The amide linkage in proteins is often referred to as a peptide bond, and the polymer (the p ­ rotein) is called a polypeptide. The amide linkage is planar (Section 23.4), and both the carbon and the nitrogen atoms are pictured as being sp2 hybridized. There is partial double-bond character in the COO and CON bonds, leading to restricted rotation about the carbon–nitrogen bond. As a consequence, each ­peptide bond in a protein is rigid and planar, a feature that plays a role in determining its structure. Proteins consist of one or more polypeptide chains that are often hundreds of amino acids long. Their molar masses are thus often thousands of grams per mole.

H

+

O

H N

H H 3C

C

C

H 2O

− O

H

alanine

H

+

H

H N

HOCH2 C

C O

H serine

H

O

−

removal of a water molecule

amino end

+

O

H N

H H 3C

C

C H

H N H

HOCH2 C

C O

O

−

carboxylate end

peptide bond

Figure 24.3  Formation of a peptide.  ​Two α-amino acids condense to form an amide linkage, often called a peptide bond. Proteins are polypeptides, polymers consisting of many amino acid units linked through peptide bonds.

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Ex am p le 24.1

Drawing Peptide Structures Problem  Peptide and protein structures are drawn starting with the free amino group on the left and ending with the free carboxylic acid group on the right. Draw the Lewis structure for the tripeptide alanine–glycine–serine at physiological pH.

What Do You Know?   You know the sequence of amino acids in the tripeptide and that they are joined by amide linkages. The structures of the amino acids are listed in Figure 24.2. Strategy   Begin with the amino group of alanine on the left. Connect it by an amide linkage to glycine and then connect glycine to serine by another amide linkage. Solution  The Lewis structure for this tripeptide is alanine (Ala)

H Peptide backbone Side chains

H

glycine (Gly)

H

O

N+ C

C

H H

H

C

serine (Ser)

H

O

N

C

C

H

H

H

H

O

N

C

C

H H

C

H

O

H

O

−

Think about Your Answer   A tripeptide contains three amino acids and two peptide bonds. At physiological pH, the dominant form of the tripeptide is the zwitterion.

Check Your Understanding  Draw the Lewis structure for the tripeptide glycine–phenylalanine–valine.

−

Protein Structure and Hemoglobin With this basic understanding of amino acids and peptide bonds, you are ready to consider some larger issues related to protein structure. One of the central tenets of biochemistry is that “structure determines function.” In other words, what a molecule can do is determined by which atoms or groups of atoms are present and how they are arranged in space. To simplify their discussions, biochemists describe proteins as having different structural levels. Each level of structure can be illustrated using hemoglobin. Hemoglobin is the molecule in red blood cells that carries oxygen from the lungs to all of the body’s other cells. It is a large iron-containing protein, made up of more than 10,000 atoms and having a molar mass of 64,500 g/mol. Hemoglobin consists of four polypeptide segments: two identical segments called the α subunits containing 141 amino acids each and two other segments called the β subunits containing 146 amino acids each. The β subunits are identical to each other but different from the α subunits. Each subunit contains an iron(II) ion locked inside an organic ion called a heme unit (Figure 24.4). The oxygen molecules transported by hemoglobin bind to these iron(II) ions. Now focus on the polypeptide part of hemoglobin (Figure 24.5). The first step in describing a structure is to identify how the atoms are covalently linked. This is called the primary structure of a protein, which is simply the sequence of amino acids linked by peptide bonds. For example, a glycine unit can be followed by an alanine, followed by a valine, and so on.

OOC

H3C

CH2

H2C

CH2 CH2 H C C C CH3 C C C C C

N

C C H

N C Fe2+

HC

H2C

COO−

C

N C

N

C

C C

CH3

HC

C H

C

C CH3 CH CH2

heme

(Fe-protoporphyrin IX)

Figure 24.4 Heme. ​The heme unit in hemoglobin (and in myoglobin, a related protein) consists of an iron ion in the center of a porphyrin ring system. (For more information about the iron-containing heme group, see “A Closer Look: Hemoglobin: A Molecule with a Tetradentate Ligand,” page 1118.) 24.1 Proteins

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1211

NH3+ CH2 O C CH2

N

C

C

H O Asparagine (Asn)

H

CH

N

C

H C

The overall three-dimensional shape of a polypeptide chain caused by the folding of various regions

CH2

OH CH3

H

Tertiary structure

Side chains

CH2

NH2

CH2

N

C

Backbone

C

H O Lysine (Lys)

H O Threonine (Thr)

Heme

Primary structure Ala

The sequence of amino acids in a polypeptide chain

Pro

1

2

1

2

Asp Asn

Thr

Lys

Val Ala

Ala

Lys

Trp Lys

Gly

Val

Secondary structure The spatial arrangement of the amino acid sequences into regular patterns such as helices (shown here) and sheets

Quaternary structure The spatial interaction of two or more polypeptide chains in a protein

Figure 24.5  The primary, secondary, tertiary, and quaternary structures of hemoglobin.

The remaining levels of structure involve intermolecular forces between amino acids in the protein. The secondary structure of a protein refers to hydrogen-­ bonding networks formed between amide linkages in the protein backbone. ­(Figure 23.19 gives an example of interchain hydrogen bonding.) Specifically, the carbonyl group of one amide linkage interacts with the amino hydrogen of another amide. Some regular patterns, such as helices and sheets, often emerge as a result of hydrogen bonding. In hemoglobin, the amino acids in large portions of the ­polypeptide chains arrange themselves into many helical regions. The tertiary structure of a protein refers to how the chain is folded, including how amino acids far apart in the sequence interact with each other. In other words, this structure deals with how the regions of the polypeptide chain fold into the overall three-dimensional structure. For proteins consisting of only one chain, the tertiary structure is the ­highest level of structure present. In proteins consisting of more than one polypeptide chain, such as hemoglobin, there is a fourth level of structure, called the quaternary structure. It is concerned with how the different chains interact. The quaternary structure of hemoglobin shows how the four subunits are related to one another in the overall protein. The subtleties of sequence, structure, and function are dramatically illustrated in the case of hemoglobin. Seemingly small changes in the amino acid sequence of hemoglobin and other molecules can be important in determining function, as is illustrated by the disease called sickle cell anemia. This disease, which is sometimes fatal, predominantly affects individuals with genetic ties to malaria-endemic regions such as the Mediterranean, Africa, and South and Central America. Persons affected by this disease are anemic; that is, they have low red blood cell counts. In addition, many of their red blood cells are elongated and curved like a sickle instead of being round disks (Figure 24.6a). These elongated red blood cells are more fragile than normal

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Dr. Stanley Flegler/Visuals Unlimited, Inc.

(a)

1 2

1

1

2

2

deoxyhemoglobin A

1 2

deoxyhemoglobin S (sickle cell)

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

Deoxyhemoglobin S forms chains. (b)

(c)

Figure 24.6  Red blood cells.  ​(a) Red blood cells are usually rounded in shape, but people afflicted with sickle cell anemia have cells with a characteristic sickle shape. (b) Sickle cell hemoglobin has a nonpolar region that can fit into a nonpolar cavity on another hemoglobin. (c) Sickle cell hemoglobins can link together to form long chainlike structures.

blood cells and thus often break, leading to anemia. The elongated cells also do not fit through the capillaries properly and so block the flow of blood, thereby decreasing the amount of oxygen that the individual’s cells receive. The cause of sickle cell anemia has been traced to a small structural difference in hemoglobin. In the β subunits of the hemoglobin in individuals carrying the sickle cell trait (hemoglobin S), a valine has been substituted for a glutamic acid present in the hemoglobin of people not affected by the condition (hemoglobin A). An amino acid in this position ends up on the surface of the protein, where it is exposed to the aqueous environment of the cell. Glutamic acid and valine are quite different from each other (Figure 24.2). The side chain in glutamic acid is ionic, whereas the side chain in valine is nonpolar. The nonpolar side chain on valine causes a nonpolar region to stick out from the molecule. When hemoglobin is in the deoxygenated state, it has a nonpolar cavity in another region. The nonpolar region around the valine on one sickle cell hemoglobin molecule fits nicely into this nonpolar cavity on another hemoglobin. The sickle cell hemoglobins thus link together, forming long chainlike structures (Figure 24.6c) that lead to the symptoms described. Just one amino acid substitution in each β subunit causes sickle cell anemia! While other amino acid substitutions may not lead to such severe consequences, sequence, structure, and function are intimately linked and of crucial importance throughout biochemistry.

Enzymes, Active Sites, and Lysozyme Because many reactions necessary for life occur too slowly on their own, biological catalysts called enzymes are used to speed them up to the appropriate level. Almost every metabolic reaction in a living organism requires an enzyme, and most of these enzymes are proteins. Enzymes are often able to speed up reaction rates tremendously, typically 107 to 1014 times faster than uncatalyzed rates. For an enzyme to catalyze a reaction, several key steps must occur: 1. A reactant (often called the substrate) must bind to the enzyme. 2. The chemical reaction must take place. 3. The product(s) of the reaction must leave the enzyme so that more substrate can bind and the process can be repeated.

Enzyme Catalysis  For an interesting example describing an experiment with the enzyme carbonic anhydrase, see Section 14.6 and “Applying Chemical Principles 14.1: Enzymes—Nature’s Catalysts,” page 716.

Typically, enzymes are very specific; that is, only a limited number of compounds (often only one) serve as substrates for a given enzyme, and the enzyme catalyzes only one type of reaction. The place in the enzyme where the substrate binds and the reaction occurs is called the active site. The active site often consists of a cavity or cleft in the structure of the enzyme into which the substrate or part of the substrate can fit. The R groups of amino acids or the presence of metal ions in an active site are often important factors in binding a substrate and catalyzing a reaction.

24.1 Proteins

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1213

Lysozyme is an enzyme obtained from human mucus and tears as well as from other sources, such as egg whites. Alexander Fleming (1881–1955) (who later discovered penicillin) is said to have discovered its presence in mucus when he had a cold. He purposely allowed some of the mucus from his nose to drip onto a dish containing a bacteria culture and found that some of the bacteria died. The chemical in the mucus responsible for this effect was a protein. Fleming called it lysozyme because it is an enzyme that causes some bacteria to undergo lysis (rupture). Lysozyme is a protein containing 129 amino acids linked together in a single polypeptide chain (Figure 24.7). Its molar mass is 14,000 g/mol. As was true in the determination of the double-helical structure of DNA (Section 8.11), X-ray crystallography and model building were key techniques used in determining lysozyme’s three-dimensional structure and method of action. Lysozyme’s antibiotic activity results from its catalysis of a reaction that breaks down a polysaccharide, a polymer of sugar molecules, found in the cell walls of some bacteria. This polysaccharide is composed of two alternating sugars: N-acetylmuramic acid (NAM) and N-acetylglucosamine (NAG). Lysozyme speeds up the reaction that breaks the bond between carbon-1 of NAM and carbon-4 of NAG (Figure 24.8), thereby destroying the cell walls of the affected bacteria. Molecules consisting of three NAG units [(NAG)3] and no NAM act as inhibitors of lysozyme. X-ray crystallography of crystals of the enzyme treated with (NAG)3 revealed that (NAG)3 binds to a cleft in lysozyme (Figure 24.9). Researchers surmised that this cleft is the active site in the enzyme and that the inhibition by (NAG)3 resulted from (NAG)3 binding to but not completely filling the active site so that the cleavage reaction does not occur.

20

His Arg Trp

Leu

Cys

S

Ile Arg Gly Glu Cys Arg

Gly

Lys

Val

Cys

129

Arg Leu

COOH

H2N

Lys

100

Ile

Lys

Ser

90 Ala

Asn Leu

Ala Lys Phe Gly Ser Asn

110 Ala

Phe

Val

Asn

Trp

Thr 40

Asn Ile

Asp Ser Gly Pro 70 Thr Arg Gly Asn

Gln Ala Thr Asn Arg

Asn Thr Asp Gly 50 Ser

Asp

Thr

80 Cys S S Cys

Ile Asp Ser

Cys

Pro

Thr

Ala

Ala Ser Asp Gly Asp Gly Asn Met Val

Ala Cys S S Asn Val

30

Trp

1

Gly Phe Lys Val

Cys

Arg Asn Arg

S

Lys Met Ala Ala 10 Ala

Gln Val Asn Ala Thr 120

Trp

S S

Asn Tyr Arg Gly Asp Tyr Ser Leu Leu Gly Asn Gly

Ser Ser Leu

Ala Leu

Asp

Trp

Tyr

Trp Arg Ser

60

Asn Ile Gln

Leu

Gly Ile

Figure 24.7  The primary structure of lysozyme.  ​The cross-​chain disulfide links (OSOSO) are links between cysteine amino acid residues.

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CH2OH O

O

HO CH2OH O

CH2OH

O O

HO

R

NH O

4

O

C

1

C

O

4

HO

CH2OH 5 3

NAM

C

H

C

O

2

CH2OH

O 1

NH O

CH3

NAG

H3C

2

NH O

CH3

R =

5 3

O

C CH3

NAG

O

R

O

C

O

C

H2O

CH2OH

H O water added

C

CH2OH

O

C CH3

site of cleavage

Cleavage occurs only after NAM.

NAM

NH O

OH

CH3

O

HO

O

NH O

NAG

CH3

NAM

R

CH3

NH O

NH

O O

O

CH2OH

O

NAG

R

O O

O

NH O

C CH3

NAM

HO

Figure 24.8  Cleavage of the bond between N-acetylmuramic acid (NAM) and N-acetylglucosamine (NAG) in the polysaccharide present in some bacterial cell walls.  ​This reaction is accelerated by the enzyme lysozyme.

This cleft in lysozyme has room for a total of six sugar units. Molecular models of the enzyme and (NAG)6 showed that five of these six sugars fit nicely into the cleft but that the fourth sugar in the sequence did not fit well. To get this fourth sugar into the active site, its structure has to be distorted. This distortion of the fourth sugar caused the researchers to focus on the bond between this sugar and the fifth sugar as being the one cleaved by lysozyme. They realized that the distortion of the fourth sugar would be advantageous for helping the cleavage reaction occur because distorting the molecule in this way causes it to more closely resemble the ­transition state of the reaction, which would speed up the reaction. In addition, they discovered that amino acids immediately around this location can also assist in the ­cleavage reaction. Finally, models showed that if an alternating sequence of NAM and NAG, such as that in the bacterial cell walls, binds to the enzyme in this cleft, NAM must bind to this fourth sugar-binding site in the cleft because it cannot fit into the third sugar-binding site, whereas NAG can. Cleavage must therefore occur only between carbon-1 of NAM in the fourth sugar-binding site and carbon-4 of the following NAG in the fifth sugar-binding site, not the other way around—and this is exactly what occurs.

(NAG)3 in active site

lysozyme

Figure 24.9 (NAG)3 binds to the active site in lysozyme by means of intermolecular forces of attraction.

24.1 Proteins

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1215

O C H

CH2OH C O

H

C

OH

HO

C

H

HO

C

H

H

C

OH

H

C

OH

H

C* OH

H

C* OH

CH2OH D-glucose

CH2OH D-fructose

Figure 24.10  d-Glucose and d-fructose.  ​d-Glucose (an

24.2 Carbohydrates Goals for Section 24.2 • Draw the Lewis structures for commonly occurring mono-, di-, and polysaccharides. • Identify the types of linkages between glucose subunits in common polysaccharides. Carbohydrates, or sugars, constitute another major class of compounds found in living things. They serve as fuel for life’s activities. Ribose and deoxyribose, two sugars, are components of the nucleic acids (Section 24.3) that store and translate the genetic information in our cells. Polymers of sugars, polysaccharides such as those in Figure 24.8, serve as structural molecules in plants and bacteria. Finally, sugar molecules linked to other biomolecules are involved in cell–cell interactions.

Monosaccharides

aldose) and d-fructose (a ketose) both have the molecular formula C6H12O6. In d-sugars the hydroxyl group attached to the chiral carbon farthest from the carbonyl (this chiral carbon is marked with an asterisk in these figures) is shown on the right in this type of structure, whereas in l-sugars it is on the left.

Carbohydrates are polyhydroxy aldehydes or ketones. The open-chain structures of glucose, a polyhydroxy aldehyde or aldose, and of fructose, a polyhydroxy ketone or ketose, are shown in Figure 24.10. Both glucose and fructose are examples of ­monosaccharides, which are the simplest sugars. Monosaccharides usually have the general formula Cx(H2O)x, and it is from this formula that this class of compounds derives its name of carbohydrate, a combination of carbon and water. Thus, the formula of glucose, C6H12O6, is equivalent to C6(H2O)6. The most abundant sugars in nature contain five or six carbon atoms and are referred to as pentoses and hexoses, respectively. Ribose and deoxyribose (­S ection 24.3), found in the nucleic acids RNA and DNA, are the most abundant pentoses in nature. By far, the most abundant hexose containing an aldehyde group is glucose, and the most abundant hexose containing a ketone group is fructose. Almost all sugars are chiral, containing one or more carbon atoms with four different groups attached (Section 23.1). In the chain structure of glucose, for example, four of the carbon atoms are bonded to four different groups. In nature, g­ lucose occurs in just one of its enantiomeric forms; thus a solution of glucose rotates polarized light. The structures shown in Figure 24.10 are the naturally occurring d-isomers of glucose and fructose. d-glucose exists in three different isomeric forms in aqueous solutions. Two of the isomers contain six-member rings, and the third is the open chain structure (Figure 24.11). Notice that the two ring structures differ by the orientation of the hydroxyl group and hydrogen atom at C-1. In α-d-glucose, the hydroxyl group is pointing down (axial), whereas in β-d-glucose, it is pointing up (equatorial). In aqueous solution, the three different forms of glucose rapidly interconvert, but the three different forms are not present in equal amounts. The straight chain i­somer makes up less than 1% of the molecules with the cyclic forms predominating and present in a 63%∶37% ratio of β-d-glucose to α-d-glucose. The predominant ­structures of most monosaccharides in aqueous solution contain rings. The structure of glucose allows you to predict some of its properties. With five polar OOH groups in the molecule, glucose is soluble in water. The aldehyde group

d and l  d and l are symbols historically used to identify absolute configurations of chiral species. They derive from dextro- (right) and levo- (left).

H HO

OH

4

5

HO

3H

H

O

2

H -D-glucose

1

OH OH

H

H HO H H

CHO 1 OH 2 H 3 OH 4 OH 5 CH2OH

open-chain form

H HO

OH

4

5

HO

3H

H

O

2

H

1

OH

OH

H

-D-glucose

Figure 24.11  The straight-chain and ring forms of d-glucose.  The predominant form in aqueous solution is β-d-glucose.

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is susceptible to chemical oxidation to form a carboxylic acid, and detection of ­glucose (in urine or blood) takes advantage of this fact. Diagnostic tests for glucose involve oxidation with subsequent detection of the products.

Disaccharides When two sugar molecules join together to form a disaccharide, they do so by a condensation reaction in which a water molecule is lost. The two sugars are connected to each other by a bridging oxygen atom; this type of bond between two sugar molecules is called a glycosidic bond. Table sugar, sucrose (Figure 24.12a), is a disaccharide formed by the reaction of the hydroxyl group at position 1 in α-d-glucose with the hydroxyl group at position 2 of β-d-fructose. This type of linkage is referred to as an α,β-1,2 glycosidic bond. Another common disaccharide is lactose (Figure 24.12b), a sugar found in milk. This sugar contains another hexose called β-d-galactose, which is combined with d-glucose by means of a β-1,4 linkage. When milk is consumed, lactose is b ­ roken down into its monosaccharides by an enzyme called lactase. As many humans grow older, their ability to produce lactase and thus digest lactose decreases; they are said to become lactose intolerant. In such individuals, the lactose is instead fermented by bacteria in the digestive tract into lactic acid, producing methane gas and ­hydrogen gas in the process. These gases can cause discomfort and lead to flatulence. In addition, the lactic acid and any undigested lactose can sometimes lead to ­diarrhea. These symptoms can certainly be avoided if the affected person abstains from dairy products, but they can also be avoided if a supplement containing the enzyme ­lactase is consumed along with the dairy products. Maltose is yet another common disaccharide (Figure 24.12c). Maltose consists of two glucose residues joined by an α-1,4 linkage. Many of the starches and OH

H HO

H2C HO

H

O

H

,-1,2 linkage

H

OH

CH2OH

O

H

H HO H glucose

-1,4 linkage

H2C

H

O OH

OH

OH

H

O

H

HO

H 2C

O

OH H

H

H

OH

H

HO

HOH2C

H H

galactose

OH

OH H

H

fructose

O

glucose

(a) Sucrose

(b) Lactose

OH

H HO

H2C HO

H

O

H

OH

HH

OH

H2C

H

H

O -1,4 linkage

HO

glucose

O

H OH

H OH

H glucose

(c) Maltose

Figure 24.12  The disaccharides sucrose, lactose, and maltose.

24.2 Carbohydrates

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1217

glycogen present in your food are broken down into maltose using enzymes present in your digestive system. The maltose is then subsequently broken down into glucose molecules using another enzyme. Maltose is also important in making beer. In one step of brewing beer, the starches in the grain used to make the beer are broken down using enzymes into smaller carbohydrates, primarily maltose. Approximately 40% of the carbohydrate content of the mixture that eventually gets fermented is maltose. Yeast is added to begin fermentation. The maltose gets transported into the yeast cells, where it is broken down into glucose, which is then converted into a two-carbon alcohol called ethanol and carbon dioxide.

Polysaccharides Long polymers of sugars, polysaccharides, form when a large number of sugar ­molecules are joined by glycosidic bonds (Figure 24.13). In animal cells, excess glucose is stored as the polysaccharide glycogen (Figure 24.13b). The basic structure of glycogen is a branched chain of glucose molecules. Within a chain, the linkages are all α-1,4 g­ lycosidic bonds, and branch points to the chain occur by means of α-1,6 glycosidic bonds. Branch points occur every 10 or so glucose units in a chain.

Figure 24.13 Polysaccharides.

OH

H

O H

H2C

O

OH

H

HH

OH

HO

O H

H2C

O

OH

H

H

OH

HO

-1,4 linkage

HH O

H

H

H2C

O

H

H

H

OH

H

OH

O

H

HO O H

O

H

-1,6 linkage

O HH

OH

HO

O

O H

H2C H

H

HO

-1,4 linkage

OH OH

HH O

O H

H2C H

H

OH

HO

H

H

(b) Amylopectin and Glycogen

OH

OH O

H

OH

H

H2C

OH

HO

(a) Amylose

H

O H

H 2C

H2C HO

H

O H

H

O

OH H

H

OH H2C

HO

-1,4 linkage

H

O H

H

O

OH H (c) Cellulose

H

OH H 2C

HO

H

O

H OH H

H

1218 Chapter 24 / Biochemistry Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Plant cells often store glucose in the form of starch. The chains in starch can range in size from molar masses of thousands to half a million or more. There are two main types of starch: amylose starch and amylopectin starch. Amylose starch contains linear chains of glucose molecules connected by means of α-1,4 glycosidic bonds (Figure 24.13a). Amylopectin starch, like glycogen, is a branched chain of glucose molecules (Figure 24.13b). The key difference is that the α-1,6 branch points in amylopectin occur less often than in glycogen, only every 30 or so glucose units. Finally, the major structural molecule in plants, cellulose, is also a ­polysaccharide with chain lengths ranging from hundreds to many thousands of glucose units (­Figure 24.13c). The linkage between glucose molecules in cellulose is β-1,4 rather than α-1,4. This seemingly small change has a tremendous effect. Humans have enzymes that break down the α-1,4 linkages in starch but do not have an enzyme that breaks down the β -1,4 linkages between glucose molecules in cellulose. Humans can thus digest starch but cannot digest cellulose, which, when ingested, passes through our systems unchanged as insoluble dietary fiber.

24.3 Nucleic Acids Goals for Section 24.3 • Draw the Lewis structures for short strands of DNA and RNA. • Predict the sequence of nucleotides complementary to a given strand of DNA. • Predict the amino acid selected by a given sequence of three nucleotides in mRNA.

Nucleic Acid Structure In the first half of the twentieth century, researchers identified deoxyribonucleic acid (DNA) as the genetic material in cells. A close relative of DNA called ribonucleic acid (RNA) was also found in cells. RNA and DNA are polymers (Figure 24.14). They consist of sugars having five carbon atoms (β-d-ribose in RNA and β-d-2-deoxyribose in DNA) that are connected by phosphodiester groups. At physiological pH, these phosphodiester ­linkages are in their ionized states with a net 1− charge per linkage. Thus, there are also cations present (not shown in Figure 24.14) in the structure to maintain electrical

DNA

RNA phosphodiester group

sugar

O

3’

H 2’ 3’

−

−

−

4’

CH2O

5’

nitrogenous bases (A, C, G, T)

5’

5’

phosphodiester group

sugar −

O O O O 4’ 4’ 4’ H CH2O + H CH2O + H CH2O + P 3’ P 3’ P P+ 3’ H H O H H O O O O O− O O− O O− O O− H H 2’ H H 2’ H H 2’ H H 1’ H 1’ H 1’ H 1’ 5’ N N N N N CH3 O O N N N N N O NH2 H NH2 O N NH2 N thymine (T) cytosine (C) H adenine (A) guanine (G) H

5’

O

3’

HO 2’ 3’

5’ 5’ O O O− O− 4’ 4’ 4’ H CH2O + H CH2O + H CH2O + P 3’ P 3’ P P+ 3’ H H H H O O O O O O− O O− O O− O O− H HO 2’ H HO 2’ H HO 2’ H HO H 1’ H 1’ H 1’ H 1’ 5’ N N N N N O O N N N N N O NH H 2 NH O N N 2 NH2 cytosine (C) uracil (U) H adenine (A) guanine (G)

H

4’

5’

CH2O

−

5’

−

nitrogenous bases (A, C, G, U)

Figure 24.14  DNA and RNA.

24.3  Nucleic Acids

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1219

NH2

NH2 N

N

N

N

N

O H

H adenine (A)

H2N

N

O

H cytosine (C)

O (a)

N

N

O− HO

H

H thymine (T)

−

base

O

O

4’

O

H

R

base H

3’

2’

5’

HO

1’

CH2

4’

H

H

R

HO

nucleoside

H uracil (U)

CH2

H

H HO

N

5’

P+ O O−

H

H

N

O

CH2 H

guanine (G) O

CH3

N

N

H

O H

N

N

O

H

−

O

1’

H

3’

2’

O

5’-nucleotide

base H

R

P+ O− O−

3’-nucleotide

(b)

Figure 24.15  Bases, nucleosides, and nucleotides. Why Use the Primes in Numbering the Sugar Positions in DNA and RNA?  In a nucleotide, there are

two ring systems present: one in the nitrogenous base and one in the sugar. To tell the difference between them, positions in the bases are numbered normally (1, 2, 3, etc.), whereas positions in the sugars are numbered using numbers followed by the prime symbol (1′, 2′, 3′, etc.).

neutrality. A phosphodiester group links the 3′ (three prime) position of one sugar to the 5′ position of the next sugar. Attached at the 1′ position of each sugar is a nitrogen-containing (nitrogenous) base. The bases in DNA are adenine (A), cytosine (C), guanine (G), and thymine (T); in RNA, the nitrogenous bases are the same as in DNA, except that uracil (U) is used rather than thymine. A single sugar with a nitrogenous base attached is called a nucleoside. If a phosphate group is also attached, then the combination is called a nucleotide (Figure 24.15). The principal chemical difference between RNA and DNA is the identity of the sugar.

HO

CH2 H

OH

O

CH2

H

H

H H

HO

HO

OH

ribose (in RNA)

OH

O H

H

H HO

H

2-deoxyribose (in DNA)

Ribose has a hydroxyl group (OOH) at the 2 position, whereas 2-deoxyribose has only a hydrogen atom at this position. This seemingly small difference turns out to have profound effects.  The polymer chain of RNA is cleaved many times faster than a corresponding chain of DNA under similar conditions due to the involvement of this hydroxyl group in the cleavage reaction. The greater stability of DNA contributes to it being a better repository for genetic information. 

Storing Genetic Information The Structure of DNA  For more

information about the structure of DNA, see the chapter opening figure and caption for Chapter 8, pages 386–387, Section 8.11: “DNA,” pages 437–441, and “A Closer Look: Hydrogen Bonding in Biochemistry,” page 551.

How does DNA store genetic information? DNA consists of a double helix (­Figure 24.16). In this structure, one strand of DNA is paired with another strand running in the opposite direction. Thus, if one strand is viewed from the 5′-end to the 3′-end, the other strand will be lined up next to it such that it runs from the 3′-end to the 5′-end. The key parts of the structure of DNA for storing genetic information are the nitrogenous bases. Adenine, A, can form two hydrogen bonds with T, and C can form three hydrogen bonds with G. The spacing between the two strands of the double helix is just right for either an A–T pair or for a C–G pair to fit, but other combinations

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S A

S

T

−

S P

S P S

P

S

G

C A

G

S P

P

S

S

P A

CH2

S P S

O C H H C H

N HC

N

C

C

C

S C S P S S T A

N

N

−

O P

+

H

N

C

C

N

CH

O

N H C H C H −

O

O 3′-end

C

C

C

C

N HC

N

N

N

O

O

H

C

N

C

CH

H

O P + O−

N H

Guanine

Figure 24.16  Base pairs and complementary strands in DNA.  ​ With the four bases in DNA, the usual pairings are adenine with thymine and cytosine with guanine. The pairing is promoted by hydrogen bonding.

−

CH

N

H

CH2 O O

H H C H C

H

O

O

CH3

H

O

O P

H

N

C

C

C H H C

C H H C

O

−

C H H C

P

T

+

O

H

O

P

A

Thymine H

O

O P

P

C T

S

O

C H H C −

T

P

−

CH2

S

S C P

Adenine

O

P

P

+

O P

P

S

C

3′-end

O

P

S

T

G

5′-end

A

A

P S

T S

P

S

P

P

Cytosine Hydrogen bond

C H H C O

CH2 O

−

O

+

−

P O O

5′-end

(such as A–G) do not fit properly. Thus, if you know the identity of a nucleotide on one strand of the double helix, you can figure out which nucleotide must be bound to it on the other strand. The two strands are referred to as complementary strands. If the two strands are separated, a new complementary strand can be constructed for each of the original strands by placing a G wherever there is a C, a T wherever there is an A, and so forth. Through this process, called replication, the cell ends up with two identical double-stranded DNA molecules for each molecule of DNA initially present. When the cell divides, each of the two resulting cells gets one copy of each DNA molecule (Figure 24.17). In this way, genetic information is passed along from one generation to the next.

Ex am p le 24.2

Complementary Strands of DNA Problem  By convention, nucleic acid sequences are usually written starting with the nucleotide on the 5′-end and proceeding to the 3′-end. A particular segment of a strand of DNA has the sequence AGTCCTCATG. What is the sequence of the complementary strand? What Do You Know?  You know the sequence of a segment of DNA and that in the complementary strand A will base pair with T, and C will base pair with G. Strategy  Write the sequence of DNA that is complementary to the given sequence by completing the base pairs (A with T and G with C). List the sequence from the 5′-end to the 3′-end. Solution  The given sequence and its complement are as follows: Given sequence: 5′-AGTCCTCATG-3′ Complementary sequence: 3′-TCAGGAGTAC-5′ In order from the 5′-end to the 3′-end, this sequence is  CATGAGGACT. 

Think about Your Answer   Knowing the sequence of one strand of DNA allows you to predict the sequence of DNA on the complementary strand.

Check Your Understanding  What is the sequence of the strand of DNA complementary to CGATACGTAC?



24.3  Nucleic Acids

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1221

A T T

T

G

A

G C

T A

C

T

G

A

G C

C G

C

Clouds Hill Imaging Ltd./Corbis

A

C

C G

G A T

Two strands of DNA. Each base is paired with its partner: adenine (A) with thymine (T), guanine (G) with cytosine (C).

The two DNA strands are separated from each other.

Two new complementary strands are built using the original strands.

G C

C G

Replication results in two identical double-stranded DNA molecules.

At this stage during cell division, the chromosomes containing the DNA have been duplicated, and the two sets have been separated.

Figure 24.17  The main steps in DNA replication.  ​The products of this replication are two identical double-helical DNA molecules.

Genetic Engineering with CRISPR-Cas9

Ever since the double helical structure of DNA was discovered, s­ cientists have looked for ways to modify the DNA of an organism. For example, they might wish to learn more about what certain portions of the DNA do, to eliminate the expression of a faulty gene, or to correct a gene to produce a healthy protein. In order to do these things, the DNA of an organism must often be cut at specific locations. Over the years, various proteins have been discovered and developed for this purpose, but recently, a new and very powerful technique to do this has been found. This method is based on a process carried out by bacterial cells to protect themselves from invading bacteria or viruses to which that species of bacteria had been exposed sometime in the past. As a result of that past invasion, portions of the invader’s DNA were introduced into the bacterial DNA. It was found that these bacteria have regions of DNA called clustered regularly interspaced palindromic repeats (CRISPRs). In these CRISPRs, there are multiple regions of 20–30 base pairs of DNA that arose from past invading species. When these segments of DNA are transcribed into RNA, the resulting short segments of RNA are joined with proteins such as the CRISPR-associated protein 9 (Cas9) and with another strand of RNA associated with the protein. When these protein–RNA hybrids come across DNA that is complementary to

Target DNA

Cas

5' 3'

Cut

Guide RNA

3' 5'

Russell, Peter J, Paul E. Hertz & Beverly McMillan. Biology: The Dynamic Science, 4ed, 2017, Cengage Learning.

A Closer Look

When a cell divides, each resulting cell gets one set.

CRISPR-Cas9.  A guide sequence of RNA is incorporated into the structure of the Cas9 protein. When a strand of DNA is encountered that is complementary to a portion of this guide sequence, both strands of the DNA are cut at the indicated locations. the CRISPR RNA, such as in an invading virus, the protein cuts both strands of the DNA, inactivating the invader and protecting the bacteria. The exact mechanism for this process was determined by a collaboration of the labs of Jennifer Doudna and Emmanuelle Charpentier. Doudna, Charpentier, and their research associates then designed a way to combine the two types of RNA into a single strand of RNA, called the guide sequence, which has one part that is complementary to the DNA to be cut and one part that binds the RNA to the Cas9 protein. The key aspect of this CRISPR-Cas9 system is that the DNA to be cut can be programmed by simply incorporating the complementary RNA into the CRISPRCas9 system. Thus, scientists can choose

to cut virtually any sequence of DNA using the appropriate RNA bound to the Cas9 protein. Once the DNA is cut, this causes DNA repair enzymes to rejoin the strands. In this process, the gene is either disrupted or replaced, perhaps with a new gene for a different protein. The 2020 Nobel Prize in Chemistry was awarded to Charpentier and Doudna “for the development of a method for genome editing.” CRISPR-Cas9 has already been used in a multitude of studies, including some successful human clinical trials for treatments based on CRISPR-Cas9 for the blood conditions sickle-cell anemia and beta thalassemia. Clinical trials for treatments of some cancers, genetic blindness, type 1 diabetes, and other conditions are either underway or being planned.

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Protein Synthesis

Table 24.1

The sequence of nucleotides in a cell’s DNA contains the instructions to make the proteins the cell needs. DNA is the information storage molecule. To use this information, the cell first makes a complementary copy of the required portion of the DNA using RNA. This step is called transcription. The molecule of RNA that results is called messenger RNA (mRNA) because it takes this message to where protein synthesis occurs in the cell. The cell uses the more rapidly cleaved RNA rather than DNA to carry out this function. It makes sense to use DNA, the more stable molecule, to store the genetic information because the cell wants this information to be passed from generation to generation intact. Conversely, it makes sense to use RNA to send the message to make a particular protein. By using the less-stable RNA, the message will not be permanent but rather will be destroyed after a certain time, thus allowing the cell to turn off the synthesis of the protein. Protein synthesis occurs in ribosomes, complex bodies in a cell consisting of a mixture of proteins and RNA, which bind to mRNA. The sequence of nucleotides in mRNA contains information about the order of amino acids in the desired protein. Beginning at the signal in mRNA to start protein synthesis, every sequence of three nucleotides provides the code for an amino acid in the protein until the ribosome reaches the signal to stop (Table 24.1). These three-nucleotide sequences in mRNA are referred to as codons, and the correspondence between each codon and its message (start, add a particular amino acid, or stop) is referred to as the genetic code. Another type of RNA is also involved in this process. Transfer RNA (tRNA) ­consists of a strand of RNA to which an amino acid can be attached (Figure 24.18). A strand of tRNA has a particular region that contains a sequence of three nucleotides that can form base pairs with a complementary (for example, A with U) codon in the mRNA. This three-nucleotide sequence in the tRNA is called the anticodon. Only if the base pairing between the codon and anticodon is complementary will the tRNA be able to bind successfully to the mRNA–ribosome complex. Not only does the anticodon determine to which codon a particular strand of tRNA can bind, but it also determines which amino acid will be attached to the 3′-end of the

Examples of the 64 Codons in the Genetic Code

Codon Base Sequence*

Amino Acid to Be Added

AAA

Lysine

AAC

Asparagine

AUG

Start

CAA

Glutamine

CAU

Histidine

GAA

Glutamic acid

GCA

Alanine

UAA

Stop

UAC

Tyrosine

*A = adenine, C = cytosine, G = guanine, U = uracil.

amino acid

3’ attachment site acceptor stem

acceptor stem

5’

amino acid 3’ attachment site

5’

anticodon loop anticodon loop

anticodon

anticodon

(a)

(b)

Figure 24.18 ​tRNA structure.  ​(a) A space-filling model of a tRNA molecule. (b) A simplified model, in

which each nucleotide is represented as a circle. A line indicates that two nucleotides are hydrogen bonded. The acceptor stem is shown in green, the anticodon in yellow, and the rest of the anticodon loop in blue.



24.3  Nucleic Acids

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1223

tRNA molecule. Thus, a codon in the mRNA is complementary to a particular tRNA ­anticodon, which in turn specifies the correct amino acid. What are the specific steps that occur to add an amino acid to a growing ­protein? In the ribosome–mRNA complex (Figure 24.19), there are three neighboring binding sites, called the A site (aminoacyl site), the P site (peptidyl site), and the E site (exit site). Each cycle that adds an amino acid to a growing protein begins with the already-constructed part of the protein located in the P site and still attached to the tRNA that brought the amino acid in the previous cycle. A new tRNA ­molecule with its amino acid attached, called an aminoacyl tRNA, binds to the A site (­Figure 24.19a). Only when the anticodon of the tRNA is complementary to the codon in the mRNA at this position does successful binding occur and the reaction proceed. The amino acid in the A site reacts with the growing protein in the P site, forming a peptide bond between the A site amino acid and the P site protein. The protein chain’s previous bond to the tRNA in the P site is broken, and the protein, now elongated by one amino acid, is attached to the tRNA in the A site (Figure 24.19b). The ribosome then moves down the mRNA chain, moving the tRNA along with the protein strand from the A site into the P site and exposing a new codon in the A site. The tRNA that had been in the P site and that no longer has an amino acid attached moves into the E site and then exits from the ribosome (Figures 24.19c and d). The process is then repeated until a Stop codon is reached in the mRNA. Converting the information from a nucleotide sequence in mRNA into an amino acid sequence in a protein is called translation. Protein synthesis thus consists of two main processes: transcription of the DNA’s information into RNA, followed by translation of the RNA’s message into the amino acid sequence of the protein. There is more involved, but the processes of transcription and translation as discussed here provide a basic introduction to this important topic. polypeptide chain

Met

Ala Val

Gln 5’

5’

new peptide bond

amino acid 3’

Met

tRNA

3’

C G U

ribosome

Gln

Val 5’

Ala 3’

3’

5’

anticodon C A A C G U

C A A G C A U G C A G G U U G C A A G C U G A U C G

G C A U G C A G G U U G C A A G C U G A U C G

5’

5’

P site

A site

mRNA

3’

P site

(a)

mRNA

A site (b)

polypeptide chain 3’

5’

Met

Gln

Val

Ala 5’

3’

Met

3’

Gln

Ser

Val

Ala 5’

C A A

ribosome movement

C G U

3’

5’

3’

U C G C G U

G C A U G C A G G U U G C A A G C U G A U C G

G C A U G C A G G U U G C A A G C U G A U C G

5’

5’

P site

A site (c)

mRNA

3’

3’

P site (d)

A site

mRNA

Figure 24.19  Protein synthesis.  ​(a) The tRNA with an anticodon complementary to the mRNA codon exposed in the A site of the ribosome brings the next amino acid to be added to the growing protein chain. (b) A new peptide bond is formed between the amino acid in the A site and the protein in the P site. (c and d) After the new peptide bond is formed, the ribosome moves down the mRNA, exposing a new codon in the A site and transferring the tRNA and the protein chain to the P site. Note that the E site, from which the previous tRNA exits the complex, is not shown in this figure.

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Ex am p le 24.3

Protein Synthesis Problem  A strand of DNA that codes for three amino acids in a protein has the following sequence: TTT TGC GTA. What is the sequence of amino acids that will result from this sequence?

What Do You Know?   The sequence of DNA is given. The information in DNA is transcribed into mRNA and then translated by tRNA into the desired amino acid sequence. You know the rules for forming base pairs. The genetic code for some of the codons in mRNA is given in Table 24.1. Strategy  From the given DNA sequence, determine the sequence of mRNA that would be transcribed. Look at the sequence of mRNA in groups of three nucleotides, which ­correspond to the codons. Determine which amino acids correspond to these codons. Solution   The given sequence of DNA and its complementary strand of mRNA are as follows: Given sequence of DNA:

5′-TTT TGC GTA-3′

Complementary sequence of mRNA: 3′-AAA ACG CAU-5′ Cells translate sequences of RNA in order, from the 5′-end to the 3′-end (Figure 24.19). This strand of mRNA has the sequence 5′-UAC GCA AAA-3′. It is divided into groups of three nucleotides each, corresponding to the codons in the mRNA. Based on Table 24.1, the amino acids selected by this sequence are  tyrosine–alanine–lysine. 

Think about Your Answer   Keep in mind that the genetic code in Table 24.1 gives the mRNA codons and the amino acids brought in by the complementary tRNAs. ­Complete listings of all possible 64 codons and their corresponding amino acids or ­messages (start, stop) can be found in many biology texts and the Internet.

Check Your Understanding  What is the sequence of amino acids encoded by the DNA sequence TTG TGC TTT?

24.4 Lipids and Cell Membranes Goals for Section 24.4 • Understand that lipids are not very soluble in water and that some lipids consist of molecules in which one end is polar and the other end is nonpolar.

• Recognize the ring system in steroids. • Understand the structure of a lipid bilayer and how materials can be transported across the bilayer.

Lipids are the principal components of cell membranes and a repository of chemical energy in the form of fat. In addition, some of the chemical messengers called hormones are lipids. Lipids include a wide range of compounds because classification of a compound as a lipid is based on its solubility rather than on a particular chemical functional group. A lipid is a compound that is at best slightly soluble in water but is



24.4  Lipids and Cell Membranes

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1225

soluble in organic solvents. Compounds in biological organisms that are nonpolar, or at least substantially nonpolar, have limited solubility in water (Section 13.2) and are therefore lipids. A major category of lipids consists of molecules that have a polar end and a nonpolar end. The polar end provides the slight water affinity necessary for it to be compatible with being in the aqueous environment of the cell, but the nonpolar end greatly limits the water solubility. Many of these lipid components are fats, oils, fatty acids, or related to these compounds. Fats and oils serve many functions in the body, a primary one being the storage of energy. Fats (solids) and oils (liquids) are triesters formed from glycerol (1,2,3-​ propanetriol) and three carboxylic acids that can be the same or different. Because they are triesters of glycerol, they are sometimes referred to as triglycerides. O H2C

O

CR O

HC

O

CR O

H2C

O

CR

Triglyceride. R = long chain fatty acid (Table 24.2)

The carboxylic acids in fats and oils are known as fatty acids and have a lengthy carbon chain, usually between 12 and 18 carbon atoms (Table 24.2). Fatty acids and triglycerides are lipids of the type where one end of the molecule is polar and the other end is nonpolar. The polar end in fatty acids is the carboxylic acid group, and the nonpolar end is the long hydrocarbon chain. The hydrocarbon chains can be saturated (Section 23.2) or may include one or more double bonds. Saturated compounds are more common in animal products while unsaturated fats and oils are more common in plants. In general, triglycerides containing saturated fatty acids are solids, and those containing unsaturated fatty acids are liquids at room temperature. The difference in melting point relates to the molecular structure. With only single bonds linking carbon atoms in saturated fatty acids, the hydrocarbon group is flexible, allowing the molecules to pack more closely together. This greater contact between the molecules leads to greater forces of attraction between the molecules and thus to higher melting points. The double bonds in unsaturated triglycerides, on the other hand, introduce kinks that make the hydrocarbon group less flexible; consequently, the molecules pack less tightly together, leading to smaller London dispersion forces and lower melting points. Steroids are another category of lipids. Steroid molecules consist of four, joined hydrocarbon rings (Figure 24.20). Three of the rings contain six carbon atoms, and one

Table 24.2 Name

Common Fatty Acids

Number of C Atoms

Formula

Saturated Acids Lauric

C12

CH3(CH2)10CO2H

Myristic

C14

CH3(CH2)12CO2H

Palmitic

C16

CH3(CH2)14CO2H

Stearic

C18

CH3(CH2)16CO2H

Unsaturated Acid Oleic

C18

CH3(CH2)7CHPCH(CH2)7CO2H

1226 Chapter 24 / Biochemistry Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

12 11 2 3

1

9

A

10 B 5

4

6

C

17 13

8

14

D

CH3

16

H3C CHCH2CH2CH2CH(CH3)2 H H

15

7

H3C

The ring structure present in all steroids: three six-member rings (A, B, and C) and one five-member ring (D).

H

HO

H

H cholesterol

Figure 24.20 Steroids.

contains five carbon atoms. Examples of steroids are the sex hormones testosterone, estradiol, and progesterone. Cholesterol (Figure 24.20) is also an important steroid. You may have heard of cholesterol because of its correlation with heart disease. While some cholesterol is necessary for humans, excess cholesterol can deposit in blood vessels, thus partially blocking them and causing the heart to work harder than it should. The most prevalent molecules in most cell membranes are phospholipids (­Figure 24.21a). These are similar to triglycerides in that they are based on ­glycerol. Two of the alcohol groups in glycerol are esterified to long-chain fatty acids. The third alcohol group is bonded to a phosphate that has another hydrocarbon chain attached. Phosphate groups are very polar. In phospholipid molecules, the phosphate end is sometimes called the head, and the nonpolar hydrocarbon chains comprise the tail. When phospholipids are placed in water, they typically arrange themselves in a bilayer structure (Figure 24.21b). This is exactly the arrangement that phospholipids have in a cell membrane. Water is present on both the inside and the outside of the bilayer, corresponding to the inside and outside of the cell. In the outside layer of the membrane, the phospholipids line up alongside each other such that their polar heads face the aqueous environment outside the cell. Moving inward, next come the tails of these lipids. The phospholipids in the second layer align themselves so that their nonpolar tails are in contact with the outer layer’s nonpolar tails. Finally, the polar heads of the second layer face the aqueous environment inside the cell. The phospholipid bilayer nicely encloses the cell and provides a good barrier between the inside and the outside of the cell, due to the different solubility characteristics of the nonpolar region in the middle of the bilayer.

Phospholipid Choline Phosphate Glycerol backbone

Fatty acid tails

(a)

CH2 N+(CH3)3 CH2 Polar O head –O P+ O– O CH2 CH CH2 O O C O O C CH2 CH2 CH2 CH2 CH CH2 CH CH2 Nonpolar CH2 CH2 tails CH2 CH2 CH2 CH2 CH3 CH3 (b)

Phospholipid bilayer cross section

Polar head Nonpolar tails Polar head

Interior aqueous compartment Exterior aqueous environment

Figure 24.21 Phospholipids. ​(a) The structure of a phospholipid. (b) A cross section of a phospholipid bilayer. The polar heads of the phospholipids are exposed to water, whereas the nonpolar tails are located in the interior of the bilayer.

24.4  Lipids and Cell Membranes

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1227

Cell Exterior (Extracellular Fluid) Phospholipid bilayer

Cholesterol

Protein

Protein

Carbohydrate chains Channel protein

Cytoskeleton filaments

Cell Interior (Cytosol)

Figure 24.22  The fluid-mosaic model of cell membranes.  ​A cell membrane is made up primarily of a phospholipid bilayer in which are embedded cholesterol, other lipids, and proteins. Movement within a layer occurs, but movement from one side of the bilayer to the other is rare.

mRNA Vaccines

The first case of COVID-19 caused by the SARS-CoV-2 virus was reported in December 2019. This virus causes mild symptoms in most people, but some get severely ill and some even die. As of this writing, there have been over 761 million reported cases worldwide (103 million U.S. cases), and 6.9 million deaths worldwide (1.1 million U.S. deaths) have been attributed to COVID-19. In December 2020, approximately one year after the first reported case, the U.S. Food and Drug Administration, issued Emergency Use Authorization for two vaccines, one produced by Pfizer/ BioNTech and the other by Moderna that were based on messenger RNA (mRNA). Full FDA approval for their use in adults was granted in August 2021 and J­ anuary 2022, respectively. In addition, these ­v accines were also fully approved or granted Emergency Use Authorization in many other countries. mRNA vaccines are different from other vaccines. Typical vaccines inject either a killed version of a germ, a weakened form

of the germ, or a subunit (for example, a protein or lipid) of the germ. When the body encounters these, it mounts an immune response so the body can protect itself if it encounters the real germ. In the case of an mRNA vaccine, however, the key material injected is mRNA that codes for a protein or portion of a protein of the germ. The mRNA is taken up by the cells,

which then use the mRNA to make the encoded protein. This protein then sets off the immune response to provide future protection. In the case of the SARS-CoV-2 virus, the protein encoded by the mRNA is a modified version of the spike protein on the virus surface (Figure A), which the virus uses to enter cells.

CDC/Alissa Eckert, MSMI; Dan Higgins, MAMS

A Closer Look

Cholesterol is an important part of animal cell membranes, helping to maintain the proper fluidity of the membrane. Proteins in the cell membrane allow select materials to cross from one side of the membrane to the other (transport proteins). Others accept chemical signals from other cells or respond to materials in the cell’s environment (receptor proteins). Finally, some enzymes are also associated with the membrane. The overall model for a cell membrane is called the fluid-mosaic model (­Figure 24.22). In this model, the membrane’s structure is largely that of a phospholipid bilayer. Embedded in this bilayer are molecules such as cholesterol and proteins. Movement of all of these components within each layer of the bilayer occurs readily; the membrane is thus fluid to a certain extent. On the other hand, there is

Figure A  The SARS-CoV-2 virus.  An illustration, created by the U.S. Centers for Disease Control and Prevention, that shows the virus. The spike proteins are shown in red on the surface of the virus.

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little movement of components, such as phospholipids, from one side of the membrane to the other. The “like dissolves like” observation provides the reason for this lack of exchange between layers. The head of a phospholipid in the outer layer, for example, would not be compatible with the very nonpolar region within the bilayer that the phospholipid would need to pass through in order to traverse from one side of the bilayer to the other. A cell membrane serves as the boundary between the cell and the rest of the u ­ niverse, but exchange of some materials between the cell and the outside world needs to occur. There are different mechanisms by which this happens (­Figure 24.23). The simplest is passive diffusion. In this process, a molecule moves through the phospholipid bilayer from a region of higher concentration to a region of lower concentration, the natural direction of flow. Because the bilayer provides such a good barrier, only a few very small uncharged molecules (such as N2, O2, CO2, and H2O) can pass through the membrane this way. Many more species enter or leave the cell through a process called facilitated diffusion. In this process, ions or molecules still travel from a region of higher concentration to a region of lower concentration, but they do not pass directly through the bilayer. Instead, they pass through channels formed by proteins embedded in the cell membrane. Sometimes, it is necessary for the cell to move species from a region of lower concentration to a region of higher concentration, a direction opposite from that normally expected. The cell accomplishes this by means of active transport. This is again mediated by transport proteins in the cell membrane. Because the species of interest must move in the opposite direction than it would normally go, the cell must expend energy (often obtained from the exothermic hydrolysis of ATP, see ­Section 24.5) in order to make this occur. Finally, cells sometimes transport materials into themselves by means of endocytosis. This process is usually mediated by a receptor protein. The species of interest

The Pfizer/BioNTech and Moderna v­ accines for COVID-19 were the first mRNA vaccines approved for use in humans, but the science underlying these vaccines goes back many years and has involved the work of ­h undreds if not thousands of researchers. mRNA was first discovered and its role in ­p rotein synthesis determined in the 1960s. RNA was not considered at that time, and for a long time afterward, as a good candidate for vaccines because it is unstable and difficult to work with. Researchers tended to focus more on DNA instead. Over the years, better methods for synthesizing and handling RNA were developed. Early studies of RNA vaccines were unsuccessful, in part, because the mRNA caused a massive inflammatory response when injected into mice. In 2005, researchers determined that if the base pseudouracil (Figure B) is used instead of uracil in the mRNA, this response is avoided. This was one of the key discoveries



O HN

NH O

Figure B Pseudouracil. that allowed for mRNA to be used in vaccines. Another major obstacle was finding a good way to encase the mRNA so that it can be transported through the body and pass through the lipid bilayer of the cell’s membrane. In 1978, liposomes, small spherical lipid bilayers, were first used to transport mRNA into mouse and human cells, and in 1987, it was discovered that if mRNA was mixed with droplets of fat, the mRNA could be absorbed by cells and result in protein production. Eventually, researchers focused on creating lipid nanoparticles (LNPs). These can be formed using special technology to mix fats dissolved in alcohol and nucleic acids dissolved in an acidic buffer. The current vaccines use LNPs composed of

four types of lipids: phospholipids, cholesterol, lipids with polyethylene glycol attached, and importantly, an ionizable lipid. When the LNPs are made, these ionizable lipids have positive charges. This enables them to interact well with the negative charges in the phosphate groups of the RNA when the LNPs form. On the other hand, when the ionizable lipids are injected into the bloodstream, which is at a higher pH, they become neutral. This is important because positively charged lipids are toxic. These LNPs can be transported through the blood and enter cells by the process of endocytosis. Finally, the mRNA vaccines also contain buffers to maintain the proper pH as well as sucrose (table sugar). The advent of mRNA vaccines for COVID-19 has saved numerous lives, and the COVID-19 vaccines are probably just the first in a wide range of vaccines using this technology.

24.4  Lipids and Cell Membranes

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1229

Figure 24.23 Transport of materials across a cell membrane.

FACILITATED DIFFUSION

PASSIV E D IFFUSIO N

High concentration

High concentration

Low concentration

Low concentration ENDOCYTOSIS

ACTIV E TRA NSPO RT

Low concentration

Receptor protein

ATP High concentration

(called the substrate) binds to the receptor protein. A portion of the cell membrane surrounds the receptor–substrate complex. This portion of the cell membrane is then broken off, bringing the complex into the cell.

24.5 Metabolism Goals for Section 24.5 • Understand that coupling of ATP hydrolysis with a reactant-favored reaction can lead to a product-favored reaction.

• Recognize that NAD+ and NADH, FAD and FADH2, and Q and QH2 are involved in biochemical redox reactions.

• Understand the key stages in the metabolism of glucose. • Contrast the overall reactions for respiration and photosynthesis. • Understand the absorption of light by chlorophyll. Why do we eat? Some components of food, such as water, are used directly in our bodies. Other chemicals are broken down to obtain the molecular building blocks required to make the compounds needed for life. Oxidation of foods also provides the energy necessary to perform the activities of life. The many different chemical reactions that foods undergo in the body to provide energy and chemical building blocks fall into the area of biochemistry called metabolism.

Energy and ATP Substances in food, such as carbohydrates and fats, are oxidized in metabolic processes. These oxidations are energetically favorable reactions, releasing large quantities of energy. For example, the oxidation of the sugar glucose (C 6H12O6) to form carbon dioxide and water is very exothermic. C6H12O6(s) + 6 O2(g) n 6 CO2(g) + 6 H2O(ℓ) ∆rH° = −2803 kJ/mol-rxn

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Figure 24.24 Adenosine-​ 5′-triphosphate (ATP).

NH2 N

N

−O

O−

O−

O−

P+

P+

P+

O

O−

O−

O

N

N O

CH2

O−

O

H

H

H

H

HO

OH

However, rather than carry out this reaction in one rapid and exothermic step, a cell carries out a more controlled oxidation in a series of steps so that it can obtain the energy in small increments. In addition, it would be inefficient if every part of a cell needed to have all the mechanisms necessary to carry out the oxidation of every type of molecule used for energy. Instead, it carries out the oxidation of compounds such as glucose in one location and stores the energy in a small set of compounds that can be used almost anywhere in the cell. The principal compound used to perform this storage and transport of energy is adenosine 5′-triphosphate (ATP). This ribonucleotide consists of a ribose molecule to which the nitrogenous base adenine is connected at the 1′ position and a triphosphate group is connected at the 5′ position (Figure 24.24). The hydrolysis of ATP to adenosine 5′-diphosphate (ADP) and the hydrogen phosphate ion (Pi) is an exothermic process that is product-favored at equilibrium (Figure 24.25). ATP + H2O n ADP + Pi  ∆rH° ≈ −24 kJ/mol-rxn (∆rG°’ = −30.5 kJ/mol-rxn)

𝚫G °  ′  For the definition of ΔG° ′, see “Applying Chemical Principles 18.1: Thermodynamics and Living Things,” page 916.

Why is this reaction exothermic? In this reaction, two bonds break, a POO bond in ATP and an HOO bond in water. But two new bonds form: a POO bond connecting the phosphate group being cleaved off the ATP with the OH of the original water and an HOO bond connecting the hydrogen from the water with the portion of the ATP that forms ADP.  In the hydrolysis reaction of ATP into ADP and the hydrogen phosphate ion, more energy is released in forming these new bonds in the products than is required to break the necessary bonds in the reactants.  In cells, many chemical processes that would be reactant-favored on their own are linked with the hydrolysis of ATP. The combination of an energetically unfavorable process with the energetically favorable hydrolysis of ATP can yield a p ­ rocess that is energetically favorable and product-favored at equilibrium. For example, most cells have a greater concentration of potassium ions and a smaller concentration of sodium ions inside them than are present outside them. The natural ­tendency, therefore, is for sodium ions to flow into the cell and for potassium ions to flow out. To maintain the correct concentrations, the cell must counteract this movement and pump sodium ions out of the cell and potassium ions into the cell. This requires energy. To accomplish this feat, the cell links this pumping process to the hydrolysis of ATP to ADP. The energy released from the hydrolysis reaction provides the energy to run a molecular pump (a transport protein) that moves the ions in the direction the cell needs. NH2

NH2 N

N −O

O−

O−

O−

P+ O

P+ O

P+ O

O−

O−

O−

CH2 H H HO

ATP

N

N O H

HPO42− + HO

N O−

O−

P+ O

P+ O

O−

O−

H

CH2 H H

OH

adenosine-5’-triphosphate

+ H2O

N

HO

N

N O

Figure 24.25 The exothermic conversion of adenosine-5′-triphosphate (ATP) to adenosine-5′-​ diphosphate (ADP).

H H OH

ADP

adenosine-5’-diphosphate 24.5 Metabolism

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1231

Common Compounds in Biochemical Oxidation–Reduction Reactions Cells also need compounds that can carry out oxidation–reduction reactions. Just as ATP is a compound used in many biochemical reactions when energy is needed, so nature uses another small set of compounds to run many redox reactions. An important example is nicotinamide adenine dinucleotide (NAD+). This compound consists of two ribonucleotides joined at their 5′ positions via a diphosphate linkage. One of the nucleotides has adenine as its nitrogenous base, whereas the other has a nicotinamide ring (Figure 24.26). Humans obtain nicotinamide from vitamin B3, also called niacin. When NAD+ is reduced, changes occur in the nicotinamide ring, such that the equivalent of a hydride ion (H−) is gained. Because this hydride ion has two electrons associated with it, the nicotinamide ring gains two electrons in the process. The resulting species, referred to as NADH, is shown on the right in Figure 24.26. In many biochemical reactions, when a particular species needs to be oxidized, it reacts with NAD+. The NAD+ is reduced to NADH, gaining two electrons in the process, and the species of interest is oxidized by losing these electrons. If a species must be reduced, the opposite process often occurs; that is, it reacts with NADH. The NADH is oxidized to NAD+, and the species of interest is reduced. Another common pair of compounds used in biochemical redox reactions is based on flavin adenine dinucleotide (FAD) (Figure 24.27a). Like NAD + and NADH, these compounds consist of two nucleotides joined at their 5′ positions by a diphosphate linkage and where one of the nucleotides has adenine as its nitrogenous base. Now, however, the other has a ring system called an isoalloxazine ring, which is derived from vitamin B2, also called riboflavin. The oxidized form is abbreviated as FAD, and the reduced form as FADH2. Once again, two electrons are transferred to convert one to the other. Finally, one more set is often encountered involving ubiquinone (Q), which is a diketone, and ubiquinol (QH2), which is a dialcohol (Figure 24.27b). Again, conversion of one to the other involves the transfer of two electrons. In the following overview of the metabolism of glucose in respiration, you will see how these redox pairs play a role in the metabolism of glucose to make ATP.

Overview of the Metabolism of Glucose in Respiration Different types of food molecules (such as carbohydrates, fats, and proteins) are metabolized through a variety of biochemical steps. The specifics of these steps are the subject of multiple chapters in more advanced biochemistry textbooks. This s­ ection will provide only an overview of the steps used in metabolizing one Figure 24.26 The structures of NAD+ and NADH.

O

H

H

C

O −O

CH2

P+ O− O

−O

P+ O

H H

O

nicotinamide

H H

HO

OH

NH2

N

O− CH2 H

N

O H

H

H

HO

OH

NAD+

N N

adenine

O C

NH2

N+

O

H

Reduction: +H+ +2 e− Oxidation: −H+ −2 e−

−O

CH2

P+ O− O

−

O

P+ O

NH2

N O

H

H

H

H

HO

OH

NH2

N

O− CH2 H

N

O

N N

H

H

H

HO

OH

NADH

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O H3C

H

N

N

N

H3C

O

H3C

NH

N

N

C

H

H

C

H H

H

C

OH

H

C

OH

H

C

OH

H

C

OH

H

C

OH

H

C

OH

C

H

C

H

H

Reduction +2 H+ +2 e− NH2

O –

O

P+

N

O–

O –

O

P+

CH2

O–

–

O

NH2

P+

N

O–

O

N –

O

O

P+

H

OH

H OH

CH2

N

O

H

H

OH

H OH

H

FAD

N

N O

O–

H H

O

O

N

N O

H

Oxidation −2 H+ −2 e−

Figure 24.27  The structures of (a) FAD and FADH2 and (b) Q and QH2.

O

N

H

H3C

N

H

FADH2

(a)

O

OH

CH3

O

CH3

CH3

O

CH2 CH

CH3 C

CH2

H

O

Q

Reduction +2 H+ +2 e−

CH3

O

CH3

Oxidation −2 H+ −2 e−

CH3

O

CH2 CH

10

(b)

CH3 C

CH2

H

OH

QH2

10

compound, the monosaccharide glucose, C 6H12O6, in the process of respiration (­Figure 24.28). Respiration is the process by which cells break down molecules, such as glucose, oxidizing them to CO2 and H2O. The first stage is called glycolysis (Figure 24.28a), a process involving ten chemical reactions that occur in the cytosol, the aqueous region inside the cell in which intracellular bodies (organelles) such as the nucleus and mitochondria are s­ uspended. The initial glucose molecule has six carbon atoms. In the first half of glycolysis, this molecule is split into two molecules containing three ­carbon atoms called ­glyceraldehyde-3-phosphate. During this process, two ATP ­molecules are used. Thus, the cell invests two ATP in this process before getting any ATP back. In the second half of glycolysis, the aldehyde group of each glyceraldehyde3-phosphate is oxidized to a carboxylic acid group. An oxidation must always be accompanied by a reduction, in this case, the reduction of NAD+ to NADH. Two ATP are formed from ADP and Pi. The eventual product of this reaction is pyruvate, the deprotonated form of pyruvic acid. Because two molecules of glyceraldehyde3-phosphate go through this process for each original molecule of glucose, there are two pyruvates, two NADH, and four ATP formed in this second half of glycolysis. Because two ATP were invested in the first half of glycolysis, the net gain of ATP directly from glycolysis is −2 (first half) + 4 (second half) = 2 ATP. After glycolysis, pyruvate is transported into the mitochondrion, an organelle in the cell with a double membrane system. The pyruvate undergoes the reactions of the pyruvate dehydrogenase complex (Figure 24.28b). In this process, each pyruvate (three carbon atoms) gets oxidized and loses a molecule of carbon dioxide (which you eventually exhale); simultaneously, NAD+ gets reduced to NADH. The remaining two-carbon atom fragment (an acetyl group) gets attached to a complex molecule called Coenzyme A (CoA) to form acetyl-CoA. Because two pyruvates were formed in glycolysis from each original molecule of glucose (six carbons), the pyruvate dehydrogenase complex results in two acetyl-CoA, two CO2, and two NADH.

24.5 Metabolism

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1233

Process and Highlights of Reaction (The equations are not balanced.)

Reaction Products for Reaction Shown CO2 NADH QH2

Reaction Products Per Molecule of Glucose

ATP

CO2 NADH QH2

ATP

(a) Glycolysis First Half

OH

H

O

H 2C

HO

H

O 2 H

H

HO

H C C

−2

2–

OH

OH

−2

OH

CH2OPO3

H

H glucose Second Half O

glyceraldehyde-3-phosphate

H C

H

C

O–

O C OH

C

CH2OPO32–

2

1

O

2

4

CH3

glyceraldehyde-3-phosphate

pyruvate

(b) Pyruvate Dehydrogenase Complex

O–

O

CoA

C C

O

O

H3C

C

+

CoA

O

C

O

1

1

2

3

2

2

4

6

2

2

6

10

2

4

CH3 pyruvate

acetyl-CoA

(c) Citric Acid Cycle

O –O

O H3C

C

O

C

C

O CH2

oxaloacetate

C

O–

O

O– CoA

CoA

–O

C C

C

C

O C

O–

2 O

C

CH2

O C

CH2

acetyl-CoA

O

oxaloacetate

CH2 HO

O

O–

O

1

1

C O

O–

citrate Subtotals for Glycolysis, Pyruvate Dehydrogenase Complex, and the Citric Acid Cycle (d) Electron Transport Chain and Oxidative Phosphorylation NADH n NAD+ and ½ O2 n H2O

2.5

−1

QH2 n Q and ½ O2 n H2O Grand Totals

−1

25

−10

1.5

−2 6

3 32

Overall Reaction C6H12O6 1 6 O2 n 6 CO2 1 6 H2O

Figure 24.28  Overview of glucose metabolism.  Note that all instances of water being a product or reactant in the reactions are not shown in this table. Overall, there are six molecules of water produced per molecule of glucose fully oxidized.

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The next stage is called the citric acid cycle, also called the tricarboxylic acid (TCA) cycle or the Krebs cycle (Figure 24.28c) that consist of six chemical reactions. In the first step of the process, acetyl-CoA reacts with oxaloacetate to form citrate, the fully deprotonated form of the TCA called citric acid. The other reactions, which include four oxidations, eventually convert the two carbons of the acetyl group to two CO2 and regenerate oxaloacetate. The four oxidations are accompanied by four reductions: three that convert NAD+ to NADH and one that converts FAD to FADH2. The FADH2 is then oxidized back to FAD by the reduction of one Q to QH2. One more ATP is also formed. Recall that two acetyl-CoAs go through the citric acid cycle per molecule of glucose, so this process yields four molecules of CO2, six NADH, two QH2, and two ATP. After the processes of glycolysis, the pyruvate dehydrogenase complex, and the citric acid cycle, all the carbon atoms of glucose have been fully oxidized to CO2, ten NAD+ have been reduced to NADH, two Q have been reduced to QH2, and four ATP have been formed. In the final stage, called the electron transport chain and oxidative phosphorylation (Figure 24.28d), the electron carriers NADH and QH2 get reoxidized to NAD+ and Q. The eventual receptor for the electrons from NADH and QH2 is molecular oxygen, O2. Six O2 molecules get reduced to 12 H2O molecules (6 H2O molecules were used in other steps of glucose metabolism, so the entire process yields a net of 6 H2O molecules). The energy from these oxidation–reduction reactions is used first to pump H+ ions across the inner membrane of the mitochondrion from the interior of the mitochondrion into the space between its two membranes, building up an excess of H+ in the space between the two membranes. This concentration and charge imbalance between the two sides of the inner membrane result in H+ ions spontaneously flowing back into the main part of the mitochondrion through a protein that uses the energy of this movement to synthesize ATP. Each NADH that goes through the electron transport chain eventually yields approximately 2.5 ATP, and each QH2 that enters the chain, yields approximately 1.5 ATP. Thus, a total of around 10 × 2.5 + 2 × 1.5 = 28 ATP are formed in oxidative phosphorylation. Added to the 4 ATP already formed in glycolysis and the citric acid cycle, this means that  a total of about 32 ATP are formed for each molecule of glucose oxidized. While different processes are used for the metabolism of food molecules other than glucose, the basic steps of breaking the molecule down into molecules consisting of two or three carbon atoms followed by the citric acid cycle often occurs. In these reactions, the electron carriers NAD+ and Q get reduced to NADH and QH2. This is then followed by the reoxidation of these electron carriers in the electron transport chain and oxidative phosphorylation to form ATP. The ATP can then be used for the cell’s purposes. Metabolic processes can be quite complex and may at first seem overwhelming to try to learn, but keep in mind that they consist of individual chemical reactions, such as oxidation–reduction reactions, that are governed by the chemical principles you have learned in this and previous chapters. What you have learned throughout your study of introductory chemistry can help you understand these and other processes occurring in living things.

Photosynthesis The sugars used in respiration can be traced back to green plants, where sugars are made by the process of photosynthesis. In photosynthesis, plants carry out the reverse of glucose oxidation—that is, the synthesis of glucose from CO2 and H2O. 6 CO2(g) + 6 H2O(ℓ) n C6H12O6(s) + 6 O2(g)

Green plants have found a way to use light to provide the energy needed to run this endothermic reaction. The key molecule involved in trapping the energy from light in photosynthesis is chlorophyll. Green plants contain two types of chlorophyll: chlorophyll a and

24.5 Metabolism

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R =

Chlorophyll a —CH3 Chlorophyll b —CHO

CH3 CH2 H II

N

H H2C CH

I

N

Mg

N N

III

OO V

IV

H CH3 H CH3

H

H

80

C OCH3

CH2 CH2

H2 C

O C

O

C H H2C

60 C CH3

Absorbance

R

CH3

Chlorophyll a

40

CH2 H2C H2C hydrophobic phytyl side chain

CH CH3 CH2

H2C CH CH3 H2C

Figure 24.29  The structure of chlorophyll and the visible absorbance spectra of chlorophylls a and b.

H2C

Chlorophyll b 20

0 400

500

600

700

Wavelength (nm)

CH2 CH CH3

H3C

chlorophyll b (Figure 24.29). The absorbance spectra of chlorophyll a and chlorophyll b are also shown in Figure 24.29. Notice that these molecules absorb best in the blue-violet and red-orange regions of the visible spectrum. Not much light is absorbed in the green region. When white light shines on chlorophyll, red-orange and blue-violet light are absorbed by the chlorophyll; green light is reflected. Your eyes see the reflected light, which is why plants appear green. The light energy absorbed by the chlorophyll is used to drive the process of photosynthesis.

Applying Chemical Principles 24.1  Polymerase Chain Reaction One night while driving to a cabin in Northern California, Kary Mullis, a chemist at the Cetus Corporation, was musing about how he might be able to develop a method to determine the sequence of a strand of DNA. As he worked on this problem, he realized that his musings had led him instead to a way to copy a sequence of DNA multiple times. Now called the polymerase chain reaction (PCR), the method allows one to amplify a segment of as little as a single copy of doublestranded DNA many times over in a relatively short period of time. Mullis shared the Nobel Prize in Chemistry in 1993 for his work in developing this technique. A piece of double-stranded DNA is heated to a temperature of around 95 °C. At this temperature, the hydrogen bonds holding the two strands together break, and the two strands separate from each other. The reaction mixture is then cooled to a

lower temperature, around 55 °C. At this lower temperature, two small segments (usually 18–30 nucleotides long) of DNA called primers bind to the DNA, one primer per strand. One primer is complementary to the 3′-region of DNA to be amplified on one strand, and the other primer is complementary to the 3′-region of DNA to be amplified on the other strand. The temperature is then raised to around 72 °C where an enzyme called Taq polymerase becomes active. This polymerase, from a species of bacteria that lives at high temperature called Thermus aquaticus (Taq), takes DNA nucleotide triphosphates (dATP, dGTP, dCTP, and dTTP) from the solution and adds them to the primers in the proper order so that new strands of DNA complementary to the original strands are produced. The process began with one double-stranded DNA, but two are now present at the end of the first cycle of PCR. The temperature is

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3’ 5’ Targeted sequence Steps 1 and 2

Heat to 95 °C, cool to 55 °C, add primers in 1000-fold excess 5’ 3’

Primer

Cycle I

Primer Step 3

Heat to 72 °C. Taq DNA polymerase, dATP, dTTP, dGTP, dCTP

2 double-stranded DNA molecules Steps 1’ and 2’

Cycle II

then raised again to 95 °C, and a new cycle is begun. After this second cycle, there will be four double-stranded DNAs present. After three cycles, the number will be eight, and so on with the amount of DNA growing exponentially. In this way, millions of copies of the original DNA can be made fairly quickly. PCR has become commonplace in forensic science where it is used to amplify DNA found at a crime scene so that it can be tested against a suspect’s DNA. It has also been used in tests for HIV infection, paternity tests, and for determining the DNA of organisms long extinct, as well as many other uses. When the COVID-19 pandemic began, there was a need for accurate tests that doctors and patients could use to determine if a person was infected with the virus. One of the most widely used and accurate tests uses a modified version of PCR. The reason for the modification is that PCR can be used to amplify only DNA, but the virus that causes COVID-19 is a retrovirus, a virus that uses RNA, instead of DNA, as its genetic material. After the sample is collected from the patient and sent to the lab for analysis, the RNA is isolated from the sample. An enzyme called reverse transcriptase (RT) is used to make a strand of DNA complementary  to  the RNA, the reverse direction of the flow of information in transcription. Once the strand of DNA has been made by RT, its complement can be synthesized to make double-stranded DNA. The double-stranded DNA can then be amplified by PCR to amounts that can be detected.

Heat to 95 °C, cool to 55 °C

Step 3’ 4 doublestranded DNA molecules

Steps 1” and 2”

1. Theoretically, how many copies of double-stranded DNA could be present after 20 cycles of PCR? 2. How does the fact that DNA molecules separate at 95 °C give evidence that the forces holding together DNA molecules are not covalent bonds? 3. Care must be exercised in selecting good primers to use in PCR. (a) Why are primers often chosen that have a high number of cytosines and guanines in their sequences? (b) Why would a primer sequence of CCCCCCCCCCGGGGGGGGGG not be a good choice? (c) Why is a vast excess of primer used for the amount of DNA to be amplified?

Cycle III

Questions

Step 3” 8 doublestranded DNA molecules etc.

PCR.  Note that at the end of three cycles of PCR, eight copies of the segment of DNA exist.

Think–Pair–Share 1. A generic amino acid with a side chain that is not ionizable can exist in aqueous solution in three different forms. This is because the amino group and the carboxylic acid group can be in either a protonated form (ONH3+, OCOOH) or a deprotonated form (ONH2, COO–). At a pH equal to the pKa of the protonated form of one of these groups, equal concentrations of the protonated form and the deprotonated form exist (recall the Henderson-Hasselbalch equation). Below the pKa, the protonated form of the group is the predominant form,

and above the pKa, the deprotonated form predominates. The pKa of the OCOOH group in a generic amino acid is typically around 2, and that of the ONH3+ group is around 9.5. (a) For a generic amino acid with a side chain that is not ionizable, predict the structure of the predominant species and its net charge at a pH (i) less than 2, (ii) between 2 and 9.5, and (iii) greater than 9.5.

Think–Pair–Share

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(b) If the side chain is ionizable, the situation is more complicated. As an example, consider the amino acid lysine. The protonated amino group on its side chain has a pKa of about 10.4. O +H N 3

CH

C

OH

CH2 CH2 CH2 CH2 NH3+ fully protonated form of lysine

Determine the pH ranges that need to be considered, the predominant structure present for each pH range, and the net charge of lysine in each pH range. 2. Carbohydrate formulas: (a) A monosaccharide is a polyhydroxy aldehyde or ketone that often has the formula Cx(H2O)x. There are no monosaccharides with fewer than three carbon atoms. Why are formaldehyde (CH2O) and acetic acid (C2H4O2) not monosaccharides even though they have formulas corresponding to C(H2O) and C2(H2O)2? (b) The molecular formula of maltose is C12H22O11, which corresponds to C12(H2O)11. Why does this have one less water molecule than would be predicted by Cx(H2O)x? (c) What should be the molecular formula of a trisaccharide formed from glucose monosaccharides? 3. Why are there three nucleotides per codon, rather than just one or two nucleotides per codon?

4. Given the following nucleotide sequence in DNA: 5′–TTGGTATGC–3′ (a) Determine the sequence of the complementary strand of DNA and report this sequence by writing it from its 5′-end to its 3′-end. (b) Determine the sequence of the strand of RNA that would be made in the process of transcription from the original strand of DNA. Report this sequence from its 5′- to its 3′-end. (c) How do your answers for parts (a) and (b) differ? (d) What sequence of amino acids is selected by the sequence of RNA determined in part (b)? All the codons in this problem are listed in Table 24.1, page 1223. 5. In the text, the steps for the complete oxidation of the monosaccharide glucose were considered. Now consider the complete oxidation of the fatty acid, palmitic acid, a ­saturated fatty acid with 16 carbons. An overview of the steps involved is: 1. Activation of the fatty acid to form palmitoyl-CoA. This requires the equivalent of two ATP invested. 2. Oxidation and cleavage of palmitoyl-CoA to form eight acetyl-CoA. In this process, seven Q get reduced to QH 2, and seven NAD+ get reduced to NADH. 3. Each acetyl-CoA enters the citric acid cycle. 4. The NADH and QH2 formed in the preceding steps enter the electron transport chain and oxidative ­phosphorylation process. (a) In what ways are the first two steps similar to the first two processes of glucose metabolism (glycolysis and the pyruvate dehydrogenase complex)? (b) Approximately how many ATP can be formed from the complete oxidation of one molecule of palmitic acid?

Chapter Goals Revisited The goals for this chapter are keyed to specific Study Questions to help you organize your review:

24.1 Proteins • Draw the Lewis structures for the most common naturally occurring amino acids. 1–4.

• Understand that under physiological conditions, amino acids exist as zwitterions in aqueous solution. 1, 2.

• Draw the Lewis structures for short polypeptides. 5–8. • Identify the categories of protein structure (primary, secondary, tertiary, and quaternary). 9, 10.

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24.2 Carbohydrates • Draw the Lewis structures for commonly occurring mono-, di-, and polysaccharides. 11–16.

• Identify the types of linkages between glucose subunits in common polysaccharides. 12, 15.

24.3  Nucleic Acids • Draw the Lewis structures for short strands of DNA and RNA. 19, 20. • Predict the sequence of nucleotides complementary to a given strand of DNA. 21, 22.

• Predict the amino acid selected by a given sequence of three nucleotides in mRNA. 21, 22.

24.4  Lipids and Cell Membranes • Understand that lipids are not very soluble in water and that some lipids consist of molecules in which one end is polar and the other end is nonpolar. 24.

• Recognize the ring system in steroids. 25, 26. • Understand the structure of a lipid bilayer and how materials can be transported across the bilayer. 23.

24.5 Metabolism • Understand that coupling of ATP hydrolysis with a reactant-favored reaction can lead to a product-favored reaction. 40.

• Recognize that NAD+ and NADH, FAD and FADH2, and Q and QH2 are involved in biochemical redox reactions. 29, 30.

• Understand the key stages in the metabolism of glucose. 29, 31, 32, 39, 40. • Contrast the overall reactions for respiration and photosynthesis. 35, 39. • Understand the absorption of light by chlorophyll. 35.

Study Questions denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully worked ­solutions in the Student Solutions Manual.

▲

Practicing Skills Proteins (See Section 24.1 and Example 24.1.) 1. (a) Draw the Lewis structure for the amino acid isoleucine, showing the amino group and the carboxylic acid group in their un-ionized forms. (b) Draw the Lewis structure for the zwitterionic form of isoleucine. (c) Which of these structures will be the predominant form at physiological pH?

2. (a) Draw the Lewis structure for the amino acid tryptophan, showing the amino group and the carboxylic group in their un-ionized forms. (b) Draw the Lewis structure for the zwitterionic form of tryptophan. (c) Which of these structures will be the predominant form at physiological pH? 3. Consider the amino acids alanine, leucine, serine, phen­yl­alanine, lysine, and aspartic acid. Which have polar R groups, and which have nonpolar R groups? Study Questions

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1239

4. Consider the amino acids arginine, cysteine, glutamine, isoleucine, threonine, and tryptophan. Which have polar R groups, and which have nonpolar R groups? 5. Draw Lewis structures for the two dipeptides that contain both alanine and glycine. 6. Do the amino acid sequences: valine–asparagine and asparagine–valine represent the same compound? Explain. 7. Draw the Lewis structure for the tripeptide serine– leucine–valine. 8. Draw the Lewis structure for the tripeptide tyrosine– histidine–glycine. 9. Identify the type of structure (primary, secondary, tertiary, or quaternary) that corresponds to the ­following statements. (a) This type of structure is the amino acid sequence in the protein. (b) This type of structure indicates how different peptide chains in the overall protein are arranged with respect to one another. (c) This type of structure refers to how the polypeptide chain is folded, including how amino acids that are far apart in the sequence end up in the overall molecule. (d) This type of structure deals with structures that arise from hydrogen bonding between amide groups in the protein’s backbone. 10. The β-pleated sheet is a structure that commonly arises in proteins. Part of a β-pleated sheet is shown in the following figure. This type of structure is an example of which level of protein structure: primary, secondary, tertiary, or quaternary? H

O

N

C

C

H H C O

C

R N

H C

C

R

H

O

R

O

H

N

C

H

C R

N H

H

O

N

C

C

H H C O

C

R N

H C

C

R

H

O

R

O

H

N H

C

C R

N H

Carbohydrates (See Section 24.2.) 11. Draw the structural formulas for α-d-glucose and β-d-glucose. 12. Draw the structural formula for two glucose molecules joined by a β-1,4 linkage.

13. The two simplest monosaccharides have the molecular formula C3H6O3. One is glyceraldehyde and the other is dihydroxyacetone. Draw the structural formulas for d-glyceraldehyde and dihydroxyacetone. 14. For each of the following disaccharides, list the monosaccharides from which they are constructed and the type of linkage that joins them: sucrose, lactose, maltose. 15. What is the difference between the chemical structures of amylose starch and cellulose? Does this difference play a role in how these compounds are digested by humans? Explain. 16. What is the major difference in the chemical structures of amylopectin starch and glycogen? Which is found in plants and which in animals?

Nucleic Acids (See Section 24.3 and Examples 24.2 and 24.3.) 17. Ribose, nucleosides, and nucleotides: (a) Draw the structural formula for the sugar β-d-ribose. (b) Draw the structural formula for the nucleoside adeno­sine (it consists of β-d-ribose and adenine). (c) Draw the structural formula for the nucleotide adeno­sine 5′-monophosphate. 18. Deoxyribose, nucleosides, and nucleotides: (a) Draw the structural formula for the sugar β-d-2-deoxyribose. (b) Draw the structural formula for the nucleoside deoxyadeno­sine (it consists of β-d-2-deoxyribose and adenine). (c) Draw the structural formula for the nucleotide deoxyadeno­sine 5′-monophosphate. 19. Draw the structural formula for the tetranucleotide AUGC. 20. Draw the structural formula for the tetradeoxy­ nucleotide CGTA. 21. Given the following nucleotide sequence in DNA: 5′OACGCGATTCO3′: (a) Determine the sequence of the complementary strand of DNA. Report this sequence by writing it from its 5′-end to its 3′-end. (b) Write the sequence (5′–3′) for the strand of mRNA that would be complementary to the original strand of DNA. (c) Assuming that this sequence is part of the coding sequence for a protein and that it is properly lined up so that the first codon of

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this sequence begins with the 5′ nucleotide of the mRNA, write the sequences for the three anticodons that would be complementary to this strand of mRNA. (d) What sequence of amino acids is selected by this mRNA? (The complete genetic code is listed in several sites on the Internet.) 22. Given the following nucleotide sequence in DNA: 5′OTCGTAGGATO3′: (a) Determine the sequence of the complementary strand of DNA. Report this sequence by writing it from its 5′-end to its 3′-end. (b) Write the sequence (5′–3′) for the strand of mRNA that would be complementary to the original strand of DNA. (c) Assuming that this sequence is part of the coding sequence for a protein and that it is properly lined up so that the first codon of this sequence begins with the 5′ nucleotide of the mRNA, write the sequences for the three anticodons that would be complementary to this strand of mRNA. (d) What sequence of amino acids is selected by this mRNA? (The complete genetic code is listed in several sites on the Internet.)

Lipids and Cell Membranes (See Section 24.4.) 23. Sketch a section of a phospholipid bilayer in which a circle represents the polar head group and curvy lines represent the hydrocarbon tails. Label the regions of the bilayer as being polar or nonpolar. 24. If a drop of oleic acid is added to a dish of water, the oleic acid will spread out and form a layer that is one molecule thick on top of the water. (a) Draw the Lewis structure for oleic acid. (b) Label the region of the oleic acid molecule that is polar and the region that is nonpolar. (c) In the layer of oleic acid that forms, which part of the oleic acid molecule points down into the water and which part points out of the water? 25. What structure do all steroids have in common? 26. The structures of the sex hormones testosterone (a male sex hormone) and estrogen (a female sex hormone) are shown in the following figures.



CH3 OH

CH3 OH CH3 O

HO testosterone

estrogen (estradiol)

(a) Compare these two structures. (b) Are these molecules steroids? Explain.

Metabolism (See Section 24.5.) 27. The section about metabolism provided a value for ∆rH° for the oxidation of one mole of ­glucose (−2803 kJ/mol-rxn). Using ∆f H° values at 25 °C, verify that this is the correct value for the equation C6H12O6(s) + 6 O2(g) n 6 CO2(g) + 6 H2O(ℓ) ∆f H°[C6H12O6(s)] = −1273.3 kJ/mol

28. The chemical equation for the fermentation of glucose into ethanol is C6H12O6(s)  n 2 C2H5OH(ℓ)  + 2 CO2(g)

Using ∆f H° values at 25 °C, calculate ∆rH° for this reaction. (See Question 27 for ∆f H° for glucose.) 29. Consider the following reaction: NADH + H+ + 1⁄2 O2 n NAD+ + H2O

(a) Which species (NADH, H+, or O2) undergoes oxidation? (b) Which species undergoes reduction? (c) Which species is the oxidizing agent? (d) Which species is the reducing agent? 30. The body processes ethanol by first converting it to acetaldehyde in a reaction catalyzed by the enzyme alcohol dehydrogenase. alcohol dehydrogenase CH3CH2OH + NAD+ 888888888888888n CH3CHO + NADH + H+

In the body, acetaldehyde can cause headaches and nausea and is one of the causes of hangovers. ­Eventually, the acetaldehyde is oxidized to the acetate ion and then converted to carbon dioxide and water. In the reaction of ethanol with NAD+, which species (ethanol or NAD+) undergoes oxidation? Which species undergoes reduction?

Study Questions

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31. ∆rG°′ for the oxidation of one mole of glucose is –2850 kJ/mol-rxn. ∆rG°′ for the formation of ATP from ADP and Pi is +30.5 kJ/mol-rxn. If the process of glucose oxidation and ATP formation were 100% efficient, how many moles of ATP could be formed from one mole of glucose? The text states that about 32 molecules of ATP are formed per molecule of glucose oxidized. Is the process of forming ATP in metabolism 100% efficient? What is the percent yield of ATP in this process?

35. Photosynthesis: (a) Calculate the enthalpy change for the ­production of one mole of glucose by the process of photosynthesis at 25 °C. ∆f H° [glucose(s)] = −1273.3 kJ/mol 6 CO2(g) + 6 H2O(ℓ) n C6H12O6(s) + 6 O2(g)

(b) What is the enthalpy change for producing one molecule of glucose by this process? (c) Chlorophyll molecules absorb light of various wavelengths. One wavelength absorbed is 650 nm. Calculate the energy of a photon of light having this wavelength.

32. Just as there are standard free energies under biochemical standard conditions (∆rG°′), there are also standard reduction potentials under biochemical standard conditions (E°′). Half-reaction 1⁄2

(d) Assuming that all of this energy goes toward providing the energy required for the photo­ synthetic reaction, can the absorption of one photon at 650 nm lead to the production of one molecule of glucose, or must multiple photons be absorbed?

E°′

O2 + 2 H + 2 e n H2O +

–

0.815 V

Q + 2 H + 2 e n QH2 +

–

0.045 V

(a) Write the overall balanced chemical equation for the reaction involving these two halfreactions that is predicted to be product-favored at equilibrium. (b) Calculate the electrochemical potential under biochemical standard conditions (E°′cell) for this reaction. (c) Calculate the value of ∆rG°′ for this reaction. (d) Do the values for E°′cell and ∆rG°′ confirm that this reaction is product-favored at equilibrium?

36. Transcription and translation: (a) According to the genetic code in Table 24.1, which amino acid is selected by the mRNA codon GAA? (b) What is the sequence in the original DNA that led to this codon being present in the mRNA? (c) If a mutation occurs in the DNA in which a G is substituted for the nucleotide at the second position of this coding region in the DNA, which amino acid will now be selected? 37. How many codons are possible? All of these codons are used in molecules of mRNA and there are only 20 amino acids normally encoded (plus start codons and stop codons). Is there exactly one codon per amino acid, must some codons select for more than one amino acid, or must there be some amino acids for which there is more than one codon?

General Questions These questions are not designated as to type or location in the chapter. They may combine several concepts. 33. Draw two Lewis structures for the dipeptide alanine–isoleucine that show the resonance ­structures of the amide linkage.

38. There are 41 = 4 mononucleotides of DNA, there are 42 = 16 possible dinucleotides, and so on. If a segment of DNA were completely random, how many nucleotides long would it need to be in order to have one possible sequence for each person on Earth (currently about 8.0 billion people)?

34. What is the equilibrium constant for the conversion of α-d-glucose to β-d-glucose in water at room temperature (assume that there is no glucose present in the open-chain structure)? H HO

H

OH

4

5

HO

3H

H 2

H

O

HO 1

OH

H

OH

-D-glucose 37%

OH

4

5

HO

3H

H 2

O 1

OH

OH

H

H -D-glucose 63%

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39. In the process of respiration C6H12O6(s) + 6 O2(g) n 6 CO2(g) + 6 H2O(ℓ)

which molecule undergoes oxidation? Which ­molecule undergoes reduction? 40. The first step of the metabolic process known as glycolysis is the conversion of glucose to glucose-6-phosphate. This process has a positive value for ∆rG°′. Glucose + Pi  n Glucose-6-phosphate + H2O  ∆rG°′ = +13.8 kJ/mol-rxn

This reaction is coupled to the hydrolysis of ATP ATP + H2O  n ADP + Pi  ∆rG°′ = −30.5 kJ/mol-rxn

What is the sum of these two equations and the value of ∆rG°′ for the coupled reaction? Is the ­coupled reaction product-favored at equilibrium?

In the Laboratory 41. The enzyme lipase catalyzes the hydrolysis of esters of fatty acids. The hydrolysis of p-nitrophenyloctanoate was followed by measuring the appearance of p-nitrophenol in the reaction mixture:

sample of insulin was dissolved in enough water to make 100. mL of solution. This solution was determined to have an osmotic pressure of 21.8 mm H2O at 25°C. What is the molar mass of insulin? (Hint: One mm Hg corresponds to 13.6 mm H2O at this temperature.)

Summary and Conceptual Questions The following questions may use concepts from this and ­previous chapters. 43. Which of the amino acids listed in Figure 24.2 is not chiral? 44. (a) What type of interaction holds DNA’s doublehelical strands together? (b) Why would it not be good for DNA’s doublehelical strands to be held together by covalent bonds? 45. Do the DNA sequences ATGC and CGTA represent the same compound? Explain.

46. For many chemical reactions in the laboratory, a percent yield of the correct product of 95% is considered very good. Many biochemical reactions, however, require a much greater percent yield of the O correct product. (a) Assume that there is a process that replicates C O (CH2)6 CH3 OH DNA with only 95% accuracy for each nucleotide O added and that you wish to make complementary copies of identical strands of DNA 10 nucleotides + H2O + C HO (CH2)6 CH3 long. What fraction of the molecules produced would have the correct sequence of nucleotides? NO2 NO2 (b) Many naturally occurring DNA polymerases, p-nitrophenol enzymes that catalyze the replication of DNA, have an accuracy much greater, often being The following data were obtained at 30 °C: 99.999999% accurate. If an enzyme with this accuracy constructed a 10-nucleotide sequence Substrate of DNA, what fraction of the molecules would Concentration Reaction Rate have the correct sequence? (mol/L) (mol/L ∙ min) 47. (a) Describe what occurs in the process of 1.0 × 10−5 6.3 × 10−6 transcription. 1.4 × 10−5 8.8 × 10−6 (b) Describe what occurs in the process of translation. 2.0 × 10−5

1.2 × 10−5

5.0 × 10−5

2.6 × 10−5

6.7 × 10−5

3.1 × 10−5

This reaction obeys Michaelis-Menten kinetics (­Section 14.6). Determine the value of Ratemax for this reaction using the method described in Study Question 14.53. 42. Insulin is a protein important in the metabolism of sugar. Its molar mass can be determined by means of an osmotic pressure experiment. A 50.0-mg



48. Which of the following statements is/are true? (a) Breaking the POO bond in ATP is an exothermic process. (b) When the phosphate group is cleaved off ATP, making a new bond between the phosphorus atom and the OH group of water is an exothermic process. (c) Breaking bonds is an endothermic process. (d) The energy released in the hydrolysis of ATP may be used to run endothermic reactions in a cell.

Study Questions

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John C. Kotz

25

Environmental Chemistry—Earth’s ­Environment, Energy, and Sustainability

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C hapt e r O ut li n e 25.1 The Atmosphere 25.2 The Aqua Sphere (Water) 25.3 Energy 25.4 Fossil Fuels 25.5 Environmental Impact of Fossil Fuels 25.6 Alternative Sources of Energy 25.7 Green Chemistry and Sustainability

We all live on this one planet, Earth. The chemistry of its environment is of ­crucial importance not only in your own daily life but in the future of all humanity. The actions of humans play a role in the health of the entire planet. It is important for you to understand the environment and its chemistry so that you can make wise environmental choices and help shape humanity’s response to environmental issues. This chapter first examines the chemistry of the atmosphere and the aqua sphere. Next, after a discussion of energy, fossil fuels (coal, petroleum, and ­natural gas) are considered. The widespread use of fossil fuels has large impacts on the environment, including one of the key environmental issues facing humans today, ­climate change. In 1988, the United Nations established the Intergovernmental Panel on Climate Change (IPCC) to provide member states with s­ cientific ­information to develop climate policies. The membership has since grown to almost 200 countries, from Andorra to Zimbabwe. Scientists worldwide evaluate a continuing stream of scientific papers “to provide a comprehensive summary of what is known about climate change, its impacts and future risks, and how adaptation and mitigation can reduce those risks.” To date, six assessment reports have been issued; the sixth was released in 2023. This chapter will provide you with the chemistry needed to understand some of the information coming from those IPCC reports. Some mitigation efforts are then considered as well as some alternatives to ­fossil fuels. The textbook concludes by revisiting a topic first introduced in Chapter 1, the effort by many chemists today to be more environmentally responsible, green chemistry.

Critical Warming Threshold­  On March 20, 2023 the IPCC reported that global average temperatures are estimated to rise 1.5 degrees Celsius above preindustrial levels sometime around “the first half of the 2030s,” as humans continue to burn coal, oil, and natural gas. This 1.5 °C increase is considered by many to be a critical threshold in Earth's warming. B. Plumer, “Earth is Nearing the Tipping Point for a Hot Future,” New York Times, March 20, 2023, A.1.

◀ Our environment.  Among other topics, this chapter is about clean air and water and about

forms of environmental pollution. This photograph was taken after a storm over a sea marsh on the coast of South Carolina. Here and around the world, people are concerned about the health of sea marshes as they protect our coasts and are the home and breeding ground for many forms of sea life. Among many concerns about the health of marshes is climate change, which has led to a significant increase in sea level in the last 100 years.

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25.1 The Atmosphere Goal for Section 25.1 • Know the important chemical components of Earth’s atmosphere, how those components affect the Earth, and how they can be affected by pollutants.

The composition of the atmosphere is being changed by human activities, and the resulting effects on climate are well established. It may be the least weighty part of the environment, but it is complex and has a disproportionately large influence on our planet. The major components of the atmosphere are listed in Table 25.1. These values are for dry air at sea level, but air also contains water vapor, which can vary considerably from place to place and from day to day. It can be as high as 40,000 ppm (parts per million) but is generally about half this value or less. There are many more components of the atmo­sphere than those listed in the table. These include trace amounts of chemicals emitted by plants, animals, and microbes as well as from burning coal, oil, and natural gas to generate electricity, from vehicle exhausts, from industry, and from our day-to-day activities. You might imagine that air is a homogeneous mixture of gases that is constantly stirred by the motion of the planet and the heat of the sun. In fact, mixing is far from complete. There is relatively little exchange of air between the north and south regions of the planet, and there are distinct layers in the atmosphere (“A Closer Look: The Earth’s Atmosphere”). For humans, the most important layer is the lowest layer—the troposphere, which goes up to about 7 km at the poles and 17 km at the equator. Above the troposphere is the stratosphere, which extends to about 30 km, getting less dense as altitude increases. There are even higher regions, but the troposphere and the stratosphere will be the focus of this chapter.

Table 25.1

The Gases of the Atmosphere*

Gases

Concentration (ppm)

Nitrogen (N2)

780,840

Oxygen (O2)

209,460

Argon (Ar)

  9340

Carbon dioxide (CO2)

   416

Neon (Ne)

    18

Helium (He)

     5.2

Methane (CH4)

     1.9

Krypton (Kr)

     1.1

Hydrogen (H2)

     0.5

Ozone (O3)

     0.4

Dinitrogen monoxide (nitrous oxide, N2O)

     0.3

Carbon monoxide (CO)

     0.1

Xenon (Xe)

     0.09

Radon (Rn)

traces

*Data on gas concentrations refer to relative numbers of particles (in parts per million) in dry air.

1246 Chapter 25 / Environmental Chemistry—Earth’s ­E nvironment, Energy, and Sustainability Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

110 0.0001 100

Thermosphere 0.001

0.01

90 Mesopause

80 70

0.1

1

Mesosphere

60 50

Stratopause

re

u rat

e mp

40

Te

10

Height (km)

Earth’s atmosphere is a mixture of gases in more or less distinct layers with widely differing temperatures. Up through the troposphere, there is a gradual decline in temperature (and pressure) with altitude. The temperature climbs again in the stratosphere due to the absorption of energy from the sun by stratospheric ozone, O3. The pressure drops steadily as altitude increases while the temperature varies dramatically. This is an illustration of the difference between temperature and thermal energy. The temperature of a gas reflects the average kinetic energy of the molecules of the gas, whereas the thermal energy present in an object is the total kinetic energy of the molecules. In the thermosphere, the molecules present have a very high temperature, but the thermal energy is exceedingly small because there are so few molecules. Gases within the troposphere are well mixed by convection. Pollutants that are produced on Earth’s surface can rise to the stratosphere, but the stratosphere acts as a thermal lid on the troposphere and prevents significant mixing of polluting gases into the stratosphere and beyond. Nonetheless, small amounts of pollutants such as chlorofluorocarbons do enter the stratosphere and contribute to destruction of ozone. The pressure of the atmosphere declines with altitude, and so the partial pressure of O 2 declines. The figure shows why climbers have a hard time breathing on Mount Everest, where the altitude is 29,028 feet (8848 m) and the O 2 partial pressure is only 29% of the sea level partial pressure. With proper training, a climber could reach the summit without supplemental oxygen. However, this

Pressure (millibars)

A Closer Look

The Earth’s Atmosphere

Ozone region

Stratosphere

30

Ozone Maximum 20 100 Tropopause

Mount Everest

10

Troposphere 1000 −100

−80 −120

−60 −80

−40

−20

−40

0

0

20 40

(°C)

0

80 (°F)

Temperature

same feat would not be possible if Mount ­ verest were farther north. Earth’s atmoE sphere thins toward the poles, and so the O 2 partial pressure would be even less if Everest’s summit were in the northern part of North America, for example.

Reference • G. N. Eby, Environmental Geochemistry, Boston, MA: Cengage Learning/Brooks/Cole, 2004.

The total mass of the atmosphere is estimated to be around 5.15 × 1015 metric tons or 5.15 × 106 gigatons. Three quarters of the mass of the atmosphere is in the troposphere, and its average temperature is 14 °C. It can be as low as −89 °C (the lowest temperature ever recorded, in Antarctica in 1983) or as high as 56.7 °C (the highest temperature ever recorded, in Death Valley, California, in 1913). Another important property of the atmosphere is the pressure it exerts, which at sea level is around 101 kPa (1 atm) and about 25 kPa (0.25 atm) at 11 km (about 35,000 feet), an altitude used by passenger aircraft. Some of the gases of the atmosphere are unreactive, notably the noble gases: helium, neon, argon, krypton, and xenon. Helium and argon are added continuously to the atmosphere because they are products of radioactive decay—helium

Gigatons  are often used in environmental studies because the masses involved are so large. A metric ton is 1000 kg, but a gigaton is a billion metric tons, that is, 109 metric tons.



25.1  The Atmosphere

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comes from alpha particles emitted by naturally occurring elements such as ­uranium, and argon is produced by decay of 40K, a long-lived radioactive isotope of potassium. The other gases of the atmosphere, particularly oxygen, nitrogen, and carbon dioxide, are chemically reactive and can be changed into other molecules by reactions driven by lightning, ultraviolet rays from the sun, or the influence of living things. The atmosphere has undergone significant changes over the lifetime of the planet, and it is still changing. The well-publicized increase of CO2 and methane in the atmosphere and the decrease of ozone in the stratosphere are probably the best-known changes, but human activity is contributing new chemicals to the environment as well.

Nitrogen and Nitrogen Oxides Nitrogen in the atmosphere, N2, arose from the out-gassing of Earth when it was simply a molten mass. Even today, some nitrogen escapes when volcanoes erupt. Every living thing on the planet has nitrogen in its cells in proteins and in DNA and RNA, among a myriad of other compounds. Gaseous N2 is not the direct source of this element in living beings, other than some soil bacteria and cyanobacteria. To be taken up by plants, nitrogen must be in the form of a usable nitrogen compound, such as ammonia or ammonium or nitrate ions. In the environment, atmospheric N2 molecules are converted to other compounds by a number of natural processes. Some of this conversion is carried out within the nitrogen cycle, shown in Figure 25.1, which charts the processes that convert between various N-containing molecules and ions in nature. Humans also influence the nitrogen cycle. On a global scale, the conversion of nitrogen to its various compounds is accomplished to the extent of around 50% by biological processes and 10% by lightning. Another 30% comes from burning fossil fuels and 10% from manufactured fertilizers. Indeed, according to recent research, nitrogen inputs to the environment from fossil fuel combustion and large-scale biomass burning accelerated greatly beginning about 1895, and another shift was seen beginning about 1970 with significant increases in nitrogen fertilizer production. Various nitrogen oxides are also present in the atmosphere, among them ­dinitrogen monoxide (nitrous oxide, N2O), nitrogen monoxide (nitric oxide, NO), and nitrogen dioxide (NO2). Collectively, nitrogen oxides are referred to as NOx compounds. Although the nitrogen cycle does produce some of these oxides in the atmosphere, the majority of NOx molecules are produced by human activities.

Figure 25.1  The nitrogen cycle.  ​The nitrogen cycle involves nitrogen fixation by soil bacteria or, in aquatic environments, by cyanobacteria. The NH4+ ions produced are converted to nitrate ions, the main form of nitrogen absorbed by plants. Nitrogen is returned to the atmosphere by denitrifying bacteria, which convert nitrate ions to N2.

Nitrogen in atmosphere (N2) Atmospheric fixation (NO2)

Animal Waste

Plant Matter

Decomposers Bacteria/Fungi A SSIM IL AT IO N

Denitrifying bacteria

A M M O N IF IC ATIO N

Ammonia (NH3) Ammonium (NH4+) N IT R IF IC ATIO N

Fertilizers (NH4+), (NH3)

Nitrogen-fixing bacteria (legume root nodules)

Nitrites (NO2–) N IT R IF Y IN G BAC T ER IA

Nitrates (NO3–)

1248 Chapter 25 / Environmental Chemistry—Earth’s ­E nvironment, Energy, and Sustainability Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

The most abundant nitrogen oxide in the atmosphere, dinitrogen monoxide (N2O), is the least reactive. Its concentration in the atmosphere has increased from a value of 280 ppb in 1900 to 330 ppb in 2019. The other nitrogen oxides make up only 0.05 ppb of the atmosphere. Although 100 million tons of NO and NO2 are generated by human processes annually, the concentration of these compounds in the atmosphere is low because they are quickly washed out of the atmosphere by rain (where they contribute to acid rain). But, NOx compounds can linger in certain situations, particularly over cities in sunny climates. There they react with hydrocarbons in the atmosphere, such as traces of unburned fuel, to produce irritating photochemical smog. Among the components of smog are PANs, peroxyacyl nitrates. PANs are formed by photochemical (lightinduced) reactions in smog involving reactions of organic compounds with O3, and NOx. They are toxic and irritating. At low concentrations, they irritate the eyes, but at higher concentrations they can cause more serious damage to both animals and vegetation. They are relatively stable; thus, it is possible they can persist for some time and travel considerable distances from where they are formed.

O H 3C

C

O

O

+

N

O

O− Peroxyacetyl nitrate (PAN), an air pollutant.  PAN is an oxidant more powerful than ozone and is a lachrymator (it makes your eyes tear). It is formed in the atmosphere when ethanol is used as a fuel. If ethanol becomes more widely as a fuel, its formation could become more of a problem.

Oxygen Over the lifetime of the planet, plant life has had a dramatic effect on the atmosphere, changing it from a reducing to an oxidizing one, that is, from there being no oxygen present to one where O2 is the second most abundant substance in the atmosphere. This change meant that life had to change. Species that could not live in the presence of oxygen either died out or were relegated to regions where oxygen is blocked out. The concentration of oxygen in the air is now midway between two extremes that would make life on Earth impossible for humans: below 17% we would suffocate, and above 25% all organic material would burn easily. The total mass of oxygen in the atmosphere is a million gigatons. Even though the burning of 7 gigatons of fossil fuel carbon per year consumes 18 gigatons of oxygen, this makes almost no discernible difference to the amount of oxygen in the atmosphere. Oxygen is a by-product of plant photosynthesis. Carbon dioxide is the source of the carbon that plants need and that they capture from the air and turn into carbohydrates such as glucose (C6H12O6) by photosynthesis. The overall chemical reaction is 6 CO2 + 6 H2O n C6H12O6 + 6 O2

and the net result is the release of an oxygen molecule into the atmosphere for each CO2 absorbed. Oxygen molecules released by photosynthesis remain in the atmosphere, on average, around 3000 years before being consumed through respiration in living beings or through other oxidation reactions. Blue-green algae, or cyanobacteria (Figure 25.2), first began producing oxygen as long ago as 3.5 billion years. But mysteriously, hundreds of millions of years

Michael Abbey/Science Source

Figure 25.2 Cyanobacteria. ​ The importance of cyanobacteria, also known as blue-green algae, lies in their production of oxygen by photosynthesis and their participation in the nitrogen cycle.



25.1  The Atmosphere

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1249

elapsed before a significant amount of oxygen was present in the atmosphere. Astrobiologists are not certain why this occurred, but it does seem plausible that the oxygen first produced did not remain in the atmosphere because it reacted with metals, especially converting abundant iron to iron(II) and iron(III) oxides. (Iron is the fourth most abundant element in Earth’s crust, where it is found primarily in the form of iron(III) oxides.) The concentration of oxygen remained low until about 2 billion years ago when it rose relatively rapidly to around 20% and the first land plants started to appear. There are three naturally occurring isotopes of oxygen: oxygen-16 accounts for 99.76% of O atoms, oxygen-17 for a mere 0.04%, and oxygen-18 for 0.2%. The ratio of oxygen-18 to oxygen-16 in the world’s oceans has varied slightly over geological time, and this has left an imprint on parts of the environment, providing ­evidence of past climates. When the world is in a cooler period, water molecules with the lighter isotope, oxygen-16, evaporate more readily from the oceans than their heavier oxygen-18 counterpart. Thus, precipitation in the form of snow is slightly richer in oxygen-16, and the water that remains in the oceans is slightly richer in oxygen-18. Marine creatures during such periods lay down shells that have more oxygen-18 than expected, and these are preserved in sediments. Analyzing the ratio of the two isotopes in such deposits reveals the cycle of global cooling and warming that has characterized the past half million years with its five ice ages.

Ozone Ozone (O3), an allotrope of oxygen, plays a key role for life on this planet, but it is also a threat. In the troposphere, it is a pollutant, whereas in the stratosphere it acts as a shield, protecting the planet from damaging ultraviolet rays from the sun. This shield is known as the ozone layer (Figure 25.3). Ozone is formed in lightning storms, so there is a natural low level of ozone in the air, about 0.02 ppm. In summer, however, the level can increase to 0.1 ppm or more as a result of sunlight acting on the nitrogen dioxide emitted by vehicles. The ultraviolet component of sunlight causes the dissociation of NO2 to NO and Figure 25.3  The ozone layer.  ​About 90% of the ozone in the atmosphere is contained in the stratosphere (roughly 15–50 km above the surface). Concentrations there range from 2 to 8 ppm.

Ozone in the Atmosphere 35

Altitude (kilometers)

30 Stratospheric ozone

25 Ozone layer 20

15

10 Ozone increases from pollution

Tropospheric ozone

5

0 Ozone concentration

1250 Chapter 25 / Environmental Chemistry—Earth’s ­E nvironment, Energy, and Sustainability Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

O atoms. The resulting O atoms then react with O2 (in the presence of a third ­molecule acting as an energy sink) to produce ozone. NO2 + energy (λ < 240 nm) n NO + O O + O2 n O3

Because ozone damages the lungs, a legal limit for exposure to ozone in workplaces, 0.1 ppm, has been established. Some growing plants are also susceptible to the gas, and even though they do not show visible signs of stress, their growth is reduced in proportion to the level of ozone in the air. In contrast to ozone in the troposphere, its presence in the stratosphere is vital to the planet because the molecules absorb high-energy ultraviolet radiation before it reaches Earth’s surface. Ozone is formed in the stratosphere when radiation with wavelengths shorter than 240 nm interacts with O2 molecules and cleaves them into two O atoms. An O atom can then combine with another O2 molecule to produce an ozone molecule. O2 + energy uv O + O O + O2 n O3

Ozone is different than oxygen in that the oxygen–oxygen bonds in ozone can be cleaved by less energetic light than the oxygen–oxygen bond in O2. (The ­oxygen–oxygen bonds in ozone have a bond order of 1.5 whereas it is 2 in the O2 molecule.) Light of 330 nm is sufficient to decompose O3 to O2 and O atoms. O3 + energy (λ < 330 nm) uv O2 + O O + O3 n 2 O2

Without the ozone layer, higher energy radiation capable of harming living cells would penetrate to Earth’s surface. An increase in this ultraviolet radiation would lead to an increased incidence of skin cancer and cataracts and suppression of the human immune response system. Damage to crops and marine phytoplankton and weathering of plastics would also result from increased levels of ultraviolet radiation.

The Ozone Hole From the late 1800s until about 1930, ammonia (NH3), chloromethane (CH3Cl), and sulfur dioxide (SO2) were widely used as refrigerants. However, leakage from refrigerators caused several fatal accidents in the 1920s, so several American corporations ­collaborated on the search for a less dangerous fluid. In 1928, Thomas Midgley, Jr., and his coworkers discovered dichlorodifluoromethane (CCl 2F2), a member of a large family of compounds called chlorofluorocarbons (CFCs). The two compounds shown here are CFC-114 (C2Cl2F4) and CFC-12 (CCl2F2).

These compounds had exactly the physical and chemical properties needed for a refrigerant: appropriate critical temperatures and pressures, no toxicity, and apparent chemical inertness. The uses of CFCs grew dramatically, not only in air conditioning and refrigeration equipment but also in applications such as propellants for aerosol cans, foaming agents in the production of expanded plastic foams, and inhalers for asthma sufferers.

Thomas Midgely, Jr., 1889–1944  Midgely was initially lauded for two important developments: leaded gasoline and chlorofluorocarbons, but both ultimately caused major environmental problems. Toxic tetraethyllead was added to gasoline for many years because it helped automobile engines run better, but it released toxic lead into the air. It was phased out starting in the 1970s when catalytic converters began to be used. Chlorofluorocarbons were introduced to replace dangerous fluids in refrigerators, but the new materials have had severe environmental consequences. S. Johnson, “The Man Who Broke the World,” New York Times Magazine, March 15, 2023, page 36.



25.1  The Atmosphere

dichlorotetrafluoroethane

dichlorodifluoromethane

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1251

Unfortunately, the properties that made CFCs so useful also led to environmental problems. CFCs are unreactive in Earth’s troposphere, which allows them to remain there for hundreds of years. Over time, however, they slowly diffuse into the stratosphere, where they are decomposed by solar radiation. CF2Cl2(g) + UV radiation n ·CF2Cl(g) + ·Cl(g)

The Cl atoms released in the decomposition of CFCs and other chlorine-­containing compounds can then destroy large numbers of ozone molecules. This occurs because Cl atoms catalyze the decomposition of O3 molecules by producing the ­radical ­species ClO (Step 1) that in turn intercepts O atoms (Step 2) from the r­ adiation-driven decomposition of O3 molecules (Step 3). Notice that chlorine atoms are regenerated (Step 2) and can start the process again with another ozone molecule. Step 1

O3(g) + ·Cl(g) n ·ClO(g) + O2(g)

Step 2

·ClO(g) + O(g) n ·Cl(g) + O2(g)

Step 3 Net reaction:

O3(g) + solar radiation n O(g) + O2(g) 2 O3(g) n 3 O2(g)

Figure 25.4  The ozone hole over the Antarctic continent.  ​ During the Antarctic winter, aerosols of HCl and ClONO2 freeze and accumulate in the polar stratospheric clouds. During the Antarctic spring, these crystals melt, and Cl and ClO radicals are rapidly formed and lead to a depletion of stratospheric ozone over the continent. Depletion is greatest in the black to deep blue zones (below 220 Dobson units). [The numbers on the color scale are Dobson units (DU), a unit of measurement for the total amount of ozone in the atmosphere above a point on the Earth’s surface. One DU is equivalent to a layer of pure ozone 0.01 mm thick at standard temperature and pressure.] https:// www.climate.gov/news-features /understanding-climate/2021antarctic-ozone-hole-larger-average

1252 Chapter 25 / Environmental Chemistry—Earth’s ­E nvironment, Energy, and Sustainability Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

NOAA Climate.gov

When repeated over and over, this sequence of reactions leads to the destruction of many ozone molecules per chlorine atom. In 1985, three British scientists of the British Antarctic Survey discovered that there was a significant reduction in the ozone concentration over the Antarctic continent in the late winter and early spring, which has since been referred to as the ozone hole (Figure 25.4). In 1986, a team from the U.S. National Center for Atmospheric Research went to Antarctica to investigate the ozone hole and found

Chemistry in Your Career

Jamil Kusiima

Jamil Kusiima Jamil Kusiima is an independent environmental consultant in Kampala, Uganda. “My work involves assessing potential environmental impacts of development projects … and proposing how to mitigate them. This includes assessing the baseline environmental conditions (air quality, water quality, soils, vegetation, ecology, etc.) and predicting using scientific models what could happen to the environment if a new project is implemented.” Kusiima draws on his scientific education (B.S. in Chemistry and M.S. in Environmental

Engineering) every day in his work. Such expertise is especially needed in low-income countries, like Uganda, where water sources can be contaminated with heavy metals such as arsenic and bacteria such as Escherichia coli. To design an effective water treatment system, you must understand the chemistry involved. “For example, if you reduce the pH, will the heavy metals become insoluble and come out as a precipitate? If yes, that’s good. But how will you then dispose [of] the precipitate, which could be harmful to the environment?”

that it had higher levels of the reaction intermediate, ·ClO, than expected in the stratosphere. The ozone hole occurrence is tied not only to ·Cl and ·ClO radicals, but also to NOx chemistry. Ozone destruction over Antarctica is partly affected in certain seasons by the combination of ·ClO and ·NO2 to give the compound ClONO2. ·ClO(g) + ·NO2(g) n ClONO2(g)

In the Antarctic winter this compound and others such as HCl are frozen in stratospheric clouds. In the spring, at warmer temperatures, the problem begins when reactions such as the following occur: ClONO2(g) + H2O(g) n HOCl(g) + HNO3(g) ClONO2(g) + HCl(g) n Cl2(g) + HNO3(g)

Both HOCl and Cl2 can be decomposed by solar radiation to give ·Cl atoms, which then lead to ozone destruction. Because of the damage caused to the stratospheric ozone layer by CFCs and related compounds, the United States banned the use of CFCs as aerosol propellants in 1978, and 68 nations had followed suit by 1987. In 1990, the United States and 140 other countries agreed to a complete halt in CFC manufacture as of December, 1995. The good news is that the ozone layer depletion is now less than observed in the 1980s. Experts estimate the layer over Antarctica will return to the pre-1980 condition around 2070.

Carbon Dioxide, CO2 Carbon is essential to life as we know it because only carbon has the ability to form stable compounds consisting of long chains and rings of atoms. This is the basis of the structures for many compounds that comprise living cells. The food you eat—carbohydrates, oils, proteins, and fiber—is made of carbon compounds, and this carbon eventually returns to the atmosphere as CO2 as part of the natural carbon cycle (Figure 25.5). This cycle moves over 200 gigatons of carbon every year between various compartments of the terrestrial ecosphere and so rules the tempo of life on Earth. Decades ago scientists began to notice increasing levels of CO 2 in the atmosphere. The Intergovernmental Panel on Climate Change (IPCC, page 1245) follows this increase closely. In the six IPCC Assessment Reports issued to date, a ­significant increase in the atmospheric carbon dioxide concentration has been

Carbon Dioxide  Since the first edition of this book was published in 1987, the global CO2 level has risen from 351 ppm to as high as 420.99 ppm. Along with methane, this has had a significant effect on the environment (Section 25.5).



25.1  The Atmosphere

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1253

Atmospheric CO2 photosynthesis

burning fossil fuels

aerobic respiration

diffusion between atmosphere and ocean

Land food webs

Dissolved carbon in ocean Fossil fuels

death, burial, compaction over millions of years

Earth’s crust

Marine organisms

sedimentation

Figure 25.5  The carbon cycle.  Through natural processes (and human activities) carbon is continuously cycled between the atmosphere, the land, and the oceans. Starr: Biology: The Unity and Diversity of Life, 14e (Cengage ©2016) ISBN 9781305073951; Figure 46.13, page 828.

reported, from around 280 ppm in 1750 to 420.99 in June 2022. (Figure 25.6 shows the increase in atmospheric CO2 over a 50-year period.) The fourth IPCC report stated that “Today’s CO2 concentration has not been exceeded during the past 420,000 years and likely not during the past 20 million years. The rate of increase over the past century is unprecedented, at least during the past 20,000 years.” As described in Section 25.5, there has been significant debate about the o ­ rigin of this increase and its effects on the planet’s climate.

Methane, CH4

140

Increase of CO2 at Mauna Loa since pre-industrial times

120

50 40

100 30

80 60

20

Percent

the graph, the left y-axis label of 0 ppm represents a 0 ppm increase over 280 ppm, and the right y-axis label of 0% represents a 0% increase from 280 ppm. The baseline of 280 ppm was chosen to represent the CO2 concentration in pre-industrial air based on a value measured in an ancient ice core in Antarctica. In June 2022, the atmospheric concentration of CO2 was 420.99 ppm (an increase of over 4 ppm from June 2020). Source: https://gml.noaa .gov/ccgg/trends/weekly.html

Methane is formed and released into the atmosphere in enormous amounts by biological processes (Figure 25.7). Some methane is produced in nature by ­anaerobic bacteria (that is, those that do not use oxygen) at the bottom of lakes and swamps, in termite mounds, and in the guts of animals such as cows and humans.

CO2 mole fraction (ppm)

Figure 25.6  Increase in CO2 concentration at Mauna Loa, Hawaii, over a 50-year period relative to a pre-industrial era concentration of 280 ppm.  On

40 10

20 Preindustrial value: 280 ppm 0

1980

1990

2000

2010

2020

0

1254 Chapter 25 / Environmental Chemistry—Earth’s ­E nvironment, Energy, and Sustainability Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Wetlands

Figure 25.7 Sources of atmospheric methane.  ​ A substantial amount of atmospheric methane comes from the extraction of fossil fuels but even more from wetlands, a natural source. Source: © nature, https://www.nature.com/articles /d41586-022-00312-2

161.6 (million metric tons per year)

Fossil-fuel extraction

129.5

Livestock

103

Landfill and agricultural waste

72.9 37.5

Natural geological seeps Rice cultivation

34.2 27.7

Burning of plant matter Termites and other wild animals

24.2 0

5

10 15 20 Proportion of total methane (%)

25

30

Landfill sites are also contributors of methane owing to the decomposition of ­organic materials in the absence of significant oxygen. Atmospheric ­methane also results from human activities such as coal mining, the increasing use of natural gas for heating and transportation, and most recently, fracking to obtain poorly ­accessible gas and oil. Regardless of the source, the atmospheric methane concentration in September 2021 was 1900 ppb (Figure 25.8). Based on studies of air bubbles trapped in ice sheets, methane is now more abundant in Earth’s atmosphere than at any time in the last 400,000 years. The causes of the rapid increase in methane (and the pause from about 2000 to 2010) are not fully understood, but carbon isotopes provide some clues. The natural abundance of the C-12 isotope is 98.9% whereas C-13 is 1.1%. Scientists have found that there is less C-13 in methane that has been generated by microbes ­digesting carbon in wetlands or in the gut of a cow than in methane from ancient underground sources such as a coal mine or natural gas well. The fact that the increase in atmospheric methane is from gas with a smaller fraction of C-13 suggests that it arises from sources such as wetlands, livestock, and agricultural sources. Concerns about methane in the atmosphere arise because it is a powerful greenhouse gas (Section 25.5). In fact, CH4 is at least 28 times better at atmospheric warming than CO2 by weight. Fortunately, however, methane survives only a decade or so in the atmosphere whereas CO2 can remain for many decades. Methane hydrates are an interesting source of methane and of possible ­commercial interest (“A Closer Look: Methane Hydrates”). There are vast amounts of methane hydrates in the depths of the ocean and in the Arctic, locked in the ­permafrost. However, there is also considerable concern that these could release additional methane directly into the atmosphere if warming of the Arctic and oceans continues.

Atmospheric methane levels (parts per billion)

Meteorological Organization reported that “global emissions [of methane] have rebounded since the COVID-related lockdowns.” Increases in methane levels in 2020 and 2021 were the largest since record keeping began in 1983. New York Times, October 26, 2022.

Figure 25.8 Atmospheric methane levels.  ​Methane levels have risen more rapidly since the Industrial Revolution. There has been a rise of almost 20% just in the past 35 years. Source: © nature, https://www.nature.com/articles /d41586-022-00312-2

1900 1850 1800 1750 1700 1650 1600

Methane Emissions Rising Rapidly  In late 2022, the World

1985

1990

1995

2000

2005

2010

2015

2020 25.1  The Atmosphere

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1255

A Closer Look

Methane Hydrates

J. Pinkston and L. Stern, U. S. Geological Survey https://www.sciencedaily.com/ releases/2009/03/090323143858.htm

Have you ever walked around a shallow lake or pond and watched as bubbles of gas rise to the surface? This is marsh gas, and it is often responsible for the characteristic smell of a marshy area. This gas is mostly methane (CH4), and it is an extremely important and possibly dangerous feature of the worldwide environment. Bodies of water are usually surrounded by vegetation, which over years or centuries falls into the water and decays. The vegetation is consumed by bacteria that release methane as a product of the digestion. Some of the methane bubbles to the surface, and in the winter the bubbles can be trapped in the ice. The white patches you see in the photo are trapped methane bubbles in a lake in Alberta, Canada. The methane can also be trapped as methane hydrate, a white solid in which

Cavan Images/Cavan/Getty Images

Methane hydrate.  When a sample is brought to the surface from the depths of the ocean, the methane oozes out of the solid and can be burned. The structure of solid methane hydrate consists of methane trapped in a lattice of water. The lattice shown here is a common structural unit of a more complex structure. Each point of the lattice is an O atom of an H2O molecule. The edges consist of an OZHOO series of atoms connected by a hydrogen bond and covalent bond. (Other, more complex structural units are known.) Such structures are often called clathrates.

Methane bubbles. Methane, produced by plant decay, is trapped in the ice in a lake in Alberta, Canada.

methane is encased in a lattice of water molecules. There are millions of tons of  methane estimated to be trapped in the hydrated form under the world’s oceans and in Arctic regions on the planet. Methane bubbles and methane hydrate are of interest as a source of needed fuel. But, in an era of climate change, likely brought on by anthropogenic excessive release of carbon dioxide (CO2) and atmospheric methane, scientists are interested

in all possible effects on the climate because methane is a far more potent greenhouse gas than CO 2. Some of the bubbles in a frozen lake come from slow methane release by methane hydrate. But what if methane is released explosively? This is of concern because the Arctic is warming, which destabilizes the buried methane hydrate. The possibility of a catastrophic, explosive methane release is hotly debated by environmental scientists.

25.2 The Aqua Sphere (Water) Goal for Section 25.2 • Know the basic steps in water purification and be aware of common water pollutants.

Although there is an abundance of water on the planet, only about 3% of it is fresh water. This water is found in lakes, rivers, and underground aquifers, locked in minerals, in ice and snow in arctic regions, and water vapor in the atmosphere. Right now, however, there are serious problems associated with this valuable resource. Some locations are deprived of sufficient fresh water, whereas there is an

1256 Chapter 25 / Environmental Chemistry—Earth’s ­E nvironment, Energy, and Sustainability Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

overabundance in other places. There are huge variations in the purity and quality of available water supplies, and evidence suggests these problems will continue to increase in the future. People need water for drinking and for their everyday needs, but agriculture and industry require even larger amounts of water. Currently, most of this water comes from available groundwater, rivers, lakes, underground aquifers, and rainwater collected and stored in reservoirs above or below ground. However, it is increasingly clear that climate change can change the amount and distribution of water resources around the world. The amount of water used will increase with continuing growth in global population. Already 500 million people worldwide are affected at some level by water shortages, although most of these problems have not yet seriously affected people in the United States. However, in the Great Plains, underground aquifers are becoming seriously depleted, and some lakes and rivers are drying up. For example, in 2021 Lake Mead, which was created by the Hoover Dam (on the Colorado River on the border between Nevada and Arizona) was 143 feet below its 2000 level, when it was considered full (Figure 25.9). The Colorado River no longer reaches the sea; much of its water is being drawn off for use in surrounding communities, especially for agriculture to irrigate crops. Water quality is always an underlying issue when talking about water. There is a lot of water on Earth, but 97% of it is in the oceans and unsuitable for most uses until it is separated from dissolved salts. Much of the remaining 3% is also unsuitable or unavailable for human consumption. For centuries humans have treated lakes, rivers, and oceans as the easiest place to dump waste. All too often, what appear to be benign human activities have unintended consequences for water supplies. For example, agricultural and industrial chemicals and drugs from discarded medications are now found in water supplies. The presence of dissolved materials in water supplies then requires complex and expensive purification processes if the water is to be used.

Drought Worst in 12 Centuries 

“The drought in the American Southwest has become so severe that it’s now the driest two decades in the region in 1200 years … and climate change is responsible.” H. Fountain, “Western Drought is the Worst in 1,200 Years,” New York Times, February 15, 2022, A.12.

Water Use  In the United States,

L. Powell/Shutterstock.com

the average daily indoor per capita water use is about 70 gallons, which is much greater than in any other country. In some developing countries, the per capita use is one tenth of this amount or less.

Figure 25.9  Lake Mead, a symbol of a crisis in water supply.  ​Lake Mead is the largest

human-made lake and reservoir in the United States. It is located on the Colorado River in the states of Nevada and Arizona. Aqueducts carry the water to Las Vegas, Los Angeles, San Diego, and other communities in the Southwest. This view of the lake from Hoover Dam shows the drop in the lake level, which used to cover the white rocks on the mountainsides.



25.2  The Aqua Sphere (Water)

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1257

The Oceans

Sea Level Rise  The following is from Climate.gov (April, 2022):

• Sea level has risen 21–24 cm since 1880. • The rate of sea level rise is accelerating: it has more than doubled from 1.4 mm per year throughout most of the twentieth century to 3.6 mm per year from 2006 to 2015. • In many locations along the U.S. coastline, high-tide flooding is now 300% to more than 900% more frequent than it was 50 years ago.

Oceans cover 71% of the area of the planet, contain about 1.3 × 109 km3 of seawater, and have an average depth of 3790 m. A multitude of life forms is known to exist in the oceans and because the depths have barely been explored, the number is likely to increase. The quantity of minerals dissolved in the oceans is vast (Table 25.2). Sodium chloride (sodium and chloride ions) is found in the largest amount, and there are sizable quantities of magnesium ions, calcium ions, potassium ions, and carbonate/bicarbonate ions. Many other elements are present in seawater in concentrations less than 1 ppm but because of the huge amounts of seawater, the amounts of these elements potentially available in seawater is striking. The lure of obtaining dissolved metals from seawater has tantalized people through the ages. Gold, for example, is present in seawater at a concentration of 10 parts per trillion (10 g in 1 trillion g of water), so the total amount in the world’s oceans is more than 13 million tons. Although some have tried, no method has been found to extract gold profitably from the oceans. Concerns have been expressed about the effect on sea levels caused by global warming. Warming leads to expansion of the volume of water in the oceans, but it also melts ice now stacked on land (such as that in Greenland). Both would cause an increase in sea level. (Melting of Arctic sea ice, which is occurring, will not raise the sea level because the volume of liquid water formed is smaller than the volume of ice.)

Water Purification You would not survive a week if you had no water to drink. Most people in developed countries have clean, safe water supplied to their homes (although curiously many prefer to buy it in small bottles). Sadly, many people, generally in low-income countries, have to make do with whatever water they can find, ­taking it from a well or even from a river, and then they are at risk of contracting a water-borne disease such as cholera, typhoid, gastroenteritis, or meningitis. The United Nations has ­estimated that 4000 children die every day from water-borne illness. Water purification began with the Egyptians 3500 years ago. They discovered that when alum—potassium aluminum sulfate, KAl(SO4) 2—was dissolved in water visible impurities were removed. In mildly basic conditions, this salt forms a

Concentrations of Some Cations and Anions in Seawater

Table 25.2

Element

Dissolved Species

Chlorine

Cl− Na

Magnesium

Mg

Calcium

Ca K

550

+

Sodium

Potassium

Concentrations (mmol/L) 460  52

2+

 10

2+

+

 10 −

Carbon

HCO3 , CO3

 30

Phosphorus

HPO42−

  B (−800.8 kJ) < C (−1029 kJ)

(a) Z* for F is 5.20; Z* for Ne is 5.85. The effective nuclear charge increases from O to F to Ne. As the effective nuclear charge increases, the atomic radius decreases, and the first ionization energy increases. (b) Z* for a 3d electron in Mn is 5.6; for a 4s electron it is only 3.6. The effective nuclear charge experienced by a 4s electron is much smaller than that experienced by a 3d electron. A 4s electron in Mn is thus more easily removed.

F

+ −

 ClF2+, 2 bond pairs and 2 lone pairs.  ClF2−, 2 bond pairs and 3 lone pairs.

(a) CN− : formal charge on C is −1; formal charge on N is 0. (b) SO32−: formal charge on S is +1; formal charge on each O is −1.

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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8.9





8.10

(a) In each resonance structure, the carbon atom has a formal charge of zero, the two oxygen atoms with single bonds to the carbon atom have formal charges of –1, and the oxygen atom with a double bond to the carbon atom has a formal charge of zero. In the carbonate ion, the carbon atom has a formal charge of zero, and each oxygen atom has a formal charge of −2⁄3. (b) In each resonance structure: carbon’s formal charge is 0; the oxygen of the OOH group and the double-bonded oxygen have a formal charge of zero; the singly bonded oxygen has a formal charge of −1. The average formal charge on the latter two oxygen atoms is −1⁄2. (c) H+ would be expected to add to one of the oxygens with a negative formal charge; that is, one of the oxygens with formal charge of −1⁄2. 16 valence electrons SqCON

Formal charges +



0

−

mn

−2

SPCPN 0

−

mn

−

0

SOCqN −

0

For each species, the electron-pair geometry and the molecular shape are the same. BF3: trigonal-planar; BF4−: tetrahedral. Adding F− to BF3 adds an electron pair to the central atom and changes the shape.

8.13

The electron-pair geometry around I is trigonalbipyramidal. The molecular geometry of the ion is linear. −

Cl Cl

Formal charges: S = +1, 0 = −1, each Cl = 0



(b) Geometry: trigonal-pyramidal. The molecule is polar. ­ egative charge The positive charge is on sulfur, the n on oxygen.

8.16

CH4(g) + 2 O2(g) n CO2(g) + 2 H2O(g)



Break 4 COH bonds and 2 OPO bonds: (4 mol COH bonds/mol-rxn)(413 kJ/mol COH bonds) + (2 mol OPO bonds/mol-rxn)(498 kJ/mol OPO bonds) = 2648 kJ/mol-rxn



Make 2 CPO bonds and 4 HOO bonds: (2 mol CPO bonds/mol-rxn)(803 kJ/mol CPO bonds) + (4 mol HOO bonds/mol-rxn) (463 kJ/mol HOO bonds) = 3458 kJ/mol-rxn



ΔrH° = 2648 kJ/mol-rxn − 3458 kJ/mol-rxn = −810. kJ/mol-rxn (value calculated using enthalpies of ­formation = −802 kJ/mol-rxn)

−

Applying Chemical Principles 8.1

Ibuprofen, A Study in Green Chemistry

1. The Lewis structures for the key portions of the molecules where the reaction takes place are the following: H H O CH

O

O + C

O

C CH

Bonds broken: 1 COO and 1 CqO (1 mol COO/1 mol-rxn)(358 kJ/mol COO) + (1 mol CqO/1 mol-rxn)(1046 kJ/mol CqO) = 1404 kJ/mol-rxn Bonds formed: 1 COC, 1 CPO, and 1 COO

ΔrH° ≈ 1404 kJ/mol-rxn − 1449 kJ/mol-rxn = −45 kJ/mol-rxn

F

The reaction is thus predicted to be exothermic.

B

Cl

NH2Cl, polar, negative side is the Cl atom. −

Cl

N H + H +

SCl2, polar, Cl atoms are on the negative side. Cl

−



Cl

BFCl2, polar, negative side is the F atom because F is the most electronegative atom in the molecule.

Cl



Cl S

(1 mol COC/1 mol-rxn)(346 kJ/mol COC) + (1 mol CPO/1 mol-rxn)(745 kJ/mol CPO) + (1 mol COO/1 mol-rxn)(358 kJ/mol COO) = 1449 kJ/mol-rxn

I



O

Tetrahedral geometry around carbon. The ClOCOCl bond angle will be close to 109.5°.

8.12

8.14

(a)

0

Formal charge considerations favor the middle structure because it has less formal charge than the left structure and, unlike the right structure, it has the negative formal charge on the most electronegative atom in the ion.

8.11

8.15

S

+

Cl

−

2. All of the atoms in ibuprofen have a formal charge of zero. 3. The most polar bond in the molecule is the OOH bond. 4. The molecule is not symmetrical and so is polar. 5. The shortest bond in the molecule is the OOH bond. 6. The CPO bond has the highest bond order. The CPC bonds in the ring have an order of 1.5. 7. Yes, there are 120° bond angles present: the bond angles around the C atoms in the ring and those around the C atom in the OCO2H group are all 120°. There are no 180° angles in this molecule.

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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A-63

8. There is one acid group in ibuprofen (OCO2H), therefore 1 mol of ibuprofen will react with 1 mol of NaOH. 200. mg ibuprofen (1 g/1000 mg)(1 mol ibuprofen/ 206.3 g ibuprofen)(1 mol NaOH/1 mol ibuprofen) (1 L/0.0259 mol NaOH)(1000 mL/1 L) = 37.4 mL Therefore 37.4 mL of the NaOH solution would be required. 8.2

8.5

C

+

8.3 Linus Pauling and the Origin of the Concept of Electronegativity 1. Using average bond-dissociation enthalpies: χCl − χH = 0.102[D(HOCl) − (D(HOH) + D(ClOCl))/2]1⁄2 = 0.102[432 kJ/mol − (436 kJ/mol + 242 kJ/mol)/2]1⁄2 = 0.98 According to Figure 8.2, χCl − χH = 1.0 2. χN − χI = 0.102[D(NOI) − (D(NON) + D(IOI))/2]1⁄2 3.0 − 2.7 = 0.102[x − (163 kJ/mol + 151 kJ/mol)/2]1⁄2



P

+

8.3

(a) Group 6A (b) Group 3A (c) Group 1A (d) Group 2A (e) GCroup 7A C (f) Group 6A C O (a) O C (b) O N (c) O N



(d) N F N F F F

A-64

Cl −

P

+

Br −

(c) B O

+

−

B S

+

−

BO is more polar



(d) B

+

F

−

B

+

I

−

BF is more polar

8.7

The greater the difference in electronegativity, the greater the ionic character of the bond.

H and H < H and C < O and H < O and Ca 8.9

(a) CH and CO bonds are polar. (b) The CO bond is most polar, and O is the most negative atom.

8.11

Group 4A (14), four bonds Group 5A (15), three bonds Group 6A (16), two bonds Group 7A (17), one

8.13 Elements from Period 3 and beyond on the periodic table can form hypervalent compounds. Thus, hypervalent compounds are possible for (b) P, (e) Cl, (g) Se, and (h) Sn. 8.15

(a) NF3, 26 valence electrons F N F F



(b) ClO3−, 26 valence electrons O

−

Cl O O

(16), six valence electrons (13), three valence electrons (1), one valence electron (2), two valence electrons (17), seven valence electrons (16), six valence electrons

−

PCl is more polar

This matches the value of 2.6 in Figure 8.2 well.

Study Questions

C N

+

(b)

x = 200 kJ/mol 3. χS = 1.97 × 10−3(IE − ΔeaH) + 0.19 = 1.97 × 10−3(1000 kJ/mol − −200.41 kJ/mol) + 0.19 = 2.55

O

−

CO is more polar

van Arkel Triangles and Bonding

1. (a) CuZn is metallic. (b) GaAs and BP are semiconductors. As and B are metalloids; Ga and P are not. In both compounds, only one element is a metalloid. (c) Mg3N2 and SrBr2 are ionic. Both are composed of a metal combined with a nonmetal. (d) Covalent bonding (e) SBr2 and C3N4 are covalent. Both elements in the compounds are nonmetallic. 2. The electronegativity difference between Be(1.6) and Cl(3.2) is 1.6. The average electronegativity of the two is 2.4. On the figure locate the point for this compound, which is on the border between ionic (red) and covalent (yellow) compounds.

8.1.

(a)

(c) HOBr, 14 valence electrons H O Br



(d) SO32−, 26 valence electrons O S O

2−

O

8.17

(a) CHClF2, 26 valence electrons H Cl C F F

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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(b) CH3CH2CO2H, 30 valence electrons

(d) There are two equivalent resonance structures.

H H O

H O N O

C O H

H C C

O

H H





(a) OH−: The formal charge on O is −1 and on H it is 0. Because O is more electronegative than H, bond polarity places the negative charge on the oxygen. Thus, for this ion, both formal charge and bond polarity considerations place the negative charge on the O. (b) BH4−: Even though the formal charge on B is −1 and on H is 0, H is slightly more electro­negative than B. The four H atoms are therefore more likely to bear the −1 charge of the ion. The BH bonds are polar with the H atom the negative end. (c) CH3CO2–: There are two equivalent resonance structures. In each, the formal charges of the C and H atoms are 0. In each, the singly bonded O has a formal charge of −1 and the doubly bonded O has a formal charge of 0, giving a net formal charge on each O of −1⁄2. The CH and CO bonds are all polar (but the COC bond is not). The most polar bonds are the carbon-oxygen bonds, with the O atoms being negative. Thus, both formal charge and bond polarity considerations predict that the negative charge will be on the O atoms.

(d) H2CCCH2, 16 valence electrons

8.19

H

H C

C C

O S



H

(a) SO2, 18 valence electrons O

O

S O

(b) HNO2, 18 valence electrons O

H O N





(c) HSCN, 16 valence electrons H S

8.21

C N

H S

C N

H S

C N

(a) BrF3, 28 valence electrons F Br F F

(b) I3−, 22 valence electrons



I

8.29

−

(a, b) N  OCl, 18 valence electrons O N Cl

I

Formal charges

I



(c) XeO2F2, 34 valence electrons F O

O

+

(a) N = 0; H = 0 (b) P = +1; O = −1

8.25

(a) N = +1; O = 0 (b) The central N is 0. In each resonance structure, the singly bonded O atom is −1, and the doubly bonded O atom is 0. The average formal charge on each oxygen is thus −1⁄2.



−

(c) N and F are both 0.

(c) B = −1; H = 0 (d) All are zero.

O

N O H O

8.23

O N O

C O H

−

O

C O H B O

O

N O H

O

F

N O

−

+1

Lewis structures: O

Xe F

0

8.31

(d) XeF3+, 28 valence electrons



−1

0

0

Cl

(c) T  he structure shown on the left is dominant because it has smaller formal charges.

Xe O

F

0

O N



F



O

8.27

N

H

O

In each, the formal charge of the central N atom is +1, the hydrogen is 0, and the oxygen with the hydrogen attached is 0. In each, the formal charge of the singly bonded O (with no hydrogen) is −1, and the doubly bonded O is 0. Thus, the average formal charge of these two oxygens is −1⁄2.

H H

N



(c) CH3CN, 16 valence electrons H C C

H O

−

O



(a) These species are isoelectronic. (b) Each has two major resonance structures. (c) In HCO3−, the H, C, and O attached to the H all have a formal charge of 0. Each of the other oxygen atoms has a formal charge of −1⁄2.



In HNO3, the same formal charges are present as in HCO3− except that the central N has a formal charge of +1. (d) HNO3 is much more acidic than HCO3−. This is due, in part, to simple electrostatics. It is much easier to remove a positively charged species (H+) from a neutral species (HNO3) than from a negatively charged one (HCO3−). Conversely, HCO3− is much more basic than HNO3; the negative charge of HCO3− will attract a positively charged H+.





Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A-65

−1

0

0

8.33 (a)  O N O

0

−

−1

0

O N O



−



(b) If an H+ ion were to attack NO2−, it would attach to an O atom because the O atoms bear the negative charge in this ion.



(c) H O N

O

O

N O

H

The structure on the left is strongly favored because all of the atoms have zero formal charge, whereas the structure on the right has a −1 formal charge on one oxygen (left) and a +1 formal charge on the other (right). 8.35

(a) Electron-pair geometry around N is tetrahedral. Molecular geometry is trigonal-pyramidal.

(b) Electron-pair geometry around Cl is trigonalbipyramidal. Molecular geometry is T-shaped. F Cl F F



(c) Electron-pair geometry around Cl is octahedral. Molecular geometry is square-planar.

F Cl F F



Cl N H

(d) Electron-pair geometry around Cl is octahedral. Molecular geometry is a square-pyramidal.

H





O Cl

(c) Electron-pair geometry around C is linear. Molecular geometry is linear. S C N



F

(b) Electron-pair geometry around O is tetrahedral. Molecular geometry is bent. Cl

−

(d) Electron-pair geometry around O is tetrahedral. The molecular geometry is bent.

F

8.41

(a) Electron-pair geometry around C is linear. Molecular geometry is linear. O C O



(b) Electron-pair geometry around N is trigonalplanar. Molecular geometry is bent. O N



O

−

(c) Electron-pair geometry around O is trigonalplanar. Molecular geometry is bent. O O O



(d) Electron-pair geometry around Cl atom is tetra­ hedral. Molecular geometry is bent. O Cl O



8.39

−

All have two atoms attached to the central atom. As the bond and lone pairs vary, the electron-pair geometries vary from linear to tetrahedral, and the molecular geometries vary from linear to bent. (a) Electron-pair geometry around Cl is trigonalbipyramidal. Molecular geometry is linear. F Cl F

A-66

−

F Cl

F F

(a) Ideal OOSOO angle = 120° (b) 120° (c) 120° (d) HOCOH = 109° and COCON angle = 180°

8.43

1 = 120°; 2 = 109°; 3 = 120°; 4 = 109°; 5 = 109°



The chain cannot be linear because the first two carbon atoms in the chain have bond angles of 109° and the final one has a bond angle of 120°. These bond angles do not lead to a linear chain.

8.45

(i) The most polar bonds are in H2O (because O and H have the largest difference in electronegativity). (ii) Not polar: CO2 and CCl4 (iii) The F atom is more negatively charged.

H O F

8.37

−

F

8.47 8.49

(a) BeCl2, nonpolar linear molecule (b) HBF2, polar trigonal-planar molecule with F atoms on the negative end of the dipole and the H atom on the positive end. (c) CH3Cl, polar tetrahedral molecule. The Cl atom is on the negative end of the dipole and the three H atoms are on the positive side of the molecule. (d) SO3, a nonpolar trigonal-planar molecule (a) Two COH bonds, bond order is 1; 1 CPO bond, bond order is 2. (b) Three SOO single bonds, bond order is 1. (c) Two nitrogen–oxygen double bonds, bond order is 2. (d) Four COH bonds, bond order is 1. One COC bond, bond order is 1. One CqN bond, bond order is 3.

8.51

(a) BOCl (b) COO (c) POO (d) CPO

8.53

Of the bonds in this problem, COC will have the longest bond length. Longer bonds result from larger atoms and smaller bond orders. Carbon atoms are larger than oxygen atoms, and a single bond (bond order = 1) between carbon atoms will be longer than a double bond (bond order = 2) between the same two atoms.

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8.55

NO bond orders: 2 in NO2+, 1.5 in NO2−; 1.33 in NO3−. The NO bond is longest in NO3− and shortest in NO2+.

8.57

The CO bond in carbon monoxide is a triple bond, so it is both shorter and stronger than the CO double bond in H2CO.

8.59

Using bond-dissociation enthalpy data, ΔrH° ≈ −44 kJ/mol-rxn. Using Δf H° data, ΔrH° = −45.9 kJ/mol-rxn. ΔrH = −126 kJ/mol-rxn

8.63

OOF bond-dissociation enthalpy = 192 kJ/mol

8.67

The NOO bonds in NO2− have a bond order of 1.5, whereas the NOO bonds in NO2+ have a bond order of 2. The shorter bonds (110 pm) are the NO bonds with the higher bond order (in NO2+), whereas the longer bonds (124 pm) in NO2− have a lower bond order.

8.77

The FOClOF bond angle in ClF2+, which has a tetrahedral electron-pair geometry, is approximately 109°. F Cl F

8.61

8.65

8.75



Element Number of Valence Electrons 1

Ti

4

Zn

2



The anion ClF2− has the greater bond angle.

Si

4

8.79

Cl

7

An H+ ion will attach to an O atom of SO32− and not to the S atom. The O atoms each have a formal charge of −1, whereas the S atom formal charge is +1.

F Cl

SeF4, BrF4−, XeF4

O

−

O H C

O

H C

O

8.81

8.71

To estimate the enthalpy change, we need bond-­ dissociation enthalpies for the following bonds: OPO, HOH, and HOO.

8.83

ΔH to break bonds ≈ (1 mol OPO bonds/1 mol-rxn)(498 kJ/mol OPO bonds) + (2 mol HOH bonds/1 mol-rxn)(436 kJ/mol HOH bonds) = 1370 kJ/mol-rxn





ΔH evolved when bonds are made ≈ (4 mol OOH bonds/1 mol-rxn)(436 kJ/mol OOH bonds) = 1852 kJ/mol-rxn



ΔrH ≈ −482 kJ/mol-rxn

8.73

All the species in the series have 16 valence electrons and all are linear. CO2 is a neutral molecule, whereas N3− and OCN− are ions.



(a) 0

0

mn

OPCPO

+1

mn

0

8.85

−1

(a)  C

N O

−1

+1

−

C N O

−1

−2

−

mn

0

NONqN

−

0

mn

+1 −2

NqNON

−



0

0

−1

OPCPN

−

−1

mn

0

0

OOCqN

−

+1

mn

0

−2

OqCON

For CO2 and N3−, the structures with two double bonds are dominant, but for OCN−, the structure with a COO single bond and a CqN triple bond is dominant.

−



+1

−

0

C

N O

−3

+1

−

+1

(b) The first resonance structure is the most reasonable because oxygen, the most electronegative atom, has a negative formal charge, and the unfavorable negative charge on the least electronegative atom, carbon, is smallest. (c) This species is so unstable because carbon, the least electronegative element in the ion, has a negative formal charge. In addition, all three resonance structures have an unfavorable charge distribution. F Xe

OqCOO

(c)

Formal charges



+1

OOCqO −2 +1

2−

(a) Calculation from bond-dissociation enthalpies: ΔrH° = −1302 kJ/mol-rxn (b) Calculation from thermochemical data: ΔrH° = −1352.3 kJ/mol-rxn

F −1 +1 −1

NPNPN



0

(b)

Formal charges





−1

0

−

O

Bond order = 3/2

Formal charges

F

S O

−

O





The ClF2− ion has a trigonal-bipyramidal electron-pair geometry with F atoms in the axial positions and the lone pairs in the equatorial positions. Therefore, the FOCOF angle is 180°.

Li

8.69



+

F 120°

F

Cl

120°

F

(a) XeF2 has three lone pairs around the Xe atom. The electron-pair geometry is trigonal-bipyramidal. Because lone pairs require more space than bond pairs, it is better to place the lone pairs in the equator of the bipyramid where the angles between them are 120°. The two F atoms are in axial positions, giving a linear molecular shape. (b) Like XeF2, ClF3 has a trigonal-bipyramidal electron-pair geometry, but with only two lone pairs around the Cl. These are again placed in the

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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A-67

equatorial plane where the angle between them is 120°. Two of the F atoms are in axial positions, and one F atom is in an equatorial position, giving a T-shaped molecule. 8.87

(a) Angle 1 = 109°; angle 2 = 120°; angle 3 = 109°; angle 4 = 109°; and angle 5 = 109° (b) The OOH bond is the most polar bond.

8.89

ΔrH = [(2 mol CON bonds/1 mol-rxn)D for CON bonds + (1 mol CPO bonds/1 mol-rxn)D for CPO bonds] − [(1 mol NON bonds/1 mol-rxn)D for NON bonds) + (1 mol CqO bonds/1 mol-rxn)D for CqO bonds] = 146 kJ/mol-rxn

8.91

(a) Two mol COH bonds and one mol OPO are broken and two mol OOC bonds and two mol HOO bonds are made in the reaction. ΔrH = −318 kJ/mol-rxn. The reaction is exothermic. (b) Both hydroxyacetone and acetone are polar. (c) The OOH hydrogen atoms are the most positive in dihydroxyacetone.



8.105 (a) Odd electron molecules: BrO (13 valence electrons) (b) Br2(g) n 2 Br(g)  ΔrH = +193 kJ/mol-rxn 2 Br(g) + O2(g) n 2 BrO(g)  ΔrH = +96 kJ/mol-rxn BrO(g) + H2O(g) n HOBr(g) + OH(g)  ΔrH = 0 kJ/mol-rxn (c) Δf H [HOBr(g)] = −101 kJ/mol (d) The first two reactions in part (b) are endothermic, the third reaction in part (b) is thermal-neutral, and the enthalpy of formation in part (c) is exothermic. 8.107 (a and b) N2O, 16 valence electrons NPNPO Formal charges

−1 +1

mn

0

NqNOO 0

+1 −1

mn

NONqO −2 +1 +1





(a) The CPC bond is stronger than the COC bond. (b) The COC single bond is longer than the CPC double bond. (c) Ethylene is nonpolar, whereas acrolein is polar. (d) The reaction is exothermic (ΔrH = −45 kJ/mol-rxn).

(c) The middle structure with a NqN triple bond and a NOO single bond is dominant. Both the left and middle structures have lower formal charges than the right structure. In addition, the middle structure has the negative formal charge on the oxygen, the more electronegative atom.



(d) ∆rH° = −96 kJ/mol-rxn

8.95

ΔrH = −211 kJ/mol-rxn



8.97

Methanol is a polar solvent. Methanol contains two bonds of significant polarity, the COO bond and the OOH bond. The COOOH atoms are in a bent configuration, leading to a polar molecule. Toluene contains only carbon and hydrogen atoms, which have similar electronegativities and which are arranged in tetrahedral or trigonal planar geometries, leading to a molecule that is largely nonpolar.

This value is likely less accurate than most ­estimates that use bond-dissociation enthalpies because, even though the middle resonance structure is dominant, the real structure of N2O is a hybrid of the three resonance structures. Thus, the NON bond is not a full triple bond, and the NOO bond is more than a single bond.

Chapter 9

8.93

8.99

(a)

H H C S H





H

Check Your Understanding

C H

9.1

The carbon and nitrogen atoms in CH3NH2 are sp3  hybridized. The COH bonds arise from overlap of carbon sp3 orbitals and hydrogen 1s orbitals. The bond between C and N is formed by overlap of sp3 orbitals from these atoms. Overlap of nitrogen sp3 and hydrogen 1s orbitals gives the two NOH bonds, and there is a lone pair in the remaining sp3 orbital on nitrogen.

9.2

(a) sp3 (b) (from left to right) sp2, sp2, sp3 (c) sp2

9.3

The two CH3 carbon atoms are sp3 hybridized, and the center carbon atom is sp2 hybridized. For each of the carbon atoms in the methyl groups, the sp3 orbitals overlap with hydrogen 1s orbitals to form the three COH bonds, and the fourth sp3 orbital overlaps with an sp2 orbital on the central carbon atom, forming a carbon–carbon sigma bond. Overlap of an sp2 orbital on the central carbon and an oxygen sp2 orbital gives the sigma bond between these ­elements. The π bond between carbon and oxygen arises by overlap of a p orbital from each element.

H

The bond angles are all approximately 109°. (b) The sulfur atom should have a slight partial negative charge, and the carbons should have slight partial positive charges. The molecule has a bent shape and is polar. (c) 1.6 × 1018 molecules

8.101 (a) CPO (Based on the difference in electronegativities, the CPO bond is slightly more polar than the NOH bond.) (b) Predicted = 120° (Actual = 124°) (c) Predicted = 120° (Actual = 113°) (d) Approximately 109.5° (slightly less, similar to the HONOH bond angles in ammonia) 8.103

(a) P4 + 6 Cl2 n 4 PCl3 (b) ΔrH = −1206.9 kJ/mol-rxn (c) Bonds broken: 6 mol POP, 6 mol ClOCl Bonds formed: 12 mol POCl  DPOCl = 322.1 kJ/mol; the value in Table 8.8 is 326 kJ/mol.

A-68

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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H2+: (σ1s)1 The ion has a bond order of 1⁄2 and is expected to exist. A bond order of 1⁄2 is predicted for He2+ and H2−, both of which are predicted to have electron configurations (σ1s)2(σ*1s)1.

9.4

Cl H C Cl

−

9.5

Li2 is predicted to have an electron configuration (σ1s)2(σ*1s)2(σ2s)2(σ*2s)1 and a bond order of 1⁄2, the positive value implying that the ion might exist.

9.6

O2+: [core electrons] (σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)1. The bond order is 2.5. The ion is paramagnetic with one unpaired electron.

Applying Chemical Principles

9.3

H N

1. Metal



Both the N and the O are sp3 hybridized. One of the sp3 orbitals on the N overlaps with one of the sp3 hybrid orbitals on the O.

9.5

O F C



2. E = hν =  hc/λ = (6.626 × 10−34 J ∙ s)(2.998 × 108 m/s)/ (58.4 × 10−9 m) = 3.401 × 10−18 J/photon (3.401 × 10−18 J/photon)(6.022 × 1023 photons/mol) = 2.05 × 106 J/mol = 2.05 × 103 kJ/mol 3. σ2p

Electron-Pair Molecular Geometry Geometry

9.7

IE =  hν – KE = 3.401 × 10−18 J − 4.23 × 10−19 J = 2.978 × 10−18 J/electron (2.978 × 10−18 J/electron)(6.022 × 1023 electrons/mol)  (1 kJ/1000 J) = 1.79 × 103 kJ/mol 5. The 15.6 eV and 16.7 eV ejected electrons came from bonding orbitals. Removing an electron from a bonding orbital weakens the bond and thus results in a longer bond length. The 18.6 eV ejected electron comes from an antibonding orbital. 9.2

Trigonal-planar

Trigonal-planar sp2

(b)

Linear

Linear

sp

(c)

Tetrahedral

Tetrahedral

sp3

(d)

Trigonal-planar

Trigonal-planar sp2

9.9

(a) C, sp3; O, sp3 (b) CH3, sp3; middle C, sp2; CH2, sp2 (c) CH2, sp3; CO2H, sp2; N, sp3

9.11

There are 32 valence electrons in both HSO3F and its anion. Both have a tetrahedral molecular geometry, so the S atom in both is sp3 hybridized. O

1. C8H4BrNO H O



S

F

O

O

  

S

F O

The C atom is sp2 hybridized. Two of the sp2 hybrid orbitals are used to form COCl σ bonds, and the third is used to form the COO σ bond. The p orbital not used in the C atom hybrid orbitals is used to form the CO pi bond.

9.15

The electron-pair and molecular geometry around the S atom are both tetrahedral. The S atom is sp3 hybridized.

9.17

(a) H3C

Nitrated butter yellow: 8

The electron-pair and molecular geometry of CHCl3 are both tetrahedral. Each COCl bond is formed by the overlap of an sp3 hybrid orbital on the C atom with a 3p orbital on a Cl atom to form a sigma bond. A COH sigma bond is formed by the overlap of an sp3 hybrid orbital on the C atom with an H atom 1s orbital.

−

O

9.13

3. Tyrian purple: 9

Study Questions

Hybrid Orbital Set

(a)

Green Chemistry, Safe Dyes, and Molecular Orbitals

2. The energy per photon is inversely proportional to wavelength. The light absorbed by butter yellow has a smaller wavelength than that absorbed by nitrated butter yellow; therefore, butter yellow absorbs higher energy light than does nitrated butter yellow.

F

The electron-pair and molecular geometries are both trigonal-planar. The carbon atom is sp2 hybridized. The σ bond between the C and O is formed by the overlap of an sp2 hybrid orbital on C with an sp2 hybrid orbital on O. The π bond is formed by the overlap of a 2p orbital on C with a 2p orbital on O.

4. E = hν = 3.401 × 10−18 J from problem 2.

(2.978 × 10−18 J)(1 eV/1.60218 × 10−19 J) = 18.6 eV

O H

H

9.1 Probing Molecules with Photoelectron Spectroscopy

9.1

Cl

CH3 C

H



CH3

(b) H C

C H

cis isomer

Cl

C H

trans isomer

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A-69

9.19

H2+ ion: (σ1s)1. Bond order is 0.5. The bond in H2+ is weaker than in H2 (bond order = 1).

9.21

MO diagram for C22− ion

9.31

The resonance structures of N2O, with formal charges, are shown here. −2

*2s

9.23

O2: (core electrons)(σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)2 O22−: (core electrons)(σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)4 O O





2−

(a) O2 is paramagnetic; O22− is diamagnetic; O2 has one net σ bond and one net π bond, and O22− has one net σ bond; O2 bond order = 2; O22− bond order = 1; the bond length in O2 is shorter than that in O22−. (b) Both valence bond and MO theories predict one σ bond and one π bond in O2 and one σ bond in O22−. Both theories predict bond orders of 2 for O2 and 1 for O22−. Notice, however, that valence bond theory does not predict the paramagnetic behavior of O2.

9.25

Shortest, N2; longest, Li2

9.27

(a) ClO has 13 valence electrons [core](σs)2(σ*s)2(πp)4(σp)2(π*p)3



(b) The HOMO is the π*p. The LUMO is the σ*p. (c) Paramagnetic (d) There are net 1 σ bond and 0.5 π bonds; bond order is 1.5.

9.29

F

−1

+1

O

N

N O

−

The electron pair and molecular geometries are both tetrahedral. The Al atom is sp3 hybridized, so the AlOF bonds are formed by overlap of an Al sp3 orbital with a p orbital on each F atom. The formal charge on each of the fluorines is 0, and that on the Al is −1. This is not a reasonable charge distribution because the less electronegative atom, aluminum, has the negative charge.

0

+1

0

N

−1

N O

B

C

The central N atom is sp hybridized in all structures. The two sp hybrid orbitals on the central N atom are used to form NON and NOO σ bonds. The two p orbitals not used in the N atom hybridization are used to form the required π bonds.

9.35

(a) All three have the formula C2H4O. (These are isomers because they have the same formula. However, they have different structures.) (b) Ethylene oxide: Both C atoms are sp3 hybridized.  Acetaldehyde: The CH3 carbon atom has sp3 hybridization, and the other C atom is sp2 hybridized. Vinyl alcohol: Both C atoms are sp2 hybridized. (c) Ethylene oxide: 109° Acetaldehyde: 109° Vinyl alcohol: 120° (d) All are polar. (e) Acetaldehyde has the strongest CO bond, and vinyl alcohol has the strongest COC bond.

9.37

(a) CH3 carbon atom: sp3 CPN carbon atom: sp2 N atom: sp2 (b) CONOO bond angle = 120°

9.39

(a) C(1) = sp2; O(2) = sp3; N(3) = sp3; C(4) = sp3; P(5) = sp3 (b) Angle A = 120°; angle B = 109°; angle C = 109°; angle D = 109° (c) The POO and OOH bonds are most polar (Δχ = 1.3).

9.41

(a) CPO bond is most polar. (b) 18 σ bonds and five π bonds (c)  CHO

H C

F

A-70

+1



F Al F



+1

N N A

2s (a) There are one net σ bond and two net π bonds. (b) 3 (c) VB and MO descriptions give the same result. (d) The bond order increases from 2 to 3 on going from C2 to C22−. (e) No, it is diamagnetic.

−

9.33 2p



N O

O

The electron-pair geometry is trigonal-planar. The molecular geometry is bent (or angular). The OONOO angle will be about 120°, the average NOO bond order is 3/2, and the N atom is sp2 hybridized.

*2p



−

N O



*2p

2p

O

H

H C

C

C CHO

H trans isomer

cis isomer

  



(d) All C atoms are sp hybridized. (e) All bond angles are 120°.

9.43

(a) The peroxide ion has a bond order of 1.

2

O O

2−

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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(b) [Core electrons](σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)4 This configuration also leads to a bond order of 1. (c) Both theories lead to a diamagnetic ion with a bond order of 1.

9.45

Paramagnetic diatomic molecules: B2 and O2



Bond order of 1: Li2, B2, F2; bond order of 2: C2 and O2; highest bond order: N2

9.47

(a) HOMO, σ2p; LUMO, π*2p (b, c)  Bond order = 2.5 (0.5 σ bond and 2 π bonds) (d) Paramagnetic

9.49

(a) All C atoms are sp3 hybridized. (b) About 109° (c) Polar (d) The six-member ring cannot be planar, owing to the tetrahedral C atoms of the ring. The bond angles are all 109°.

9.51

(a) The geometry about the boron atom is trigonalplanar in BF3 but tetrahedral in H3NOBF3. (b) Boron is sp2 hybridized in BF3 but sp3 hybridized in H3NOBF3. (c) The ammonia molecule is polar with the N atom partially negative. While the BF3 molecule is nonpolar overall, each of the BOF bonds is polarized such that the B has a partial positive charge. The partially negative N in NH3 is attracted to the partially positive B in BF3. (d) One of the lone pairs on the oxygen of H2O can form a coordinate covalent bond with the B in BF3. The resulting compound would be (the lone pairs on the F’s not shown):





H H

(b)

H H N S



COC bond: overlap of sp2 hybrid orbital on the carbon of OCO2− with sp 3 hybrid orbital on the other C.



COO σ bond of CPO double bond: overlap of sp2 hybrid orbital on C with sp2 hybrid orbital on O



COO π bond of CPO double bond: overlap of p orbital on C with p orbital on O



COO single bond: Each Lewis structure by itself would make it appear this bond is formed by the overlap of an sp2 hybrid orbital on C with an sp 3 hybrid orbital on O. In the resonance hybrid, however, this oxygen would also have sp2 hybridization.

9.57





MO theory pictures one net σ bond for each SOO linkage plus a contribution from π bonding. The π bonding in this molecule will be similar to that in O3 discussed in the text. There will be two electrons in a π bonding MO and two electrons in a π nonbonding MO. This gives an overall bond order of 1 for the π bonding in the entire molecule and therefore a net π bond order of 0.5 for each SOO linkage. The total bond order for each SOO linkage is therefore 1.5 (1 from σ bonding, 0.5 from π bonding).

9.59

A C atom may form, at most, four hybrid orbitals (sp3). The minimum number is two, for example, the sp hybrid orbitals used by carbon in CO. Carbon has only four valence orbitals, so it cannot form more than four hybrid orbitals.

9.61

(a) C, sp2; N (predicted based on Lewis structure), sp3 (b) The amide or peptide link has two resonance structures (shown here with formal charges on the O and N atoms). Structure B is less favorable, owing to the separation of charge.





O S O

O

0

O

−

O O

The bond angles around the N and the S are all approximately 109°. (c) The N does not undergo any change in its hybridization; the S changes from sp2 to sp3. (d) The SO3 is the acceptor of an electron pair in this reaction. The electrostatic potential map confirms this to be reasonable because the sulfur has a partial positive charge.

H



O B F

O

C C O

Hybridization of C in OCO2− is sp2

SO3: electron-pair geometry = molecular geometry = trigonal-planar, hybridization of S = sp2

H

C C O

−



F

(a) NH2−: electron-pair geometry = tetrahedral, molecular geometry = bent, hybridization of N = sp3

H O

H

F

9.53

−

H O H

CN has nine valence electrons. [core electrons](σ2s)2(σ*2s)2(π2p)4(σ2p)1



9.55

R

C A



O 0

N R

H

R

C B

S O

−

+

N

H

R

(c) The fact that the amide link is planar indicates that structure B has some importance and that the N of the peptide linkage is sp2 hybridized.

The principal sites of positive charge are the nitrogen in the amide linkage and the hydrogen of the OOOH group. The principal regions of negative charge are oxygen atoms and the nitrogen of the free ONH2 group.

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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A-71

9.63

9.65

MO theory is better to use when explaining or understanding the effect of adding energy to molecules. A molecule can absorb energy and an electron can thus be promoted to a higher level. Using MO theory, one can see how this can occur. Additionally, MO theory is a better model to use to predict whether a molecule is paramagnetic.

10.4

V2 = V1(P1/P2)(T2/T1) = (22 L)(150 atm/0.9934 atm)(295 K/304 K) = 3200 L



At 5.0 L per balloon, there is sufficient He to fill 640 balloons.

10.5

44.8 L of O2(g) is required; 44.8 L of H2O(g) and 22.4 L of CO2(g) are produced.

Lowest energy = orbital C < orbital B < orbital A = highest energy

10.6

PV = nRT



(750/760 atm)(V) =   (1300 mol)(0.08206 L ∙ atm/mol ∙ K)(296 K)

9.67

B surrounded by three electron pairs: electron-pair geometry = molecular geometry = trigonal planar; hybridization = sp 2; formal charge = 0



V = 3.2 × 104 L

10.7



B surrounded by four electron pairs; electron-pair geometry = molecular geometry = tetrahedral; hybridization = sp 3; formal charge = −1

According to Equation 10.5, density is inversely proportional to T (K).



d (at 55° C ) = (1.18 g/L)(298 K/328 K) = 1.07 g/L

10.8

PV = (m/M)RT; M = mRT/PV



M = (0.105 g)(0.08206 L ∙ atm/mol ∙ K)(296.2 K)/  [(561/760) atm (0.125 L)] = 27.7 g/mol

10.9

2 Na(s) + 2 H2O(ℓ) n 2 NaOH(aq) + H2(g)



Amount H2 = (15.0 g Na)(1 mol Na/22.99 g Na)  (1 mol H2/2 mol Na) = 0.3262 mol Volume of H2 = V = nRT/P = (0.3262 mol) (0.08206 L ∙ atm/mol ∙ K)(298.2 K)/1.10 atm = 7.26 L

9.69

9.71

HF has a bond order of 1. HF2− has a bond order of 1 for the entire three-center-four-electron bond and thus a 0.5 bond order per HOF linkage. It should therefore be easier to break the bond in HF2−, leading to a smaller bond enthalpy for HF2− than for HF. (a) empirical formula = Br2O



Br O Br



The O should have sp 3 hybrid orbitals. (b) 13 valence electrons; (σs)2(σ*s)2(πp)4(σp)2(π*p)3

9.73

HOMO = π*p

(a) Electron dot structure for SO3. O O



S

O

(b) Formal charges: S atom = +2; O atom with double bond = 0; O atoms with single bond = −1 (c) Structure is planar and trigonal. (d) S atom hybridization is sp2. (e) Cyclic trimer O O

S O O

O S

O S

O

O O

Chapter 10 Check Your Understanding 10.1

0.29 atm(760 mm Hg/1 atm) = 220 mm Hg 0.29 atm(1.01325 bar/1 atm) = 0.29 bar 0.29 bar(105 Pa/1 bar)(1 kPa/103 Pa) = 29 kPa

10.2

V2 = P1V1/P2 = (745 mm Hg)(65.0 L)/[0.70(745 mm Hg)] = 93 L

10.3

V1 = 45 L and T1 = 298 K; V2 = ? and T2 = 263 K



V2 = V1(T2/T1) = (45 L)(263 K/298 K) = 40. L

A-72

10.10 Phalothane (5.00 L) =   (0.0760 mol)(0.08206 L ∙ atm/mol ∙ K) (298.2 K)

Phalothane = 0.3719 atm (or 282.7 mm Hg) = 0.372 atm (or 283 mm Hg)



Poxygen (5.00 L) =   (0.734 mol)(0.08206 L ∙ atm/mol ∙ K)(298.2 K)



Poxygen = 3.592 atm (or 2730 mm Hg) = 3.59 atm (or 2730 mm Hg)



Ptotal = Phalothane + Poxygen =   282.7 mm Hg + 2730 mm Hg = 3010 mm Hg

10.11 For He: Use Equation 10.9, with M =  4.00 × 10−3 kg/mol, T = 298 K, and R =  8.314 J/mol ∙ K to calculate the rms speed of 1360 m/s. A similar calculation for N2, with M = 28.01 × 10−3 kg/mol, gives an rms speed of 515 m/s. 10.12 The molar mass of CH4 is 16.0 g/mol. Rate for CH4 n molecules/1.50 min   Rate for unknown n molecules/4.73 min

Munknown 16.0

Munknown = 159 g/mol

Applying Chemical Principles 10.1 The Atmosphere and Altitude Sickness 1. (a) A  tmospheric pressure = 0.29(760 mm Hg) = 220. mm Hg = 220 mm Hg. (b) Partial pressure of O2 = 0.21(220. mm Hg) = 46 mm Hg.

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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2.

Atmospheric Pressure vs. Elevation 800

Pressure (mm Hg)

700 600

2.70 × 102 mm Hg

10.7

61.6 mL

10.9

3.7 L

10.11 −47 °C

500

10.13 250 mm Hg

400

10.15 3.2 × 102 mm Hg

300

10.17 9.72 atm

200 100

10.5

0

1000 2000 3000 4000 5000 6000 7000 8000 9000 10,000 Elevation (m)

As altitude increases, atmospheric pressure decreases. Although the relationship at lower altitudes is close to linear, at the highest altitude there is a significant deviation from linearity.

10.19 (a) 75 mL O2 (b) 150 mL NO2 10.21 0.919 atm 10.23 V = 2.9 L 10.25 1.9 × 106 g He 10.27 3.7 × 10−4 g/L

3. Atmospheric pressure = 90 mm Hg/0.21 = 429 mm Hg Assuming the relationship between altitude and pressure is close to linear between 3000 and 5000 meters, you can estimate a value of about 4400 m for x (elevation). There are not enough data to give a truly accurate and precise answer.

10.35 0.096 atm; 73 mm Hg

10.2 The Chemistry of Airbags

10.37 170 g NaN3

1. (a) The decomposition of NaN3 produces 3 mol N2 from 2 mol NaN3. From the reaction that captures the sodium by-product, there will be an additional 1 mol N2 produced from 10 mol Na (or 0.2 mol of N2 per 2 mol Na, which corresponds to 2 mol NaN3). Adding these amounts of N2 from the two reactions shows that 3.2 mol of N2 will be produced per 2 mol of NaN3. First calculate the amount of N2 needed using PV = nRT, and then complete the stoichiometry problem.  n = (3.0 atm)(75 L)/[(0.08206 L ∙ atm/mol ∙ K) (298.2 K)] = 9.19 mol N2 9.19 mol N2(2 mol NaN3/3.2 mol N2)(65.01 g NaN3/1 mol NaN3) = 374 g NaN3 = 370 g NaN3 (b) 374 g NaN3(1 mol NaN3/65.01 g NaN3) (2 mol Na/2 mol NaN3)(2 mol KNO3/10 mol Na) (101.1 g KNO3/1 mol KNO3) = 116 g KNO3 2. (a) 9.19 mol of gases are needed (see Question 1), and 3 mol (1 mol N2O and 2 mol H2O) of gas are produced per mol of NH4NO3. (9.19 mol gases)(1 mol NH4NO3/3 mol gases) (80.04 g NH4NO3/1 mol NH4NO3) = 250 g NH4NO3 (b) 1.0 atm of N2O(g) and 2.0 atm of H2O(g)

10.39 1.7 atm O2

10.29 34.0 g/mol 10.31 57.5 g/mol 10.33 Molar mass = 74.9 g/mol; (d) B6H10

10.41 4.1 atm H2; 1.6 atm Ar; total pressure = 5.7 atm 10.43 (a) 0.30 mol halothane/1 mol O2 (b) 3.0 × 102 g halothane 10.45 (a) CO2 has the higher kinetic energy. (b) The root mean square speed of the H2 molecules is greater than the root mean square speed of the CO2 molecules. (c) The number of CO2 molecules is greater than the number of H2 molecules [n(CO2) = 1.8n(H2)]. (d) The mass of CO2 is greater than the mass of H2. 10.47 Average speed of CO2 molecules = 3.65 × 104 cm/s 10.49 Average speed increases (and molar mass decreases) in the order CH2F2 < Ar < N2 < CH4. 10.51 (a) F2 (38 g/mol) effuses faster than CO2 (44 g/mol). (b) N2 (28 g/mol) effuses faster than O2 (32 g/mol). (c) C2H4 (28.1 g/mol) effuses faster than C2H6 (30.1 g/mol). (d) CFCl3 (137 g/mol) effuses faster than C2Cl2F4 (171 g/mol). 10.53 36 g/mol

Study Questions

10.55 27.7 g/mol

10.1

(a) 481 mm Hg (b) 0.641 bar (c) 64.1 kPa

10.57 (c) 10 atm, 0 °C (high pressure and low temperature)

10.3

(a) 0.754 bar (b) 650 kPa (c) 934 kPa



10.59 P from the van der Waals equation = 26.0 atm

P from the ideal gas law = 30.6 atm

10.61 (a) As an ideal gas, P = 162 atm. Using the van der Waals equation, P = 111 atm. (b) a(n/V)2

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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A-73

10.63 

Standard atmosphere N2 partial pressure H2 pressure Air

atm

mm Hg

kPa

bar

1

760

101.325

1.013

0.780

593

79.1

0.791

131

9.98 × 104

1.33 × 104

133

0.333

253

33.7

0.337

10.65 T = 290. K or 17 °C 10.67 2 C4H9SH(g) + 15 O2(g) n  8 CO2(g) + 10 H2O(g) + 2 SO2(g)

Total pressure = 37.3 mm Hg. Partial pressures: CO2 = 14.9 mm Hg, H2O = 18.6 mm Hg, and SO2 = 3.73 mm Hg.

10.69 4 mol 10.71 V = 44.8 L O2 10.73 Ni is the limiting reactant; 1.31 g Ni(CO)4 10.75 (a, b)  Sample 4 (He) has the largest number of molecules and sample 3 (H2 at 27 °C and 760 mm Hg) has the fewest number of molecules. (c) Sample 2 (Ar) 10.77 8.54 g Fe(CO)5 10.79 S2F10 10.81 (a) 28.7 g/mol ≃ 29 g/mol (b) X of O2 = 0.17 and X of N2 = 0.83 10.83 Molar mass = 86.4 g/mol. The gas is probably ClO2F. 10.85 Calculate molar mass (44 g/mol). Gas is probably CO2. 10.87 (d) is not correct. The rate of effusion is really inversely proportional to the square root of a gas’s molar mass. 10.89 1.3 × 102 g/mol 10.91 n(He) = 0.0127 mol 10.93 Weight percent KClO3  = 69.1% 10.95 (a) NO2 < O2 < NO (b) P(O2) = 75 mm Hg (c) P(NO2) = 150 mm Hg 10.97 P(NH3) = 69 mm Hg and P(F2) = 51 mm Hg

Pressure after reaction = 17 mm Hg

10.99 (a) At 20 °C, there is 7.8 × 10−3 g H2O/L. (b) At 0 °C, there is 4.6 × 10−3 g H2O/L. The mass of water is greater at 20 °C and 45% relative humidity. 10.101 The mixture contains 0.22 g CO2 and 0.77 g CO.

P(CO2) = 0.22 atm; P(O2) = 0.12 atm; P(CO) = 1.22 atm

10.103 The formula of the iron compound is Fe(CO)5.

A-74

10.105 (a) P(B2H6) = 0.0160 atm (b) P(H2) = 0.0320 atm, so Ptotal = 0.0480 atm 10.107

Amount of Na2CO3  = 0.00424 mol Amount of NaHCO3  = 0.00951 mol Amount of CO2 produced = 0.0138 mol Volume of CO2 produced = 0.343 L

10.109 Decomposition of 1 mol of Cu(NO3)2 should give 2 mol NO2 and 1⁄2 mol of O2. Total actual amount = 4.72 × 10−3 mol of gas. Average molar mass = 41.3 g/mol Mole fractions: X(NO2) = 0.666 and X(O2) = 0.334 Amount of each gas: 3.13 × 10−3 mol NO2 and 1.57 × 10−3 mol O2. The ratio of these amounts is 1.99 mol NO2/mol O2. This is different from the 4 mol NO2/mol O2 ratio expected from the reaction. If some NO2 molecules combine to form N2O4, the apparent mole fraction of NO2 would be smaller than expected (= 0.8). As this is the case, it is apparent that some N2O4 has been formed (as is observed in the experiment). 10.111 (a) M = 138 g/mol; the unknown compound is P2F4. (b) M = 1.4 × 102 g/mol; this is consistent with the result obtained in part (a). 10.113 (a) 10.0 g of O2 represents more molecules than 10.0 g of CO2. Therefore, O2 has the greater partial pressure. (b) The average speed of the O2 molecules is greater than the average speed of the CO2 molecules. (c) The gases are at the same temperature and so have the same average kinetic energy. 10.115 (a) P(C2H2) > P(CO) (b) There are more molecules in the C2H2 container than in the CO container. 10.117 (a) Not a gas. A gas would expand to an infinite volume. (b) Not a gas. A density of 8.2 g/mL is typical of a solid. (c) Insufficient information (d) Gas 10.119 (a) There are more molecules of H2 than atoms of He. (b) The mass of He is greater than the mass of H2. 10.121 The speed of gas molecules is related to the square root of the absolute temperature, so a doubling of the temperature will lead to an increase of about (2)1/2 or 1.4.

Chapter 11 Check Your Understanding 11.1

Because F− is the smaller ion, water molecules can approach more closely and interact more strongly. Thus, F− should have the more negative enthalpy of hydration.

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11.2

% N in cyanuric acid = 42/129 × 100 = 33%

H H3C

Both of these compounds have a greater percentage of nitrogen than the average protein.

O H

O CH3



Hydrogen bonding in methanol entails the attraction of the hydrogen atom bearing a partial positive charge (δ+) on one molecule to the oxygen atom bearing a partial negative charge (δ−) on a second molecule. The strong attractive force of hydrogen bonding will cause the boiling point and the enthalpy of vaporization of methanol to be quite high.

11.3

11.4

11.5 11.6

Water is a polar solvent, while hexane and CCl4 are nonpolar. London dispersion forces are the primary forces of attraction between all pairs of dissimilar solvents. For mixtures of water with the other solvents, dipole–induced dipole forces will also be important.

Study Questions 11.1

11.3

(a) O2: induced dipole–induced dipole forces only. (b) CH3OH: strong hydrogen bonding (dipole–dipole forces) as well as induced dipole–induced dipole forces. (c) Forces between water molecules: strong hydrogen bonding and induced dipole–induced dipole forces. Between N2 and H2O: dipole–induced dipole forces and induced dipole–induced dipole forces. (1.00 × 103 g)(1 mol/32.04 g)(35.2 kJ/mol) =  1.10 × 103 kJ



(a) At 40 °C, the vapor pressure of ethanol is about 120 mm Hg. (b) The equilibrium vapor pressure of ethanol at 60 °C is about 320 mm Hg. At 60 °C and 600 mm Hg, ethanol is a liquid. If vapor is present, it will condense to a liquid.

Applying Chemical Principles 11.1 Chromatography 1. (a) The 1,5-pentanediol is most attracted to the mobile phase. The primary attractive force is hydrogen bonding. Dipole–dipole forces and dispersion forces are also present. (b) The ethyl propyl ether is most attracted to the stationary phase by dispersion forces but there are also some dipole–induced dipole interactions. (c) The molecules will elute in the following order (first to last): 1,5-pentanediol, 1-pentanol, and propyl ethyl ether. 2. Order of elution from first to last: pentane, hexane, heptane, octane. The forces of attraction between the hydrocarbons and the stationary phase increase in this order. 11.2 A Pet Food Catastrophe 1. % N in melamine = 84/126 × 100 = 67%



2. 454 g sample (0.14 g melamine/1,000,000 g sample) = 6.4 × 10−5 g melamine = 0.064 mg melamine = 64 μg melamine

(a) Dipole–dipole interactions (and hydrogen bonds) and induced dipole–induced dipole interactions (b) Induced dipole–induced dipole forces (c) Dipole–dipole interactions (and hydrogen bonds) and induced dipole–induced dipole interactions (a) Induced dipole–induced dipole forces (b) Dipole–dipole and induced dipole–induced dipole forces (c) Induced dipole–induced dipole forces (d) Dipole–dipole forces (and hydrogen bonding) and induced dipole–induced dipole forces

11.5

(a), (b), and (c). The boiling points of these liquids are: Ne, (−246 °C), CO (−192 °C), CH4 (−162 °C), and CCl4 (77 °C).

11.7

(c) HF; (d) acetic acid; (f) CH3OH

11.9

(a) MgCl2. The Mg2+ ion is smaller than the Ba2+ ion (Figure 7.12), which makes the ion–dipole forces of attraction stronger for MgCl2. (b) Ca(NO3)2. The Ca2+ ion is smaller than the K+ ion (Figure 7.12), and the Ca2+ ion has a 2+ charge (as opposed to 1+ for potassium). Both of these effects lead to stronger ion–dipole forces of attraction for calcium nitrate. (c) ZnI2. The zinc(II) ion has a larger charge than Cs+ and is considerably smaller. Both effects mean that there are stronger ion–dipole forces of attraction for zinc(II) iodide.



11.11 q = +90.1 kJ 11.13 (a) Water vapor pressure is about 150 mm Hg at 60 °C. (Appendix G gives a value of 149.4 mm Hg at 60 °C.) (b) 600 mm Hg at about 93 °C (c) At 70 °C, ethanol has a vapor pressure of about 520 mm Hg, whereas that of water is about 225 mm Hg. 11.15 At 30 °C, the vapor pressure of ether is about 590 mm Hg. (This pressure requires 0.23 g of ether in the vapor phase at the given conditions, so there is sufficient ether in the flask.) At 0 °C, the vapor pressure is about 160 mm Hg, so some ether condenses when the temperature declines. 11.17

(a) O2 (−183 °C) (bp of N2 = −196 °C) (b) SO2 (−10 °C) (CO2 sublimes at −78 °C) (c) HF (+19.7 °C) (HI, −35.6 °C) (d) GeH4 (−90.0 °C) (SiH4, −111.8 °C)

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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A-75

11.19 (a) CS2, about 620 mm Hg; CH3NO2, about 80 mm Hg (b) CS2, induced dipole–induced dipole forces; CH3NO2, dipole–dipole forces and induced dipole–induced dipole (c) CS2, about 46 °C; CH3NO2, about 100 °C (d) About 39 °C (e) About 95 °C

enthalpies of vaporization here are given at the boiling point of the liquid.) 11.39 5.49 × 1019 atoms/m3 11.41 (a) 70.3 °C (b)  7 6 ln (Vapor pressure)

11.21 (a) 80.1 °C (b) At about 48 °C, the liquid has a vapor pressure of 250 mm Hg. The vapor pressure is 650 mm Hg at 75 °C. (c) 33.5 kJ/mol (from slope of plot) 11.23 No, CO cannot be liquefied at room temperature because the critical temperature is lower than room temperature. 11.25 An application of surface tension is that it is possible to float a paperclip on the surface of water, but if the paperclip breaks through the surface, it will sink. Surface tension results from intermolecular forces because molecules in the interior of a liquid interact through intermolecular forces with molecules all around them, but molecules on the surface have intermolecular forces only with molecules at or below the surface layer and thus feel a net inward force of attraction. 11.27 This phenomenon results from capillary action. The water is attracted to OOH groups in the paper. As a result of these adhesive forces, some water molecules begin to move up the paper. Other water molecules remain in contact with these water molecules by means of cohesive forces. A stream of water molecules thus moves up the paper. 11.29 These three molecules differ in the attachment of the hydroxyl, OOH, group relative to the carboxyl, OCOOH, group. In 2-hydroxybenzoic acid the hydroxyl group is adjacent to the carboxyl group. The relative closeness of the two groups interferes with the formation of strong, multiple hydrogen bonds from one 2-hydroxybenzoic acid molecule to other molecules in the solid state. This lowers the melting point relative to the other two molecules. 11.31 (a) Induced dipole–induced dipole, dipole–dipole, hydrogen bonding (b) Induced dipole–induced dipole, dipole–dipole (c) Induced dipole–induced dipole, dipole–dipole 11.33

11.35

Li+

+

ions are smaller than Cs ions (78 pm and 165 pm, respectively; see Figure 7.12). Thus, there will be a stronger attractive force between Li+ ions and water molecules than between Cs+ ions and water molecules. (a) 350 mm Hg (b) Ethanol (lower vapor pressure at every temperature) (c) 84 °C (d) CS2, 46 °C; C2H5OH, 78 °C; C7H16, 99 °C (e) CS2, gas; C2H5OH, gas; C7H16, liquid

11.37 Molar enthalpy of vaporization increases with increasing intermolecular forces: C2H6 (14.69 kJ/mol; induced dipole) < HCl (16.15 kJ/mol; dipole) < CH3OH (35.21 kJ/mol, hydrogen bonds). (The molar

A-76

5 4 3 2 1 0 0.002



0.003 1/T (K)

0.004

Using the equation for the straight line in the plot



ln P = −3885 (1/T) + 17.949

we calculate that T = 311.6 K (39.5 °C) when P = 250 mm Hg. When P = 650 mm Hg, T = 338.7 K (65.5 °C). (c) Calculated ΔvapH = 32.3 kJ/mol 11.43 (a) When the can is inverted in cold water, the water vapor pressure in the can, which was approximately 760 mm Hg, drops rapidly—say, to 9 mm Hg at 10 °C. This creates a partial vacuum in the can, and the can is crushed because of the difference in pressure inside the can and the pressure of the atmosphere pressing down on the outside of the can. (b) 

Before heating

After heating

11.45 Acetone and water can interact by hydrogen bonding. hydrogen bond −

H3C

O C

+

O H

+

−

H

+

CH3

11.47 Ethylene glycol’s viscosity will be greater than ethanol’s, owing to the greater hydrogen-bonding capacity of ethylene glycol. 11.49 (a) Water has two OH bonds and two lone pairs, whereas the O atom of ethanol has only one OH bond (and two lone pairs). More extensive hydrogen bonding is likely for water.

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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(b) Water and ethanol interact extensively through hydrogen bonding, so the volume is expected to be slightly smaller than the sum of the two volumes.

11.51 Two pieces of evidence for H2O(ℓ) having considerable intermolecular attractive forces: (a) Based on the boiling points of the Group 6A (16) hydrides (Figure 11.4), the boiling point of water should be approximately −80 °C. The actual boiling point of 100 °C reflects the significant hydrogen bonding that occurs. (b) Liquid water has a specific heat capacity that is higher than almost any other liquid. This reflects the fact that a relatively larger amount of energy is necessary to overcome intermolecular forces and raise the temperature of the liquid. 11.53 (a) HI, hydrogen iodide (b) The large iodine atom in HI leads to a significant polarizability for the molecule and thus to a large dispersion force. (c) The dipole moment of HCl (1.07 D, Table 8.6) is larger than for HI (0.38 D). (d) HI. See part (b). 11.55 A gas can be liquefied at or below its critical temperature. The critical temperature for CF4 (−45.7 °C) is below room temperature (25 °C), so it cannot be liquefied at room temperature.

Chapter 12 Check Your Understanding 12.1

(a) The strategy to solve this problem is given in Example 12.1.



Step 1. Mass of the unit cell = (197.0 g/mol)(1 mol/6.022 × 1023 atom) (4 atoms/unit cell) = 1.3085 × 10−21 g/unit cell



Step 2. Volume of unit cell = (1.3085 × 10−21 g/unit cell)(1 cm3/19.32 g) = 6.7730 × 10−23 cm3/unit cell

Step 3. Length of side of unit cell = [6.7730 × 10−23 cm3/unit cell]1/3  = 4.0762 × 10−8 cm Step 4. Calculate the radius from the edge dimension. Diagonal distance = 4.0762 × 10−8 cm (2½) =  4 (rAu) rAu = 1.441 × 10−8 cm (= 144.1 pm) (b) To verify a body-centered cubic structure, calculate the mass contained in the unit cell. If the structure is bcc, then the mass will be the mass of 2 Fe atoms. (Other possibilities: fcc − mass of 4 Fe; primitive cubic − mass of 1 Fe atom.) This calculation uses the four steps from the previous exercise in reverse order.

11.57 Hydrogen bonding is most likely at the OOH group at the “right” end of the molecule, and at the CPO and NOH groups in the amide group (ONHOCOO).

Step 1. Use radius of Fe to calculate cell dimensions. In a body-centered cube, atoms touch across the diagonal of the cube.

11.59 Boiling point, enthalpy of vaporization, volatility, surface tension

Side dimension of cube =  4 (1.26 × 10−8 cm)/( 3 ) = 2.910 × 10−8 cm

11.61 The more branching in the hydrocarbons, the lower the boiling point. This implies weaker induced dipole–induced dipole forces. Greater branching results in a more compact shape with less surface area available for contact and therefore smaller induced dipole–induced dipole forces. 11.63 F2 < Cl2 < Br2 < I2 He < Ne < Ar < Kr < Xe

Molar mass and boiling point correlate with these orders.

11.65 The boiling point of the water in the pressure cooker is 121 °C. 11.67 Using the equation of the line from Question 11.66, P = 8.4 atm. 11.69 (a) There will be water in the tube in equilibrium with its vapor, but almost all of the water will be present as liquid water. (b) 760. mm Hg (c) 10.4 cm3 (d) 0.0233 g H2O







Diagonal distance  = side dimension ( 3 ) = 4 rFe

Step 2. Calculate unit cell volume

Unit cell volume = (2.910 × 10−8 cm)3 =  2.464 × 10−23  cm3 Step 3. Combine unit cell volume and density to find the mass of the unit cell. Mass of unit cell = (2.464 × 10−23  cm3) × (7.8740 g/cm3) = 1.94 × 10−22 g Step 4. Calculate the mass of 2 Fe atoms, and compare this to the answer from step 3. Mass of 2 Fe atoms = 55.85 g/mol (1 mol/6.022 × 1023 atoms)(2 atoms) = 1.85 × 10−22 g. This is a fairly good match, and clearly much better than the two other possibilities, primitive and fcc. 12.2

M2X. In a face-centered cubic unit cell, there are four anions and eight tetrahedral holes in which to place metal ions. All of the tetrahedral holes are inside the unit cell, so the ratio of atoms in the unit cell is 2∶1.

12.3

Calculate the mass and volume of the unit cell. The density of KCl will then be mass/volume. Select units so the density is calculated as g/cm3.



Step 1. Mass: The unit cell contains 4 K+ ions and 4 Cl− ions.

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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A-77





Unit cell mass = (39.10 g/mol)(1 mol/6.022 × 1023 K+ ions)(4 K+ ions) + (35.45 g/mol)(1 mol/6.022 × 1023 Cl− ions)(4 Cl− ions) = 2.5971 × 10−22 g +  2.3547 × 10−22 g = 4.9518 × 10−22 g Step 2. Volume: Assuming K+ and Cl− ions touch along one edge of the cube, the side dimension =  2 r K+ + 2 rCl−. The volume of the cube is the cube of this value. (Convert the ionic radius from pm to cm.)



V = [2(1.33 × 10−8 cm) + 2(1.81 × 10−8 cm)]3 =  2.477 × 10−22 cm3



Step 3. Density = mass/volume =  4.9518 ×10−22 g/2.477 × 10−22 cm3) = 2.00 g/cm3

Applying Chemical Principles 1. 73,000,000 metric tons Li2CO3 (13.88 metric tons Li/ 73.89 metric tons Li2CO3) = 1.4 × 107  metric tons Li (= 14 million metric tons Li) 2. The unit cell for lithium metal is body-centered cubic. (There are atoms at each of the corners of a cube and one atom embedded in the middle of the cube.) 3. 351 pm (1 m/1012 pm)(100 cm/1 m) = 3.51 × 10−8 cm V of unit cell = (3.51 × 10−8 cm)3 = 4.324 × 10−23 cm3 Mass of unit cell = 6.94 g Li/mol (1 mol/6.022 × 1023 atoms)(2 atoms/unit cell) = 2.305 × 10−23 g/unit cell Density = mass/volume = 2.305 × 10−23 g/4.324 × 10−23 cm3 = 0.533 g/cm3 4. One method is to carry out the following gas-forming acid–base reaction: Li2CO3(aq) + 2 HCl(aq) n  2 LiCl(aq) + H2O(ℓ) + CO2(g) 5. MnO2 unit cell Volume of unit cell = 5.56706 × 10−23 cm3 Mass of unit cell (with 2 Mn and 4 O atoms) = 2.88721 × 10−22 g Density of MnO2 = 5.1862 g/cm3 To three significant figures, this is 5.19 g/cm3. 12.2 Nanotubes and Graphene—Network Solids 1. We want to know the distance x in the diagram of the hexagon. The interior angles in a hexagon are all 120°. To find x, we can find the distance y in the figure and then double it (x = 2y). The angle bounded by the side of the hexagon and y is 30°, and, from geometry, cos 30° = y/(hexagon side) so y = (cos 30°)(139 pm) = (0.866)(139 pm) = 120.4 pm. Finally, x = 240.8 pm = 241 pm. 30°

3. The thickness would be approximately 150 pm. This ­corresponds to the diameter of a carbon atom, found by multiplying the radius of a carbon atom (given in ­Figure 7.6) by 2 and rounding off to two significant figures. 4. The six-sided benzene ring (with sp2-hybridized carbon atoms) is the basic structural feature (page 472). 12.3 Tin Disease 1. β-Tin has four atoms in the unit cell. α-Tin has eight atoms in the unit cell. 2. Volume of unit cell = (5.83 × 10−8 cm)(5.83 × 10−8 cm) (3.18 × 10−8 cm) = 1.081 × 10−22 cm3

12.1 Lithium and Electric Vehicles

139 pm

2. 1.0 μm (1 m/106 μm)(1012 pm/1 m)(1 C6-ring/240.8 pm) = 4.2 × 103 C6-rings

Mass of unit cell = (118.7 g Sn/1 mol Sn) (1 mol/6.022 × 1023 atoms)(4 atoms/unit cell) = 7.8844 × 10−22 g Density =  7.8844 × 10−22 g/1.081 × 10−22 cm3 = 7.29 g/cm3 3. There are eight atoms per unit cell (8 corner atoms × 1/8 + 6 face atoms × 1/2 + 4 body atoms × 1 = 8). Mass of unit cell = (118.7 g Sn/1 mol Sn) (1 mol/6.022 × 1023 atoms)(8 atoms/unit cell) = 1.5769 × 10−21 g Volume = mass/density = 1.5769 × 10−21 g/5.769 g/cm3 = 2.7334 × 10−22 cm3 Length of one side = (2.7334 × 10−22 cm3)1/3 = 6.490 × 10−8 cm = 649.0 pm 4. Tetragonal: The unit cell of β-tin contains 4 atoms. The volume of the unit cell (from question 2) is 1.081 × 10−22 cm3 = 1.081 × 108 pm3. The volume of space occupied by four atoms is V = 4(4/3)(πr3) = 4[(4/3) π(141 pm)3] = 4.697 × 107 pm3. The percent of space occupied is (4.697 × 107 pm3/1.081 × 108 pm3) × 100% = 43.5%. Cubic: The unit cell of gray tin contains 8 atoms. The volume of the unit cell (from question 3) is 2.7334 × 10−22 cm3 = 2.7334 × 108 pm3. The volume of space occupied by eight atoms is V = 8(4/3)(πr3) = 8[(4/3)π(141 pm)3] = 9.394 × 107 pm3. The percent of space occupied is (9.394 × 107 pm3/2.7334 × 108 pm3) × 100% = 34.4%.

Study Questions 12.1

Two possible unit cells are illustrated here. The simplest formula is AB8.

y x

120°

A-78

  

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12.23 In carbon the band gap is too large for electrons to move up in energy from the valence band to the conduction band, whereas in silicon the band gap is small enough to permit this.

12.3

(a) Body-centered cubic



(b) Mass of unit cell = (39.10 g K/1 mol K) (1 mol/6.022 × 1023 atoms)(2 atoms/unit cell) = 1.2986 × 10−22 g



Volume of unit cell = (5.33 × 10−8 cm)3 = 1.514 × 10−22 cm3



Density = 1.2986 × 10−22 g/1.514 × 10−22 cm3 = 0.858 g/cm3

12.25 An intrinsic semiconductor (such as Si or Ge) is one that naturally can have electrons move across the band gap, whereas an extrinsic semiconductor (such as Ge with added As) requires the addition of dopants in order for it to conduct.

12.5

Ti4+: 8 corners (1⁄8 Ti4+ per corner) = 1 net Ti4+ ion

12.27 (c) Buckyballs (made up of C60 molecules)



O2−: 12 edges (1⁄4 O2− per edge) = 3 net O2− ions



Ca2+: 1 Ca2+ in center of unit cell = 1 net Ca2+ ion



Formula = CaTiO3

12.7



(a) There are eight O2− ions at the corners and one in the center for a net of two O2− ions per unit cell. There are four Cu ions in the interior in tetrahedral holes. The ratio of ions is Cu2O. (b) The oxidation number of copper must be +1.

12.29 (a) Eight C atoms per unit cell. There are eight corners (= 1 net C atom), six faces (= 3 net C atoms), and four internal C atoms. (b) Face-centered cubic (fcc) with C atoms in the tetrahedral holes

12.9

Calcium atom radius = 197 pm

12.11 There are three ways the edge dimensions can be calculated: (a) Calculate mass of unit cell (= 1.103 × 10−21 g/uc) Calculate volume of unit cell from mass (= 3.53 × 10−22 cm3/uc) Calculate edge length from volume (= 707 pm) (b) Assume I− ions touch along the cell diagonal (Check Your Understanding 12.1) and use I− radius to find the edge length. Radius I− = 220 pm Edge = 4(220 pm)/21/2 = 622 pm (c) Assume the I− and K+ ions touch along the cell edge (page 591)

Edge = 2 × I− radius + 2 × K+ radius = 706 pm

Methods (a) and (c) agree. It is apparent that the sizes of the ions are such that the I− ions cannot touch along the cell diagonal.

12.13 Least to most negative lattice energy: RbI, LiI, LiF, CaO 12.15 As the ion–ion distance decreases, the force of attraction between ions increases. This should make the lattice more stable, and more energy should be required to melt the compound.

12.31 Substance: type of solid, particles, forces, property (a) Gallium arsenide: network, covalently bonded atoms, covalent bonds, semiconductor (b) Polystyrene: amorphous, covalent bonds within the polymer molecules and dispersion forces between the polymer molecules, thermal insulator (c) Silicon carbide: network, covalently bonded atoms, covalent bonds, very hard material (d) Perovskite: ionic, Ca2+ and TiO32− ions, ion–ion interactions, high melting point 12.33 q (for fusion) = −1.97 kJ; q (for melting) = +1.97 kJ 12.35 (a) The density of liquid CO2 is less than that of solid CO2. (b) CO2 is a gas at 5 atm and 0 °C. (c) Critical temperature = 31 °C, so CO2 cannot be liquefied at 45 °C. 12.37 q (to heat the liquid) = 9.42 × 102 kJ

q (to vaporize NH3) = 1.64 × 104 kJ



q (to heat the vapor) = 8.79 × 102 kJ



qtotal = 1.8 × 104 kJ

12.39 O2 phase diagram. (i) Note the slight positive slope of the solid–liquid equilibrium line. It indicates that the density of solid O2 is greater than that of liquid O2. (ii) Using the diagram here, the vapor pressure of O2 at 77 K is between 150 and 200 mm Hg. 800

12.21 In metals, thermal energy causes some electrons to occupy higher-energy orbitals in the band of molecular orbitals. For each electron promoted, two singly occupied levels result: a negative electron above the Fermi level and a positive hole below the Fermi level. Electrical conductivity results because in the presence of an electric field, these negative electrons will move toward the positive side of the field and the positive holes will move toward the negative side.

600 Pressure (mm Hg)

12.19 The 1000 2s orbitals will combine to form 1000 molecular orbitals. In the lowest energy state, 500 of these will be populated by pairs of electrons, and 500 will be empty.

SOLID

12.17 Δf H° = −607 kJ/mol

Normal freezing point

Normal boiling point

LIQUID

400

GAS

200 Triple point 0 50

60

70 80 Temperature (K)

90

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

100

A-79

12.41 Radius of silver = 145 pm

Pb2+ ions and four S2− ions per unit cell, a 1:1 ratio that matches the compound formula.

12.43 Nitinol density = 5.84 g/cm3

12.63 4 formula units/unit cell; mass of unit cell = 7.905 × 10−22 g

12.45 1.356 × 10−8 cm (literature value is 1.357 × 10−8 cm) 12.47 Mass of 1 CaF2 unit calculated from crystal data = 1.29633 × 10−22 g. Divide molar mass of CaF2 (78.077 g/mol) by mass of 1 CaF2 to obtain A ­ vo­gadro’s number. Calculated value = 6.0230 × 1023 CaF2/mol. 12.49 Diagram A leads to a surface coverage of 78.5%. Diagram B leads to 90.7% coverage. 12.51 (a) The lattice can be described as an fcc lattice of Si atoms. (b) Si atoms are located in one half of the tetrahedral holes. (c) There are eight Si atoms in the unit cell. (d) Mass of unit cell = 3.7311 × 10−22 g Volume of unit cell = 1.6019 × 10−22 cm3



Density = 2.329 g/cm3 (which is the same as the literature value) (e) In the Si unit cell we cannot assume the atoms touch along the edge or along the face diagonal. Instead, we know that the Si atoms in the tetrahedral holes are bonded to the Si atoms at the corner. Si atom in tetrahedral hole Si atom in middle of face

Si

109.5°/2

384 pm/2

Si

Si atom at cell corner

Si

Distance = 1/2 (cell diagonal) = 384 pm



Distance across cell face diagonal = 768.1 pm



Sin (109.5°/2) = 0.8166 = (384.0 pm/2)/(Si-Si distance)



Distance from Si in tetrahedral hole to face or corner Si = 235.1 pm



Si radius = 118 pm



Figure 7.6 gives Si radius as 117 pm.



Dimension of one side of unit cell = 2r(K+) + 2r(Br−) = 6.58 × 10−8 cm; so volume of the unit cell = side3 = 2.849 × 10−22 cm3



Density = 7.905 × 10−22 g/2.849 × 10−22 cm3 = 2.77 g/cm3

12.65 (a) BBr3(g) + PBr3(g) + 3 H2(g) n BP(s) + 6 HBr(g) (b) If B atoms are in an fcc lattice, then the P atoms must be in 1⁄2 of the tetrahedral holes. (In this way it resembles Si in Question 12.51.) (c) Unit cell volume = 1.092 × 10−22 cm3 Unit cell mass = 2.775 × 10−22 g Density = 2.54 g/cm3 (d) The solution to this problem is identical to Question 12.51. In the BP lattice, the cell face diagonal is 676.0 pm. Therefore, the calculated BP distance is 207 pm. 12.67 (a) There are 6 Ni atoms in the cube faces, each ­contributing 1⁄2 an atom for a net of 3 Ni atoms. The 8 corner Al atoms contribute a net of 1 atom, for a formula of Ni3Al. (b) The mass of the unit cell is 3.37 × 10−22 g. Using the density the calculated volume is 4.373 × 10−23 cm3, and so the cell dimension is 3.52 × 10−8 cm (the ­literature value). 12.69 Step 1. There are four atoms within a ccp unit cell. Calculate the volume of these four atoms (V of a sphere = (4/3)π r3).

Step 2. Calculate the total volume of the unit cell (Vcell = side3).



Step 3. Percent occupancy = (Vatoms/Vcell)100%

12.71 (a) Two triple points. One with diamond, graphite and liquid C; the second with liquid C, graphite, and carbon vapor. (b) No (c) Diamond (d) Graphite is the stable form of carbon at room temperature and 1 atmosphere of pressure. 12.73 Assuming the spheres are packed in an identical way, the water levels are the same. A face-centered cubic lattice, for example, uses 74% of the available space, regardless of the sphere size.

12.53 (a) Mg2+ ions are in 1⁄8 of the eight possible tetrahedral holes, and Al3+ ions are in 1⁄2 of the four available octahedral holes. (b) Fe2+ ions are in 1⁄8 of the eight possible tetrahedral holes, and Cr3+ ions are in 1⁄2 of the four available octahedral holes.

Chapter 13

12.55 8.5 × 10 nm

13.1

2

12.57 Germanium has a smaller band gap, so it has a higher conductivity. 12.59 Because boron is electron deficient in comparison to carbon, this is a p-type semiconductor. 12.61 Lead(II) sulfide has the same structure as sodium chloride, not the same structure as ZnS. There are four

A-80

Check Your Understanding (a) 10.0 g sucrose = 0.02921 mol; 250. g H2O =  13.88 mol

Xsucrose = (0.02921 mol)/(0.02921 mol + 13.88 mol) = 0.00210 

csucrose = (0.02921 mol sucrose)/(0.250 kg solvent) = 0.117 m



Weight % sucrose =  (10.0 g sucrose/260.0 g soln)(100%) = 3.85%

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

(b) 1.08 × 104 ppm = 1.08 × 104 mg Na+ per 1000 g soln = (1.08 × 104 mg Na+/1000 g soln) (1050 g soln/1 L) = 1.134 × 104 mg Na+/L = 11.34 g Na+/L



 13.2

13.3 13.4

(11.34 g Na+/L)(58.44 g NaCl/22.99 g Na+) =  28.8 g NaCl/L

ΔsolnH° = Δf H°[NaOH(aq)] − Δf H°[NaOH(s)] = −469.2 kJ/mol − (−425.9 kJ/mol) = −43.3 kJ/mol Solubility of CO2 = kHPg = 0.034 mol/kg ∙ bar × 0.33 bar = 1.1 × 10−2 mol/kg The solution contains sucrose [(10.0 g)(1 mol/342.3 g)  = 0.02921 mol] in water [(225 g)(1 mol/18.02 g) =  12.49 mol].



Xwater = (12.49 mol H2O)/(12.49 mol +  0.02921 mol) = 0.9977



Pwater = XwaterP°water = 0.9977(149.4 mm Hg) =  149 mm Hg

13.5

cglycol = ΔTbp/Kbp = 1.0 °C/(0.512 °C/m) = 1.95 m  = 1.95 mol/kg

Applying Chemical Principles 13.1 Distillation 1. X(hexane) = 0.59 2. Four plates 3. Mass hexane = 0.20 mol C6H14(86.18 g C6H14/1 mol  C6H14) = 17.2 g C6H14 Mass heptane = 0.80 mol C7H16(100.21 g C7H16/1 mol  C7H16) = 80.2 g C7H16 Mass percent hexane = 17.2 g C6H14/(17.2 g C6H14 + 80.2 g C7H16) × 100% = 18% hexane 4. ln(P2/P1) = −(ΔvapH/R)(1/T2 − 1/T1) ln(361.5 mm Hg/760 mm Hg) = −(ΔvapH/ 0.0083145 kJ/ mol ∙ K)(1/348.2 K − 1/371.6 K) ΔvapH = 34.2 kJ/mol 13.2 Henry’s Law and Exploding Lakes 1. PV = nRT 4.0 atm(0.025 L) = n(0.08206 L ∙ atm/mol ∙ K)(298 K); n = 4.1 × 10−3 mol



massglycol = (1.95 mol/kg)(0.125 kg)(62.07 g/mol)  = 15 g

13.6

cglycol = (525 g)(1 mol/62.07 g)/(3.00 kg) = 2.819 m

4.0 atm(0.025 L) = 3.7 × 10−4 atm (V2); V2 = 270 L



ΔTfp = Kfp × m = (−1.86 °C/m)(2.819 m)  = −5.24 °C

The gas expanded by a factor of 11,000 (= 270 L/0.025 L).



You will be protected only to about −5 °C and not to −25 °C.

13.7

c (mol/L) = (0.033 g bradykinin)(1 mol bradykinin/1060 g bradykinin)/0.0500 L = 6.23 × 10−4 M



Π = cRT = (6.23 × 10−4 mol/L) (0.082057 L ∙ atm/mol ∙ K)(293 K) = 0.015 atm

13.8

ΔTfp = 5.265 °C − 5.50 °C = −0.235 °C



ΔTfp = Kfp × msolute



msolute = −0.235 °C/−5.12 °C/m = 0.0459 mol/kg



nsolute = (0.0459 mol solute/kg benzene) (0.02346 kg benzene) = 0.00108 mol solute



M = 0.448 g/0.00108 mol = 416 g/mol



M(compound)/M(empirical formula) = 416 g/mol/104.1 g/mol = 4.0



Molecular formula = (C2H5)8Al4F4

13.9

c (mol/L) = Π/RT = [(1.86 mm Hg)(1 atm/ 760 mm Hg)]/[(0.08206 L ∙ atm/mol ∙ K)(298 K)] = 1.001 × 10−4 M



(1.001 × 10−4 mol/L)(0.100 L) = 1.001 × 10−5 mol



Molar mass = 1.40 g/1.001 × 10−5 mol =  1.40 × 105 g/mol



(Assuming the polymer is composed of CH2 units, the polymer is about 10,000 units long.)

13.10 cNaCl =  (25.0 g NaCl)(1 mol/58.44 g)/(0.525 kg)  = 0.8148 m



ΔTfp =  Kfp × m × i = (−1.86 °C/m)(0.8148 m)(1.85)  = −2.80 °C

2. P1V1 = P2V2

3. Solubility of CO2 = kHPg = 0.034 mol/kg ∙ bar (3.7 × 10−4 bar) = 1.26 × 10−5 mol/kg = 1.3 × 10−5 mol/kg 4. Before opening: Solubility of CO2 before opening = kHPg = 0.034 mol/kg ∙ bar (4.0 bar) = 0.136 mol/kg Assume the density of the solution is 1.0 g/cm3, therefore 1.0 L corresponds to 1.0 kg. The amount of CO2 dissolved is 0.136 mol. After opening: Solubility of CO2 = 1.26 × 10−5 mol/kg as calculated in part 3 above. The amount of CO2 in 1.0 kg is 1.26 × 10−5 mol. Mass of CO2 released = (0.136 mol CO2 − 1.26 × 10−5 mol CO2)(44.01 g CO2/1 mol CO2) = 6.0 g CO2 13.3 Narcosis and the Bends 1. Density of water in SI units = 1.00 g/cm3(1 kg/1000 g) (100 cm/1 m)3 = 1.00 × 103 kg/m3 Pressure of water in pascals = density × acceleration due to gravity × depth = 1.00 × 103 kg/m3(9.81 m/s2) (10.0 m) = 9.81 × 104 Pa Pressure of water in atm = 9.81 × 104 Pa(1 atm/ 101325 Pa) = 0.9682 atm Total pressure = 1.000 atm + 0.9682 atm = 1.9682 atm = 1.968 atm Partial pressure of N2 = XN2(Ptotal) = 0.7808(1.9682 atm) = 1.537 atm Partial pressure of O2 = XO2(Ptotal) = 0.2095(1.9682 atm) = 0.4123 atm

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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A-81

2. Partial pressure of O2 = XO2(Ptotal) = 0.2095(3.0 atm) = 0.629 atm Solubility of O2 = kHPg = (1.3 × 10−3 mol/kg ∙ bar) (0.629 atm)(1 bar/0.98692 atm) = 8.28 × 10−4 mol/kg = 8.3 × 10−4 mol/kg Assume the density of water is 1.00 g/cm3, so 1.0 L corresponds to 1.0 kg of water. 1.0 kg(8.28 × 10−4 mol O2/kg)(32.00 g O2/mol O2) = 0.026 g O2 3. Partial pressure of N2 = XN2(Ptotal) = 0.7808(1.0 atm) = 0.781 atm Solubility of N2 = kHPg = (6.0 × 10−4 mol/kg ∙ bar) (0.781 atm)(1 bar/0.98692 atm) = 4.75 × 10−4 mol/kg Solubility of O2 = kHPg = (1.3 × 10−3 mol/kg ∙ bar) [(0.2095)(1.0 atm)(1 bar/0.98692 atm)] = 2.76 × 10−4 mol/kg In 1.0 L (= 1.0 kg) of H2O, there are 4.75 × 10−4 mol N2 ​ and 2.76 × 10−4 mol O2 that will be expelled. XO2 = 2.76 × 10−4 mol/(2.76 × 10−4 mol + 4.75 × 10−4  mol) = 0.37

Compound

13.33 (a) Molality = 8.60 m; (b)  28.4% 13.35 Molality = 0.1948 m; ΔTfp = fp = −0.362 °C 13.37 Solvent flows by osmosis from the solution less concentrated in solute to the solution more concentrated in solute. Thus, water will flow through the semipermeable membrane from the 0.01 M sucrose solution to the 0.1 M sucrose solution, diluting the more concentrated sucrose solution. 13.39 Π = 5.31 atm 13.41 (a) ΔTfp = −0.3482 °C; fp = −0.348 °C (b) ΔTbp = +0.09587 °C; bp = 100.0959 °C (c) Π = 4.58 atm

The osmotic pressure is large and can be measured with a small experimental error.

13.43 Molar mass = 360 g/mol; C20H16Fe2 13.47 Molar mass = 170 g/mol 13.49 Molar mass = 6.0 × 103 g/mol

Concentration (m) = 0.0254 m Mole fraction of acid = 4.57 × 10−4 Weight percent of acid = 0.299%

13.3

13.31 ΔTbp = 0.8082 °C; solution boiling point = 62.51 °C

13.45 Molar mass = 150 g/mol

Study Questions 13.1

13.29 Calculated boiling point = 84.15 °C

Molality

13.51 Freezing point = −24.6 °C

Weight Percent

Mole Fraction

13.53 0.08 m CaCl2 < 0.1 m NaCl < 0.04 m Na2SO4 < 0.1 m sugar 13.55 Π = 7.25 atm

NaI

0.15

2.2

2.7 × 10

13.57 This is a sol, a solid dispersed in a liquid.

C2H5OH

1.1

5.0

0.020

4.9

2.7 × 10

13.59 (a) BaCl2(aq) + Na2SO4(aq) n BaSO4(s) + 2 NaCl(aq) (b) Initially, the BaSO4 particles form a colloidal suspension. (c) Over time, the particles of BaSO4(s) grow and precipitate.

C12H22O11

0.15

−3

13.5

2.65 g Na2CO3; X(Na2CO3) = 3.59 × 10−3

13.7

210 g glycerol; 5.5 m

13.9

(a) 16.2 m; (b)  37.1%; (c)  XHCl = 0.226

−3

13.11 Molality = 2.6 × 10−5 m (assuming that 1 kg of seawater is equivalent to 1 kg of solvent) 13.13 (b) and (c) 13.15 (a) +0.8 kJ/mol (b) +25.7 kJ/mol 13.17 Raise the temperature of the solution, and add some NaCl. Above 40 °C the solubility increases with temperature. 13.19 2 × 10−3 g O2 13.21 1100 mm Hg or 1.5 bar 13.23 (a) S = 2.0 × 10−3 mol/kg (b) Raising the pressure to 1.7 atm will result in an increased concentration of dissolved O2; lowering the pressure to 1.0 atm will lead to a lower concentration of dissolved O2.

13.61 Molar mass = 110 g/mol 13.63 (a) Increase in vapor pressure of water 0.20 m Na2SO4 < 0.50 m sugar < 0.20 m KBr < 0.35 m ethylene glycol (b) Increase in boiling point 0.35 m ethylene glycol < 0.20 m KBr < 0.50 m sugar < 0.20 m Na2SO4 13.65 (a) 0.4564 mol DMG and 11.40 mol ethanol; X(DMG) = 0.0385 (b) 0.869 m (c) VP ethanol over the solution at 78.4 °C = 730.7 mm Hg (d) bp = 79.5 °C 13.67 For ammonia: 23 m; X(NH3) = 0.29; 28% 13.69 0.592 g Na2SO4 13.71 (a) 0.20 m KBr; (b) 0.10 m Na2CO3

13.25 35.0 mm Hg

13.73 Freezing point = −11 °C

13.27 X(H2O) = 0.8692; 16.71 mol glycol; 1040 g glycol

13.75 4.0 × 102 g/mol

A-82

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13.77 4.7 × 10−4 mol/kg 13.79 (a) Molar mass = 4.9 × 104 g/mol (b) ΔTfp = −3.8 × 10−4 °C It would not be easy to determine the molar mass of starch by measuring the freezing point depression because the freezing point depression is so small.



(page 550), dimer formation is more prevalent at the lower temperature. In this process two molecules become one entity, lowering the number of separate species in solution and lowering the molality.

13.97

Molar mass in benzene = 1.20 × 102 g/mol; molar mass in water = 62.4 g/mol. The actual molar mass of acetic acid is 60.1 g/mol. In benzene, the molecules of acetic acid form “dimers.” That is, two molecules form a single unit through hydrogen bonding (page 550).

13.99

The empirical formula, calculated from the percent composition data, is C7H6N2. The calculated molar mass of the compound is 118 g/mol, which agrees with the molar mass calculated from the vapor pressure data.

13.81 ΔsolnH° [Li2SO4] = −28.0 kJ/mol

ΔsolnH° [LiCl] = −36.9 kJ/mol



ΔsolnH° [K2SO4] = +24.7 kJ/mol



ΔsolnH° [KCl] = +17.2 kJ/mol



Both lithium compounds have exothermic enthalpies of solution, whereas both potassium compounds have endothermic values. Consistent with this is the fact that lithium salts (LiCl) are often more water-soluble than potassium salts (KCl) (see Figure 13.11).

13.83 X(benzene in solution) = 0.667 and X(toluene in solution) = 0.333

Ptotal = Ptoluene + Pbenzene =  7.33 mm Hg +  50.0 mm Hg = 57.3 mm Hg = 57 mm Hg X(toluene in vapor) =

7.33 mm Hg = 0.13 57.3 mm Hg

X(benzene in vapor) =

50.0 mm Hg = 0.87 57.3 mm Hg

13.85 i = 1.7. That is, there is 1.7 mol of ions in solution per mole of compound. 13.87 (a) Calculate the number of moles of ions in 106 g H2O: 550.1 mol Cl−; 469.8 mol Na+; 53.08 mol Mg2+; 9.422 mol SO42−; 10.28 mol Ca2+; 9.719 mol K+; 0.839 mol Br−. Total moles of ions = 1103.2 per 106 g water. This gives ΔTfp of −2.05 °C. (b) Π = 27.0 atm. This means that a minimum pressure of 27 atm would have to be used in a reverse osmosis device. 13.89 (a) i = 2.06 (b) There are approximately two particles in solution, so H3O+ + HSO4− best represents H2SO4 in aqueous solution. 13.91 Mass of N2O = 0.53 g; concentration = 1.1 × 103 ppm 13.93 The best method would be to shine a laser through the liquid and look for the Tyndall effect. Some other properties of colloids that could be utilized are that the dispersed material in a colloid will either not crystallize or crystallize only with difficulty and will either not diffuse across a membrane or do so much more slowly than a true solute. 13.95 The calculated molality at the freezing point of benzene is 0.47 m, whereas it is 0.99 m at the boiling point. A higher molality at the higher temperature indicates more molecules are dissolved. Therefore, assuming benzoic acid forms dimers like acetic acid

13.101 The strength of the interactions is related to the size of the ion. Thus, Be2+ is most strongly hydrated, and Ca2+ is least strongly hydrated. 13.103 Colligative properties depend on the number of ions or molecules in solution. Each mole of CaCl2 provides 1.5 times as many ions as each mole of NaCl. 13.105 Benzene is a nonpolar solvent. Thus, ionic substances such as NaNO3 and NH4Cl will certainly not dissolve. However, naphthalene is also nonpolar and resembles benzene in its structure; it should dissolve very well. (A chemical handbook gives a solubility of 33 g naphthalene per 100 g benzene.) Diethyl ether is weakly polar but is miscible to some extent with benzene. Water is a polar solvent. The ionic compounds, NaNO3 and NH4Cl, are both soluble in water (NaNO3 solubility (at 25 °C) = 91 g per 100 g water; NH4Cl solubility (at 25 °C) = 40 g per 100 g water). The slightly polar compound diethyl ether is soluble to a small extent in water (about 6 g per 100 g water at 25 °C). Nonpolar naphthalene is not soluble in water (solubility at 25 °C = 0.003 g per 100 g water). 13.107 The COC and COH bonds in hydrocarbons are nonpolar or weakly polar and tend to make such dispersions hydrophobic (water-hating). The COO and OOH bonds in starch present opportunities for hydrogen bonding with water. Hence, starch is expected to be more hydrophilic. 13.109 [NaCl] = 1.00 M and [KNO3] = 0.878 M. The KNO3 solution has a higher solvent concentration, so solvent will flow from the KNO3 solution to the NaCl solution. 13.111 (a) X(C2H5OH) = 0.051; X(H2O) = 0.949 (b) P(C2H5OH) over mixture = 38 mm Hg; P°(H2O) calculated by linear interpolation using the vapor pressures at 78 °C and 79 °C = 334.2 mm Hg; P(H2O) over mixture = 317 mm Hg (c) X(C2H5OH) = 0.11; X(H2O) = 0.89 (d) The mole fraction of ethanol increased from 0.051 to 0.11, a factor of 2.1 times.



Weight percent of C2H5OH = 24%

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A-83

13.113 The molality of the 5.00% NaCl solution is 0.9006 m. Assuming no dissociation, this leads to ΔTfp(calculated) = −1.675 °C and i = 1.82. The molality of the 5.00% Na2SO4 solution is 0.3705 m, ΔTfp(calculated) = −0.6892 °C, and i = 1.97. These values are consistent with the trends seen in Table 13.4. As the concentration of the ionic solute increases, the value of i continues to decrease for both solutes, and rate of decrease is falling for both solutes. In the case of NaCl, this value is leveling off quite a bit. In the case of Na2SO4, it is continuing to decrease by significant amounts. Interestingly, the value of i for Na2SO4 has now fallen below 2 and is getting closer to that for NaCl. 13.115 This does not lend credence to this story. The reported lowest temperature for a NaCl solution ­corresponds to a temperature of 2.372 °F. The ­temperature of 0 °F is still below this.

For 125I, t1/2 = 0.693/(0.011 d−1) = 63 d (b) 125I decays much faster. (c) ln [(n)/(1.6 × 1015 atoms)] = −(0.011 d−1)(2.0 d)

Chapter 14

n/1.6 × 1015 atoms = 0.9782; n = 1.57 × 1015 atoms

14.7

1/[HI] − 1/[HI]0 = kt



1/[HI] − 1/[0.010 M] = (30. L/mol ∙ min)(12 min)



[HI] = 0.0022 M

14.8

(0.060 M/0.24 M) = 0.25; thus 1⁄4 of the original material remains and two half-lives have transpired. t1/2 = 141 min.



k = ln 2/t1/2 = (ln 2)/141 min = 4.916 × 10−3 min−1 = 4.92 × 10−3 min−1



Initial rate = k[H2O2]0 = 4.916 × 10−3 min−1 (0.24 mol/L) = 1.2 × 10−3 mol/L ∙ min

14.9

(a) For 241Am, t1/2 = 0.693/k = 0.693/(0.0016 y−1) =  430 y



Check Your Understanding 14.1

For the first 2 hours:



−Δ[sucrose]/Δt = −[(0.033 − 0.050) mol/L]/(2.0 h) = 0.0085 mol/L ∙ h



For the last 2 hours:



− Δ[sucrose]/Δt = −[(0.010 − 0.015) mol/L]/(2.0 h) = 0.003 mol/L ∙ h

14.2

−1⁄2(Δ[NOCl]/Δt) = 1⁄2(Δ[NO]/Δt) = Δ[Cl2]/Δt

14.3

Compare experiments 1 and 2: Doubling [O2] causes the rate to double, so the rate is first order in [O2]. Compare experiments 2 and 4: Doubling [NO] causes the rate to increase by a factor of 4, so the rate is second order in [NO]. Thus, the rate law is

Since the answer should have two significant figures, we should round this off to 1.6 × 1015 atoms. The approximately 2% that has decayed is not discernable within the limits of accuracy of the data presented.

14.10 An Arrhenius plot was constructed by plotting ln k on the y-axis and 1/T on the x-axis. Using Microsoft Excel, the equation of the best-fit line is y = −22336x + 27.304.

Ea = −R ∙ (slope) = −(0.0083145 kJ/mol ∙ K) (−22336 K) = 1.9 × 102 kJ/mol

14.11 ln (k2/k1) = (−Ea/R)(1/T2 − 1/T1)

ln [(1.00 × 104)/(4.5 × 103)] = −(Ea/8.3145 × 10−3 kJ/mol ∙ K)(1/283 K − 1/274 K)



Ea = 57 kJ/mol



Rate = −(∆[NO]/∆t) = k[NO]2[O2]

14.12 All three steps are bimolecular.



Using the data in experiment 1 to determine k:



For step 1: Rate = k[NO]2



0.028 mol/L ∙ s = k[0.020 mol/L]2[0.010 mol/L]



For step 2: Rate = k[N2O2][H2]

k = 7.0 × 10 L /mol  ∙ s



For step 3: Rate = k[N2O][H2]

14.4

Rate =  k[Pt(NH3)2Cl2] = (0.27 h−1)(0.020 mol/L) = 0.0054 mol/L ∙ h



There are two intermediates, N2O2(g) and N2O(g).



14.5

ln ([sucrose]/[sucrose]0) = −kt



ln ([sucrose]/[0.010]) = −(0.21 h−1)(5.0 h)

When the three equations are added, N2O2 (a product in the first step and a reactant in the second step) and N2O (a product in the second step and a reactant in the third step) cancel, leaving the net equation:



[sucrose] = 0.0035 mol/L



2 NO(g) + 2 H2(g) n N2(g) + 2 H2O(g).

14.6

(a) The fraction remaining is [CH3N2CH3]/[CH3N2CH3]0.

14.13 (a) 2 NH3(aq) + OCl−(aq) n N2H4(aq) + Cl−(aq) + H2O(ℓ) (b) The second step is the rate-determining step. (c) Rate = k[NH2Cl][NH3] (d) NH2Cl, N2H5+, and OH− are intermediates.



3

2

2

ln ([CH3N2CH3]/[CH3N2CH3]0) =  −(3.6 × 10−4 s−1)(150 s) [CH3N2CH3]/[CH3N2CH3]0= 0.95 (b) After the reaction is 99% complete [CH3N2CH3]/[CH3N2CH3]0 = 0.010 −4



ln (0.010) = −(3.6 × 10



t = 1.3 × 104 s (210 min)

A-84

−1

s )(t)

14.14 Overall reaction: 2 NO2Cl(g) n 2 NO2(g) + Cl2(g)

Rate = k′[NO2Cl]2/[NO2] (where k′ = k1k2/k−1)



Increasing [NO2] causes the reaction rate to decrease.

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Applying Chemical Principles 14.1 Enzymes—Nature’s Catalysts 1. To decompose an equivalent amount of H2O2 catalytically would take 1.0 × 10−7 years; this is equivalent to 3.2 seconds. 2. (100. mL)(2 mg/mL)(1 g/1000 mg)(1 mol/29,000 g) = 6.9 × 10−6 mol enzyme = 7 × 10−6 mol enzyme (6.9 × 10−6 mol enzyme)(1 × 106 mol CO2/mol enzyme ∙ s)​(44 g CO2/mol CO2) = 3 × 102 g CO2/s 14.2 Kinetics and Mechanisms: A 70-Year-Old Mystery Solved 1. (a)  E = hc/λ = (6.626 × 10−34 J ∙ s)(2.998 × 108 m/s)/ (5.78 × 10−7 m) = 3.437 × 10−19 J/photon Energy per mole of photons = (3.437 × 10−19 J/photon)(6.022 × 1023 photons/mol photons) (1 kJ/1000 J) = 207 kJ/mol photons (b) (151 kJ/mol photons)(1000 J/1 kJ)(1 mol photons/ 6.022 × 1023 photons) = 2.507 × 10−19 J/photon λ = hc/E = (6.626 × 10−34 J ∙ s)(2.998 × 108 m/s)/ (2.507 × 10−19 J) = 7.922 × 10−7 m(109 nm/1 m) = 792 nm 2. For the fast equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, thus k1[I2] = k−1[I]2 where k1 and k−1 are the forward and reverse rate constants, respectively. The rate limiting step determines the rate law, Rate = k2[H2][I]2. Substituting for [I]2 gives Rate = (k1k2/k−1)[H2][I2] = k[H2][I2]. 3. The probability of three particles colliding simultaneously with the correct geometry for a successful reaction is low. 4. A graph of ln(k) versus 1/T, where T is in kelvins, yields a straight line with a slope of −2.235 × 104 K. So, −2.235 × 104 K = −Ea/R where R = 8.3145 × 10−3 kJ/K ∙ mol. Ea = 186 kJ/mol.

Study Questions 14.1

1 [O3] 1 [O2] (a)   2 t 3 t

[A]

(b) 





14.7

The reaction is second order in A, first order in B, and third order overall.

14.9

For the rate equation: Rate = k[A][B]2, if [A] is doubled, the rate doubles. If [B] doubles, the rate ­quadruples. If both [A] and [B] are doubled, the rate increases by 2 × 4 = 8.

t





14.3

1 [O2] 1 [O3] [O3] 2 [O2]  or  3 t 2 t t 3 t so [O3]/t  1.0  103 mol/L ·s.

14.5



(a) The graph of [B] (product concentration) versus time shows [B] increasing from zero. The line is curved, indicating the rate changes with time; thus the rate depends on concentration. Rates for the four 10-s intervals are as follows: 0–10 s, 0.0326 mol/L ∙ s; from 10–20 s, 0.0246 mol/L ∙ s; 20–30 s, 0.0178 mol/L ∙ s; 30–40 s, 0.0140 mol/L ∙ s.

2 t

throughout the reaction

In the interval 10–20 s,

[A] t

 0.0123

mol L⋅s

14.11 (a) Rate = k[NO2][O3] (b) If [NO2] is tripled, the rate triples. (c) If [O3] is halved, the rate is halved. 14.13 (a) The reaction is second order in [NO] and first order in [O2]. [NO]

 k[NO]2[O2]



(b) Rate =



(c) k = 25 L2/mol2 ∙ s (d) Rate = 2.8 × 10−5 mol/L ∙ s (e) When −Δ[NO]/Δt = 1.0 × 10−4 mol/L ∙ s, Δ[O2]/Δt = 5.0 × 10−5 mol/L ∙ s and Δ[NO2]/Δt = 1.0 × 10−4 mol/L ∙ s.

t

14.15 (a) Rate = −∆[NO]/∆t = k[NO]2[O2] (b) 50. L2/mol2 ∙ h (c) Rate = 8.4 × 10−9 mol/L ∙ h 14.17 k = 3.73 × 10−3 min−1 14.19 5.0 × 102 min 14.21 (a) 153 min (b) 1790 min 14.23 44 s 14.25 140 s 14.27 6.3 s 14.29 0.0016 mol/L 14.31 t1/2 = 135 s; k = 0.0051 s−1 14.33 (a) t1/2 = 1.0 × 104 s

(b) 34,000 s

14.35 1.0 × 10  min 3

14.37 Fraction of 1 [HOF] 1 [HF] [O2] (b)    2 t 2 t t

1 [B]



64

Cu remaining = 0.030

14.39 The straight line obtained in a graph of ln[N2O] versus time indicates a first-order reaction.

k = (−slope) = 0.0128 min−1



The rate when [N2O] = 0.035 mol/L is 4.5 × 10−4 mol/L ∙ min.

14.41 The graph of 1/[NO2] versus time gives a straight line, indicating the reaction is second order with respect to [NO2] (see Table 14.1 on page 690). The slope of the line is k, so k = 1.1 L/mol ∙ s. 14.43 −Δ[C2F4]/Δt = k[C2F4]2  = (0.04 L/mol ∙ s)[C2F4]2 14.45 Activation energy = 102 kJ/mol 14.47 k = 0.3 s−1

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A-85

14.49

Energy

H2 + F

14.61 Doubling the concentration of A will increase the rate by a factor of 4 because the concentration of A appears in the rate law as [A]2. Halving the concentration of B will halve the rate The net result is that the rate of the reaction will double.

Ea = 8 kJ

∆rH = −133 kJ

14.63 After measuring pH as a function of time, one could then calculate pOH and then [OH−]. Finally, a plot of 1/[OH−] versus time would give a straight line with a slope equal to k.

HF + H

14.65 72 s represents two half-lives, so t1/2  = 36 s.

Reaction Progress

14.51 The lock-and-key model assumes a rigid structure for the enzyme in which the key (the substrate) fits precisely in the active site (the lock). The induced-fit model suggests that there is flexibility in the enzyme that allows the enzyme active site to change to fit the substrate. The induced-fit model is currently favored.

1/Rate

14.53

4.00 3.75 3.50 3.25 3.00 2.75 2.50 2.25 2.00 1.75 1.50 1.25 1.00 0.75 0.50 0.25 0

14.67 (a) A plot of 1/[C2F4] versus time indicates the reaction is second order with respect to [C2F4]. The rate law is Rate = k[C2F4]2. (b) The rate constant (= slope of the line) is about 0.045 L/mol ∙ s. (The graph does not allow a very accurate calculation.) (c) Using k = 0.045 L/mol ∙ s, the concentration after 600 s is 0.03 M (to one significant figure). (d) Time = 2000 s [using k from part (b)]. 14.69 (a) A plot of 1/[NH4NCO] versus time is linear, so the reaction is second order with respect to NH4NCO. (b) Slope = k = 0.0109 L/mol ∙ min. (c) t1/2 = 200. min (d) [NH4NCO] = 0.0997 mol/L 14.71 Mechanism 2 14.73 k = 0.0176 h−1 and t1/2 = 39.3 h 14.75 (a) After 125 min, 0.250 g remains. After 145, 0.144 g remains. (b) Time = 43.9 min (c) Fraction remaining = 0.016

0

.5

1

1.5

2 2.5 1/[S]

3

3.5

4

4.5

From the graph, we obtain a value of 1/Rate = 1.47 when 1/[S] = 0. From this, Ratemax = 0.68 mmol/min.

14.77 Plotting 1/concentration versus the time gives a ­reasonably good linear correlation. The reaction is second order.

14.57 (a) The second step; (b)  Rate = k[O3][O]



14.59 (a) NO2 is a reactant in the first step and a product in the second step. CO is a reactant in the second step. NO3 is an intermediate, and CO2 is a product. NO is a product. (b) Reaction coordinate diagram

14.81 The slope of the ln k versus 1/T plot is −6370. From slope = −Ea/R, we derive Ea = 53.0 kJ/mol.

Energy

14.55 (a) Rate = k[NO3][NO] (b) Rate = k[Cl][H2] (c) Rate = k[(CH3)3CBr]

14.79 The rate equation for the slow step is Rate = k[O3][O]. The equilibrium constant, K, for step 1 is K = [O2][O]/[O3]. Solving this for [O], we have [O] = K[O3]/[O2]. Substituting the expression for [O] into the rate equation we find

NO + NO3 Ea step 2 NO2 + CO Ea step 1

∆rH

NO + CO2

Rate = k[O3]{K[O3]/[O2]} = kK[O3]2/[O2]

14.83 Step 2 is the rate-determining step, and N2O2 is an intermediate.

Rate = −1⁄2 Δ[NO]/Δt = k[NO]2[O2]

14.85 (a) k = 3.41 L/mol ∙ min (b) The rate constant (k′) for the rewritten equation is 1 ⁄2 the value of the rate constant, k, for the original equation. The rate equation for the original equation is −∆[NO2]/∆t = k[NO2]. For the rewritten equation the rate equation is −(1⁄2)∆[NO2]/∆t = k′[NO2] or −∆[NO2]/∆t = 2k′[NO2]. Therefore, k = 2k′ or k′ = (1⁄2)k. 14.87 Estimated time at 90 °C = 4.76 min

Reaction Progress

A-86

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14.89 After 30 min (one half-life), PHOF = 50.0 mm Hg and Ptotal = 125.0 mm Hg. After 45 min, PHOF = 35.4 mm Hg and Ptotal = 132 mm Hg.



14.91 (a) Reaction is first order in NO2NH2 and −1 for H3O+. In a buffered solution, [H3O]+ is constant, so the reaction has an apparent order of 1. (b, c)  Mechanism 3

14.103 (a) True (b) True (c) False. As a reaction proceeds, the reactant concentration decreases and the rate decreases. (d) False. It is possible to have a one-step mechanism for a third-order reaction if the slow, ratedetermining step is termolecular.

In step 1, K = k4/k4′ = [NO2NH−][H3O+]/[NO2NH2] Rearrange this and substitute into the rate law for the slow step.

Rate = k5[NO2NH ] = k5K[NO2NH2]/[H3O ] −

+

This is the same as the experimental rate law, where the overall rate constant k = k5K. (d) Addition of OH− ions will shift the equilibrium in step 1 (by reacting with H3O+) to produce a larger concentration of NO2NH−, the reactant in the rate-determining step, thus the reaction rate will increase. 14.93 (a) Average rate for t = 0 s to t = 15 s is about 4.7 × 10−5 M/s. For t = 100 s to 125 s, the average rate is about 1.6 × 10−5 M/s. The rate slows because the rate of the reaction is dependent on the concentration of reactant and this concentration is declining with time. (b) A plot of ln (concentration) versus time is a straight line with an equation of y = −0.010x −5.30. This indicates the reaction is a first-order reaction; Rate = k[phenolphthalein]. The slope, which is equal to −k, is −0.010, so k = 0.010 s−1. (c) From the data the half-life is 69.3 s, and the same value comes from the relation t1/2 = ln 2/k.

(e) False. If the concentrations of both reactants are doubled, the rate will increase by a factor of 4. (f) True



14.105 (a) Decrease (b) Increase (c)  No change

1/Rate = 94 (1/[S]) + 7.6 × 104



so Ratemax = 1/(7.6 × 104) = 1.3 × 10−5 M min−1.

14.97 (a) The reaction is first order in [ClO−] (experiments 1 and 3), first order in [I−] (experiments 2 and 3), and −1 order in [OH−] (experiments 3 and 4). Therefore, Rate = k[ClO−][I−]/[OH−]. (b) Step 2 is the rate determining step. For this step, rate = k[I−][HOCl]. From the first step, Keq = [HOCl][OH−]/[ClO−], so [HOCl] = Keq [ClO−]/[OH−]; substituting for [HOCl] in the rate equation for the second step gives the observed rate law.

14.109 (b) The atoms are lined up so that the new NOO bond can form and the OOO bond can break concurrently.

Chapter 15 Check Your Understanding 15.1

(a) K = [CO]2/[CO2] (b) K = [Cu2+][NH3]4/[[Cu(NH3)4]2+] (c) K = [H3O+][CH3CO2−]/[CH3CO2H]

15.2

(a) Q = 0.00218/0.00097 = 2.2. The system is not at equilibrium; Q < K. To reach equilibrium, [isobutane] will increase and [butane] will decrease. (b) Q = 0.00260/0.00075 = 3.5. The system is not at equilibrium; Q > K. To reach equilibrium, [butane] will increase and [isobutane] will decrease.



15.3



(a) Equation

C6H10I2 uv

Initial (M)

0.050

Change (M)



C6H10

+

I2

0

0

−0.035

+0.035

+0.035

0.015

0.035

0.035

Equilibrium (M)

(b) K = (0.035)(0.035)/(0.015) = 0.082

15.4 Equation

14.99 The finely divided rhodium metal will have a significantly greater surface area than the small block of metal. This leads to a large increase in the number of reaction sites and vastly increases the reaction rate. 14.101 (a) False. The reaction may occur in a single step but this does not have to be true. (b) True (c) False. Raising the temperature increases the value of k. (d) False. Temperature has no effect on the value of Ea.

(d) No change (e) No change (f ) No change

14.107 (a) There are three mechanistic steps. (b) The overall reaction is exothermic.

14.95 A plot of 1/Rate versus 1/[S] gives the equation



+

H2

I2

uv 2 HI

Initial (M)

6.00 × 10

6.00 × 10

Change (M)

−x

−x

+2x

Equilibrium (M)

0.00600 − x

0.00600 − x

+2x

−3

−3

0

(2x)2 (0.00600  x)2



Kc = 33 = 



x = 0.00445 M, so [H2] = [I2] = 0.0015 M and [HI] = 0.0089 M.

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A-87

2. (a) For CH4(g) + H2O(g) n CO(g) + 3 H2(g)

15.5 Equation

PCl5(g)

Initial (M)

0.1000

Change (M)

−x

Equilibrium (M)

0.1000 − x

uv PCl3(g) + Cl2(g) 0

0

+x

+x

x

x



Kc = [PCl3][Cl2]/[PCl5]



33.3 = x2/0.1000 − x



We cannot use the simplifying assumption in this case (K is > 1 and 100 ∙ K > 0.1000), so we must solve using the quadratic formula.



x2 + 33.3x − 3.33 = 0



Using the quadratic formula, x = 0.09970 (the other root, x = −33.40, is not possible because it leads to negative concentrations).



[PCl3] = [Cl2] = 0.09970 M = 0.0997 M



[PCl5] = 0.1000 M − 0.09970 M = 0.0003 M

15.6

(a) K′ = K2 = (2.5 × 10−29)2 = 6.25 × 10−58 = 6.3 × 10−58 (b) K″ = 1/K2 = 1/(6.25 × 10−58) = 1.6 × 1057

15.7

Equation

Butane

Initial (M)

0.020

uv Isobutane

For CO(g) + H2O(g) n CO2(g) + H2(g)

ΔrH° = (1 mol CO2/mol-rxn)[Δf H°(CO2)] − (1 mol CO/mol-rxn)[Δf H°(CO)] − (1 mol H2O/mol-rxn)[Δf H°(H2O)] ΔrH° = (1 mol CO2/mol-rxn)(−393.5 kJ/mol) − (1 mol CO/mol-rxn)(−110.5 kJ/mol) − (1 mol H2O/mol-rxn)(−241.8 kJ/mol) = −41.2 kJ/mol-rxn (exothermic) (b) (15 billion kg = 1.5 × 1013 g) Add the two equations: CH4(g) + 2 H2O(g) n CO2(g) + 4 H2(g) CH4 required = (1.5 × 1013 g NH3) (1 mol NH3/17.03 g NH3)(3 mol H2/2 mol NH3) (1 mol CH4/4 mol H2)(16.04 g CH4/1 mol CH4) = 5.3 × 1012 g CH4

0.050

CO2 formed = (1.5 × 1013 g NH3) (1 mol NH3/17.03 g NH3)(3 mol H2/2 mol NH3) (1 mol CO2/4 mol H2)(44.01 g CO2/1 mol CO2) = 1.5 × 1013 g CO2

0.020

0.050 + 0.0200

15.2 Trivalent Carbon

Change (M)

+x

−x

Equilibrium (M)

0.020 + x

0.070 − x



K = 2.50 = (0.070 − x)/(0.020 + x)



Solving for x gives x = 0.00571 M. Therefore, [isobutane] = 0.070 − 0.00571 = 0.064 M and [butane] = 0.020 + 0.00571 = 0.026 M.

Applying Chemical Principles 15.1 Applying Equilibrium Concepts— The Haber-Bosch Ammonia Process 1. (a) NH3 + HNO3 n NH4NO3 (b)  ΔrH° = (1 mol (NH2)2CO/mol-rxn) [Δf H°{(NH2)2CO}] + (1 mol H2O/mol-rxn)​ [Δf H°(H2O)] − (2 mol NH3/mol-rxn)[Δf H°(NH3)] − (1 mol CO2/mol-rxn)[Δf H°(CO2)] ΔrH° = (1 mol (NH2)2CO/mol-rxn)(−333.1 kJ/mol) + (1 mol H2O/mol-rxn)(− 241.8 kJ/mol) − (2 mol NH3/mol-rxn)(−45.90 kJ/mol) − (1 mol CO2/mol-rxn)(−393.5 kJ/mol) ΔrH° = −89.6 kJ/mol-rxn.

The reaction as written is exothermic, so the equilibrium will be more favorable for product formation at a low temperature. The reaction converts three moles of gaseous reactants to one mole of gaseous products; thus, high pressure will be more favorable to product formation.

A-88

ΔrH° = (1 mol CO/mol-rxn)[Δf H°(CO)] − (1 mol CH4/mol-rxn)[Δf H°(CH4)] − (1 mol H2O/mol-rxn)[Δf H°(H2O)]

ΔrH° = (1 mol CO/mol-rxn)(−110.5 kJ/mol) − (1 mol CH4/mol-rxn)(−74.87 kJ/mol) − (1 mol H2O/mol-rxn)(−241.8 kJ/mol) = 206.2 kJ/mol-rxn (endothermic)

After adding more isobutane (M)





1. (a) Concentration (molality) = −0.542 °C/−5.12 °C/m = 0.1059 mol dimer/kg benzene Amount dimer = (0.1059 mol dimer/kg benzene) (0.0100 kg benzene) = 1.059 × 10−3 mol dimer Molar mass of dimer = 0.503 g/1.059 × 10−3 mol = 475 g/mol (b) Each molecule of the dimer that decomposes produces two monomer particles, increasing the total number of moles of particles in solution. When the mass of the dimer is divided by the moles of particles, the calculated molar mass is too low. 2. Kc = 4.1 × 10−4 = (2x)2/(0.015 − x) Solving the quadratic equation gives x = 0.00119 M. The concentration of the monomer = 2(0.00119 M) = 0.0024 M. The concentration of the dimer = 0.015 M − 0.00119 M = 0.014 M. 3. [Dimer]0 = ((0.64 g)(1 mol/486.658 g))/0.0250 L = 0.0526 M Kc = 4.1 × 10−4 = (2x)2/(0.0526 − x) Solving the quadratic equation gives x = 0.00227 M. [Monomer] = 2(0.00227 M) = 0.0045 M [Dimer] = 0.0526 M − 0.00227 M = 0.050 M 4. The monomer, the product of the reaction, is yellow in color. Heating produces more of the monomer so the reaction is endothermic. 5. (b) Triphenylmethyl radical

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Study Questions

15.43 To obtain the second equation, the first equation has been reversed and multiplied by 2. (c) K2 = 1/K12

15.1

(a) K c 

[H2O]2[O2 ] [H2O2 ]2



(b) K c 

[CO2 ] [CO][O2 ]1/2



(c) K c 

[CO]2 [CO2 ]



(d) K c 

[CO2 ] [CO]

15.3

(a) 2 SO2(g) + O2(g) st 2 SO3(g)



(b) 2 CO(g) + O2(g) st 2 CO2(g)

15.47 (a) The equilibrium will shift to the left on adding more Cl2. (b) Kc is calculated (from the quantities of reactants and products at equilibrium) to be 0.004709. After Cl2 is added, the concentrations are: [PCl5] = 0.00198 M, [PCl3] = 0.00231 M, and [Cl2] = 0.00404 M.



(c) PbCl2(s) st Pb2+(aq) + 2 Cl−(aq)

15.49 Kp = 0.215



(d) HF(aq) + H2O(ℓ) st F−(aq) + H3O+(aq)

15.5

Q = (2.0 × 10−8)2/(0.020) = 2.0 × 10−14 Q < Kc, so the reaction proceeds to the right.

15.51 (a) Fraction dissociated = 0.15 (b) Fraction dissociated = 0.189. If the pressure decreases, the equilibrium shifts to the right, increasing the fraction of N2O4 dissociated.

15.7

Q = 1.0 × 103, so Q > Kc and the reaction is not at equilibrium. It proceeds to the left to convert products to reactants.

15.9

Kc = 1.2

15.45 (a) No change (b) Shifts left (c) No change

15.11 (a) Kc = 0.025 (b) Kc = 0.025 (c) The amount of solid does not affect the equilibrium.



(d) Shifts right (e) Shifts right

15.53 [NH3] = 0.67 M; [N2] = 0.57 M; [H2] = 1.7 M; Ptotal = 180 atm 15.55 (a) [NH3] = [H2S] = 0.013 M (b) [NH3] = 0.027 M and [H2S] = 0.0067 M 15.57 P(NO2) = 0.096 atm and P(N2O4) = 0.066 atm; P(total) = 0.16 atm

15.19 [COBr2] = 0.0026 M; [CO] = [Br2] = 0.0224 M 89.6% of the COBr2 has decomposed.

15.59 (a) Kp = Kc = 56. Because 2 mol of reactant gases gives 2 mol of product gases, Δn does not change and Kp = Kc (page 742). (b) Total pressure before reaction is 0.52 atm. After reaction the total pressure is the same because the amount of gas present has not changed. (c) After reaction, at equilibrium, P(H2) = P(I2) = 0.052 atm and P(HI) = 0.42 atm.

15.21 (b) K2 = K12

15.61 P(CO) = 0.0010 atm

15.23 (d) K2 = 1/K12

15.63 3.9 × 1017 O atoms

15.25 (e) K2 = 1/(K1)2

15.65 Glycerin concentration should be 1.7 M

15.13 (a) [COCl2] = 0.00308 M; [CO] = 0.0071 M (b) Kc = 140 15.15 [isobutane] = 0.024 M; [butane] = 0.010 M 15.17 [I2] = 6.14 × 10−3 M; [I] = 4.80 × 10−3 M

15.27 Kc = 13.7 15.29 (a) Kc = Kp/(RT) = 0.16/[(0.08206)(298)] = 6.5 × 10−3 (b) ∆n = 0, therefore Kc = Kp = 1.05 ∆n

15.31

(a) Equilibrium (b) Equilibrium (c) Equilibrium (d) Equilibrium

shifts shifts shifts shifts

to to to to

the the the the

right left right left

15.33 Equilibrium concentrations are the same under both cir­ cumstances: [butane] = 1.1 M and [isobutane] = 2.9 M. 15.35 Kc = 3.9 × 10−4 15.37 For decomposition of COCl2, Kp = 1/(Kp for COCl2 formation) = 1/(6.5 × 1011) = 1.5 × 10−12 15.39 Kc = 3.9 15.41 Q is less than Kc, so the system shifts to form more isobutane.

15.67 (a) Kp = 0.20 (b) When initial [N2O4] = 1.00 atm, the equilibrium pressures are [N2O4] = 0.80 atm and [NO2] = 0.40 atm. When initial [N2O4] = 0.10 atm, the equilibrium pressures are [N2O4] = 0.050 atm and [NO2] = 0.10 atm. The percent dissociation is now 50.%. This is in accord with Le Chatelier’s principle: If the initial pressure of the reactant is smaller, the equilibrium shifts to the right, increasing the fraction of the reactant dissociated. [This might be clearer if you imagine beginning with the equilibrium system in the case when the initial pressure of N2O4 was 1.00 atm and then increasing the volume tenfold (obtaining the same equilibrium system as starting with 0.10 atm N2O4). Le Chatelier’s principle predicts a shift in the direction that has a greater number of gas molecules.]

At equilibrium, [butane] = 0.86 M and [isobutane] = 2.14 M. Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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A-89

15.69 (a) The flask containing (H3N)B(CH3)3 will have the largest partial pressure of B(CH3)3. (b) P[B(CH3)3] = P(NH3) = 0.23 and P[(H3N)B(CH3)3] = 0.012 atm

16.4

From the pH, we can calculate [H3O+] =  1.91 × 10−3 M. Also, [butanoate−] = [H3O+] =  1.91 × 10−3 M. Use these values along with [butanoic acid] to calculate Ka.

Ptotal = 0.48 atm



Ka =  [1.91 × 10−3] [1.91 × 10−3]/(0.055 − 1.91 × 10−3) = 6.8 × 10−5

16.5

Ka = 1.8 × 10−5 = [x][x]/(0.10 − x)



x = 1.34 × 10−3 M



[H3O+] = [CH3CO2−] = 1.3 × 10−3 M; [CH3CO2H] = 0.099 M; pH = 2.87

16.6

HF(aq) + H2O(ℓ) st H3O+(aq) + F−(aq)



Ka = 7.2 × 10−4  = [x][x]/(0.00150 − x)



The x in the denominator cannot be dropped. This equation must be solved with the quadratic formula or by successive approximations. Because this problem follows the section dealing with the method of successive approximations, that method will be used here.



7.2 × 10−4  = x2/(0.00150); x = 1.04 × 10−3



7.2 × 10−4  = x2/(0.00150 − 1.04 × 10−3); x = 5.76 × 10−4



7.2 × 10−4  = x2/(0.00150 − 5.76 × 10−4); x = 8.16 × 10−4



7.2 × 10−4  = x2/(0.00150 − 8.16 × 10−4); x = 7.02 × 10−4



7.2 × 10−4  = x2/(0.00150 − 7.02 × 10−4); x = 7.58 × 10−4



7.2 × 10−4  = x2/(0.00150 − 7.58 × 10−4); x = 7.31 × 10−4



7.2 × 10−4  = x2/(0.00150 − 7.31 × 10−4); x = 7.44 × 10−4



7.2 × 10−4  = x2/(0.00150 − 7.44 × 10−4); x = 7.38 × 10−4



7.2 × 10−4  = x2/(0.00150 − 7.38 × 10−4); x = 7.41 × 10−4



7.2 × 10−4  = x2/(0.00150 − 7.41 × 10−4); x = 7.39 × 10−4



7.2 × 10−4  = x2/(0.00150 − 7.39 × 10−4); x = 7.40 × 10−4



7.2 × 10−4  = x2/(0.00150 − 7.40 × 10−4); x = 7.40 × 10−4



The result has converged to three digits.



[H3O+] = [F−] = 7.40× 10−4 M = 7.4 × 10−4 M



[HF] = 0.00150 M − 7.40 10−4 M = 7.6 × 10−4 M



pH = 3.13

16.7

ClO−(aq) + H2O(ℓ) st HClO(aq) + OH−(aq)



Kb = 2.9 × 10−7 = [x][x]/(0.015 − x)



x = 6.60 × 10−5 M



[OH−] = [HClO] = 6.6 × 10−5 M



pOH = 4.181; pH = 9.82



Percent dissociation = 95%

15.71 (a) As more KSCN is added, Le Chatelier’s principle predicts more of the red complex ion [Fe(H2O)5(SCN)]+ will form. (b) Adding Ag+ ions leads to a precipitate of AgSCN, thus removing SCN− ions from solution. The equilibrium shifts left, dropping the concentration of the red complex ion. 15.73 (a) True (b) False. The equilibrium constant for the reverse reaction is the reciprocal of the original equilibrium constant. (c) True (d) False. Δn = 1, so Kp = Kc(RT). (e) True 15.75 (a) Product-favored at equilibrium, K >> 1 (b) Reactant-favored at equilibrium, K > 1 15.77 Begin with an equilibrium system containing 14N2, H2, and 14NH3. Introduce some 15N2 and allow the system to equilibrate. The presence of 15NH3 indicates that the forward reaction has occurred. In addition, the presence of 15N14N indicates that the reverse ­reaction has occurred. Further evidence of the reverse ­reaction occurring could be obtained by carrying out another trial in which 15NH3 is added to the initial equilibrium mixture. The presence of 15N14N or 15N2 indicates that the reverse reaction is occurring. (A similar set of experiments could be run using 2 H instead of 15N.) 15.79 Kp = 3.2 × 10

−7

Chapter 16 Check Your Understanding 16.1

[H3O+] = 4.0 × 10−3 M; [OH−] = Kw/[H3O+] =  2.5 × 10−12 M

16.2

(a) pH = 7 (b) pH < 7 (NH4+ is an acid) (c) pH < 7 [Al(H2O)6]3+ is an acid (d) pH > 7 (HPO42− is a stronger base than it is an acid)

16.3

(a) NH4+ is a stronger acid than HCO3−. CO32−, the conjugate base of HCO3−, is a stronger base than NH3, the conjugate base of NH4+. (b) Reactant-favored at equilibrium; the reactants are the weaker acid and base. (c) Reactant-favored at equilibrium; the reactants are the weaker acid and base, so reaction lies to the left.



A-90

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16.8

Equivalent amounts of acid and base react to form water, CH3CO2− and Na+. Acetate ion hydrolyzes to a small extent, giving CH3CO2H and OH−. We need to determine [CH3CO2−] and then solve a weak base equilibrium problem to determine [OH−].

(c) HCl is a stronger acid than NH4+, so HCl will be completely ionized. The solution will be a strong conductor. (d) O2− + NH3 st OH− + NH2− The reaction is product-favored at equilibrium.



Amount CH3CO2− = moles base = 0.12 mol/L × 0.015 L = 1.80 × 10−3 mol

Study Questions

[CH3CO2−] = 



Total volume = 0.030 L, so (1.80 × 10−3 mol)/0.030 L = 0.0600 M = 0.060 M



CH3CO2−(aq) + H2O(ℓ) st  CH3CO2H(aq) + OH−(aq)



Kb = 5.6 × 10−10 = [x][x]/(0.0600 − x)



16.1

(a) CN−, cyanide ion (b) SO42−, sulfate ion (c) F−, fluoride ion

x = 5.80 × 10−6 M

16.3

(a) HCO3−, hydrogen carbonate ion (b) HClO2, chlorous acid (c) H3PO4, phosphoric acid



[OH−] = [CH3CO2H] = 5.8 × 10−6 M

16.5



pOH = 5.237; pH = 8.76

16.9

H2C2O4(aq) + H2O(ℓ) st H3O+(aq) + HC2O4−(aq)



Ka1 = 5.9 × 10−2 = [x][x]/(0.10 − x)



The x in the denominator cannot be dropped. This equation must be solved with the quadratic formula or by successive approximations.



(a) H3O+(aq) + NO3−(aq); H3O+(aq) is the conjugate acid of H2O, and NO3−(aq) is the conjugate base of HNO3. (b) H3O+(aq) + SO42−(aq); H3O+(aq) is the conjugate acid of H2O, and SO42−(aq) is the conjugate base of HSO4−. (c) H2O + HF; H2O is the conjugate base of H3O+, and HF is the conjugate acid of F−.



x = 5.28 × 10−2 M

16.7



[H3O+] = [HC2O4−] = 5.3 × 10−2 M

Brønsted acid: HC2O4−(aq) + H2O(ℓ) st  H3O+(aq) + C2O42−(aq)



pH = 1.28





Ka2 = [H3O+][C2O42−]/[HC2O4−]; because [H3O+] =  [HC2O4−],

Brønsted base: HC2O4−(aq) + H2O(ℓ) st  H2C2O4(aq) + OH−(aq)



[C2O42−] = Ka2 = 6.4 × 10−5 M

16.9



(a)

Applying Chemical Principles 16.1 Would You Like Some Belladonna Juice in Your Drink? 1. (100. mg C17H23NO3)(1 g/1000 mg)(1 mol C17H23NO3/ 289.4 g C17H23NO3) = 3.46 × 10−4 mol C17H23NO3

Conjugate Conjugate Acid (A) Base (B) Base of A Acid of B H2O

HCO2−

H3O+

(b) H2S

NH3

HS−

NH4+

(c)

OH−

SO42−

H2O

HCO2H

HSO4−

2. The proton will attach to the N.

16.11 [H3O+] = 1.8 × 10−4 M; [OH−] = 5.6 × 10−11 M; acidic

3. The pKa of protonated atropine (4.35) is less than that of the ammonium (pKa = 9.26), methylammonium (pKa = 10.70), and ani­linium (pKa = 4.60) ions.

16.13 HCl is a strong acid, so [H3O+] = concentration of the acid. [H3O+] = 0.0075 M and [OH−] = 1.3 × 10−12 M. pH = 2.12.

16.2 The Leveling Effect, Nonaqueous Solvents, and Superacids

16.15 Ba(OH)2 is a strong base, so [OH−] = 2 × concentration of the base.

1. HClO4 (K = 5 × 10−6) > H2SO4 (K = 2 × 10−7) > HCl (K = 2 × 10−9)



2. (a) 2 CH3CO2H st CH3CO2H2+ + CH3CO2− (b) Let x = [CH3CO2H2+] = [CH3CO2−]

x2 = 3.2 × 10−15



x = [CH3CO2H2+] = [CH3CO2−] = 5.7 × 10−8 M

3. NH2−(aq) + H2O(ℓ) st NH3(aq) + OH−(aq) The reaction is product-favored at equilibrium. 4. HClO4 in glacial acetic acid is a weak conductor of electricity as it only partially ionizes. 5. (a) 2 NH3 st NH4+ + NH2− (b) The strongest acid is NH4+; the strongest base is NH2−.

[OH−] = 3.0 × 10−3 M; pOH = 2.523; and pH = 11.48

16.17 pH = 14.00 − 2.55 = 11.45; [H3O+] = 10−pH = 3.5 × 10−12 M 16.19 Ka = [H3O+][NH3]/[NH4+] 16.21 (a) The strongest acid is HCO2H (largest Ka) and the weakest acid is C6H5OH (smallest Ka). (b) The strongest acid (HCO2H) has the weakest ­conjugate base. (c) The weakest acid (C6H5OH) has the strongest ­conjugate base. 16.23 (c) HClO, the weakest acid in this list (Table 16.2), has the strongest conjugate base.

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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A-91

16.25 (a) HCO3−. Decide based on the base strength (Table 16.2). The strongest base of the three listed examples has the weakest conjugate acid. 16.27 CO32−(aq) + H2O(ℓ) st HCO3−(aq) + OH−(aq) 16.29 [Fe(H2O)6]3+(aq) + H2O(ℓ) st [Fe(H2O)5(OH)]2+(aq) + H3O+(aq) 16.31 Highest pH, (a) Na2S; lowest pH, (f) AlCl3 (which gives the weak acid [Al(H2O)6]3+ in solution)

16.73 (a) The reaction produces acetate ion, the conjugate base of acetic acid. The solution is weakly basic. pH is greater than 7. (b) The reaction produces NH4+, the conjugate acid of NH3. The solution is weakly acidic. pH is less than 7. (c) The reaction mixes equal molar amounts of strong base and strong acid. The solution will be neutral. pH will be 7. 16.75 H2C2O4(aq) + H2O(ℓ) st HC2O4−(aq) + H3O+(aq) HC2O4−(aq) + H2O(ℓ) st C2O42−(aq) + H3O+(aq)

16.33 pKa = 8.08



16.35 Ka = 3.0 × 10−10; in Table 16.2, this acid falls between hexaaquairon(II) ion and the hydrogen carbonate ion.

16.77 H2C2O4(aq) + H2O(ℓ) st HC2O4−(aq) + H3O+(aq)  Ka1 = 5.2 × 10−2

16.37 (b) 2-Chlorobenzoic acid is the stronger acid; it has the smaller pKa value. 16.39 Kb = 7.1 × 10−12 16.41 Kb = 6.3 × 10−5 16.43 HNO2(aq) + HCO3−(aq) st  NO2−(aq) + H2CO3(aq)

Equilibrium lies predominantly to the right because HNO2 is a stronger acid than H2CO3.

16.45 (a) Left; CH3CO2− and HBr are the stronger base and acid, respectively. (b) Right; H3PO4 and F− are the stronger acid and base, respectively. (c) Left; [Ni(H2O)5(OH)]+ and H2S are the stronger base and acid, respectively. 16.47 (a) OH−(aq) + HPO42−(aq) st H2O(ℓ) + PO43−(aq) (b) OH− is a stronger base than PO43−, so the equilibrium will lie to the right. 16.49 (a) CH3CO2H(aq) + HPO42−(aq) st  CH3CO2−(aq) + H2PO4−(aq) (b) CH3CO2H is a stronger acid than H2PO4−, so the equilibrium will lie to the right.



HC2O4−(aq) + H2O(ℓ) st H2C2O4(aq) + OH−(aq)  Kb2 = 1.9 × 10−13



Sum: 2 H2O(ℓ) st H3O+(aq) + OH−(aq)  Kw = Ka1 × Kb2 = 1.0 × 10−14

16.79 (a) pH = 1.17; (b) [SO32−] = 6.2 × 10−8 M 16.81 (a) [OH−] = [N2H5+] = 9.2 × 10−5 M; [N2H62+] = 8.9 × 10−16 M (b) pH = 9.96 16.83 HOCN should be a stronger acid than HCN because the H atom in HOCN is attached to the more electronegative O atom. The electron attachment enthalpy of OCN is thus more negative than that of CN, which stabilizes the conjugate base that forms and makes the ionization of HOCN more product-favored at equilibrium. 16.85 The S atom is surrounded by three highly electronegative O atoms. These help stabilize the conjugate base that needs to form so that the negative charge is more readily accepted. 16.87 (a) Lewis base (b) Lewis acid (c) Lewis base (owing to lone pair of electrons on the N atom)

16.51 (a) 2.1 × 10−3 M; (b) Ka = 3.6 × 10−4

16.89 CO is a Lewis base in its reactions with transition metal atoms. It donates a lone pair of electrons on the C atom.

16.53 Kb = 6.6 × 10−9

16.91 pH = 2.671

16.55 (a) [H3O+] = 1.6 × 10−4 M (b) Moderately weak; Ka = 1.1 × 10−5

16.93 Both Ba(OH)2 and Sr(OH)2 dissociate completely in water to provide M2+ and OH− ions. 2.50 g Sr(OH)2 in 1.00 L of water gives [Sr2+] = 0.021 M and [OH−] = 0.041 M. This concentration of OH− is reflected in a pH of 12.61.

16.57 [CH3CO2−] = [H3O+] = 1.9 × 10−3 M and [CH3CO2H] = 0.20 M 16.59 [H3O+] = [CN−] = 3.2 × 10−6 M; [HCN] = 0.025 M; pH = 5.50

16.95 H2S(aq) + CH3CO2−(aq) st  CH3CO2H(aq) + HS−(aq)

16.61 [NH4+] = [OH−] = 1.6 × 10−3 M; [NH3] = 0.15 M; pH = 11.22



16.63 [OH−] = 0.010 M; pH = 12.01; pOH = 1.99 16.65 pH = 3.25

16.97 [Χ−] = [H3O+] = 3.0 × 10−3 M; [HΧ] = 0.007 M; pH = 2.52

16.67 [H3O+] = 1.1 × 10−5 M; pH = 4.98

16.99 Ka = 1.37 × 10−5; pKa = 4.86

16.69 [HCN] = [OH−] = 3.3 × 10−3 M; [H3O+] = 3.0 × 10−12 M; [Na+] = 0.441 M

16.101 pH = 5.84

16.71 [H3O+] = 1.5 × 10−9 M; pH = 8.81

A-92

The equilibrium lies to the left and favors the reactants.

16.103 (a) Ethylamine is a stronger base than ethanolamine. (b) For ethylamine, the pH of the solution is 11.82.

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

16.105 pH = 7.66 16.107

Acidic: NaHSO4, NH4Br, FeCl3 Neutral: KClO4, NaNO3, LiBr Basic: Na2CO3, (NH4)2S, Na2HPO4 Highest pH: (NH4)2S, lowest pH: NaHSO4

16.109 Knet = Ka1 × Ka2 = 3.8 × 10−6

16.135 (a) For the weak acid HA, the concentrations at equilibrium are [HA] = C0 − αC0, [H3O+] = [A−] = αC0. Putting these into the usual expression for Ka we have Ka = α2C0/(1−α). (b) For 0.10 M NH4+, α = 7.5 × 10−5 (reflecting the fact that NH4+ is a much weaker acid than acetic acid).

16.111 For the reaction HCO2H(aq) + OH−(aq) n  H2O(ℓ) + HCO2−(aq),  Knet = Ka (for HCO2H) × [1/Kw] = 1.8 × 1010

16.137 (a) Add the three equations.

NH4+(aq) + H2O(ℓ) st NH3(aq) + H3O+(aq)  K1 = Ka

16.113 To double the percent ionization, you must dilute 100 mL of solution to 400 mL.



CN−(aq) + H2O(ℓ) st HCN(aq) + OH−(aq)  K2 = Kb

16.115  H2O > H2C2O4 > HC2O4− = H3O+ > C2O42− > OH−



H3O+(aq) + OH−(aq) st 2 H2O(ℓ) K3 = 1/Kw



NH4+(aq) + CN−(aq) st NH3(aq) + HCN(aq)  Knet = K1K2K3 = KaKb/Kw



(b) The salts NH4CN, NH4CH3CO2, and NH4F have Knet values of 1.4, 3.1 × 10−5, and 7.8 × 10−7, respectively. Only in the case of NH4CN is the base (the cyanide ion) strong enough to remove a proton from the ammonium ion and produce a significant concentration of products. (c) NH4CN: basic, Kb of CN− > Ka of NH4+

16.117  H2O > Na+ = Cl− > HF > F− = H3O+ >> OH− 16.119  Measure the pH of 0.1 M solutions of the three bases. The solution containing the strongest base will have the highest pH. The solution having the weakest base will have the lowest pH. 16.121  The possible cation–anion combinations are NaCl (neutral), NaOH (basic), NH4Cl (acidic), NH4OH (basic), HCl (acidic), and H2O (neutral).

 A = H solution; B = NH4 solution; C = Na+ solution; Y = Cl− solution; Z = OH− solution



NH4CH3CO2: neutral, Kb of CH3CO2− = Ka of NH4+



NH4F: acidic, Kb of F− < Ka of NH4+

+

+

16.123  Ka = 3.0 × 10−5 16.125  (a) Aniline is both a Brønsted and a Lewis base. As a proton acceptor it gives C6H5NH3+. The N atom can also donate an electron pair to give a Lewis acid–base adduct, F3B m NH2C6H5.  (b) pH = 7.97 16.127  Water can both accept a proton (a Brønsted base) and donate a lone pair (a Lewis base). Water can also donate a proton (Brønsted acid), but it cannot accept a pair of electrons (and act as a Lewis acid). 16.129  (a) HOCl is the strongest acid (smallest pKa and largest Ka), and HOI is the weakest acid.  (b) Cl is more electronegative than Br or I, so the OCl− anion is more stable than the other two oxoanions.

Chapter 17 Check Your Understanding 17.1

pH of 0.30 M HCO2H:



Ka = [H3O+][HCO2−]/[HCO2H]



1.8 × 10−4 = [x][x]/[0.30 − x]



x = 7.35 × 10−3 M; pH = 2.13



pH of 0.30 M formic acid + 0.10 M NaHCO2



Ka = [H3O+][HCO2−]/[HCO2H]



1.8 × 10−4 = [x][0.10 + x]/(0.30 − x)



x = 5.40 × 10−4 M; pH = 3.27

17.2

16.131  (a) HClO4 + H2SO4 st ClO4− + H3SO4+  (b) The O atoms on sulfuric acid have lone pairs of electrons that can be used to bind to an H+ ion. O H O S



HCO2H + H2O st H3O+ +

HCO2−

Initial (M)

0.50

0.70

Change (M)

−x

Equilibrium (M)

0.50 − x

0 +x x

+x 0.70 + x

O



Ka = 1.8 × 10−4 = (x)(0.70 + x)/(0.50 − x)



The value of x will be insignificant compared to 0.50 M and 0.70 M.

(b) I−(aq) [Lewis base] + I2(aq) [Lewis acid] n  I3−(aq)



1.8 × 10−4 = (x)(0.70)/(0.50)



x = [H3O+] = 1.29 × 10−4 M



pH = −log[H3O+] = 3.89

16.133   (a)  I I I

O H

Equation

−

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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A-93

17.3

(15.0 g NaHCO3)(1 mol/84.01 g) = 0.1786 mol NaHCO3, and (18.0 g Na2CO3)(1 mol/106.0 g) =  0.1698 mol Na2CO3

17.7

75.0 mL acid will partially neutralize the base.



Initial amount NH3 =  (0.100 mol/L)(0.1000 L)  = 0.01000 mol



Amount HCl added =  (0.100 mol/L)(0.0750 L)  = 0.007500 mol



Amount NH3 after reaction =  0.01000 − 0.007500 = 0.00250 mol



Amount NH4+ after reaction = 0.007500 mol



Solve using the Henderson–Hasselbalch equation; use Ka for the weak acid NH4+:



pH = pKa + log {[base]/[acid]}



pH = −log (4.8 × 10−11) + log {[0.1698]/[0.1786]}



pH = 10.319 − 0.022 = 10.30

17.4

pH = pKa + log {[base]/[acid]}



5.00 = −log (1.8 × 10−5) + log {[base]/[acid]}



5.00 = 4.745 + log {[base]/[acid]}



[base]/[acid] = 1.8





To prepare this buffer solution, the ratio [base]/[acid] must equal 1.8. For example, you can dissolve 1.8 mol (148 g) of NaCH3CO2 and 1.0 mol (60.05 g) of CH3CO2H in some amount of water.

pH = pKa + log {[base]/[acid]}



pH = −log (5.6 × 10−10) + log {[0.00250]/ [0.007500]}



pH = 9.252 − 0.477 = 8.77

17.8

BaF2(s) st Ba2+(aq) + 2 F−(aq)



[F−] = 2[Ba2+] = 2(3.6 × 10−3 M) = 7.2 × 10−3 M



Ksp = [Ba2+][F−]2 = (3.6 × 10−3)(7.2 × 10−3)2 = 1.9 × 10−7

17.5

Initial pH (before adding acid):



pH =  pKa + log {[base]/[acid]} = −log (1.8 × 10−4) + log {[0.70]/[0.50]} = 3.745 + 0.146 = 3.89



After adding acid, the added HCl will react with the weak base (formate ion) and form more formic acid. The net effect is to change the ratio of [base]/[acid] in the buffer solution.

17.9

AgCN(s) st Ag+(aq) + CN−(aq)



Ksp = [Ag+][CN−]



Let x = solubility of AgCN in mol/L



Initial amount HCO2H =  0.50 mol/L × 0.500 L  = 0.250 mol



6.0 × 10−17 = x2



x = 7.75 × 10−9 mol/L = 7.7 × 10−9 mol/L



Initial amount HCO2− =  0.70 mol/L × 0.50 L  = 0.350 mol



(7.75 × 10−9 mol AgCN/L)(133.9 g AgCN/1 mol AgCN) = 1.0 × 10−6 g/L



Amount HCl added =  1.0 mol/L × 0.010 L  = 0.0100 mol

17.10 Ca(OH)2(s) st Ca2+(aq) + 2 OH−(aq)



Amount HCO2H after HCl addition = 0.250 mol +  0.0100 mol = 0.260 mol



Ksp = [Ca2+][OH−]2; Ksp = 5.5 × 10−6



5.5 × 10−6 = [x][2x]2 (where x = solubility in mol/L)



Amount HCO2− after HCl addition = 0.350 mol − 0.0100 mol = 0.340 mol



x = 1.11 × 10−2 mol/L = 1.1 × 10−2 mol/L



Solubility in g/L = (1.11 × 10−2 mol/L)(74.1 g/mol) =  0.82 g/L



pH = pKa + log {[base]/[acid]}



pH = −log (1.8 × 10−4) + log {[0.340]/[0.260]}



pH = 3.745 + 0.117 = 3.86

17.6

35.0 mL base will partially neutralize the acid.



Initial amount CH3CO2H =  (0.100 mol/L)(0.1000 L) = 0.01000 mol

17.11 (a) In pure water: K sp = [Ba2+][SO42−]; 1.1 × 10−10 = [x][x]; x = 1.0 × 10−5 mol/L (b) In 0.010 M Ba(NO3)2, which furnishes 0.010 M Ba2+ in solution:



K sp = [Ba2+][SO42−] 1.1 × 10−10 = [0.010 + x][x] x = 1.1 × 10−8 mol/L



Amount NaOH added = (0.100 mol/L)(0.0350 L) =  0.003500 mol



Amount CH3CO2H after reaction = 0.01000 − 0.003500 = 0.00650 mol



Amount CH3CO2− after reaction = 0.003500 mol



[CH3CO2H] after reaction =  0.00650 mol/0.1350 L = 0.0481 M



[CH3CO2−] after reaction =  0.00350 mol/0.1350 L = 0.02593 M



Ka = [H3O+][CH3CO2−]/[CH3CO2H]

17.13 Ksp = [Pb2+][I−]2. Let x be the concentration of I− required at equilibrium.



1.8 × 10−5 = [x][0.02593 + x]/[0.0481 − x]



9.8 × 10−9 = [0.050][x]2



x = [I−] = 4.4 × 10−4 mol/L. A concentration greater than this value will result in precipitation of PbI2.



+

−5

x = [H3O ] = 3.34 × 10

A-94

M; pH = 4.48

17.12 When [Pb2+] = 1.1 × 10−3 M, [I−] = 2.2 × 10−3 M.

Q = [Pb2+][I−]2 = [1.1 × 10−3][2.2 × 10−3]2 =  5.3 × 10−9



This value is less than Ksp, which means that the system has not yet reached equilibrium and more PbI2 will dissolve.

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.



Let x be the concentration of Pb2+ in solution, in equilibrium with 0.0015 M I−.



9.8 × 10−9 = [x][1.5 × 10−3]2



x = [Pb2+] = 4.4 × 10−3 M

17.14 First, determine the concentrations of Ag+ and Cl−; then calculate Q, and see whether it is greater than or less than Ksp. Concentrations are calculated using the final volume, 105.0 mL, in the equation Cdil × Vdil =  Cconc × Vconc. [Ag ](0.1050 L) = (0.0010 mol/L)(0.1000 L) +

−4



[Ag ] = 9.52 × 10



[Cl−](0.1050 L) = (0.025 M)(0.0050 L)



[Cl−] = 1.19 × 10−3 M



Q = [Ag+][Cl−] = [9.52 × 10−4][1.19 × 10−3] =  1.1 × 10−6



Because Q > K sp, precipitation occurs.

M

[Ag(CN)2]− st AgCN + CN−

Initial (M)

0.0071

0

Change

−x

+x

Equilibrium (M)

0.0071 − x

x

1/K2 = [CN−]/[[Ag(CN)2]−] 1/(7.80 × 104) = x/(0.0071 − x) x = 9.10 × 10−8 M [CN−] = 9.1 × 10−8 M; [[Ag(CN)2]−] = 0.0071 M

17.2 Take a Deep Breath

Equation

[Ag(NH3)2]+ st Ag+ +

Initial (M)

0.0050

Change

−x

Equilibrium (M)

0.0050 − x

0 +x x

2 NH3

+

7.40 = 7.20 + log[HPO42−]/[H2PO4−]

+2x

[HPO42−]/[H2PO4−] = 1.58 = 1.6

0.99 + 2x

2. Assign x = [HPO42−], then [H2PO4−] = (0.020 − x)

−10

x = [Ag ] = 4.6 × 10

 mol/L

1.58 = x/(0.020 − x); x = 0.0123 [HPO42−] = x = 0.012 mol/L [H2PO4−] = 0.020 − x = 0.008 mol/L

−

17.16 Cu(OH)2(s) st Cu (aq) + 2 OH (aq)

Ksp = [Cu2+][OH−]2



Cu2+(aq) + 4 NH3(aq) st [Cu(NH3)4]2+(aq)



Kf = [Cu(NH3)4 ]/[Cu ][NH3]



Net: Cu(OH)2(s) + 4 NH3(aq) st  [Cu(NH3)42+](aq) + 2 OH−(aq)



Knet = K sp × Kf = (2.2 × 10−20)(2.1 × 1013) =  4.6 × 10−7

2+

1. pH = pKa + log[HPO42−]/[H2PO4−]

1.00 − 2(0.0050)

K = 1/Kf = 1/1.1 × 107 = [x][0.99]2/0.0050 2+

2+

Study Questions 17.1

(a) Decrease pH; (b) increase pH; (c) no change in pH

17.3

pH = 9.25

17.5

pH = 4.74

17.7

pH = 4.38

17.9

pH = 9.12; pH of buffer is lower than the pH of the original solution of NH3 (pH = 11.17).

4

Applying Chemical Principles 17.1 Everything That Glitters . . . 1. (0.10 g NaCN)(100 g solution/0.035 g NaCN) (1 mL solution/1 g solution) = 2.9 × 102 mL 2. (1.0 × 103 kg ore)(1000 g/1 kg)(0.012 g Au/100 g ore) (1 mol Au/196.97 g Au)(8 mol NaCN/4 mol Au) (1 L/0.0071 mol NaCN) = 1.7 × 102 L 3. Au+(aq) + 2 CN−(aq) st [Au(CN)2]−(aq) Kf = [[Au(CN)2]−]/([Au+][CN−]2) 2.0 × 1038 = 1.1 × 10−4/(x(0.0071)2) x = [Au+] = 1.1 × 10−38 M. Yes, the conclusion is reasonable; less than one free Au+ ion is present per liter of solution. 4. (a)  K1 = 1/Ksp = 1/(6.0 × 10−17) = 1.67 × 1016 = 1.7 × 1016



Equation

5. 2 NaAu(CN)2(aq) + Zn(s) n 2 Au(s) + Na2Zn(CN)4(aq)

17.15



(c) Because K2 >> 1, assume that the reaction to form the complex goes to completion and then calculate the amount of dissociation of the complex.

+





(b)  K3 = K1K2 1.3 × 1021 = (1.67 × 1016)(K2) K2 = 7.80 × 104 = 7.8 × 104

17.11 4.7 g 17.13 pH = 4.92 17.15 [CH3CO2H]/[CH3CO2−] = 1.8 17.17 (a) pH = 3.59; (b) [HCO2H]/[HCO2−] = 0.45 17.19 (c) CH3CO2H + NaCH3CO2 17.21 The buffer must have a ratio of 0.51 mol NaH2PO4 to 1 mol Na2HPO4. For example, dissolve 0.51 mol NaH2PO4 (61 g) and 1.0 mol Na2HPO4 (140 g) in some amount of water. 17.23 46 mL of 1.0 M NaOH 17.25 (a) pH = 4.95; (b) pH = 5.05 17.27 (a) pH = 9.55; (b) pH = 9.50 17.29 (a) pH = 7.21 (b) 2.7 L of 0.15 M NaOH

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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A-95

17.31 (a) Original pH = 5.62 (b) [Na+] = 0.0323 M, [OH−] = 1.5 × 10−3 M, [H3O+] = 6.5 × 10−12 M, and [C6H5O−] = 0.0308 M (c) pH = 11.19 17.33 (a) Original NH3 concentration = 0.0154 M (b) At the equivalence point [H3O+] = 1.9 × 10−6 M, [OH−] = 5.3 × 10−9 M, [NH4+] = 6.25 × 10−3 M. (c) pH at equivalence point = 5.73 17.35 The titration curve begins at pH = 13.00 and drops slowly as HCl is added. Just before the equivalence point (when 30.0 mL of acid has been added), the curve falls steeply. The pH at the equivalence point is exactly 7. Just after the equivalence point, the curve flattens again and begins to approach the final pH of just over 1.0. The total volume at the equivalence point is 60.0 mL. 17.37

(a) Starting pH = 11.12 (b) pH at equivalence point = 5.28 (c) pH at midpoint (half-neutralization point) = 9.25 (d) Methyl red, bromocresol green



(e) Acid (mL) Added pH

17.39

 5.00

9.85

15.0

9.08

20.0

8.65

22.0

8.39

30.0

2.04

See Figure 17.11 on page 852. (a) Thymol blue or bromophenol blue (b) Phenolphthalein (c) Methyl red; thymol blue

17.41 (a) Silver chloride, AgCl; lead(II) chloride, PbCl2 (b) Zinc carbonate, ZnCO3; zinc sulfide, ZnS (c) Iron(II) carbonate, FeCO3; iron(II) oxalate, FeC2O4 17.43 (a) and (b) are soluble, (c) and (d) are insoluble. 17.45 (a) AgCN(s) st Ag+(aq) + CN−(aq), Ksp = [Ag+][CN−] (b) NiCO3(s) st Ni2+(aq) + CO32−(aq), Ksp = [Ni2+][CO32−] (c) AuBr3(s) st Au3+(aq) + 3 Br−(aq), Ksp = [Au3+][Br−]3 17.47 Ksp = (1.9 × 10−3)2 = 3.6 × 10−6 17.49 Ksp = 4.37 × 10−9 17.51 Ksp = 1.4 × 10−15 17.53 (a) 9.2 × 10−9 M; (b) 2.2 × 10−6 g/L 17.55 (a) 2.4 × 10−4 M; (b) 0.018 g/L 17.57 Only 2.1 × 10−4 g dissolves. 17.59 (a) PbCl2; (b) FeS; (c) Fe(OH)2 17.61 (a) PbBr2; (b) Hg2Cl2; (c) TlI

A-96

17.63 Solubility in pure water = 1.0 × 10−6 mol/L; solubility in 0.010 M SCN− = 1.0 × 10−10 mol/L 17.65 (a) Solubility in pure water = 2.2 × 10−6 mg/mL (b) Solubility in 0.020 M AgNO3 = 1.0 × 10−12 mg/mL 17.67 [Fe2+] = 4.9 × 10−3 M 17.69 (a) PbS (b) Ag2CO3 (c) Al(OH)3 17.71 (a) AgCN (b) Ag2C2O4 (c) Hg2CO3 17.73 Q < Ksp, so no precipitate forms. 17.75 Q > Ksp; Zn(OH)2 will precipitate. 17.77 [OH−] must exceed 1.0 × 10−5 M. 17.79 Using Ksp for Zn(OH)2 and Kf for [Zn(OH)4]2−, Knet for Zn(OH)2(s) + 2 OH−(aq) uv [Zn(OH)4]2−(aq)

is 1 × 101. This indicates that the reaction is productfavored at equilibrium.

17.81 Knet for AgCl(s) + 2 NH3(aq) uv [Ag(NH3)2]+(aq) + Cl−(aq)is 1.98 × 10−3. When all the AgCl dissolves, [[Ag(NH3)2]+] = [Cl−] = 0.050 M. To achieve these concentrations, [NH3] must be 1.124 M. Therefore, the amount of NH3 added must be 2 × 0.050 mol/L (to react with the AgCl) plus 1.124 mol/L (to achieve the proper equilibrium concentration). The total is 1.22 mol/L NH3. 17.83 (a) Solubility in pure water = 1.3 × 10−5 mol/L or 0.0019 g/L. (b) Knet for AgCl(s) + 2 NH3(aq) uv [Ag(NH3)2]+(aq) + Cl−(aq) is 1.98 × 10−3. When using 1.0 M NH3, the concentrations of species in solution are [[Ag(NH3)2]+] = [Cl−] = 0.0409 M = 0.041 M and so [NH3] = 1.0 − 2(0.0409) M or about 0.9 M. The amount of AgCl dissolved is 0.041 mol/L or 5.9 g/L. 17.85 (a) NaBr(aq) + AgNO3(aq) n NaNO3(aq) + AgBr(s) (b) 2 KCl(aq) + Pb(NO3)2(aq) n  2 KNO3(aq) + PbCl2(s) 17.87 Q > Ksp, so BaSO4 precipitates. 17.89 [H3O+] = 1.9 × 10−10 M; pH = 9.73 17.91 BaCO3 < Ag2CO3 < Na2CO3 17.93 Original pH = 8.62; dilution will not affect the pH. 17.95 (a) 0.100 M acetic acid has a pH of 2.87. Adding sodium acetate slowly raises the pH. (b) Adding NaNO3 to 0.100 M HNO3 has no effect on the pH. In part (a), adding the conjugate base of a weak acid creates a buffer solution. In part (b), HNO3 is a strong acid, but its conjugate base (NO3−) is so weak that the base has no effect on the complete ionization of the acid.

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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17.97

(a) pH = 4.13 (b) 0.5 g of C6H5CO2H (c) 0.82 mL of 2.0 M HCl should be added.

17.99

K = 2.1 × 106; the equilibrium lies to the right; yes, AgI forms.

17.117 110 mL NaOH

17.103 (a) [F−] = 1.3 × 10−3 M (b) [Ca2+] = 2.9 × 10−5 M

17.119 Add dilute HCl, say 1 M HCl, to a solution of the salts. Both AgCl and PbCl2 will precipitate, but Cu2+ ions will stay in solution (as CuCl2 is water-soluble). Decant off the copper-containing solution to leave a precipitate of white AgCl and PbCl2. Lead(II) chloride (Ksp = 1.7 × 10−5) is much more soluble than AgCl (Ksp = 1.8 × 10−10). Warming the precipitates in water will dissolve the PbCl2 and leave the AgCl as a white solid.

17.105 (a) PbSO4 will precipitate first. (b) [Pb2+] = 5.1 × 10−6 M

17.121 Cu(OH)2 will dissolve in a nonoxidizing acid such as HCl, whereas CuS will not.

17.107 (a) Knet = 16; (b)  1.9 g Na2S2O3.

17.123 When Ag3PO4 dissolves slightly, it produces a small concentration of the phosphate ion, PO43−. This ion is a strong base and hydrolyzes to HPO42−. As this reaction removes the PO43− ion from equilibrium with Ag3PO4, the equilibrium shifts to the right, producing more PO43− and Ag+ ions. Thus, Ag3PO4 dissolves to a greater extent than might be calculated from a Ksp value (unless the Ksp value was actually determined experimentally).

17.101 (a) 0.030 L (30. mL) (b) 0.48 g CaC2O4

17.109 (a) Add H2SO4, precipitating BaSO4 and leaving Na+(aq) in solution. (b) Add HCl or another source of chloride ion. PbCl2 will precipitate, but NiCl2 is water-soluble. 17.111 (a) BaSO4 will precipitate first. (b) [Ba2+] = 3.2 × 10−5 M 17.113

(a) pH = 2.81 (b) pH at equivalence point = 8.72 (c) pH at the midpoint = pKa = 4.62 (d) Phenolphthalein (e) After 10.0 mL, pH = 4.39. After 20.0 mL, pH = 5.07. After 30.0 mL, pH = 11.84. (f) A plot of pH versus volume of NaOH added would begin at a pH of 2.81, rise slightly to the midpoint at pH = 4.62, and then begin to rise more steeply as the equivalence point is approached (when the volume of NaOH added is 27.0 mL). The pH rises vertically through the equivalence point, and then begins to level off above a pH of about 11.0.

17.115 The Kb value for ethylamine (4.27 × 10−4) is found in Appendix I. (a) pH = 11.89 (b) Midpoint pH = 10.63 (c) pH = 10.15 (d) pH = 5.93 at the equivalence point (e) pH = 2.13 (f) Titration curve 14

pH

Chapter 18 18.1

8



(a) O3; larger molecules generally have higher entropies than smaller molecules. (b) SnCl4(g); gases have higher entropies than liquids.

6

18.2

(a) ΔrS° = ΣnS°(products) − ΣnS°(reactants)

4





ΔrS° = (l mol/mol-rxn) S°[NH4Cl(aq)]) − (l mol/mol-rxn) S°[NH4Cl(s)]





ΔrS° =  (l mol/mol-rxn)(169.9 J/mol ∙ K)  − (1 mol/mol-rxn)(94.85 J/mol ∙ K)  = 75.1 J/K ∙ mol-rxn





A gain in entropy for the formation of a mixture (solution) is expected.

10

2 0

0

20

40

60

80

100

120

140

Titrant Volume (mL)



17.127 (a) COCOC angle, 120°; OOCPO, 120°; COOOH, 109°; COCOH, 120° (b) Both the ring C atoms and the C in CO2H are sp2 hybridized. (c) Ka = 1 × 10−3 (d) 10% (e) pH at half-way point = pKa = 3.0; pH at equivalence point = 7.3

Check Your Understanding

12



17.125 (a) Base is added to increase the pH. The added base reacts with acetic acid to form more acetate ions in the mixture. Thus, the fraction of acid declines and the fraction of conjugate base rises (that is, the ratio [CH3CO2H]/[CH3CO2−] decreases) as the pH rises. (b) At pH = 4, acid predominates (85% acid and 15% acetate ions). At pH = 6, acetate ions predominate (95% acetate ions and 5% acid). (c) At the point the lines cross, [CH3CO2H] = [CH3CO2−]. At this point pH = pKa, so pKa for acetic acid is 4.74.

(g) Alizarin or bromocresol purple (see Figure 17.11)

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

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A-97





(b) ΔrS° = (2 mol CO2/mol-rxn) S°(CO2) +  (3 mol H2O/mol-rxn) S°(H2O) − [(1 mol C2H5OH/mol-rxn) S°(C2H5OH) +  (3 mol O2/mol-rxn) S°(O2)]

ΔrS° = (2 mol/mol-rxn)(213.74 J/mol ∙ K) + (3 mol/mol-rxn)(188.84 J/mol ∙ K) − [(1 mol/mol-rxn)(282.70 J/mol ∙ K) + (3 mol/mol-rxn)(205.07 J/mol ∙ K)]

18.5

SO2(g) + 1⁄2 O2(g) n SO3(g)



ΔrG° = ΣnΔf G°(products) − ΣnΔf G°(reactants)



ΔrG° = (1 mol/mol-rxn)Δf G°[SO3(g)] − {(1 mol/mol-rxn)Δf G°[SO2(g)]  + (0.5 mol/mol-rxn)Δf G°[O2(g)]}



ΔrG° = −371.04 kJ/mol-rxn − (−300.13 kJ/mol-rxn + 0 kJ/mol-rxn)



ΔrS° = 96.09 J/K ∙ mol-rxn







An increase in entropy is expected because there is an increase in the number of moles of gases.

18.6

18.3

ΔS°(system) =  ΔrS°  = ΣnS°(products) − ΣnS°(reactants)



ΔrS° = ( 2 mol HCl/mol-rxn) S°[HCl(g)]  − {(1 mol H2/mol-rxn) S°[H2(g)] + (1 mol Cl2/mol-rxn) S°[Cl2(g)]}





= ( 2 mol HCl/mol-rxn)(186.2 J/K ∙ mol HCl)  − {(1 mol H2/mol-rxn)(130.7 J/K ∙ mol H2)  + (1 mol Cl2/mol-rxn) (223.08 J/K ∙ mol Cl2(g)} = 18.62 J/K ∙ mol-rxn = 18.6 J/K ∙ mol-rxn



ΔrH° = ΣnΔf H°(products) − ΣnΔf H°(reactants)



ΔrH° = ( 2 mol HCl/mol-rxn) Δf H°[HCl(g)]  − {(1 mol H2/mol-rxn) Δf H°[H2(g)]  + (1 mol Cl2/mol-rxn) Δf H°[Cl2(g)]}





= ( 2 mol HCl/mol-rxn)(−92.31 kJ/mol HCl)  − {(1 mol H2/mol-rxn)(0 kJ/mol H2)  + (1 mol Cl2/mol-rxn) (0 kJ/mol Cl2(g)} = −184.62 kJ/mol-rxn

= −70.91 kJ/mol-rxn

HgO(s) n Hg(ℓ) + 1⁄2 O2(g); determine the temperature at which ΔrG° = ΔrH° − TΔrS° = 0. T is the unknown in this problem.



ΔrH° = [−Δf H° for HgO(s)] = 90.83 kJ/mol-rxn



ΔrS° = S°[Hg(ℓ)] + 1⁄2 S°[O2(g)] − S°[HgO(s)]



ΔrS° = (1 mol/mol-rxn)(76.02 J/mol ∙ K) + [(0.5 mol/mol-rxn)(205.07 J/mol ∙ K) − (1 mol/mol-rxn)(70.29 J/mol ∙ K)] = 108.27 J/K ∙ mol-rxn



ΔrH° − T(ΔrS°) = 0 = 90,830 J/mol-rxn − T(108.27 J/K ∙ mol-rxn)



T = 839 K (566 °C)

18.7

Pb2+(aq) + 2 I–(aq) uv PbI2(s)



K = 1/Ksp = 1/(9.8 × 10–9) = 1.02 × 108

∆rG° = −RT ln Ksp ∆rG° = (−0.0083145 kJ/K ∙ mol-rxn)(298.15 K) ln(1.02 × 108) ∆rG° = – 45.7 kJ/mol-rxn 18.8

ΔrG° = ΣnΔf G°(products) − ΣnΔf G°(reactants)



Both ΔrH° (< 0) and ΔrS° (> 0) are favorable, so this reaction is predicted to be spontaneous under standard conditions.



ΔrG° = (2 mol/mol-rxn)Δf G°[H2(g)] + (1 mol/mol-rxn)Δf G°[O2(g)] − (2 mol/mol-rxn)Δf G°[H2O(ℓ)] 



ΔS°(surroundings) = −ΔrH°/T = −(−184.62 kJ/mol-rxn)/298.15 K = 0.619219 kJ/K ∙ mol-rxn = 619.219 J/K ∙ mol-rxn



ΔrG° = (2 mol/mol-rxn)(0 kJ/mol) + (1 mol/mol-rxn)(0 kJ/mol) − (2 mol/mol-rxn)(−237.15 kJ/mol)



ΔS°(universe) = ΔS°(system) + ΔS°(surroundings) = 18.62 J/K ∙ mol-rxn + 619.219 J/K ∙ mol-rxn = 637.8 J/K ∙ mol-rxn



ΔrG° = 474.30 kJ/mol-rxn



 e−ΔrG°/(RT) Kp = = e−(474.30 kJ/mol-rxn)/[(0.0083145 kJ/ K ∙ mol-rxn)(298.2 K)] = 8 × 10−84

18.9

(a) ΔrG° = 2 Δf G°(NO) = 2 mol/mol-rxn × 86.58 kJ/mol = 173.16 kJ/mol-rxn = 173.2 kJ/mol-rxn



ΔS°(universe) > 0, so the reaction is spontaneous under standard conditions.

18.4

For the reaction N2(g) + 3 H2(g) n 2 NH3(g):



ΔrH° = (2 mol/mol-rxn) Δf H° for NH3(g) =  (2 mol/mol-rxn)(−45.90 kJ/mol) = −91.80 kJ/mol-rxn



The reaction is reactant-favored at equilibrium.

(b) ΔrG = ΔrG° + RT ln Q = 173.16 kJ/mol-rxn + (0.0083145 kJ/K ∙ mol-rxn)(298.15 K) ln[P(NO)]2/[P(N2)][P(O2)]



ΔrS° = (2 mol/mol-rxn) S°(NH3) − [(1 mol/mol-rxn) S°(N2) + (3 mol/mol-rxn) S°(H2)]



ΔrS° = (2 mol/mol-rxn)(192.77 J/ mol ∙ K) − [(1 mol/mol-rxn)(191.56 J/mol ∙ K)  + (3 mol/mol-rxn)(130.7 J/mol ∙ K)]





ΔrS° = −198.12 J/K ∙ mol-rxn = −198.1 J/K ∙ mol-rxn (= −0.1981 kJ/K ∙ mol-rxn)

Applying Chemical Principles



 rH° − TΔrS° = −91.80 kJ/mol-rxn − ΔrG° = Δ (298 K)(−0.19812 kJ/K ∙ mol-rxn)



ΔrG° = −32.8 kJ/mol-rxn

A-98

ΔrG = 173.16 kJ/mol-rxn − 11.42 kJ/mol-rxn = 161.7 kJ/mol-rxn. The reaction is not spontaneous.

18.1 Thermodynamics and Living Things 1. Creatine phosphate + H2O n creatine + HPi  ΔrG°′ = −43.3 kJ/mol-rxn Adenosine + HPi n adenosine monophosphate + H2O  ΔrG°′ = + 9.2 kJ/mol-rxn

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Net reaction (sum of the two reactions):

Study Questions

Creatine phosphate + adenosine n creatine + adenosine monophosphate

18.1

(a) Endothermic (b) Yes (c) Yes (d) Yes

18.3

All four processes are spontaneous.

ΔrG°′ = ΔrG° − 39.9 kJ/mol

18.5

(a) Decrease in the entropy of the system. (b) Decrease in the entropy of the system. (c) Increase in the entropy of the system. (d) Increase in the entropy of the system.

18.2 Are Diamonds Forever?

18.7

(c) is reversible; (a), (b), and (d) are not reversible.

1. C(diamond) n C(graphite) (a)  ΔrG° = ΣnΔf G°(products) − ΣnΔf G°(reactants) ΔrG° = (1 mol/mol-rxn)Δf G°[C(graphite)] − (1 mol/mol-rxn)Δf G°[C(diamond] ΔrG° = (1 mol/mol-rxn)(0 kJ/mol) − (1 mol/mol-rxn)(2.900 kJ/mol) ΔrG° = −2.900 kJ/mol-rxn Kp = e−ΔrG°/(RT) = e−(−2.900 kJ/mol-rxn)/[(0.0083145 kJ/ K ∙ mol-rxn)(298.2 K)] = 3.22 (b) ΔrH° = ΣnΔf H°(products) − ΣnΔf H°(reactants) ΔrH° = (1 mol/mol-rxn)Δf H°[C(graphite)] − (1 mol/mol-rxn)Δf H°[C(diamond] ΔrH° = (1 mol/mol-rxn)(0 kJ/mol) − (1 mol/mol-rxn)(1.8 kJ/mol) ΔrH° = −1.8 kJ/mol-rxn  ΔrS° = ΣnS°(products) − ΣnS°(reactants) ΔrS° = (1 mol/mol-rxn)S°[C(graphite)] − (1 mol/mol-rxn)S°[C(diamond] ΔrS° = (1 mol/mol-rxn)(5.6 kJ/mol ∙ K) − (1 mol/mol-rxn)(2.4 kJ/mol ∙ K) ΔrS° = 3.2 kJ/mol-rxn ∙ K = 0.0032 kJ/ mol ∙ K ΔrG° = ΔrH° − TΔrS° = −1.8 kJ/mol-rxn − (1000 K)(0.0032 kJ/K ∙ mol-rxn) ΔrG° = −5.0 kJ/mol-rxn Kp = e−ΔrG°/(RT) = e−(−5.0 kJ/mol-rxn)/[(0.0083145 kJ/ K ∙ mol-rxn)(1000 K)] = 1.8 Though still favoring graphite, the equilibrium has shifted more toward diamond. (c) Greater pressures favor the formation of diamond, which is more dense than graphite. (d) The rate of reaction is negligibly slow at room temperature. 2. C60(s) n 60 C(diamond) (a) ΔrH° = ΣnΔf H°(products) − ΣnΔf H°(reactants) ΔrH° = (60 mol/mol-rxn)Δf H°[C(diamond)] − (1 mol/mol-rxn)Δf H°[C60(s)] ΔrH° = (60 mol/mol-rxn)(1.8 kJ/mol) − (1 mol/mol-rxn)(2320 kJ/mol) ΔrH° = −2212 kJ/mol-rxn = −2210 kJ/mol-rxn (b) ΔrS° ≈ 0 kJ/mol-rxn ∙ K ΔrG° = ΔrH° − TΔrS° = −2212 kJ/mol-rxn − (298.2 K)(0 kJ/K ∙ mol-rxn) ΔrG° = −2210 kJ/mol-rxn Thus, the formation of diamond from C60 is productfavored at equilibrium.

18.9

qrev = (333 J/g)(18.015 g/mol)(0.50 mol) = 3.00 × 103 J ΔS° = qrev/T = 3.00 × 103 J/373 K = 8.0 J/K

For this, ΔrG°′ = −43.3 kJ/mol-rxn + 9.2 kJ/mol-rxn = −34.1 kJ/mol-rxn. The negative value indicates that the transfer of phosphate from creatine phosphate to adenosine is product-favored at equilibrium. 2. ΔrG°′ = ΔrG° + RT ln[C][H3O+]/[A][B] = ΔrG° + (8.31 × 10−3 kJ/mol ∙ K)(298 K) ln[1][1 × 10−7]/[1][1]



18.11 ΔS =  k (ln Wfinal – ln Winitial) = k ln(Wfinal/Winitial) = (1.381 × 10−23 J/K)ln(30/5) = 2.47 × 10−23 J/K 18.13 The specific heat capacity for H2O(s) and the enthalpy of fusion for water are needed. 18.15

(a) CO2(g) at 0 °C (b) Liquid water at 50 °C (c) Ruby (d) One mole of N2 at 1 bar

18.17 (a) ΔrS° = +12.7 J/K ∙ mol-rxn. Entropy increases. (b) ΔrS° = −102.56 J/K ∙ mol-rxn. Significant decrease in entropy. (c) ΔrS° = +93.3 J/K ∙ mol-rxn. Entropy increases. (d) ΔrS° = −129.7 J/K ∙ mol-rxn. The solution has a smaller entropy (with H+ forming H3O+ and hydrogen bonding occurring) than HCl in the gaseous state. 18.19 (a) ΔrS° = +9.3 J/K ∙ mol-rxn (b) ΔrS° = −294.0 J/K ∙ mol-rxn 18.21 (a) ΔrS° = −507.3 J/K ∙ mol-rxn; entropy declines as a gaseous reactant is incorporated in a solid compound. (b) ΔrS° = +313.25 J/K ∙ mol-rxn; entropy increases as five molecules (three of them in the gas phase) form six molecules of products (all gases). 18.23 ΔS°(system) = −134.2 J/K ∙ mol-rxn; ΔH°(system) = −662.75 kJ/mol-rxn; ΔS°(surroundings) = +2222.9 J/K ∙ mol-rxn; ΔS°(universe) = +2088.8 J/K ∙ mol-rxn. The reaction is spontaneous under standard conditions. 18.25 ΔS°(system) = +163.3 J/K ∙ mol-rxn; ΔH°(system) = +285.83 kJ/mol-rxn; ΔS°(surroundings) = −958.68 J/K ∙ mol-rxn; ΔS°(universe) = −795.4 J/K ∙ mol-rxn

The reaction is not spontaneous under standard conditions because the overall entropy change in the universe is negative. The reaction is disfavored by energy dispersal.

18.27 (a) Type 2 (Table 18.1, page 901). The reaction is enthalpy-favored but entropy-disfavored. It is more favorable at low temperatures.

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A-99



(b) Type 4 (Table 18.1, page 901). This endothermic reaction is not favored by the enthalpy change nor by the entropy change. It is not spontaneous under standard conditions at any temperature.

18.29 (a) ΔrH° = −438 kJ/mol-rxn; ΔrS° = −201.7 J/K ∙ mol-rxn; ΔrG° = −378 kJ/mol-rxn. The reaction is product-favored at equilibrium and is enthalpy-driven. (b) ΔrH° = −86.61 kJ/mol-rxn; ΔrS° = −79.4 J/K ∙ mol-rxn; ΔrG° = −62.9 kJ/mol-rxn. The reaction is product-favored at equilibrium. The enthalpy change favors the reaction. 18.31 (a) ΔrH° = +116.7 kJ/mol-rxn; ΔrS° = +168.0 J/K ∙ mol-rxn; Δf G° = +66.6 kJ/mol-rxn. (b) ΔrH° = −425.93 kJ/mol-rxn; ΔrS° = −154.6 J/K ∙ mol-rxn; Δf G° = −379.82 kJ/mol-rxn. (c) ΔrH° = +17.51 kJ/mol-rxn; ΔrS° = +77.95 J/K ∙ mol-rxn; Δf G° = −5.73 kJ/mol-rxn.

The reactions in (b) and (c) are predicted to be product-favored at equilibrium under standard conditions.

18.33 (a) ΔrG° = −817.54 kJ/mol-rxn; productfavored at equilibrium (b) ΔrG° = +256.6 kJ/mol-rxn; reactantfavored at equilibrium (c) ΔrG° = −1101.14 kJ/mol-rxn; productfavored at equilibrium 18.35 Δf G°[BaCO3(s)] = −1134.4 kJ/mol 18.37 (a) ΔrH° = +66.2 kJ/mol-rxn; ΔrS° = −121.62 J/K ∙ mol-rxn; ΔrG° = +102.5 kJ/mol-rxn. Both the enthalpy and the entropy changes are unfavorable. There is no temperature at which the reaction will be product-favored at equilibrium. This is a case like that in the right panel in Figure 18.10 and is a Type 4 reaction (Table 18.1). As the temperature increases, the reaction becomes even more reactant-favored (the value of ΔrG° becomes more positive). (b) ΔrH° = −221.05 kJ/mol-rxn; ΔrS° = +179.1 J/K ∙ mol-rxn; ΔrG° = −274.45 kJ/mol-rxn. The reaction is favored by both enthalpy and entropy and is product-favored at equilibrium at all temperatures. This is a case like that in the left panel in Figure 18.10 and is a Type 1 reaction. As the temperature increases, the reaction becomes even more product-favored (the value of ΔrG° becomes more negative). (c) ΔrH° = −179.0 kJ/mol-rxn; ΔrS° = −160.2 J/K ∙ mol-rxn; ΔrG° = −131.23 kJ/mol-rxn. The reaction is favored by the enthalpy change but disfavored by the entropy change.



The reaction becomes less product-favored as the temperature increases; it is a case like the blue line in the middle panel of Figure 18.10. (d) ΔrH° = +822.2 kJ/mol-rxn; ΔrS° = +181.28 J/K ∙ mol-rxn; ΔrG° = +768.19 kJ/mol-rxn.

The reaction is not favored by the enthalpy change but favored by the entropy change. The reaction becomes more product-favored as the temperature increases; it is a case like the red line in the middle panel of Figure 18.10. 18.39 (a) ΔrS° = +174.75 J/K ∙ mol-rxn; ΔrH° = +116.94 kJ/mol-rxn; ΔrG° = +64.84 kJ/mol-rxn.

(b) The reaction is not product-favored at equilibrium under standard conditions at 298.15 K. (c) As the temperature increases, ΔrS° becomes more important, so ΔrG° can become negative at a sufficiently high temperature and the reaction is product-favored at equilibrium.

18.41 ΔrG° =  −RTlnK = −(8.3145 J/K ∙ mol)(298 K)ln(1.8 × 10−5) = 2.71 × 104 J/mol = 27.1 kJ/mol

The reaction is reactant-favored at equilibrium.

18.43 Kp = 6.8 × 10−16. Note that Kp is very small and that Δf G° is positive. Both indicate a process that is reactant-favored at equilibrium. 18.45 ΔrG° = −100.24 kJ/mol-rxn and Kp = 4 × 1017. Both the free energy change and K indicate a process that is product-favored at equilibrium. 18.47 CaCO3(s) uv CaO(s) + CO2(g)

ΔrG° = +106.1 kJ/mol; Kp = 7 × 10−13; A positive ΔrG° (and a Kp < 1) indicate a reactant-favored reaction at equilibrium.

18.49 (a) ΔrG° = −32.74 kJ/mol-rxn. The reaction is product-favored at equilibrium. (b) ΔrG = −21.33 kJ/mol-rxn. The reaction is spontaneous in the forward direction. 18.51

(a) HBr(g) (b) NH4Cl(aq) (c) C2H4(g) (d) NaCl(g)

18.53 ΔrG° = −98.9 kJ/mol-rxn. The reaction is productfavored at equilibrium. It is enthalpy-driven. 18.55 ΔrH° = −1560.66 kJ/mol-rxn; ΔrS° = −309.6 J/K ∙ mol-rxn; ΔS°(universe) = +4930 J/K ∙ mol-rxn. Combustion reactions are spontaneous, and this is confirmed by the sign of ΔS°(universe). 18.57 (a) The reaction occurs spontaneously and is product-favored at equilibrium. Therefore, ΔS°(universe) is positive and ΔrG° is negative. The reaction is likely to be exothermic, so ΔrH° is negative, and ΔS°(surroundings) is positive.

A-100 Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

ΔS°(system) is expected to be negative because two moles of gas form one mole of solid. The ­calculated values are as follows:

ΔS°(system) = −284.1 J/K ∙ mol-rxn



ΔrH° = −176.34 kJ/mol-rxn



ΔS°(surroundings) = +591.45 J/K ∙ mol-rxn



ΔS°(universe) = +307.4 J/K ∙ mol-rxn

ΔrG° = −91.64 kJ/mol-rxn (b) Kp = 1.1 × 1016

Rhombic sulfur is more stable than monoclinic sulfur at 80 °C, but the reverse is true at 110 °C. (b) T = 370 K or about 96 °C. This is the temperature at which the two forms are at equilibrium under standard conditions. 18.79 ΔrG° at 298 K = 22.64 kJ/mol; reaction is not product-favored at equilibrium. It does become product-favored above 469 K (196 °C). 18.81 Δf G°[HI(g)] = −10.9 kJ/mol

18.59 Kp = 1.3 × 1029 at 298 K (ΔrG° = −166.1 kJ/mol-rxn). The reaction is already extremely product-favored at equilibrium at 298 K. A higher temperature would make the reaction less productfavored because ΔrS° has a negative value (−242.3 J/K ∙ mol-rxn).

18.83 (a) ΔrG° = +194.8 kJ/mol-rxn and K = 6.7 × 10−11 (b) The reaction is reactant-favored at equilibrium at 727 °C. (c) Keep the pressure of CO as low as possible (by removing it during the course of the reaction).

18.61 At the boiling point, ΔrG° = 0 = ΔrH° − TΔrS°.



Kp = 1 at 620.3 K or 347.2 °C when PHg(g) = 1.000 bar



T when PHg(g) = (1/760) bar is 398.3 K or 125.2 °C.



Here ΔrS° = ΔrH°/T = 112 J/K ∙ mol-rxn at 351.15 K.

18.63 ΔrS° is +137.2 J/K ∙ mol-rxn. A positive entropy change means that raising the temperature will increase the product favorability of the reaction (because −TΔS° will become more negative). 18.65 The reaction is exothermic, so ΔrH° should be negative. Also, a gas and an aqueous solution are formed, so ΔrS° should be positive. The calculated values are ΔrH° = −183.32 kJ/mol-rxn (with a negative sign as expected) and ΔrS° = −7.7 J/K ∙ mol-rxn.

The entropy change is slightly negative, not positive as predicted. The reason for this is the negative entropy change upon dissolving NaOH. Apparently the OH− ions in water hydrogen-bond with water molecules, an effect that also leads to a small, negative entropy change.

18.67 ΔrH° = +126.03 kJ/mol-rxn; ΔrS° = +78.2 J/K ∙ mol-rxn; and ΔrG° = +103 kJ/mol-rxn. The reaction is not product-favored at equilibrium. 18.69 ΔrG° from K value = 4.87 kJ/mol-rxn

ΔrG° from free energies of formation = 4.73 kJ/mol-rxn

18.71 ΔrG° = −2.27 kJ/mol-rxn 18.73 (a) ΔrG° = +141.82 kJ/mol-rxn, so the reaction is not product-favored at equilibrium. (b) ΔrH° = +197.86 kJ/mol-rxn; ΔrS° = +187.95 J/K ∙ mol-rxn T = ΔrH°/ΔrS° = 1052.7 K or 779.6 °C (c) ΔrG° at 1500 °C (1773 K) = −135.4 kJ/mol-rxn

Kp at 1500 °C = 1 × 10

4

18.75 ΔrS° = −459.0 J/K ∙ mol-rxn; ΔrH° = −793 kJ/mol-rxn; ΔrG° = −657 kJ/mol-rxn

The reaction is product-favored at equilibrium. It is enthalpy-driven.

18.77 (a) ΔrG° at 80.0 °C = +0.14 kJ/mol-rxn



ΔrG° at 110.0 °C = −0.12 kJ/mol-rxn

18.85 Kp = PHg(g) at any temperature

18.87 (a) True (b) False. Whether an exothermic system is spontaneous also depends on the entropy change for the system. (c) False. Reactions with +ΔrH° and +ΔrS° are product-favored at equilibrium at higher temperatures. (d) True 18.89 Dissolving a solid such as NaCl in water is a productfavored process. Thus, ΔG° < 0. If ΔH° = 0, then the only way the free energy change can be negative is if ΔS° is positive. Generally the entropy change is the important factor in forming a solution. 18.91 2 C2H6(g) + 7 O2(g) n 4 CO2(g) + 6 H2O(g) (a) Not only is this an exothermic combustion reaction, but there is also an increase in the number of molecules of gases from reactants to products. Therefore, we would predict a positive value for ΔS° for both the system and the surroundings and thus for the universe as well. (b) The exothermic reaction has ΔrH° < 0. Combined with a positive ΔS°(system), the value of ΔrG° is negative. (c) The value of Kp is likely to be much greater than 1. Further, because ΔS°(system) is positive, the value of Kp will be even larger at a higher temperature. (See the left panel of Figure 18.10.) 18.93 Reaction a: Δr S°a = −80.7 J/K ∙ mol-rxn Reaction b: Δr S°b = −161.60 J/K ∙ mol-rxn

Reaction c: Δr S°c = −242.3 J/K ∙ mol-rxn Δr S°a + Δr S°b = Δr S°c; entropy is a state function.

18.95 (a) ΔrH° = −352.88 kJ/mol-rxn and ΔrS° = +21.31 J/K ∙ mol-rxn. Therefore, at 298 K, ΔrG° = −359.23 kJ/mol-rxn. (b) 4.8 g of Mg is required.

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions A-101

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18.97

(a) N2H4(ℓ) + O2(g) n 2 H2O(ℓ) + N2(g)

19.2

O2 is the oxidizing agent and N2H4 is the ­reducing agent. (b) ΔrH° = −622.29 kJ/mol-rxn and ΔrS° = +4.87 J/K ∙ mol-rxn. Therefore, at 298 K, ΔrG° = −623.74 kJ/mol-rxn. (c) 0.0027 K (d) 7.5 mol O2 (e) 4.8 × 103 g solution (f) 7.5 mol N2(g) occupies 170 L at 273 K and 1.0 atm of pressure. 18.99

Iodine dissolves readily, so the process is productfavored at equilibrium and ΔG° must be less than zero. Because ΔH° = 0, the process is entropy-driven.



Al(s) + 3 OH−(aq) n Al(OH)3(s) + 3 e−



Reduction half-reaction:



S(s) + H2O(ℓ) + 2 e− n HS−(aq) + OH−(aq)



Overall reaction:

2 Al(s) + 3 S(s) + 3 H2O(ℓ) + 3 OH−(aq) n  2 Al(OH)3(s) + 3 HS−(aq) (b) Aluminum is the reducing agent and is oxidized; sulfur is the oxidizing agent and is reduced. 19.3

Construct two half-cells, the first with a silver electrode and a solution containing Ag+(aq), and the second with a nickel electrode and a solution containing Ni2+(aq). Connect the two half-cells with a salt bridge. When the electrodes are connected through an external circuit, electrons will flow from the anode (the nickel electrode) to the cathode (the silver e­ lectrode). The overall cell reaction is Ni(s) + 2 Ag+(aq) n Ni2+(aq) + 2 Ag(s). To maintain ­ egative ions electrical neutrality in the two half-cells, n will flow from the Ag  ∙ Ag+ half-cell to the Ni  ∙ Ni2+ halfcell, and positive ions will flow in the opposite direction.

19.4

Zn(s) ∙ Zn2+(aq) ∙ SO42− ∙ PbSO4(s) ∙ Pb(s)

19.5

(a) Using Appendix M, the order determined for these metals from least strong reducing agent to strongest reducing agent is Hg < Pb < Sn. (The farther down the table, the stronger the metal is as a reducing agent.) (b) F2, Cl2, and Br2 all can oxidize mercury to mercury(II); Hg is located “southeast” of them on the table. I2 cannot oxidize mercury to mercury(II); mercury is located “northeast” rather than “southeast” of it.

18.101 CH3OH(g) n C(s, graphite) + 2 H2(g) + 1⁄2 O2(g) (a) The reaction becomes more product-favored at equilibrium as temperature increases. (b) There is no temperature between 400 K and 1000 K at which the decomposition is productfavored at equilibrium. 18.103 (a) For the electrolysis of water: ΔrG° = 474.3 kJ. For the formation from methane: ΔrG° = 142.2 kJ. (b) For the electrolysis of water: ΔrG° = 237.2 kJ/mol H2 For the formation from methane: ΔrG° = 47.4 kJ/mol H2 (c) The free energy change for the formation of H2 from methane is less than that to obtain H2 from water electrolysis.



18.105 (a, b) Temperature (K) 𝚫rG° (kJ/mol)

19.6

Overall reaction: 2 Al(s) + 3 Fe2+(aq) n  2 Al3+(aq) + 3 Fe(s)



(E°cell = 1.22 V, n = 6)



Ecell = E°cell − (0.0257/n) ln {[Al3+]2/[Fe2+]3} = 1.22 − (0.0257/6) ln {[0.025]2/[0.50]3} = 1.22 V − (−0.0227) V = 1.24 V

19.7

Overall reaction: 2 Ag+(aq) + H2(g) n  2 Ag(s) + 2 H+(aq)

K

  298 K

  −32.8

5.5 × 105

  800. K

  +73

  2 × 10−5

1300. K

+184.0

4.06 × 10−8



(c) The largest mole fraction of NH3 in an equilibrium mixture will be at 298 K.

18.107

(a) False (b) True (c) False (d) False

Chapter 19

(a) Oxidation half-reaction:



(E°cell = 0.7994 V, n = 2)



Ecell = E°cell − (0.0257/n) ln {[H+]2/([Ag+]2 PH2)}



0.902 V = 0.7994 − (0.0257/2) ln {x2/([1.0]2(1.0)}



x = [H+] = 0.0185 M = 0.018 M



pH = −log(0.0185) = 1.73

19.8

 −nFE°  ΔrG° = = −(2 mol e−)(96,500 C/mol e−) (−0.76 V)(1 J/1 C ∙ V) = 150,000 J = 150 kJ



The negative value of E° and the positive value of ΔrG° both indicate a reactant-favored reaction at equilibrium.

Check Your Understanding 19.1

Oxidation (Fe2+, the reducing agent, is oxidized):



Fe2+(aq) n Fe3+(aq) + e−



Reduction (MnO4−, the oxidizing agent, is reduced)



MnO4−(aq) + 8 H+(aq) + 5 e− n Mn2+(aq) + 4 H2O(ℓ)

19.9

E°cell = E°cathode − E°anode = 0.799 V − 0.855 V =  −0.056 V; n = 2



Overall reaction:



nE°/0.0257 = ln K



MnO4−(aq) + 8 H+(aq) + 5 Fe2+(aq) n Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(ℓ)



2(−0.056)/0.0257 = ln K



K = 0.013

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19.10 Cathode: 2 H2O(ℓ) + 2 e− n 2 OH−(aq) + H2(g)

Equation 3: Cu2+(aq) + 2 OH−(aq) n Cu(OH)2(s) K = 1/Ksp = 1/(2.2 × 10−20) = 4.55 × 1019; ΔrG° = −RTlnK = −(8.3145 J/K ∙ mol)(298 K) ln(4.55 × 1019) = −1.121 × 105 J



E°cathode = −0.83 V



Anode: 4 OH−(aq) n O2(g) + 2 H2O(ℓ) + 4 e−



E°anode = 0.40 V



Overall: 2 H2O(ℓ) n 2 H2(g) + O2(g)



E°cell = E°cathode − E°anode = −0.83 V − 0.40 V = −1.23 V

Total ΔrG° = 6.503 × 104 J/mol-rxn + −7.72 × 104 J/mol-rxn + −1.121 × 105 J/mol-rxn = −1.243 × 105 J/mol-rxn



The minimum voltage needed under standard conditions to cause this reaction to occur is 1.23 V.



K=  e−ΔrG°/(RT) = e−(−1.243 × 105 J/mol-rxn)/[(8.3145 J/ K ∙ mol-rxn)(298 K)] = 6.1 × 1021

19.11 (1) O2 is formed at the anode, by the reaction 2 H2O(ℓ) n 4 H+(aq) + O2(g) + 4 e−. (0.445 A)(45 min)(60 s/min)(1 C/1 A ∙ s) (1 mol e−/96,500 C)(1 mol O2/4 mol e−) (32.0 g O2/1 mol O2) = 0.10 g O2 (2) The cathode reaction (electrolysis of molten NaCl) is



Na+(melt) + e− n Na(ℓ). (25 × 10 A)(60 min)(60 s/min)(1 C/1 A ∙ s) (1 mol e−/96,500 C)(1 mol Na/mol e−)(23.0 g Na/1 mol Na) = 21,000 g Na = 21 kg



3

Applying Chemical Principles 19.1 Electric Batteries versus Gasoline  −nFE° 1. (a) ΔrG° = = −(1 mol e−)(96,485 C/mol e−)(3.6 V) = −3.47 × 105 C ∙ V ΔrG° = (−3.47 × 105 C ∙ V)(1 J/1 C ∙ V) (1 kJ/1000 J) = −347 kJ = −350 kJ (b) ( 1000 g)(1 mol Li/6.94 g Li)(−347 kJ/mol) = −5.00 × 104 kJ = −5.0 × 104 kJ 2. (15 gallons)(4 qt/1 gallon)(1 L/1.057 qt)(0.70 kg/L)  (46 MJ/kg)(1000 kJ/1 MJ) = 1.83 × 106 kJ = 1.8 × 106 kJ 3. (1.83 × 106 kJ)/(5.00 × 104 kJ/kg Li) = 37 kg Li

5. (a) Cu(OH)2(s) + 2 e− n Cu(s) + 2 OH−(aq) (b) Zn(s) + 2 OH−(aq) n Zn(OH)2(s) + 2 e− (c) Cu(OH)2(s) + Zn(s) n Cu(s) + Zn(OH)2(s) (d)  E = −0.36 V − (−1.245 V) = 0.89 V

Study Questions 19.1

(a) Cr(s) n Cr3+(aq) + 3 e−

Cr is a reducing agent; this is an oxidation reaction. (b) AsH3(g) n As(s) + 3 H+(aq) + 3 e− AsH3 is a reducing agent; this is an oxidation reaction. (c) VO3−(aq) + 6 H+(aq) + 3 e− n  V2+(aq) + 3 H2O(ℓ) VO3−(aq) is an oxidizing agent; this is a reduction reaction. (d) 2 Ag(s) + 2 OH−(aq) n Ag2O(s) + H2O(ℓ) + 2e− Silver is a reducing agent; this is an oxidation reaction. 19.3

(a) Ag(s) n Ag+(aq) + e−

e− + NO3−(aq) + 2 H+(aq) n NO2(g) + H2O(ℓ) Ag(s) + NO3−(aq) + 2 H+(aq) n Ag+(aq) + NO2(g) + H2O(ℓ) (b) 2[MnO4−(aq) + 8 H+(aq) + 5 e− n Mn2+(aq) + 4 H2O(ℓ)]

19.2 Sacrifice!



1. An insulator will prevent the flow of electrons from the zinc to copper, preventing zinc from keeping the copper reduced.

5[HSO3−(aq) + H2O(ℓ) n SO42−(aq) + 3 H+(aq) + 2 e−]

2. (a) tin, (c) iron, (d) nickel, and (e) chromium

2 MnO4−(aq) + H+(aq) + 5 HSO3−(aq) n  2 Mn2+(aq) + 3 H2O(ℓ) + 5 SO42−(aq)

3. (e) chromium



(c) 4[Zn(s) n Zn2+(aq) + 2 e−]

4. (a) Oxidation half-reaction: Cu(s) n Cu2+(aq) + 2 e−

2 NO3−(aq) + 10 H+(aq) + 8 e− n N2O(g) + 5 H2O(ℓ)





Reduction half-reaction: O2(g) + H2O(ℓ) + 2 e− n 2 OH−(aq)

1⁄2

(b) Equation 1: Cu(s) n Cu2+(aq) + 2 e− E° = −0.337 V; ΔrG° = −nFE° = −(2 mol e−) (96,485 C/mol e−)(−0.337 V)(1 J/1 C ∙ V) = 6.503 × 104 J Equation 2: 1⁄2 O (g) + H O(ℓ) + 2 e− n 2 OH−(aq) 2 2 E° = 0.40 V; ΔrG° = −nFE° = −(2 mol e−)(96,485 C/mol e−)(0.40 V)(1 J/1 C ∙ V) = −7.72 × 104 J





4 Zn(s) + 2 NO3−(aq) + 10 H+(aq) n 4 Zn2+(aq) + N2O(g) + 5 H2O(ℓ)

(d) Cr(s) n Cr3+(aq) + 3 e−



3 e− + NO3−(aq) + 4 H+(aq) n NO(g) + 2 H2O(ℓ)



Cr(s) + NO3−(aq) + 4 H+(aq) n Cr3+(aq) + NO(g) + 2 H2O(ℓ)

19.5

(a) 2[Al(s) + 4 OH−(aq) n Al(OH)4−(aq) + 3 e−]



3[2 H2O(ℓ) + 2 e− n H2(g) + 2 OH−(aq)]



2 Al(s) + 2 OH−(aq) + 6 H2O(ℓ) n 2 Al(OH)4−(aq) + 3 H2(g)

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions A-103

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(b) 2[CrO42−(aq) + 4 H2O(ℓ) + 3 e− n  Cr(OH)3(s) + 5 OH−(aq)]



3[SO32−(aq) + 2 OH−(aq) n SO42−(aq) + H2O(ℓ) + 2 e−]



2 CrO42−(aq) + 3 SO32−(aq) + 5 H2O(ℓ) n 2 Cr(OH)3(s) + 3 SO42−(aq) + 4 OH−(aq)



(c) Zn(s) + 4 OH−(aq) n [Zn(OH)4]2−(aq) + 2 e−



Cu(OH)2(s) + 2 e− n Cu(s) + 2 OH−(aq)



Zn(s) + 2 OH−(aq) + Cu(OH)2(s) n [Zn(OH)4]2−(aq) + Cu(s)



(d) 3[HS−(aq) + OH−(aq) n S(s) + H2O(ℓ) + 2 e−]



ClO3−(aq) + 3 H2O(ℓ) + 6 e− n Cl−(aq) + 6 OH−(aq)



3 HS−(aq) + ClO3−(aq) n 3 S(s) + Cl−(aq) + 3 OH−(aq)

19.7

Electrons flow from the Cr electrode to the Fe electrode. Negative ions move via the salt bridge from the Fe|Fe2+ half-cell to the Cr|Cr3+ half-cell (and positive ions move in the opposite direction).



Anode (oxidation): Cr(s) n Cr3+(aq) + 3 e−



Cathode (reduction): Fe2+(aq) + 2 e− n Fe(s)

19.9

(a) Oxidation: Fe(s) n Fe2+(aq) + 2 e− Reduction: O2(g) + 4 H+(aq) + 4 e− n 2 H2O(ℓ) Overall: 2 Fe(s) + O2(g) + 4 H+(aq) n  2 Fe2+(aq) + 2 H2O(ℓ) (b) Anode, oxidation: Fe(s) n Fe2+(aq) + 2 e− Cathode, reduction: O2(g) + 4 H+(aq) + 4 e− n  2 H2O(ℓ) (c) Electrons flow from the negative anode (Fe) to the positive cathode (site of the O2 half-reaction). Negative ions move through the salt bridge from the cathode compartment in which the O2 reduction occurs to the anode compartment in which Fe oxidation occurs (and positive ions move in the opposite direction).



19.11 (a) Reduction of Fe3+ ions by copper.

Reduction: Fe3+(aq) + e− n Fe2+(aq)



Oxidation: Cu(s) n Cu2+(aq) + 2 e−

Overall: 2 Fe3+(aq) + Cu(s) n  2 Fe2+(aq) + Cu2+(aq) (b) Reduction of Fe3+ ions by lead

Reduction: Fe (aq) + e n Fe (aq)



Oxidation: Pb(s) + SO42−(aq) n PbSO4(s) + 2 e−



Overall: Pb(s) + SO42−(aq) + 2 Fe3+(aq) n PbSO4(s) + 2 Fe2+(aq)

3+

−

2+

19.13 Cu(s) ∙ Cu2+(aq) ∙ Cl−(aq) ∙ Cl2(g) ∙ Pt 19.15 (a) All are primary batteries, not rechargeable. (b) Dry cells and alkaline batteries have Zn anodes and are primary batteries. Ni-Cd batteries have a cadmium anode and are rechargeable. (c) Dry cells have an acidic environment, whereas the environment is alkaline for alkaline and Ni-Cd cells.

19.17 (a) E°cell = −1.298 V; not product-favored at equilibrium (b) E°cell = −0.51 V; not product-favored at equilibrium (c) E°cell = −1.02 V; not product-favored at equilibrium (d) E°cell = +0.028 V; product-favored at equilibrium 19.19 (a) Sn2+(aq) + 2 Ag(s) n Sn(s) + 2 Ag+(aq) E°cell = −0.94 V; not product-favored at equilibrium (b) 3 Sn4+(aq) + 2 Al(s) n 3 Sn2+(aq) + 2 Al3+(aq) E°cell = +1.81 V; product-favored at equilibrium (c) 2 ClO3−(aq) + 10 Ce3+(aq) + 12 H+(aq) n  Cl2(aq) + 10 Ce4+(aq) + 6 H2O(ℓ) E°cell = −0.14 V; not product-favored at equilibrium (d) 3 Cu(s) + 2 NO3−(aq) + 8 H+(aq) n  3 Cu2+(aq) + 2 NO(g) + 4 H2O(ℓ)

E°cell = +0.62 V; product-favored at equilibrium

19.21 (a) Al (b) Zn and Al (c) Fe2+(aq) + Sn(s) n Fe(s) + Sn2+(aq); reactant-favored at equilibrium (d) Zn2+(aq) + Sn(s) n Zn(s) + Sn2+(aq); reactant-favored at equilibrium 19.23 Best reducing agent, (e) Cr(s) (Use Appendix M.) 19.25 (d) Ag+ 19.27 See Example 19.5 (a) F2, most readily reduced (b) F2 and Cl2 19.29 E°cell = +0.3923 V. When [Zn(OH)42−] = [OH−] = 0.025 M and P(H2) = 1.0 bar, Ecell = 0.34 V. 19.31 E°cell = +1.562 V and Ecell = +1.59 V 19.33 E°cell = +1.563 V. When Ecell = 1.48 V, n = 2, and [Zn2+] = 1.0 M, the concentration of Ag+ = 0.040 M. 19.35 (a) ΔrG° = −29.0 kJ; K = 1 × 105 (b) ΔrG° = +89 kJ; K = 3 × 10−16 19.37 E°cell for AgBr(s) n Ag+(aq) + Br−(aq) is −0.7281 V.

Ksp = 4.9 × 10−13

19.39 Kformation = 2 × 1025 19.41 See Figure 19.20. Electrons from the battery or other source enter the cathode where they are transferred to Na+ ions, reducing the ions to Na metal. Chloride ions move toward the positively charged anode where an electron is transferred from each Cl− ion, and Cl2 gas is formed. 19.43 O2 from the oxidation of water is more likely than F2. See Example 19.10. 19.45 See Example 19.10. (a) Cathode: 2 H2O(ℓ) + 2 e− n H2(g) + 2 OH−(aq) (b) Anode: 2 Br−(aq) n Br2(ℓ) + 2 e−

A-104 Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

19.47 Mass of Ni = 0.0334 g 19.49 Time = 2300 s or 38 min 19.51 Charge = 5.9 × 105 C



19.53 (b) Only Mg, among the four choices. 19.55 E°cell = E°cathode − E°anode = 0.40 V − (−0.44 V) = 0.84 V. A positive E°cell indicates that the reaction is product-favored at equilibrium. Decreasing the pH decreases [OH−]. Because OH− is one of the products of the net reaction, decreasing its concentration will result in more product favorability. 19.57 (a) UO2+(aq) + 4 H+(aq) + e− n  U4+(aq) + 2 H2O(ℓ) (b) ClO3−(aq) + 6 H+(aq) + 6 e− n  Cl−(aq) + 3 H2O(ℓ) − (c) N2H4(aq) + 4 OH (aq) n  N2(g) + 4 H2O(ℓ) + 4 e− (d) ClO−(aq) + H2O(ℓ) + 2 e− n  Cl−(aq) + 2 OH−(aq) 19.59 (a, c)  The electrode at the right is a magnesium anode. (Magnesium metal supplies electrons and is oxidized to Mg2+ ions.) Electrons pass through the wire to the silver cathode, where Ag+ ions are reduced to silver metal. Nitrate ions move via the salt bridge from the AgNO3 solution to the Mg(NO3)2 solution (and Na+ ions move in the opposite direction). A salt bridge is needed to maintain electrical neutrality in each half-cell and to complete the electrical circuit. (b) Anode: Mg(s) n Mg2+(aq) + 2 e−

Cathode: Ag+(aq) + e− n Ag(s)

Net reaction: Mg(s) + 2 Ag+(aq) n Mg2+(aq) + 2 Ag(s) 19.61 (a) For 1.7 V: Use chromium as the anode to reduce Ag+(aq) to Ag(s) at the cathode. The cell potential is +1.71 V. (b) For 0.5 V, several choices are possible: (i) One possibility involving silver as the cathode is to oxidize Cu(s) to Cu2+(aq) at the anode (E° = 0.46 V). (ii) Another possibility involving silver as the cathode is to oxidize Hg(ℓ) to Hg2Cl2(s) at the anode (E° = 0.53 V). (iii) One possibility involving silver as the anode is to reduce N2H5+(aq) to NH4+(aq) at the cathode (E° = 0.44 V). (iv) Another possibility involving silver as the anode is to reduce Cr2O72−(aq) at to Cr3+(aq) at the cathode (E° = 0.53 V). 19.63

(a) Zn2+(aq) (b) Au+(aq) (c) Zn(s) (d) Au(s) (e) Yes, Sn(s) will reduce Cu2+. (f) No, Ag(s) can only reduce Au+(aq). (g) Cu2+, Ag+, and Au+ (h) Ag+(aq) can oxidize Cu, Sn, Co, and Zn.

19.65 (a) The cathode is the site of reduction, so the halfreaction must be 2 H+(aq) + 2 e− n H2(g).

This is the case with the following half-reactions: Cr3+(aq) n Cr(s), Fe2+(aq) n Fe(s), and Mg2+(aq) n Mg(s). (b) Choosing from the half-cells in part (a), the reaction of Mg(s) and H+(aq) would produce the most positive potential (2.37 V), and the reaction of H2 with Cu2+ would produce the least positive potential (+0.337 V).

19.67 8.1 × 105 g Al 19.69 (a) E°anode = −0.268 V (b) Ksp = 2 × 10−5 19.71 ΔrG° = −409 kJ 19.73 6700 kWh; 820 kg Na; 1300 kg Cl2 19.75 Ru2+, Ru(NO3)2 19.77 9.5 × 106 g Cl2 per day 19.79 Anode: 2 H2O(ℓ) n O2(g) + 4 H+(aq) + 4 e− Cathode: Cu2+(aq) + 2 e− n Cu(s) 19.81 The rate of the reaction with H2O is more rapid than the rate of the reaction with NO3−.

Products formed at anode: O2(g) and H+(aq)



Anode half-reaction: 2 H2O(ℓ) n  O2(g) + 4 H+(aq) + 4 e−

19.83 Under standard conditions, the spontaneous reaction is Hg2+(aq) + 2 Fe2+(aq) n Hg(ℓ) + 2 Fe3+(aq) with E° = 0.084 V.

Under the conditions present, E = −0.244 V. The reaction is not spontaneous in the same direction as under standard conditions but in the opposite direction: Hg(ℓ) + 2 Fe3+(aq) n Hg2+(aq) + 2 Fe2+(aq). Under these conditions, the anode for the spontaneous reaction is the Hg(ℓ) ∙ Hg2+(aq, 0.020 M) electrode and the measured voltage will be 0.244 V.

19.85 E° = 0.771 V. When [H+] = 1.0 × 10−7 M, E = 1.185 V. The reaction is more favorable at a lower [H+] (higher pH). 19.87 (a) See Figure 19.5. In this case, both electrodes are made of silver metal. On one side, the solution contains 1.0 × 10−5 M Ag+ and on the other the solution contains 1.0 M Ag+. The cathode is the electrode on the side that has the 1.0 M Ag+ solution [half-reaction: Ag+(aq) + e− n Ag(s)], and the anode is the electrode on the side that has the 1.0 × 10−5 M Ag+ solution [half-reaction: Ag(s) n Ag+(aq) + e−]. Connecting the two electrodes is a wire. Electrons flow through the wire from the anode to the cathode. A salt bridge also connects the two compartments. (b) E = 0.30 V 19.89 (a) K = 3.4 × 10−10; reactant-favored at equilibrium (b) K = 3.0; product-favored at equilibrium 19.91 (a) ΔrG° = −3.6 × 102 kJ/mol-rxn (b) ΔrG° = −79 kJ/mol-rxn

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions A-105

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19.93 (a) 2[Ag+(aq) + e− n Ag(s)]





C6H5CHO(aq) + H2O(ℓ) n C6H5CO2H(aq) + 2 H+(aq) + 2 e−



2 Ag+(aq) + C6H5CHO(aq) + H2O(ℓ) n C6H5CO2H(aq) + 2 H+(aq) + 2 Ag(s)



(b) 3[CH3CH2OH(aq) + H2O(ℓ) n  CH3CO2H(aq) + 4 H+(aq) + 4 e−]



2[Cr2O72−(aq) + 14 H+(aq) + 6 e− n 2 Cr3+(aq) + 7 H2O(ℓ)]

3 CH3CH2OH(aq) + 2 Cr2O72−(aq) + 16 H+(aq) n  3 CH3CO2H(aq) + 4 Cr3+(aq) + 11 H2O(ℓ) 19.95 (a) 0.974 kJ/g (b) 0.60 kJ/g (c) The silver-zinc battery produces more energy per gram of reactants. 19.97 (a) 2 NO3−(aq) + 3 Mn2+(aq) + 2 H2O(ℓ) n  2 NO(g) + 3 MnO2(s) + 4 H+(aq) 3 MnO2(s) + 4 H+(aq) + 2 NH4+(aq) n  N2(g) + 3 Mn2+(aq) + 6 H2O(ℓ) (b) E° for reduction of NO3− with Mn2+ is −0.27 V. E° for oxidation of NH4+ with MnO2 is +1.50 V. 19.99 19.101

(a) Fe2+(aq) + 2 e− n Fe(s) 2[Fe2+(aq) n Fe3+(aq) + e−] 3 Fe2+(aq) n Fe(s) + 2 Fe3+(aq) (b) E°cell = −1.21 V; not product-favored at equilibrium (c) K = 1 × 10−41

wire e− Cd −

+ Ni

salt bridge

Cd2+(aq)

Ni2+(aq)

Anode



19.107 0.054 g Au 19.109 I− is the strongest reducing agent of the three halide ions. Iodide ion reduces Cu2+ to Cu+, forming insoluble CuI(s).

Cathode

(b) Anode: Cd(s) n Cd (aq) + 2 e Cathode: Ni2+(aq) + 2 e− n Ni(s) Net: Cd(s) + Ni2+(aq) n Cd2+(aq) + Ni(s) (c) The anode is negative and the cathode is positive. (d) E°cell = E°cathode − E°anode = (−0.25 V) − (−0.40 V) = +0.15 V (e) Electrons flow from anode (Cd) to cathode (Ni). 2+

−

2 Cu2+(aq) + 4 I−(aq) n 2 CuI(s) + I2(aq)



19.111 (a) 92 g HF required; 230 g CF3SO2F and 9.3 g H2 isolated (b) H2 is produced at the cathode. (c) 48 kWh 19.113 290 h 19.115

(a) 3.6 mol glucose and 22 mol O2 (b) 86 mol electrons (c) 96 amps (d) 96 watts

Check Your Understanding

19.105 (a) 





Chapter 20

19.103 No, both Na and K would react vigorously and rapidly with seawater.





20.1

(a) Emission of six α particles leads to a decrease of 24 in the mass number and a decrease of 12 in the atomic number. Emission of four β particles increases the atomic number by 4 but doesn’t affect the mass. The final product of this process has a mass number of 232 − 24 = 208 and an atomic number of 90 − 12 + 4 = 82, identifying it as 20882Pb.



(b) Step 1:

(a) +4 in CoO2 and +3 in LiCoO2 (b) Cathode reaction: CoO2(s) + Li+(solv) + e− n LiCoO2(s) Anode reaction: Li(on carbon) n Li+(solv) + e− (c) No, because Li, an alkali metal, would react ­directly with water (giving H2 and LiOH).

NO3− Na+

(f) Positive ions move from the anode compartment to the cathode compartment. Anions move in the opposite direction. (g) K = 1 × 105 (h) Ecell = 0.21 V; yes, the net reaction is still the same. (i) 480 h



Step 2:



Step 3:

232 90Th 228 88Ra 228 89Ac

n n n

228 4 88Ra + 2α 228 0 89Ac + −1β 228 0 90Th + −1β

20.2

22 0 (a) +01β  (b)  4119K  (c)  −1β  (d)  10Ne

20.3



32 0 15P + −1β 45 45 (b) 22Ti n 21Sc + +01β or 4252Ti + −01e (c) 23994Pu n 42α + 23952U (d) 4129K n 4220Ca + −01β

20.4

Δm = 0.034345 g/mol



ΔE =  (3.4345 × 10−5 kg/mol)(2.998 × 108 m/s)2 = 3.0869 × 1012 J/mol (= 3.0869 × 109 kJ/mol)



Eb = 5.145 × 108 kJ/mol nucleons

20.5

(a) 49.2 years is exactly four half-lives; quantity remaining = 1.5 mg(1/2)4 = 0.094 mg (b) Three half-lives, 36.9 years (c) 1% is between six half-lives, 73.8 years (1/64 remains), and seven half-lives, 86.1 years (1/128 remains). [Using the integrated first-order rate equation with [R]/[R]0 = 0.010 and k = (ln 2)/t1/2 = 0.05635 y−1, the amount of time is calculated to be 81.7 years.]





(a) 3124Si n

n

45 21Sc

A-106 Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

20.6

(a) ln ([A]/[Ao]) = −kt 3

3



ln ([3.18 × 10 ]/[3.35 × 10 ]) = −k(2.00 d)



k = 0.02604 d−1



t1/2 = 0.693/k = 0.693/(0.02604 d−1) = 26.6 d

(b) k = 0.693/t1/2 = 0.693/200. y = 3.465 × 10−3 y−1





ln ([A]/[Ao]) = −kt

Mass Tc = (5.41 × 10−9 mol NaTcO4) (1 mol Tc/mol NaTcO4)(97.9 g Tc/mol) = 5.3 × 10−7 g Tc 4. 24 hours is 4 half-lives for 99mTc. Thus, mass remaining is 1/16th of 1.0 μg, or 0.063 μg. 5. Beta particle



ln (4.615 × 10−10) = −(3.465 × 10−3 y−1)t

6. The attractive forces binding an anion to the column are related to the anion charge. The anion MoO42− has a higher charge than TcO4−.



t = 6.20 × 103 y

20.3 The Age of Meteorites

ln ([3.00 × 103]/[6.50 × 1012]) =  −(3.465 × 10−3 y−1)t

20.7

ln ([A]/[Ao]) = −kt

1. −4

−1

y )t



ln ([9.32]/[13.4]) = −(1.21 × 10



t = 3.00 × 103 y



This compares quite well with the estimated date.

20.8

98 42Mo



99 42Mo

20.9

4.17 × 10−5(0.0100 g Pb2+) = 4.17 × 10−7 g Pb2+



Solubility = [4.17 × 10−7 g Pb2+ (1 mol Pb2+/ 207.2 g Pb)(1 mol PbCrO4/1 mol Pb2+)]/ 0.01000 L = 2.01 × 10−7 mol PbCrO4/L

+ 10n n n

99 43 Tc

99 42Mo

+

n

87 38Sr

+

0 −1β

2. (b) Beta emission

+γ

3. k = ln (2)/t1/2 = 0.6931/(4.965 × 1010 y) = 1.3961 × 10−11 y−1 = 1.396 × 10−11 y−1 4. ln([87Rb]t/[87Rb]0) = −(1.3961 × 10−11 y−1)(4.5 × 109 y)

0 −1β

Applying Chemical Principles 20.1 A Primordial Nuclear Reactor 1. In a 100-g sample today, there are 0.720 g of 99.274 g of 238U.

87 37Rb

235

U and

For 235U, k = ln (2)/(7.038 × 108 y) = 9.8486 × 10−10 y−1 ln (0.720) = −(9.8486 × 10−10 y−1)(2.0 × 109 y) + ln(x) x = 5.2 g

[87Rb]t/[87Rb]0 = 0.939 The fraction of 87Rb that remains is 0.939, so the fraction that has decayed is 0.061. 5. A graph of [87Sr/86Sr] versus [87Rb/86Sr] was constructed using Microsoft Excel. The resulting equation of the line was y = 0.0406x + 0.785. ekt − 1 = 0.0406 ekt = 1.0406 kt = 0.039797 t = 0.039797/(1.420 × 10−11 y−1) = 2.80 × 109 y 6. Begin with the integrated rate law for the first-order decomposition of 87Rb. ln([87Rb]/[87Rb]0) = −kt This can be rearranged to give an equation for [87Rb]0, [87Rb]0 = [87Rb]ekt

For U, k = ln (2)/(4.468 × 109 y) = 1.5514 × 10−10 y−1 ln (99.274) = −(1.5514 × 10−10 y−1)(2.0 × 109 y) + ln(y) y = 135 g

The [87Sr] equals its initial concentration plus the concentration created by the decomposition of 87Rb: [87Sr] = [87Sr]0 + ([87Rb]0 − [87Rb])

Percent abundance of 235U = 5.2 g/(5.2 g + 135 g) × 100% = 3.7%

This can also be solved for [87Rb]0: [87Rb]0 = [87Sr] + [87Rb] − [87Sr]0

2. The ratio of rates will be given by the ratio of the two rate constants.

Setting the two equations for [87Rb]0 equal to each other, we obtain [87Sr] + [87Rb] − [87Sr]0 = [87Rb]ekt [87Sr] = [87Rb]ekt − [87Rb] + [87Sr]0 [87Sr] = [87Rb](ekt − 1) + [87Sr]0

238

k for 235U/k for 238U = (9.8486 × 10−10 y−1)/(1.5514 × 10−10 y−1 = 6.348 so 235U decomposes 6.348 times faster than 238U. 3.

235 9 2U

n

231 90Th

+ 42α

4. (0.00720)(235.0409) + (0.99274)(238.0508) = 238.01 20.2 Technetium-99m and Medical Imaging 1.

99 42Mo

n

99 43 Tc

+

0 −1β

2. Oxidation number = 7 Tc7+ electron configuration = [Kr] Diamagnetic 3. Amount NaTcO4 = (1.0 μg)(1 g/106 μg)(1 mol/184.9 g) = 5.41 × 10−9 mol = 5.4 × 10−9 mol

Finally, dividing both sides of the equation by [86Sr] gives the desired equation: [87Sr]/[86Sr] = ([87Rb]/[86Sr])(ekt − 1) + [87Sr]0/[86Sr]

Study Questions 20.1

(a) Increasing mass: γ < β < α (b) Increasing penetrating power: α < β < γ

20.3

Binding energy per nucleon is determined by first calculating the mass defect for an isotope, converting the mass defect to binding energy (using E = mc2), and then dividing the binding energy by the mass number of the nucleus.

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions A-107

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20.5



20.7

20.9

An element of fairly high atomic number is bombarded with a beam of high-energy particles. Initially, neutrons were used; later, helium nuclei and then larger nuclei such as 11B and 12C were used; and, more recently, highly charged ions of elements such as calcium, chromium, cobalt, and zinc have been chosen. The bombarding particle fuses with the nucleus of the target atom, forming a new nucleus that lasts for a short time before decomposing.



Neutrons are effective when used as a bombarding particle because they have no charge. They are therefore not repelled by the positively charged nuclear particles.



While an organism lives, the percentage of carbon that is 14C in the organism will equal the percentage in the atmosphere. When the organism dies, the 14C is no longer replenished. By measuring the activity of 14C in the organism, comparing it to the ambient 14C activity, and using first-order kinetics, it is possible to determine how long ago the organism stopped taking in 14 C (died). Limitations: (1) The method assumes that the amount of 14C in the atmosphere has remained constant, whereas it has varied by as much as 10%. (2) 14C dating cannot be used to date an object less than 100 or more than about 40,000 years old. (3) The accuracy of 14C dating is about ±100 years. 8 α and 6 β

20.11 The initiation step involves the addition of a neutron to 235U and its subsequent fission. The fission produces neutrons that react with further 235U atoms in the propagation step. The chain reaction is terminated when either there is no 235U remaining or when the neutrons escape without reacting or when they are absorbed by a moderator. 20.13 A moderator is present in a nuclear reactor to control the rate of a fission reaction and are moved in and out of the reactor to slow or speed up the reaction. 20.15 (a) Curie: radioactive decomposition per second (1 curie = 3.7 × 1010 disintegrations per second) (b) Rad: energy absorbed per mass of tissue (1 rad = 0.01 J per kilogram of tissue) 20.17

14 7N 15 8O 0 +1β

+ 21H n 158O + 10n n 157N + +01β + −01e n 2 γ

20.19 (a) 5268Ni; (b) 10n; (c) 3125P; (d) 9473 Tc; (e) −01β; 0 1e

(positron)



219 4 223 88Ra n 86Rn + 2α 215 4 219 86Rn n 84Po + 2α 211 4 215 84Po n 82Pb + 2α 211 0 211 82Pb n 83Bi + −1β 211 0 211 83Bi n 84Po + −1β 207 4 211 84Po n 82Pb + 2α 198 0 80Hg + −1β (b) 22826Rn n 21884Po + 42α (c) 13575Cs n 13576Ba + −01β (d) 11409In n 11408Cd + 01e

20.25 (a) 19789Au n

20.27 (a) 8305Br has a high neutron/proton ratio of 45/35. Beta decay will allow the ratio to decrease:

80 35Br

+ 42α

(c) Cobalt-61 has a high n/p ratio, so beta decay is likely:



61 27Co

n

+

61 28Ni

n

0 −1β

(d) Carbon-11 has only five neutrons, so K-capture or positron emission may occur:



11 6C 11 6C

+

0 −1e

n

n

11 5B

+

11 5B 0 e 1

20.29 Generally beta decay will occur when the n/p ratio is high, whereas positron emission will occur when the n/p ratio is low.

(a) Beta decay:



3 1H

n

3 2He

20 9F

+

n

20 10Ne

+

0 −1β

0 −1β

(b) Positron emission



22 11Na

n

22 10Ne

+ 01β

20.31 Binding energy per mole of nucleons for 11 B = 6.70 × 108 kJ

Binding energy per mole of nucleons for 10 B = 6.25 × 108 kJ

20.33 8.253 × 108 kJ/mol nucleons. This value is consistent with the value shown in Figure 20.4. 20.35 7.700 × 108 kJ/mol nucleons 20.37 (a) 32He + 42He n 74Be + γ

(b)  74Be +

0 −1e

n

7 3Li

20.39 0.781 micrograms 20.41 (a) 13513I n

131 54Xe

20.21

20.43 9.5 × 10−4 mg

20.23

235 231 4 92U n 90 Th + 2α 231 0 231 90 Th n 91Pa + −1β 227 4 231 91Pa n 89Ac + 2α 227 0 227 89Ac n 90 Th + −1β 223 4 227 90 Th n 88Ra + 2α

20.45 (a) 22826Rn n



236 96Cm



(a) −01β; (b) 8377Rb; (c) 42α; (d) 22868Ra; (e) −01β; (f) 2141Na



240 98Cf

(b) Alpha decay is likely:

(f)



0 −1β.







+

80 36Kr

n





+

0 −1β

(b) 0.075 micrograms 218 84Po

+ 42α

(b) Time = 8.87 d

20.47 (a) 15.8 y; (b) 88% 20.49

239 94Pu

+ 42α n

240 95Am

+ 11H + 2 10n

A-108 Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

20.51

48 20Ca

+

242 94Pu

n

Chapter 21

+ 3 10n

287 114Fl

20.53 (a) 11458Cd;   (b) 74Be;   (c) 42α;   (d) 6239Cu 20.55

10 5B

+ 10n n 73Li + 42α

20.57 Time = 4.6 × 1010 y 20.59 If t1/2 = 14.28 d, then k = 4.854 × 10−2 d−1. If the original disintegration rate is 3.2 × 106 dpm, then (from the integrated first order rate equation), the rate after 365 d is 0.065 dpm. The plot will resemble Figure 20.6. 20.61 (a) 23982U + 10n n

239 92U

+γ

239 0 93Np + −1β 239 239 0 93Np n 94Pu + −1β 239 1 1 94Pu + 0n n 2 0n +



(b) 23992U n



(c)



(d)

energy + other nuclei

20.63 (a) Calcium-48. Possible reaction is 248 96Cm

+

48 20Ca







(b) Three neutrons

n

296 116Lv

Check Your Understanding 21.1

(a) 2 Na(s) + Br2(ℓ) n 2 NaBr(s) (b) Ca(s) + Se(s) n CaSe(s) (c) 2 Pb(s) + O2(g) n 2 PbO(s) (d) 2 Al(s) + 3 Cl2(g) n 2 AlCl3(s)

21.2

(a) H2Te (b) Na3AsO4 (c) SeCl6 (d) HBrO4

21.3

(a) NH4+ (ammonium ion) (b) O22− (peroxide ion) (c) N2H4 (hydrazine) (d) NF3 (nitrogen trifluoride)

21.4

In Na2Cl, chlorine would have the unlikely charge of 2− (to balance the two positive charges of the two Na+ ions). The formula CaCH3CO2 would require either the calcium ion to have the formula Ca+ or the acetate ion to have the formula CH3CO22−. In all of its compounds, calcium occurs as the Ca2+ ion. The acetate ion, formed from acetic acid by loss of H+, has a 1− charge. In Mg2O, the magnesium ions would need to have the incorrect charge of 1+ to balance charge of the O2− ion or else the oxygen would need to have the incorrect charge of 4− to balance the charge of the two Mg2+ ions. Neither of these possibilities is acceptable.



20.65 About 2700 years old 20.67 Plot ln (activity) versus time. The slope of the plot is −k, the rate constant for decay. Here, k = 0.0050 d−1, so t1/2 = 140 d.



20.69 Time = 1.9 × 109 y 20.71 Energy obtained from 1.000 lb (453.6 g) of 235 U = 4.05 × 1010 kJ

Mass of coal required = 1.6 × 103 ton (or about 3 million pounds of coal)

20.73 The percentage of tagged fish in the lake is (27/5250)100% = 0.51%. Therefore, 1000 tagged fish is 0.51% of the total or approximately 190,000 fish. 20.75 (a) 20893Bi + 10n n



210 83Bi n 210 84Po n

210 83Bi 210 0 −1e + 84Po 4 206 2α + 82Pb –1



235

U decay series.

(c) Pa-231 is present to the extent of 1 part per million. Therefore, 1 million grams of pitchblende need to be used to obtain 1 g of Pa-231. (d) 23911Pa n

227 89Ac

+ 42α

20.79 Pitchblende contains 23982U and 23952U. Thus, both radium and polonium isotopes must belong to either the 4n + 2 or 4n + 3 decay series. Furthermore, the isotopes must have sufficiently long half-lives in order to survive the separation and isolation process. These criteria are satisfied by 226Ra and 210Po.



1. 50. ppb is 50. g in 1 × 109 g of blood. Assume the density of blood is 1.0 g/mL. In 1.0 × 103 mL (i.e., 1.0 L) of blood, there will be 50. × 10−6 g of Pb. From this:

2. (750 mL wine)(1.0 g wine/mL wine) (2000 g Pb/1,000,000 g wine) = 1.5 g Pb

(b) 23952U n 23910Th + 42α 231 231 0 90Th n 91Pa + −1β

21.1 Lead in the Environment

(50. × 10−6 g) (1 mol Pb/207.2 g Pb) (6.022 × 1023 atoms Pb/mol Pb) = 1.5 × 1017 atoms Pb

(b) k = 0.00502 day Amount remaining = 9 μg

20.77 (a) 231Pa isotope belongs to the

Applying Chemical Principles

3. (0.15 g paint)(12 g Pb/100 g paint) (1 mol Pb/207.2 g Pb)/4.7 L = 1.8 × 10−5 M 21.2 Hydrogen Storage 1.

H H H

N

B

H

H H

Formal charges: N, 1+; B, 1−; H, 0. 2. Percent H = (6.047 g/30.865 g) × 100% = 19.59% Mass H in 1 kg = (1.00 × 103 g)(0.1959) = 196 g 3. (1.00 L)(1000 cm3/1 L)(0.780 g/cm3) = 780. g Mass hydrogen in 780. g of ammonia borane = (780. g)(0.1959) = 152.8 g; therefore the hydrogen density = 152.8 g/L = 153 g/L 4. (140 g/152.8 g) × 100% = 92%

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions A-109

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

5.

H

H H H

N B

B N

N B

H

H

H

H

N B

H

B N

H N B

H

H

H

H

N

H

The compound is isoelectronic with benzene.

B

B N

N B

H H

H

Study Questions

The alkali metal halides are white, crystalline solids. They have high melting and boiling points, and they are soluble in water.

21.31 (a) 2 NaCl(aq) + 2 H2O(ℓ) n  Cl2(g) + H2(g) + 2 NaOH(aq) (b) If this were the only process used to produce chlorine, the mass of Cl2 reported for industrial production would be 0.88 times the mass of NaOH produced (2 mol NaCl, 117 g, would yield 2 mol NaOH, 80 g, and 1 mol Cl2, 70 g). The amounts quoted indicate a Cl2 to NaOH mass ratio of 0.96. Chlorine is presumably also prepared by other routes than this one.

21.1

(d) Ca2O3

21.3

(b) +3

21.5

4 Li(s) + O2(g) n 2 Li2O(s)



Li2O(s) + H2O(ℓ) n 2 LiOH(aq)



2 Ca(s) + O2(g) n 2 CaO(s)



CaO(s) + H2O(ℓ) n Ca(OH)2(s)

21.7

These are the elements of Group 3A(13): boron, B; aluminum, Al; gallium, Ga; indium, In; and thallium, Tl.

21.33 (c) Gypsum, CaSO4 ∙ 2 H2O

21.9

2 Na(s) + Cl2(g) n 2 NaCl(s)

21.35 2 Mg(s) + O2(g) n 2 MgO(s)



The reaction is exothermic and the product is ionic (see Figure 1.2).



21.11 The product, NaCl, is a colorless solid and is soluble in water. Other alkali metal chlorides have similar properties. 21.13 Calcium will not exist in the Earth’s crust because the metal reacts with water to form Ca(OH)2 and oxidizes to CaO in air. 21.15 Increasing basicity: CO2 < SiO2 < SnO2 21.17

(a) 2 Na(s) + Br2(ℓ) n 2 NaBr(s) (b) 2 Mg(s) + O2(g) n 2 MgO(s) (c) 2 Al(s) + 3 F2(g) n 2 AlF3(s) (d) C(s) + O2(g) n CO2(g)

21.37 CaCO3 is used in agriculture to neutralize acidic soil, to prepare CaO for use in mortar, and in steel production.

21.41 (c) Third



3 H2(g) + N2(g) n 2 NH3(g)

21.23 CH4(g) + H2O(g) n CO(g) + 3 H2(g)

ΔrH° = +206.2 kJ/mol-rxn; ΔrS° = +214.7 J/K ∙ mol-rxn; ΔrG° = +142.2 kJ/mol-rxn (at 298 K).

21.25 Step 1: 2 SO2(g) + 4 H2O(ℓ) + 2 I2(s) n  2 H2SO4(ℓ) + 4 HI(g)

−

21.43

O O −

O

B

B

−

O

O

O

B

−

O

−

O

−

O

−

B O B O B2O54−

B3O63−

21.21 2 H2(g) + O2(g) n 2 H2O(g) H2(g) + Cl2(g) n 2 HCl(g)

CaCO3(s) + H2O(ℓ) + CO2(g) n  Ca2+(aq) + 2 HCO3−(aq)

21.39 1.4 × 106 g SO2

21.19 (a) Neon

3 Mg(s) + N2(g) n Mg3N2(s)

21.45 (a) 2 B5H9(g) + 12 O2(g) n 5 B2O3(s) + 9 H2O(ℓ) (b) Enthalpy of combustion of B5H9 = −4539.2 kJ/mol. This is more than double the enthalpy of combustion of B2H6. (c) Enthalpy of combustion of C2H6(g) [to give CO2(g) and H2O(ℓ)] = −1560.7 kJ/mol. C2H6 produces 51.9 kJ/g, whereas diborane produces much more (78.4 kJ/g).



Step 2: 2 H2SO4(ℓ) n 2 H2O(ℓ) + 2 SO2(g) + O2(g)



Step 3: 4 HI(g) n 2 H2(g) + 2 I2(g)



Net: 2 H2O(ℓ) n 2 H2(g) + O2(g)



2 Al(s) + 3 Cl2(g) n 2 AlCl3(s)

21.27 (b) Has a high melting point (> 400 °C)



4 Al(s) + 3 O2(g) n 2 Al2O3(s)

21.29 2 Na(s) + F2(g) n 2 NaF(s)

21.49 2 Al(s) + 2 OH−(aq) + 6 H2O(ℓ) n  2 Al(OH)4−(aq) + 3 H2(g)



2 Na(s) + Cl2(g) n 2 NaCl(s)



2 Na(s) + Br2(ℓ) n 2 NaBr(s)



2 Na(s) + I2(s) n 2 NaI(s)

21.47 2 Al(s) + 6 HCl(aq) n  2 Al3+(aq) + 6 Cl−(aq) + 3 H2(g)



Volume of H2 obtained from 13.2 g Al = 18.4 L

A-110 Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

21.51 Al2O3(s) + 3 H2SO4(aq) n Al2(SO4)3(s) + 3 H2O(ℓ)

Mass of H2SO4 required = 860. g and mass of Al2O3 required = 298 g.

21.53 (c) SiO2 21.55 Pyroxenes have as their basic structural unit an extended chain of linked SiO4 tetrahedra. The ratio of Si to O is 1∶3. 21.57 This structure has a six-member ring of Si atoms with O atom bridges. Each Si also has two O atoms attached. The basic unit is SiO32−, and the overall charge is −12 in [(SiO3)6]−12. (Electron lone pairs are omitted in the following structure.) −O −O −O

O

O− O−

Si O

Si

O −O

O−

Si

O Si

−O

O

Si O

−O

S S

disulfide ion

21.75 (b) Cl2 21.77 E°cell = E°cathode − E°anode = +1.44 V − (+1.51 V) = −0.07 V

The reaction is not product-favored under standard conditions.

21.79 Cl2(aq) + 2 Br−(aq) n 2 Cl−(aq) + Br2(ℓ)

Cl2 is the oxidizing agent, Br− is the reducing agent; E°cell = 0.28 V.

21.81 The reaction consumes 4.32 × 108 C to produce 8.51 × 104 g F2. 21.83 (c) XeF3+ 21.85 Xe-F bond-dissociation enthalpy = 132 kJ/mol

O−

21.89

O−

21.59 (c) Between 2 and 3

Element

Appearance

State

Na, Mg, Al

Silvery metal

Solids

Si

Black, shiny metalloid

Solid

P

White, red, and black allotropes; nonmetal

Solid

S

Yellow nonmetal

Solid

Cl

Pale green nonmetal

Gas

Ar

Colorless nonmetal

Gas

21.61 Consider the general decomposition reaction: NxOy n x/2 N2 + y/2 O2

For all NxOy molecules ΔrG° = −Δf G°. These data show that the decomposition reaction is spontaneous for all of the nitrogen oxides. All are unstable with respect to decomposition to the elements. Compound −𝚫fG° (kJ/mol) NO(g)

  −86.58

NO2

  −51.23

N 2O

−104.20

N2O4

  −97.73

21.63 ΔrH° = −114.4 kJ/mol-rxn; exothermic ΔrG° = −70.7 kJ/mol-rxn, product-favored at equilibrium 21.65 (a) N2H4(aq) + O2(g) n N2(g) + 2 H2O(ℓ) (b) 1.32 × 103 g 21.67 Diphosphorous acid (H4P2O5) should be a diprotic acid (losing the two H atoms attached to O atoms). O

O

H P

O P

H

O

21.69 (c) +3 21.71 (a) 3.5 × 103 kg SO2 (b) 4.1 × 103 kg Ca(OH)2



H

O H

2−

21.87 0.015 g/L of argon; 2.6 × 103 L for 1.0 mol of argon. O−

Si

21.73

21.91

(a) 2 KClO3(s) n 2 KCl(s) + 3 O2(g) (b) 2 H2S(g) + 3 O2(g) n 2 H2O(g) + 2 SO2(g) (c) 2 Na(s) + O2(g) n Na2O2(s) (d) P4(s) + 3 KOH(aq) + 3 H2O(ℓ) n  PH3(g) + 3 KH2PO2(aq) (e) NH4NO3(s) n N2O(g) + 2 H2O(g) (f) 2 In(s) + 3 Br2(ℓ) n 2 InBr3(s) (g) SnCl4(ℓ) + 2 H2O(ℓ) n SnO2(s) + 4 HCl(aq)

21.93 1.4 × 105 metric tons 21.95 Mg: ΔrG° = +64.9 kJ/mol-rxn Ca: ΔrG° = +131.38 kJ/mol-rxn Ba: ΔrG° = +219.67 kJ/mol-rxn

Relative tendency to decompose: MgCO3 > CaCO3 > BaCO3

21.97 (a) Δf G° should be more negative than (−95.1 kJ) × n. (b) Ba, Pb, Ti 21.99

OOF bond enthalpy = 190 kJ/mol

21.101 (a) N2O4 is the oxidizing agent (N is reduced from +4 to 0 in N2), and H2NN(CH3)2 is the reducing agent. (b) 1.3 × 104 kg N2O4 is required. Product masses: 5.7 × 103 kg N2; 4.9 × 103 kg H2O; 6.0 × 103 kg CO2.

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions A-111

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

21.103 ΔrH° = −257.78 kJ/mol-rxn. This reaction is entropydisfavored, however, with ΔrS° = −963 J/K ∙ mol-rxn because of the decrease in the number of moles of gases. Combining these values gives ΔrG° = +29.19 kJ/mol-rxn, indicating that under standard conditions at 298 K the reaction is not spontaneous. (The reaction has a favorable ΔrG° at temperatures less than 268 K, indicating that further research on this system might be worthwhile. Note that at that temperature water is a solid.) 21.105 A = B2H6; B = B4H10; C = B5H11; D = B5H9; E = B10H14 21.107 x = 1.5 21.109 ΔrG° = −834.28 kJ/mol-rxn, therefore the reaction is product-favored at equilibrium at 298 K. ΔrS° = −149.9 J/K ∙ mol-rxn at 298 K. Because this is negative, the reaction will be less product-favored at high temperatures. 21.111 (a) 2 CH3Cl(g) + Si(s) n (CH3)2SiCl2(ℓ) (b) 0.823 atm (c) 12.2 g 21.113 When A is heated it gives 0.00506 mol of gas and white solid B. When the gas reacts with Ca(OH)2 it gives C, a white solid. This would be the behavior of CO2, and it is likely that C is solid, white CaCO3. When B is placed in water it gives a basic solution, a clue that B is a metal oxide. When B is treated with HCl, the ultimate solid product, 1.055 g of D, gives a green flame. This suggests D is a barium compound, likely BaCl2. To further confirm that D is BaCl2, 1.055 g is equivalent to 0.00507 mol, which is also equal to the amount of CO2 produced on heating 1.00 g of A, BaCO3, to give B, BaO. The identity of B is strengthened by knowing that treating BaO with H2SO4 will give a white precipitate of BaSO4, compound E. A = BaCO3, B = BaO, C = CaCO3, D = BaCl2, and E = BaSO4 21.115 (a) Br2O3 (b) The structure of Br2O is reasonably well known. Several possible structures for Br2O3 can be imagined, but experiment confirms the structure below. bent

BrOOOBr

trigonal pyramid

21.117 (a) The NO bond with a length of 114.2 pm is a double bond. The other two NO bonds (with a length of 121 pm) have a bond order of 1.5 (as there are two resonance structures involving these bonds).

N

N

121 pm

O



(b) K = 1.90; ΔrS° = 141 J/K ∙ mol-rxn (c) Δf H° = 82.9 kJ/mol

21.119 Generally, a sodium fire can be extinguished by smothering it with sand. The worst choice is to use water (which reacts violently with sodium to give H2 gas and NaOH). 21.121 Nitrogen is a relatively unreactive gas, so it will not participate in any reaction typical of hydrogen or oxygen. The most obvious property of H2 is that it burns, so attempting to burn a small sample of the gas would immediately confirm or deny the presence of H2. If O2 is present, it can be detected by allowing it to react as an oxidizing agent. There are many reactions known with low-valent metals, especially transition metal ions in solution, that can be detected by color changes. 21.123 The reducing ability of the Group 3A (13) metals declines considerably on descending the group, with the largest drop occurring on going from Al to Ga. The reducing ability of gallium and indium are similar, but another large change is observed on going to thallium. In fact, thallium is most stable in the +1 oxidation state. This same tendency for elements to be more stable with lower oxidation numbers is seen in Groups 4A (14) (Ge and Pb) and 5A (15) (Bi). 21.125 (a) CH4(g) + 2 H2O(ℓ) n CO2(g) + 4 H2(g)

SiH4(g) + 2 H2O(ℓ) n SiO2(s) + 4 H2(g) (b) Reaction of CH4:



 Δr G° = Δf G°(CO2) − Δf G°(CH4) − 2 Δf G°(H2O) = +130.7 kJ/mol-rxn







 Δr G° = Δf G°(SiO2) − Δf G°(SiH4) − 2 Δf G°(H2O) = −439.51 kJ/mol-rxn



The reaction of silane with water is productfavored at equilibrium. This is an important difference between methane and silane. (c) Electronegativities: C = 2.5, Si = 1.9, H = 2.2. Polarities are in opposite directions: in CH4, Cδ−Hδ+ (H is positive); in SiH4, Siδ+Hδ− (H is negative). (d) Si prefers to form four single bonds, rather than form a double bond to O, similar to what is seen in acetone. We predict that a molecular species Si2H4, analogous to ethene, will not exist; a compound of this formula will be a polymer O(SiH2SiH2)xO



bent

O A BrOOOBrOO

O

O

114.2 pm



Reaction of SiH4:

O H3C

C

CH3 CH3

Acetone

O Si

CH3 O

Si

CH3 O

Si

O

CH3 CH3 CH3 [(CH3)2SiO]n polymer

A-112 Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

21.127 (a) HXeO4− (Xe oxidation number = +6) is both oxidized (to XeO64−, Xe oxidation number = +8) and reduced (to Xe). In addition, O is oxidized to O2. (b) One Xe atom in HXeO4− loses 2 electrons to give the Xe atom in XeO64−. A second Xe atom in HXeO4− gains 6 electrons to give Xe. Thus, 2 HXeO4− ions require a net of 4 electrons. These electrons can be supplied by the oxidation of 2 O2− ions to give O2.

22.6

Chapter 22

1. Preparation and formula of blue: (4.5392 g Y2O3)(1 mol/225.809 g)(2 mol Y/1 mol Y2O3) = 0.0402039 mol Y (5.302 g In2O3)(1 mol/277.64 g)(2 mol In/1 mol In2O3) = 0.038193 mol In (0.1585 g MnO2)(1 mol/86.936 g)(1 mol Mn/1 mol MnO2) = 0.0018232 mol Mn Mol In/mol Y = 0.9500 = x Formula of the compound is YIn0.95Mn0.05

Check Your Understanding 22.1

(a) Co(NH3)3Cl3 (b) Fe(H2NCH2CH2NH2)2Br2 = Fe(en)2Br2

22.2

(a)   (i) K3[Co(NO2)6]: a complex of cobalt(III) with a coordination number of 6 (ii) Mn(NH3)4Cl2: a complex of manganese(II) with a coordination number of 6 (b) NH4[Co(EDTA)]: a complex of cobalt(III) with a coordination number of 6

22.3 22.4

22.5

(a) hexaaquanickel(II) sulfate (b) dicyanobis(ethylenediamine)chromium(III) chloride (c) potassium amminetrichloroplatinate(II) (d) potassium dichlorocuprate(I) (a) Geometric isomers are possible (with the NH3 ligands in cis and trans positions). (b) Only a single structure is possible. (c) Only a single structure is possible. (d) This compound is chiral; there are two optical isomers. (e) Only a single structure is possible. (f) Two structural isomers are possible based on coordination of the NO2− ligand through oxygen or nitrogen. (a) [Ru(H2O)6]2+: an octahedral complex of ruthenium(II) (d 6). A low-spin complex has no unpaired electrons and is diamagnetic. A highspin complex has four unpaired electrons and is paramagnetic. dx 2−y 2 dxy

dxz

dz 2

dx 2−y 2

dyz

dxy

high-spin Ru2+



dyz

low-spin Ru2+

(b) [Ni(NH3)6]2+: an octahedral complex of nickel(II) (d 8). Only one electron configuration is possible; it has two unpaired electrons and is paramagnetic. dx 2−y 2 dxy

dxz

dz 2 dyz

Ni2+ion (d 8)



dxz

dz 2



1. A wavelength of 500 nm corresponds to green light being absorbed. The complex ion will appear magenta. 2. The complex appears yellow because blue light is being absorbed. The high energy of blue light indicates that Δo is large and the complex is therefore low spin.

Applying Chemical Principles 22.1 Blue!

2. (a) Mn2+ has 5 d electrons. One electron is placed in each of the 5 orbitals. (b) The dz2 orbital is highest in energy as it points directly at the oxide ions on the z axis. The dxz and dyz orbitals do not lie in the plane of the three ligands in the trigonal plane and so are less affected and are the lowest energy orbitals. 3. In Prussian blue a CN− ligand bridges an iron(II) and an iron(III) ion. 22.2 C  isplatin: Accidental Discovery of a Chemotherapy Agent 1. k = 0.693/t1/2 = 0.277 hr−1 ln (fraction remain) = −(0.277 hr−1)(24 hr) = −6.65 fraction remain = e−6.65 = 0.0013 Mass remaining = 0.01 mg 2. Cis-diamminedichloroplatinum(II) 3. The d-orbitals split into four groups, in order of increasing energy: dxz = dyz < dz2 < dxy< dx2−y2. The four pairs of electrons occupy the four lowest energy orbitals. 22.3 The Rare Earths 1. Nd: [Xe]4f 46s2; Eu: [Xe]4f  76s2 2. Ce3+: [Xe]4f  1 Nd3+: [Xe]4f  3 3. Fe2+(aq) + [Ce(NO3)6]2−(aq) n  Fe3+(aq) + Ce3+(aq) + 6 NO3−(aq) C = [0.181 g Fe(1 mol Fe/55.85 g Fe)  (1 mol [Ce(NO3)6]2−/mol Fe)]/0.03133 L = 0.103 M 4. 2 CeO2(s) + CO(g) n Ce2O3(s) + CO2(g) 2 Ce2O3(s) + O2(g) n 4 CeO2(s) 5. Perovskite structure. If there are N3− ions in the faces, this accounts for a total of 9− in ion charges. The lanthanide ions at the 8 corners must be La3+ ions (for a total charge of +3 for the unit cell). The tungsten ion in the center of the cell is a 6+ ion. This structure is analogous to the one for the perovskite in Example 12.2, page 591, where La3+, W6+, and N3− ions are used in place of Ca2+, Ti4+, and O2−, respectively.

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions A-113

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Study Questions 22.1



(b) H3N

Curium (Cm) and californium (Cf)

22.3

Rhodium (Rh) and ruthenium (Ru)

22.5

Chemical property: (a) Physical property: (b), (c), (d), (e)

H3N

22.9

3+

(a) Cr : [Ar]3d  , paramagnetic (b) V2+: [Ar]3d 3, paramagnetic (c) Ni2+: [Ar]3d 8, paramagnetic (d) Cu+: [Ar]3d 10, diamagnetic

22.11

(a) Fe3+: [Ar]3d 5, isoelectronic with Mn2+ (b) Zn2+: [Ar]3d 10, isoelectronic with Cu+ (c) Fe2+: [Ar]3d 6, isoelectronic with Co3+ (d) Cr3+: [Ar]3d 3, isoelectronic with V2+

3

22.17 C and CO 22.19 Monodentate: CH3NH2, CH3CN, N3−, Br−

Bidentate: en, phen (see Figure 22.12)

22.21

(a) H2NCH2CH2NH2 (b) C2O42− (c) ammonia (d) thiocyanate ion

22.37

22.27

(a) Ni(en)2Cl2 (en = H2NCH2CH2NH2) (b) K2[PtCl4] (c) K[Cu(CN)2] (d) [Fe(NH3)4(H2O)2]2+

22.29

(a) Diaquabis(oxalato)nickelate(II) ion (b) Dibromobis(ethylenediamine)cobalt(III) ion (c) Amminechlorobis(ethylenediamine)cobalt(III) ion (d) Diammineoxalatoplatinum(II)

22.31

(a) [Fe(H2O)5OH]2+ (b) Potassium tetracyanonickelate(II) (c) Potassium diaquabis(oxalato)chromate(III) (d) (NH4)2[PtCl4]

22.33 Cis and trans geometric isomers. The cis isomer is chiral giving rise to two optical isomer.

H3N

NH3 cis

NH3 Co NO2

(d)

N

Cl Co

SCN

NO2

O2N

NO2

H3N

NH3 Co NO2

NH3 NO2

mer −

Cl

Only one structure possible. (N N is the bidentate ethylenediamine ligand.)

(a) Fe2+ is a chiral center. (b) Co3+ is not a chiral center. (c) Co3+ is not a chiral center. (d) Pt2+ is not a chiral center. Square-planar complexes are never chiral.

22.39 In ligand field theory, the bonds between ligands and the metal are described as ionic, attraction being between the positively charged metal ion and a negatively charged anion or polar molecule. 22.41 The lower energy group is made up of the dxy, dyz, and dxz orbitals.



(b) [Co(NH3)6]3+: d 6, low-spin Co3+ complex is diamagnetic.



(c) [Fe(H2O)6]3+: d 5, low-spin Fe3+ complex is paramagnetic [one unpaired electron; same as part (a)]. (d) [Cr(en)3]2+: d 4, Cr2+ complex is paramagnetic (two unpaired electrons).

22.25 [Ni(en)(NH3)3(H2O)]2+

Fe

H3N

NH3

22.43 (a) [Mn(CN)6]4−: d 5, low-spin Mn2+ complex is paramagnetic (one unpaired electron).

22.23 (a) Mn2+; (b) Co3+; (c) Co3+; (d) Cr2+

H3N

H3N

Pt trans

N Cl Cl

22.15 (b) Iron oxide

NH3

SCN

fac

22.13 (b) A  tomic radii similar to 5th period transition elements.

22.35 (a)

(c)

H3N

(a) Os (b) Hg (c) Tc (d) Fe, Co, Mo

Br

Br

cis



22.7

Pt

Cl

Cl

Cl

H3N

NH3 Fe NH3 trans

NH3 Cl



22.45

(a) Fe2+, d 6, paramagnetic, four unpaired electrons (b) Co2+, d 7, paramagnetic, three unpaired electrons (c) Mn2+, d 5, paramagnetic, five unpaired electrons (d) Zn2+, d 10, diamagnetic, zero unpaired electrons

22.47

(a) 6 (b) Octahedral (c) +2 (d) Four unpaired electrons (high spin) (e) Paramagnetic

A-114 Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

22.49 With four ligands, complexes of the d 8 Ni2+ ion can be either tetrahedral or square-planar. The CN− ligand is at one end of the spectrochemical series and leads to a large ligand field splitting, whereas Cl− is at the opposite end and often leads to complexes with small orbital splitting. With ligands such as CN− the complex will be square-planar (and for a d 8 ion it will be diamagnetic). With a weak field ligand (Cl−) the complex will be tetrahedral and, for the d8 ion, two electrons will be unpaired, giving a paramagnetic complex.

22.69 N H2O

N H3N

N H2O

22.55 Fe2+ has a d 6 configuration. Low-spin octahedral complexes are diamagnetic, whereas high-spin octahedral complexes of this ion have four unpaired electrons and are paramagnetic.

Co

H3N

NH3

Cl

x2-y2

Cl

H3N

cis

NH3 Co NH3

xy

+



Cl

22.65 [Co(en)2(H2O)Cl]

2+

N Cr

Cl

Cl

Cl

Cl

N

Cl

mer

N Cl

N



N Cr N

Cr N

N N

fac

N Cr Br

Cl

N

N

Cl

trans chlorides

N

Cl

3+

N

N

N

N

N Cr N

N Cr Cl

N Co NH3

3+

OH2 NH3

z2

xz

yz

xy

xz

yz

[Mn(CN)6]4− paramagnetic, −    1 unpaired e

This shows that Δo for CN− is greater than for H2O.

22.75 (a) The light absorbed is in the orange region of the spectrum (page 1134). Therefore, the light transmitted (the color of the solution) is blue or cyan. (b) Using the cobalt(III) complexes in Table 22.3 as a guide, we might place CO32− between F− and the oxalate ion, C2O42−. Carbonate ion is a weak-field ligand. (c) Δo is small, so the complex should be high spin and paramagnetic.

22.67 N

OH2

22.73 (a) Ammonium tetrachlorocuprate(II) (b) [Cr(H2O)4Cl2]Cl (c) [Co(H2O)(NH2CH2CH2NH2)2(SCN)](NO3)2

trans

  

OH2

NH3

z2

[Mn(H2O)6]2+ paramagnetic, 5 unpaired e−

NH3

Co

3+

x2-y2

22.63 Name: tetraamminedichlorocobalt(III) chloride +

N

22.71 In [Mn(H2O)6]2+ and [Mn(CN)6]4−, Mn has an oxidation number of +2 (Mn is a d 5 ion).

22.61 (a) Mn2+; (b) 6; (c) octahedral; (d) 5; (e) para­ magnetic; (f) cis and trans isomers exist.

Cl

NH3

H2O trans and NH3 cis, not chiral

22.59 Two geometric isomers are possible.

NH3

OH2

H2O cis and NH3 trans, not chiral

22.53 Determine the magnetic properties of the complex. Square-planar Ni2+ (d 8) complexes are diamagnetic, whereas tetrahedral complexes are paramagnetic.

H3N

Co

3+

NH3

H2O and NH3 cis, chiral

22.51 The light absorbed is in the green region of the spectrum (page 1134). Therefore, the light transmitted— which is the color of the solution—is magenta.

22.57 Square-planar complexes most often arise from d 8 transition metal ions. Therefore, it is likely that (b) [Ni(CN)4]2− (Ni2+) and (d) [Pt(CN)4]2− (Pt2+) are square planar. (See also Study Question 22.49.)

N

Br N

cis chlorides 3+

N

N

N

N

N Cr N

3+

N N

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions A-115

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22.77

N

CO2−

O = H2N CH2 H2O O O Cu N N H2O O

N

Cu H2O H 2O O

N Cu

H2O H 2O N

O

Cu H2O H 2O

2−

2−

H2O N O Cu O N H2O

2−

O

O

N

2−

O

Cu OH2 N H2O

N 2−

N

N

N

O

O H2O 2−

O

O

Cu

O

enantiomeric pair

2−

O

enantiomeric pair

OH2 2−

N

enantiomeric pair

22.79 (a) In complexes such as M(PR3)2Cl2 the metal is Ni2+ or Pd2+, both of which are d 8 metal ions. If an Ni2+ complex is paramagnetic it must be tetrahedral, whereas the Pd2+ must be square planar. (A d 8 metal complex cannot be diamagnetic if it has a tetrahedral structure.)

xy xz x2−y2

yz z2 2+

Tetrahedral Ni paramagnetic



complex,

UO22+(aq) + 4 H+(aq) + 2 e− n U4+(aq) + 2 H2O(ℓ)



UO22+(aq) + 4 H+(aq) + Zn(s) n U4+(aq) + 2 H2O(ℓ) + Zn2+(aq)



2[MnO4−(aq) + 8 H+(aq) + 5 e− n Mn2+(aq) + 4 H2O(ℓ)]



5 U4+(aq) + 2 MnO4−(aq) + 2 H2O(ℓ) n 5 UO22+(aq) + 4 H+(aq) + 2 Mn2+(aq)

22.87 (a) Define length of the side of the cube as x, then the length of the diagonal across the cube is x 3 . This is set equal to: 2 rTi + 2 rNi, i.e., x 3 = 2 rTi + 2 rNi = 540 pm; x = 311.8 pm = 312 pm (a = b = c = 3.118 × 10−8 cm = 3.12 × 10−8 cm) (b) Calculated density:

Volume of the unit cell is x3 = (3.118 × 10−8 cm)3 = 3.030 × 10−23 cm3

z2 xz

(c) 5[U4+(aq) + 2 H2O(ℓ) n  UO22+(aq) + 4 H+(aq) + 2 e−]

Mass of one unit cell is the mass of one Ti and one Ni atom = (47.87 g/mol)(1 mol/6.022 × 1023 atoms Ti) + (58.69 g/mol)(1 mol/6.022 × 1023 atoms Ni) = 1.770 × 10−22 g

x2−y2

xy





Cu OH2 N H2O

N

22.85 (a) There is 5.406 × 10−4 mol of UO2(NO3)2, and this provides 5.406 × 10−4 mol of Un+ ions on reduction by Zn. The 5.406 × 10−4 mol Un+ requires 2.157 × 10−4 mol MnO4− to reach the equivalence point. This is a ratio of 5 mol of Un+ ions to 2 mol MnO4− ions. The 2 mol MnO4− ions require 10 mol of e− (to go to Mn2+ ions), so 5 mol of Un+ ions provide 10 mol e− (on going to 5 UO22+ ions, with a uranium oxidation number of +6). This means the Un+ ion must be U4+. (b) Zn(s) n Zn2+(aq) + 2 e−

yz

Square-planar Pd2+ complex, diamagnetic

(b) A tetrahedral Ni2+ complex cannot have isomers, whereas a square-planar complex of the type M(PR3)2Cl2 can have cis and trans isomers (see page 1123).

22.81 (a) Ce [Xe]4f 15d 16s2 Ce3+ [Xe]4f 1 Ce4+ [Xe] (b) Ce3+ is paramagnetic with 1 unpaired electron. Ce4+ is diamagnetic. (c) A unit cell of CeO2 consists of Ce4+ ions as a facecentered cube (4 Ce4+ ions) with O2− ions in all of the tetrahedral holes (8 O2− ions). The ratio of the two types of ions give an empirical formula of CeO2. 22.83 A, dark violet isomer: [Co(NH3)5Br]SO4

B, violet-red isomer: [Co(NH3)5(SO4)]Br



[Co(NH3)5Br]SO4(aq) + BaCl2(aq) n [Co(NH3)5Br]Cl2(aq) + BaSO4(s)

Calculated density = 1.770 × 10−22 g/ 3.030 × 10−23 cm3 = 5.84 g/cm3 The agreement is not very good, probably because atoms don’t pack together as tightly as is assumed. (c) As free atoms, both Ti and Ni are paramagnetic.

Chapter 23 Check Your Understanding 23.1

(a) Isomers of C7H16 CH3CH2CH2CH2CH2CH2CH3    heptane CH3 CH3CH2CH2CH2CHCH3

2-methylhexane

CH3 CH3CH2CH2CHCH2CH3

3-methylhexane

CH3 CH3CH2CHCHCH3

2,3-dimethylpentane

CH3

A-116 Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

23.5

CH3 CH3CH2CH2CCH3

1,4-diaminobenzene

2,2-dimethylpentane

NH2

3,3-dimethylpentane

NH2

CH3 CH3 CH3CH2CCH2CH3 CH3

23.6



OH

CH3 CH3CHCH2CHCH3

CH3CHCH2OH

CH3CH2CHCH2CH3

3-ethylpentane

OH CH3CCH3

CH3 CHCH3

23.7



(b) Two isomers, 3-methylhexane and 2,3-dimethylpentane, are chiral.

23.2

The names accompany the structures in the “Check Your Understanding” answer in Example 23.1.

23.3

Isomers of C6H12 in which the longest chain has six C atoms: H

H C C

H

CH2CH2CH2CH3 H

H C

C CH2CH2CH3

H

CH2CH2CH3

(a) CH3CH2CH2OH: 1-propanol, has an alcohol (OOH) group

CH3CO2H: ethanoic acid (acetic acid), has a carboxylic acid (OCO2H) group CH3CH2NH2: ethylamine, has an amino (ONH2) group (b) 1-propyl ethanoate (propyl acetate) (c) Oxidation of this primary alcohol first gives propanal, CH3CH2CHO. Further oxidation gives propanoic acid, CH3CH2CO2H. (d) N-ethylacetamide, CH3CONHCH2CH3 (e) The amine is protonated by hydrochloric acid, forming ethylammonium chloride, [CH3CH2NH3]Cl. 23.8

H3C

Kevlar is a condensation polymer, prepared by the reaction of terephthalic acid and 1,4-diaminobenzene.

n HO2CC6H4CO2H + n H2NC6H4NH2 ( COC6H4CONHC6H4NH )n + 2n H2O

C C H3C

H H

H C C

H3CCH2

CH2CH3

H

CH2CH3 C

H C C



H

  (b)

H H Br

23.1 An Awakening with l-DOPA 1.

l-DOPA

is chiral. Its chiral center is indicated in the ­following structural formula. chiral center

H

Names (in order, top to bottom): 1-hexene, cis-2-hexene, trans-2-hexene, cis-3-hexene, trans-3-hexene. None of these isomers is chiral. (a)

Applying Chemical Principles

C

H3CCH2

23.4

2-methyl-2-propanol

CH3

2,2,3-trimethylbutane

CH3



2-methyl-1-propanol

CH3

CH2CH3

CH3C

2-butanol

CH3CH2CHCH3

2,4-dimethylpentane

CH3

H3C

CH3CH2CH2CH2OH    1-butanol

Br Br

H3C

H O H O

C C

C C

H C C

H

O

H N H H H

H

C C CH3

H H

H H

bromoethane

2,3-dibromobutane

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions A-117

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2. Dopamine is not chiral. Epinephrine is chiral. Its chiral center is indicated in the following structural formula. chiral center

H

C

H

H H H H

C

C

C

23.1



H



(c) H



(d) H

H

C C

C

O

H C C

H H

H C

H

C

H C

N

C

O N

H H

H

C

H

H

C

C

urea

formaldehyde

2. The electron-pair and molecular geometries are both trigonal-planar. The COH σ bonds are each formed by the overlap of an sp2 hybrid orbital on the C atom with the 1s orbital of the H atom. The σ bond between the C and O is formed by the overlap of an sp2 hybrid orbital on the C with an sp2 hybrid orbital on the O. The π bond between the C and O is formed by the overlap of a 2p orbital on C with a 2p orbital on O.

H

23.3

Energy = 299 kJ/mol of photons This energy is not sufficient to break a COC bond.

23.5

(b) Structural isomers

23.7

Two structural isomers (butane and 2-methylpropane) exist. There are no geometric or optical isomers.

23.9

Heptane

H phenol

H C

H

1. Structural formulas: H

(b) H H

23.2 Green Adhesives

C

C H

C

3. 5.0 g l-DOPA (1 mol l-DOPA/197.2 g l-DOPA) = 0.025 mol l-DOPA

H

H H

H O

O

(a) H

C N C H

O H C C C H H H O

Study Questions

23.11 (b) C5H12 and (c) C14H30 are alkanes. 23.13 3-Ethyl-2-methylhexane CH3 CH2CH3 CH3 CH CH CH2 CH2 CH3



2,3,3,4-Tetramethylpentane CH3 CH3 CH3

3. Similarity: Both nylon-6,6 and proteins are polyamides.

CH3 CH

Differences: 1. In proteins there is one direction for the amide linkage: CONH. In nylon-6,6 two orientations are present: CONH and NHCO.



23.15 2,3-Dimethylbutane 23.17 (a) 2,3-Dimethylhexane CH3 CH3 CH



CH2

CH3

(b) 2,3-Dimethyloctane CH3

Atom economy = (228/246) × 100% = 92.7%

CH3

2. Both are condensation polymers.

CH

CH CH2

CH2

CH2 CH2

CH3

CH3



(c) 3-Ethylheptane CH2CH3

4. (156 lb)(0.454 kg/lb)(1.5 μg/kg/day) = 1.1 × 102 μg/day 5. Volume of base needed = (0.300 g BPA) (1 mol BPA/228.3 g BPA)(2 mol NaOH/1 mol BPA) (1.00 L/0.050 mol NaOH) = 0.053 L (or 53 mL)

CH CH2 CH3

23.3 Bisphenol A (BPA)

3. (15 lb)(0.454 kg/1 lb)(13 μg/kg/day) = 89 μg/day. Yes, the infant ingests over 50 μg/day.

CH3

Other possibilities include 2,2,3,4-tetramethylpentane, 3-ethyl-2,2-dimethylpentane, 3-ethyl-2,3-dimethylpentane, 3-ethyl-2,4-dimethylpentane, and 3,3-diethylpentane.

4. Proteins are chiral, whereas nylon-6,6 is not.

1. Reactants: 15 C, 18 H, 3 O, molar mass = 246; Products: 15 C, 16 H, 2 O, molar mass = 228.

CH

CH3

2. In proteins there is only one C between the amide linkages. In nylon-6,6 there are four or six carbons between amide linkages. 3. Proteins have numerous R groups that can be attached to the carbon in between the amide groups, whereas nylon-6,6 has only hydrogen atoms attached to the carbons in between the amide groups.

C

CH3 CH2

CH CH2 CH2

CH2 CH3

A-118 Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.



23.31 (a) H

(d) 3-Ethyl-2-methylhexane

C

CH2CH3 CH3

CH CH

CH2 CH2

H3C

C

C CH2CH2CH2CH2CH3

2-methylheptane

CH3CH2CH2

C CH2CH2CH3

3-methylheptane. The C atom with an asterisk is chiral.

CH3

H

H

C

C H

H

H



C

H H

H

H



CH3 C H3C



CH2CH3

trans -4-methyl-2-hexene

H

H

H

CH2Cl C

C H

3-chloro-1-propene

H C

H

C

CH3

2-chloropropene

CH3

trans-1-chloropropene

cis -4-methyl-2-hexene

Cl C

H

C

Cl

H

CH

H

C

H C

H

23.29

H

CH3 C

H

23.39

H

CH2CH3

cis-1-chloropropene

CH3

CH3CHCH2CH3

+ HBr

3-ethylheptane. Not chiral.

23.27 Alkane: C5H12; alkene: C5H10; alkyne: C5H8.

C

C

C

C12H26, dodecane: a colorless liquid at room temperature. Expected to be insoluble in water but quite soluble in nonpolar solvents.

CH2CH3

Br

23.37 Four isomers are possible.

23.25 C4H10, butane: a low-molar-mass fuel gas at room temperature and pressure. Slightly soluble in water.

CH

CH2 CH2

H C

CH2CH3

H3C

H

trans-2-pentene

CH2

Cl

C CH2CH2CH2CH3

C

H3C

H

H

H



C

23.35 The three alkenes are 1-butene, cis-2-butene, and trans-2-butene. The reaction with 1-butene is shown below:

H 4-ethylheptane. The compound is not chiral.

CH2CH3 CH3CH2

CH2CH3

H

23.33 (a) 1,2-Dibromopropane, CH3CHBrCH2Br (b) Pentane, CH3CH2CH2CH2CH3

H CH3CH2CH2 C CH2CH2CH3

H

cyclopentane

Axial hydrogens are shown in red; equatorial hydrogens are shown in blue.

23.23

C

cis-2-pentene

C

(b)  H2C

H C

C H

C H

3-methyl-1-butene

H2C

H

CH2CH3

CH3 A CH CH3 C

23.21 Chair form of cyclohexane:

C

H3C

CH3

H

H C* CH2CH2CH2CH3

H

CH3

C

H

4-methylheptane

CH3 CH3CH2

H3C

2-methyl-1-butene

H

C

2-methyl-2-butene

CH2CH3

H

CH3



H

H

H C

1-pentene

H

H3C

C

H

CH3

CH3

23.19

CH2CH2CH3

C

+ H2

CH3CH2CH2CH2CH2CH3

CH2CH2CH2CH3

Hydrogenation is often carried out in the presence of a metal catalyst (Pt, Pd, Rh). Hydrogenation is used in the food industry to convert liquid oils to solids and to make them less susceptible to spoilage.

C H

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions A-119

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23.41 H C H

H C

+ HBr CH2

CH2

CH3

1-pentene

H Br H C C

Br H CH2

CH3 + H C C

CH2

H H

CH2

CH3

H H

2-bromopentane major product

23.43 (a)

CH2



Cl

1-bromopentane minor product

(b)

CH3

Cl Br m-dichlorobenzene

p-bromotoluene

23.55

(a) Diethylamine (b) Methyldipropylamine (c) Butylamine (d) Tripropylamine

23.57 (a) 1-Butanol, CH3CH2CH2CH2OH (b) 2-Butanol OH

23.45 (a) 1,2-Dichlorobenzene (b) 1,3,5-Trinitrobenzene (c) 1,2,4-Tribromobenzene 23.47

CH3CH2

C CH3 H

CH3



CH3

(c) 2-Methyl-1-propanol H

CH3

CH3Cl/AlCl3

CH3 C

CH2OH

CH3

CH3

23.49

CH3 1,2,4-trimethylbenzene

CH3

(d) 2-Methyl-2-propanol OH CH3 C CH3

CH3 CH3

CH3

CH3

NO2 NO2 1,2-dimethyl-3-nitrobenzene

23.51



1,2-dimethyl-4-nitrobenzene

23.59 (a) C6H5NH2(ℓ) + HCl(aq) n (C6H5NH3)Cl(aq) (b) (CH3)3N(aq) + H2SO4(aq) n [(CH3)3NH]HSO4(aq) 23.61

(a) 1-Propanol, primary (b) 1-Butanol, primary (c) 2-Methyl-2-propanol, tertiary (d) 2-Methyl-2-butanol, tertiary

(a)

(b)

23.53 (a) Ethylamine, CH3CH2NH2 (b) Dipropylamine, (CH3CH2CH2)2NH CH3CH2CH2 N CH2CH2CH3

(c)

H



(c) Butyldimethylamine CH3CH2CH2CH2

N CH3 CH3



(d)

(d) Triethylamine

CH3CH2CH2 CH C

CH3CH2 CH C O

(b)

CH3 O

HO C CH2CH2 C

OH

O NH2CH2CH2 C

OH

O CH3



OH

CH3 O

23.63 (a)

CH3CH2 N CH2CH3 CH2CH3

CH3 O

C CH2CH2CH3

O H C CH2CH2CH2CH2CH3

A-120 Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.



CH3CH2CH2CH2 23.65



O

(c)

C OH

(a) Acid, 3-methylpentanoic acid (b) Ester, methyl propanoate (c) Ester, butyl acetate (or butyl ethanoate) (d) Acid, p-bromobenzoic acid

O CH3CH2CH2CH2

H H

H H

H H

C

C

C C

C

H O

H O

H O

C

O

23.67 (a) Pentanoic acid



(b) Three units of PVA:

C OH

C

CH3 O

23.81 Polyacrylonitrile from acrylonitrile. H CN H CN n

(c) 2-Octanol

CH2

C C C C

CHCN

H H H H

OH C CH2CH2CH2CH2CH2CH3

23.83

Cl

23.69 Step 1: Oxidize 1-propanol to propanoic acid. H CH3CH2

O

C OH

oxidizing agent

CH3CH2

C OH

CH3CH2

C OH + CH3CH2

C OH

C

+H2O

C CH3

H

OH

−H2O

H C

C H

CH3 CH3 H Br +HBr

H C

C H

CH3 CH3 Cl Cl

O O CH2CH2CH3

+Cl2

23.71 Sodium acetate, NaCH3CO2, and 1-butanol, CH3CH2CH2CH2OH

H C

C H

CH3 CH3

23.73 (a) Trigonal-planar (b) 120° (c) The molecule is chiral. There are four different groups around the carbon atom marked 2. (d) The acidic H atom is the H attached to the CO2H (carboxyl) group. 23.75

C

Cl

H

H

H CH3CH2 C

C

trans isomer

H3C

H

Cl

H

H

Step 2: Combine propanoic acid and 1-propanol. O

(b)

C

H

23.85 Five isomers: 1-hexene, cis- and trans-2-hexene, and cis- and trans-3-hexene. 23.87

H



H

cis isomer

n

Cl

H C

C

H

(d) No reaction.

Cl

Cl C

(a)

H

CH3

(c) Hydrolysis of polyvinyl acetate will yield polyvinyl alcohol.

CH2CH2CH2CH2CH2OH

H3C

C

CH3 O

(b) 1-Pentanol



C

23.89 (a) O H3C



O

C OH + NaOH

−

H3C C O

Na+ + H2O

H

(b)  H3C

[CH3NH3]+Cl−

N H + HCl

O

23.91

CH3CH2CH2CNHCH3



This compound is an amide.



CH3CH2CH2CO2H + CH3NH2 n  CH3CH2CH2CONHCH3 + H2O

H

H C

n

23.77 (a) Alcohol   (b) amide   (c) acid   (d) ester

H

C

C H

23.79 (a) Prepare polyvinyl acetate (PVA) from vinylacetate. C

n

H

H H

H

H

n HOCH2CH2OH + n HO

C C

C O C O

CH3

H O O

C

n

CH3

O

C

H H

n

O

O

C

C OH

O

O

C

C OCH2CH2O

+ 2 n H2O n



Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions A-121

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23.93 (a) 2, 2-Dimethylpentane

23.103

O

CH3 H3C

CH2CH2CH3

C

CH3

(b) 3, 3-Diethylpentane CH2CH3 CH3CH2

H2C

O C(CH2)10CH3 O

HC

O C(CH2)10CH3

H2C

O C(CH2)10CH3 O

(a)

C CH2CH3

H2C

O C(CH2)10CH3 O

CH2CH3

H CH2CH3 CH3 C

H2C

O C(CH2)10CH3

C CH2CH3

CH3 H

O C(CH2)10CH3 + 3 NaOH O

HC

(c) 3-Ethyl-2-methylpentane

H2C OH

(d) 3-Ethylhexane

HC OH + 3

CH2CH3 CH3CH2 C CH2CH2CH3

H2C

O C(CH2)10CH3 O

HC

O C(CH2)10CH3 O

H2C

O C(CH2)10CH3

H C CH2CH3 Cl Cl Cl

1,2-Dichloropropane

H C C

CH3

H H H C C

C

H H H H C C

H2C OH

C

H

23.105

H CH2OH

H Cl H add H2

23.97 CH3

CH3 CH3

H C C H H H

CH3

H

CH3

oxidize

CH3

H 3C

CH3

CH3 1,2,3-trimethylbenzene

O

HC OH + 3 CH3(CH2)10COCH3

H

H Cl H 2,2-Dichloropropane

1,2,4-trimethylbenzene

1,3,5-trimethylbenzene

H

CH2OH C

H

C

H

H

H C H3C

+H2

C CH3

H C

H CH2OH H CH2OH

CH3

(b)  H

C CH3 CH3

C C

C C

H H

H H

n

O CH3CO2H

H

CH3 CH3 butane (not chiral)



H

polymerize

H C

C

H

C H

CO2H

23.99 Replace the carboxylic acid group with an H atom. 23.101 (a)

Na+

+ 3 CH3OH

H2C OH

Cl H Cl 1,3-Dichloropropane

CH3(CH2)10CO

O

(b) Cl

1,1-Dichloropropane

−

H2C OH

H 23.95

O

23.107 (a)

H H C

H3C

H C

H H

C H

C O CH2CH

CH2

H Br H +HBr

H C

C

C H

H H H 2-bromopropane

A-122 Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.



H CH3 H

(b)

H3C C C

H CH3 H

C H

+H2O

H3C C C

H

C H

H CH3 H

(c)

H3C C C

C

H2C C

C

H

H3C C C

C

H

The product is the same for parts (b) and (c).

CH3 H

OH CH3 H H3C C

C

H

23.109 (a) The only structural difference between theobromine and caffeine occurs on the N in the sixmember ring that is between the two CPO groups. In theobromine, there is an H atom attached to this N. In caffeine, there is a CH3  group attached. (b) 5.00 g sample (2.16 g theobromine/100 g sample) = 0.108 g theobromine

C

H H

C H



C

C

H

H C H C H H

C

H H

+ Br2

C H

H

C

C

H

Br C H C H

H

H

Benzene, however, needs much more stringent conditions to react with bromine; then Br2 will substitute for H atoms on benzene and not add to the ring.

23.115 (a) The compound is either propanone, a ketone, or propanal, an aldehyde. H O H H C C H



H C C

C H

H H

H

propanone (a ketone)



H H O

C H

    

CH3

oxidizing agent

O H3C C

CH3 H

CH3 H C

C CH3

CH3 H 3,3-dimethyl-2-pentanone

23.121 The 18O label will be found in dimethyl terephthalate. 23.123

H H C H

methane

four single bonds

formaldehyde

one double bond and two single bonds

allene

two double bonds

acetylene

one single bond and one triple bond

H O

23.113 Cyclohexene, a cyclic alkene, will add Br2 readily (to give C6H10Br2). H

C

Y = 3,3-dimethyl-2-pentanol

23.111 Compounds (b), acetaldehyde, and (c), ethanol, produce acetic acid when oxidized.

Br H

CH3

+H2O

H OH H

H

C

X = 3,3-dimethyl-1-pentene

H CH3 H +H2O

H

CH3 H

H

H OH H 2-methyl-2-butanol



23.119

propanal (an aldehyde)

(b) The ketone will not undergo oxidation, but the aldehyde will be oxidized to the acid, CH3CH2CO2H. Thus, the unknown is likely propanal. (c) Propanoic acid

H

C

H

H

H C C C

H

H

H

C C H

23.125 (a) Cross-linking makes the material very rigid and inflexible. (b) The OH groups give the polymer a high affinity for water. (c) Hydrogen bonding allows the chains to form coils and sheets with high tensile strength. 23.127 C2H6(g) + 7/2 O2(g) n 2 CO2(g) + 3 H2O(ℓ) C2H5OH(ℓ) + 3 O2(g) n 2 CO2(g) + 3 H2O(ℓ) (a) Enthalpy of combustion of ethane = −1560.66 kJ/mol C2H6; enthalpy change per gram = −51.901 kJ/g C2H6 Enthalpy of combustion of ethanol = −1367.5 kJ/mol C2H5OH; enthalpy change per gram = −29.684 kJ/g C2H5OH (b) The enthalpy change for the combustion of ethanol is less negative than for ethane, so partially oxidizing ethane to form ethanol decreases the amount of energy per gram available from the combustion of the substance.

23.117 2-Propanol will react with an oxidizing agent such as KMnO4 (to give the ketone), whereas methyl ethyl ether (CH3OC2H5) will not react. In addition, the alcohol should be more soluble in water than the ether.



Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions A-123

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Study Questions

23.129 (a) Empirical formula, CHO (b) Molecular formula, C4H4O4 (c) O O H

O

C

C

C

24.1

H



N

H



(b)

H

N

+

C

C

O

C

H

H H

H

C

C

N

H

C

H C

H H

C

C

H

H

H

H

H

H

O

C

C

O

H

H

−

C

H

C

O

−

H

H H

C

C

H

C

H H

H

H C

C H

H

H

C

N

O

H H

O

H

H N

C

C

+

H

O

C

H

Check Your Understanding H

O

H

C

Chapter 24 24.1 H

H

H

O

H H More experiments would need to be performed to determine that the cis- configuration is correct for maleic acid. (d) All four C atoms are sp2 hybridized. (e) 120°



(a)

H

C

C

C

H

H

H

H



(c) The zwitterionic form is the predominant form at physiological pH.

24.3

Polar: serine, lysine, aspartic acid Nonpolar: alanine, leucine, phenylalanine

24.5

C C

H

C

H

O

N

C

C

H

H

H

H H

+

H

O

N

C

C

H H

C

H

O

−

H

24.2

5′-GTACGTATCG-3′

24.3

mRNA: 5′-AAA GCA CAA-3′ amino acid sequence: lysine–alanine–glutamine

H

H

24.1 Polymerase Chain Reaction 1. 220 = 1,048,576 2. Hydrogen bonds are weaker than covalent bonds. At 95 °C, the amount of thermal energy is not enough to break covalent bonds. 3. (a) One reason is that cytosine and guanine bind with three hydrogen bonds per base pair, whereas adenine and thymine (uracil) bind with only two. Sequences with a high number of cytosines and guanines thus bind more strongly. [There are additional reasons for stronger binding between G and C than between A and T (or U) than just the number of hydrogen bonds per base pair.] (b) This sequence could fold over and base pair to itself instead of to the target strand. (c) Using a large excess of primers ensures that the target strands will bind to the primers instead of having the complementary target strands bind to each other.

H

O

N

C

C

H H

C

H

H

Applying Chemical Principles

+

H

O

N

C

C

H

H

O

−

H

24.7 H H

N

+

H

O

H

O

C

C

N

C

C

H

H H

C

H H

H

C

C

N

H

O

C

C

O

−

H H H

C

H

O

H H

C

C

H

C

H H

H

H H

H

C

H H

H

A-124 Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

24.9



(a) Primary (b) Quaternary (c) Tertiary (d) Secondary

Adenosine-5′-monophosphate

(c)

NH2 N

24.11 H HO

H

OH

4

5

HO

H

3H

2

O

HO 1

H

OH

4

5

HO

OH

H

24.13

H

3H

2

O 1

OH

OH

C

H

H

H

C

OH

H

C

OH

H

H

C

OH

C

O

C

OH

N

dihydroxyacetone

24.17 (a) O

H

N

H

−O

O

H OH

P

+

−O

O

CH2

O

N

H

H

O

H OH

P

+

O

O−

C H

H

−O

H

H C

O

H OH

+

O

O

H

H Adenosine

(b)

NH2 N N HO

5’ 4’

O H

H 3’



H

1’

N

24.21 24.23

N

CH2

H

O

NH2 NH2 N

H

P

N

O

H

O−

C

NH

N

CH2

H

C

O

N

O O

O−

H

O NH

H

H

H O

C

OH

O

H

-d-Ribose H

OH

N

HO CH2

24.15 In amylose starch, the glucose subunits are joined by α-1,4 linkages. In cellulose, the glucose subunits are joined by β-1,4 linkages. Whereas humans can digest amylose starch, they cannot digest cellulose. Ingested cellulose is dietary fiber.

H

NH2 N

H

d-glyceraldehyde

H H

24.19

H

N

O

O−

-D-glucose

O

H

P+ O

−O

H

H

-D-glucose

N

O−

OH

N

O

O

H

H

OH

H OH

(a) 5′-GAATCGCGT-3′ (b) 5′-GAAUCGCGU-3′ (c) 5′-UUC-3′, 5′-CGA-3′, and 5′-ACG-3′ (d) glutamic acid, serine, and arginine polar

N nonpolar

H OH

OH 2’

polar

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions A-125

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24.25 T  he 4-ring structure present in all steroids is given in Figure 24.20. 24.27 C6H12O6(s) + 6 O2(g) n 6 CO2(g) + 6 H2O(ℓ) ∆rH° = Σn∆f H°(products) − Σn∆f H°(reactants) ∆rH° = (6 mol CO2/mol-rxn)[∆f H°(CO2)] + (6 mol H2O/mol-rxn)[∆f H°(H2O)] − (1 mol C6H12O6/mol-rxn)[∆f H°(C6H12O6)] ∆rH° = (6 mol CO2/mol-rxn)(−393.5 kJ/mol CO2) + (6 mol H2O/mol-rxn)(−285.8 kJ/mol H2O) − (1 mol C6H12O6/mol-rxn)(−1273.3 kJ/mol C6H12O6) ∆rH° = −2,803 kJ/mol-rxn 24.29 (a) NADH (b) O2



(c) O2 (d) NADH

24.31 If the process were 100% efficient, then 93 mol of ATP could be formed. The percent yield is 34%. 24.33

H H

+

N

H H

H

O

C

C

N

H

H H

C

H

O

C

C H

O

C

C

H

H

C

H H

H

C

H

H

−

H

O

H

N

C

C

H H

C

H

+

−

O

N

C

C H

O

H H

C

C

H

H

C

H H

H

C

H

H



24.37 T  here are four nucleotide bases, and each codon is three nucleotides long, so there are 43 = 64 codons possible. Some amino acids have more than one codon. 24.39 oxidation: C6H12O6 reduction: O2 24.41 A  graph of y = 1/Rate and x = 1/[S] yields a straight line with the equation y = 1.5x + 9200. The y-intercept corresponds to 1/Ratemax, so Ratemax = 1/9200 = 1.1 × 10−4 mol/L ∙ min. 24.45 T  he sequences differ in the positions of attachments of the phosphate to deoxyribose on adjacent units. Consider the A–T attachments. In ATGC, the phosphate links the 3′ position on A to the 5′ position on T. In CGTA, the phosphate links the 5′ position on A to the 3′ position on T.

−

Chapter 25 Applying Chemical Principles

H

24.35 (a) 6 CO2(g) + 6 H2O(ℓ) n C6H12O6(s) + 6 O2(g) ∆rH° = Σn∆f H°(products) − Σn∆f H°(reactants) ∆rH° = (1 mol C6H12O6/mol-rxn)[∆f H°(C6H12O6)] − {(6 mol H2O/mol-rxn)[∆f H°(H2O)] + (6 mol CO2/mol-rxn)[∆f H°(CO2)]} ∆rH° = (1 mol C6H12O6/mol-rxn) (−1273.3 kJ/mol C6H12O6) − [(6 mol H2O/mol-rxn)(−285.8 kJ/mol H2O) + (6 mol CO2/mol-rxn)(−393.5 kJ/mol CO2)] ∆rH° = +2,802.5 kJ/mol-rxn = +2803 kJ/mol-rxn

E = hc/λ = (6.626 × 10−34 J ∙ s) (3.00 × 108 m ∙ s−1)/(6.5 × 10−7 m) =  3.1 × 10−19 J (d) The energy per photon is less than the amount required per molecule of glucose, therefore multiple photons must be absorbed.



24.47 (a) In transcription, a strand of RNA complementary to the segment of DNA is constructed. (b) In translation, an amino acid sequence is constructed based on the information in an mRNA sequence.

H +



(b) (2802.5 kJ/mol)(1 mol/6.022 × 1023 molecules) (1000 J/1 kJ) = 4.654 × 10−18 J/molecule (c) λ = 650 nm(1 m/109 nm) = 6.5 × 10−7 m

24.43 glycine

H

H



25.1 Chlorination of Water Supplies 1. (a) One chlorine atom is oxidized (oxidation number increases from 0 to +1), and one chlorine atom is reduced (oxidation number decreases from 0 to −1). (b) Reduction, cathode: Cl2 + 2 e− n 2 Cl− Oxidation, anode: Cl2 + 2 H2O n 2 HClO + 2 H+ + 2e− E°cell = E°cathode − E°anode = 1.36 V − 1.63 V = −0.27 V (c) In the overall reaction, H+ appears on the right side of the equation. Increasing pH (decreasing [H+] and making the solution more basic) shifts the equilibrium to the right, making the reaction more favorable. Decreasing pH (making the solution more acidic) will have the opposite effect. 2. lnK = nE°/0.0257 = 2(−0.27)/0.0257 = −21.0; K = 7 × 10−10 3. Kb = Kw/Ka = 2.86 × 10−7 = x2/(0.010 − x) x = 5.35 × 10−7 M [HClO] = [OH−] = 5.3 × 10−5; pH = 9.73

A-126 Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

4. H2O2(aq) + ClO−(aq) n Cl−(aq) + H2O(ℓ) + O2(g) 5. The reaction of chlorine bleach (which contains hypochlorite ion, ClO−) and ammonia in basic solution produces toxic vapors of chloramine, NH2Cl(g). 25.2 Hard Water

25.15 NH4+(aq) + NO2−(aq) n N2(g) + 2 H2O(ℓ) 2 NH4+(aq) + 3 O2(g) n  2 NO2−(aq) + 2 H2O(ℓ) + 4 H+(aq) 25.17 (b) hydroelectric power 25.19 From methane: H2O(g) + CH4(g) n 3 H2(g) + CO(g)

1. For Mg2+: (50. mg)(1 mmol Mg2+/24.31 mg) (1 mmol CaO/mmol Mg2+)(56.08 mg CaO/ 1 mmol CaO) = 115 mg CaO



For Ca2+: (150 mg)(1 mmol Ca2+/40.08 mg) (1 mmol CaO/mmol Ca2+)(56.08 mg CaO/ 1 mmol CaO) = 210 mg CaO

28.7 g H2 produced From coal: H2O(g) + C(s) n H2(g) + CO(g)

37.7 g of H2 produced From petroleum: H2O(g) + CH2(ℓ) n  2 H2(g) + CO (g)

16.8 g H2 produced

Total CaO = 115 mg + 210 mg = 330 mg (two significant figures)

25.21 70. lb(453.6 g/lb)(33 kJ/g) = 1.0 × 106 kJ

There are 2 mol CaCO3 per mole Ca2+ and 1 mol each of CaCO3 and MgCO3 per mole Mg2+.

25.23 Assume burning oil produces 43 kJ/g (the value for crude petroleum in Table 25.5).

CaCO3 from Ca2+ reaction: (0.15 g Ca2+) (1 mol/40.08 g Ca2+)(2 mol CaCO3/1 mol Ca2+) (100.1 g CaCO3/1 mol CaCO3) = 0.749 g



CaCO3 from Mg2+ reaction: (0.050 g Mg2+) (1 mol/24.31 g Mg2+)(1 mol CaCO3/1 mol Mg2+) (100.1 g CaCO3/1 mol CaCO3) = 0.206 g MgCO3 from Mg reaction: (0.050 g Mg ) (1 mol/24.31 g Mg2+)(1 mol MgCO3/1 mol Mg2+) (84.31 g MgCO3/1 mol MgCO3) = 0.173 g 2+

2+

Total mass of solids = 0.749 g + 0.206 g + 0.173 g = 1.13 g 2. CaCO3(s) + 2 CH3CO2H(aq) n  Ca(CH3CO2)2(aq) + H2O(ℓ) + CO2(g) This is a gas-forming reaction.



Energy per liter = −1.785 × 104 kJ/L.



Finally, use the kW-h to kJ conversion factor to obtain the answer.



(1.785 × 104 kJ/L)(1 kW-h/3600 kJ) = 4.96 kW-h/L.

25.27 (a) Area of parking lot = 1.625 × 104 m2 (2.6 × 107 J/m2)(1.625 × 104 m2) = 4.2 × 1011 J (b) 1.3 × 107 g C

25.1

(b) CH4

25.3

For gases, ppm refers to numbers of particles, and hence to mole fractions. Gas pressure exerted is directly proportional to mole fraction. Thus, 40,000 ppm water vapor would exert a pressure of 40,000/1,000,000 of one atmosphere, or 30.4 mm Hg (0.0400 × 760 mm Hg). This would be the case at a little over 29 °C, at 100% humidity.



(a) H O N O Electron-pair geometry around N is trigonal planar and around O (between H and N) is tetrahedral. (b) 5.95 × 10−7 m (or 595 nm)

25.7

(d) Al2(SO4)3

25.9

[HCO3−]/[CO3−] = 170

25.11 The amount of NaCl is limited by the amount of sodium present. From a 1.0-L sample of seawater, a maximum of 0.460 mol NaCl could be obtained. The mass of this amount of NaCl is 26.9 g [(0.460 mol/L) (1.00 L)(58.44 g NaCl/1 mol NaCl) = 26.9 g]. 25.13 Ca(OH)2(s) + Mg (aq) n Mg(OH)2(s) + Ca (aq) Mg(OH)2(s) + 2 H3O+(aq) n Mg2+(aq) + 4 H2O(ℓ) MgCl2(ℓ) n Mg(s) + Cl2(g) 2+



25.25 Δ  rH° for the reaction CH3OH(ℓ) + 1.5 O2(g) n CO2(g) + 2 H2O(ℓ) is −726.8 kJ/mol-rxn.

25.29 Energy per gallon of gas = 1.339 × 105 kJ/gal

Study Questions

25.5

7.0 gal(3.785 L/gal)(1000 cm3/L)(0.8 g/cm3)(43 kJ/g) =  0.9 × 106 kJ. Uncertainty in the numbers is one significant figure. This value is close to the value for the energy obtained by burning from 70 pounds of coal (calculated in Question 25.21).

2+

Energy to travel 1 mile = 2430 kJ

25.31 (a) The cage is a dodecahedron, so it has 20 vertices, each of which is the O atom of a water molecule (the red spheres in the model are O atoms). (b) 30 hydrogen bonds. (This estimate is made by assuming there is an H bond on each of the edges.) (c) A dodecahedron has 12 faces. 25.33 (d) O2 25.35 (a) The most significant resonance structure is the following with formal charges as shown. The molecular geometry is linear. 0 1+ 1– N N O



(b) If the connectivity were NOOON, the central oxygen would have a formal charge of 2+, a highly unfavorable situation.



(c) (1.00 L)(800 nmol/L)(1 mol/109 nmol) (44.01 g/mol) = 4 × 10−5 g

25.37 (c) Hydrogen is explosive 25.39 (a) C13H27CO2CH3(ℓ) + 43/2 O2(g) n  15 CO2(g) + 15 H2O(g) (b) ∆rH° = −8759.1 kJ/mol-rxn

Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions A-127

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

(c) Hexadecane (ΔH° = −9951.2 kJ/mol) provides more energy per mole of fuel than methyl myristate (ΔH° = −8759.1 kJ/mol).



Hexadecane (ΔH° = −3.4 × 104 kJ/L) also provides more energy per liter of fuel than methyl myristate (ΔH° = −3.1 × 104 kJ/L).





25.41 (a)

H H H

C

C

H H O H

H

C

C

N H

H H ethanol

H H H ethylamine

Both carbon atoms are tetrahedral; oxygen atom is bent.

Both carbon atoms are tetrahedral. Nitrogen atom is pyramidal.

H H



C

C

N

H

C

C

C

N

H acetonitrile

H H acrylonitrile

CH3 carbon atom is tetrahedral. Carbon atom of CN group is linear.

CH2 and CH carbon atoms are trigonal planar. Carbon atom of CN group is linear.

(b) CH3CH2OH + NH3 n CH3CH2NH2 + H2O CH3CH2NH2 + O2 n CH3CN + 2 H2O (c) Reactants (CH3CH2OH, NH3, and O2) have 2 C, 9 H, 1 N, and 3 O; molar mass = 95. Product (CH3CN) has 2 C, 3 H, and 1 N; molar mass = 41. Atom economy = (41/95) × 100% = 43%

25.43 Using a COCl bond enthalpy of 339 kJ/mol, the wavelength of the required radiation is 3.53 × 10−7 m or 353 nm. This is in the ultraviolet region. 25.45 NH4+(aq) + 2 H2O(ℓ) n NO2−(aq) + 8 H+(aq) + 6 e−

NO2−(aq) + H2O(ℓ) n NO3−(aq) + 2 H+(aq) + 2 e−

25.47 (a) Isooctane:

C8H18(ℓ) + 25/2 O2(g) n 8 CO2(g) + 9 H2O(ℓ)

(b) Isooctane = 70.0 mol CO2 and ethanol = 43.4 mol CO2. Ethanol produces less CO2 per kilogram. (c) On the basis of this simplistic comparison, there is not a clear winner. Isooctane releases more energy per kilogram but also releases more CO2 per kilogram.

25.49 (a) Only 92% of the ice is submerged, and the water displaced by ice (the volume of ice under the surface of water) is 23 cm3 (0.92 × 25 cm3 =  23 cm3). Thus, the liquid level in the graduated cylinder will be 123 mL. (b) Melting 25 cm3 of ice will produce 23 mL of liquid water [25 cm3 ice (0.92 g H2O/cm3 ice) (1.0 cm3 liquid H2O/g H2O) = 23 cm3 liquid H2O]. The water level will be 123 mL (the same as in (a); that is, the water level won’t rise as the ice melts). 25.51 N  onrenewable resources are not replenished when used. Renewable resources are, for the foreseeable future, not depleted when used. Nonrenewable: coal, natural gas; renewable: solar, geothermal energy, wind power. 25.53 Mercury: from coal burning power plants Lead: paint residues (before 1970), soil (from the use of tetraethyllead in gasoline), water supplies (from use of lead in pipes and plumbing) Arsenic: some ground water sources, residual from some common chemicals (use in wood treatment, for example) 25.55 T  he main pollutant is SO2, arising from the burning of coal. Some SO3 in the atmosphere arises from oxidation of SO2, and this combines with water to give H2SO4, a primary component of acid rain. Removing SO2 is best done at the source, by extracting it from the flue gases of coal burning plants. This can be done by passing the gases from coal oxidation through a scrubber containing a CaCO3 slurry.

ΔrH° = −5461.2 kJ/mol-rxn (or −47,809 kJ/kg C8H18) Ethanol:

C2H5OH(ℓ) + 3 O2(g) n 2 CO2(g) + 3 H2O(ℓ)

ΔrH° = −1367.5 kJ/mol-rxn (or −29,684 kJ/kg C2H5OH) Isooctane releases more energy per kilogram.

A-128 Appendix N / Answers to Study Questions, Check Your Understanding, and Applying Chemical Principles Questions Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Index of Names

Note: Italicized page numbers indicate pages containing figures/illustrations.

A Agutter, Alexandra, 816 Aizenberg, Joanna, 36 Ampere, A. M., 1082 Amundsen, Roald, 616 Anastas, Paul, 7 Anderson, Carl, 998 Armbruster, Peter, 1020 Arnold, J. R., 1013 Arrhenius, Svante, 156, 157, 158 Aspdin, Joseph, 1058 Autumn, Kellar, 556 Avogadro, Amedeo, 100, 506

B Balmer, Johann, 313 Banavar, Shubha, 1107 Bartlett, Neil, 1089 Becquerel, Henri, 73, 82, 994, 1022 Bednorz, Georg, 177 Blackett, Patrick, 1014 Bohr, Christian, 1118 Bohr, Niels, 313–317, 314, 319, 321–323, 326, 328–329, 1020, 1118 Boltzmann, Ludwig, 518, 892, 894, 1020 Born, Max, 321–322, 595 Bosch, Carl, 762 Boyle, Robert, 502, 508 Bragg, Lawrence, 440, 593 Bragg, William Henry, 593

Brandt, Hennig, 1072 Branly, Édouard, 82 Brønsted, Johannes N., 157, 159, 812 Brown, Darius Z., 7 Bunsen, Robert, 332

C Callmann, Cassandra, 783 Carey, Tiffany, 66 Carothers, Wallace, 1190 Chadwick, James, 74, 1015 Chapman, Eddie Nathaniel, 938 Charles, Jacques Alexandre César, 503, 508, 510, 1049 Charpentier, Emmanuelle, 1222 Cielo, César, 52 Clapeyron, B. P. E., 564 Clausius, R., 564 Cockcroft, J. D., 1014 Coulson, C. A., 459 Crick, Francis, 405, 439–441, 551 Crookes, William, 72 Curie, Irène, 82, 1020 Curie, Marie, 12, 73, 81, 82, 993, 994, 997, 1022, 1081 Curie, Pierre, 12, 73, 81, 82, 1022, 1081 Cusumano, James, 7

Davy, Sir Humphry, 970, 978–979, 1050, 1074 Debierne, André-Louis, 82 de Broglie, Louis Victor, 320–321 Debye, Peter, 426 DeNiro, Robert, 1192 Doudna, Jennifer, 1222

E Ehrlich, Paul, 118 Eiffel, Gustave, 42 Einstein, Albert, 12, 308–311, 993, 1004, 1020 Emsley, John, 1072 Ertem, S. Piril, 543

F Faraday, Michael, 970, 1167 Feliú-Mójer, Mónica, 307 Fermi, Enrico, 12, 1015, 1020 Fleming, Alexander, 1214 Fowler, Joanna, 1024 Franklin, Benjamin, 508 Franklin, Rosalind, 439–441, 551 Fraunhofer, Joseph von, 332 Frayn, Michael, 314 Frisch, Otto Robert, 1020 Fuller, R. Buckminster, 79

D

G

Dalton, John, 64, 72, 514 David, Jacques-Louis, 141 Davisson, C. J., 320

Galvani, Luigi, 940 Gay-Lussac, Joseph, 506 Geiger, Hans, 74

Index of Names I-1

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Geim, Andre, 615 Gerlach, Walther, 330 Germer, L. H., 320 Gibbs, J. Willard, 903 Gomberg, Moses, 763 Goodyear, Charles, 1188 Gosling, Raymond, 440–441 Graham, Thomas, 521, 656–657 Gray, Louis H., 1022 Gray-Scully, Uraina, 629

H Haber, Fritz, 595, 762 Hahn, Otto, 1018, 1020 Hall, Charles Martin, 1050, 1060 Hanna-Attisha, Mona, 1260 Harriss, Zachary, 346 Heisenberg, Werner, 314, 321 Heroult, Paul, 1050, 1060 Hitler, Adolph, 1020 Hoffmann, R., 443

I Ichoku, Charles, 1043 Isaac, Jessica N., 194

J Jahren, Hope, 646 Jimenez, Ederly, 501 Joliot, Frédéric, 82 Joule, James P., 37, 257

K Kekulé, August, 472, 1168 Kelvin. See Thomson, William Ketelaar, J. A. A., 442 Kirchhoff, Gustav Robert, 332 Kohlrausch, Friedrich, 780 Kusiima, Jamil, 1253

L Lavoisier, Antoine Laurent, 140–141 Lavoisier, Marie-Anne Pierrette Paulze, 141 LeClanche, George, 946 Lee, H., 1274 Lewis, Gilbert Newton, 388, 389, 459, 777, 812 Li, Kaichang, 1193 Li, Tao, 1208 Libby, W. F., 1013 Lockyer, Sir Joseph Norman, 332 Long, Crawford Williamson, 1172 Lowry, Thomas M., 157, 159, 812

I-2

M

R

Macintosh, Charles, 1188 Manuel, Simone, 52 Markel, Howard, 441 Markovnikov, Vladimir, 1165 Marsden, Ernest, 74 Matsche, Alex, 675 Maxwell, James Clerk, 306, 518 McGrayne, Sharon B., 1020 McMillan, Edwin, 1014 Meitner, Lise, 12, 381, 1018, 1020 Mendeleev, Dmitri Ivanovich, 12, 78, 345, 1016, 1040 Menten, Maud L., 706 Michaelis, Leonor, 706 Midgely, Thomas, Jr., 1251 Millikan, Robert, 73 Miolati, A., 1140 Moisson, Henri, 1082 Mond, Ludwig, 1140 Moore, Tiffany, 832 Morton, William T. G., 1172 Moseley, H. G. J., 78 Müller, Karl, 177 Mulliken, Robert S., 443, 459 Mullis, Kary, 1236

Rahm, M., 443 Ramsay, Sir William, 118, 762 Rayleigh. See Strutt, John William Robert, Nicolas-Louis, 508 Roentgen, Wilhelm, 1022 Roosevelt, Franklin D., 993 Rosenberg, Barnett, 1138 Rubin, A., 1052 Rumford. See Thompson, Benjamin Rutherford, Ernest, 63, 74, 78, 994–995, 1014, 1020 Rydberg, Johannes, 313

N Navedo, Thalia I., 38 Newton, Isaac, 508, 1134 Niepce de Saint Victor, Abel, 82 Nisbet, Euan, 1273 Nobel, Alfred, 12, 1172 Novoselov, Kostya, 615

O Oganessian, Yuri, 1017 Olah, George, 817 Oleksiak, Penny, 52 Ostwald, Wilhelm, 100

P Paul, Sonali, 743 Pauli, Wolfgang, 346 Pauling, Ava Helen, 405 Pauling, Linus, 390, 404, 405, 414, 439–440, 443, 459, 462 Pauling, Peter, 440 Paulze, Marie-Anne, 141 Perkin, William, 486 Planck, Max, 308–311, 1020 Prabhune, Nandan, 141 Priestley, Joseph, 141

S Sacks, Oliver, 508, 1192 Sagan, Carl, 312 Santos, Randy, 260 Scerri, E., 361 Scheele, Karl Wilhelm, 141, 1082 Schrödinger, Erwin, 321 Scott, Robert, 616 Seaborg, Glenn T., 1014, 1017 Sievert, Rolf, 1022 Skłodowska, Bronya, 82 Skłodowska, Maria. See Curie, Marie Smalley, Richard, 1263 Soddy, Frederick, 995 Stern, Otto, 330 Stock, Alfred, 1260 Stokes, A.R., 441 Strassmann, Fritz, 1018, 1020 Strutt, John William, 118–119 Sullivan, John H., 717

T Thenard, Louis, 203 Thompson, Benjamin, 141, 257 Thomson, Sir Joseph John, 63, 72, 320 Thomson, William, 33, 504 Tibbetts, Kaitlyn, 891 Torricelli, Evangelista, 499 Tyson, Neil deGrasse, 4

V van Arkel, A. E., 442 van der Waals, Johannes, 525 van’t Hoff, Jacobus Henricus, 654 Villard, Paul, 994 Volta, Alessandro, 940 von Laue, Max, 593

Index of Names

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W Walton, E. T. S., 1014 Wang, Yan, 584 Warner, John, 7 Watson, James D., 439–441, 551 Weir, Andy, 234 Werner, Alfred, 1140



Wilkins, Maurice, 439–441, 551 Williams, Robin, 1192 Wilson, H.R., 441 Windham, Ian, 393 Witte, Kristen, 464 Wrublewski, Donna, 1159

Z Zahn, Sophia, 999 Zeng, T., 443

Index of Names

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I-3

Index and Glossary

Italicized page numbers indicate pages containing illustrations, and those followed by “t” indicate tables. Glossary terms, printed in boldface, are defined here as well as in the text abbreviations, A-9–A-10 absolute temperature scale. See Kelvin temperature scale absolute zero The lowest possible temperature, equivalent to −273.15 °C, used as the zero point of the Kelvin scale, 33 Charles’s law and, 504 zero entropy at, 894 absorbance The negative logarithm of the transmittance, 228–229 concentration and, 230 absorbed dose, 1022 absorption spectrophotometer, 228, 229 absorption spectrum A plot of the intensity of light absorbed by a sample as a function of the wavelength of the light, 228 of chlorophyll a and b, 1236, 1236 spectrophotometric analysis and, 229, 231 absorptivity, molar, 229 abundance of elements, 1039–1040, 1040 of elements in Earth’s crust, 1040t of isotopes, 69t, 70–72 acceptor level, in semiconductor, 601 accuracy The agreement between the measured quantity and the accepted value, 38, 38 and error, 39–40 measurements, 38 precision and, 37–41, 38 acetaldehyde, 700, 1171t, 1177, 1178t, 1180, 1186 acetaminophen, structure of, 1183

acetate ion buffer solution of, 832, 836–837, 840 reaction with phosphoric acid, 792 acetic acid, 1179 bonding in, 469–470 in bread, 1179 buffer solution of, 836 glacial, 817 hydrogen bonding in, 550 ionization of, 147, 158 partial charges in, 810 reaction with sodium bicarbonate, 792–793 reaction with sodium hydroxide, 164, 202 reaction with water, 147, 160 titration with sodium hydroxide, 846–848 as weak acid, 158, 164, 795 as weak electrolyte, 149 acetic anhydride, 201 acetone, 1151, 1176, 1178t hydrogenation of, 436–437 structure of, 1176 acetylacetonate ion, as ligand, 1116 acetylene bond length in, 432 orbital hybridization in, 470–471 structure of, 1152 valence bond description of bonding in, 471 N-acetylglucosamine (NAG), 1215 N-acetylmuramic acid (NAM), 1214, 1215 N-acetyl-p-aminophenol (acetaminophen), 1183 acetylsalicylic acid, 140 acid(s) A substance that, when dissolved in pure water, increases the concentration of hydrogen ions, 156, 156–163, 157t, 777, 831, 1182t. See also Brønsted-Lowry acid(s), Lewis acid(s) acid–base properties of salts, 789–791 anion and salt solubility in, 863

Arrhenius definition for, 158–159 bases and, 156–163 (See also acid–base reaction(s)) bonding and acid–base behavior, 807–812 Brønsted, 161 Brønsted-Lowry definition, 159–161, 777–780 calculations with equilibrium constants, 796–805 carboxylic (See carboxylic acid(s)) common, 157t conjugate acid–base pairs, 779–780 determining molar mass by titration, 224–225 direction of acid–base reactions, 791–793 equilibrium constants for, 783–788 ionization constants for, 788 ions and corresponding, 158t Lewis definition of, 812–816 molecular structure and acid–base behavior, 807–812 naming common, 157–158 oxides of nonmetals and metals, 161–163 pH as concentration scale for, 215–217 polyprotic, 805–807 properties of, 156 properties of ions in aqueous solution, 789t standardizing, 223–224 standardizing by titration, 223–224 strong (See strong acid(s)) types of acid–base reactions, 794–795 water and pH scale, 780–782 weak (See weak acid(s)) acid–base adduct The product of a Lewis acid–base reaction in which the lone pair donated by the Lewis base is shared with an atom in the Lewis acid, 812 acid–base behavior acid strength of hydrogen halides, HX, 807–808

I-4 Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

anions Brønsted bases, 812 and Brønsted acids, 810–812 and carboxylic acids, 810–812 hydrated metal cations as Brønsted acids, 812 oxoacids, 808–810 acid–base indicator(s), 216, 220–222, 508, 832, 851, 852, 975 acid–base reaction(s) An exchange reaction between an acid and a base producing a salt and water, 163–167, 164, 166t, 174–175 Brønsted-Lowry, 163–167 characteristics of, 794t gas-forming, 165–167, 166t net ionic equation for, 165 strong acid with a strong base, 794 strong acid with a weak base, 795 weak acid with a strong base, 795 weak acid with a weak base, 795 acid–base titrations, 221–223, 843–853 determining purity by, 222–223 pH indicators, 851–853 strong acid with strong base, 843–844 weak acid with strong base, 845–848 weak base with strong acid, 849–851 of weak polyprotic acids, 848–849 acidic oxide(s) An oxide of a nonmetal that acts as an acid, 163 acidic solution A solution in which the concentration of hydronium ions is greater than the concentration of hydroxide ions, 935–940 acid ionization constant (Ka) The equilibrium constant for the ionization of an acid in aqueous solution, 787 acidosis metabolic, 872 respiratory, 873 acoustic energy The energy of compression and expansion of spaces between molecules, 20 actinide(s) The series of elements between actinium and rutherfordium in the periodic table, 83, 357, 358, 1105 activation energy (Ea) The minimum amount of energy that must be absorbed by a system to cause it to react, 696, 697 and collision theory, 696–698 and temperature, 698 active site The location on an enzyme where the substrate binds and the reaction occurs, 705, 1213 and enzymes, 1213–1215 and lysozyme, 1213–1215 activity (A) the number of disintegrations observed in a radioactive sample per unit time, 1010, 1022 variation of atmospheric carbon-14, 1012



actual yield The measured amount of product obtained from a chemical reaction, 201 addition polymer(s) A synthetic organic polymer formed by directly joining monomer units, 1185–1188 natural and synthetic rubber, 1188 polyethylene and polyolefins, 1185–1187 addition reaction A reaction in which a compound with a double or triple bond reacts with a material with the general formula X–Y such that X ends up attached to one of the atoms that was originally involved in the double or triple bond and Y ends up attached to the other, 1164 adduct, acid–base, 812 adenine in DNA and RNA, 437 hydrogen bonding to thymine, 438 adenosine 5’-diphosphate (ADP), 1231, 1232, 1234 adenosine 5’-triphosphate (ATP) A ribonucleotide that acts as an energy storage and transport molecule in cells, 1231, 1231–1232, 1233–1235 adhesive force A force of attraction between molecules of two different substances, 567 adhesives chemistry of, 1193 on gecko’s toes, 556 adipoyl chloride, 1190 adrenaline, 1192 adsorption, of gas molecules on solid surface, 944 aerosol propellants, 1253 Agency for Toxic Substances and Disease Registry (ATSDR), 1261 air argon in, 1247–1248 components of, 1246 environmental concerns, 1271–1272 fractional distillation of, 1080 noble gases in, 83, 87, 91, 354, 370, 388–389, 900, 1041 pollution of, 1271–1272 air bags, 509–510 Air Canada Flight 51–52, 52 air pollution, 1271–1272 alanine, structure of, 1210–1211 albite, dissolved by rain water, 234 alchemy, 1054 alcohol(s) Any of a class of organic compounds characterized by the presence of a hydroxyl group bonded to a saturated carbon atom, 287–288, 1170, 1171t, 1182t, A-17 and amines, 1170–1176 and ethers, 1170–1176 methanol as simplest, 1170 miscibility in water, 632

oxidation to carbonyl compounds, 1176 primary, 1177 properties of, 1174–1175, 1174 secondary, 1177 structural isomers of, 1173 tertiary, 1177 uses of, 1174 alcohol tester, 1177 aldehyde(s) Any of a class of organic compounds characterized by the presence of a carbonyl group, in which the carbon atom is bonded to at least one hydrogen atom, 1177–1178, A-17 and ketones, 1178, 1178t naming of, 1179, A-17 and odors, 1178 reduction to primary alcohols, 1177 aldose, 1216 algae bloom from excess phosphate in a pond, 1260 oxygen production by, 1249 phosphates and, 1260 alkali metal ions, 544t alkali metal(s) The metals in Group 1A (1) of the periodic table, 77, 1049–1054 lithium, 1051–1054 potassium compounds, 1051–1054 preparation of sodium and potassium, 1050–1051 properties of sodium and potassium, 1051 reaction with halogens, 1054 reaction with oxygen, 1051 reaction with water, 345, 1054 reactivity of, 1054 sodium, 1051–1054 alkaline batteries, 946–947, 947 alkaline battery A battery that uses the oxidation of zinc and the reduction of MnO2 in a basic (alkaline) environment, 947 alkaline earth elements, 1054–1058 beryllium, 1055 calcium minerals and their applications, 1056–1058 production of magnesium, 1056 properties of magnesium and calcium, 1056 alkaline earth metal(s) The elements in Group 2A (2) of the periodic table, 77 biological uses of, 1057 and biology, 1057 electron configuration of, 77 alkalosis, 872 metabolic, 872 respiratory, 872 alkane(s) Any of a class of hydrocarbons in which each carbon atom is bonded to four other atoms, 1155–1156, 1156, A-15–A-16 chirality in, 1158 cycloalkanes, CnH2n, 1161 derivatives of, 1159, 1173

Index and Glossary

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I-5

alkane(s) (Continued) naming of, 1159–1160, A-15–A-16 properties of, 1160–1161 structural isomers, 1156–1158 Alka-Seltzer tablet, 173 alkene(s) Any of a class of hydrocarbons in which there is at least one carbon– carbon double bond, 1161–1167, A-16 naming of, 1162, A-16 properties of, 1164–1167 alkyl group A part of a molecular structure derived by removing a hydrogen atom from an alkane molecule, 1159 alkyne(s) Any of a class of hydrocarbons in which there is at least one carbon– carbon triple bond, 1161–1167, 1164t, A-16 naming of, A-16 properties of, 1164–1167 allotrope(s) Different forms of the same element that exist in the same physical state under the same conditions of temperature and pressure, 75 alloy A mixture that includes a metal and one or more additional elements, done with the aim of improving the properties of the metal but still retaining the metallic behavior, 606–608, 607 alpha particle(s) A helium nucleus ejected from certain radioactive substances, 1008, 1022 alpha radiation, 994 characteristics of, 994t penetrating ability of, 994 alpha ray(s). See alpha particle(s) aluminum compounds, 1063–1065 and Group 3A (13) elements, 1058–1065 American Chemical Society, 13, 1085 amide(s) Any of a class of organic compounds characterized by the presence of a carbonyl group, in which the carbon atom is bonded to an amino group, 1182–1184 amine(s) A derivative of ammonia in which one or more of the hydrogen atoms are replaced by organic groups, 1175, 1175 primary, 1175 properties of, 1175–1176 secondary, 1175 tertiary, 1175 amino acid An organic molecule that contains an amino functional group (ONH2) and a carboxylic acid functional group (OCOOH), 1208 amino group A functional group related to ammonia, in which some or all of the hydrogen atoms are replaced by organic groups, 1208, 1210

I-6

ammonia, 159, 1072–1073, 1278–1279 hydrogen and, 1278–1279 molecular geometry of, 420 oxidation of, 196 reaction with silver chloride in aqueous solution, 755 amorphous solid(s) A solid that lacks long-range regular structure and displays a melting range instead of a specific melting point, 604, 604–606 and crystalline solids, 605t amount, of pure substance, 10. See also mole (mol) amount of substance and thermodynamic calculations, 897 amounts table A table used in solving stoichiometry problems that summarizes the mole relationships of reactants and products in a reaction, 193, 197, 199 ampere (A) The unit of electric current, 973–974 amphiprotic substance A substance that can behave as either a Brønsted acid or a Brønsted base, 161, 779 amphoteric metal hydroxides, 814t amphoteric substance A substance, such as a metal hydroxide, that can behave as either an acid or base, 814 amplitude The height of a wave, as measured from the axis of propagation, 306 analysis, chemical. See chemical analysis analytical laboratory balance and significant figures, 43 ångstrom units, 35 anhydrous compound The substance remaining after the water has been removed (usually by heating) from a hydrated compound, 98 anion(s) An ion with a negative electric charge, 91, 360–361 electron configurations, 360–361 and Lewis structures, 402 as Brønsted bases, 812 as ligands, 1116 monatomic, 91, 92 oxoacids and, 402 oxoanions, 92 polyatomic, 92, 398 and salt solubility, 861–863 and salt solubility in acid, 863 in seawater, 1258t anode The electrode of an electrochemical cell at which oxidation occurs, 941 of lead storage battery, 948 of lithium battery, 949 reactions in iron corrosion, 975 sacrificial, 978 anodic inhibition The process of painting or allowing a thin oxide film to form on a metal surface to prevent corrosion, 977

antibonding molecular orbital A molecular orbital in which the energy of the electrons is higher than that of the parent orbital electrons, 474 anticodon The three-nucleotide sequence in tRNA that is complementary to a codon in mRNA, 1223 antilogarithm, A-3 antimony, metalloid, 71, 71 antineutrinos, 999 applications and calcium minerals, 1056–1058 of nuclear chemistry, 1023–1027 aqua sphere (water), 1256–1263 aqueous equilibria acid–base titrations, 843–853 buffers, 831–842 equilibria involving complex ions, 867–871 precipitation reactions, 863–867 solubility of salts, 853–863 aqueous solution A solution in which the solvent is water, 147–151, 149 classifying reactions in, 173–176 electrolysis of, 969–973 ions in, 147–149 molecules in, 147–149 solubility of ionic compounds in water, 149–151 standard reduction potentials in, 956t, A-32–A-35 stoichiometry of reactions in, 217–219, 220–227 aqueous weak acids ionization constants for, 785t, A-20–A21 aqueous weak bases ionization constants for, 785t, A-22 aragonite (CaCO3), 1039 in seawater, 1276 argon, 118–119 aromatic compound(s) Any of a class of hydrocarbons characterized by the presence of a benzene ring or related structure, 1167–1170, 1167t benzene derivatives, 1168–1169 properties of, 1169–1170 structure of benzene, 1168 aromatic rings, 1168 Arrhenius definition for acids and bases, 158–159 Arrhenius equation A mathematical expression that relates the rate constant and temperature, 699–701 Arrhenius plot, 700 artificial nuclear reactions, 1014–1017 search for new elements, 1016–1017 asbestos, 1067, 1068 aspartame (C14H18N2O5), 1167 aspirin, 1181 molecular formula of, 104

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

atmosphere, 1246–1256, 1246t. See also air carbon dioxide, CO2, 1253–1254 CO2 levels, 1273 methane, 1255 methane, CH4, 1254–1256 methane levels, 1255 nitrogen and nitrogen oxides, 1248–1249 NO, 1272 oxygen, 1249–1250 ozone, 1250–1253 atom(s) The smallest particle of an element that retains the characteristic chemical properties of that element, 13 Bohr theory and spectra of excited, 316–319 components of, 72–74 core electrons of, 352 electron configurations of, 350–360 formal charge of, 410 and mass, 102–103 and molar mass, 100–103 in molecular models, 86 and molecules, 99–106 and moles, 102–103 nature of, 72–74 sharing at cube corners and faces, 585 size of, 364–367 subshell energies and filling order in multielectron, 348 atom economy, 232 atomic bomb tests, 1022 atomic composition, 65 atomic line spectra and Niels Bohr, 312–319 atomic mass The experimentally determined mass of an atom of one isotope, 63–67 determining, 68 relative, 64 atomic mass unit (u) A non-SI unit of a scale of relative atomic masses of the elements; 1 u = 1/12 of the mass of a carbon atom with six protons and six neutrons, 64 atomic nuclei band of stability and radioactive decay, 1002–1003 nuclear binding energy, 1003–1005 stability of, 1000–1005 atomic number (Z) The number of protons in the nucleus of an atom of an element, 64 of beta particles, 995 even, 1006 odd, 1006 variation in orbital energies, 359 atomic orbital(s) The matter wave for an allowed energy state of an electron in an atom, 323, 325–329, 327, 474 d orbitals, 327, 327–328 f orbitals, 328, 328 p orbitals, 326–327, 327 shapes of, 325–329 s orbitals, 325–326

atomic properties and periodic trends, 364–374 atomic radii, 365–367 atomic size, 364–367 atomic structure, 63–64, 64 atomic line spectra and Niels Bohr, 312–319 electromagnetic radiation, 305–308 electronic structure, 321–325 electron spin, 330 modern view of, 321–325 quantization, 308–312 shapes of atomic orbitals, 325–329 wave or quantum mechanics, 321–325 wave–particle duality, 319–321 atomic subshell energies effective nuclear charge, Z*, 348–350 and electron assignments, 347–350 order of, 347–348 atomic theory of matter A theory that describes the structure and behavior of substances in terms of ultimate chemical particles called atoms and molecules, 8–9 atomic weight The average mass of an atom in a natural sample of the element, 67–74, 69t atomic mass, determining, 68 calculating, from isotope abundance, 70–71 isotopic abundances and, 68, 70 ATP, and energy, 1231–1232 Aufbau principle The “building up” principle, which states that electrons fill the orbitals with the lowest energy level first, 347 autoionization of water Proton transfer between two water molecules to produce a hydronium ion and a hydroxide ion, 780–782 autoionization reaction, 780 Avogadro’s hypothesis Equal volumes of gases under the same conditions of temperature and pressure have equal numbers of particles, 506–508 Avogadro’s number The number of particles in one mole of any substance (6.02214076 × 1023), 100 axis of propagation A line representing the average height of a wave and defined as having a height of zero, 306 azurite, 859, 1105, 1137 bacon fat and addition reactions, 1165 balanced chemical equation A chemical equation showing the relative amounts of reactants and products, 139, 142–144 in acid solution, 935–938 in basic solution, 939–940 ball-and-stick model A way of representing a molecule in which spheres of different colors represent the atoms, and sticks represent the bonds holding them together, 86

Balmer equation An equation that reproduces the wavelengths of the visible portion of the emission spectrum of hydrogen atoms, 313 Balmer series A series of lines in the visible portion of the emission spectrum of hydrogen atoms that correspond to the electron moving from higher energy levels down to the second energy level, 313, 318, 318 band gap, 600–602 band of stability The narrow range of neutron-to-proton ratios that lead to stable nuclei, 1001, 1002–1003 band theory The theory of metals that treats the electron energy states as overlapping in a continuous band, 597 and electrical conductivity in metals, 597–599 for insulators, 599 for metals, 599 for semiconductors, 599 bar A unit of pressure; 1 bar = 100 kPa, 500, A-7 barium sulfate, 856 barometer An apparatus used to measure atmospheric pressure, 499 base(s) A substance that, when dissolved in pure water, increases the concentration of hydroxide ions, 156, 156–163, 157t, 1220. See also Brønsted base(s), Lewis base(s) acid–base properties of salts, 789–791 acids and, 156–163 (See also acid–base reaction(s)) Arrhenius definition for, 158–159 Brønsted, 161 Brønsted-Lowry definition, 159–161, 777–780 calculations with equilibrium constants, 796–805 common, 157t conjugate acid–base pairs, 779–780 equilibrium constants for, 783–788 Lewis concept of acids and bases, 812–816 molecular structure, bonding and acid– base behavior, 807–812 pH as concentration scale for, 215–217 polyprotic, 805–807 predicting the direction of acid–base reactions, 791–793 properties of, 156 properties of ions in aqueous solution, 789t standardizing, 223–224 strong (See strong base(s)) titration of acid in aqueous solution with, 220 types of acid–base reactions, 794–795 water and pH scale, 780–782 weak (See weak base(s)) base ionization constant (Kb) The equilibrium constant for the ionization of a base in aqueous solution, 787 Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I-7

base pairs, 1221 basic oxide(s) An oxide of a metal that acts as a base, 163 basic oxoanions, 812t basic solution A solution in which the concentration of hydronium ions is less than the concentration of hydroxide ions, 935–940 battery A device consisting of two or more electrochemical cells, 946 alkaline, 946–947 dry cells, 946–947 lead storage, 948 lithium, 949–950 nickel-cadmium, 949 nickel-metal hydride, 949 primary, 946–947 rechargeable, 947–950 secondary, 947–950 becquerel The SI unit of radioactivity, 1 decomposition per second, 1022 Beer–Lambert law The absorbance of a sample is proportional to the path length and the concentration, 228–229 Bell Laboratories, 36 benzene derivatives of, 1168–1169, A-17 and pi bonding, 472–473 structure of, 1168 valence bond theory, 472–473 beryllium (Be), 353, 353, 1055 beta particle(s) An electron ejected from certain radioactive substances, 994 beta radiation, 994 characteristics of, 994t penetrating ability of, 994 beta rays. See beta particle(s) bidentate ligand A ligand that coordinates to a metal ion by means of two atoms and therefore two coordinate covalent bonds, 1116 bidentate ligands, 1116, 1116, 1117 Big Bang theory, 1006–1007 bilayer structure The arrangement of phospholipids in water and in cell membranes where the non-polar hydrocarbon chains (tails) associate in the interior structure away from the water, and the polar phosphate groups (heads) position themselves at the exterior of the structure near the water, 1227 bimolecular process An elementary step involving two molecules (or ions, atoms, or free radicals), 708 binary compound(s) A compound formed from two elements, 86 binary molecular compounds, 87–88 binding energy The energy required to separate a nucleus into individual protons and neutrons, 1003–1005 nuclear, 1003–1005 per mole of nucleons, 1004

I-8

biochemistry An area of study that draws from biology and chemistry to focus on molecules involved in biological processes and on how chemical reactions occur in an organism, 1207 carbohydrates, 1216–1219 defined, 551, 1207 lipids and cell membranes, 1225–1230 metabolism, 1230–1236 metals in, 376 nucleic acids, 1219–1225 proteins, 1208–1215 biodiesel An alternative to petroleumbased fuels used in diesel engines that is made from a mixture of esters of fatty acids, 1280 bus, 1281 bioethanol Ethanol made from fermenting materials derived from biological sources, typically used as a gasoline additive, 1280 biofuels, 1279–1281 biology alkaline earth metals and, 1057 dimensions in, 36 bituminous coal, 1269 black gunpowder, 286–287, 287 black smokers metal sulfides from, 178 and volcanoes, 178 blast furnace, 1112 molten iron being poured from a basic oxygen, 1112 bleach, 232–233 blue apatite, 1057 body-centered cubic (bcc) unit cell A unit cell consisting of the primitive cubic unit cell with an additional particle positioned at the center of the cube that is identical to the particles positioned at the corners of the cube, 583 Bohr, Niels atomic line spectra and, 312–319 Bohr model, 315 of the hydrogen atom, 313–314, 315 and the spectra of excited atoms, 316–319 boiling, 564 boiling point The temperature at which the vapor pressure of a liquid is equal to the external pressure on the liquid, 547t, 560, 564 for common substances, 559t elevation, 643–645 elevation and freezing point depression constants, 644t of liquids, 564–565 Boltzmann’s constant The proportionality constant that relates the entropy (S) of a system to the natural logarithm of the number of microstates (W) in the relationship, S = k lnW, 892

bond(s) An interaction between two or more atoms that holds them together by reducing the potential energy of their electrons energies, 434–437 bond angle The angle between two atoms bonded to a central atom, 420 bond-dissociation enthalpy The enthalpy change for breaking a bond in a molecule, with the reactants and products in the gas phase at standard conditions, 434–437, 435t, 715 calculating enthalpy of reaction from bond energies, 434–437 of halogen compounds, 1085 bonding, 387 in acetic acid, 469–470 in acetylene, 471 acid strength of hydrogen halides, HX, 807–808 anions as Brønsted bases, 812 in benzene, 472 bond properties, 432–437 and Brønsted acids, 810–812 and carboxylic acids, 810–812 deoxyribonucleic acid (DNA), 387, 437–441, 438, 1219–1221 electronegativity and bond polarity, 390–393 electron pair, 389. See also bond pair(s). formal charges in covalent molecules and ions, 410–416 in formaldehyde, 469 HNO2, 808–810 HNO3, 808–810 hydrated metal cations as Brønsted acids, 812 in ionic compounds, 594–596 Lewis electron dot symbols and chemical bond formation, 388–389 Lewis structures of molecules and polyatomic ions, 393–410 in metals, 596–602 in methanol, 464–465 molecular polarity, 424–431 molecular shapes, 416–424 and octet rule, 406–410 orbital overlap model of, 459–461 oxoacids, 808–810 in ozone, 482 resonance, 404–406 in semiconductors, 596–602 molecular orbital model for ozone, 482 bonding molecular orbital A molecular orbital in which the energy of the electrons is lower than that of the parent orbital electrons, 474 bond length The distance between the nuclei of two bonded atoms, 432–433, 433t bond order The number of bonding electron pairs shared by two atoms in a molecule, 432

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

bond pair(s) Two electrons, shared by two atoms, that contribute to the bonding attraction between the atoms, 394 bond polarity, 390–393 bond properties bond-dissociation enthalpy (D), 434–437 bond length, 432–433 bond order, 432–433, 475–483 borax (mineral source), 1062 Born–Haber cycle A thermodynamic cycle based on Hess’s law and used to determine the lattice enthalpy of ionic solids, 595 boron (B), 353 compounds, 1061–1063 and Group 3A (13) elements, 1058–1065 and other elements of Group 3A (13), 353, 353 boron carbide in fireworks, 1062 boron halides (molecular compounds), 1042 boron icosahedron, 1060 boron minerals and production of element, 1059–1060 boundary surface The depiction of an atomic orbital as a region centered on the nucleus that is drawn in such a way that there is a 90% probability of finding the electron somewhere inside the sphere, 326 Boyle’s law A law stating that the pressure and volume of a gas are inversely proportional (under conditions of constant pressure and amount of gas), 502, 502, 505–506 bromine, 81 bromobenzene, 115 Brønsted-Lowry acid(s) A proton donor, 159–161, 777–780 Brønsted-Lowry acid–base reactions, 163–167 Brønsted-Lowry base(s) A proton acceptor, 159–161, 777–780 buffer capacity, 838 buffer solutions(s) A solution that resists a change in pH when hydroxide or hydronium ions are added, 831–842, 832, 836t determining pH of, 832–836 general expressions for concentration of hydronium ions or pH of buffer solutions, 836–838 and maintaining pH, 840–842 preparing solutions, 832, 836, 838–840 Bureau Intérnational des Poids et Mésures (BIPM, International Bureau of Weights and Measures), 34, 100 butene isomers, properties of, 1162t



calcite in seawater, 1276 calcium, 1055 minerals and applications, 1056–1058 properties of, 1056 calcium carbonate (CaCO3) decomposition, 909 calibration curve, 230 calibration plot In spectrophotometric analysis, a plot of absorbance as a function of concentration for a series of standard solutions whose concentrations are accurately known, also known as a calibration curve, 230 caloric fluid, 257 calorie (cal) The quantity of energy required to raise the temperature of 1.00 g of pure liquid water from 14.5 °C to 15.5 °C, 37, A-8t calorimeter, 274 constant-pressure (coffee-cup), 274 constant-volume (bomb), 277 calorimetry The experimental determination of the energy transferred as heat in reactions, 274, 274, 274–278 constant-pressure calorimetry, measuring ∆H, 274–276 constant-volume calorimetry, measuring ∆U, 276–278 capillary action The spontaneous movement of a liquid up a narrow tube, 566–567 carbohydrate(s) An aldehyde or ketone containing multiple hydroxyl groups; sugar(s), 1216–1219 disaccharides, 1217–1218 monosaccharides, 1216–1217 polysaccharides, 1218–1219 carbon (C) alcohols, ethers, and amines, 1170–1176 allotropes of, 80 bonding patterns, 1152 compounds, 1154–1155 compounds with a carbonyl group, 1176–1184 hydrocarbons, 1155–1170 importance of, 1151–1155 isomers, 1153–1154 nanotubes, 614–615 and organic compounds, 402–403 and other elements of Group 4A (14), 353 polymers, 1184–1191 stability of carbon compounds, 1154–1155 structural diversity, 1152 carbon-14 as a tracer, 1025–1026 carbon cycle, 1254 carbon dating (See radiocarbon dating) carbon dioxide (CO2), 177–178, 1253–1254 in the oceans, 21–22 phase diagrams, 613 carbonyl compounds, 1176

carbonyl group The functional group consisting of a carbon atom doubly bonded to an oxygen atom, 1176 aldehydes, 1177–1178 amides, 1182–1184 carboxylic acids, 1178–1180 and compounds, 1176–1184 esters, 1181–1182 ketones, 1177–1178 carboxylic acid(s) Any of a class of organic compounds characterized by the presence of a carbonyl group, in which the carbon atom is bonded to another carbon atom and to a hydroxyl group, 810–812, 1178–1180, 1180t, A-17 naming of, 1179 carotene, 1164 CAS database, 13 catalyst(s) A substance that increases the rate of a reaction while not being consumed in the reaction, 679, 702–706, 1165 and chemical reactions, 673 in chemical reactions, 681 enzymes, 705–706 and reaction rate, 702–705 catalytic converter, 1272 catalyzed decomposition, 679 cathode rays, 72 cathode The electrode of an electrochemical cell at which reduction occurs, 941 reactions in iron corrosion, 975 cathodic protection A method of inhibiting metal oxidation by forcing the metal to become a cathode instead of an anode in an electrochemical cell by connecting it with another metal that is more easily corroded, 978 cation(s) An ion with a positive electrical charge, 89–91, 360–361 electron configurations, 360–361 in seawater, 1258t cave chemistry, 145 cell membranes, 1225–1230 cells electrolytic, 968–973 fuel, 950–951 voltaic (or galvanic), 940–945, 946–951, 960–963 Celsius temperature scale A scale defined by the freezing and boiling points of pure water, defined as 0 °C and 100 °C, 32–33, 33 central science, 3 cerium, 375 cesium chloride (CsCl) unit cell, 589, 589 chain reaction A reaction sequence where a reaction product acts as a reactant in a reaction occurring earlier in the sequence, 716, 1018 chalcogens, 81

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I-9

change chemical, 18–19, 968–973 chemistry and, 4 physical, 18–19 properties and, 15–19 change of state, 258, 265–266 defined, 262 energy and, 262–266 characteristic The part of a logarithm to the left of the decimal point, A-3 charge distribution, 410 identifying, on transition metal cations, 95 charge distribution The way in which electrons are distributed in a molecule, 410 Charles’s law A law stating that the volume and absolute temperature of a gas are directly proportional (under conditions of constant pressure and moles of gas, 504, 504, 504t, 505–506 chelating ligand A ligand that forms more than one coordinate covalent bond with the central metal ion in a complex, 1116 chemical analysis The determination of the amounts or identities of the components of a mixture, 106 chemical equations and, 202–208 defined, 106 determining compound formulas, 106–114 determining formula from mass data, 111–114 empirical formula and, 204 empirical formulas from percent composition, 108–111 molecular formulas from percent composition, 108–111 percent composition, 106–108 titration as method of, 220–221 using spectrophotometry in, 230–231 chemical bond formation, 388–389 Lewis electron dot symbols and, 388–389 and valence electrons, 388–389 chemical bonds, 13, 387, 389. See bond(s), bonding theories of, 483–485 chemical change(s) A change that involves the transformation of one or more substances into one or more different substances, 18–19 See also reaction(s) chemical compound(s). See compound(s) chemical energy The energy stored in chemical compounds, 20, 20, 256 in chemical reaction, 273

I-10

chemical equation(s) A written representation of a chemical reaction, showing the reactants and products, their physical states, and the direction in which the reaction proceeds, 19, 139–142, 754–756 balancing, 142–144, 934–940 and chemical analysis, 202–208 chemical equilibrium, 145–147, 737–738 addition or removal of a reactant or product, 757–759 chemical equilibrium, See equilibrium. disturbing, 757–762 equilibrium constants, 738–748, 754–756 and Gibbs free energy ∆G, 903–905 reaction quotient, 738–745 temperature changes and equilibrium composition, 760–762 using equilibrium constants in calculations, 748–753 volume changes and gas-phase equilibria, 759–760 chemical formula A sequence of chemical symbols that indicates the component atoms of a compound and their relative proportions, 14 chemical kinetics The study of the rates of chemical reactions under various conditions and of reaction mechanisms, 673 catalysts, 702–706 concentration–time relationships, 686–694 defined, 673 effect of concentration on reaction rate, 680–685 integrated rate laws, 686–694 interplay between thermodynamics and, 914–915 microscopic view of reaction rates, 694–701 rates of chemical reactions, 673–678 reaction conditions and rate, 678–679 reaction mechanisms, 706–716 chemical property A property that indicates whether, and how readily, a substance will undergo a chemical change in reaction with another substance, 19 of hydrogen, 1046–1047 periodic trends and, 374–375 chemical reaction(s). See reaction(s) chemistry The scientific study of the composition, structure, and properties of matter and the changes in both composition and energy that matter undergoes during reactions, 3 and change, 4 defined, 3 dimensions in, 36 and its methods, 3–6 chemistry, green, 7–8, 1281–1282

chiral compound A molecule that is not superimposable on its mirror image, 1124, 1154. See also enantiomers chirality and death, 1154 and elephants, 1155 chiral metal complexes, 1126 chlorine, 13, 1082–1083 electrolysis of molten NaCl, 969 in fireworks, 1087 oxoacids of, 1086–1087 preparation, 1083 reaction with sodium, 5 in water treatment, 1259 chlorine compounds, 1086–1087 hydrogen chloride, 1086 oxoacids of chlorine, 1086–1087 cinnabar, 4, 4 cis–trans isomers Isomers involving either octahedral or square planar complexes in which the two groups of interest are directly across from each other (trans) or in adjacent positions at 90° (cis) or in compounds containing carbon–carbon double bonds in which the two groups of interest are on opposite sides of the double bond (trans) or on the same side of the double bond (cis), 471–472, 1123, 1153, 1162 Clausius–Clapeyron equation The value expressing the linear relationship produced when the reciprocal of the Kelvin temperature (1/T) is plotted as a function of the logarithm of the vapor pressure (ln P), 563–564 clay, 80, 1068 climate change The gradual warming of the surface air and oceans of the Earth caused by the accumulation of greenhouse gases in the atmosphere and the consequences of that warming, 4, 287, 1272–1274 and water resources, 1274 closed-circuit breathing apparatus, 1051 coal, 1268–1269 mercury from, 1268 types of, 1268t codon A three-nucleotide sequence in mRNA that corresponds to a particular amino acid in protein synthesis, 1223, 1223t coffee-cup calorimeter, 274, 275–276 cohesive force A force of attraction between molecules of a single substance, 567 coke (reducing agent), 1111 colligative properties The properties of a solution that depend only on the number of solute particles per solvent molecule, 627, 640–655 boiling point elevation, 643–645

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

changes in vapor pressure, 640–642 freezing point depression, 645–647 and molar mass determination, 651–653 osmotic pressure, 647–651 Raoult’s law, 640–642 of solutions containing ions, 653–655 collision theory A theory of reaction rates that assumes that molecules must collide in order to react, 695–696 and activation energy, 696–698 and concentration, 695–696 and molecular orientation on reaction rate, 699 and reaction rate, 695–696 and temperature, 698 colloid(s) A state of matter intermediate between a solution and a suspension, in which solute particles are large enough to scatter light but too small to settle out, 655–659, 656t surfactants, 658–659 types of, 657–658 colors color disks to analyze, 1134 of coordination compounds, 1134–1135 in fireworks, 331–332 primary, 1134, 1134 combined gas law, 506 combustion analysis A method of chemical analysis in which the empirical formula of an unknown substance is determined through combustion, 204–208 empirical formula of compound containing C, H, and O, 207–208 empirical formula of hydrocarbon, 205–206 of hydrocarbon, 205 combustion reaction The reaction of a compound with molecular oxygen to form products in which all elements are combined with oxygen, 142, 143 balancing an equation for, 143–144 of hydrogen gas, 887 Commission on Isotopic Abundances and Atomic Weights (CIAAW), 70 common ion effect The limiting of acid (or base) ionization caused by addition of its conjugate base (or conjugate acid) or of salt solubility by addition of anther salt containing one of its ions, 832, 859–861 common logarithms, A-2–A-3 common substituent groups in organic compounds, A-16t complementary strands Two nucleic acid strands where each nitrogenous base in one strand can be paired with the complementary base on the other strand, 1221 complete ionic equation For an exchange reaction in aqueous solution, the equation that includes all of the dissociated ions, including spectator ions, 154

complex(es), 1114–1117. See also coordination complex(es); coordination compound(s) complex ions, 813, 867–871, 868, 1114 and equilibria, 867–871 and solubility, 869–871 compound formulas, determining, 106–117 compound(s) Matter that is composed of two or more kinds of atoms chemically combined in definite proportions, 13–14 aluminum, 1063–1065 aromatic, 1167–1170 in biochemical oxidation–reduction reactions, 1232–1233 boron, 1061–1063 carbon, 1154–1155 and carbonyl group, 1176–1184 chlorine, 1086–1087 coordination, 1114–1122 electron-deficient, 406–407 elements and, 14 fluorine, 1084–1086 hypervalent, 407–409 with indefinite proportions, 14 intensive property used to distinguish, 17 measuring concentrations in solution, 209–215 and molar mass, 103–106 and molecules, 103–106 nitrogen, 1072–1076 paramagnetism of, 362 properties of, 13 specific heat capacities of some, 259 sulfur, 1081–1082 xenon, 1089 compressibility The change in volume with change in pressure, 502–503 concentration(s) The amount of solute dissolved in a given amount of solution absorbance and, 230 molality (m), 628 molarity (M), 209–212 mole fraction (X), 628 ppm (parts per million), 629 units of, 627–630 weight percent, 628 concentration–time relationships first-order reactions, 686–688 graphical methods for determining rate constant, 690 graphical methods for determining reaction order, 690 half-life and first-order reactions, 690–694 second-order reactions, 688–689 zero-order reactions, 689–690 concentrations of CO2 at Mauna Loa, Hawaii, 1254 condensation The state change from gas to liquid, 558 of liquids, 558–561

condensation polymer(s) An organic polymer formed by combining monomer units in such a way that a small molecule, usually water, is eliminated, 1185, 1188–1191 polyamides, 1190–1191 polyesters, 1189–1190 condensation reaction(s) A reaction in which parts of the reactant molecules bond together and a small molecule, usually water, is eliminated, 1188 condensed formula A variation of a molecular formula that shows groups of atoms, 85 condition(s), standard. See standard state conduction band, 600 conductivity The extent to which a solution or material conducts electricity, 148 conjugate acid–base pair(s) A pair of compounds or ions that differ by the presence of one hydrogen ion, 779–780 conjugate acid ionization constants for, 788 conjugate base ionization constants for, 788 constant-pressure (coffee-cup) calorimeter, 274 constant-pressure calorimetry, 274–276 constant-volume (bomb) calorimeter, 277 constant-volume calorimetry, 276–278 continuous spectrum The spectrum of white light emitted by a heated object, consisting of light of all wavelengths, 312 control, as goal of science, 6 conversion factor(s) A multiplier that relates the desired unit to the starting unit, 32, 47 common, 32 and dimensional analysis, 47–48 coordinate covalent bond(s) Interatomic attraction resulting from the sharing of a lone pair of electrons from one atom with another atom, 407, 1116 coordination chemistry, 1114 coordination complex(es) A species in which a metal ion or atom is bonded to one or more molecules or anions to compose a structural unit, 813–814, 1114, 1116, 1136 coordination compound(s) A compound containing a coordination complex, 1114–1122 bonding in, 1128–1133 colors of, 1133–1137 common coordination geometries, 1122 complexes and ligands, 1114–1117 d orbitals, 1128–1129 Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I-11

coordination compound(s) (Continued) electron configurations and magnetic properties, 1130–1133 formulas of, 1117–1120 isomerism, 1122–1127 ligand field theory, 1128–1129 naming of, 1120–1122 of Ni2+ ion, 1115 spectrochemical series, 1135–1137 structures of, 1112–1127 coordination geometry The arrangement in space of the central metal atom or ion in a coordination compound and the ligands attached to it, 1116, 1122 coordination isomerism, 1123 coordination isomers Two or more complexes in which a coordinated ligand and a noncoordinated counterion are exchanged, 1123 coordination number The number of ligands attached to the central metal atom or ion in a coordination compound, 1116 copolymer A polymer formed by combining two or more different monomers, 1188 copper metallurgy, 1113 production, 1113–1114 core electrons The electrons in an atom’s completed set of shells, 352, 388 corrosion The deterioration of metals by oxidation–reduction reactions, 975–978 of aluminum, 1060 economic costs of, 975 electrochemical process, 975–977 in environment, 975–978 inhibitor, 977 of iron, 975, 976 protecting metal surfaces from, 977–978 corundum, 1063 Coulomb’s law The force of attraction between oppositely charged ions of an ionic compound is directly proportional to their charges and inversely proportional to the square of the distance between them, 97, 98, 313, 544 and electrostatic forces, 98 coulomb (C) The quantity of charge that passes a point in an electric circuit when a current of 1 ampere flows for 1 second, 1022 counterion(s) An ion in a compound that is present for charge balance, 1119–1120, 1123 covalent bond(s) An interatomic attraction resulting from the sharing of electrons between atoms, 389, 461 covalent compound(s) A compound formed by atoms that are covalently bonded to each other, 1041

I-12

covalent molecules formal charge, 416 formal charges in equivalent resonance structures, 412–413 importance of structures using formal charge, 414–416 Lewis structures, 393–409 polarity, 424–431 shapes, 416–424 covalent to ionic bonding continuum, 391 COVID-19 pandemic, 4, 232, 233, 1228–1229 CRISPR-Cas9 Complex, 1207 critical point The upper end of the curve of vapor pressure versus temperature, 565 critical pressure The pressure at the critical point, 565 critical temperature The temperature at the critical point; above this temperature the vapor cannot be liquefied at any pressure, 565 for CO2, 565 for common compounds, 566t of liquids, 565–566 crystal lattice A solid, regular array of particles, 98, 581–588 cubic unit cells, 583–584 lattice points of, 583 number of atoms per cubic unit cell, 584–588 unit cells for metals, 584–588 crystalline solids, 604, 605t cubic close-packed (ccp) unit cell, 588 cubic hole The empty space that exists between the packing of atoms in a primitive cubic unit cell, 589 cubic unit cell A unit cell having eight identical points at the corners of a cube, 583–584 curie A unit of radioactivity, 1022 curium, 12 current Referring to electricity, the amount of charge flowing per unit time, 973 cyanobacteria, 1249 cycloalkane A hydrocarbon in which the carbon atoms are connected by only COC single bonds and which contains a ring of carbon atoms, 1161 cycloalkenes, 1164 Dalton’s law of partial pressures The total pressure of a mixture of gases is the sum of the pressures of the components of the mixture, 514, 515 d-block elements The group of elements coinciding with the filling of d orbitals, 1105 de Broglie’s equation, 320–321 Debye forces. See dipole–induced dipole attraction. deciliter (dL), 36

decomposition reaction(s) A reaction in which a compound is broken down into smaller constituents, 174 degenerate, 347 degenerate orbital(s) Electron orbitals in an atom that have the same energy, 347 density The ratio of the mass of an object to its volume, 15, 16, 1110 deoxyribonucleic acid (DNA) A biopolymer with monomers consisting of deoxyribose sugar molecules linked by phosphodiester groups and having nitrogenous bases attached that is responsible for encoding genetic information, 1219 base pairs in, 1221 complementary strands in, 1221 discovery of double helix in, 439–441 hydrogen bonding in, 551 replication of, 1221–1222 structure of, 1220 deposition The process that occurs when a gas changes directly to a solid, 611, 611 detergent A surfactant used for cleaning, 658 and surface tension of water, 659, 660 determinate errors, 40 deuterium (D), 66 diamagnetic, 362 diamagnetism The physical property of being repelled by a magnetic field, 361–363 diamond, 13, 600, 603 diatomic elements, 80 diborane (B2H6), 484, 1062–1063 diene(s) A hydrocarbon containing two double bonds, 1164 dietary Calorie (Cal), 37 diethyl ketone, 1178t diffusion The gradual mixing of the molecules of two or more substances by random molecular motion, 521–524 digit term, 42 dilemmas, in science, 6 dilutions preparing solution by, 213–214 serial, 215 dimensional analysis A general problem solving approach that uses the dimensions or units of each value to guide you through calculations, 47 conversion factors and, 47–48 problem solving by, 47–48 dimethyl ether condensed formula, 85 molecular formula, 85 molecular models, 85 structural formula, 85 dipolar bond. See polar covalent bond

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

dipole, 425 dipole–dipole attraction The electrostatic force between two neutral molecules that have permanent dipole moments, 546 dipole–dipole forces (interaction), 546–547 dipole–induced dipole attraction (Debye forces) The electrostatic force between two neutral molecules, one having a permanent dipole and the other having an induced dipole, 553 dipole moment The product of the magnitude of the partial charges in a molecule and the distance by which they are separated, 425, 425t disaccharide Two sugar molecules connected to each other by a bridging oxygen molecule, 1217–1218 discovery of neutrons, 1015 dispersion forces Intermolecular attractions involving induced dipoles, 553–555 disproportionation reaction, 1086 dissociation The process by which a substance is broken into smaller components, 147 distances on molecular scale, 35 DNA. See deoxyribonucleic acid (DNA) d orbital(s), 327, 327–328, 1128, 1128–1129. See also atomic orbital(s) double bond A bond formed by sharing two pairs of electrons, one pair in a sigma bond and the other in a pi bond, 394 double displacement reaction(s). See exchange reaction(s) drought, 1257 dry cell batteries, 946–947 d-to-d transition The excitation of an electron from one d orbital in a metal to a d orbital that has a different energy due to the presence of a ligand, 1135 dynamic equilibrium The condition of a reaction system in which both the forward and reverse reactions occur at equal rates such that the net concentrations of all substances remain constant, 146, 561 dynamite, 1172 École Supérieure de Physique et de Chimie Industrielles de la Ville de Paris (ESPCI), 82 EDTA42 (hexadentate ligand), 1117 effective nuclear charge (Z*) The nuclear charge experienced by an electron in a multielectron atom, as modified by the other electrons, 348–350, 349t, 350 effusion The movement of gas molecules through a membrane or other porous barrier by random molecular motion, 521 gases, 521–524

Eiffel Tower (Paris, France), 42, 42 eka-silicon, 78 elastomer(s) An organic polymer with very high elasticity, 1188 electrical conductivity in metals, 599 electrical energy The energy due to the motion of electrons in a conductor, 20 electrochemical cell(s) A device that produces an electric current as a result of an electron transfer reaction, 942 and Nernst equation, 960–963 under nonstandard conditions, 960–963 electrochemical cell notations, 945 electrochemical processes, 941 electrochemical units, 951 electrochemistry, 964–967 E° and equilibrium constant, 965–967 work and free energy, 964–965 electrode(s) A device such as a metal plate or wire for conducting electrons into and out of solutions in electrochemical cells, 148 electrolysis The use of electrical energy to produce chemical change, 940, 968, 968, 968–973 of aqueous NaI, 970 of aqueous solutions, 969–973 of aqueous tin(II) chloride, 971 of molten salts, 968–969 using electrical energy, 968–973 of water, 893, 940, 965, 965–967 electrolyte(s) A substance that dissociates or ionizes in water to form an electrically conducting solution, 148 electromagnetic radiation Radiation that consists of wave-like electric and magnetic fields, including light, microwaves, radio signals, and X-rays, 305–308, 306 electromagnetic spectrum, 307 electromotive force (emf) The difference in potential (voltage) between two electrodes in an electrochemical cell (with zero current), 751, 951–952 electron(s) (e2) A negatively charged subatomic particle found in the space about the nucleus, 71, 973–975 assigning quantum numbers to, 355 diffraction of, 320 molecules with odd number of, 409–410 number in electron shells and subshells, 347t octet of, 388 wave nature of, 320 electron affinity The internal energy change occuring when an anion loses an electron, 369–370 and electron attachment enthalpy, 369 and sign conventions, 369 electron assignments atomic subshell energies and, 347–350

electron attachment enthalpy The enthalpy change occurring when an atom of the element in the gas phase gains an electron, 369–370, 370 and electron affinity, 369 for the halogens, 370 and sign conventions, 369 electron capture A nuclear process in which an inner-shell electron is captured, 998 electron configurations, 351t, 359, 1107 anions and cations, 360–361 of atoms, 350–360 of coordination compounds, 1130–1133 diamagnetism and paramagnetism, 361–363 elements of period 3, 354–355 of fourth-period transition elements, 1108t ground state, 350–351, 351t for heteronuclear diatomic molecules, 480–481 for homonuclear diatomic molecules and ions, 474–480 of ions, 360–364 lanthanides and actinides, 358 of main group elements, 352–354 and periodic table, 350 and quantum numbers, 356–357 of transition elements, 357–360, 1107 writing, 355 electron dot symbols. See Lewis electron dot symbol/structure(s) electronegativity A measure of the ability of an atom in a molecule to attract electrons to itself, 390–393 electronic transitions in atoms, 318, 316–319 electron-pair geometry The geometry determined by all the bond pairs and lone pairs in the valence shell of the central atom, 418, 418 electron sea model of metallic bonding, 597 electron shell All atomic orbitals having the same value of the principal quantum number, n, 323 electrons in, 347t electron spin quantum number A number describing the apparent “spin” of an electron, 330 electron subshells, 347t electron transfer reaction(s). See also oxidation–reduction reaction(s) chemical change using electrical energy, 968–973 commercial voltaic cells, 946–951 corrosion, 975–978 counting electrons, 973–975 electrochemical cells under nonstandard conditions, 960–963 electrochemistry and thermodynamics, 964–967 electrolysis, 968–973 Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I-13

electron transfer reaction(s) (Continued) oxidation–reduction reactions, 168–169, 933–940 redox reactions in environment, 975–978 standard electrochemical potentials, 951–960 voltaic (or galvanic) cells, 940–945 electron volt (eV), A-8 electrostatic energy The energy due to the separation of electrical charges, 20, 20 electrostatic force(s) Forces of attraction or repulsion caused by electric charges, 97, 98 Coulomb’s law and, 98 electrostatic potential surfaces, 428 element(s) Matter that is composed of only one kind of atom, 12, 12, 12–13 abundance of, 1039–1040 brief overview of periodic table and, 76, 77–84 chlorine, 1082–1083 and compounds, 14 as diatomic or triatomic molecules, 80 in the Earth’s Crust, 77 fluorine, 1082 of Group 4A (14), 1070 of Group 5A (15), 1076 of Groups 7A and 8A (17 and 18), 354 iodine, 1083–1084 most abundant elements in Earth’s crust, 77t names of, 12–13 origin of, 1006–1008 Pauling electronegativity values for, 390 of period 3, 354–355 periodic table as guide to properties, 1040–1045 preparation of, 1080–1081, 1082–1084 production of, 1059–1060 properties of, 1080–1081 sources of, 1111 specific heat capacities of some, 259 transition, 1107–1110 elemental nitrogen, properties of, 1071–1072 elementary step A simple event in which some chemical transformation occurs; one of a sequence of events that form the reaction mechanism, 707 Elementary Treatise on Chemistry (Lavoisier), 141 emission lines from excited hydrogen atoms, 318–319 emission spectra of hydrogen, 313 of mercury, 313 of neon, 313 empirical formula A chemical formula showing the simplest possible ratio of atoms in a compound, 109 and chemical analysis, 204 determining, 109 from percent composition, 108–111

I-14

emulsifying agent A substance that allows two otherwise immiscible liquids to form an emulsion, 658 emulsion A colloidal dispersion of one liquid in another, 658 enantiomers A stereoisomeric pair consisting of a chiral compound and its mirror image isomer, 1124, 1154 endothermic process A thermodynamic process in which energy is transferred as heat into a system from its surroundings, 257, 258 end point. See equivalence point energy The capacity to do work and/or be transferred as heat, 20, 255, 308–312, 1263–1267 absorption of, 317 activation (See activation energy) and ATP, 1231–1232 for automobiles, 950 balance sheet on energy, 1263–1265 basic principles, 20–21, 255–258 calorimetry, 274–278 and chemistry, 311 conservation of, 21 conversions of, 20 dispersal of, 887–889, 890, 891, 892–893 electrical, 968–973 energy and changes of state, 262–266 energy resources, 1265–1266 energy usage, 1266–1267 enthalpy calculations, 278–285 enthalpy changes for chemical reactions, 271–274 first law of thermodynamics, 266–271 and food, 37 free, 964–965 and frequency, 310 ground state, 315–316 and Hund’s Rule, 353 ionization (See ionization energy) in a physical process, 267 and power, A-7–A-8 and product- or reactant-favored reactions, 285–286 profile for iodine-catalyzed reaction of cis-2-butene to form trans-2-butene, 704 and quantum numbers, 323 rainstorms release an enormous quantity of, 561 sources of, 1277–1281 specific heat capacity, 258–262 supply and demand, 1263–1265 transfer of, 261, 264 transferred as heat, 270 and wavefunctions, 322 and wavelength, 310 and work, 269 world energy consumption, 255 energy consumption by some countries, 1267t energy distribution curve, 698

energy levels diagrams, 281–282, 282 radiation emitted on changes in, 317 energy resources, 1265–1266 energy transfer, 256 energy units, 37 conversions, 37, A-8t energy usage, 1266–1267 enthalpy calculations, 278–285 energy level diagrams, 281–282 enthalpy change for reactions, 283 Hess’s law, 279–281 standard enthalpies of formation, 282–283 enthalpy change Energy transferred as heat at constant pressure, 271–274 for chemical reactions, 271–274, 283 enthalpy of fusion The energy required to convert one mole of a substance from a solid to a liquid at constant temperature, 609 of elements and compounds, 610t of fifth period metals, 610 of fourth period metals, 610 of sixth-period metals, 610 enthalpy of hydration, 544 enthalpy of solution The amount of heat involved in the process of solution formation, 633–636, 636t and ionic solids, 634 and thermodynamic data, 636–637 enthalpy of sublimation The energy required to convert one mole of a substance from a solid to a vapor, 611 enthalpy of vaporization The quantity of heat required to convert 1 mol of a liquid to a gas at constant temperature, 279, 553, 560, 563–564 enthalpy (H) The sum of the internal energy of the system and the product of its pressure and volume, 270–271 of formation, 284–285 and internal energy, 270 and non-expansion work, 270 of solvation, 544 of sublimation, 611 entropy (S) A measure of the dispersal of energy in a system, 889 changes, 893, 898–902 changes in physical and chemical processes, 896–897 dispersal of energy, 890–893 dispersal of matter, 892–893 Gibbs free energy, 902–906 kinetics and thermodynamics, 914–915 measurement and values, 894–897 microscopic understanding, 889–893 and solution process, 634 and spontaneity, 887–889, 898–902 standard entropy values, S°, 895–896 and states of matter, 894 values, 894–897 entropy changes, 893, 898–902 in chemical processes, 896–897 on gas expansion, 893

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

how temperature affects ∆S°(universe), 902 in physical processes, 896–897 and spontaneity, 901–902 environment, 1245 air pollution, 1271–1272 corrosion in, 975–978 of fossil fuels, impact of, 1271–1276 global warming/climate change, 4, 287, 1272–1274 greenhouse effect, 1272–1274 ocean acidification, 1274–1276 problems with SO2, 1081 redox reactions in, 975–978 environmental chemistry aqua sphere (water), 1256–1263 atmosphere, 1246–1256 energy, 1263–1267 environmental impact of fossil fuels, 1271–1276 fossil fuels, 1267–1271 green chemistry and sustainability, 7–8, 1281–1282 sources of energy, 1277–1281 enzyme(s) A biological catalyst, 705–706, 1213 and active sites, 1213–1215 enzyme catalysis, 705–706, 1213 equation(s) Arrhenius, 699–701 balancing chemical, 142–144 Balmer, 313 and chemical equilibrium, 754–756 Clausius–Clapeyron, 563–564, 564 complete ionic, 154 de Broglie’s, 320–321 half-life, 692 Henderson–Hasselbalch, 837 Nernst, 960–963 for nuclear reactions, 995–996 Planck’s, 308–310 quadratic, 751, A-4–A-5 rate (See rate equations) van der Waals, 525–526 writing, for precipitation reaction, 152–153 equilibrium A condition in which the forward and reverse reaction rates in a physical or chemical system are equal, 146 and complex ions, 867–871 temperature and, 761 equilibrium composition effect of temperature changes on, 760–762 effects of disturbance on, 762t equilibrium constant (K) The constant in the equilibrium constant expression, 714, 738–745, 754–756 for acids and bases, 783–788 adding two chemical equations, 755–756 in calculations, 748–753 chemical equation, reversing, 754–755 determining, 746–748 determining K from initial concentrations, 796–798 and E°, 965–967

and Gibbs free energy, 905 ionization constants for an acid, 785t, 783–788, A-20–A-22 Ka and Kb values for polyprotic acids, 785t, 787, A-20–A-22 logarithmic scale of relative acid strength, pKa, 788 magnitude of, 742–743 measured pH, 796–798 pH of an aqueous solution of a weak acid or base, 798–805 and reaction quotient, 738–745 reversing chemical equation, 754–755 solution involves quadratic expression, 750–753 stoichiometric coefficients, 754 and temperature, 833 writing equilibrium constant expressions, 740–742 using different stoichiometric coefficients, 754 equilibrium constant expression A mathematical expression that relates the concentrations of the reactants and products at equilibrium at a particular temperature to a numerical constant, 740, 854 Kc and Kp, 741–742 reactions in solution, 741 reactions involving gases, 741–742 reactions involving solids, 740–741 equilibrium vapor pressure The pressure of the vapor of a substance at equilibrium in contact with its liquid or solid phase in a sealed container, 561, 640 equivalence point The point in a titration at which one reactant has been exactly consumed by addition of the other reactant, 221, 843 error The difference between the measured quantity and the accepted value accuracy and, 39–40 determinate, 40 indeterminate (random), 40 percent, 39 Escherichia coli, 376 ester(s) Any of a class of organic compounds characterized by the presence of a carbonyl group, in which the carbon atom is bonded to another carbon atom and to an oxygen atom that is attached to another organic group, 1181–1182, 1182t esterification reaction A reaction between a carboxylic acid and an alcohol in which a molecule of water and an ester is formed, 1181 ethanol, 115, 287, 287 condensed formula, 85 energy content of, 1280 molecular formula, 85 molecular models, 85 structural formula, 85 ethene, 1152 See also ethylene.

ether(s) Any of a class of organic compounds characterized by the presence of an oxygen atom singly bonded to two carbon atoms, 1172–1173 ethylene derivatives, 1187t ethylene glycol properties of, 1174 uses of, 645, 1174 ethyne, 1152 See also acetylene. evaporation, 546. See also vaporization Evonik, 232 excess reactant A reactant present in a greater amount than required by stoichiometry, 196 exchange reaction(s) A chemical reaction that proceeds by the interchange of reactant cation–anion partners, 151 excited state The state of an atom or molecule in which at least one electron is not in the lowest possible energy level, 314 of the H atom, 315–316 exclusion principle. See Pauli exclusion principle exothermic process A thermodynamic process in which energy is transferred as heat from a system to its surroundings, 257, 258 experimental error, 38–39 explaining, as goal of science, 6 exponential notation, 42 exponential term, 42 exposure In reference to radiation, the amount of radiation that one is exposed to, measured in coulombs/ kilogram (C/kg) or roentgens (R), 1022 extensive properties Physical properties that depend on the amount of matter present, 17 extrinsic semiconductor A semiconductor whose properties are due to the presence of foreign atoms (dopants), 601 face-centered cubic (fcc) unit cell A unit cell consisting of the primitive cubic unit cell with additional particles positioned at the center of each face of the cube that are identical to the particles positioned at the corners of the cube, 583 factor-label method, 47. See also dimensional analysis Fahrenheit temperature scale A scale defined by the freezing and boiling points of pure water, defined as 32 °F and 212 °F, 33, 33 families, 74–75, 75 Faraday constant (F) The proportionality constant that relates standard free energy of reaction to standard potential; the charge carried by one mole of electrons, 960, 974 Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I-15

fat A lipid that is solid at physiological temperatures, has a primary function of energy storage, and consists of a triester of glycerol and three carboxylic acids, 1226 fatty acid A long-chain carboxylic acid contained in a fat or oil, 1226 f-block elements The group of elements coinciding with the filling of f orbitals, 1105 Fermi level The highest filled electron energy level in a metal at absolute zero temperature, 599 ferromagnetic, 364 ferromagnetism The physical property of certain solids in which the spins of unpaired electrons in a cluster of atoms are aligned in the same direction, 364, 364 filtration A means of separation in which a porous barrier is used to remove an undissolved solid from a solution, 11 fireworks, 305 boron carbide in, 1062 colors in, 331–332 first law of thermodynamics The total energy of the universe is constant; the internal energy change for a system is equal to the sum of the energy transferred as heat and the energy transferred as work between the system and the surroundings, 266–271 first-order reactions, 686–688 and half-life, 690–694 fission The highly exothermic process by which very heavy nuclei split to form lighter nuclei, 1018–1021 Flint water crisis, 1260, 1261 fluid-mosaic model A cell membrane model where the phospholipid bilayer molecules can move within one side of the bilayer but in which they do not move from one side to the other and in which molecules like proteins and cholesterol embedded in the bilayer readily move within it, 1228 of cell membranes, 1228 fluorine (F), 354, 1082 fluorine compounds, 1043t, 1084–1086 food and energy, 37 f orbital(s), 328, 328. See atomic orbital(s) force defined, A-6 electromotive, 951–952 intermolecular (See intermolecular forces) and weight, A-6–A-7 formal charge The charge on an atom in a molecule or ion calculated by assuming equal sharing of the bonding electrons, 410 in covalent molecules and ions, 410–416

I-16

in equivalent resonance structures, 412–413 importance of structures using, 414–416 formaldehyde, 1178t formation constant An equilibrium constant for the formation of a complex ion, 868, A-24 formic acid (HCO2H), 1180, 1180t formula(s) of binary nonmetal compounds, 87 calculating, from percent composition, 109–111 of common polyatomic ions, 93t of compound from combining masses, 112–113 condensed, 85 of coordination compounds, 1117–1120 deriving, 108 determining, by mass spectrometry, 114 determining from mass data, 111–114 of hydrated compound, 113–114 for ionic compounds, 94–95 for ions, 94 structural, 85 formula unit The simplest ratio of ions in an ionic compound, 590 fossil fuels, 1267–1271 air pollution, 1271–1272 coal, 1268–1269 energy released by combustion of, 1267t environmental impact of, 1271–1276 global warming/climate change, 1272–1274 greenhouse effect, 1272–1274 methane/natural gas, 1269–1270 ocean acidification, 1274–1276 petroleum, 1270–1271 fracking, 1266 free energy changes in course of a reaction, 904 and equilibrium, 903 Gibbs free energy, 902–914 kinetics and thermodynamics, 914–915 and spontaneity, 903–905 standard free energy of formation, 906 free energy under conditions other than standard conditions, 904–905, 912–914 free energy change under standard conditions, 903, 905–912 and equilibrium constant, 905, 911–914 and standard free energies of formation, 906–908 and temperature, 908–911 and work, 905 free radical(s) A chemical species containing an unpaired electron, 409 chain reaction, 715–716 chemistry of NO2, 410 freezing point depression, 645–647 of solutions containing ionic solutes, 654t French Academy of Sciences, 34

French National Assembly, 34, A-10 French Revolution, 34, 42 frequency (ν) The number of complete waves passing a point in a given amount of time, 306 and energy, 310 and wavelength, 310 frequency factor The pre-exponential term in the Arrhenius equation that is related to the number of molecular collisions having geometry that allows the reaction to proceed, 699 fructose d-fructose structure of, 1216 fuel cell A voltaic cell in which reactants are continuously added, 950–951, 1277, 1277–1278 fuel cell design, 951 functional group A structural fragment found in all members of a class of compounds, 1170, 1171t fusion The state change from solid to liquid, 1021 gallium arsenide (GaAs), 602 galvanic cell An electrochemical cell that uses chemical reactions to produce an electric current, 940. See also voltaic cell and electrochemical cell(s) galvanized iron Iron that has been coated with a film of zinc to protect it from corrosion, 978 gamma radiation, 994 characteristics of, 994t penetrating ability of, 994 gamma ray(s) High-energy electromagnetic radiation, 1022, 1025 gangue A mixture of sand and clay in which a desired mineral is usually found, 1111 gas(es) The phase of matter in which a substance has no definite shape and a volume defined only by the size of its container, 9, 499–501, 628 chemical reactions, 513–514 density of, 510–511 diffusion, 521–524 dissolving in liquids, 637–639 effusion, 521–524 gas laws, 502–508, 513–514 gas mixtures, 514–517 and gas pressure, 499–501 ideal gas law, 508–513 kinetic-molecular theory of, 517–521 and molecular speed, 517–520 molecular view of, 517 movement of, 517 nonideal behavior of, 524–526 partial pressures, 514–517 gas constant (R) The proportionality constant in the ideal gas law, 0.082057 L·atm/mol·K or 8.314510 J/mol·K, 509

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

gas density, 510, 511 gaseous diffusion, 521, 522 gaseous effusion, 522 gas-forming acid–base reactions, 165–167, 166t, 173,175 gas laws, 502–508, 513–514 Avogadro’s hypothesis, 506–508 Boyle’s law, 502–503 Charles’s law, 503–505 compressibility of gases, 502–503 effect of temperature on gas volume, 503–505 and general gas law, 505–506 and kinetic-molecular theory, 520–521 and stoichiometry, 513–514 gas mixtures, 514–517 gasoline, 287–288, 1270–1271 gas-phase equilibria effect of volume changes on, 759–760 gas pressure, 499–501, 520 gas solubility and pressure, 638 gas volume effect of temperature on, 503–505 Geiger–Müller counter, 1010 gel A colloidal dispersion with a structure that prevents it from flowing, 657 gelatinous precipitates, 657 gems, solubility of, 859 general gas law An equation that allows calculation of pressure, temperature, and volume when a given amount of gas undergoes a change in conditions, 505–506 genetic code The specific correspondences between the codon and its corresponding amino acid in the context of protein synthesis, 1223 geometric isomers Isomers in which the atoms of the molecule are arranged in different geometric relationships, 471–472, 1123, 1124, 1153, 1162 cis-trans isomers, 471–472, 1123–1124, 1153, 1162 fac–mer isomers, 1124 geometry(ies) coordination, 1116 electron-pair, 418 and hybridization, 473 hybrid orbitals for molecules and ions with linear electron-pair geometry, 465–466, 483 for molecules and ions with octahedral electron-pair geometry, 466–467 for molecules and ions with tetrahedral electron-pair geometry, 462–465, 483 for molecules and ions with trigonalbipyramidal geometry, 466–467 for molecules and ions with trigonalplanar electron-pair geometry, 465–466, 483 molecular, 420–422

Gibbs free energy (G) A thermodynamic state function relating enthalpy, temperature, and entropy, 902–906. See also free energy. change in, 903 and chemical equilibrium, 903–905 and equilibrium constant (K), 905 and free energy, 905–906 and reaction favorability, 905 and reaction quotient (Q), 905 and spontaneity, 903–905 gigatons, 1247 glass A non-crystalline, usually transparent solid that has many uses in our lives and commerce, 605 glassware and significant figures, 44 global warming, 1272–1274 glucose d-glucose structure, 1216 ring forms of, 1216 straight-chain forms of, 1216 glucose, metabolism of, 1234 glycosidic bond The bridging oxygen bond connecting two sugar molecules, 1217 gold colloid, 655 Graham’s law A gas law relating the rate of effusion of two gases, 522, 523–524 graphite (network solid), 79–80, 604 graphs and graphing, 48–49 gravimetric analysis A method of chemical analysis in which the mass of a precipitate is used to determine the composition of an unknown sample, 203 gravitational energy The energy due to the attraction between masses, 20, 20 gray The SI unit of radiation dosage, 1022 green chemistry An area of chemistry focused on reducing environmental impacts of chemicals, including reducing pollution and increasing safety, 7–8, 1281–1282 examples of, 441, 486, 1189, 1190, 1193, 1282 principles of, 7–8, 1281–1282 sustainability and, 7–8 Green Chemistry: Theory and Practice (Anastas and Warner), 7, 1281 greenhouse effect The process by which an accumulation of insulating gases in the troposphere traps energy in Earth’s atmosphere, warming it, 1272–1274, 1273 greenhouse gases, 1273 green jade, 1135 ground state The state of an atom, ion, or molecule in which all electrons are in the lowest possible energy levels, 314 electron configurations, 350–351, 351t

Group 1A (1) elements, 77, 352, 1049–1059. See also alkali metal(s) Group 2A (2) elements, 77, 1054–1058. See also alkaline earth metal(s) Group 3A (13) elements, 79, 90, 1058–1065 and aluminum, 1058–1065 aluminum compounds, 1063–1065 boron compounds, 1061–1063 boron minerals and production of, 1059–1060 chemistry of, 1058–1059 chemistry of Group 3A (13) elements, 1058–1059 metallic aluminum, 1060–1061 Group 4A (14) elements, 79, 1065–1070 Group 5A (15) elements, 80, 1070–1079 hydrogen compounds of, 1076 Group 6A (16) elements, 81, 1079–1082 Group 7A (17) elements, 81, 83, 1082–1087. See also halogens Group 8A (18) elements, 83, 1087–1089. See also noble gas(es) group names, 81 group(s) The vertical columns in the periodic table of the elements, 74–75, 75. See also families similarities, 1044–1045 gunpowder, 286–287, 1053 half-cell A compartment of an electrochemical cell in which a halfreaction occurs, 941, 941 half-life (t1/2) The time required for the concentration of one of the reactants to reach half of its initial value, 690, 1008–1010 equations, 692 and first-order reactions, 690–694, 692 and radioactivity, 692, 1008–1010 of a reaction, 690 half-reaction method A systematic procedure for balancing oxidation– reduction reactions, 934–935 half-reactions The two chemical equations into which the equation for an oxidation–reduction reaction can be divided, one representing the oxidation process and the other the reduction process, 934–935 halide ions Ions of the elements of Group 7A (17), 91 halitosis (bad breath), 1081 halogens The elements in Group 7A (17) of the periodic table, 83, 1082–1087 chlorine compounds, 1086–1087 fluorine compounds, 1084–1086 preparation of elements, 1082–1084 hard water, 1056 Harry Potter and the Philosopher’s Stone, 1072 Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I-17

heat The process by which energy is transferred from an area with higher temperature to an area with lower temperature, 256 of crystallization, 631 directionality and extent of transfer of, 256–257 quantitative aspects of energy transferred as, 260–262 heat capacity The quantity of energy required to change the temperature of a substance by 1 kelvin, 258–260 heating and cooling, 258–266 heat of fusion The quantity of heat required to convert 1 g of a solid to a liquid at constant temperature, 263. See also enthalpy of fusion heat of solution. See enthalpy of solution heat of vaporization The quantity of heat required to convert 1 g of a liquid to a gas at constant temperature, 263. See also enthalpy of vaporization heats of transformation, A-14t heavy metals, 1261 heavy water, 1046 Heisenberg’s uncertainty principle It is impossible to determine both the position and the momentum of an electron in an atom simultaneously with great certainty, 323 helium, 75, 83, 83, 1087 electron configuration of, 352 elemental, 475 and MRI units, 83 as noble gas, 83 nonflammable, 1049 in nucleosynthesis, 1006–1008 in nuclear fusion, 1021 photoelectron spectra of, 371 and stars, 332 as UV radiation source, 485 in weather balloons, 83, 506 heme An iron(II) ion held in the center of a porphyrin ring (for example, in the binding site for O2 in hemoglobin), 1118, 1211 hemoglobin, 1118, 1211–1213 primary structure of, 1212 quaternary structure of, 1212 secondary structure of, 1212 tertiary structure of, 1212 Henderson–Hasselbalch equation An equation used to determine the pH of a buffer, 837 Henry’s law The concentration of a gas dissolved in a liquid at a given temperature is directly proportional to the partial pressure of the gas above the liquid, 637–639, 637t limitations of, 638 hertz The unit of frequency, or cycles per second; 1 Hz = 1 s21, 306

I-18

Hess’s law If a reaction is the sum of two or more other reactions, the enthalpy change for the overall process is the sum of the enthalpy changes for the constituent reactions, 278, 279–281 heterogenous alloy An alloy formed by two or more different phases, leading to regions of different composition and crystal structure, 607, 608 heterogeneous mixture A mixture in which the properties in one region or sample are different from those in another region or sample, 10–12, 11 heteronuclear diatomic molecule(s) A molecule composed of two atoms of different elements, 480–481 hexagonal close-packed (hcp) unit cell, 588 highest occupied molecular orbital (HOMO) The highest occupied molecular orbital in a molecule (or polyatomic ion), 480 high-spin configuration The electron configuration for a coordination complex with the maximum number of unpaired electrons, 1130 hippuric acid, 111 HOMO (highest occupied molecular orbital), 480 homogenous alloy A class of alloys that consist of a single phase, including interstitial and substitutional alloys, 607, 608 homogeneous catalyst A catalyst that is in the same phase as the reaction mixture, 681 homogeneous mixture A mixture in which the properties are the same throughout, regardless of the optical resolution used to examine it, 10–12, 11 homonuclear diatomic ion, electron configuration for, 480 homonuclear diatomic molecule(s) A molecule composed of two identical atoms, 477 molecular orbitals in, 474–480 hormone A molecule that acts as a chemical messenger in biological systems, 1225 Hund’s rule The most stable arrangement of electrons in a subshell is that with the maximum number of unpaired electrons, all with the same spin direction, 353, 474–475 hybrid orbital(s) An orbital formed by mixing two or more atomic orbitals, 462 for molecules and ions with linear electron-pair geometry, 465–466, 483

for molecules and ions with octahedral electron-pair geometry, 466–467 for molecules with tetrahedral electronpair geometry, 461–465, 483 for molecules with trigonal-planar electron-pair geometry, 465–466, 483 for molecules with trigonal-bipyramidal electron-pair geometry, 466–467 hydrated compound A compound in which molecules of water are associated with ions, 98 complex ions in, 1114–1115 determining formula of, 113–114 molar mass of, 104 hydrated metal cations as Brønsted acids, 812 hydration The process of water molecules surrounding particles of solute in an aqueous solution, 634 hydrazine, 87, 88, 400, 902, 1072–1073 hydrocarbon(s) A compound that contains only carbon and hydrogen, 11, 88, 1155–1170, 1156t, A-15–A-17 of the alkane family, 1156t alkanes, 1155–1156, A-15–A-16 alkenes, 1161–1167, A-16 alkynes, 1161–1167, A-16 aromatic compounds, 1167–1170 combustion analysis determining empirical formula of, 205–206 combustion analysis of, 205 derivatives of, A-17 hydrochloric acid, 149, 157, 158, 780, 784, 785, 1086 in acid–base titrations, 220–227, 843–844, 849–851 in aqua regia, 1076 addition to buffer, 831–832, 840–842 catalyst for esterification reactions, 704, 1181 in double displacement reaction, 174 factors affecting strength, 807–808 in gas-forming acid–base reactions, 165–166, 175–176, 218 as monoprotic acid, 778 oxidation of metals, 1107–1108 in pH electrode, 963 reaction with aluminum, 1063 reaction with aluminum hydroxide, 815 reaction with ammonia, 138, 165, 174, 517, 521–522, 803–805, 849–851 reaction with calcium carbonate, 165–166, 218 reaction with magnesium, 275–276, 1048 reaction with sodium hydroxide, 163, 164, 794, 843–844 reaction with zinc, 156, 219 in stomach acid, 872 as strong acid, 157–160, 783, 784, 786, 791, 794–795 as strong electrolyte, 149, 783 hydrogen, 1046–1049, 1048, 1278–1279 and ammonia, 1278–1279 Bohr model of atom, 313–314, 315

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

bond, 548 chemical properties of, 1046–1047 combustion of, 887 electrodes, 944 by electrolysis, 1048 electron configuration of, 352 emission spectrum of, 313 excited states of, 315–316 forming binary compounds, 87 line emission spectrum of, 313 physical properties of, 19, 1046–1047 preparation of, 1047 production from natural gas, 1047 visible emission spectrum of, 312 water electrolysis, 1047–1049 hydrogenation An addition reaction in which the reagent is molecular hydrogen, 1165 hydrogen atoms Bohr model of, 313–314, 315 emission lines from excited, 318–319 hydrogen bonding Attraction between a hydrogen atom that is bonded to a very electronegative atom, and another atom to produce an unusually strong dipole–dipole attraction, 548 in acetic acid, 550 between adenine and thymine, 551 in biochemistry, 551 in DNA, 551 effect of, 549–550 H atom charge density in, 549 between HF molecules, 549 between molecules with permanent dipole, 548–552 between polyamide chains, 1191 in protein secondary structure, 1212 and unusual properties of water, 550–552 in water and the structure of ice, 550 hydrogen chemistry, 1046–1049 hydrogen chloride, 1086 hydrogen compounds, 548 of nitrogen, 1072–1073 hydrogen electrode, 944 hydrolysis reaction A reaction with water, 862 and hydrolysis constants, 789 hydrometallurgy Recovery of metals from their ores by reactions in aqueous solution, 1111, 1113–1114 hydronium ion concentration and pH, 215–217, 780–782 hydrophilic colloid A substance that is strongly attracted to water, 657 hydrophobic colloid A substance that is not strongly attracted to water or that is insoluble in water, 657 hydrosulfuric acid, 157 hygroscopic compound A compound that absorbs water from the air, 331, 1053 hypertonic solution A solution with higher solute concentration than a cell, 651

hypervalent compound(s) A compound in which the central atom is surrounded by more than four valence electron pairs, 407 electron-pair geometry of, 418, 422–423, 446 Lewis structures for, 407–409 molecular geometry of, 418, 422–423, 446 orbital hybridization in, 466–467 and three-center bonds, 484 hypophosphorous acid, 1078t hypothesis A tentative explanation or prediction based on experimental observations, 4 hypotonic solution A solution with lower solute concentration than a cell, 651 ICE table A table used to organize reaction data that shows what the initial (I) concentrations are, how those concentrations change (C) on proceeding to equilibrium, and what the concentrations are at equilibrium (E), 739 ideal gas A simplification of gas behavior in which it is assumed that there are no forces between the molecules and that the molecules occupy no volume, 508–513 ideal gas law A law that relates pressure, volume, number of moles, and temperature for an ideal gas, 508–513 density of gases, 510–511 determining molar mass of a gas, 511–513 ideal solution A solution that obeys Raoult’s law, 640 immiscible liquids Liquids that are not soluble in each other, 632 indeterminate (random) errors, 40 indicator(s) A substance used to signal the equivalence point of a titration by a change in some physical property such as color, 851 induced dipole(s) Separation of charge in a normally nonpolar molecule, caused by the approach of a polar molecule, 554 induced dipole–induced dipole forces The electrostatic force between two neutral molecules, both having induced dipoles, 554, 554. See also London dispersion forces induced-fit model A model for enzyme binding in which binding of the substrate causes the structure of the enzyme to change so that it fits the substrate, 705 industrial steam-reforming plant, 1279 inert gas(es). See noble gas(es) inertia, A-6 initial rate The instantaneous reaction rate at the start of the reaction, 682–684

inner transition elements, 357. See also actinide(s) and lanthanide(s) insoluble compounds, 149–156, 853–857, 853, 854t Institute of Theoretical Physics, 314 instrumental analysis determining compound formulas, 114–117 determining formula by mass spectrometry, 114 determining structure by nuclear magnetic resonance spectroscopy, 208 isotopes in mass spectrometry, 114–117 molar mass in mass spectrometry, 114–117 integrated rate equation An equation relating the concentration of a reactant to its initial concentration and the elapsed time, 686 integrated rate laws first-order reactions, 686–688 graphical methods for determining rate constant, 690 graphical methods for determining reaction order, 690 half-life and first-order reactions, 690–694 nuclear decay, 1011 second-order reactions, 688–689 zero-order reactions, 689–690 integrity, in science, 6 intensive properties Physical properties that do not depend on the amount of matter present, 17 intercalation In chemistry, the ability to insert atoms, ions, and even small molecules into layered compounds such as in the formation of lithiumion batteries, 949 Intergovernmental Panel on Climate Change (IPCC), 1245 intermediate. See reaction intermediate intermetallic alloy (Ni3Al), 608 intermetallic compound A class of alloy with a definite stoichiometry and formula where the atoms are ordered rather than random, which often leads to higher melting points and better structural stability, 607 intermolecular forces Interactions between molecules, between ions, or between molecules and ions, 542, 554, 556 capillary action, 567 formation of a meniscus, 567 involving nonpolar molecules, 552–555 involving ions and molecules with permanent dipole, 543–545 hydrogen bonding, 548–552 in a liquid and surface tension, 566 involving molecules with a permanent dipole, 546–552 and properties of liquids, 557–567 states of matter and, 541–543 van der Waals forces, 555–557 Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I-19

intermolecular forces involving nonpolar molecules Debye forces, 552–553 dipole–induced dipole forces, 552–553 induced dipole–induced dipole forces, 553–555 London dispersion forces, 553–555 internal energy The sum of the potential and kinetic energies of the particles in the system, 267, 271 and enthalpy, 270 and non-expansion work, 270 International Prototype Meter, 34 International System of Units, A-7, A-11 International Union of Pure and Applied Chemistry (IUPAC), 64, 70, A-15 International Union of Pure and Applied Physics (IUPAP), 64 interstitial alloy A type of homogenous alloy in which the solute atoms occupy the interstices of the lattice, 608 intrinsic semiconductor A semiconductor whose properties occur naturally for the pure substance, 601 iodine, 81, 1083–1084 preparation of, 1083 iodine clock reaction, 678 ion(s) An atom or group of atoms that has lost or gained one or more electrons so that it is no longer electrically neutral, 14, 89–93, 373 in aqueous solutions, 147–149 colligative properties of solutions containing, 653–655 and corresponding acids, 158, 158t drawing Lewis structures for, 398–399 electron configurations of, 360–364 formation of, 89–90 formation constants for complex, A-24 formulas for, 94 isoelectronic, 373–374 and isoelectronic molecules, 402 monatomic ions, 89–91 polyatomic ions, 92–93 sizes, 372–374 ion–dipole attraction The electrostatic force between an ion and a neutral molecule that has a permanent dipole moment, 544 ionic bond(s) The attraction between a positive ion and a negative ion resulting from the complete (or nearly complete) transfer of one or more electrons from one atom to another, 389 ionic compound(s) A compound formed by the combination of positive and negative ions, 89, 150, 210, 389 bonding in, 594–596 calculating lattice enthalpy from thermodynamic data, 595–596 formulas for, 93–99 guidelines to predict aqueous solubility of, 150 hydrated, 98 lattice energies of, 594t

I-20

of main group elements, 1041–1042 naming of, 93–99 properties of, 93–99 solubility in water, 149–151 ionic solid(s) A solid formed by the condensation of anions and cations, 99 and enthalpy of solution, 634 formulas of, 589–593 structure of, 589–593 in water, 636 ionization constant(s) The equilibrium constant for an ionization reaction, 784 for aqueous weak acids, 785, A-20–A21 for aqueous weak bases, 785, A-22 ionization energy The energy change required to remove an electron from an atom or ion in the gas phase, 367–369, 368t, 369, 371 iron corrosion, anode and cathode reactions in, 975 production of, 1111–1113 irreversible process A process that involves nonequilibrium conditions, 890 isoelectronic ions Ions that have the same number of electrons but different numbers of protons, 373–374 isoelectronic molecules, 402t isoelectronic species Molecules or ions that have the same number of valence electrons and similar Lewis structures, 401–402 isolated system A system that can transfer neither energy nor matter to or from its surroundings, 257 isomerism, 1122–1127, 1153–1155 cis-trans, 471–472, 1123, 1153–1154 coordination, 1123 fac-mer, 1124 geometric isomerism, 1123–1124, 1153–1155 linkage, 1123 optical, 1124–1127, 1154–1155 structural isomerism, 1123 isomer(s) Two or more compounds with the same molecular formula but different arrangements of atoms, 471, 1122, 1153–1154 geometric, 1123–1124, 1153–1155 linkage, 1123 optical, 1124–1127, 1153–1155, 1154 stereoisomers, 1122, 1153 structural, 1122, 1153 isotonic saline solution, 651, 651 isotonic solution A solution with a solute concentration equal to that of a cell, 651 isotope(s) Atoms with the same atomic number but different mass numbers, because of a difference in the number of neutrons, 66 in mass spectrometry, 114–117

Ötzi, the Iceman of the Alps, 117–118 separation by gas centrifuge, 1019 isotope composition, 66–67 isotope dilution, 1026–1027 isotopic abundance, 69t and atomic weights, 69t, 70 calculating, 71–72 calculating atomic weight from, 70–71 determining, 68 by mass spectrometry, 68, 116–117 isotopic masses and mass defect, 68 jade (network solid), 603 James Webb Space Telescope, 993, 1055 joule (J) The SI unit of energy, 37, A-7 Journal of Chemical Education, 7 K capture. See electron capture Kelvin temperature scale A scale in which the unit is the same size as the Celsius degree but the zero point is the lowest possible temperature, 32–33, 33. See also absolute zero ketone(s) Any of a class of organic compounds characterized by the presence of a carbonyl group, in which the carbon atom is bonded to two other carbon atoms, 1177–1178, A-17 naming of, 1179 kilogram (kg) The SI base unit of mass, 34, A-10t kilojoule (kJ), 37, 558 kilopascal (kPa), 500 kinetically stable A state that is longlasting because its reaction is slow, even though it may not be at the lowest energy possible, 914 kinetic control The reaction condition in which the product ratio is influenced by the rates of the chemical reactions, 915 kinetic energy The energy of a moving object, dependent on its mass and velocity, 9, 20 of a gas, 517–519 kinetic-molecular theory A theory of the behavior of matter at the molecular level, 8–9 of gases, 517–521 molecular speed, 517–520 and gas laws, 520–521 of matter, 8 kinetics. See chemical kinetics of nuclear decay, 1010–1012 krypton, photoelectron spectrum of, 371 lactose, 1217, 1154 Lake Mead (symbol of crisis in water supply), 1257 lanthanide(s) The series of elements between lanthanum and hafnium in the periodic table, 83, 357, 358, 1105

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

lanthanide contraction The decrease in ionic radius that results from the filling of the 4f orbitals, 1109 lanthanum, 375, 1109, 1139 lattice energy The energy of formation of one mole of a solid crystalline ionic compound from ions in the gas phase, 594–596, 595 calculating a lattice enthalpy from thermodynamic data, 595–596 lattice enthalpy The enthalpy of formation of one mole of a solid crystalline ionic compound from ions in the gas phase, 594 calculating from thermodynamic data, 595–596 lattice point(s) The locations that define a particular type of unit cell, 583 law A concise verbal or mathematical statement of a relation that is always the same under the same conditions, 5 Beer–Lambert law, 228–229 Boyle’s law, 502, 502, 505–506 Charles’s law, 504, 504, 504t, 505–506 Coulomb’s law, 97–98, 313–314 general gas law, 505–506 Graham’s law, 522–524 Henry’s law, 637–639 Hess’s law, 278, 279–281 ideal gas law, 508–513 Raoult’s law, 640–642 rate law, 680 of thermodynamics first law, 266–271 second law, 889 third law, 894 law of chemical periodicity The law stating that the properties of the elements are periodic functions of their atomic number, 78 law of conservation of energy The law stating that energy can neither be created nor destroyed, 21, 21, 256 law of conservation of matter The law stating that matter is neither created nor destroyed in ordinary chemical reactions, 6, 140 law of definite proportions The law stating that a compound has a definite composition (by mass) of its combining elements, also called the law of constant composition, 13 laws of electrolysis, 970 lead, 12, 15, 70, 74, 79–80, 102, 1041, 1070 lead poisoning, 36, 1070, 1089, 1251, 1260–1261 lead storage battery (rechargeable battery), 948, 948 Le Chatelier’s principle A change in any of the factors determining an equilibrium will cause the system to adjust to reduce the effect of the change, 639–640, 757–762 Le Grand K, 34, 34

length, 34–37 common units of, A-12 leveling effect, 783, 817 Lewis acid(s) A substance that can accept a pair of electrons to form a new bond, 812–813 Lewis acid–base complex, 813 Lewis base(s) A substance that can donate a pair of electrons to form a new bond, 812–813 Lewis concept of acids and bases coordination complexes, 813–814 molecular Lewis acids, 814–815 molecular Lewis bases, 815–816 Lewis electron dot symbol/structure(s) A notation for the electron configuration of an atom or molecule, 388–409, 388t, 393 and chemical bond formation, 388–389 and Lewis symbols for atoms, 388–389 and valence electrons, 388–389 carbon-based organic compounds, 402–403 and isoelectronic species, 401–402 of oxoacids and anions, 402 patterns of bonding in, 399–403 ligand field model, 1128 ligand(s) The molecules or anions bonded to the central metal atom in a coordination compound, 868, 1114–1117, 1115 anions as, 1116 bidentate, 1116, 1116, 1117 chelating, 1116 polydentate, 1116 ligand field splitting The difference in potential energy between sets of d orbitals in a metal atom or ion surrounded by ligands, 1129 ligand field theory A theory of metal-ligand bonding in coordination compounds, 1128–1129 light absorption, 229 and color, 1135 light-emitting diodes (LEDs), 600 LiMA (Leading in Methacrylates), 232 lime, 163 limestone (calcium carbonate) placed in vinegar, 165 Limited Test Ban Treaty, 1026 limiting reactant The reactant present in limited supply that determines the amount of product formed, 196–200 line emission spectrum The spectrum of light emitted by excited atoms in the gas phase, consisting of discrete wavelengths, 312, 313 linkage isomerism, 1123 linkage isomers Two or more complexes in which a ligand is attached to the metal atom through different atoms, 1123

lipid(s) Any of a class of biological compounds that are poorly soluble in water but soluble in organic solvents, 1225–1230 liquid(s) The phase of matter in which a substance has no definite shape but a definite volume, 9 conversion of solid into, 609–610 dissolving gases in, 637–639 dissolving in liquids, 632 molecular view of, 517 properties of, 557–567 solids dissolving in, 632–633 states of matter and intermolecular forces, 541–543 liquid gallium, 79 liquid oxygen, 459 liter (L) A unit of volume convenient for laboratory use; 1 L = 1000 cm3, 36, A-12t lithium, 352 and green cars, 950 compounds of, 1051–1054 lithium batteries, 949–950, 1052 children and, 949 lock-and-key model A model of enzyme binding in which the shape of the substrate is pictured as fitting into the enzyme’s active site precisely, much like a key (the substrate) fits a lock (the enzyme), 705 logarithms, 215, 216, 782, 967, A-2–A-3 characteristic in, A-3 common, A-2–A-3 mantissa in, A-3 mathematical operations using, A-3–A-4 natural, A-2–A-3 London dispersion forces, 542, 553–554, 554, 560, 609, 634. See also induced dipole–induced dipole forces intermolecular forces involving nonpolar molecules, 553–555 lone pair(s) Pairs of valence electrons that do not contribute to bonding in a covalent molecule, 394 bond angles and, 420–422, 446 central atoms with more than four valence electron pairs and, 422–423, 446 low-density polyethylene (LDPE), 1186 lowest unoccupied molecular orbital (LUMO) The lowest unoccupied molecular orbital in a molecule (or polyatomic ion), 480 low-spin configuration The electron configuration for a coordination complex with the minimum number of unpaired electrons, 1130 Lucite, 232, 1187 LUMO (lowest unoccupied molecular orbital), 480 lycopodium powder, 679 Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I-21

Lyman series A series of lines in the ultraviolet portion of the emission spectrum of hydrogen atoms that correspond to transitions of the electron from higher energy levels down to the first energy level, 318, 318 lysozyme, 1213–1215 macroscopic change, 146 macroscopic level Processes and properties on a scale large enough to be observed directly, 9–10 magic numbers, of protons and neutrons, 1016 magnesite, 1054, 1056 magnesium production of, 1056 properties of, 1056 magnesium carbonate, 1057 magnesium chloride, 1056, 1065 magnesium oxide, 591 magnetic properties of coordination compounds, 1130–1133 magnetic quantum number (mℓ), 324 magnetic resonance imaging (MRI), helium and, 83 magnetism. See also diamagnetism; paramagnetism in high-spin and low-spin complexes, 1130 main group element(s) Elements from the A groups on the periodic table, which belong to the s- and p-blocks, 74, 1038–1092 abundance of, 1039–1040 alkali metals, Group 1A (1), 1049–1054 alkaline earth elements, Group 2A (2), 1054–1058 aluminum, 1058–1065 beryllium (Be) and elements of Group 2A (2), 353, 353 boron (B), 353, 353, 1058–1065 carbon (C) and other elements of Group 4A (14), 353 electron configurations of, 352–354 first, second, and third ionization energies for, 368t fluorine (F), 354 Group 3A (13) elements, 1059–1065 Group 4A (14) elements, 1065–1070 Group 5A (15) elements, 1070–1079 Group 6A (16) elements, 1079–1082 Group 7A (17) elements, 1082–1087 Group 8A (18) elements, 1087–1089 halogens, 1082–1087 hydrogen, 1046–1049 ionic compounds of, 1041–1042 molecular compounds of, 1042–1044 neon (Ne), 354 nitrogen (N), 354, 1070–1079 noble gases, Group 8A (18), 1087–1089 oxygen (O), 354, 1079–1082 periodic table as guide to elements, 1040–1045 phosphorus, 1070–1079 silicon and the Group 4A (14) elements, 1065–1070

I-22

sulfur, 1079–1082 valence electrons for, 1041 maltose, 1217 mantissa The part of a logarithm to the right of the decimal point, A-3 Markovnikov’s rule A rule stating that when a reagent HX adds to an unsymmetrical alkene, the hydrogen atom in the reagent becomes attached to the carbon that already has the larger number of hydrogens, 1165 Mars Climate Orbiter, 31 Mars Oxygen In-situ Resource Utilization Experiment (MOXIE), 234 The Martian (Andy Weir), 234–235 mass A measure of the quantity of matter in a body, 34–37 and atoms, 102–103 common units of, A-11 formula of a compound from combining, 112–113 and moles, 102–103 and weight, A-11 mass defect The difference in mass between an atomic nucleus and the same number of separated protons and neutrons, 1003 mass number (A) The sum of the number of protons and neutrons in the nucleus of an atom of an element, 64–65 mass percent. See percent composition mass relationships in chemical reactions, 191–195 mass spectrometer, 68 mass spectrometry determining a formula by, 114 isotopic abundance by, 116–117 molar mass and isotopes in, 114–117 matches, phosphorus sulfide in, 1077 mathematics of chemistry, 41–46 operations using logarithms, A-3–A-4 matter Anything that has mass and occupies space, 3, A-6 classification of, 8–12, 10 dispersal of, 892–893 gases, 9 heterogeneous mixtures, 10–12 homogeneous mixtures, 10–12 kinetic-molecular theory, 8–9 law of conservation of, 6, 140 levels of, 9 liquids, 9 at macroscopic levels, 9–10 at particulate levels, 9–10 pure substances, 10 solids, 8 states of, 8–9, 541–543 Maxwell-Boltzmann curves, 518 mean square speed, of gas molecules, 519 measurements accuracy, 38 concentrations of compounds in solution, 209–215 energy units, 37

experimental error, 38–39 length, volume, and mass, 34–37 precision, 37 SI base units, 31–32, 32t, 34 standard deviation, 40–41 temperature scales, 32–33, 33 uncertainties in, 37–41 units of, 30–37, 32t mechanical energy The energy due to the motion of macroscopic objects, 20 mechanism, reaction. See reaction mechanism medical imaging, 1024–1025 melamine, pet food adulterated with, 568–569 melting point The temperature at which the crystal lattice of a solid collapses and solid is converted to liquid, 1110 of elements and compounds, 610t of transition metals, 1110 membrane cell, 1227–1228 proton exchange, 950–951 semipermeable, 647–648, 649 memory metal, 1148 meniscus The curved boundary at the interface between a liquid and another phase, 566, 567 mercury from cinnabar, 4, 4 emission spectra of, 313 line emission spectrum of, 313 mercury barometer, 499 mercury electrode, 943 messenger RNA (mRNA) The RNA formed in transcription to transport the genetic information from DNA to where protein synthesis will occur in the cell, 1223 metabolic acidosis, 872 metabolic alkalosis, 872 metabolism The entire set of chemical reactions that take place in the body, 1230–1236 compounds in biochemical oxidation– reduction reactions, 1232–1233 energy and ATP, 1231–1232 of glucose in respiration, 1233–1235 photosynthesis, 1235–1236 metal(s) An element characterized by a tendency to give up electrons and by good thermal and electrical conductivity, 75 and band theory, 597–599 in biochemistry, 376 bonding in, 596–602 electrical conductivity in, 599 and electron sea model, 597 mixtures of, 606–608 oxides of, 161–163 radii, 1109–1110 unit cells for, 584–588 metal atom radii, 1109–1110

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

metal bonding band theory, 597 electron sea model, 597 metal cations in water, 814 metallic aluminum and production, 1060–1061 metallic bonding forces The coulombic forces of attraction between the electrons and the charged nuclei in metals, 389 metalloid(s) An element with properties of both metals and nonmetals, 75 antimony, 71 metallurgy The process of obtaining metals from their ores, 1111 copper production, 1113–1114 hydrometallurgy, 1113–1114 iron production, 1111–1113 pyrometallurgy, 1111–1113 metal sulfide solubility, 862 metaphosphoric acid, 1078t metathesis reaction(s). See exchange reaction(s) meter (m) The SI base unit of length, 32, A-10t methane (CH4), 1254–1256 bubbles trapped in ice, 3 emissions, 1255 hydrates, 1256, 1269 molecular geometry of, 420 molecules, 86 and natural gas, 1269–1270 methanol (CH3OH) boiling point of, 1174 bonding in, 464–465 hydrogen bonding involving, 557 methanol fuel cell car, 199 as polar molecule, 1174 properties and uses of, 1174 as simplest alcohol, 1170 valence bond description of bonding in, 464–465 and zeolites, 1069 methanol fuel cell car, 199, 199 method of successive approximations, A-5 N-methylacetamide (amide), 1182 methyl ethyl ketone, 1178t methyl methacrylate, 232 methyl propanoate, A-17 metric system A decimal system for recording and reporting measurements, in which all units are expressed as powers of 10 times some basic unit, 31 selected prefixes used in, 32t metric units, prefixes used with, A-11 mica (sheet silicate), 1068 microplastics, 1262, 1262–1263 microstate A specific configuration describing how energy can be distributed throughout a thermodynamic system with a certain probability, 891 Millikan’s experiment, 73

milliliter (mL) A unit of volume equivalent to one thousandth of a liter; 1 mL = 1 cm3, 36, A-12t millimeter of mercury (mm Hg) A common unit of pressure, defined as the pressure that can support a 1-millimeter column of mercury; 760 mm Hg = 1 atm, 500 mineral oil, density of, 50–51 minerals, 859, 863 analysis of, 203 boron, 1059–1060 calcium, 1056–1057 clay, 80 containing calcium, 1055 pyroxenes, 1067 silicate, 1067–1068 solubility of, 859 sulfide-containing, 1079 miscible liquids Two liquids that are soluble in each other to an appreciable extent, 632 mixture(s) A combination of two or more substances that can be separated by physical techniques, 13 heterogeneous, 10–12 homogeneous, 10–12 quantitative analysis of, 202–203 separating by selective precipitation, 155–156 mobile phase, in chromatography, 567–568 model for energy changes on dissolving KF, 635 models, molecular, 14, 85–86 models of substrate binding, 706 moderator, nuclear, 1027 Mohr method, 215, 234 molal boiling point elevation constant (Kbp) The proportionality constant relating the raising of boiling point to the molality of the solution, 643 molal freezing point depression constant (Kfp) The proportionality constant relating the lowering of freezing point to the molality of the solution, 644, 645 molality (m) The number of moles of solute per kilogram of solvent, 628 molar absorptivity, 229 molar enthalpy of vaporization (ΔvapH°) The energy required to vaporize 1 mole of a liquid, 558, 559t molar heat capacity The quantity of heat required to raise the temperature of 1.00 mol of a substance by 1.00 kelvin, 258 molarity (M) The number of moles of solute per liter of solution, 209–212 amount and, 212 defined, 209 molar mass (M) The mass in grams of one mole of particles of any substance, 100, 547t atoms and, 100–103

and compounds, 103–106 determination, and colligative properties, 651–653 determining by titration, 224–225 of gas from pressure (P), temperature (T), and volume (V), 511–513 Graham’s law of effusion to calculate, 523–524 in mass spectrometry, 114–117 and molecules, 103–106 and moles, 105–106 molar volume, standard, 509, A-13 mole (mol) The SI base unit for amount of substance, 100, 101, 104 and atoms, 102–103 and mass, 102–103 molar mass and, 105–106 of reaction, 272, 677 of reaction and limiting reactants, 200 molecular collisions effect of concentration on frequency of, 696 molecular composition, 107 molecular compound(s) A compound formed by the combination of atoms without significant ionic character, 89. See also covalent compound(s) of main group elements, 1042–1044 naming, 86–87 molecular formula A written formula that expresses the number of atoms of each type within one molecule of a compound, 84–85, 85, 109 determining, 109 dimethyl ether, 85 from percent composition, 108–111 molecular geometry The arrangement in space of the central atom and the atoms directly attached to it, 418, 420–422 and lone pairs, 420–422 and multiple bonds, 424 molecular ion, 114 molecularity The number of particles colliding in an elementary step, 708 of elementary steps, 708 molecular Lewis acids, 814–815 molecular Lewis bases, 815–816 molecular models, 14, 85–86 molecular orbital energy level diagram, 483 molecular orbital(s) An orbital created by the combination of atomic orbitals, 473 in benzene, 482, 483 for heteronuclear diatomic molecules, 480–481 for homonuclear diatomic molecules, 474–480 in ozone, 481–482, 482 molecular orbital theory, 473–483 theories of chemical bonding, 483–485 Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I-23

molecular orbital theory A model of bonding in which pure atomic orbitals combine to produce molecular orbitals that are delocalized over two or more atoms, 459, 473–483 principles of, 474–480 resonance, 481–483 molecular polarity, 424–431 molecular scale, 35 molecular shapes, 416–424 central atoms surrounded only by single-bond pairs, 418–419 central atoms with single-bond pairs and lone pairs, 420–423 multiple bonds and molecular geometry, 424 molecular solid(s) A solid formed by the condensation of covalently bonded molecules, 602 molecular speed, 518 kinetic-molecular theory of gases, 517–520 molecular structure acid-base properties and, 807–812 acid strength of hydrogen halides, HX, 807–808 anions as Brønsted bases, 812 and bonding, 807–812 bond properties, 432–437 and Brønsted acids, 810–812 and carboxylic acids, 810–812 chemical bond formation, 388–389 deoxyribonucleic acid (DNA), 437–441, 1219–1221 electronegativity and bond polarity, 390–393 formal charges in covalent molecules and ions, 410–416 HNO2, 808–810 HNO3, 808–810 hydrated metal cations as Brønsted acids, 812 Lewis electron dot symbols, 388–389 Lewis structures of, 393–399 molecular polarity, 424–431 molecular shapes, 416–424 octet rule, 406–410 oxoacids, 808–810 patterns of bonding in Lewis structures, 399–403 resonance, 404–406 VSEPR model of, 416–424 molecular weight, 103. See also molar mass molecule(s) A neutral particle consisting of more than one atom in which the atoms are covalently bonded to each other, 13, 84–88 in aqueous solutions, 147–149 binary molecular compounds, 87–88 and compounds, 103–106 diatomic elements, 80 electron-pair geometries for, 422 formulas, 84–85 heteronuclear diatomic, 480–481 homonuclear diatomic, 477

I-24

models, 85–86 and molar mass, 103–106 molecular geometries for, 422 naming molecular compounds, 86–87 with odd number of electrons, 409–410 with permanent dipole dipole–dipole forces, 546–547 hydrogen bonding, 548–552 properties of water, 550–552 mole fraction (X) The ratio of the number of moles of one substance to the total number of moles in a mixture of substances, 515, 628 molten ionic compounds, 609 molten salts, electrolysis of, 968–969 monatomic ion(s) An ion consisting of one atom bearing an electric charge, 89–91, 170 monatomic anions, 91–92 monatomic cations, 89–91 naming, 91 monodentate ligand A ligand that coordinates to a metal ion by means of one atom and therefore one coordinate covalent bond, 1116 monomer(s) The small units from which a polymer is constructed, 1184 monoprotic acid(s) A Brønsted acid that can donate one proton, 778 monosaccharide The simplest type of sugar, usually having the general formula Cx(H2O)x, 1216–1217 motion, A-6 MRI units, 83 multiple bonds drawing Lewis structures with, 397–399 and molecular geometry, 424 and valence bond theory, 468–472 multiple ion charges, 91 municipal water treatment plant, 1259 Mylar balloons, 1189 naming of aldehydes and ketones, 1179 of alkanes, 1159–1160, A-15–A-16 of alkenes, A-16 of alkynes, A-16 of carboxylic acids, 1179 of common acids, 157–158 of coordination compounds, 1120–1122 of ionic compounds, 96–97 molecular compounds, 14, 86–87 monatomic ions, 91 of polyatomic ions, 92–93, 93t of substituted alkanes nanotechnology, 32 nanotubes, carbon, 614–615 National Institute of Standards and Technology (NIST), 38 natural acids and bases, 787

natural gas, 1269, 1270 as energy resource, 1263–1265 hydrogen production from, 1047 natural logarithms, A-2–A-3 natural radioactivity, 994 natural rubber, 1188, 1188 negative entropy values, 894 neon (Ne), 354 emission spectrum of, 313 photoelectron spectrum of, 371 Nernst equation An equation that relates the potential of an electrochemical cell to the concentrations of the cell reactants and products, 960–963 net ionic equation(s) A chemical equation involving only those substances undergoing chemical changes in the course of the reaction, 153–155 for an acid–base reaction, 165 balancing, 154–155 writing, 154–155, 164 network solid(s) A solid composed of a network of covalently bonded atoms, 602–604 neurotransmitter, 1192 neutralization reaction(s) An acid– base reaction that produces a neutral solution of a salt and water, 164 neutral solution A solution in which the concentrations of hydronium ion and hydroxide ion are equal, 790–791 neutrino(s) A massless, chargeless particle emitted by some nuclear reactions, 999 neutron(s) An electrically neutral subatomic particle found in the nucleus, 1003, 1015 new elements, search for, 1016–1017 newton (N) The SI unit of force; 1 N = 1 kg·m/s2, 500, A-6 nickel-cadmium batteries, 949 nickel(II) complexes and magnetism, 1131 nickel-metal hydride batteries, 949 nicotinamide adenine dinucleotide (NAD+) A compound used in biological oxidation–reduction reactions that consists of two ribonucleotides (one containing the base adenine and the other nicotinamide) connected at their 5’ positions by a phosphate linkage, 1232 nicotine, 1176 Niels Bohr Institute, 314 night vision goggles, 309 nitinol, 617, 622, 1148 nitric acid (HNO3) bonding in, 808–810 factors affecting acid strength of, 808–810

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

molecular structure in, 808–810 preparation of, 1075 properties of, 1075 nitrogen, 354, 1070–1079 ammonia and hydrazine, 1072–1073 and Group 5A (15) elements, 1070–1079 hydrogen compounds of, 1072–1073 in Lewis and Brønsted bases, 816 nitrogen compounds, 1072–1076 and nitrogen oxides, 1248–1249 oxidation numbers for, 1072 oxides of, 1073–1076, 1073t oxoacids of, 1073–1076 properties of elemental nitrogen, 1071–1072 nitrogen cycle, 1248 nitroglycerin, 1172 nitrous acid (HNO2) acid-base behavior in, 808–810 bonding in, 808–810 molecular structure in, 808–810 nitrous oxide, N2O (laughing gas), 1074 NMR spectrum, 208, 208 noble gas(es) The elements in Group 8A (18) of the periodic table, 83, 83, 1087–1089 xenon compounds, 1089 noble gas notation An abbreviated form of spdf notation that replaces the completed electron shells with the symbol of the corresponding noble gas in brackets, 352 orbital box notation with, 355 nodal surface A surface on which there is zero probability of finding an electron, 327 of d orbitals, 327 of p orbitals, 327 node(s) A point of zero amplitude of a wave, 322 nonbonding electrons, 394. See also lone pair(s) nonelectrolyte A substance that dissolves in water to form an electrically nonconducting solution, 148, 149 non-expansion work and enthalpy, 270 and internal energy, 270 nonmetal(s) An element characterized by a lack of metallic properties, 75, 161–163 nonpolar molecules, 424–431 intermolecular forces involving, 552–555 properties of some, 547t, 553t solubility in polar and nonpolar solvents, 547 nonspontaneous reaction. See reactantfavored reaction(s) at equilibrium normal boiling point The boiling point of a liquid when the external pressure is 1 atm, 564

northwest-southeast rule A productfavored reaction involves a reducing agent below and to the right of the oxidizing agent in the table of standard reduction potentials, 955 n-type semiconductor An extrinsic semiconductor whose dopant atoms have more valence electrons than the host atoms, 602 nuclear binding energy The energy required to separate the nucleus of an atom into protons and neutrons, 1003–1005 nuclear chemistry The study and use of radioactive elements and their compounds, 993 analytical methods, 1026–1027 applications of, 1023–1027 artificial nuclear reactions, 1014–1017 carbon-14 as tracer, 1025–1026 defined, 993 isotope dilution, 1026–1027 medical imaging, 1024–1025 natural radioactivity, 994 nuclear fission and nuclear fusion, 1018–1021 nuclear medicine, 1024–1025 nuclear reactions and radioactive decay, 995–1000 nucleosynthesis, 1006–1008 origin of the elements, 1006–1008 radiation health and safety, 1021–1023 radiation therapy, 1025 radioactive tracers, 1026 rates of nuclear decay, 1008–1013 stability of atomic nuclei, 1000–1005 nuclear decay half-life, 1008–1010 kinetics of, 1010–1012 radiocarbon dating, 1012–1013 rates of, 1008–1013 nuclear electricity generated by some countries, 1265t nuclear energy, 996, 1021 nuclear fission, 1018, 1018, 1018–1021 nuclear fusion The process by which several small nuclei react to form a larger nucleus, 1018–1021 nuclear magnetic resonance (NMR) spectroscopy, 208, 208 nuclear medicine medical imaging, 1024–1025 radiation therapy, 1025 nuclear moderator, 1027 nuclear reaction(s) A reaction involving one or more atomic nuclei, resulting in a change in the identities of the isotopes, 995–1000 artificial, 1014–1017 equations for nuclear reactions, 995–996 radioactive decay series, 996–1000 radon as environmental problem, 1000 nuclear reactor A container in which a controlled nuclear reaction occurs, 1018–1019

nucleic acid(s) A class of polymers, including RNA and DNA, that are the genetic material of cells, 1219–1225 protein synthesis, 1223–1225 structure, 1219–1220 nucleon A nuclear particle, either a neutron or a proton, 1004 nucleoside A single sugar with its attached nitrogenous base from DNA or RNA, 1220 nucleosynthesis, 1006–1008 Big Bang theory, 1006–1007 in stars, 1007–1008 nucleotide The combination of a nucleoside and an attached phosphate group, 1220 nucleus The core of an atom, made up of protons and neutrons, 64 numbers with trailing zeroes and significant figures, 44 Nylon-6,6, 1190 oceans, 1258 acidification, 1274–1276 CO2 in, 21–22 octadecane, 568 octahedral complexes high spin, 1130–1133, 1131 low-spin, 1130–1133, 1131 octahedral electron-pair geometry, 418, 422–423 octahedral hole The empty space at the center of a face-centered cubic unit cell, 589 octet rule When forming bonds, atoms of main group elements gain, lose, or share electrons to achieve a stable configuration having eight valence electrons, 389, 394 atom with fewer than eight valence electrons, 406–407 atom with more than eight valence electrons, 407–409 molecules with odd number of electrons, 409–410 oil A lipid that is liquid at physiological temperatures, has a primary function of energy storage, and consists of a triester of glycerol and three carboxylic acids, 1226 Omega-3 tablets, 1180 optical isomers Isomers that are nonsuperimposable mirror images of each other, 1124–1127, 1125, 1154, 1154 orbit, 323 orbital(s) The matter wave for an allowed energy state of an electron in an atom or molecule, 323–324, 324t, 346 atomic (See atomic orbital(s)) for carbon, 462 molecular (See molecular orbital(s)) and quantum numbers, 328 Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I-25

orbital angular momentum quantum number (ℓ), 323 orbital box diagram A notation for the electron configuration of an atom in which each orbital is depicted as a box and the number and spin direction of the electrons are shown by arrows, 346, 355, 358 for elements Ca through Zn, 358t orbital box notation, 355 with noble gas notation, 355 orbital energies, Z*, 359 orbital hybridization The combination of atomic orbitals to form a set of equivalent hybrid orbitals that minimize electron-pair repulsions, 462 orbital overlap Partial occupation of the same region of space by orbitals from two atoms, 459–461 order, bond. See bond order ore(s) A sample of matter containing a desired mineral or element, usually with large quantities of impurities, 167–168 organic chemistry The chemistry of compounds containing carbon (except the oxides of carbon, carbides, and carbonates), 1151 organic compounds, 1151–1205 and carbon, 402–403 orthophosphoric acid, 1078t osmosis The movement of solvent molecules through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration, 647, 648, 650 and living cells, 651 at particulate level, 648 osmotic pressure (Π) The pressure exerted by osmosis in a solution system at equilibrium, 647–651 Ötzi, the Iceman of the Alps, 117, 117–118 overall reaction order, 681 oxidation The loss of electrons by an atom, ion, or molecule, 167 of ammonia, 196 of copper by silver ions, 934 of transition elements, 1107–1109 of transition elements in first transition series, 1109 oxidation number(s) A number assigned to each element in a compound in order to keep track of the electrons during a reaction, 170, 171, 933 determining, 170–171 guidelines for assigning, 170 oxidation–reduction reactions, 169–171 oxidation–reduction equations balancing equations in acidic or basic solutions, 935–940 oxidation–reduction reaction(s) A reaction involving the transfer of one or more electrons from one species to another, 138, 167–173, 172, 175, 226, 933–940

I-26

balancing equations, 934–940 common molecules used in biochemical, 1232–1233 common oxidizing and reducing agents, 172t and electron transfer, 168–169 oxidation numbers, 169–171 recognizing, 169, 169, 171–172, 172t titrations using, 225–227 oxides nitrogen, 1248–1249 of nitrogen, 1073–1076 of nonmetals and metals, 161–163 phosphorus, 1076, 1078t oxidizing agent(s) The substance that accepts electrons and is reduced in an oxidation–reduction reaction, 168, 169, 172, 172t, 173, 958–960 relative strengths of, 958–960 oxoacid(s) A type of acid in which an atom (usually a nonmetal) is bonded to one or more oxygen atoms, 808, 808t of chlorine, 1086–1087 and Lewis structures, 402 of nitrogen, 1073–1076 oxoanion(s) Polyatomic anions containing oxygen, 92 oxy-acetylene torch, 1164 oxygen, 81, 354, 1249–1250 and Group 6A (16) elements, 1079–1082 liquid, 458 molecular orbitals, 478 orbital hybridization in, 469 preparation and properties of, 1080–1081 reactions of metal and two nonmetals with, 142 and sulfur, 1080 sulfur compounds, 1081–1082 oxygen binding to hemoglobin, 1118 ozone, 1250–1253 in the atmosphere, 404 molecular orbital model for, 482 ozone hole, 1251–1253, 1252 ozone layer The accumulation of ozone (O3) in the stratosphere that acts as a shield protecting the planet from damaging ultraviolet rays from the sun, 1250 pairing energy The additional potential energy due to the electrostatic repulsion between two electrons in the same orbital, 1130 pandemic and air pollution, 1272 paraffin wax and mineral oil, 1160 paramagnetic, 362 paramagnetism The physical property of being attracted by a magnetic field, 361–363, 364, 364, 473, 1130 parent ion, 114 Parkinson’s disease, 1192

partial charge(s) The charge on an atom in a molecule or ion calculated by assuming sharing of the bonding electrons proportional to the electronegativity of the atom, 390, 428 for acetic acid, 810 partial pressure(s) The pressure exerted by one gas in a mixture of gases, 514, 741 of gases, 514–517 particulate level Representations of chemical phenomena in terms of atoms and molecules; also called submicroscopic level, 9 partition chromatography, 568 parts per million (ppm) The number of grams of solute per million grams of solution, 629 pascal (Pa) The SI unit of pressure; 1 Pa = 1 N/m2, 500, A-7 p atomic orbitals, 326–327 in π (pi) molecular orbitals, 478 in σ (sigma) molecular orbitals, 478 Pauli exclusion principle No two electrons in an atom can have the same set of four quantum numbers, 345–347 p-block elements Elements from Group 3A (13) through Group 8A (18), which have the outer shell configuration ns2npx, where x varies from 1 to 6, 353 peer review, 6 peptide bond The amide linkage connecting two amino acids in an amino acid polymer, 1210, 1210 percent abundance The percentage of the atoms of a natural sample of the pure element represented by a particular isotope, 68 percent composition The percentage of the mass of a compound represented by each of its constituent elements, 106–108 calculating formula from, 109–111 chemical analysis, 106–108 empirical formula from, 108–111 molecular formula from, 108–111 percent error The difference between the measured quantity and the accepted value, expressed as a percentage of the accepted value, 38, 39 percent yield The actual yield of a chemical reaction as a percentage of its theoretical yield, 200–202, 201 calculating, 201–202 defined, 201 periodic groups, similarities within, 1041t periodicity The periodic repetition of the properties of elements, 78 periodic table of the elements A listing of the elements in order of atomic number where elements with similar chemical and physical properties are placed in the same column of the table, 13 and electron configurations, 350

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

of the elements, 76 features of, 74–77 groups or families, 74–75, 75 as guide to elements, 1040–1045 ionic compounds of main group elements, 1041–1042 main group elements, 74 Mendeleev and, 78 metalloids, 75 metals, 75 molecular compounds of main group elements, 1042–1044 nonmetals, 75 overview, 77–84 semimetals, 75 transition elements, 74 using group similarities, 1044–1045 valence electrons for main group elements, 1041 periodic trends atomic properties and, 364–374 atomic size, 364–367 and chemical properties, 374–375 in the d-block, 1109–1110 electron attachment enthalpy and electron affinity, 369–370 ionization energy (IE), 367–369 ion sizes, 372–374 periodic trends in d-block elements density, 1110 melting point, 1110 metal atom radii, 1109–1110 periods The horizontal rows in the periodic table of the elements, 75 in the periodic table, 75 permanent dipole dipole–dipole forces, 546–547 hydrogen bonding, 548–552 and molecules, 546–552 properties of water, 550–552 peroxides, 170 peroxyacetyl nitrate (PAN), 1249 petroleum, 1155, 1270–1271 chemistry of, 1271 fractional distillation of, 660 reserves of, 1270 tar sands (oil sands), 1270, 1270 PET scans, 1025 PET technology, 1024 PFOA (perfluorooctanoic acid, C8HF15O2), 1263 PFOS (perfluorooctane sulfonic acid C8HF17O3S), 1263 pH The negative of the base-10 logarithm of the hydrogen ion concentration; a measure of acidity, 782, 963 of aqueous solution of weak acid, 798–805 of aqueous solution of weak base, 798–805 of buffer, 831–842 as concentration scale for acids and bases, 215–217 determining, 216 and equilibrium constants, 796–798

indicators, 851–853 meter, 216, 963, 963 scale, 780–782 of solutions, 217 during titrations, 843–853 values of common substances, 216 phase changes, 609–613 conversion of solid into liquid, 609–610 conversion of solid into vapor, 611 melting, 609–610 phase diagrams, 611–613 sublimation, 611 phase diagram A graph showing which phases of a substance exist at various temperatures and pressures, 611–613 for carbon dioxide, 613, 613 and thermodynamics, 612–613 for water, 611–612, 612 phenolphthalein (acid–base indicator), 851 phosphate rock uses of, 1077 phosphating The application of a thin metal phosphate coating on the surface of a metal to inhibit corrosion, 977 phospholipids A major component of cell membranes consisting of a triglyceride where one carboxylic acid esterified to glycerol is replaced by a phosphate which in turn is bound to another hydrocarbon chain, 1227–1228 phosphoric acid, uses of, 1077 phosphorous acid, 1078t phosphorus, 80–81, 1071–1072 and Group 5A (15) elements, 1070–1079 hydrogen compounds of, 1076 phosphorus oxides and sulfides, 1076–1077 phosphorus oxoacids and salts, 1077–1079 properties of, 1071–1072 red allotrope of, 1071 uses of, 1077 white allotrope of, 1071 photoelectric cell, 310 photoelectric effect The ejection of electrons from a metal bombarded with light of at least a minimum frequency, 310–311 photoelectron spectroscopy (PES), 371, 485–486 photon(s) A “particle” of electromagnetic radiation having zero mass and an energy given by Planck’s law, 308–312 photosynthesis A biological process taking place in plants where the energy in sunlight is used to synthesize sugars from CO2 and H2O, 1235–1236, 1249 physical change A change that involves only physical properties, 18–19, 18

physical constants, A-13–A-14, A-13t physical properties Properties of a substance that can be observed and measured without changing the composition of the substance, 15, 15t of hydrogen, 19, 1046–1047 temperature dependence of, 16 pi bond(s) The second (and third, if present) bond in a multiple bond; results from sideways overlap of p atomic orbitals, 468 in benzene, 472–473 pKa The negative of the base-10 logarithm of the acid ionization constant, 788 planar node. See atomic orbital(s) and nodal plane Planck’s constant (h) The proportionality constant that relates the frequency of radiation to its energy, 308–312 Planck’s equation, 308–310 plasma A gas-like phase of matter that consists of charged particles, 1021 Plaster of Paris, 98 pOH The negative of the base-10 logarithm of the hydroxide ion concentration; a measure of basicity, 782 poisoning arsenic, 1261–1262 atropine, 816 carbon monoxide, 1111 hydrogen sulfide, 1081 lead, 36, 1070, 1089, 1251, 1260–1261 polar covalent bond A covalent bond in which there is unequal sharing of the bonding electron pair, 390 polarizability The extent to which the electron cloud of an atom or molecule can be distorted by an external electric charge, 553 polarization, 553 of OOH bonds, 812 polar molecules, 391, 427 polar substances, 547t pollution air, 1271–1272 water, 1260–1262 polyamide(s) A condensation polymer containing amide linkages between what were carboxylic acid and amine groups in the monomers, 1190–1191 polyatomic anions, 92, 398 polyatomic ion(s) An ion consisting of more than one atom, 92–93, 93t, 393–399 formulas and names of some common, 93t and Lewis electron dot structures, 394–399 naming, 92–93 polydentate ligand A ligand that coordinates to a metal ion by means of more than one atom, 1116 Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I-27

polyester(s) A condensation polymer containing ester linkages between what were carboxylic acid and alcohol groups in the monomers, 1189, 1189, 1189–1190 polyethylene, 1185–1187, 1186 polylactic acid (PLA), 1189 polymer(s) A large molecule composed of many smaller repeating units, usually arranged in a chain, 1184–1191 addition polymers, 1185–1188 classifying, 1184–1185 condensation polymers, 1188–1191 defined, 1184 polymer-based consumer products, 1185 polymethyl methacrylate (PMMA), 232, 232 polyolefins, 1185–1187 polypeptide An amino acid polymer, 1210 polyprotic acid(s) A Brønsted acid that can donate more than one proton, 778, 779t, 805–807, 848–849 Ka and Kb values for, 787 polyprotic base(s) A Brønsted base that can accept more than one proton, 779, 779t, 805–807 polysaccharide(s) A polymer in which the monomers are sugar molecules, 1214, 1218, 1218–1219 p orbital(s), 326–327, 327. See atomic orbital(s) positron emission, 998 positron(s) A particle having the same mass as an electron but a positive charge, 998 positron emission tomography (PET), 1024–1025 potassium compounds, 1051–1054 and lithium and sodium compounds, 1051–1054 preparation of, 1050–1051 properties of, 1051 potassium permanganate (KMnO4), 210, 230 dissolving in water, 893 potential energy The energy that results from an object’s position, 20 power The amount of energy delivered per unit time energy and, A-7–A-8 precipitate A water-insoluble solid product of a reaction, usually of watersoluble reactants, 151 precipitation reaction(s) An exchange reaction that produces an insoluble salt, or precipitate, from soluble reactants, 151–156, 174, 863–867 net ionic equations, 153–155 and reaction quotient, Q, 864–867 and solubility product constant, Ksp, 864–867 writing equation for, 152–153

I-28

precision The agreement of repeated measurements of a quantity with one another, 37, 38 and accuracy, 38 measurements, 37 and standard deviation, 40–41 prediction, as goal of science, 6 prefixes used in the metric system, 32t used with traditional metric units and SI units, A-11 pressure The force exerted on an object divided by the area over which the force is exerted, 499, A-7 conversions, 500–501, A-7 critical, 565 and gas solubility, 637–638, 638 and ideal gas law, 508–512 and kinetic-molecular theory, 520–521 molar mass of gas from, 511–513 osmotic, 647–651 partial, 514–517 vapor pressure of liquid, 561–565 and volume of a gas, 502–503 primary alcohols, 1177 primary amines, 1175 primary battery A battery that cannot be returned to its original state by recharging, 946–947 primary colors, 1134, 1134 primary standard A pure, solid acid or base that can be accurately weighed for preparation of a titrating reagent, 223 primary structure The specific sequence of amino acids that are linked by peptide bonds in a protein, 1211 primitive cubic (pc) unit cell A unit cell consisting of eight identical particles positioned at the corners of a cube, 583 principal quantum number (n) A positive, unitless number represented by n that is used to describe electron orbital size, 314, 323 principle of electroneutrality The observation that electrons in a molecule or polyatomic ion will be distributed in such a way that the charges on all atoms are as close to zero as possible, 414 probability density The probability of finding an atomic electron within a given region of space, related to the square of the electron’s wavefunction, 322 problem solving and chemical arithmetic, 49–51 by dimensional analysis, 47–48 Problem Solving Tip balanced equations and equilibrium constants, 755 balancing equations in basic solution, 940

balancing oxidation–reduction equations, 938 buffer solutions, 842 calculating ΔT, 261 choosing the central atom in a dot structure, 396 concepts of thermodynamics, 888 determining a rate equation, 685 determining strong and weak acids, 786 drawing structural formulas, 1158 electrochemical conventions for voltaic cells and electrolysis cells, 969 entropy-favored processes, 895 finding empirical and molecular formulas, 109 formulas for ions and ionic compounds, 94 general approach to finding an empirical formula by chemical analysis, 204 ligand field theory, 1133 moles of reactions with a limiting reactant, 200 naming aldehydes, ketones, and carboxylic acids, 1179 pH after mixing equal amounts (moles) of an acid and a base, 805 preparing a solution by dilution, 214 recognizing gas-forming reactions, 166 relating rate equations and reaction mechanisms, 716 stoichiometry calculations, 193 stoichiometry calculations involving solutions, 218 using a calculator, 45 using the quadratic formula, 753 writing electron configurations, 355 writing net ionic equations, 164 product(s) A substance formed in a chemical reaction, 19, 140 product-favored reaction(s) at equilibrium A reaction in which reactants are completely or largely converted to products at equilibrium, 146, 285–286 and the equilibrium constant, 742–743 and standard cell potential, 954 and standard free energy change, 904 production of aluminum, 1060–1061, 1061 and boron minerals, 1059–1060 of nitric acid, 196 of water gas, 1047 properties of liquids boiling point, 564–565 capillary action, 566–567 Clausius–Clapeyron equation, 563–564 condensation, 558–561 critical pressure, 565 critical temperature, 565–566 enthalpy of vaporization, 563–564 surface tension, 566–567 vaporization, 558–561 vapor pressure, 561–564 viscosity, 566–567

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

protein(s) A polymer formed by con­ densation of amino acids, 1208–1215 amino acids are building blocks of, 1208–1211 drawing structures of, 1211–1213 as emulsifying agent, 658 enzymes, active sites, and lysozyme, 1213–1215 hemoglobin, 1211–1213 primary structure, 1211 protein structure and hemoglobin, 1211–1213 quaternary structures of, 1212 secondary structures of, 1212 structural levels of, 1212–1213 synthesis, 1223–1225, 1224 tertiary structures of, 1212 protium, 66 proton(s) A positively charged subatomic particle found in the nucleus, 159 proton exchange membrane, 950–951 pseudo–first order reaction, 682 p-type semiconductor An extrinsic semiconductor whose dopant atoms have fewer valence electrons than the host atoms, 601 pure substance A form of matter that cannot be separated into two different species by any physical technique, and that has a unique set of properties, 10 purification heterogeneous mixture by filtration, 11 of silicon by zone refining, 1066 P–V (pressure-volume) work, 268 pyrometallurgy Recovery of metals from their ores by high-temperature processes, 1111–1113 pyrophosphoric acid, 1078t quadratic equations, 751, A-4–A-5 quadratic formula, A-4 qualitative chemical analysis, 156 qualitative information Nonnumerical experimental observations, such as descriptive or comparative data, 5 quantitative analysis of a mixture, 202–203 quantitative chemical analysis The determination of the amount or concentration of a substance in a sample, 202–203 quantitative chemistry, 30–61 experimental error, 37–41 graphs and graphing, 48–49 making measurements, 37–41 mathematics of chemistry, 41–46 precision and accuracy, 37–41 problem solving and chemical arithmetic, 49–51 problem solving by dimensional analysis, 47–48



standard deviation, 37–41 units of measurement, 30–37, 32t quantitative information Numerical experimental data, such as measurements of changes in mass or volume, 5 quantization The observation that only certain energy levels are allowed in a system, 309 Einstein, 308–312 of electron spin, 330 of energy, 308–312, 890 of matter, 890 photons, 308–312 Planck’s assumption of, 308–312 quantum mechanics A general theoretical approach to atomic behavior that describes the electron in an atom as a matter wave, 319–321 quantum numbers and orbitals, 323–324 shells and subshells, 324–325 quantum number(s) A set of numbers that define the properties of an atomic orbital, 322, 323–324 allowed values of, 323–325 assigning to electrons, 355 electron configurations and, 356–357 and energy, 323 interrelationships of, 324t magnetic, 324 and orbital, 324t orbital angular momentum, 323 orbital information conveyed, 324t orbitals and, 328 Pauli exclusion principle and, 345 principal, 323 spin, 330 quartz, 604 forms of, 1066 quaternary structure A structure in proteins containing more than one polypeptide chain that indicates spatial interactions of the different chains, 1212 quicklime, 163 rad A unit of radiation dosage, 1022 radial distribution plot, 325 radiation doses and effects, 1022–1023 effects of a single dose of, 1023t exposure from natural and artificial sources, 1023t health and safety, 1021–1023 units for measuring radiation, 1021–1022 radiation therapy, 1025 radioactive decay, 995–1000, 1002–1003 band of stability and, 1002–1003 radioactive decay series A series of nuclear reactions by which a radioactive isotope decays to form a stable isotope, 996–1000 actinium (4n + 3), 997 neptunium (4n + 1), 997

thorium (4n), 997 uranium-238 (4n + 2), 996 radioactive tracers, 1026 radioactivity, 73, 73, 994, 1022 discovery of, 73, 82, 994 radiocarbon dating, 1012–1013 radioisotopes used in medical diagnostic procedures, 1024t Radium Institute in Paris, 82 radius atomic, 364−365 ionic, 372−373 radon detector, 1000 as environmental problem, 1000 Raoult’s law The vapor pressure of the solvent is proportional to the mole fraction of the solvent in a solution, 640–642 rare earth elements, 83, 84, 375. See lanthanide(s) rare gas(es). See noble gas(es) Raschig reaction, 712, 1072 rate. See reaction rate(s) rate constant (k) The proportionality constant in the rate equation, 680, 681–682 Arrhenius equation, 699 graphical methods for, 690 half-life and, 691 units of, 682 rate-determining step The slowest elementary step of a reaction mechanism, 711 rate equation(s) The mathematical relationship between reactant concentrations and reaction rate, 680 determining, 682–685 for elementary steps, 708–710 and reaction mechanisms, 710–716 reaction mechanisms involving an initial equilibrium step, 713–715 reaction mechanisms involving freeradical chain reaction, 715–716 and reaction rate, 680–681 stoichiometry and, 695 rate law, 680 rate of enzyme-catalyzed reaction, 706 rate of reaction, 674 reactant(s) A starting substance in a chemical reaction, 19, 140 moles of reaction and limiting, 200 stoichiometry calculation with limiting, 196–199 reactant-favored reaction(s) at equilibrium A reaction in which only a small amount of reactants is converted to products at equilibrium, 147, 285–286 and the equilibrium constant, 743 and standard cell potential, 954 and standard free energy change, 904

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I-29

reaction(s) A process in which substances are changed into other substances by rearrangement, combination, or separation of atoms, 139 (n, ), 1015 acid–base, 163–167. See also acid–base reaction(s). addition, 1064 autoionization, 780 balancing chemical equations, 142–144 chemical equilibrium, 145–147 classifying, in aqueous solution, 173–176 chain, 715, 1018, 1236 condensation, 1088, 1210, 1217 disproportionation, 1086 effect of concentration on, 680–685 electron transfer, 167–173. See also oxidation–reduction reaction(s). enthalpy change for, 271–274, 283 enthalpy of, 434–437 esterification, 1181 exchange, 151, 174 free energy change for, 904 gas-forming, 165–167 hydrogenation, 1165 hydrolysis, 1181 neutralization, 164 neutron capture, 1125 nuclear, 995–1000 oxidation–reduction, 167–173, 933–940. See also oxidation– reduction reaction(s). precipitation, 151–156, 863–867 Raschig, 712 rates of, 673–679 spontaneity of, predicting, 898 substitution, 1170 synthesis, 174 transesterification, 1189 reaction coordinate diagram A diagram on which the y-axis is energy (or enthalpy), and the x-axis is a measure of a reaction’s progress, 696–697 comparing a catalyzed and uncatalyzed reaction, 704 for the addition of HBr to 1,3-butadiene, 915 for a two-step reaction, 697 reaction favorability and Gibbs free energy ∆G, 905 reaction intermediate A species that is produced in one step of a reaction mechanism and completely consumed in a later step, 703 reaction mechanism(s) The sequence of events at the molecular level that control the speed and outcome of a reaction, 673, 706–716 involving free-radical chain reaction, 715–716 involving initial equilibrium step, 713–715 molecularity of elementary steps, 708 and rate equations, 710–716 rate equations for elementary steps, 708–710

I-30

reaction order The exponent of a concentration term in the reaction’s rate equation, 690, 691 reaction quotient (Q) The product of concentrations of products divided by the product of concentrations of reactants, each raised to the power of its stoichiometric coefficient in the chemical equation, 738–745. See also equilibrium constant and equilibrium constant, 738–745 and Gibbs free energy, 905 and relation to cell potential, 960 and solubility product constant, Ksp, 864–865 and standard free energies, 912–914 reaction rate(s) The change in concentration of a reagent per unit time and activation energy, 696–698 Arrhenius equation, 699–701 of chemical reactions, 674–676 and collision theory, 695–698, 699 and concentration, 695–696 determining a rate equation, 682–685 effect of catalysts on, 702–705 effect of molecular orientation on, 699 order of reaction, 681 and rate constant, k, 681–682 and rate equations, 680–681 and temperature, 698, 699 rechargeable batteries, 946, 947–950 recycling, 1190 red blood cells, 1213 redox reaction(s), 169, 933. See also oxidation–reduction reaction(s) electrochemical process, 975–977 in environment, 975–978 reducing agent(s) The substance that donates electrons and is oxidized in an oxidation–reduction reaction, 167, 168, 172, 172, 173, 958–960 relative strengths of, 958–960 reduction The gain of electrons by an atom, ion, or molecule of Cu2+ by Al, 935 half-reactions, potential ladder of, 954 of vanadium (V) with zinc, 936 relative atomic mass The ratio of the average mass per atom of an element to 1/12 of the mass of an atom of 12C, 64–67. See also atomic weight relative rates, 677–678 rem A unit of radiation dosage to biological tissue, 1022–1023 renewable resources, 1265, 1280 replication The process whereby two identical double-stranded DNA molecules are produced from a single double-stranded DNA molecule, 1221 resonance, 404–406 molecular orbital theory and, 481–483 resonance stabilization The additional stability a compound gains from the delocalization of electrons, 1169

resonance structure(s) The possible structures of a molecule for which more than one Lewis structure can be written, differing by the number of bond pairs between a given pair of atoms, 404 bond length in, 433 bond order in, 432 formal charge in equivalent, 412 formal charge in nonequivalent, 414–415 resources energy, 1265–1266 respiration metabolism of glucose in, 1233–1235 respiratory acidosis, 873 respiratory alkalosis, 872 reverse osmosis The application of pressure greater than the osmotic pressure of impure solvent to force solvent through a semipermeable membrane to the region of lower solute concentration, 649, 1259 reversible chemical reactions, 145 and dynamic equilibrium, 737 reversible process A process for which it is possible to return to the starting conditions along the same path without altering the surroundings, 890 ribonucleic acid (RNA) A biopolymer similar to DNA that plays a role in protein synthesis and other processes, 1219, 1219 messenger RNA (mRNA), 1223–1225 mRNA vaccines, 1228–1229 transfer RNA (tRNA), 1223–1225 ribosome The place in a cell where protein synthesis takes place, consisting of a complex body containing proteins and RNA, 1223 Ritz-Paschen series, 318 rms speed. See root-mean-square speed RNA. See ribonucleic acid roentgen A unit of radiation dosage, 1022 root-mean-square (rms) speed The square root of the average of the squares of the speeds of the molecules in a sample, 519 roses, (indicator), 851 ROY G BIV spectrum of colors, 1134 Rutherford’s experiment, 74 Rydberg constant (R) A constant used to predict the wavelength of light that results from an electron moving between energy levels, 313 Saccharin (C7H5NO3S), 1167 sacrificial anode A more readily oxidized metal that is attached to a metal surface to protect it from corrosion using cathodic protection, 978 salt(s) An ionic compound whose cation comes from a base and whose anion comes from an acid, 1050 acid-base properties of, 789–791 molten, 968–969

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

and phosphorus oxoacids, 1077–1079 in seawater, 233–234 solubility of, 149–151, 853–863 salt bridge A device for maintaining the balance of ion charges in the compartments of an electrochemical cell, 941 SARS-CoV-2 virus, 1229 saturated compound(s) A hydrocarbon containing only single bonds, 630, 854, 1155. See also alkane(s) saturated solution(s) A stable solution in which the maximum amount of solute has been dissolved, 630 and Ksp, 864 s-block elements Elements of Group 1A (1) and Group 2A (2), which have the valence electron configurations ns1 and ns2, respectively, 353 science control, 6 dilemmas and integrity in science, 6 goals of, 6 methods of, 4–6 prediction, 6 scientific instruments and glassware, 31 scientific notation A way of presenting very large or very small numbers in which the number is represented as the product of a number between 1 and 9.999... and a power of ten, 42 screening constant (S), 349 scuba diving, 627 sea level rise, 1258 sea urchins, 1276 seawater anions in, 1258t salt in, 233–234 second law of thermodynamics The total entropy of the universe is continually increasing, 889 secondary alcohols, 1177 secondary amines, 1175 secondary battery A battery in which the redox reaction can be reversed and the battery can thus be recharged, 946, 947–950 lead storage battery, 948 lithium batteries, 949–950 nickel-cadmium batteries, 949 nickel-metal hydride, 949 secondary structure A type of protein structure signifying the spatial arrangement of the amino acid sequence in regular patterns, 1212 second-order reactions, 688–689 seesaw molecular geometry, 422, 422 selective precipitation separating a mixture by, 155–156 selenium, 81, 1080

semiconductor(s) Substances that can conduct small quantities of electric current, 600–602 and band gap, 600–602 bonding in, 596–602 semimetals, 75. See also metalloid(s) semipermeable membrane A thin sheet of material through which only certain types of molecules can pass, 647–648, 649 serial dilutions, 215 sevoflurane, 1084–1085 shells and subshells, 324–325 SI Abbreviation for Système International d’Unités, a uniform system of measurement units in which a single base unit is used for each measured physical quantity, 31–32, 31t, 32t, 34 base units of, 32, 34, A-10 derived units, A-11 prefixes used with, 32, A-11 sievert The SI unit of radiation dosage to biological tissue, 1022 sigma bond(s) A bond formed by the overlap of orbitals head to head, and with bonding electron density concentrated along the axis of the bond, 460 significant digits, 43. See also significant figures significant figure(s) The digits in a measured quantity that are known exactly, plus one digit that is inexact, 43–46, 195. See also significant digits defined, 43 determining, 43–44 ties in swimming and, 52 using, 46 using in calculations, 44–45 silicate minerals and chain structures, 1067–1068 and ribbon structures, 1067–1068 silicates and aluminosilicates, 1068–1069 and sheet structures, 1068–1069 silicon, 355, 1065–1066 and Group 4A (14) elements, 1065–1070 metalloid, 75 silicate minerals with chain and ribbon structures, 1067–1068 silicates with sheet structures and aluminosilicates, 1068–1069 silicon dioxide, 1066–1067 silicone polymers, 1069–1070 silicon dioxide, 1066–1067 silicone polymers, 1069–1070 silt, formation of, 657 silver chloride in aqueous ammonia, 755 precipitation of, 152 single bond A bond formed by sharing one pair of electrons; a sigma bond, 420–421

single displacement reaction A reaction in which one element is substituted for another element in a compound, 174 slaked lime, 163 slime from polyvinyl alcohol, 1186 slope of a line, 690 soap A salt produced by the hydrolysis of a fat or oil by a strong base, 658–659 cleaning action of, 658 and hard water, 659, 1283–1284 and surfactants, 658 soda ash, 1053 sodium, 13 electrolysis of molten NaCl, 969 and lithium compounds, 1051–1054 and potassium compounds, 1051–1054 preparation of, 1050–1051 properties of, 1051 reaction with chlorine, 5 reaction with water, 6, 344 sodium bicarbonate acetic acid reaction with, 792–793 sodium carbonate (Na2CO3), 223–224, 807, 1052 sodium chloride (common salt, NaCl), 13, 99, 589 sodium hypochlorite (NaClO), 232–233 sodium propanoate, A-17 sodium silicate forms of, 1067 sol A colloidal dispersion of a solid substance in a fluid medium, 657 solid(s) The phase of matter in which a substance has both definite shape and definite volume, 8 amorphous, 604–606 bonding in ionic compounds, 594–596 bonding in metals and semiconductors, 596–602 conversion into liquid, 609–610 conversion into vapor, 611 crystal lattices and unit cells, 581–588 dissolving in liquids, 632–633 formulas of ionic, 589–593 molecular, 602 network, 602–604 phase changes, 609–613 structure of ionic, 589–593 structures and formulas of ionic solids, 589–593 solid materials, types of, 602–608 alloys, 606–608 amorphous solids, 604–606 mixtures of metals, 606–608 molecular solids, 602 network solids, 602–604 properties of, 603t structures of, 603t solubility The concentration of solute in equilibrium with undissolved solute in a saturated solution, 631, 855–859 and complex ions, 869–871 dissolving gases in liquids, 637–639 factors affecting, 637–640 Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I-31

solubility (Continued) guidelines, 150–151 Henry’s law, 637–639 Le Chatelier’s principle, 639–640 and pressure, 637–640 of salts, 853–863 and solubility product constant, Ksp, 855–859 and solutions, 637–640 and temperature, 637–640 solubility of salts and common ion effect, 859–861 effect of basic anions on, 861–863 guidelines, 150–151 relating solubility and solubility product constant, Ksp, 855–859 solubility product constant, Ksp, 854–855 solubility product constant (Ksp) An equilibrium constant relating the concentrations of the ionization products of a dissolved substance, 854–855, A-23 and precipitation reactions, 864–867 and reaction quotient, Q, 864–865 and solubility, 855–859 solute The substance dissolved in a solvent to form a solution, 147 solution(s) A homogeneous mixture in a single phase, 11, 147, 210, 213 colligative properties, 640–655 colloids, 655–659 concentration, 209–212, 627–630 diluting a more concentrated, 213–214 equilibrium constant for reactions in, 741 general expressions for buffer, 836–838 molarity, 209–212 pH of, 217 preparing by combining a weighed solute with solvent, 212–213 preparing by dilution, 214 and pressure, 637–640 and solubility, 637–640 solution process, 630–637 stoichiometry calculations involving, 217–227 and temperature, 637–640 units of concentration, 209–212, 627–630 solution process, 630–637 enthalpy of solution, 633–637 entropy, 633 liquids dissolving in liquids, 632 solids dissolving in liquids, 632–633 solvent The medium in which a solute is dissolved to form a solution, 147 combining weighed solute with, 212–213 s orbitals, 325–326 sources of energy biofuels, 1279–1281 fossil fuels, 1267–1276 fuel cells, 1277–1278 hydrogen and ammonia, 1278–1279 nuclear reactions, 1018–1021

I-32

space-filling models Representations of molecules in which the atoms are partial spheres that have diameters proportional to those of the atoms and are joined directly to one another, 86 spdf notation A notation for the electron configuration of an atom in which the number of electrons assigned to a subshell is shown as a superscript after the subshell’s symbol, 352, 355 with noble gas notation, 355 specific heat capacity (c) The quantity of heat required to raise the temperature of 1.00 g of a substance by 1.00 kelvin, 258 of compounds, 259, A-14 of elements, 259, A-14 heating and cooling, 258–262 quantitative aspects of energy transferred as heat, 260–262 spectator ion(s) An ion that is present in a solution in which a reaction takes place, but that is not involved in the net process, 154 spectrochemical series An ordering of ligands by the magnitudes of the splitting energies they cause, 1136 of coordination compounds, 1135–1137 spectrophotometer, 229 spectrophotometric analysis, 229–231 spectrophotometry An analytical method based on the absorption and transmission of specific wavelengths of light, 227–231, 228 using in chemical analysis, 230–231 spherical nodal surface, 329 splitting of d orbitals in octahedral complexes, 1129 in square-planar complexes, 1129 in tetrahedral complexes, 1129 spontaneity, 898–902 and chemical equilibrium, 743–745, 903–905 and dispersal of energy, 887–889 and entropy, 887–889 and Gibbs free energy, 903–905 spontaneous or not, 901–902 and standard conditions, 901 spontaneous expansion of a gas, 888 stable isotopes, 1001 standard atmosphere (atm) A unit of pressure; 1 atm = 760 mm Hg, 500, A-7 standard conditions In an electrochemical cell, all reactants and products are pure liquids or solids, or 1.0 M aqueous solutions, or gases at a pressure of 1 bar, 953 standard deviation A measure of precision, calculated as the square root of the sum of the squares of the deviations for each measurement from the average, divided by one less than the number of measurements, 40–41 precision and, 40–41 standard electrochemical cell potentials (E°cell). See standard potential (E°cell)

standard enthalpies of formation, 282–283 standard entropy values, S°, 895–896 standard free energy change of reaction (Δr   G°) The free energy change for a reaction in which all reactants and products are in their standard states, 903–914 and the equilibrium constant, 904–905, 911–912 and standard electrochemical cell potential, 964–965 calculating from standard free energy of formations, 906, 908 calculating from temperature and standard enthalpy and entropy changes, 903, 906 temperature and, 908–911 standard free energy of formation of a compound (Δf   G°) The free energy change when one mole of a compound is formed from its component elements with all of the reactants and products in their standard states, 906 table of, A-25–A-31 using to calculate standard free energy change of reaction, 906, 908 standard hydrogen electrode (SHE) An electrode with [H+] = 1.0 M and P(H2) = 1.0 bar, 952, 953 standardization The accurate determination of the concentration of an acid, base, or other reagent for use in a titration, 223–224 standard molar enthalpy of formation The enthalpy change of a reaction for the formation of one mole of a compound directly from its elements, all in their standard states, 282 table of, A-25–A-31 using to calculate standard enthalpy change for a reaction, 283–285 standard molar enthalpy of fusion The quantity of heat required to convert 1 mol of a solid to a liquid at 1 bar and constant temperature, 609 standard molar enthalpy of vaporization The quantity of heat required to convert 1 mol of a liquid to a gas at 1 bar and constant temperature, 558–559 standard molar entropy (S°) The entropy of a substance in its most stable form at a pressure of 1 bar, 895 table of, A-25–A-31 using to calculate standard entropy change for a reaction, 896–897 standard molar free energy of formation The free energy change for the formation of one mole of a compound from its elements, all in their standard states, 906 tables of, A-25–A-31 using to calculate standard free energy change of a reaction, 906, 908

Index and Glossary

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standard molar volume The volume occupied by 1 mol of gas at standard temperature and pressure; 22.414 L, 509 standard potential (E°cell) The potential of an electrochemical cell measured under standard conditions, 951–960 calculating from equilibrium constants, 964–967 calculating from standard free energy changes, 964–965 calculating from standard reduction potentials, 957 electromotive force, 951–952 measuring, 952–954 standard reaction enthalpy The enthalpy change of a reaction that occurs with all reactants and products in their standard states, 272 standard reduction potential The potential that would be measured when a half-cell of interest is set up as the cathode in an electrochemical cell and the standard hydrogen electrode is set up as the anode, 951, 954–955 calculating standard cell potentials, E°cell, 957 relative strengths of oxidizing and reducing agents, 958–960 tables of, 956t, A-32–A-35 standard state The most stable form of an element or compound in the physical state in which it exists at 1 bar and the specified temperature, 272 standard temperature and pressure (STP) A temperature of 0 °C and a pressure of exactly 1 atm, 509 standing wave A single-frequency wave having fixed points of zero amplitude, 321, 322 starch-iodine, 233 stars nucleosynthesis in, 1007–1008 state(s), 8 change of, 258, 263, 558–566, 609–613 energy and changes of, 262–266 state function A quantity whose value is determined only by the state of the system, 271, 271 state of matter An observable physical property of matter being either a solid, liquid, or gas, 8–9 statistical thermodynamics, 892 stereoisomerism in chemistry and nature, 1125 stereoisomers Two or more compounds with the same molecular formula and the same atom-to-atom bonding, but with different arrangements of the atoms in space, 1122, 1153 steroids A category of lipids consisting of four joined hydrocarbon rings with three rings being 6-member and one ring being 5-member, 1227

stoichiometric coefficients The multiplying numbers assigned to the species in a chemical equation in order to balance the equation, 140, 272, 283, 754, 955 stoichiometric factor(s) A conversion factor relating moles of one species in a reaction to moles of another species in the same reaction, 192 stoichiometry The study of the quantitative relations between amounts of reactants and products, 140, 191–200, 217–227, 513–514, 677–678 chemical analysis, 202–208 concentrations of compounds in solution, 209–215 defined, 191 and gas laws, 513–514 limiting reactant, 195–200 mass relationships in chemical reactions, 191–195 percent yield, 200–202 pH, a concentration scale for acids and bases, 215–217 of reactions in aqueous solution, 217–227 spectrophotometry, 227–231 titrations, 220–227 storage batteries, 946 STP. See standard temperature and pressure strategy maps, 50 strong acid(s) An acid that ionizes completely in aqueous solution, 158, 217, 784 strong base(s) A base that ionizes completely in aqueous solution, 158, 784, 844 strong electrolyte A substance that dissolves in water to form a good conductor of electricity, 148, 148 structural formula A variation of a molecular formula that expresses how the atoms in a compound are connected, 85 structural isomers Two or more compounds with the same molecular formula but with different atoms bonded to each other, 1122–1123, 1153, 1156–1158 of butane C4H10, 1156 coordination isomers, 1123 linkage isomers, 1123 of pentane C5H12, 1157 structure The arrangement of atoms of a molecule (or ion) in space, 387 subatomic particles A collective term for protons, neutrons, and electrons, 65, 65t sublimation The direct conversion of a solid to a gas, 266, 611 submicroscopic level Representations of chemical phenomena in terms of atoms and molecules; also called particulate level, 9

subshell A group of orbitals that share the same value of the principal quantum number n and the orbital angular momentum quantum number l, 323, 324, 348 electrons accommodated in, 347t energies and filling order in multielectron atom, 348 substance(s), pure A form of matter that cannot be separated into two different species by any physical technique, and that has a unique set of properties, 10 substituent group(s) A group of atoms attached to an organic molecule, replacing a hydrogen atom, 1159 substituted alkanes nomenclature of, 1165 substitutional alloy A type of homogenous alloy in which the solute atoms replace solvent atoms in the original crystal structure, 608 substrate The substance on which an enzyme acts, 705, 1213 sulfide-containing minerals, 1079 sulfides, 1076–1077 sulfur, 81, 81, 162, 356, 740 compounds, 1081–1082 and Group 6A (16) elements, 1079–1082 and oxygen, 1080 sulfuric acid, 162 sun, 332 sunburn, 330–331 sunscreens, 330–331 superconductivity, 177 superconductor(s) A material that has no resistance to the flow of electric current, 177 supercritical fluid A substance at or above the critical temperature and pressure, 565 superoxides, 170 supersaturated solution(s) A solution that temporarily contains more than the saturation amount of solute, 631 surface density plot A plot representing the probability of finding an electron at a particular distance from the nucleus, 325 1s electrons of lithium, 349 2s electrons of lithium, 349 surface tension The energy required to disrupt the surface of a liquid, 566–567 surfactant(s) A substance that changes the properties of a surface, typically in a colloidal dispersion, 658–659 and soaps, 658 surroundings Everything outside the system in a thermodynamic process, 256 calculating, 898 Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I-33

suspension A distribution of solid particles within a solvent, which gradually separate and settle to the bottom of the container, 655 sustainability, 7–8, 1281–1282 sustainable development, 7 symbol(s), in chemistry, A-9–A-10 synthesis reaction(s) A reaction in which a compound is prepared from elements or other compounds, 174 synthetic rubber, 1188 Système International d’Unités (International System of Units, SI). See SI system The object or collection of objects being studied for energy content in a thermodynamic process, 256 tar sands Sandy deposits that contain bitumen and are a source of petroleum, also called oil sands, 1270, 1270 tartaric acid, 779 temperature A physical property that determines the direction of heat transfer in an object on contact with another object and activation energy, 698 dependence of physical properties, 16 dependence of solubility of ionic compounds in water, 640 dependence of water density, 15, 15t and equilibrium, 761 and Kw, 780 and pressure, 637–640 and reaction rates, 699 and solubility, 637–640 and solutions, 637–640 and standard free energies, 908–911 vapor pressure of water at various, A-19 variation in global mean surface, 1274 temperature scales, 32–33, 33 termolecular process An elementary step involving three molecules (or atoms, ions, or free radicals), 708 tertiary alcohols, 1177 tertiary amines, 1175 tertiary structure An aspect of protein structure that takes into account the overall three-dimensional shape of a polypeptide chain caused by the folding and spatial arrangements of different regions, 1212 tetrahedral hole In a lattice, the empty space formed by four particles arranged in a tetrahedral geometry, 590, 590 tetrahedral molecules polarity of, 428 thenardite, 203 theoretical yield The maximum amount of product that can be obtained from the given amounts of reactants in a chemical reaction, 201

I-34

theory A unifying principle that explains a body of facts and the laws based on them, 6 atomic (See atomic theory of matter) kinetic-molecular, 520–521 molecular orbital (See molecular orbital theory) quantum (See quantum mechanics) valence bond (See valence bond theory) VSEPR (See valence shell electron-pair repulsion (VSEPR) model) thermal decomposition of NH4Cl(s), 902 thermal energy The energy due to the motion of atoms, molecules, or ions, 20, 255 thermal equilibrium A condition in which the system and its surroundings are at the same temperature and net heat transfer stops, 256–257 thermite reaction, 199 thermodynamic, 272 thermodynamically stable The state of a reaction in which it is at its lowest energy state, 914 thermodynamic control The state of a reaction in which the ratio of products is defined by the equilibrium between the two species, 915 thermodynamics The science of heat or energy flow in chemical reactions, 255, 256, 612, 914–915, 964–967 data, calculating lattice enthalpy from, 595–596 defined, 234, 255 directionality and extent of transfer of heat, 256–257 E°cell and equilibrium constant, 965–967 first law of, 266–271 and phase diagrams, 612–613 product- or reactant-favored reactions and, 285–286 second law of, 889 systems and surroundings, 256 thermal equilibrium, 256–257 third law of, 894 work and free energy, 964–965 thermodynamic values, A-25–A-31 thermoplastic polymer(s) A polymer that softens but is unaltered on heating, 1185 thermosetting polymer(s) A polymer that is initially soft but sets to a solid when heated and that cannot then be resoftened, 1185 third law of thermodynamics The entropy of a pure, perfectly formed crystal at 0 K is zero, 894 thorium decay series, 997 three-center bond, 484 Three Mile Island, 1021 threshold frequency The minimum frequency at which electrons are ejected when light strikes the surface of a light-sensitive material, 310

thyroxine, 1084 ties in swimming and significant figures, 52 titrant The substance being added during a titration, 844 titrations, 220–227, 843–853 acid–base, 221–223 of an acid in aqueous solution with a base, 220 determining molar mass by, 224–225 and limiting reactants, 843 as method of chemical analysis, 220–221 standardizing an acid by, 223–224 of strong acid with strong base, 843–844 using oxidation–reduction reactions, 225–227 of weak acid with strong base, 845, 845–848 of weak base with strong acid, 849, 849–851 of weak polyprotic acids, 848–849 torr A unit of pressure equivalent to one millimeter of mercury, 499, A-7 transcription A step in biological protein synthesis where a sequence complementary to the required region of DNA is made using RNA, 1223 transesterification A reaction where one ester is converted to another, 1189 transfer of heat directionality and extent of, 256–257 thermal equilibrium, 256–257 transfer RNA (tRNA) A type of RNA that functions to convert the mRNA genetic code into a specific sequence of amino acids, 1223 transition elements (or transition metals) Some elements that lie in rows 4 to 7 of the periodic table, comprising scandium through zinc, yttrium through cadmium, and lanthanum through mercury, 74, 83, 357 atomic radii for, 367 bonding in coordination compounds, 1128–1133 cations, 95–96 colors of coordination compounds, 1133–1137 electron configurations of, 357–360, 1107 metallurgy, 1111–1114 overview, 1105–1107 oxidation states, 1107–1109 periodic properties of, 1107–1110 structures of coordination compounds, 1112–1127 transition state The arrangement of reacting molecules and atoms at the point of maximum potential energy, 696 translation The conversion of the information in a nucleotide sequence in mRNA to a specific amino acid sequence in a protein, 1224

Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

transmittance (T) The ratio of the amount of light transmitted by the sample to the amount of incident light, 228–229 transmutation The conversion of atoms of one element to another element by a nuclear reaction, 1014. See also nuclear reaction(s) transport proteins, 1228 transuranium elements An element with an atomic number greater than that of uranium (92), 1051 in nature, 1014 Treaty of the Meter, 34 triglycerides A term applied to fats and oils that signifies that they are triesters of glycerol, 1226 triple point The temperature and pressure at which the solid, liquid, and vapor phases of a substance are in equilibrium, 611 tritium (T), 66 Tyndall effect The scattering of light that occurs when a beam of light is passed through a colloid, 656, 656 ultraviolet radiation, 330–331 uncertainty principle, 323. See also Heisenberg’s uncertainty principle understanding, as goal of science, 6 unified atomic mass units, 64 unimolecular process An elementary step involving only one molecule (or atom, ion, or free radical), 708 unit cell(s) The smallest repeating unit in a crystal lattice that has the characteristic symmetry of the solid, 581–588, 583, 585 cubic unit cells, 583–584 for metals, 584–588 number of atoms per cubic, 584–588 for solid made from circular atoms, 582 units of amount, 100 of density, 15 of energy, 20 for enthalpy of formation, 282 of frequency, 306 of length, 34–35, A-12 of mass, 36–37 of measurement, 30–37, 32t for measuring radiation, 1021–1022 for pressure, 500 of radiation, 1022 SI, 31–37 of temperature, 32–33 of volume, 36, A-12 universe, entropy, calculating, 899, 901 unsaturated compound(s) A hydrocarbon containing double or triple carbon–carbon bonds, 630, 1164 unstable isotopes, 1001 uranium-238 radioactive decay series, 996

U.S. Centers for Disease Control and Prevention (CDC), 36, 233 U.S. Food and Drug Administration (FDA), 37, 1194, 1228 U.S. primary energy production by source (1950–2019), 1264 valence band, 600 valence bond theory A model of bonding in which a bond arises from the overlap of atomic orbitals on two atoms to give a bonding orbital with electrons localized between the atoms, 459–473 benzene, 472–473 hybridization using s and p atomic orbitals, 461–466, 473 hybrid orbitals for molecules and ions with linear electron-pair geometries, 465–466 for molecules or ions with octahedral electron-pair geometries, 466–467 for molecules and ions with tetrahedral electron-pair geometries, 461–465 for molecules with trigonal-planar electron-pair geometries, 465–466 for molecules and ions with trigonalbipyramidal geometries, 466–467 and multiple bonds, 468–472 orbital overlap model of bonding, 459–461 and pi bonding, 472–473 possible d-orbital participation, 466–467 valence electron(s) the electrons beyond the core electrons and that determine the chemical properties of an element, 352, 388, 1041 valence shell electron-pair repulsion (VSEPR) theory A model for predicting the shapes of molecules, in which structural electron pairs are arranged around each atom to maximize the angles between them, 416–424 van Arkel–Ketelaar diagram, 392, 422 van der Waals constants, 525t van der Waals equation A mathematical expression that describes the behavior of nonideal gases, 525–526 van der Waals force(s) A general term for intermolecular forces that includes the attraction and repulsion arising from dipole–dipole forces, dipole–induced dipole forces, and induced dipole–induced dipole forces, 542, 555–557 vanillin (C8H8O3), 1026 van’t Hoff factor The ratio of the experimentally measured freezing point depression of a solution to the value calculated from the apparent molality, 654–655 vaporization The state change from liquid to gas, 558–561 molar enthalpies of, 558, 559t

vapor pressure The pressure of the vapor of a substance in contact with its liquid or solid phase in a sealed container, 561, 561, 564 of benzene by addition of nonvolatile solute, 643 colligative properties changes in, 640–642 of water at various temperatures, A-19 viscosity The resistance of a liquid to flow, 567, 567 volatility The tendency of the molecules of a substance to escape from the liquid phase into the gas phase, 561 voltaic cell An electrochemical cell that uses chemical reactions to produce an electric current, 940–945, 942, 946, 952. See also galvanic cell and electrochemical cell(s) commercial, 946–951 dry cells and alkaline batteries, 946–947 electrochemical cell notations, 945 fuel cells, 950–951 with hydrogen electrodes, 944 with inert electrodes, 943–944 primary batteries, 946–947 secondary or rechargeable batteries, 947–950 volt (V) The electric potential through which 1 coulomb of charge must pass in order to do 1 joule of work, 952, A-8 volume (V), 36 common units of, 36, A-12 volumetric flask, 211, 212 VSEPR theory. See valence shell electron pair repulsion (VSEPR) theory water, 611–612, 780–782 autoionization constant, Kw, for, 780 autoionization of, 780–782 contamination in supply, 1262 detergent and surface tension of, 659 distribution of species for a solution of CO2 in, 1275 electrolysis of, 1047–1049 hard, 1056 hydrogen bonding and unusual properties of water, 550–552 interactions between molecules with permanent dipole, 550–552 microplastics in, 1262–1263 molecular geometry of, 420 in oceans, 1258 phase diagram, 611–612 pollution of, 1260–1262 properties of, 550–552 purification of, 1258–1260 reaction of P4O10, 1078 reaction with sodium, 6 reaction with Na/K alloy, 1054 solubility of ionic compounds in, 149–151, 853–863 Index and Glossary

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I-35

water (Continued) temperature dependence of density of, 15t use, per capita in U.S., 1257 vapor pressure of, A-19 water pollution, 1260–1263 wavefunction(s) A set of equations that characterize the electron as a matter wave, 322 for 1s orbital, 329 for 2p orbital, 329 for 2px orbital, 329 for 2s orbital, 329 and energy, 322 H atom orbital shapes and, 329 wavelength-frequency conversions, 307–308 wavelength The distance between successive crests (or troughs) in a wave, 306 and energy, 310 and frequency, 306, 310 wave mechanics. See quantum mechanics wave or quantum mechanics. See quantum mechanics wave–particle duality The idea that electromagnetic radiation and electrons each have properties of both a wave and a particle, 312, 320 prelude to quantum mechanics, 319–321

I-36

waves, electromagnetic amplitude, 306 axis of propagation, 306 frequency, 306 wavelength, 306 weak acid(s) An acid that is only partially ionized in aqueous solution, 157, 158, 217, 784 aqueous, A-20–A-21 ionization constants for aqueous, A-20–A-21 pH of aqueous solution of, 798–805 titration with a strong base, 845–848 titration with a weak base, 843 weak base(s) A base that is only partially ionized in aqueous solution, 784 examples of, 801 ionization constants for aqueous, A-22 pH of aqueous solution of, 798–805 titration with a strong acid, 849–851 weak electrolyte A substance that dissolves in water to form a poor conductor of electricity, 148, 149 weak acid as, 158 weak base as, 159 weight common units of, A-11 force and, A-6–A-7 mass and, A-11 weight percent The mass of one component of a solution or mixture divided by the total mass, multiplied by 100%, 628, 629

Willamette iron meteorite, 365 work Energy transfer that occurs as a mass is moved through a distance against an opposing force, 266–269 energy and, 20 and free energy, 964–965 non-expansion, 270 pressure-volume (P–V), 268–269 sign conventions, 266 world energy consumption, 255, 1264 World War I, 82 World War II, 314, 993, 1020, 1022, 1025, 1049, 1190 xenon compounds, 1089 X-ray crystallography The science of using x-rays to determine crystal structures and the underlying structures of atoms and molecules, 439, 585, 593, 1214 zeolites, 1069 zero entropy, 894 zeroes and common laboratory mistakes, 44 and significant figures, 43–44 zero-order reactions, 689–690 zwitterion An amino acid in which both the amino group and the carboxylic acid group are ionized, 1210

Index and Glossary

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Notes

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Notes

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Notes

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Notes

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Notes

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Notes

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Notes

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Notes

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Notes

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Notes

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Physical and Chemical Constants NA = 6.02214076 × 1023/mol

Avogadro’s number

−19

Electronic charge

e

= 1.60217663 × 10

Faraday’s constant

F

= 96,485.3 C/mol electrons

Gas constant

R

= 8.31446 J/K ∙ mol = 0.082057 L ∙ atm/K ∙ mol

C

π

π = 3.1415926536

Planck’s constant

h

= 6.62607 × 10−34 J ∙ s

Speed of light (in a vacuum)

c

= 2.99792458 × 108 m/s

Useful Conversion Factors and Relationships Length

Volume

Pressure

SI unit: Cubic meter (m3) 1 liter (L) = 1.00 × 10−3 m3 = 1000 cm3 = 1.056688 quarts 1 gallon = 4.00 quarts

SI unit: Pascal (Pa) 1 pascal = 1 N/m2 = 1 kg/m ∙ s2 1 atmosphere = 101.325 kilopascals = 760 mm Hg = 760 torr = 14.70 lb/in2 = 1.01325 bar 1 bar = 105 Pa (exactly)

Mass

Energy

Temperature

SI unit: Kilogram (kg) 1 kilogram = 1000 grams 1 gram = 1000 milligrams 1 pound = 453.59237 grams = 16 ounces 1 ton = 2000 pounds

SI unit: Joule (J) 1 joule = 1 kg ∙ m2/s2 = 0.23901 calorie

SI unit: Meter (m) 1 kilometer = 1000 meters = 0.62137 mile 1 meter = 100 centimeters 1 centimeter = 10 millimeters 1 nanometer = 1.00 × 10−9 meter 1 picometer = 1.00 × 10−12 meter 1 inch = 2.54 centimeter (exactly) 1 Ångstrom = 1.00 × 10−10 meter

= 1 C × 1 V 1 calorie = 4.184 joules

SI

unit: kelvin

(K) 0 K = −273.15 °C K = °C + 273.15°C ? °C = (5 °C/9 °F)(°F − 32 °F) ? °F = (9 °F/5 °C)(°C) + 32 °F

Location of Useful Tables and Figures Atomic and Molecular Properties Atomic electron configurations Table 7.3 Atomic radii Figures 7.6, 7.9 Bond-dissociation enthalpies Table 8.8 Table 8.7 Bond lengths Electron attachment enthalpy Figure 7.11, Appendix F Electronegativity Figure 8.2 Elements and their unit cells Figure 12.5 Ionic radii Figure 7.12 Figure 7.10, Table 7.5, Appendix F Ionization energies

Acids, Bases, and Salts Acid and base properties of some   ions in aqueous solution Common acids and bases Formation constants Ionization constants for weak   acids and bases Names and composition of   polyatomic ions Solubility guidelines Solubility product constants

Thermodynamic Properties Enthalpy, free energy, entropy Appendix L Lattice energies Table 12.1 Specific heat capacities Figure 5.4, Appendix D

Miscellaneous Charges on common monoatomic    cations and anions Figure 2.19 Melting points and enthalpies of   fusion of some elements and   compounds Table 12.4 Oxidizing and reducing agents Table 3.4 Polymers Table 23.12 Selected alkanes Table 23.2 Standard reduction potentials Table 19.1, Appendix M Structures and properties of various   types of solid substances Table 12.2

Table 16.3 Tables 3.1, 3.2 Appendix K Table 16.2, Appendices H, I Tables 2.4, 3.2 Figure 3.10 Appendix J

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 tandard Atomic Weights S of the Elements 2021 Name

Symbol

Based on relative atomic mass of 12C = 12, where 12C is a neutral atom in its nuclear and electronic ground state.

Atomic Number

Atomic Weight

Actinium* Ac 89 (227) Aluminum Al 13 26.9815384(3) Americium* Am 95 (243) Antimony Sb 51 121.760(1) Ar 18 39.95 Argon† Arsenic As 33 74.921595(6) Astatine* At 85 (210) Barium Ba 56 137.327(7) Berkelium* Bk 97 (247) Beryllium Be 4 9.0121831(5) Bismuth Bi 83 208.98040(1) Bohrium* Bh 107 (270) B 5 10.81 Boron† Bromine† Br 35 79.904 Cadmium Cd 48 112.414(4) Cesium Cs 55 132.90545196(6) Calcium Ca 20 40.078(4) Californium* Cf 98 (251) † C 6 12.011 Carbon Cerium Ce 58 140.116(1) Chlorine† Cl 17 35.45 Chromium Cr 24 51.9961(6) Cobalt Co 27 58.933194(3) Cn 112 (285) Copernicium* Copper Cu 29 63.546(3) Curium* Cm 96 (247) Darmstadtium* Ds 110 (281) Dubnium* Db 105 (268) Dysprosium Dy 66 162.500(1) Einsteinium* Es 99 (252) Erbium Er 68 167.259(3) Europium Eu 63 151.964(1) Fermium* Fm 100 (257) Fl 114 (289) Flerovium* Fluorine F 9 18.998403162(5) Francium* Fr 87 (223) Gadolinium Gd 64 157.25(3) Gallium Ga 31 69.723(1) Germanium Ge 32 72.630(8) Gold Au 79 196.966570(4) Hafnium Hf 72 178.486(6) Hassium* Hs 108 (277) Helium He 2 4.002602(2) Ho 67 164.930329(5) Holmium Hydrogen† H 1 1.008 Indium In 49 114.818(1) Iodine I 53 126.90447(3) Iridium Ir 77 192.217(2) Iron Fe 26 55.845(2) Krypton Kr 36 83.798(2) Lanthanum La 57 138.90547(7) Lawrencium* Lr 103 (262) Pb 82 207.2 Lead† Lithium† Li 3 6.94 Livermorium* Lv 116 (293) Lutetium Lu 71 174.9668(1) Magnesium† Mg 12 24.305 Manganese Mn 25 54.938043(2) Meitnerium* Mt 109 (276) Source of atomic weights: https://ciaaw.org/atomic-weights.htm The uncertainties in atomic weight values are given in parentheses following the last significant figure to which they are attributed. For example, the atomic weight of 192.217(2) for iridium indicates that the atomic weight determined for iridium in normal materials should be between 192.215 and 192.219. † IUPAC recommends a range of atomic weights for this element because the variation in atomic weight obtained from different samples is so great. See p. 70 for the actual range. The value shown in this table is a conventional value often used.

Name

Symbol

Atomic Number

Atomic Weight

Mendelevium* Md 101 (258) Mercury Hg  80 200.592(3) Molybdenum Mo 42 95.95(1) Moscovium* Mc 115 (290) Neodymium Nd 60 144.242(3) Neon Ne 10 20.1797(6) Neptunium* Np 93 (237) Nickel Ni 28 58.6934(4) Nihonium* Nh 113 (286) Niobium Nb 41 92.90637(1) † 7 14.007 Nitrogen N Nobelium* No 102 (259) Oganesson* Og 118 (294) Osmium Os 76 190.23(3) † 8 15.999 Oxygen O Palladium Pd 46 106.42(1) Phosphorus P 15 30.973761998(5) Platinum Pt 78 195.084(9) Plutonium* Pu 94 (244) Polonium* Po 84 (209) Potassium K 19 39.0983(1) Praseodymium Pr 59 140.90766(1) Pm 61 (145) Promethium* Protactinium* Pa 91 231.03588(1) Radium* Ra 88 (226) Radon* Rn 86 (222) Rhenium Re 75 186.207(1) Rhodium Rh 45 102.90549(2) Roentgenium* Rg 111 (282) Rubidium Rb 37 85.4678(3) Ruthenium Ru 44 101.07(2) Rutherfordium* Rf 104 (267) Samarium Sm 62 150.36(2) Scandium Sc 21 44.955907(4) Seaborgium* Sg 106 (269) Selenium Se 34 78.971(8) Silicon† Si 14 28.085 Silver Ag 47 107.8682(2) Sodium Na 11 22.98976928(2) Strontium Sr 38 87.62(1) Sulfur† S 16 32.06 Tantalum Ta 73 180.94788(2) Technetium* Tc 43 (98) Tellurium Te 52 127.60(3) Tennessine* Ts 117 (294) Terbium Tb 65 158.925354(7) † 81 204.38 Thallium Tl Thorium* Th 90 232.0377(4) Thulium Tm 69 168.934219(5) Sn 50 118.710(7) Tin Titanium Ti 22 47.867(1) Tungsten W 74 183.84(1) Uranium* U 92 238.02891(3) Vanadium V 23 50.9415(1) Xenon Xe 54 131.293(6) Ytterbium Yb 70 173.045(10) Yttrium Y 39 88.905838(2) Zinc Zn 30 65.38(2) Zirconium Zr 40 91.224(2) *Elements with no stable nuclide; the value given in parentheses is the atomic mass number of the isotope of longest known half-life. However, three such ele­ments (Th, Pa, and U) have a characteristic terrestrial isotopic composition, and the atomic weight is tabulated for these. https://ciaaw.org/radioactive-elements.htm

Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.