Quantum : chemistry : third year of secondary cycle two : student textbook [2-3] 9782765214380, 2765214387

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Third Year of Secondary Cycle Two

CHEMISTRY

Student Textbook

Ivan Couture Marie-Ève Lacombe-Harvey Geneviève Levasseur-Thériault

Detailed Periodic Table

1

Phases of elements in ambient conditions

(IA) 1

1

2.20 1312 259.3 252.9

0.98 520 180.5 1342

0.93 496 97.8 882.9

0.82 419 63.5 759

K

0.82 403 39.31 688

0.79 376 28.5 671

4

1.57 899 1287 2471

Rb

Cs

0.7 ~375 27 677

Fr

Metalloids 2 1.40 1.85

Be

12

1.31 738 650 1090

Np

Synthetic elements

C

Solids

Liquids

He

Non-metals

Mg

Gases

2 1.72 1.74

3

Magnesium 24.305 1 2.77 0.86

20

1.00 590 842 1484

1 2.98 1.53

1 3.34 1.87

2 2.23 1.55

Ca Calcium 40.078

38

0.95 549 777 1382

21

1.36 631 1541 2832

39

1.22 616 1522 3345

56

0.89 503 727 1897

3 2.09 2.99

Sc

3 2.27 4.46

Y

22

1.54 658 1668 3287

1.33 660 1855 4409

Francium (223)

88

0.9 509 700 1137

1.63 650 1910 3407

41

1.6 664 2477 4744

1.3 642 2233 4603

Hf

73

1.5 761 3017 5458

Hafnium 178.490 104

2 — 5.0

— — — —

89-103

Radium (226)

24

1.66 653 1907 2671

5 3 2.08 8.57

Nb

Ta

42

2.16 685 2623 4639

Rutherfordium (267)

105 — — — —

74

1.7 770 3422 5555

6 2.01 10.22

6 2.02 19.3

W

25

1.55 717 1246 2061

43

2.10 702 2157 4265

Dubnium (268)

106 — — — —

26

1.83 759 1538 2861

7 1.95 11.50

75

1.9 760 3186 5596

Re

7 1.97 21.02

107 — — — —

Seaborgium (271)

3 2 1.72 7.89

Fe

27

1.88 760 1495 2927

44

2.2 711 2334 4150

Ru

3 4 1.89 12.44

45

2.28 720 1964 3695

Ruthenium 101.070 76

2.2 840 3033 5012

Os

108 — — — —

Bohrium (272)

2 3 1.67 8.9

Co Cobalt 58.933

4 1.92 22.57

Rh

3 1.83 12.41

Rhodium 102.906 77

2.2 880 2446 4428

Osmium 190.230 — —

Bh

9

(VIIIB)

Iron 55.845

Rhenium 186.207 — —

Sg

Tc

Technetium (98)

Tungsten 183.840 — —

2 4 1.79 7.3

Mn

Manganese 54.938

Molybdenum 95.960

5 2.09 16.65

Db

3 2 1.85 7.19

Cr

Mo

8

(VIIB)

Chromium 51.996

Tantalum 180.948 — —

Rf

5 4 1.92 6.11

V

Niobium 92.906

4 2.16 13.29

7

(VIB)

Vanadium 50.942 4 2.16 6.51

Zr

72

57-71

Ra

23

Zirconium 91.224

Barium 137.327 1 — 1.87

4 3 2.00 4.55

Ti

40

6

(VB)

Titanium 47.867

Yttrium 88.906

2 2.78 3.5

Ba

5

(IVB)

Scandium 44.956 2 2.45 2.54

Sr

4

(IIIB)

Strontium 87.620

Cesium 132.905 87

7

1 2.23 0.97

Rubidium 85.468 55

6

2

(IIA)

Potassium 39.098 37

5

Na

Natural elements

Beryllium 9.012

Sodium 22.990 19

4

1 2.05 0.53

Lithium 6.941 11

3

Li

S

Metals

Hydrogen 1.008 3

2

H

1 1 0.79 0.09

Ir

4 1.87 22.42

Iridium 192.217 — —

Hs

Hassium (270)

109 — — — —

— —

Mt

Meitnerium (276)

Lanthanides 57

6

1.10 538 920 3469

La

58

1.12 527 798 3443

Lanthanum 138.905 89

7

3 2.74 6.17

1.1 499 1050 3200

Ac

3 — 10.07

3 2.70 6.77

Ce Cerium 140.116

90

1.3 587 1750 4788

Actinium (227) Actinides

Th

4 — 11.72

Thorium 232.038

59

1.13 523 930.8 3512.8

Pr

3 2.67 6.77

Praseodymium 140.908 91

1.5 568 1572 4030

Pa

5 4 — 15.37

Protactinium 231.036

60

1.14 530 1021 3074

3 2.64 6.80

Nd

Neodymium 144.242 92

1.7 584 1132.2 3927

U

6 4 — 19.05

Uranium 238.029

61

— 536 1100 3000

Pm

3 2.62 6.47

Promethium (145) 93

1.3 597 637 4000

Np

5 — 20.25

Neptunium (237)

62

1.17 543 1074 1794

Sm

3 2 2.59 7.54

Samarium 150.360 94

1.3 585 639.4 3230

Pu

4 6 — 19.84

Plutonium (244)

63

— 547 822 1529

Eu

3 2 2.56 5.28

Europium 151.964 95

— 578 1176 2607

Am

3 4 — 13.67

Americium (243)

8

Atomic number

3.44 1314 218.3 182.9

Electronegativity Energy of the 1st ionization (kJ/mol) Melting point (°C) Boiling point (°C)

Ionic charge (if there is more than one, the first is the one that is most commonly used)

2 0.65 1.43

O

Atomic radius ( 1012 m) Density (g/L or g/cm3) Symbol

Oxygen 15.999

Name Atomic weight (u) Molar mass (g/mol)

18 (VIIIA) 2

5

2.04 800 2075 4000

13

14

15

16

17

(IIIA)

(IVA)

(VA)

(VIA)

(VIIA)

1.17 2.87

B

6

2.55 1086 3527 4027

C

Boron 10.811 13

10 28

1.91 737 1455 2913

11 2 3 1.62 8.90

29

1.90 745 1084.6 2562

Nickel 58.693 46

2.20 805 1554.9 2963

Pd

2.2 870 1768.4 3825

Pt

2 4 1.79 12.02

47

1.93 731 961.8 2162

— — — —

Ds

1.20 593 1313 3273

Gd

4 2 1.83 21.45

1 1.75 10.5

Ag

79

2.4 890 1064.2 2856

Au

— —

111 — — — —

3 2.54 7.90

— 581 1340 3110

Cm Curium (247)

1.69 868 321.1 767

3 — 13.51

— —

3 2.51 8.23

Tb Terbium 158.925

— 601 986 —

Bk

2 1.71 8.65

3 4 — 14.00

Berkelium (247)

— —

Uub

Ununbium (285)

66

1.22 572 1412 2567

Dy

1.78 558 156.6 2072

1.8 589 304 1473

— 608 900 —

Cf

Tl

32

2.01 761 938.3 2833

3 2.49 8.54

1 3 2.08 11.85

3 — 15.1

Californium (251)

— —

Ununtrium (284)

1.23 581 1470 2720

Ho

1.96 708 231.9 2602

82

1.8 715 327.5 1749

— 619 860 —

Es

4 1.52 5.32

4 2 1.72 7.29

Sn

Pb

3 2.47 8.78

P

33

2.18 947 817 (28 atm) 614 (sublime)

Uuq

68

Er

Einsteinium (252)

100 — 627 1527 —

3.44 1314 218.3 182.9

As

51

2.05 834 630.6 1587

Sb

83

1.9 703 271.3 1564

Bi

3 1.23 1.82

16

2.58 999 119.6 444.6

115 — — — —

Uup

3 2.45 9.04

69

1.25 597 1545 1950

Fm

Fermium (257)

Tm

3 1.33 5.72

101 — 635 827 —

Md

9

3.98 1681 219.6 188.1

S

34

2.55 941 220.5 684.9

Se

3 5 1.53 6.61

52

2.1 869 449.5 988

Te

2 1.09 2.07

17

3.16 1256 101.5 34.04

84 2.0 813 254 962

Po

2 1.22 4.79

3 2.42 9.29

2 1.42 6.24

53

2.66 1009 113.7 184.4 (35 atm)

2 4 1.53 9.32

I

85

2.2 (926) 302 337

Yb

Mendelevium (258)

— 1520 189.3 185.8

At

3 2 2.40 6.97

No

Nobelium (259)

2 3 — —

0.88 1.78

Ar

Argon 39.948 36

— 1351 157.4 153.2

Kr

1.03 3.741

Krypton 83.798 1 1.32 4.93

54

— 1170 111.8 108.1

Xe

1.24 5.90

Xenon 131.293 1 1.43 7

86

— 1037 71 61.8

Rn

1.34 9.73

Radon (222) 118

— —

102 — 642 827 —

18

Astatine (210) — — — —

Uuo

— —

Ununoctium (294)

71

1.0 524 1663 3402

Ytterbium 173.054 2 3 — —

1 0.97 3.21

1 1.12 3.12

2.96 1143 7.2 58.78

0.51 0.90

Ne

Neon 20.180

Ununhexium (293)

— 603 824 1196

— 2080 248.6 246.1

Iodine 126.904

Uuh

70

10

Bromine 79.904

116 — — — —

Cl

35

Polonium (209) — —

1 0.57 1.70

Chlorine 35.453

Tellurium 127.600

3 5 1.63 9.74

F

0.49 0.18

He

Helium 4.003

Fluoride 18.998

Selenium 78.960

Thulium 168.934 3 — —

2 0.65 1.43

Sulfur 32.065

Bismuth 208.980 — —

O

Oxygen 15.999

Ununquadium Ununpentium (289) (288)

1.24 589 1529 2868

8

Arsenic 74.922

Erbium 167.259 3 — —

3 0.75 1.25

Antinomy 121.760

2 4 1.81 11.35

114 — — — —

N

Phosphorus 30.974

Ge

50

Holmium 164.930 99

2.19 1012 44.2 277

Lead 207.200

Uut

67

15

Tin 118.710

113 — — — —

1.46 2.33

Si

Germanium 72.640 3 2.00 7.29

In

81

Dysprosium 162.500 98

1.90 786 1414 3265

Thallium 204.383

112

— — — —

14

Indium 114.818

2 1 1.76 13.55

1.9 1107 38.83 356.7

3 1.81 5.90

Ga

49

Mercury 200.590

Rg

97

1.81 579 29.76 2204

3.04 1402 209.9 195.8

Nitrogen 14.007

Silicon 28.086

Gallium 69.723

Cd

80

3 1.82 2.70

Al

31

Cadmium 112.411

3 1 1.79 19.32

65

— 565 1356 3230

2 1.53 7.11

Zn

48

Gold 196.967

Gadolinium 157.25 96

1.65 906 419.6 907

Zinc 65.380

Darmstadtium Roentgenium (281) (280)

64

30

Silver 107.868

Platinum 195.084 110

2 1 1.57 8.96

Cu

7

Carbon 12.011

Aluminum 26.982

(IIB)

Copper 63.546

Palladium 106.420 78

12

(IB)

Ni

1.61 577 660.3 2519

4 2 0.91 1.8-3.5

— 2372 272.2 268.9

Lu

2 2.25 9.83

Lutetium 174.967 103 — — 1627 —

Lr

3 — —

Lawrencium (262)

Notes : The elements presented in this periodic table are those recognized by the National Research Council Canada (NRC) in 2008. The names of the elements and their atomic weights are derived from the 2007 update of the International Union of Pure and Applied Chemistry.

Third Year of Secondary Cycle Two

CHEMISTRY

Student Textbook

Ivan Couture Marie-Ève Lacombe-Harvey Geneviève Levasseur-Thériault

Translators: Jacquie Charlton, Cristina Cusano, Natasha DeCruz, Joann Egar, Gwen Schulman

Quantum – Chemistry Third Year of Secondary Cycle Two

Acknowledgements

Student Textbook Ivan Couture, Marie-Ève Lacombe-Harvey, Geneviève Levasseur-Thériault © 2011 Chenelière Education Inc. Editors: Marie-Eve Robitaille, Isabel Rusin Coordinators: Geneviève Blanchette, Dominique Lapointe, Christiane Gauthier Copy-editors: Ginette Gratton, Nicole Blanchette Proofreaders: Renée Bédard, Caroline Bouffard Graphic design and layout: Dessine-moi un mouton Cover design: Josée Brunelle Illustrations: Michel Rouleau Visual research: Marie-Chantal Laforge Printer: Imprimeries Transcontinental English Version Editor: Colleen Ovenden Scientific Advisor: Michael Quinn Copy-editor: Eileen Travers Proofreader: Piera Palucci Layout: Interscript Printer: Transcontinental Printing

The Editor would like to thank the following people for their valuable consulting work: Abderrahmane Amane, École internationale St-Edmond, CS Marie-Victorin; Bouthaïna Bouzid, École PèreMarquette, CS de Montréal; Claudie Chartré, Collège Jean de la Mennais; Somali Keuk, École secondaire des Pionniers, CS du Chemin-du-Roy; Christian Lamotte, École secondaire DanielJohnson, CS de la Pointe-de-l’Île; Mona Langlois, École JacquesRousseau, CS Marie-Victorin; Danielle Lanneluc-Sanson, École secondaire De Mortagne, CS des Patriotes; Nathalie Lefebvre, École Félix-Leclerc, CS des Affluents; Mourad Meziane, École secondaire Mont-Royal, CS Marguerite-Bourgeoys; Mireille Paris, École secondaire Mont-Bleu, CS des Portages-de-l’Outaouais; Gilles Roy, Collège Jean de la Mennais; Anne-Marie Talbot-Fournier, Collège de Montréal; Rénald Veilleux, École secondaire Joseph-François Perrault, CS de Montréal; Mary Zarif, École secondaire Saint-Luc, CS de Montréal. The Editor would also like to thank the following people for their valuable expertise and scientific revisions: Jeffrey Keillor, Ph. D., Full Professor, Université de Montréal (all chapters); Pierre Baillargeon, Ph. D., teacher, Cégep de Sherbrooke (Chapters 2, 11, 12); Hugo Bélisle, teacher, École de technologie supérieure de Montréal (Chapter 2); Mohammed Mecif, Chemist (Chapter 1); Dana F. Simon, postdoctoral fellow, Université de Montréal (Chapter 1). The Editor of the English version would like to thank Michael Quinn for his valuable expertise and constructive review during the translation of this textbook. The Editor would like to thank the following people for their valuable copy-editing: Patrick Germain, teacher, Cégep de Sherbrooke (Redox Reaction supplement: p. 349 to 369); Serge Rodrigue, teacher, Cégep de Sorel-Tracy (exercises); Évelyne Beaubien (features: p. 142 and 143 and index), Mourad Meziane, teacher, École secondaire Mont-Royal, CS Marguerite-Bourgeoys (feature: p. 164), Agence Science-Presse (features: p. 43, 49, 114, 165, 224, 229, 268, 269, 300 and 366), Guy Lapointe, lecturer, UQAM. The Editor would also like to thank Collège Mont-Sacré-Cœur for having graciously agreed to release Mr. Ivan Couture from his teaching duties during the writing of this book.

7001 Saint-Laurent Blvd. Montréal (Québec) Canada H2S 3E3 Telephone: 514 273-1066 Fax: 450 461-3834 / 1 888 460-3834 [email protected]

ALL RIGHTS RESERVED. No part of this book may be reproduced in any form or by any means without written permission from the Publisher. ISBN 978-2-7652-1438-0 Legal deposit: 1st quarter 2011 Bibliothèque et Archives nationales du Québec Library and Archives Canada Printed in Canada 1

2

3

4

5 ITIB

14 13

12

11

10

We acknowledge the financial support of the Government of Canada through the Book Publishing Industry Development Program (BPIDP) for our publishing activities. Government of Québec – Tax credit program for book publishing – SODEC These programs are funded by Québec’s Ministère de l’Éducation, du Loisir et du Sport and through contributions from the Canada-Québec Agreement on Minority-Language and Second-Language Instruction.

List of Laboratories and Demonstrations . . . . . . . . . . VIII

8.2

Synthesis, decomposition and precipitation. . . . . . . . . . . . . . . . . . . . . . . 25

REVIEW . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

8.3

Endothermic and exothermic reactions . . . . . 26

1.

Organization of matter . . . . . . . . . . . . . . . . . . . . . . . . . 4

8.4

Oxidation and combustion . . . . . . . . . . . . . . . 27

1.1

Atoms and molecules . . . . . . . . . . . . . . . . . . . 4

8.5

Photosynthesis and respiration . . . . . . . . . . . 27

1.2

Chemical formulas and ions . . . . . . . . . . . . . . . 4

Organization of the Textbook . . . . . . . . . . . . . . . . . . . . . IX

2.

3.

4.

5.

6.

7.

8.

9.

Nature of chemical bonds. . . . . . . . . . . . . . . . . . . . . . 28

Representations of atoms. . . . . . . . . . . . . . . . . . . . . . . 5

9.1

Ionic bonding . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.1

Rutherford-Bohr atomic model . . . . . . . . . . . . 5

9.2

Covalent bonding . . . . . . . . . . . . . . . . . . . . . . 29

2.2

Simplified atomic model and the neutron . . . . 5

2.3

Lewis notation . . . . . . . . . . . . . . . . . . . . . . . . . 7

Periodic classification . . . . . . . . . . . . . . . . . . . . . . . . . . 8

10. Forms of energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 10.1

Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . 30

10.2

Potential energy . . . . . . . . . . . . . . . . . . . . . . . 30

10.3

The law of conservation of energy. . . . . . . . . 32

10.4

3.1

Periodic table . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3.2

Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.3

Periods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Relationship between thermal energy, specific heat capacity, mass and temperature change . . . . . . . . . . . . . . . . . . . . 32

Nomenclature and notation rules. . . . . . . . . . . . . . . . 11

11. Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

4.1

Naming a compound from its chemical formula . . . . . . . . . . . . . . . . . . . . . . 11

11.1

Compressible and incompressible fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

4.2

Writing chemical formulas of compounds . . . 12

11.2

Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

Enumeration of matter . . . . . . . . . . . . . . . . . . . . . . . . 13 5.1

The Mole concept . . . . . . . . . . . . . . . . . . . . . . 13

5.2

Molar mass . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

Physical changes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

GASES

6.1

Phase changes . . . . . . . . . . . . . . . . . . . . . . . . 15

6.2

Dissolution and solubility . . . . . . . . . . . . . . . . 15

6.3

Concentration and dilution . . . . . . . . . . . . . . . 16

6.4

Electrolytes and electrolysis. . . . . . . . . . . . . . 19

INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

6.5

Measuring physical change using the pH scale . . . . . . . . . . . . . . . . . . . . . 20

ORGANIZATION CHART . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Chemical changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 7.1

The law of conservation of mass . . . . . . . . . . 21

7.2

Balancing chemical equations . . . . . . . . . . . . 22

7.3

Stoichiometry . . . . . . . . . . . . . . . . . . . . . . . . . 23

Examples of types of chemical reactions. . . . . . . . . . 25 8.1

Acid-base reactions . . . . . . . . . . . . . . . . . . . . 25

CHAPTER 1 CHEMICAL PROPERTIES OF GASES . . . 37 1.1 Daily use of gases. . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 1.1.1 Gases and natural phenomena . . . . . . . . . . . 38 1.1.2 Gases and their technological applications. . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Applications – Compressed gases. . . . . . . . . . . . . . . . . . . . 43 1.2 Chemical reactivity of gases. . . . . . . . . . . . . . . . . . . . 44

Table of Contents

III

1.2.2 Combustible gases . . . . . . . . . . . . . . . . . . . . . 46

2.4.4 Relationship between the volume and quantity of gas expressed in number of moles 9 and 10 . . . . . . . . . . . . . . . . . . . . . . . . . 88

1.2.3 Oxidizing gases . . . . . . . . . . . . . . . . . . . . . . . . 48

2.4.5 Molar volume of gases. . . . . . . . . . . . . . . . . . 92

A brief history of… Anesthesia. . . . . . . . . . . . . . . . . . . . . . 49

2.4.6 Relationship between the pressure and the quantity of gas expressed in number of moles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

1.2.1 The causes of the chemical reactivity of gases . . . . . . . . . . . . . . . . . . . . . 45

Review – Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 Practice makes perfect - Chapter 1. . . . . . . . . . . . . . . . 51

Practice makes perfect – Section 2.4 . . . . . . . . . 97 2.5 Ideal gas law. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

CHAPTER 2 PHYSICAL PROPERTIES OF GASES . . . . . . . . . . . . . . . . . . . . . . . . . . 53 2.1 Kinetic theory of gases . . . . . . . . . . . . . . . . . . . . . . . . 54 2.1.1 Particle behaviour in various 1 phases of matter . . . . . . . . . . . . . . . 54 2.1.2 Kinetic energy of gas particles and temperature . . . . . . . . . . . . . . . . . . . . . . . 58 2.1.3 Hypotheses of the kinetic theory of gases . . . . . . . . . . . . . . . . . . . . . . . . 61

2.5.1 Molar mass of a gas

11

. . . . . . . . . . . 103

Practice makes perfect – Section 2.5 . . . . . . . . 104 2.6 General gas law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Practice makes perfect – Section 2.6 . . . . . . . . 107 2.7 Stoichiometry of gases . . . . . . . . . . . . . . . . . . . . . . . 108 Practice makes perfect – Section 2.7 . . . . . . . . 110 2.8 Dalton's law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 Practice makes perfect – Section 2.8 . . . . . . . . 113

Practice makes perfect – Section 2.1 . . . . . . . . . 62

Applications – Gasoline engines and diesel engines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

2.2 Behaviour of gases . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

A brief history of… Dirigible balloons . . . . . . . . . . . . . . . 115

2.2.1 Compressibility 2.2.2 Expansion

. . . . . . . . . . . . . . . . 63

Review – Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

. . . . . . . . . . . . . . . . . . . . . 63

Practice makes perfect – Chapter 2 . . . . . . . . . . . . . . 119

2 3

2.2.3 Diffusion and effusion

4

. . . . . . . . . . . 64

Practice makes perfect – Section 2.2 . . . . . . . . . 68 2.3 Pressure of gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 2.3.1 Atmospheric pressure. . . . . . . . . . . . . . . . . . . 70 2.3.2 Measuring the pressure of gases. . . . . . . . . . 71 Practice makes perfect – Section 2.3 . . . . . . . . . 74 2.4 Simple gas laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 2.4.1 Relationship between pressure and volume 5 and 6 . . . . . . . . . . . . . . . . . . . . . . . . . 76 2.4.2 Relationship between volume 7 and absolute temperature

IV

ENERGY CHANGES IN REACTIONS INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 ORGANIZATION CHART . . . . . . . . . . . . . . . . . . . . . . . . . 126

. . . . . . . . 80

CHAPTER 3 ENERGY TRANSFER . . . . . . . . . . . . . . . . 127

2.4.3 Relationship between pressure and 8 temperature . . . . . . . . . . . . . . . . . . . 85

3.1 Difference between heat and temperature . . . . . . . 128

Table of Contents

Practice makes perfect – Section 3.1 . . . . . . . . 130

For a detailed list of laboratories and demonstrations see page VIII

3.2 The law of conservation of energy. . . . . . . . . . . . . . 131

A brief history of… The use of chemical energy . . . . . . . 165

3.2.1 Types of systems . . . . . . . . . . . . . . . . . . . . . 131

Review – Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

3.2.2 Calorimetry and the calorimeter . . . . . . . . . 132

Practice makes perfect – Chapter 4 . . . . . . . . . . . . . . 168

Practice makes perfect – Section 3.2 . . . . . . . . 133 3.3. Relationship between thermal energy, specific heat capacity, mass and 12 . . . . . . . . . . . . . . . . . 134 temperature change Practice makes perfect – Section 3.3 . . . . . . . . 137 3.4 Calculating energy transfer

13

. . . . . . . . . . . . 138

Practice makes perfect – Section 3.4 . . . . . . . . 141 Applications – The ground-source heat pump. . . . . . . . . . 142 A brief history of… Calorimetry . . . . . . . . . . . . . . . . . . . 143 Review – Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 Practice makes perfect – Chapter 3 . . . . . . . . . . . . . . 145

CHAPTER 5 GRAPHICAL REPRESENTATION OF ENTHALPY CHANGE . . . . . . . . . . . . 171 5.1 Activated complex, activation energy 15 . . . . . . . . . . . . . . . 172 and the energy diagram 5.1.1 Activated complex . . . . . . . . . . . . . . . . . . . . . 172 5.1.2 Activation energy. . . . . . . . . . . . . . . . . . . . . . 173 5.1.3 Energy diagram . . . . . . . . . . . . . . . . . . . . . . . 174 5.2 Observing the use of an energy diagram to plot the progress of a conversion. . . . . . . . . . . . . 176 5.2.1 Height of the activation energy barrier and the spontaneous reaction . . . . . . . . . . . . 176 5.2.2 Direct and reverse reactions . . . . . . . . . . . . . 178

CHAPTER 4 ENTHALPY CHANGE . . . . . . . . . . . . . . . . 147

Review – Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

4.1 Enthalpy and enthalpy change . . . . . . . . . . . . . . . . . 148

Practice makes perfect – Chapter 5 . . . . . . . . . . . . . . 183

4.1.1 Standard molar enthalpy change . . . . . . . . . 149 4.2 Endothermic and exothermic 14 . . . . . . . . . . . . . . . . . . . . . . . . . . 150 reactions

CHAPTER 6 MOLAR HEAT OF REACTION . . . . . . . . . 185 6.1 Molar heat of dissolution

. . . . . . . . . . . . . 186

16

4.2.1 Endothermic and exothermic physical changes. . . . . . . . . . . . . . . . . . . . . . 150

6.2 Molar heat of neutralization

4.2.2 Endothermic and exothermic chemical reactions. . . . . . . . . . . . . . . . . . . . . 152

Review – Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

4.2.3 Enthalpy diagram of endothermic and exothermic reactions . . . . . . . . . . . . . . . 153 Practice makes perfect – Section 4.1 and 4.2 . . 155 4.3 Energy balance. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 4.3.1 Performing an energy balance. . . . . . . . . . . . 156 Practice makes perfect – Section 4.3 . . . . . . . . 160

17

. . . . . . . . . . . 192

A brief history of… Chewing gum. . . . . . . . . . . . . . . . . . . 194 Practice makes perfect – Chapter 6 . . . . . . . . . . . . . . 195 CHAPTER 7 HESS’S LAW. . . . . . . . . . . . . . . . . . . . . . . 197 7.1 Reaction mechanism

18

. . . . . . . . . . . . . . . . . 198

7.1.1 Graphical representation of a reaction mechanism . . . . . . . . . . . . . . . . 200 7.2 Summation of enthalpies

19 and 20

. . . . . . . 201

4.4. Calculating enthalpy change using stoichiometry. . . . . . . . . . . . . . . . . . . . . . . . . . 161

7.2.1 Presentation of Hess's Law . . . . . . . . . . . . . . 202

Practice makes perfect – Section 4.4 . . . . . . . . 162

Review – Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

Applications – Phase-changing materials . . . . . . . . . . . . . 164

Practice makes perfect – Chapter 7 . . . . . . . . . . . . . . 208

For a detailed list of laboratories and demonstrations see page VIII

7.2.2 Applying Hess’s law . . . . . . . . . . . . . . . . . . . 203

Table of Contents

V

REACTION RATE INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 ORGANIZATION CHART . . . . . . . . . . . . . . . . . . . . . . . . . 212 CHAPTER 8 MEASURING REACTION RATE . . . . . . 213 8.1 Expressing reaction rate . . . . . . . . . . . . . . . . . . . . . 214 8.2 Reaction rate in terms of the stoichiometric coefficients of a balanced chemical equation . . . . 216 8.2.1 General reaction rate . . . . . . . . . . . . . . . . . . 217 Practice makes perfect – Section 8.2 . . . . . . . 219 8.3 Ways to measure reaction rate

21

. . . . . . . . 220

8.3.1 Measuring reaction rate in terms of the state of the reactant or the product and the ease of using the measuring method . . . . . . . . . 220 8.3.2 Measuring reaction rate in terms of the intended use of the results . . . . . . . . . . 221 Applications – Galvanization . . . . . . . . . . . . . . . . . . . . . . . 224 Practice makes perfect – Section 8.3 . . . . . . . . 225 8.4 Average reaction rate and instantaneous 22 . . . . . . . . . . . . . . . . . . . . . . . 226 reaction rate

CHAPTER 10 FACTORS THAT AFFECT REACTION RATES . . . . . . . . . . . . . . . . 23 . . . . . . . . . . . . . . . . . . 10.1 Nature of reactants 10.1.1 Phase of the reactants . . . . . . . . . . . . . . . . . 10.1.2 Number and strength of the bonds of the reactants to be broken . . . . . . . . . . . . Practice makes perfect – Section 10.1 . . . . . . . 24 . . . . . . . . . . . . . . 10.2 Surface area of reactants 10.3 Concentration of reactants 25 . . . . . . . . . . . . . . . . . . . . and the rate law

245 246 247 248 250 251 253

10.3.1 Concentration of reactants . . . . . . . . . . . . . . 253 10.3.2 Rate law. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 Practice makes perfect – Section 10.3 . . . . . . . 258 10.4 Temperature of the reaction 26 . . . . . . . . . . . . . . . . . . . . . . . . 260 environment 27 . . . . . . . . . . . . . . . . . . . . . . . . . . 262 10.5 Catalysts 10.5.1 How a catalyst functions . . . . . . . . . . . . . . . 262 10.5.2 Homogeneous catalysts. . . . . . . . . . . . . . . . . 264 10.5.3 Heterogeneous catalysts . . . . . . . . . . . . . . . 266 Practice makes perfect – Section 10.4 and Section 10.5. . . . . . . . . . . . . . . . . . . . . . . . . . . 267 Applications – What is hydrogenation? . . . . . . . . . . . . . . 268 A brief history of… Catalysis . . . . . . . . . . . . . . . . . . . . . . 269 Review – Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 Practice makes perfect – Chapter 10 . . . . . . . . . . . . . 272

Practice makes perfect – Section 8.4 . . . . . . . . 228 A brief history of… Food preservation . . . . . . . . . . . . . . . 229 Review – Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 Practice makes perfect – Chapter 8 . . . . . . . . . . . . . . 231

CHAPTER 9

COLLISION THEORY . . . . . . . . . . . . . . . 235

9.1 Types of collisions. . . . . . . . . . . . . . . . . . . . . . . . . . . 236

CHEMICAL EQUILIBRIUM

9.1.1 Orientation of reactants . . . . . . . . . . . . . . . . 237 9.1.2 Activation energy . . . . . . . . . . . . . . . . . . . . . 237

INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274

9.2 Reaction mechanism explained by collision theory. . . . . . . . . . . . . . . . . . . . . . . . . . . 239

ORGANIZATION CHART . . . . . . . . . . . . . . . . . . . . . . . . . 276

9.2.1 Simple reaction using collision theory. . . . . . . . . . . . . . . . . . . . . . . . 239 9.2.2 Complex reaction using collision theory . . . . . . . . . . . . . . . . . . . . . . . 240 Review – Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 Practice makes perfect – Chapter 9 . . . . . . . . . . . . . . 243

VI

Table of Contents

CHAPTER 11 THE QUALITATIVE ASPECT OF CHEMICAL EQUILIBRIUM . . . . . . 277 11.1 Static equilibrium and 28 . . . . . . . . . . . . . . . . . 278 dynamic equilibrium 11.1.1 Phase or "physical" equilibrium. . . . . . . . . . . 278 11.1.2 Solubility equilibrium. . . . . . . . . . . . . . . . . . . 279

For a detailed list of laboratories and demonstrations see page VIII

12.2.1 Theories of acids and bases. . . . . . . . . . . . . 320

11.1.3 Chemical equilibrium . . . . . . . . . . . . . . . . . . . 279 11.2 Irreversible and reversible reactions . . . . . . . . . . . . 280 11.2.1 Irreversible chemical reactions . . . . . . . . . . . 280 11.2.2 Reversible chemical reactions. . . . . . . . . . . . 281

12.2.2. Ionization constant of water. . . . . . . . . . . . . 325 12.2.3 Relationship between the pH and the molar concentration of hydronium and hydroxide ions . . . . . . . . . . . . . . . . . . . . 328

11.3 Necessary conditions 29 . . . . . . . . . . . . . . . 283 for attaining equilibrium 11.3.1 Change is reversible . . . . . . . . . . . . . . . . . . . 283 11.3.2 Change occurs in a closed system . . . . . . . . 284 11.3.3 Macroscopic properties are constant . . . . . . 285 Practice makes perfect – Section 11.2 and Section 11.3 . . . . . . . . . . . . . . . . . . . . . . . . . . 287 11.4 Le Chatelier’s principle . . . . . . . . . . . . . . . . . . . . . . . 288 11.5 Factors that affect the state of equilibrium . . . . . . 289 11.5.1 Changes in concentration

30

11.5.2 Changes in temperature 11.5.3 Changes in pressure

31 32

. . . . . . . 289 . . . . . . . . 292

12.2.4 Acidity constant and basicity constant 34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 12.2.5 Solubility product constant 35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336 Practice makes perfect – Section 12.2 . . . . . . . 338 Applications – Saltwater pools . . . . . . . . . . . . . . . . . . . . . 341 A brief history of… Sunglasses . . . . . . . . . . . . . . . . . . . 342 Review – Chapter 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343 Practice makes perfect – Chapter 12 . . . . . . . . . . . . . 345

. . . . . . . . . . . 295

11.5.4 Adding a catalyst . . . . . . . . . . . . . . . . . . . . . 297

SUPPLEMENT: REDOX REACTIONS . . . . . . . . . . . . . . 349

Practice makes perfect – Section 11.4 and Section 11.5 . . . . . . . . . . . . . . . . . . . . . . . . . . 298

1.

Oxidation and reduction . . . . . . . . . . . . . . . . . . . . . . 350

2.

Oxidation number . . . . . . . . . . . . . . . . . . . . . . . . . . . 353

11.6 Chemical equilibrium in everyday life . . . . . . . . . . . 299

3.

Reducing power of metals . . . . . . . . . . . . . . . . . . . . 357

Applications – Halogen light bulbs . . . . . . . . . . . . . . . . . . 300

4.

Electrochemical cell . . . . . . . . . . . . . . . . . . . . . . . . . 358

A brief history of… The Nobel Prize . . . . . . . . . . . . . . . . 301

5.

Redox potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360

Review – Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302

Applications – Domestic batteries. . . . . . . . . . . . . . . . . . . 366

Practice makes perfect – Chapter 11 . . . . . . . . . . . . . 304 APPENDICES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370 CHAPTER 12 THE QUANTITATIVE ASPECT OF EQUILIBRIUM . . . . . . . . . . . . . . . . . 307 12.1 Equilibrium constant

33

. . . . . . . . . . . . . . . . . 308

12.1.1 Expressing the equilibrium constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 12.1.2 Interpreting the value of the equilibrium constant . . . . . . . . . . . . . 312 12.1.3 The effect of temperature on the value of the equilibrium constant . . . . . . . . . . . . . 313

1.

Labratory Safety . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372

2.

Methods Used in Chemistry . . . . . . . . . . . . . . . . . . . 377

3.

Labratory Instruments and Techniques. . . . . . . . . . . 383

4.

Presenting Scientific Results . . . . . . . . . . . . . . . . . . 388

5.

Interpreting Measurement Results. . . . . . . . . . . . . . 394

6.

Mathematics in Science . . . . . . . . . . . . . . . . . . . . . 398

7.

Units of Measure in Chemistry. . . . . . . . . . . . . . . . . 408

8.

Reference Tables. . . . . . . . . . . . . . . . . . . . . . . . . . . . 410

12.1.4 Calculating concentrations at equilibrium . . . . . . . . . . . . . . . . . . . . . . . . 314

Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424

Practice makes perfect – Section 12.1 . . . . . . . 318

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428

12.2 Ionic equilibrium in solutions . . . . . . . . . . . . . . . . . . 320

Sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435

For a detailed list of laboratories and demonstrations see page VIII

Table of Contents

VII

CHAPTER 2

PHYSICAL PROPERTIES OF GASES

DEMO 1

Observing particle movement

DEMO 2

Compressibility of gases

DEMO 3

Expansion of gases

DEMO 4

Observing gas diffusion

LAB 5 DEMO 6

CHAPTER 8 DEMO 21 LAB 22

The relationship between the pressure and volume of a gas Boyle's law

MEASURING REACTION RATE Changing different parameters in time to calculate reaction rate Determining the mean rate and the different instantaneous reaction rates between magnesium (Mg) and hydrochloric acid (HCl)

CHAPTER 10 FACTORS THAT AFFECT REACTION RATES

LAB 7

The relationship between the volume and temperature of a gas

DEMO 8

The relationship between the pressure and temperature of a gas

LAB 23

The effect of the nature of reactants on reaction rate

LAB 9

The relationship between the volume and number of moles of a gas

LAB 24

The effect of surface area of reactants on reaction rate

Evaluating Avogadro’s law

LAB 25

Determining the molar mass and molar volume of an unknown gas

The effect of reactant concentration on reaction rate

LAB 26

The effect of the temperature of the reaction environment on reaction rate

LAB 27

The effect of a catalyst on reaction rate

DEMO 10 LAB 11

CHAPTER 3

ENERGY TRANSFER

LAB 12

Observing energy transfer

LAB 13

Calculating energy transfer

CHAPTER 4 LAB 14

CHAPTER 5 DEMO 15

CHAPTER 6

ENTHALPY CHANGE Chemical reactions and thermal energy GRAPHICAL REPRESENTATION OF ENTHALPY CHANGE Observing the analogue model of chemical reactions MOLAR HEAT OF REACTION

LAB 16

Determining the molar heat of dissolution of a substance

LAB 17

Determining the molar heat of neutralization

CHAPTER 7

HESS’S LAW

CHAPTER 11 THE QUALITATIVE ASPECT OF CHEMICAL EQUILIBRIUM DEMO 28

Representing the dynamic microscopic aspect of a system at equilibrium

DEMO 29

Observing the behaviour of a given reaction in closed and open systems

LAB 30

The effect of a change in concentration on the state of equilibrium

LAB 31

The effect of a change in temperature on the state of equilibrium

DEMO 32

The effect of a change in pressure on the state of equilibrium

CHAPTER 12 THE QUANTITATIVE ASPECTS OF EQUILIBRIUM

LAB 18

Determining the molar heat of an overall reaction

LAB 19

Hess’s law

LAB 33

Expressing the equilibrium constant

LAB 20

Determining the molar heat of formation of magnesium oxide (MgO)

LAB 34

Determining the acidity constant

LAB 35

Determining the solubility product constant

VIII

Laboratories and Demonstrations

The QUANTUM textbook contains four units: Gases, Energy Changes in Reactions, Reaction Rate and Chemical Equilibrium. These units are subdivided into 12 chapters, each with a certain number of sections depending on the concepts being addressed. There is a supplement on oxidation-reduction following the chapters. Appendices provide information on scientific strategies, methods and techniques, and elaborate on problem-solving techniques. The appendices also contain useful reference tables. At the end of the textbook, there is a glossary with definitions of the concepts under study and important terms used in science. An index at the end of the textbook provides a detailed list of keywords, with the page numbers on which the words appear.

Review The Review suggests reflections on important concepts introduced in the first and second years of Secondary Two. These concepts relate to those in the Third Year of Secondary Cycle Two Chemistry Program.

Each concept is clearly identified.

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At the beginning of each unit The title of the unit.

An organization chart gives a quick view of all the chapters and sections in the unit.

An introduction presents the general concepts discussed in the unit. Organization of the Textbook

IX

L’ORGANISATION DU MANUEL

Within a chapter

The title of the chapter. The introductory text presents the general concepts addressed in the chapter. The Review box provides a quick reference to the concepts presented in the Review section at the beginning of the textbook.

A summary introduces the sections of the chapter.

Numerous diagrams, photographs and illustrations make it easier to understand the concepts.

The names of the concepts and scientific terms are printed in blue in the text. Definitions of these terms are found in the glossary. Notions that are important for understanding the concepts appear in the subheadings.

Difficult words are marked with an asterisk* and are defined in the margin.

This pictogram indicates that there is an appendix with more extensive information on the subject being discussed.

The History highlights feature presents important historical figures from the worlds of science and technology.

The Info-science feature offers supplementary information related to the material under study. Formulas and equations are highlighted in a framed box. Calculation examples are always presented in a distinctive box.

X

Organization of the Textbook

Important elements are highlighted in a shaded blue box. References to other concepts or other scientific terms provide more extensive information when needed.

The Furthering your understanding feature introduces additional ideas beyond what is prescribed in the program.

At the end of a chapter A brief history of… presents the history of a subject related to the concepts addressed in the chapter.

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The Review summarizes the most important aspects of each of the concepts studied in the chapter.

The Practice makes perfect pages for the chapter propose questions and exercises that help to better understand and apply all of the knowledge acquired in the chapter. Pictograms indicate the degree of difficulty: easy, medium, difficult,  challenging.

The Applications feature presents a technological application related to a scientific concept addressed in the chapter. Appendices provide information on scientific strategies, methods and techniques, and elaborate on problem-solving techniques. The appendices also contain useful reference tables.

Organization of the Textbook

XI

2

TABLE OF CONTENTS 1

Organization of matter . . . . . . . . . . . . . . . . . . . . . . . . 4

7

Chemical changes . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.1 Atoms and molecules . . . . . . . . . . . . . . . . . . . . . . . 4

7.1 The law of conservation of mass . . . . . . . . . . . . . 21

1.2 Chemical formulas and ions . . . . . . . . . . . . . . . . . . 4

7.2 Balancing chemical equations . . . . . . . . . . . . . . . . 22 7.3 Stoichiometry. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2

Representations of atoms . . . . . . . . . . . . . . . . . . . . . 5 2.1 Rutherford-Bohr atomic model . . . . . . . . . . . . . . . . 5

3

8

Examples of types of chemical reactions . . . . . . 25

2.2 Simplified atomic model and the neutron . . . . . . . 5

8.1 Acid-base reactions . . . . . . . . . . . . . . . . . . . . . . . . 25

2.3 Lewis notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

8.2 Synthesis, decomposition and precipitation . . . . . 25 8.3 Endothermic and exothermic reactions. . . . . . . . . . 26

Periodic classification . . . . . . . . . . . . . . . . . . . . . . . . 8

8.4 Oxidation and combustion . . . . . . . . . . . . . . . . . . . 27

3.1 Periodic table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

8.5 Photosynthesis and respiration . . . . . . . . . . . . . . . 27

3.2 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 3.3 Periods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4

9

Nature of chemical bonds . . . . . . . . . . . . . . . . . . . . 28 9.1 Ionic bonding. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

Nomenclature and notation rules . . . . . . . . . . . . . 11

9.2 Covalent bonding . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.1 Naming a compound from its chemical formula . . 11 4.2 Writing chemical formulas of compounds . . . . . . 12

5

10

10.1 Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

Enumeration of matter . . . . . . . . . . . . . . . . . . . . . . . 13

10.2 Potential energy. . . . . . . . . . . . . . . . . . . . . . . . . . . 30

5.1 The Mole concept . . . . . . . . . . . . . . . . . . . . . . . . . 13

10.3 The law of conservation of energy . . . . . . . . . . . . 32

5.2 Molar mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

6

Forms of energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

10.4 Relationship between thermal energy, specific heat capacity, mass and temperature change . . . 32

Physical changes . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 6.1 Phase changes . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 6.2 Dissolution and solubility . . . . . . . . . . . . . . . . . . . 15 6.3 Concentration and dilution . . . . . . . . . . . . . . . . . . 16 6.4 Electrolytes and electrolysis. . . . . . . . . . . . . . . . . . 19

11

Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 11.1 Compressible and incompressible fluids . . . . . . . 33 11.2 Pressure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

6.5 Measuring physical change using the pH scale . . . . . . . . . . . . . . . . . . . . . . . . 20

3

REVIEW

1

1 Organization of matter Matter, whether it is living or not, is composed of atoms and molecules. Ions are generated by a process in which atoms gain or lose electrons.

1.1

Oxygen (O) Hydrogen (H)

Figure 1 The water molecule can be represented by balls and rods that indicate, respectively, the atoms and the bonds, or without rods when it is not necessary to show the bonds.

Atoms and molecules

Matter is composed of atoms. In nature, the atoms that make up most substances are assembled in molecules. The simplest molecules can be composed of only two atoms, while the most complex can have thousands. The atoms within molecules are joined together by chemical bonds. Molecules can be represented with or without their chemical bonds (see Figure 1). Molecules are also represented by their chemical formulas. These indicate the number of each of the atoms that make up the molecules (see Figure 2). Symbol for hydrogen

Symbol for oxygen

H2O

The number 2 indicates that there are two hydrogen atoms in the molecule.

The absence of a number indicates that there is just one oxygen atom in the molecule.

Figure 2 Chemical formula for the water molecule

1.2

Chemical formulas and ions

Ions are particles that carry a positive or negative electrostatic charge because the atoms from which they were formed either lost or gained one or more electrons. Table 1 Common polyatomic ions Formula

Name of ion

Polyatomic cations H3O

Hydronium

NH4

Ammonium

Polyatomic anions OH

Hydroxide

NO3

Nitrate

HCO3

Hydrogen Carbonate

SO4

Sulphate

PO43

Phosphate

2

Carbonate

CO3

4

2

REVIEW 1 Organization of matter

• A positive ion is a cation; a negative ion is an anion. • Generally, metals form positive ions, while non-metals form negative ions. For example, the atoms of sodium (Na), a metal, tend to lose an electron and form (Na) ions. When this happens, their electron configuration resembles that of neon (Ne). • Polyatomic ions are groups of atoms with strong bonds, carrying an electric charge. They behave like ions made of one single atom (see Table 1).

In Enpractice pratique pr 1. What ions do the following atoms tend to form? a) Nitrogen (N) d) Potassium (K) b) Iodine (I) e) Sulphur (S) c) Calcium (Ca) f) Fluorine (F)

2 Representations of atoms

REVIEW

2

There are different ways to represent atoms. The Rutherford-Bohr atomic model, the simplified atomic model and the neutron, and Lewis notation are three theoretical representations.

2.1

Electron inner shell Lowest energy level

Rutherford-Bohr atomic model

In the Rutherford-Bohr atomic model, the atom is shown with two types of elementary particles arranged in a particular way: protons (p), with a positive charge, and electrons (e), with a negative charge. The protons are contained in a very small, very dense nucleus. The electrons, equal in number to the protons, revolve around the nucleus in shells called energy levels. According to this model, the atom is made up largely of empty space (see Figure 3). • Rutherford helped to create this model by discovering a dense, heavy nucleus that contains a number of protons equal to the atomic number (represented by the letter Z) of the element in the periodic table. He also formulated the hypothesis that electrons move in a space close to the nucleus. • Bohr’s contribution to the model was to define the precise energy levels (electron shells) at which the electrons rotate around the nucleus. • The number of electron shells in an element corresponds to the number of the element’s period in the periodic table. For example, magnesium (Mg) is located in the third period of the table as it has three electron shells (see Figure 4).

Electron outer shell Highest energy level

Figure 3 The Rutherford-Bohr atomic model (1913), also called the planetary model

• Electron configuration refers to the distribution of electrons, with a maximum of 2, 8 and 18 electrons in the first three energy levels or “shells.” A lower level/shell must be full before the next energy level begins to be filled, and the outermost shell never contains more than 8 electrons, which are called valence electrons. • Valence electrons are located in an atom’s highest energy level (outermost shell). For example, magnesium has two valence electrons: the two electrons in the outermost shell. The valence electrons are the electrons involved in chemical reactions and that form bonds between atoms.

Figure 4 Magnesium (Mg) electrons are distributed in three electron shells in this electron configuration: 2, 8, 2.

In Enpractice pratique pr 2. a) What are the electron shells of an atom? b) What is the electron configuration of an atom?

2.2

Neutron

Simplified atomic model and the neutron

Proton

Electron

The simplified atomic model has the same characteristics as the RutherfordBohr model, but it includes one additional elementary particle in the nucleus: the neutron (n0). • The neutron is an uncharged particle that bonds with the proton and ensures that the nucleus of the atom is held together (see Figure 5).

Nucleus

Electron shell

Figure 5 Simplified atomic model

REVIEW 2 Representations of atoms

5

REVIEW

2

• The simplified atomic model can be represented in a more schematic way. The element’s chemical symbol can represent the nucleus and arcs of circles can represent the electron shells, under which is indicated the number of electrons found in the shells (see Figure 6).

Be

Na 2

2

Ar 2

8

1

2

b) Sodium (Na)

a) Beryllium (Be) Z4 Configuration: 2, 2

8

8

c) Argon (Ar)

Z  11 Configuration: 2, 8, 1

Z  18 Configuration: 2, 8, 8

Figure 6 Electron configurations of beryllium (Be), sodium (Na) and argon (Ar)

• The nucleons (neutrons and protons) are particles found in the nucleus. They make up the bulk of an atom’s mass (see Table 2). The sum of the protons and neutrons in the nucleus is called the mass number.

Table 2 Properties of protons, neutrons and electrons Particle

Position in the atom

Symbol

Charge

Mass

Proton

P

1

1.672  1024 g

In the nucleus

Neutron

n0

0

1.674  1024 g

In the nucleus

Electron

e

1

9.109  1028 g

Around the nucleus

• Isotopes are atoms of an element whose nuclei do not have the typical number of neutrons (see Figure 7).

Mass number (A)

Atomic number (Z)

Oxygen 16 8 protons – 8 neutrons

Oxygen 17 8 protons – 9 neutrons

Oxygen 18 8 protons – 10 neutrons

Figure 7 Oxygen has three natural isotopes. In this notation, the mass number (A) is written in superscript to the left of the element’s symbol, and the atomic number (Z) is written in subscript to the left of the symbol.

6

REVIEW 2 Representations of atoms

To find out how many neutrons an isotope has, subtract the atomic number (Z) from the mass number (A). For example, the number of neutrons in one of oxygen’s isotopes, oxygen 17, is 9.

REVIEW

2

16O 8 17O 8 18O 8

Number of neutrons  A  Z Number of neutrons in oxygen 17  17  8  9 neutrons

In Enpractice pratique pr 3. Draw a schematic diagram of the electron configuration of the following elements: a) Fluorine (F) d) Helium (He) b) Hydrogen (H) e) Potassium (K) c) Chlorine (Cl) f) Magnesium (Mg) 4. How many neutrons does a fluorine 19 atom have? North

2.3

West

Lewis notation

Lewis notation is a way of schematically representing the valence electrons of an atom. • The rules of notation are simple: valence electrons are represented by dots placed around the atom’s symbol (see Figure 8). • If the atom has more than four valence electrons, the number of dots is doubled. Dots representing the first four valence electrons are entered. Then a second line of dots is entered to indicate additional valence electrons, making pairs of dots. These are called electron pairs. The dots that are not in pairs represent odd electrons (see Figure 9). Electron pairs Odd electron

N

O

Cl

Nitrogen

Oxygen

Chlorine

Chemical symbol

East

South a) Arrangement of valence electrons around an imaginary square.

Na

Ca

Al

Sodium

Calcium

Aluminum

b) Arrangement of dots around chemical symbols.

Figure 8 Lewis notation for elements with one to three valence electrons

Figure 9 Lewis notation for elements with more than four valence electrons

In Enpractice pratique pr 5. Use Lewis notation to illustrate the following atoms: a) Hydrogen (H) d) Boron (B) b) Sodium (Na) e) Oxygen (O) c) Helium (He) f) Neon (Ne)

REVIEW 2 Representations of atoms

7

3 Periodic classification

REVIEW

3

Periodic classification is a way of listing all of the currently known elements within a system that takes into account a variety of factors, including the groups and periods of the elements.

3.1

Periodic table

The periodic table is a classification tool that groups the elements into groups and periods according to properties they share. Each box shows the symbol of an element and supplies useful information about the element (see Figure 10).

8

Atomic number

3.44 1314 218.3 182.9

Electronegativity Energy of the first ionization (kJ/mol) Melting point (°C) Boiling point (°C)

O

Ionic charge (if there is more than one, the first is the one that is most commonly used)

2 0.65 1.43

Atomic radius (1012m) Density (g/L or g/cm3)

Symbol

oxygen 15.999

Name Atomic weight (u) Molar mass (g/mol)

Figure 10 Each box in the periodic table supplies information about that element.

• The periodic table contains the three major categories of elements: metals, non-metals and metalloids (see Figure 11).

Cd Cu Cr

Pb

Bi

Metals

Cl As Si

S

Sb

B

Te

Br

C

Metalloids

I Non-metals

Figure 11 Some examples from the three categories of elements in the periodic table

• Metals tend to combine with non-metals to form compounds. All metals have these properties: – Metals are good conductors of heat and electricity.

8

REVIEW 3 Periodic classification

– Metals are shiny, malleable and ductile.

REVIEW

3

– When metals form compounds, they are usually electron donors, and many metals react chemically with acids. • Non-metals tend to combine with metals to form compounds. Elements in this category also have many properties in common: – Solid non-metals are not shiny or ductile and they are brittle. – Non-metals are usually good insulators. – When non-metals form compounds, they are electron acceptors. • Metalloids have the following properties: – Metalloids are usually breakable and not ductile. – Metalloids conduct electricity to varying degrees, but they are not good heat conductors.

3.2

Groups

The periodic table organizes the elements into chemical groups. These groups are families of elements that share similar physical and chemical properties, arranged in columns because they have the same number of valence electrons. In addition to being numbered in two ways (from 1 to 18 and with Roman numerals), some groups have specific names: 1 (IA) alkali metals, 2 (IIA) alkaline earth metals, 17 (VIIA) halogens and 18 (VIIIA) noble gases, also called inert gases (see Figure 12).

Like all alkali metals, lithium (Li) has just one valence electron.

Like all halogens, iron (Fe) has seven valence electrons.

Hydrogen (H) does not belong to any chemical group. Like all alkaline earth metals, magnesium (Mg) has two valence electrons.

1 (1A) 1

2 (IIA)

Like all inert gases, neon (Ne) has eight valence electrons.

18 (VIIIA) 13 14 (IIIA) (IVA)

15 (VA)

16 17 (VIA) (VIIA)

2 3

3 (IIB)

4 (IVB)

5 6 (VB) (VIB)

7 (VIIB)

8

9 (VIIIB)

10

11 (IB)

12 (IIB)

4 5 6 7

Figure 12 The four chemical families each have specific names: alkali metals, alkaline earth metals, halogens and noble gases. The Roman numerals for the numbers of the families from IA to VIIIA indicate the number of valence electrons of these elements.

REVIEW 3 Periodic classification

9

In Enpractice pratique pr

REVIEW

3

6. What is the connection between the number of the chemical group and the electron configuration of the elements in that group?

3.3

Periods

The periodic table is also comprised of periods, which are rows of elements arranged in order of their atomic number. The number of the period indicates the number of electron shells contained in the elements in that row (see Figure 13). 18 (VIIIA)

1 (IA)

1 2 (IIA)

13 (IIIA)

14 (IVA)

15 (VA)

16 (VIA)

17 (VIIA)

2

3

4

Figure 13 The numbers of the periods at the left of the table provide the number of electron shells in the elements in that row.

In Enpractice pratique pr 7. What information is provided by the number of the period? 8. Using the periodic table, find each of the elements corresponding to the following descriptions: a) A metal of the alkaline earth metal group that is found in the third period. b) A metalloid of group 15 (VA) in the fifth period. c) A non-metal of the oxygen group, located in the fourth period. d) A metal that is liquid in ambient conditions. e) Gaseous elements in the second period.

10

REVIEW 3 Periodic classification

4 Nomenclature and notation rules

REVIEW

4

Nomenclature and notation rules allow us to name compounds and write their chemical formulas, according to established conventions, in order to differentiate among them.

4.1

Naming a compound from its chemical formula

Use the following method to name a compound from its chemical formula: 1. First, name the element on the left hand side of the chemical formula. 2. Then, name the second element on the right hand side of the chemical formula by adding the suffix “ide” to the radical of the name of the original element. For example, the name of the element chlorine (Cl) becomes chloride in this way and the compound NaCl is named “sodium chloride” (see Figure 14). Some suffixes for elements on the right hand side in a chemical formula are formed in a special way. For example, hydrogen (H) does not become “hydrogenide,” but rather “hydride” (see Table 3).

Table 3 Specific formation of some suffixes

Name of the element

Name of the negative ion

Sulphur

Sulphide

Nitrogen

Nitride

Hydrogen

Hydride

Oxygen

Oxide

Carbon

Carbide

Chlorine n chloride

NaCl Na

Cl

Element at the left Element at the right

Sodium chloride

Table 4 List of prefixes

Figure 14 To name a compound, we must rename the element on the right hand side of its chemical formula.

1

Di

2

Tri

3

Tetra

4

Penta

5

Hexa

6

Potassium nitrate

Hepta

7

Ammonium chloride

Octa

8

Sodium carbonate

Nona

9

Magnesium hydroxide

Deca

10

Table 5 Nomenclature of polyatomic ions Cation

Anion

KNO3

K

NO3

NH4Cl

NH4

Cl

Na2CO3

Na

CO32

Mg(OH)2

Mg

OH

Quantity

Mono

In the case of a compound containing a polyatomic ion, we follow the nomenclature rules, using the name of the polyatomic ion without the suffix “ide” (see Table 5). For example, the compound KNO3 is called “potassium nitrate.”

Chemical formula

Prefix

Nomenclature of the ionic compound

REVIEW 4 Nomenclature and notation rules

11

4.2

REVIEW

4

Writing chemical formulas of compounds

Use the following method to write a chemical formula of compounds: 1. When the name of the compound is known In this case, write the chemical formula using the name of the compound. For example, sulphur fluoride is written “SF2” (see Figure 15). Sulphur fluoride

S

F2

Element at the left Element at the right

SF2 Figure 15 Chemical formula for s-ulphur fluoride

2. When the name of the compound is unknown In this case, the first element in the compound’s chemical formula is usually the one that is farthest to the left in the periodic table. For example, when a compound containing the elements fluorine (F) and calcium (Ca) forms, it is the element calcium, in the IIA group, which is farthest left on the table. That tells us that the compound’s formula begins with Ca and is followed by F. However, this information is not enough to write the correct formula for the compound: the proportions of the calcium and fluorine atoms in the compound must also be checked. Since calcium atoms tend to lose two electrons to resemble the closest inert gas and each fluorine atom must gain an electron to do the same thing, the compound will be formed of two fluorine atoms, which will each gain one electron from the calcium atom while the ionic bonds are forming. The chemical formula for this compound, calcium fluoride, is then CaF2. If the two elements that form a compound are from the same chemical group, the element that is lowest in the group must be written first. For example, oxygen (O) and sulphur (S) atoms bind to form a compound, sulphur dioxide, whose formula is SO2. Hydrogen does not belong to any group and is therefore the only exception to these rules. So hydrogen must be viewed as straddling the VA and VIA families to be able to apply these rules of formula writing. For example, “lithium hydride” would be written LiH since lithium (Li) is farther to the left than hydrogen (H) in the periodic table.

In Enpractice pratique pr 9. Find the name or the formula for the following compounds: a) CsCl e) Sulphur hexachloride b) AgF f) Compound formed by strontium and chlorine c) Carbon tetrabromide g) Compound formed by chlorine and aluminum d) Ca(NO3)2 h) Potassium iodide

12

REVIEW 4 Nomenclature and notation rules

5 Enumeration of matter

REVIEW

5

Matter can be enumerated, which means that the number of elements in a substance can be determined using the concept of the mole and molar mass.

5.1

The Mole concept

The mole is used to count atoms, ions, molecules and elementary particles. • The value of the mole is 6.02  1023 particles and corresponds to Avogadro’s number, NA (see Figure 16).

Helium (He) 4.0 g

Iron (Fe) 55.8 g

Water (H2O) 18.0 g

Ethanol (C2H5OH) 46.1 g

Carbon (C) 12.0 g

Salt (NaCl) 58.5 g

Mercury (Hg) 200.6 g

Sulphur (S) 32.0 g

Glucose (C6H12O6) 180.2 g

Figure 16 Samples of a mole of different substances

In Enpractice pratique pr 10. How many atoms or molecules are there in the following quantities? a) 3 mol of Fe (s) atoms c) 100 mol of N2 (g) b) 0.5 mol of water molecules d) 2.7 mol of NaCl (s) 11. How many moles correspond to each of the numbers of the following atoms or molecules? a) 7.525  1024 atoms of Cu c) 4.816  1024 molecules of CH4 26 b) 2.408  10 molecules of O2 d) 3.311  1024 molecules of He

REVIEW 5 Enumeration of matter

13

REVIEW

5

5.2

Molar mass

Molar mass is the mass of one mole of atoms of an element. • Molar mass (M) is expressed in grams per mole (g/mol). It corresponds to the same value as the atomic weight in the periodic table. • The molar mass of a substance is calculated by adding the molar mass of each of the atoms that form it, as in the following example: Example A What is the molar mass of methane (CH4)? Data: MC  12.011 g/mol

Calculation: MCH  MC  4MH 4

MH  1.008 g/mol

 12.011 g/mol  4 (1.008 g/mol)

MCH  ?

 16.043 g/mol

4

Answer: The molar mass of methane (CH4) is 16.043 g/mol. • Use the molar mass to find the number of moles that correspond to the given mass of a substance, as the following example shows: Example B What is the number of moles equivalent to 80.175 g of sulphur (S)? Data: m  80.175 g MS  32.065 g/mol n?

Calculation: m  n  MS m n MS 

80.175 g 32.065 g/mol

 2.5 mol

Answer: The number of moles of sulphur (S) is 2.5.

In Enpractice pratique pr 12. How many moles are there in each of the following quantities? a) 48 g of potassium (K) d) 0.27 g of sodium nitrate (NaNO3) b) 27 g of nitrogen (N2) e) 150 g of calcium hydroxide (Ca(OH)2) c) 345 g of carbon dioxide (CO2) 13. What is the mass of each of the following samples? a) 1 mol of lead (Pb) d) 18 mol of magnesium chloride (MgCl2) b) 1 mol of ozone (O3) e) 0.000 23 mol of aluminum sulfide (Al2S3) c) 15 mol of water (H2O) 14. How many moles are there in each of the following samples? a) 500 g of table salt (NaCl) d) 200 g of gold (Au) b) 250 g of road salt (CaCL2) e) 500 g of glucose (C6H12O6) c) 500 g of sodium hydroxide (NaOH)

14

REVIEW 5 Enumeration of matter

6 Physical changes

REVIEW

6

A physical change takes place when a substance undergoes a modification in its appearance that does not alter its nature or its characteristic properties.

6.1

Phase changes

Phase changes take place when a substance moves from one phase (solid, liquid or gas) to another. Only the appearance of the substance is altered (see Figure 17).

co

S nd olid en sa tio n

n tio ma bli Su

uid on Liq nsati e nd

co ion

zat

ori

p Va

Melting Solid phase

Solidification

Liquid phase

Figure 17 Phase changes of matter

6.2

Dissolution and solubility

In dissolution, one or more solutes are mixed in a solvent to obtain a solution. • During dissolution: – the mass of the substances before and after dissolution is the same. – the total volume of the solution that is formed is slightly less than the sum of the volumes of its components. – when the solvent cannot dissolve any more solute, the solution is said to be “saturated.” • Solubility is the property that indicates the maximum amount of solute that can be dissolved in a given volume of solvent at a given temperature. It is usually expressed in grams of solute per 100 mL (g/100 mL) of solvent. • A substance’s solubility varies according to temperature and, in the case of a gaseous solute, pressure. – The solubility of solids in water usually increases as the temperature rises. – The solubility of gases decreases as water temperature rises. – The greater the pressure exerted on a gas, the more soluble the gas is in the solvent.

REVIEW 6 Physical changes

15

In Enpractice pratique pr

REVIEW

6

15. a) If 50 g of sugar is dissolved in 1000 g of water, what is the mass of the solution? b) If 80 mL of alcohol is dissolved in 200 mL of water, what is the volume of the solution? 16. a) If a maximum of 5 g of powdered solute is dissolved in 25 mL of water at 25°C, what is the solubility of the substance expressed in g/100 mL? b) Given that solid substances are usually more soluble in a hot solvent, what would happen to the prepared solution in a if it were cooled? 17. Examine the graph below. a) Which solute’s solubility varies little with temperature? b) The solubility of solids usually increases with a rise in temperature. According to this graph, which solute is the exception to this rule? c) How much sodium nitrate (NaNO3) can be dissolved at 10°C? Solubility curves of different substances in water

140 130 120

Potassium iodide (KI)

110 Potassium nitrate (KNO3)

Solubility (g/100 mL)

100 90 80 70

Sodium nitrate (NaNO3)

60 50 40 Sodium chloride (NaCI)

30 20 10 0

Cerium sulfate (CeSO4)

10 20 30 40 50 60 70 80 90 100 Temperature of the solvent (°C)

6.3

Concentration and dilution

The concentration of a solution is the ratio of the amount of dissolved solute to the total amount of the solution. • Concentration is expressed by the following general equation: Concentration of solution Concentration 

16

REVIEW 6 Physical changes

Amount of solute Amount of solution

• Concentration is expressed using a variety of units (g/mL, %, ppm and mol/L). • The molar concentration of a solution is the number of moles of solute dissolved in 1 L of solution. It is expressed as moles per litre (mol/L). It is represented in two ways:

REVIEW

6

– by the symbol for molar concentration (C) – by brackets on either side of the chemical formula of the substance in question. For example, [HCl] = 2 mol/L means that the molar concentration of a solution of hydrochloric acid is 2 mol/L • The molar concentration is described by the following formula:

Molar concentration C

n V

where C  Molar concentration, expressed in moles per litre (mol/L) n  Amount of solute, expressed in moles (mol) V  Volume of solution, expressed in litres (L)

The following example shows how to calculate molar concentration: Example What is the molar concentration of an aqueous solution containing 5 g of ethanol (C2H5OH) in 0.750 L of solution? Data: m(solute)  5.00 g

1. Calculation of the molar mass of ethanol: MC H OH  2MC  6MH  MO 2 5

V(solution)  0.750 L

 2(12.011 g/mol)  6(1.008 g/mol)  15.999 g/mol

M(solute)  46.08 g/mol

 46.069 g/mol

n? C?

2. Conversion of the mass of ethanol into number of moles: m MC H OH  2 5 n m n MC H OH 2 5



5.00 g 46.069 g/mol

 0.109 mol 3. Calculation of the molar concentration: n C V 0.109 mol  0.750 L  0.145 mol/L Answer: The molar concentration of ethanol (C2H5OH) is 0.145 mol/L.

REVIEW 6 Physical changes

17

REVIEW

6

• Dilution is a physical change that results in a final solution that has a lower concentration than the original solution (see Figure 18). Strong solution

Dilute solution

Figure 18 A strong solution contains more particles of dissolved solute than a dilute solution of the same volume.

• This is the equation for calculating the concentration of solutions:

Concentration of solutions C1V1  C2V2 where C1  Concentration of the original solution V1  Volume of the original solution C2  Concentration of the original solution V2  Volume of the original solution When this equation is used, it is important to express the data in the same units of measurement, as shown in the following example: Example We want to prepare a dilute solution of vinegar to wash windows. If we have a bottle containing 250 mL of commercial vinegar, with a concentration of 4% V/V, what volume of water has to be added to obtain a solution of 1% V/V? Data: C1  4 % V/V V1  250 mL C2  1 % V/V V2  ?

1. Calculation of the final volume of the solution: C1V1  C2V2 V2  

C1V1 C2 4 % V/V  250 mL  1000 mL 1 % V/V

2. Calculation of the volume of water to add: Volume of water to add  Final volume (V2)  Original volume (V1)  1000 mL  250 mL  750 mL Answer: We have to add 750 mL of water to the 250 mL of the original solution.

18

REVIEW 6 Physical changes

In Enpractice pratique pr

REVIEW

6

18. Express the concentration of the following solutions in grams per litre (g/L), and then arrange them in ascending order of concentration: a) 3.5 g/75 mL c) 35 g/450 mL b) 5.6 kg/2000 L 19. a) What is the concentration in parts per million (ppm) of 1 g of carbon dioxide (CO2) in 1000 g of air? b) The calcium (Ca) content of spring water is 35 ppm. Express this content in grams per litre (g/L) and in milligrams per litre (mg/L). 20. Calculate the molar concentration of the solutions formed by the following quantities: a) 20 g of sodium hydroxide (NaOH) dissolved in a total volume of 1000 mL of solution b) 250 g of calcium carbonate (CaCO3) dissolved in a total volume of 5 L of solution c) 30 g of copper sulfate (CuSO4) dissolved in a total volume of 2500 mL of solution 21. How many grams of sodium hydroxide (NaOH) are dissolved in a total volume of 100 mL of solution with a sodium hydroxide concentration of 3 mol/L? 22. A 350 mL solution of potassium nitrate (KNO3), with a concentration of 36 g/L, is diluted with 250 mL of water. What is the concentration of the diluted solution? 23. A base solution with a concentration of 660 g/L of hydrochloric acid is used to prepare a solution of hydrochloric acid (HCl). How much should be used to obtain 1500 mL of solution with a concentration of 36g/L?

6.4

Electrolytes and electrolysis

Electrolytes are substances which, when dissolved in water, enable an electrical current to pass through the water, because of the presence of ions (see Figure 19).

Battery

Electrodes

Ions

Figure 19 The ions liberated by electrolytic dissociation enable a current to pass between the two electrodes inserted into the solution.

• Electrolytic dissociation or “electrolysis” occurs when a solute dissociates into ions with opposite charges at the moment of its dissolution in a solvent. • When acids, bases and salts are dissolved in water, they become electrolytes with particular characteristics (see Table 6 on page 20).

REVIEW 6 Physical changes

19

REVIEW

6

Table 6 Types of electrolytes Types of electrolytes Characteristics

Acids

Bases

• They release H ions.

• They release OH– ions.

• They neutralize bases.

• They neutralize acids.

• Their pH  7.

• Their pH  7.

• They turn blue litmus paper red.

• They turn red litmus paper blue.

Salts • They release different metal and non-metal ions of H and OH–. • They are one of the products of an acid-base neutralization. • Their pH is variable.

General chemical formula

H  Non-metal

Metal  OH

Metal – Non-metal

Sample equation for electrolytic dissociation

Hydrochloric acid: HCl (g) n H(aq)  Cl(aq)

Sodium hydroxide: NaOH (s) n Na(aq)  OH(aq)

Sodium chloride: NaCl (s) n Na(aq)  Cl(aq)

In Enpractice pratique pr 24. Which of these beakers contain electrolyte solutions? a) b)

c)

25. Which of the following dissolution equations represent electrolytic dissociations? Explain your answer. a) CH3COOH (l) n CH3COO(aq)  H(aq) b) C12H22O11 (s) n C12H22O11 (aq)

c) C2H5OH (l) n C2H5OH (aq) d) KNO3 (s) n K(aq)  NO3(aq)

26. What do acids, bases and salts have in common? 0

Battery acid

< 1.0

Gastric acid

2.0

Lemon juice

2.4

Cola (soft drink)

2.5

Vinegar

2.9

Orange or apple juice

3.5

Beer

4.5

Coffee

5.0

Tea

5.5

Acid rain

More acidic

Hydrochloric acid

6.5

6.5

Pure water (neutral) Blood

7.0 7.34 - 7.45

Seawater

8.0 9.0 - 10.0

Ammonia

11.5

Lime

12.5

Sodium hydroxide

14.0

More basic

Soap

Figure 20 The potential of hydrogen (pH) is expressed by a range of values from 0 to 14.

20

Measuring physical change using the pH scale

The pH (potential of hydrogen) scale is used to indicate the acidity or alkalinity of a solution. It is calibrated from 0 to 14 (see Figure 20). Solutions with a pH less than 7 are acids, those with a pH of 7 are neutral, and those with a pH of more than 7 are basic (or alkaline).

< 5.6

Milk

27. Indicate whether the following substances belong to the category of acids, bases or salts by looking at their chemical formulas: a) Copper sulfate (CuSO4) d) Nitric acid (HNO3) b) Perchloric acid (HClO4) e) Magnesium chloride (MgCl2) c) Potassium hydroxide (KOH) f) Cesium hydroxide (CsOH)

REVIEW 6 Physical changes

In Enpractice pratique pr 28. a) Is a solution with a pH of 5.6 an acid, a base or neutral? Explain your answer. b) Some soaps have a pH of 10. How many times more acidic than soap is distilled water with a pH of 7?

7 Chemical changes

REVIEW

7

A chemical change takes place when substances, the reactants, react with each other to form new substances, the products. • Indications that a chemical change has taken place: – release of a gas – change in colour – formation of a precipitate – change of energy in the form of heat, light or an explosion • Chemical equations show these changes (see Figure 21). Chemical equation Reactants

4 Fe (s)  Coefficient Chemical symbol

Product

3 O2 (g)

n

Index of the number of atoms

2 Fe2O3 (s)

changes to

Chemical Phase formula

Phase (s) (l) (g) (aq)

Solid Liquid Gaseous In an aqueous solution (the solvent is water)

Figure 21 Symbols used in chemical equations

7.1

The law of conservation of mass

The law of conservation of mass states that in a chemical change the total mass of the reactants is always equal to the total mass of the products. • In chemical reactions, the atoms in the elements and compounds formed by the reactants are rearranged to form the atoms of the products. Since the atoms of the reactants are all conserved, the products are formed only of atoms coming from the reactants (see Figure 22). Molecules of the reactants

Molecules of the products

2 C2H2 (g)  5 O2 (g) n 4 CO2 (g)  2 H2O(g) 

Number of atoms Mass Total mass

4 4 52 g

 212 g

Carbon (C) Oxygen (O) Hydrogen (H)



10 160 g

n

4 8



176 g



In Enpractice pratique pr

N2 (g)  3 H2 (g) n 2 NH2 (g)

4 2 

29. This equation describes the synthesis of ammonia:

36 g

212 g

Figure 22 When acetylene (C2H2) burns, the number of atoms of each element remains unchanged and the mass stays the same. However, since the atoms are combined differently, new compounds are produced.

If 56 g of nitrogen (N2) is combined with hydrogen (H2), and 68 g of ammonia (NH3) is produced, what mass of hydrogen was used?

REVIEW 7 Chemical changes

21

7.2

REVIEW

7

Balancing chemical equations

Balancing chemical equations means adding coefficients to the beginning of the chemical formulas of the reactants and the products to conform to the law of conservation of mass. • A skeleton equation is a chemical equation that presents the reagents and products of a chemical change without taking into account the law of conservation of mass. Skeleton equation for octane (C8H18) combustion C8H18 (g)  O2 (g) n CO2 (g)  H2O (g) • To observe the law of conservation of mass, a skeleton equation must be balanced to ensure equal numbers of atoms in the reactants and the products. These rules must be followed to balance a chemical equation: – Only coefficients may be added; the indexes in the chemical formulas of compounds cannot be changed. – Do not write the coefficient “1”; it is understood. – When the equation is balanced, the coefficients (“balancing numbers”) that are used must be whole numbers reduced to the lowest terms possible. – When the equation is balanced, the number of atoms of each element must be the same in the reagents and the products. Balanced equation for octane (C8H18) combustion 2 C8H18 (g)  25 O2 (g) n 16 CO2 (g)  18 H2O (g) The following example shows the balancing of a skeleton equation that presents the reagents and products in the combustion of methane (CH4):

Example Balance the following skeleton equation: REACTANTS CH4  O2

n

Solution: Number of atoms of reactants: C: 1 atom H: 4 atoms O: 2 atoms

PRODUCTS CO2  H2O

3. Balance the equation: H: 4 atoms

CH4 C: 1 atom

Number of atoms of products: C: 1 atom H: 2 atoms O: 3 atoms

1. Begin by balancing the most complex reactant or product: Underline methane (CH4). CH4  O2 n CO2  H2O 2. Finish by balancing the least complex reactant or product: Circle oxygen (O2). CH4  O2 n CO2  H2O



2 O2

B n

C: 1 atom

CO2

A O: 2  1 atom  4 atoms

REVIEW 7 Chemical changes

2 H2O

C O: 2  1 atom  2 atoms O: 2 atoms

A First check the carbon (C) in methane (CH4) and carbon dioxide (CO2). Since there is one atom of carbon on either side of the equation, it is possible to conclude that the carbon is balanced. B Write the coefficient “2” in front of the water (H2O) to balance the 4 hydrogen (H) atoms in methane. C This coefficient “2” in front of H2O brings the total number of oxygen atoms (0) to 4 in the products (2 in CO2 and 2 in H2O). B a l a n c e by adding a coefficient “2” in front of the oxygen (O2), the least complex reagent. The equation is now balanced. Answer: CH4  2 O2 n CO2  2 H2O

22

H: 4 atoms



In Enpractice pratique pr

REVIEW

7

30. Balance the following chemical equations: a) NO2 n N2O4 c) FeCl3  NaOH n Fe(OH)3  NaCl b) CO  O2 n CO2 d) Fe2O3  CO n Fe  CO2

7.3

Stoichiometry

Stoichiometry is the study of the quantitative relationships between the amounts of matter (reactants and products) that react in a chemical change. • The coefficients before the chemical formulas of reactants and products in a chemical reaction indicate the proportions of the reactants that combine to form the products. • Stoichiometric calculations determine the quantities of substances (reactants and products) and the energy involved in a chemical change. • The mole ratios are the proportions of the moles of reagents and products in a balanced chemical equation. The following two examples provide a method for making stoichiometric calculations to find the number of moles or the mass of a reactant or a product: Example A The combustion of gaseous ethane (C2H6) is shown in the following skeleton equation: C2H6  O2 n CO2  H2O How many moles of oxygen (O2) are needed to burn 16 mol of gaseous ethane? Data: Number of moles of C2H6  16 mol Number of moles of O2  ?

1. Balancing the chemical equation: Skeleton equation: C2H6  O2 n CO2  H2O Balanced equation: 2 C2H6  7 O2 n 4 CO2  6 H2O

2. Mole ratios of the reagents and the products, and amounts carried forward: n 2 C2H6  7 O2 4 CO2  6 H2O 2 mol 7 mol 4 mol 6 mol 16 mol ? 3. Calculation of moles of oxygen: 7 mol of O2 ?  16 mol of C2H6 2 mol of C2H6 ?

7 mol of O2  16 mol of C2H6 2 mol of C2H6

 56 mol Answer: It would take 56 mol of oxygen (O2) to burn 16 mol of gaseous ethane (C2H6).

REVIEW 7 Chemical changes

23

REVIEW

7

Example B Aboard a space shuttle such as the Endeavour, special measures must be taken to prevent the levels of carbon dioxide (CO2), produced by the astronauts’ breathing, from rising too high. This involves passing the cabin air through filters containing lithium hydroxide granules (LiOH). The lithium hydroxide reacts with the carbon dioxide to form lithium carbonate (Li2CO3) and water. What mass of lithium hydroxide per person per day is needed to maintain air quality in the cabin, if we know that each astronaut exhales an average of 1056.0 g of carbon dioxide a day? Data: mCO  1056.0 g 2 mLiOH  ?

1. Balancing the chemical equation: Skeleton equation: LiOH  CO2 n Li2CO3  H2O Balanced equation: 2 LiOH  CO2 n Li2CO3  H2O

2. Mole ratios of the reagents and products, and conversion using molar mass: n Li2CO3  2 LiOH  CO2 H2O 2 mol 1 mol 2 mol  MLiOH 1 mol  MCO 2

2 mol  23.948 g 1 mol  44.009 g 1 mol 1 mol 47.896 g

44.009 g

3. Calculation of the mass of lithium hydroxide: ? 47.896 g of LiOH  1056.0 g of CO2 44.009 g of CO2 ?

47.896 g of LiOH  1056.0 g of CO2 44.009 g of CO2  1149.27 g

Answer: It would take 1149.3 g of lithium hydroxide (LiOH) per person per day to maintain air quality in the cabin.

Figure 23 Canadian astronauts Julie Payette and Robert Thirsk aboard the space shuttle Endeavour

In Enpractice pratique pr 31. A car’s airbag fills with nitrogen (N2) produced in the following decomposition reaction: 2 NaN3 (s) n 3 N2 (g)  2 Na (s) In a small car, it usually takes 2 mol of N2 (g) to inflate the driver’s airbag. How many moles of sodium azide (NaN3) are required to produce the nitrogen? 32. Gaseous ammonia (NH3) reacts with oxygen (O2) to form water and nitrogen oxide (NO) according to this equation: 4 NH3 (g)  5 O2 (g) n 4 NO (g)  6 H2O (g) a) What mass of oxygen will react with 34 g of ammonia? b) How many moles of nitrogen oxide will be formed if 320 g of oxygen react with a sufficient amount of ammonia?

24

REVIEW 7 Chemical changes

8 Examples of types of chemical reactions

REVIEW

There are different types of chemical reactions that can produce chemical changes. These are the most common:

8.1

8

Acid-base reactions

Acid-base neutralization is a chemical change by which an acid and a base react with each other to form a salt and water. • If an acidic solution is mixed with a basic solution, the hydrogen (H) ions in the acid react with the hydroxide (OH) ions in the base to form water and a salt whose nature depends on the reagents. • When the original solutions (the reactants) contain equal amounts of hydrogen and hydroxide ions, the acid-base neutralization reaction is complete and the final solution that is produced is neutral. When the quantities of the reactants are different, the acid-base neutralization reaction is incomplete: the final solution will have an acidic pH if it contains a surplus of hydrogen ions, and it will have a basic pH if it contains a surplus of hydroxide ions. • This is how the general equation for acid-base neutralization is written: General equation for acid-base neutralization acid (aq)  base (aq) n water (l)  salt (aq)

In Enpractice pratique pr 33. Pour a solution of hydrochloric acid (HCl) into a beaker and insert the tip of a pH meter into the solution. The pH meter reads 1.9. a) What will happen if we add a solution of potassium hydroxide (KOH) to the acid, one drop at a time? b) What is the complete chemical equation for this reaction? c) When the quantity of hydroxide (OH) ions is higher than the quantity of hydrogen (H) ions, what will happen to the pH?

8.2

Synthesis, decomposition and precipitation

Synthesis, decomposition and precipitation are all types of chemical change. • Synthesis is a chemical change in which elements or simple compounds react and form a more complex compound. This is how the general equation for synthesis is written: General equation for synthesis A  B n AB • Decomposition is a chemical change in which a compound separates into elements or simple compounds. It is the opposite of synthesis.

REVIEW 8 Examples of types of chemical reactions

25

REVIEW

This is how the general equation for decomposition is written:

8

General equation for decomposition AB n A  B • Precipitation is the formation of a slightly soluble or insoluble solid produced from a mixture of two electrolytic solutions. The solid thus formed is called a precipitate (see Figure 24). Pb(NO3)2 (aq)  2KI (aq) n PbI2 (s)  2KNO3 (aq) Precipitate

In Enpractice pratique pr 34. Indicate which type of reaction (synthesis, decomposition or precipitation) takes place in each of the following chemical changes: a) CaCl2 (s) n Ca (s)  Cl2 (g) b) CaBr2 (aq)  Pb(NO3)2 (aq) n PbBr2 (s)  Ca(NO3)2 (aq) c) 2 Li (s)  Cl2 (g) n 2 LiCl (s)

8.3

Endothermic and exothermic reactions

All chemical changes cause energy changes in a system and an environment. The energy changes depend on the type of change: energy can be absorbed or released. Figure 24 The yellow precipitate that forms is lead iodide (PbI2)

• Endothermic reactions are chemical reactions that absorb energy. This is the general equation for an endothermic reaction: General equation for an endothermic reaction Reactants  Energy n Products

• Exothermic reactions are chemical reactions that release energy. This is the general equation for an exothermic reaction: General equation for an exothermic reaction Reactants n Products  Energy

In Enpractice pratique pr 35. Determine whether the following chemical changes are endothermic or exothermic: a) When water is decomposed by electrolysis, as soon as the electrical current is cut, the decomposition reaction stops. b) When hydrogen (H2) is subjected to a flame test, an explosion can be heard. c) 2 H2 (g)  C (s) n CH4 (g)  75 kJ 36. Look at the following reaction: H2 (g)  I2 (g)  173 kJ n H2I2 (g) a) Calculate the amount of energy involved in a reaction that forms 768 g of hydrogen iodide (H 2I2). b) Indicate whether heat is absorbed or released (an endothermic or exothermic reaction).

26

REVIEW 8 Examples of types of chemical reactions

8.4

Oxidation and combustion

REVIEW

Oxidation reactions are very common on Earth: almost 21 percent of the atmosphere is made up of oxygen (O2), the reactant that causes the great majority of oxidation reactions. Combustion, cellular respiration, the formation of rust, and corrosion are all oxidation reactions.

8

• Oxidation is a chemical change in which oxygen or another substance with similar properties combine with a reactant to form an oxide. For example, the oxidation of copper (Cu) causes copper oxide (CuO) to form, as shown in this chemical equation: 2 Cu (s)  O2 (g) n 2 CuO (s) • Combustion is an oxidation reaction that releases energy. A fuel, an oxidant and thermal energy are required elements for combustion. • Octane that burns in the presence of oxygen can be used to illustrate the combustion equation. 2 C8H18 (g)  25 O2 (g) n 16 CO2 (g)  18 H2O (g)

In Enpractice pratique pr 37. Determine the chemical equation for the oxidation reaction of magnesium (Mg).

8.5

Photosynthesis and respiration

Almost all life on Earth depends on two chemical reactions: photosynthesis (a synthesis reaction) and cellular respiration (a combustion reaction). • Photosynthesis is a chemical change through which living organisms transform the Sun’s radiant energy into chemical energy. This is the chemical equation for photosynthesis: Chemical equation for photosynthesis 6 CO2 (g)  6 H2O (l)  Energy n C6H12O6 (s)  6 O2 (g) • Respiration is a chemical change through which energy contained in sugars is released in living cells. It is the reverse of the photosynthesis process, and this is its chemical equation: Chemical equation for respiration C6H12O6 (s)  6 O2 (g) n 6 CO2 (g)  6 H2O (l)  Energy

In Enpractice pratique pr 38. Explain why respiration and photosynthesis are inverse reactions.

REVIEW 8 Examples of types of chemical reactions

27

9 Nature of chemical bonds

REVIEW

9

A chemical bond is an exchange or sharing of electrons between two atoms, which causes the formation of a compound or a diatomic element. The force with which a valence electron is held by the nucleus is called electronegativity.

9.1

Ionic bonding

An ionic bond is formed when electrons are transferred from one atom to another (see Figure 25). • This type of bond is formed when one of the atoms has much greater electronegativity than the other atom. This bond usually forms between a metal and a non-metal. It causes the formation of an ionic compound.

Neutral atoms

Loss

Gain

Loss and gain of electrons

Ions that are formed

Na Number of protons

of  Number electrons

Cation (positive ion)

Cl Number of protons

of  Number electrons

Anion (negative ion)

Figure 25 When the sodium (Na) and chlorine (Cl) ions form, the electron transfer takes place from the sodium to the chlorine. Ions with opposite charges attract each other and form ionic bonds.

28

REVIEW 9 Nature of chemical bonds

• Lewis notation is a way of representing the ionic bond (see Figure 26).

Na

Cl

Na



Cl

REVIEW

9



a) A single bond between two ions

Mg

O

Mg

2

O

2

b) Two ionic bonds between two ions

F Ca

Ca

F



F



2

F c) Two ionic bonds between three ions

Figure 27 Each oxygen atom (O) provides two electrons to form two electron pairs shared with the carbon atom (C). The carbon atom provides four electrons, two for each oxygen atom.

Figure 26 The formation of ionic bonds shown in Lewis notation

9.2

Covalent bonding

A covalent bond is formed when electrons are shared by two atoms. • This type of bond usually forms when there is only a small difference in the electronegativity of the two elements. This bond generally forms between two non-metals, between two identical atoms or between a non-metal and hydrogen (H). • The covalent bond causes the formation of a covalent compound in which there are one or more single, double or triple covalent bonds (see Figure 27). • Lewis notation provides a way of representing the covalent bond (see Figure 28).

Cl

Cl

a) The diatomic chlorine (Cl2) molecule has a single covalent bond.

H H

C

H

H

In Enpractice pratique pr 39. What is the difference between ionic bonds and covalent bonds in terms of valence electrons? 40. What categories of elements usually form ionic bonds with each other? 41. With regards to the following compounds: CO2, H2O, CaCl2, CCl4, LiF, CaCO3, PBr3, FeSO4 a) Which ones contain one or more ionic bonds? b) Use Lewis notation to represent those which contain one more covalent bonds.

b) The methane (CH4) molecule has four single covalent bonds.

H

O

H

c) The water (H2O) molecule has two single covalent bonds.

Figure 28 Lewis notation is a way of representing the covalent bonds, by showing the shared electron pairs.

REVIEW 9 Nature of chemical bonds

29

REVIEW

10

10 Forms of energy There are several forms of energy including kinetic energy and potential energy.

10.1

Kinetic energy

Kinetic energy (Ek) is concerned with the movement of a body. • Kinetic energy is the product of the mass (m) and the square of the velocity (v) of a body in motion. • The following equation can be used to find the kinetic energy of a body in motion: Kinetic energy 1 Ek  mv 2 2 where Ek  Kinetic energy, expressed in joules (J) m  Mass of an object, expressed in kilograms (kg) v  Velocity of an object, expressed in metres per second (m/s) • Kinetic energy grows in proportion to the mass of an object and the square of its velocity.

In Enpractice pratique pr 42. What is the kinetic energy of a 45-gram golf ball that is moving at: a) 70 m/s? b) 45 m/s?

10.2

Potential energy

Potential energy is stored in a body and can be transformed into another form of energy, for example in gravitational potential energy. • Gravitational potential energy stored in a body is the product of the mass (m), the gravitational acceleration (g) and the height (h) of the body in relation to a reference point. • The following equation can be used to determine the gravitational potential energy of a body in motion: Gravitational potential energy Ep  mgh where Ep  Gravitational potential energy, expressed in joules (J) m  Mass of an object, expressed in kilograms (kg) g  Gravitational acceleration, which is 9.8 m/s2 on Earth h  Height of an object in relation to a reference point, expressed in metres (m)

30

REVIEW 10 Forms of energy

The following table lists the main forms of energy (see Table 7). The potential energies are associated with a force: electric force for electric potential energy, nuclear force for nuclear potential energy, etc.

REVIEW

10

Table 7 Various forms of energy Form of energy

Example

Kinetic energy Energy associated with an object’s movement.

A car driving along a road.

Gravitational potential energy Energy associated with an object’s position above the ground.

Water at the top of a waterfall. Because of its height, this water has more gravitational potential energy than does the water at the bottom of the waterfall.

Electric potential energy Energy present when electric charges interact.

An electrically charged storm cloud.

Elastic potential energy Energy stored in materials that have been compressed or stretched.

A spring that has been compressed or stretched.

Nuclear potential energy Energy stored in the nucleus of an atom.

Uranium inside the core of a nuclear reactor.

Chemical energy Energy stored in chemical bonds between atoms.

Energy stored in glucose, gasoline, etc.

Thermal energy Energy related to the agitation of the molecules and atoms that make up an object or a substance. Radiant energy Energy carried by an electromagnetic wave.

Boiling water. The thermal energy of the water molecules increases and they become more agitated.

Light from the Sun or an electric light bulb, microwaves that reheat food, radio waves, waves emitted and captured by cell phones.

In Enpractice pratique pr 43. A person lifts an 8-kg box off the ground and puts it on top of a table that is 1.5 m high. What is the potential energy stored in the box once it is placed on the table? 44. A diver weighing 65 kg is on a diving board. The diving board is 10 m above the surface of the water, which is in a pool 5 m deep. How much potential energy is stored in the diver: a) in relation to the surface of the water? b) in relation to the bottom of the pool?

REVIEW 10 Forms of energy

31

REVIEW

10

10.3

The law of conservation of energy

The law of conservation of energy states that energy can be neither created nor destroyed, but can be changed from one form to another. • An isolated system is one that does not exchange matter or energy with its surroundings.

In Enpractice pratique pr 45. A stone falls from the top of a cliff that is 54 m high. Assuming that there is no friction, determine whether the kinetic and potential energy are at their maximum, non-existent or equal: a) when the stone is at the top of the cliff. b) In Enpractice pratique pwhen r the stone is at a height of 27 m during its fall. 46. The specific heat capacity of water is 4.19 J/g°C. Calculate the amount of heat required to raise the temperature of 240 g of water by 30°C. 47. The specific heat capacity of water is 4.19 J/g°C and that of copper (Cu) is 0.39 J/g°C. If we heat 10 g of water and 10 g of copper to 25°C: a) which of the two substances will store more thermal energy? Explain your answer. b) which of the two substances will have a more rapid rise in temperature?

32

REVIEW 10 Forms of energy

• Mechanical energy (Em ) is the sum of the potential energy (Ep) and kinetic energy (Ek ) in a system. The following equation expresses the relationship between these forms of energy:

Mechanical energy Em  Ep  E k where Em  Mechanical energy, expressed in joules (J) Ep  Potential energy, expressed in joules (J) Ek  Kinetic energy, expressed in joules (J)

10.4

Relationship between thermal energy, specific heat capacity, mass and temperature change

Thermal energy is a form of energy possessed by a substance due to the agitation of its particles. It depends on the number of particles (mass) in the substance, the degree of agitation (temperature) of the particles and the nature of the substance. • The change in thermal energy (Q) of a substance is the product of the substance’s mass, its capacity to store heat and the change in its temperature. • Specific heat capacity (c) is the amount of thermal energy that must be transferred to one gram of a substance to raise its temperature by 1°C. It is a physical property that is characteristic of the substance. • This equation is used to find the amount of heat needed to raise the temperature of a substance.

Thermal energy Q  mcT where Q  Amount of heat, expressed in joules (J) m  Mass of a substance, expressed in grams (g) c  Specific heat capacity of a substance, expressed in joules per gram per degree Celsius (J/g°C) T  Variation in temperature (Tf  Ti), expressed in degrees Celsius (°C)

11 Fluids

REVIEW

Substances in a liquid or gaseous state are fluids, and they have the ability to diffuse in all directions.

11.1

11

Compressible and incompressible fluids

Fluids can be either compressible or incompressible. They have the distinctive characteristic of exerting equal pressure in all directions. • Gaseous fluids are fluids that are compressible because their volume can decrease under pressure. The particles that make up gases are far apart and can move closer together when compressed (see Figure 29).

Figure 29 A gas is a compressible fluid.

• Liquid fluids are fluids that are incompressible because their particles are too close to each other to be forced closer under pressure.

11.2

Pressure

Pressure is the measurement of a force exerted on a surface. • Pressure is calculated by the following formula:

Pressure P

F A

where P  Pressure, expressed in newtons per square meter (N/m2) or in pascals (Pa) F  Force, expressed in newtons (N) A  Area where force is applied, expressed in square meters (m2)

• The unit of measurement of pressure used in the International System of Units (SI) is the pascal (Pa). It is the equivalent of 1 N/m2. The kilopascal (kPa) is used to measure major quantities like atmospheric pressure, which has an average value of 101.3 kPa.

REVIEW 11 Fluids

33

34

CONTENTS CHAPTER 1

Chemical Properties of Gases . . . . 37 CHAPTER 2

Physical Properties of Gases . . . . . 53

The scientific study of gases allows a better understanding of the world around us. For example, by understanding the gas composition of the upper at mos phere, scientists have been able to explain the phe nomenon of aurora borealis. Knowledge of how gases behave has also made the development of aviation possible. In addition to the field of technology, where gases are used for many different purposes, gases play a key role in many natural phenomena, such as volcanoes.

In this unit, you will learn about the different uses of gases in daily life, which draw on the chemical properties of gases, such as chemical reactivity, and on their physical properties, such as compressibility. Moreover, after explaining the kinetic theory of gases, you learn the laws that establish relationships between the various physical properties of gases, such as pressure, volume, temperature and quantity.

35

CHAPTER

UNIT

1

CHEMICAL PROPERTIES OF GASES

1

1.1 Daily use of gases 1.2 Chemical reactivity of gases

2.1 Kinetic theory of gases

GASES

2.2 Behaviour of gases 2.3 Pressure of gases CHAPTER

2

PHYSICAL PROPERTIES OF GASES

2.4 Simple gas laws 2.5 Ideal gas law 2.6 General gas law 2.7 Stoichiometry of gases 2.8 Dalton’s law

36

Chemical Properties of Gases

G

ases play a role in our everyday lives, from our comforts to our jobs and leisure activities. In all spheres of society, various technological applications of gases are essential. For example, natural gas is used to heat our homes and cook our food. In fact, many of these technological applications are made possible by the chemical reactivity of gases. Gases also play a key role in many natural phenomena.

In this chapter, you will learn about the different uses of gases in daily life and study the chemical reactivity of certain types of gases that play a role in natural phenomena or are used in various technological applications.

Review Periodic table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Nomenclature and notation rules . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Oxidation and combustion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Photosynthesis and respiration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.1 1.2

Daily use of gases ........................38 Chemical reactivity of gases .............44 CHAPTER 1 Chemical Properties of Gases

37

1.1 Daily use of gases The daily use of gases is widespread in many areas of human activity. Gases play an important role in the environment and in certain technological applications. Gases are present almost everywhere on the Earth: in the atmosphere, dissolved in the oceans and sometimes even trapped in solid materials such as porous rocks (see Figure 1).

Figure 1 Pumice is a volcanic rock that cools quickly and traps gas bubbles trying to escape from the lava, which is what gives it a spongy texture.

Most gases are invisible, but many have an odour that allows us to detect them. Gases are fluids that do not have a defined shape or volume. They flow like liquids but, contrary to liquids and solids, they will not remain in an open container. Instead, they diffuse in all directions. However, gases will spread throughout a closed container, filling it completely and taking on its shape. Gases play an important role in many natural phenomena such as the respiration of living organisms and the process of food production for plants through photosynthesis. Gases are also used in many technological applications such as the manufacturing of energy-efficient windows, where they improve our quality of life without our even knowing of their presence.

1.1.1

Gases and natural phenomena

The Earth is surrounded by a gaseous layer called the atmosphere. The terrestrial atmosphere is composed of a mixture of various gases that come into play in many natural phenomena (see Figure 2). Nitrogen 78%

Oxygen 21%

1%

Argon (0.93%) Carbon dioxide (0.03%) Neon Helium Methane Krypton (traces) Xenon Hydrogen Ozone Radon

Figure 2 The composition of the atmosphere. Since the water that is present in the atmosphere is generally considered a component of the hydrosphere, it does not appear in the diagram. The quantity of water vapour in the atmosphere varies according to meteorological conditions.

Nitrogen (N2) is the most abundant gas in the atmosphere. This gas is vital to plant growth. However, plants are not able to directly assimilate the nitrogen present in the atmosphere. Nitrogen must first be transformed into nitrogen compounds by bacteria that live in soil or in the nodules of certain leguminous plants (see Figure 3). Plants are then able to absorb it.

Figure 3 The bacteria that fix nitrogen (N) live in symbiosis in the nodules that dot the roots of leguminous plants.

38

UNIT 1 Gases

Oxygen (O2) is the second most abundant gas in the terrestrial atmosphere. It is vital to the respiration of living organisms and is the most common oxidizer in a combustion reaction (see Figure 4).

*

Oxygen is primarily produced by photosynthesis (see Figure 5). Photosynthesis also enables the oxidation of metals such as iron (Fe) and some of its alloys, including steel and cast iron. In addition to nitrogen (N2) and oxygen, approximately 1% of the atmosphere is made up of a mixture of several gases. This mixture includes carbon dioxide (CO2) and methane (CH4), two greenhouse gases. Plants use carbon dioxide as a source of carbon (C) during photosynthesis.

Radiant energy

A substance that causes * Oxidizer the combustion of a fuel during a combustion reaction.

Figure 4 A glowing wood splint reignites in the presence of oxygen (O2).

Carbon dioxide (CO2) is present in the air. Oxygen production (O2)

Water originates in the roots of the plant.

6 CO2 (g)  6 H2O (l)  Energy n C6H12O6 (s)  6 O2 (g)

Glucose production (C6H12O6) (aq)

Figure 5 A schematic diagram of photosynthesis

Methane present in the atmosphere is the main component of biogas resulting from the anaerobic fermentation of organic matter. This type of fermentation occurs in marshlands and landfills. Methane is also produced in mammals’ stomachs, particularly in ruminants, and is released into the environment through the process of flatulence. Large quantities of methane are also trapped in permafrost and the ocean floor (see Figure 6). Melting permafrost, due to climate change, could cause very large quantities of methane to be released, thereby exacerbating global warming.

*

Relating to a process that * Anaerobic can occur without oxygen (O ). 2

Figure 6 Methane (CH4) is released in the form of bubbles on the surface of a lake in the Mackenzie River Delta in the Northwest Territories.

CHAPTER 1 Chemical Properties of Gases

39

Figure 7 A radon (Rn) detector

Radon (Rn), a radioactive gas in the group of noble gases, is produced by the natural disintegration of uranium (U) present in soil and rock. In the open air and in well-ventilated areas, the quantity of radon is minimal and harmless. However, radon can seep through cracks in cement floors and accumulate in the basements of houses. In these confined spaces, where air circulation is limited, radon can reach toxic concentrations. Exposure to high concentrations of radon increases the risk of lung cancer. However, it is difficult to predict the concentration of radon in a particular house. In fact, on the same street, the concentration can vary from one house to another. A radon detector can be used so that corrective measures can be taken if concentrations exceed acceptable levels (see Figure 7). Certain natural phenomena, such as volcanic eruptions, emit large quantities of gases into the atmosphere (see Figure 8). These gases include carbon dioxide (CO2), water vapour (H2O), sulphur dioxide (SO2) and carbon monoxide (CO).

Figure 8 Eruption of Guagua Pichincha in Quito, Ecuador, on October 7, 1999. When a volcano like this erupts, it releases tonnes of gases and ash into the atmosphere. wind A stream of highly * Solar charged particles constantly projected by the Sun into space. These particles include mostly ions and electrons.

Gases present in the upper atmosphere play a role in a natural phenomenon that can be observed at night near the Earth’s poles: aurora borealis, in the northern hemisphere, and aurora australis, in the southern hemi sphere (see Figure 9). Aurora borealis and australis are produced by solar wind particles that penetrate the atmosphere close to the magnetic poles and ionize gas particles in the upper atmosphere. These ionized gas particles then release surplus energy they have absorbed by emitting light that is visible in the Figure 9 Aurora borealis form of multicoloured veils.

*

40

UNIT 1 Gases

The buoyancy of fish Gases are involved in natural aquatic phenomena. Some fish species have a pouch called a swim bladder in their abdomen just below the spine. This pouch is filled primarily with oxygen (O2) and helps to ensure the buoyancy of fish. Depending on the quantity of gas it contains, the swim bladder determines the depth at which a fish swims. This pouch allows a fish to maintain the same density as the water at various depths. In many species, oxygen travels to the swim bladder through the blood, where the gas is dissolved. The dissolved oxygen penetrates the blood vessels that line the swim bladder and enters the bladder where it returns to its gaseous state. This way, the fish does not have to swim to the surface to fill its swim bladder with gas.

Spine

Stomach

Swim bladder

Intestines

Caudal fin

Kidneys

Figure 10 The fish stores gas in its swim bladder, which allows it to float at various depths in the water.

The movement of air on the planet caused by prevailing winds regulates the climate in each region. It also helps to disperse plant pollen. The Earth’s atmosphere contains a variable quantity of water in gaseous form which condenses into tiny droplets and tiny ice crystals, forming clouds. These more or less humid air masses displaced by radiant energy from the Sun form weather systems such as cyclones and anticyclones. These systems determine the weather in a region at a particular time (see Figure 11).

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

GEORGES CLAUDE French chemist and inventor (1870–1960)

Figure 11 Hurricane Ike over Cuba, September 10, 2008

1.1.2

Gases and their technological applications

Many technological applications, such as neon (Ne) signs, require a variety of gases, and their supply is essential. For some of these applications, gases serving as raw materials are drawn directly from the atmosphere. This is the case for the production of ammonia (NH3) using nitrogen (N2) in the air. For other applications, the necessary gases are synthesized. For example, the raw material petroleum is used to synthesize propane (C3H8), which is used as a fuel. But regardless of the origin of these gases, various technological applications use them in multiple and sometimes unexpected ways (see Table 1 on the following page).

In 1902, Georges Claude developed a process for liquifying air, which improved the efficiency of the process used at the time. His research allowed him to separate certain components of air, such as nitrogen (N2), oxygen (O2) and a few noble gases. He subjected these gases to electric discharges in gas tubes coated in fluorescent powder, which he designed, and obtained tubes that emitted light in various colours, depending on the fluorescent powder and gas used. This led, in 1910, to the invention of neon (Ne) light, which is still used today in illuminated signs.

CHAPTER 1 Chemical Properties of Gases

41

Table 1 The use of gases in various technological applications

Gas Nitrogen (N2)

Examples of technological applications • It is used as an unreactive agent to preserve the freshness of foods in modified atmosphere packaging. • It constitutes the raw material used in the manufacture of fertilizers with an ammonia (NH3) base. • It is used to inflate tires in the aviation industry and, increasingly, in the car industry.

Oxygen (O2)

• It is administered to people with respiratory problems such as asthma, and used to ensure the respiratory function of patients during certain operations under general anesthesia. • It enables pressurization inside commercial planes. • It acts as an oxidizer to ignite the flame of an oxyacetylene welding torch.

Compressed air

• Scuba divers carry tanks of compressed air so that they can remain under water for long periods of time. • It is used to operate pneumatic tools (percussion drill, dentist’s drill, etc.). • It is used to produce bubbles in certain industrial processes for manufacturing light or insulating materials, such as polystyrene.

Combustible gases

• Acetylene (also called ethene) (C2H2) is widely used as a fuel in certain welding procedures. • Natural gas and propane (C3H8) are used for heating homes and cooking. • They fuel internal combustion engines, such as gasoline-powered cars.

Noble gases

• They are enclosed in the tubes of certain illuminated signs that emit white or coloured light when they are turned on. For example, neon (Ne) emits an orangey-red colour. • They are found in incandescent bulbs and between the panes of energyefficient windows. • They fill probe balloons used to gather meteorological data.

Carbon dioxide (CO2)

• It is used to put out fires in computer rooms without damaging the computers. • It is used to preserve foods in controlled-atmosphere warehouses. • Tanks of compressed carbon dioxide (CO2) are used for paintball.

42

UNIT 1 Gases

APPLICATIONS Compressed gases Since gases are compressible, a mechanical compressor can be used to reduce the volume of a gas or to increase its pressure in a given container. The energy contained in air that is compressed in this way can be used for multiple applications in a wide range of activities, including housekeeping, cosmetics, dentistry, construction and deep-sea diving. Compressed gases are used to make the various aerosol products present in our daily lives. These products are prepared by mixing a propellant gas with cream or paint, for example, in the form of liquid or solid particles in a pressurized can. The most commonly used propellant gases are propane (C3H8), butane (C4H10) and nitrogen (N2). When the valve on the can is pressed, the gas escapes, reducing the pressure inside the can. The volume of the gas therefore increases, and the mixture of gas and other substances flows out of the bottle in aerosol form, that is, in tiny droplets suspended in the air. This is the case for aerosol paint, hairspray and shaving cream. The phenomenon in which the volume of a gas increases due to a drop in pressure is called expansion. Expansion is the opposite of compression.

Figure 12 A dentist’s drill is equipped with a rotating head that, among

Figure 13 A nail gun is a pneumatic tool used to drive nails like a stapler.

A dentist’s drill turns at up to 200 000 revolutions per minute using the energy stored in compressed gases (see Figure 12). This energy is called pneumatic energy. It powers various pneumatic tools such as staple guns, hammers, screwdrivers and nail guns (see Figure 13). These tools, often used in construction, are much more efficient. They are powerful and greatly reduce the amount of effort required by construction workers. For deep-sea diving, the gas contained in tanks is usually air. However, for diving in very deep waters, enriched mixtures of oxygen (O2), hydrogen (H2) and nitrogen (N2) are used. The pressure of the mixture in the tank must be carefully controlled so that it does not affect the lungs. Moreover, divers must come back up to the surface slowly, decompressing level by level, to prevent air bubbles from forming in their blood vessels, causing the bends.

other things, polishes tooth enamel.

CHAPTER 1 Chemical Properties of Gases

43

1.2 Chemical reactivity of gases The chemical reactivity of a gas is its tendency to undergo a chemical change due to various factors such as heat, light or contact with other substances. Certain physical properties, such as colour, mass, volume and length, do not allow the substances that compose matter to be identified with certainty. These are non-characteristic properties. The use of certain indicators, for example, litmus paper, allows substances to be classified into various categories according to the characteristic chemical properties they share. In this respect, chemical reactivity is often specific to each gas or to a particular group of gases. This makes it possible to classify gases in a group when they share similar chemical reactivity, or to distinguish one gas from others. For example, several gases, such as butane (C4H10), are flammable and react to heat (see Figure 14). Figure 14 Butane (C4H10) in a lighter is ignited on contact with a spark.

The classification adopted by the periodic table of elements includes 11 elements that are gaseous under ambient conditions (see Figure 15). Six of these elements, in monatomic form (He, Ne, Ar, Kr, Xe and Rn), are classified in the group of noble gases, also called inert gases or rare gases, due to their low chemical reactivity. The five other elements are gaseous under ambient conditions in the form of diatomic molecules (H2, N2, O2, F2 and Cl2).

Figure 15 Eleven elements classified in the periodic table are gaseous under ambient conditions.

Knowing the chemical reactivity of gases allows us to, among other things, determine in which technological applications they can be used (see Table 1 on page 42) and how they can be used safely. The gases used in various technological applications can be combustible, oxidizing, toxic, flammable or corrosive. Before using a gas, it is important to read the label on the container to find out about its properties.

Figure 16 Carbon monoxide (CO) detectors warn residents in the event of a leak in the heating system.

44

UNIT 1 Gases

All gases, whether they are toxic or not, can have an asphyxiating or suffocating effect because they replace the oxygen (O2) in the air when they are inhaled. This is the case of carbon monoxide (CO), a toxic gas that can be found in homes with defective gas or wood heating systems. To prevent carbon monoxide poisoning, a detector should be installed to alert residents before the toxicity threshold is reached (see Figure 16).

The use of compressed gases in pressurized containers is subject to strict regulations in which warning labels must be used to indicate the danger associated with the use of these substances. Moreover, at home and at school, many household products contain hazardous substances for which symbols must be used by the manufacturers to indicate the danger associated with their use (see Figure 17). For example, pressurized containers must be stored away from heat and flames and must not be perforated, since they sometimes contain extremely flammable gaseous substances. Since many compressed gases are corrosive, skin and respiratory tracts must be protected when using them. The same applies to products with the poison symbol, which obviously must not be inhaled or ingested.

1.2.1

APPENDIX 1 Danger symbols, p. 372.

The causes of the chemical reactivity of gases

The chemical reactivity of a gas largely depends on the electron configuration of its atoms and, more specifically, the strength of attraction between the nucleus and the valence electrons in the outermost shell. The group of noble gases, for example, contains elements whose atoms all have an outermost shell filled to capacity with valence electrons that are strongly attracted by their nucleus (see figures 18 and 19). This is what explains their low chemical reactivity. In fact, unless they are subjected to exceptional conditions in a lab, noble gases do not form chemical compounds, and their reactivity is almost nonexistent.

Figure 18 An atom of helium (He) has only one electron shell. This is why two valence electrons are enough to fill it completely.

Figure 17 The symbols on this insecticide indicate that the pressurized container must be handled and stored with caution.

Figure 19 An atom of neon (Ne) has eight valence electrons that completely fill its outermost shell.

In contrast, fluorine (F2) and chlorine (Cl2), which are elements that belong to the halogen group, have very high chemical reactivity, primarily because they are missing only one electron to complete the outermost shell of their atoms (see Figure 20).

Figure 20 An atom of chlorine (Cl) has seven valence electrons, which explains its high chemical reactivity. When this atom accepts an additional electron during a chemical reaction, its outermost shell is complete.

CHAPTER 1 Chemical Properties of Gases

45

See Energy balance, p. 156.

The chemical reactivity of gases is also governed by the strength of the bonds between the atoms of the gas molecules. In fact, for a chemical reaction to occur, the bonds within the molecules of the reactants must break and then form new bonds, thereby forming new products. The energy balance of this process corresponds to the difference between the energy absorbed to break the bonds and the energy released when the new bonds are formed. Molecular nitrogen (N2), for example, is a very nonreactive gas because a large quantity of energy is required to break the triple covalent bond that unites the two atoms of nitrogen (N) in the molecule. However, molecular fluorine (F2) is very reactive, since the bond between the two atoms of fluorine (F) is relatively weak and little energy is required to break it.

1.2.2

Combustible gases

Combustion is a very common oxidation reaction that releases energy. Usually, the energy is released in the form of heat that can be used for various purposes including heating homes and cooking. The three components of the fire triangle must be present for combustion to occur: the fuel, the oxidizer and the source of ignition (see Figure 21). The fuel is the substance that burns during combustion. The oxidizer is the substance that supports the combustion of the fuel, while the source of ignition is the temperature that must be attained for combustion to be triggered. The fuel can be solid, liquid or gas. Combustible gases include hydrocarbons and hydrogen (H2). Figure 21 The three components of the fire triangle must be present for combustion to occur.

Hydrocarbons are organic compounds which come from living organisms. They are formed of molecules mainly composed of carbon (C) and hydrogen (H) atoms. Gaseous hydrocarbons under ambient conditions are methane (CH4), ethane (C2H6), propane (C3H8) and butane (C4H10). Like all hydrocarbons, when they burn they release water (H2O) and carbon dioxide (CO2), a greenhouse gas. Methane, which is the main component of natural gas, burns with a flame whose colour varies depending on whether or not the combustion is complete (see Figure 22). When the mixture of fuel and oxidizer is optimal, the flame takes on a bluish colour, since it only contains carbon dioxide and water, the products of the combustion reaction. However, when the reactants are not completely burned, the residue colours the flame yellow.

Figure 22 The combustion of methane (CH4) using a Bunsen burner produces a flame of many colours, depending on the quality of the combustion.

46

UNIT 1 Gases

The equation for the combustion of methane (CH4) is as follows: CH4 (g)  2 O2 (g) → CO2 (g)  2 H2O (g) Traces of hydrogen (H2), a very reactive gas, are present very high up in the atmosphere, from where it escapes fairly easily due to its low density. Consequently, hydrogen used as a combustible gas must be produced industrially through the electrolysis of water (H2O) molecules, or through extraction using methane. The flame test can be used to detect the presence of hydrogen. However, its high reactivity, when put in contact with a lit wood splint, means that this test must be conducted with very small quantities of gas (see Figure 23). Hydrogen combustion produces water vapour according to the following equation:

Figure 23 A lit wooden splint introduced into a test tube filled with hydrogen (H2) causes a small explosion.

2 H2 (g)  O2 (g) → 2 H2O (g) It is due to this reaction, producing only water vapour, that we call hydrogen a “clean fuel.” Much hope has been pinned on this clean fuel, which could become the fuel of the future and replace oil. However, producing hydrogen, which involves extracting it from methane, consumes a lot of energy and, ultimately, produces a considerable amount of greenhouse gases. Water electrolysis is the inverse reaction of hydrogen combustion (see Figure 24). Water electrolysis occurs according to the following equation: 2 H2O (l) → 2 H2 (g)  O2 (g) By combining the equation above and the one for hydrogen combustion, we can see that the combustion of hydrogen produced by electrolysis is a clean cycle that releases only water in its gas phase. While hydrogen production through water electrolysis also consumes a lot of energy, it is possible to reduce its impact on the environment by using renewable sources of energy that do not produce greenhouse gases, such as hydroelectricity or wind energy. Researchers are seeking ways to optimize electrolysis production methods that draw on renewable and less polluting energy sources.

Figure 24 Water electrolysis using an electric current makes it possible to separate water into its two constituent gases: hydrogen (H2) and oxygen (O2).

CHAPTER 1 Chemical Properties of Gases

47

1.2.3

Oxidizing gases

Combustion is a reaction that requires an oxidizer to react with a fuel. The most common oxidizer on Earth is oxygen (O2), which is present in the atmosphere at a proportion of approximately 21% (see Figure 2 on page 38). Oxygen is therefore present in most combustion reactions, from rapid ones, such as the explosion of gasoline fumes, to slow ones, such as cellular respiration. a) An ozone molecule (O3)

b) A diatomic oxygen molecule (O2) Oxygen (O)

Figure 25 An ozone molecule (O3) contains an oxygen (O) atom and a diatomic oxygen (O2) atom.

Ozone (O3) is a very toxic oxidizer. Its molecule is composed of three atoms of oxygen (see Figure 25). It has some of the distinct properties of diatomic oxygen, which can make it useful but also hazardous depending on its altitude in the atmosphere or how it is used. Ozone is useful when it is in the stratosphere and forms the layer that protects living organisms from the Sun’s ultraviolet rays. Ozone is also useful in treatment plants to purify water. However, ozone is toxic when it is closer to the ground, in the first few metres of the troposphere where it is a component of smog (see Figure 26). Some people experience respiratory problems when they come into contact with it.

Figure 26 Some cities, such as Linfen in China, experience smog so often that pedestrians have to use masks to protect themselves.

There are other very reactive oxidizing gases, but they are used much less often than oxygen. Of these oxidizers, two elements from the halogen group are extremely reactive, namely, fluorine (F2) and chlorine (Cl2). Fluorine gas is very toxic and extremely corrosive. It must be handled with great caution as it can cause serious skin burns. Inhaling it is very corrosive to the mucous membranes in the respiratory system. It is used in many compounds in the form of a fluoride ion (F−), for example in sodium fluoride (NaF) added to toothpaste. Chlorine gas is used in many compounds in the form of a chloride ion (Cl−) for water purification, the production of plastic (including PVC) and the manufacture of disinfectants.

Figure 27 Halogen bulbs provide brighter light and last longer than conventional incandescent bulbs.

48

UNIT 1 Gases

Halogens are not used exclusively as oxidizers. For instance, halogen bulbs contain one of the elements from the halogen group in a gas state. When the electric current passes through the tungsten (W) filament, a chemical reaction occurs between the filament and the gas present in the bulb (see Figure 27). This reaction extends the lifespan of the filament and produces very bright light.

Anesthesia Anesthesia helps to reduce or eliminate a patient’s pain during a delicate medical procedure like surgery. The discovery of anesthetic agents dates back thousands of years. Almost 5000 years ago, the Chinese discovered that inhaling substances derived from plants could reduce pain. It was not until the mid-19th century, however, that more effective anesthesia methods involving the inhalation of gas were invented. In 1800, English chemist Humphry Davy studied the chemical and medical properties of nitrous oxide (N2O), also known as laughing gas. He recommended its use during surgery after testing the product on himself. Forty-four years later, American den tist Horace Wells reco gnized the analgesic properties of nitrous oxide and performed many tooth extractions on patients who had inhaled this gas (see Figure 28). Ether (C4H10 O), a very volatile and flammable liquid with a distinct odour, was also one of the first anesthetics to be disco- Figure 28 A patient inhales nitrous vered. In 1842, Ameri- oxide (N2O) in 1874. can surgeon Crawford Long performed the first surgery under ether anesthetic. A handkerchief soaked in ether was placed over the mouth and nose of the patient who then brea thed in the vapours (see Figure 29). However, scientists discovered that ether was difficult to administer be cause its vapours caused vomiting and choking. Figure 29 The first public demonstration of anesthesia using ether was conducted by dentist William Morton in 1846 at Boston’s Massachusetts General Hospital.

To avoid these problems, physicians turned to another volatile anesthetic, chloroform (CHCl3). Scottish physician James Young Simpson was the first to use it in 1847 and administered it in the same way as ether. One of his patients was Queen Victoria, who delivered her eighth child in 1853 under chloroform anesthetic. Non-flammable, chloroform acted faster than ether and caused less agitation in patients. In 1862, English physician Joseph Thomas Clover invented a device to administer chloroform to his patients. However, the bag, which contained a large quantity of gas, could cause overdoses (see Figure 30). Methods for administering anesthetics have chan- Figure 30 Joseph Thomas Clover ged over time. The use of shows how chloroform is admincompresses impregnated istered using his device, Clover’s drop by drop was followed Chloroform Apparatus. by the use of immense tanks filled with liquid anesthetic. Valves were used to administer the vapours to the patients through a tube. Subsequently, less cumbersome, more mobile machines were invented. Today, inhalers are mechanical and are programmed to administer the correct quantity of anesthetic gas (see Figure 31). Nitrous oxide is still used, but it is now mixed with oxygen (O2) to prevent asphyxiation. Since nitrous oxide is a weak anesthetic, it is combined with other anesthetizing gases, particularly powerful halogen compounds with a fluorine (F) base. Figure 31 An anesthetist administers anesthetizing gases with a modern inhaler during surgery.

CHAPTER 1 Chemical Properties of Gases

49

CHAPTER

1

Chemical Properties of Gases

1.1 Daily use of gases • The daily use of gases is very common in many areas of human activity. Gases play an important role in the environment and in several technological applications. • Gases are present in many natural phenomena including photosynthesis, respiration of living organisms, the greenhouse effect, aurora borealis and australis, and meteorological phenomena. • Gases are used in many technological applications including welding, and in the areas of food, agriculture and health, among others.

1.2 Chemical reactivity of gases • The chemical reactivity of a gas is its tendency to undergo a chemical change under the effect of various factors such as heat, light and contact with other substances. • Chemical reactivity makes it possible to distinguish one gas from another or to classify gases in a group based on similar chemical reactivity. • Knowing the chemical reactivity of gases makes it possible to handle them safely. Danger symbols displayed on the containers of dangerous gases indicate how to handle them safely. • The chemical reactivity of a gas depends on: – the electron configuration of its atoms – the force of attraction between the nucleus and valence electrons in its outermost shell – the strength of the bonds between the atoms that make up the gas molecules • Combustible gases, such as hydrocarbons and hydrogen (H2), are used in the combustion reaction to produce energy. • Oxidizing gases, such as oxygen (O2), are essential to the combustion reaction, in which they react with the fuel.

50

UNIT 1 Gases

CHAPTER 1

Chemical Properties of Gases

1. Identify various uses of the halogens.

7. Give two examples of gas in the living world.

2. Name two gases that are used in the food industry to create favourable conditions for preserving food.

8. Which gas is most present in the atmosphere?

3. How is oxygen (O2) used in medicine and industry? 4. What precautions must be taken when handling a household product that displays the following warning symbols on its label? a) b) c) 5. Match each atmospheric gas with its corresponding description. Gas a) Carbon dioxide (CO2) b) Nitrogen (N2)

c) Methane (CH4)

d) Oxygen (O2)

Description 1) A constituent of the atmosphere essential for respiration. 2) A very powerful greenhouse gas that is the main component of biogas. 3) A greenhouse gas used as a source of carbon (C) by plants in photosynthesis. 4) The main component of the atmosphere.

9. Compare the effects of ozone (O3) close to the ground and in the upper atmosphere. 10. a) Explain why hydrogen (H2) is referred to as a “clean fuel.” b) Explain the production methods that must be used for hydrogen to truly merit the label “clean fuel.” 11. Give examples of natural phenomena in which gases play an essential role. 12. Name the most common oxidizing gas on Earth. 13. To keep potato chips intact during transport, they are packaged in bags filled with gas. The gas used is not air, but rather nitrogen (N2). Why is nitrogen rather than air used to inflate bags of chips? 14. Ozone (O3) is an oxidizing gas with numerous properties. Which properties of ozone make it useful? Which ones make it harmful? 15. When a plane flies in the upper atmosphere, it sometimes leaves a white trail behind. Why does this phenomenon occur at this altitude but not at takeoff?

6. a) What are the products resulting from the complete combustion of hydrocarbons? b) What is the difference between a yellow flame and a blue flame during the combustion of methane? c) What is the impact of the combustion of hydrocarbons on the environment?

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51

16. How do the electron configurations of noble gases and halogens describe their particular chemical reactivity? 17. What gas is responsible for browning the surface of an apple when it is sliced? 18. How can the photosynthesis reaction help to reduce greenhouse gases? 19. A space shuttle is equipped with two tanks that give it powerful propulsion during takeoff: one contains oxygen (O2) and the other, hydrogen (H2). a) What is the chemical role of each of these gases? b) What is the chemical equation that describes the chemical reaction between the gases in these tanks?

20. At the beginning of the last century, dirigible balloons were built to transport many passengers and a lot of material, much like a cruise ship. They were equipped with tanks containing gas that was lighter than air, which allowed them to fly. Sadly, on May 6, 1937, the Hindenburg airship exploded and crashed, killing 36 people. Why is it said that the designers made a mistake by using hydrogen (H2) instead of helium (He) to inflate the balloon?

Tank containing oxygen (O2)

21. Why is it beneficial to vacuum pack food? Intertank

Tank containing hydrogen (H2)

52

UNIT 1 Gases

22. The smoke that comes out of the exhaust pipe of a car is often visible in the winter but practically invisible in the summer. Why?

Physical Properties of Gases G ases surround us and are essential to life, but their behaviour is often a mystery to us. The fluidity of this air bubble, for example, makes us forget that once it is released by a diver in the ocean’s depths, it will likely double or triple in volume before reaching the surface.

In this chapter, you will learn how different phases of matter behave at the particle level, with a focus on the gaseous phase. The kinetic theory of matter will help you to understand the behaviour of gases and to apply the

Review Organization of matter. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Representations of atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 The Mole concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Phase changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Stoichiometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Compressible and incompressible fluids . . . . . . . . . . . . . . . . . . . . . 33 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

laws derived from this theory in order to accurately predict how gases will react at different pressures, temperatures and volumes.

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

Kinetic theory of gases ...................54 Behaviour of gases .......................63 Pressure of gases ........................69 Simple gas laws .........................75 Ideal gas law ............................100 General gas law .........................105 Stoichiometry of gases ..................108 Dalton's law .............................111 CHAPTER 2 Physical Properties of Gases

53

2.1 Kinetic theory of gases The kinetic theory of gases seeks to explain the similarities observed in the behaviour of gases based on the movement of the particles that compose them. The kinetic theory of gases was developed on the basis of observations made by many scientists during the 18th century. This theory provides a better understanding of the physical properties of gases using a hypothetical model of gas known as the ideal gas. This model attempts to explain, at the particle level, the experimental results observed for most gases. Kinetic theory allows us to understand how gases behave and to apply these gas laws to predict their behaviour under various conditions. To study the kinetic theory of gases, we must first understand the particle behaviour of matter in various phases, as well as the relationship between the kinetic energy of gas particles and temperature.

2.1.1

Particle behaviour in various phases of matter

Matter on Earth is generally composed of substances in a solid, liquid or gas phase. Depending on the conditions in which it is found, a substance can move from one phase to another without any change in its nature. These physical transformations, in which only the appearance of the substance changes, are called phase changes (see Figure 1). Gas phase

n

ion

De p

os

itio

n

tio ma Su bli

n

zat

tio

sa

en

nd Co ri po Va

Fusion (also refered to as “melting”)

Solid phase

Solidification

Liquid phase

Figure 1 Phase changes of matter

Figure 2 The plasma ball provides a close look at the fourth phase of matter: plasma.

54

UNIT 1 Gases

Plasma is the fourth phase of matter that exists under particular conditions of temperature and pressure. Plasma can be closely observed using plasma screens or a decorative plasma ball (see Figure 2). In the latter case, the ball is filled with a noble gas, and the high electric voltage that is applied partially ionizes the particles of the gas, which emit light in a variety of colours.

The particle model of matter can explain some of the properties of the liquid, solid and gas phases of matter. Particle model of matter According to this model: • All matter is composed of particles (ions, atoms or molecules) that are infinitely small with more or less space between them depending on their phase. • Particles of matter attract or repel each other, and the force that attracts or repels them from each other varies depending on the distance that separates them. • Particles of matter are always moving. The movements of a particle vary depending on whether it is in the solid, liquid or gas phase. Particles of matter exhibit three types of movement: vibrational motion, rotational motion and translational motion (see Figure 3). Vibration

Rotation

Translation

x z y a) A water molecule can vibrate according to three modes: symmetrical stretching, bending and asymmetrical stretching.

b) A water molecule can turn on its own axis, along axes x, y and z.

c) By translation, a water molecule moves in a straight line from one collision to the next.

Figure 3 Vibrational, rotational and translational motion of a water molecule

Vibrational motion is present in the three phases of matter. The vibrational motion of a particle is the constant oscillation of its atoms around a fixed point. This oscillation can occur due to the compression or stretching of the bonds between the atoms or by a change in the angles between them. For example, a water molecule can adopt three modes of vibration: bending, symmetrical stretching and asymmetrical stretching. The three modes of vibration occur around the molecule’s centre of gravity, which constitutes its fixed point. Rotational motion is present in liquids and gases. In rotational motion, a particle can turn on its own axis in three ways, along axes x, y and z. Translational motion occurs mostly in gases, and, to a lesser degree, in liquids. In translational motion, a particle moves in a straight line. This motion produces the longest movements of particles.

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55

Particle behaviour in solids

Sodium (Na) Chloride (Cl)

Figure 4 The cubic shape of table salt crystals (NaCl)

The only type of motion exhibited by solids is vibration. This is one reason why solids have a definite form. Indeed, particles that compose solids cannot move in relation to each other. They are maintained in a particular structure and orientation due to forces that exist between them. The position of particles results from an equilibrium between these forces. Particles cannot slide against each other and change the shape of the solid, since the forces that bind them are very difficult to modify. This is why solids, such as sodium chloride (NaCl), are considered virtually incompressible, that is, they cannot be compressed. The cubic shape of table salt crystals (NaCl) can be explained by the alternating of Na+ and Cl− ions, which are bound by forces that are difficult to modify, rendering the crystals virtually incompressible (see figures 4 and 5). The particles of solids are constantly vibrating, but this vibrational motion is extremely weak compared to the size of the object. For example, a grain of salt vibrates, but this motion is so small compared to its size that it is imperceptible. Consequently, there is very little disorder between the molecules of solids (see Figure 6).

Figure 5 Table salt crystals

Vibration

Figure 6 The particles of solids vibrate weakly Rotation

Particle behaviour in liquids In liquids, particles move more easily than in solids, but their movement is not completely free. In addition to vibrational motion, particles in liquids are displaced by rotational motion and weak translational motion.

Translation

Weak bonds Oxygen (O) Hydrogen (H)

Figure 7 Liquid water molecules are held together by weak bonds that exist between one atom of hydrogen (H) and one atom of oxygen (O).

56

UNIT 1 Gases

Particles of liquids are able to move because the strong bonds that existed between them in the solid phase have been broken. What remains are relatively weak forces of attraction, which provide a degree of cohesion between the particles of liquids while allowing them to slide against each other (see Figure 7). Since the particles of liquids are in rotational and vibrational motion, they can change position with each other. They are also displaced through translational motion, but this movement occurs to a lesser degree than rotation and vibration. In part, it is this combination of rotational, vibrational and translational motion that allows liquids to be poured and to change shape while maintaining their volume.

In a liquid, the distance between particles is usually greater than in a solid (with the notable exception of water). Nevertheless, the forces of attraction between particles make liquids, like solids, virtually incompressible. However, the freedom of movement of the particles of liquids is greater than that of the particles of solids. Consequently, the degree of disorder in the liquid phase is much greater than in the solid phase.

Particle behaviour in gases

Ice that floats When liquid water transforms into ice, the molecules move away from each other. They form a three-dimensional arrangement comprised of hexagons.

Particles of most gases exhibit three types of motion: vibrational, rotational and translational (see Figure 8). Translational motion predominates in gases since there are no interactions between particles. This contrasts sharply with solids, where there is no translational motion, and liquids, where this type of motion is very limited. In translational motion, gas particles follow random trajectories that remain linear until they collide with other particles or other objects. Figure 9 The alveolar structure of frozen water molecules Translation Rotation Vibration

This particular organization of water molecules in ice means that there are fewer molecules in a litre of ice than in a litre of liquid water. Consequently, ice has a lower density than water, which makes it lighter than water. This is why ice floats on water. In most other substances, such as benzene (C6H6), matter in the solid phase sinks to the bottom of the matter in the liquid phase. This unique property of water allows ice to form on the surface of lakes in the winter.

Figure 8 In the steam from a kettle, the water molecules in the gas phase are primarily displaced by translation although they also turn on their own axis by vibration.

Gas particles move much faster than liquid particles. Contrary to liquids that usually flow downwards, gases flow in all directions, including upward, despite gravitational force, until all of the available space is occupied. Gases expand to completely fill their container. Since gas particles move freely, the gas phase is extremely disorderly compared to the liquid and solid phases.

Figure 10 Cubes of benzene (C6H6) sink to the bottom of liquid benzene (beaker on the left) while ice cubes float on water (beaker on the right).

CHAPTER 2 Physical Properties of Gases

57

In gases, the distance between particles is much greater than in solids or liquids. This is why gases are compressible. When gases are compressed, the particles move closer together, but the distance between them remains great. This distance is reduced considerably when the gas transforms into a liquid through condensation. The following table summarizes macroscopic properties and particle behaviour in the different phases of matter: Table 1 Macroscopic properties and particle behaviour in the various phases of matter Properties and behaviour Macroscopic properties

Particle behaviour

2.1.2

Phases of matter Solid state

Liquid state

Gas state

Volume

Determinate

Determinate

Indeterminate

Shape

Determinate

Indeterminate

Indeterminate

Compressibility

Negligible

Negligible

High

Relative position of particles

Very close together

Close together

Very far apart

Types of motion

Vibrational

Vibrational Rotational Weak translation

Vibrational Rotational Strong translational

Amount of motion

Very small

Small

Very large

Strength of interparticle bonds

Strong

Weak

Nil

Kinetic energy of gas particles and temperature

The temperature of a substance corresponds to the amount of movement in its particles. In a solid, the only possible movement is vibration. The particles in a warm solid vibrate faster than those in a cold solid. Similarly, the particles in a warm gas move more than those in a cold gas. Since gases exhibit three types of motion, increasing their temperature speeds up the rate of the vibration, rotation and translation of their particles. Since translation is a linear displacement, increasing the rate of translation also increases particle velocity. The energy of an object in motion, such as a moving particle of gas, is called kinetic energy (Ek). The following formula is used to determine the kinetic energy of a gas particle in motion. Kinetic energy

1 Ek  mv 2 2

where Ek  Kinetic energy, expressed in joules (J) m  Mass of the particle, expressed in kilograms (kg) v  Velocity of the object, expressed in metres per second (m/s)

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UNIT 1 Gases

The formula shows that the kinetic energy of a gas particle is determined by its mass (m) and its velocity (v). However, the second variable has a greater effect on kinetic energy since it increases in proportion to the square of the velocity. Consequently, the greater the mass of the gas particle and the faster it moves, the greater its quantity of kinetic energy.

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

In a sample of a given gas, the particles do not all have the same kinetic energy. Therefore, they do not all move at the same velocity. In fact, when two particles collide, kinetic energy is exchanged, which generally slows down one particle while accelerating the other. Consequently, some particles move quickly, while others move more slowly. This is why it is best to consider the mean velocity of the particles and, thus, their mean kinetic energy when analyzing a gas sample at a given temperature. In the 19th century, James Clerk Maxwell analyzed the behaviour of gas particles to determine the velocity of each of the particles in a sample of gas at a given temperature. The results of his work allowed him to establish the velocity distribution curve of gas particles, also known as Maxwell’s distribution curve (see Figure 11).

JAMES CLERK MAXWELL Scottish physicist (1831–1879) In the 19th century, James Clerk Maxwell tried to explain the behaviour of molecules and their collisions within gaseous systems. Based on the research of his predecessors, he developed the kinetic theory of gases in 1866. He postulated that each molecule is discernable before or after a collision and possesses its own kinetic energy. Maxwell’s law is used to determine the distribution of velocity and kinetic energy of molecules in a gas at a given temperature. This theory, while incomplete, is still used today as a reference in chemistry.

Number of particles of a gas sample in relation to their velocity at a given temperature vmp

Number of particles

vx—

Particle velocity

Figure 11 Maxwell’s velocity distribution curve for a gas sample at a given temperature

The shape of the curve shows that at a given temperature, many of the particles move at the same velocity. This velocity is the most probable velocity (vmp). The curve also shows that the mean velocity (vx) of the gas particles is just above the most probable velocity. Therefore, few particles move very slowly or very quickly; most move at a velocity close to the mean velocity.

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59

Maxwell pursued his work by observing the effect of increasing temperature on the velocity distribution of particles (see Figure 12). He noted that the curves on the graph had the same general appearance but that their peaks moved to the right as the temperature increased. This indicated that the most probable velocity and the mean velocity increase with temperature. In other words, the higher the temperature of a gas, the faster its particles move. Number of particles of nitrogen gas in relation to their velocity at three different temperatures vmp

Number of particles

Low temperature

vmp Intermediate temperature vmp High temperature

500

1000

1500

Particle velocity (m/s)

Figure 12 The distribution curve of the velocities of particles of nitrogen (N2) gas shows that at higher temperatures, the most probable velocity (vmp ) of the particles is greater.

Since the kinetic energy of molecules depends primarily on their velocity, an increase in temperature has the same effect on the distribution curve of kinetic energies as on the velocity distribution curve. In fact, the curves rise until they reach a maximum, then they drop to zero when the kinetic energy increases. In addition, as the temperature increases, the peaks of the curves move to the right, which means that the mean kinetic energy of the particles is greater. Consequently, the following generalization can be made for all gases: the mean kinetic energy of particles increases as the temperature of the gas increases (see Figure 13).

Number of particles

Number of particles of a sample of gas in relation to their kinetic energy at an intermediate or high temperature

Intermediate temperature High temperature

5

10 15 20 Kinetic energy (10−21 J per molecule)

Figure 13 The distribution of kinetic energy (Ek ) of the particles of a gas at

two temperatures. These curves show that an increase in the temperature of a gas increases its mean kinetic energy.

60

UNIT 1 Gases

2.1.3

Hypotheses of the kinetic theory of gases

Beginning in the 18th century, many scientists conducted research in an attempt to explain the behaviour of gases and to better understand their properties. The combined results of this research led to the development of a theory that explained some of the facts observed. Since the theory is based on the movement of gas particles and on the kinetic energy that causes this movement, it is called the kinetic theory of gases. The kinetic theory of gases describes a hypothetical gas, called an ideal gas, enclosed in a rigid container. In an ideal gas, the particles do not exert any attraction on each other and occupy virtually no space, which means that the gas is essentially empty space. The kinetic theory of gases also explains the behaviour of most real gases, except when they are subjected to extreme conditions.

See Ideal gas law, p. 100.

The kinetic theory of gases is based on the following hypotheses: Hypothesis 1 The particles of a gas are infinitely small and the size of a particle is negligible compared to the volume of the container that holds the gas. The particles are considered points that have mass, but whose volume is negligible. This means that the particles of a gas are extremely far from each other and that most of the container is empty. Hypothesis 2 The particles of a gas are in constant motion and move in a straight line in all directions. The translational motion of the particles of a gas never stops, so the particles of a gas never pile up the way they do in solids and liquids. The particles collide with each other randomly and hit the sides of the container. These collisions are perfectly elastic, which means that they occur without any loss of energy. This is why the mean kinetic energy of all particles remains constant when conditions are stable. Hypothesis 3 The particles of a gas do not exert any force of attraction or repulsion on each other. Aside from the elastic collisions they undergo, the particles of a gas are completely independent of each other. Hypothesis 4 The mean kinetic energy of the particles of a gas is directly proportional to the absolute temperature. The higher the temperature of a gas, the more kinetic energy its particles possess (see Figure 12 on page 60 ), which means heightened agitation or a faster movement of particles. At a given temperature, the mean kinetic energy of particles is the same for all gases, regardless of their nature.

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61

Kinetic theory of gases

1. Explain each of the following phenomena using the particle model of matter: a) The smell of skunk spray seeps into a house even when the windows are closed. b) The smell of toast fills every room in the house. c) Fishing on a lake in the winter requires a little more effort since the ice must be broken. d) A hydroplane can easily fly through a cloud, and can also land on the surface of a lake. e) Milk spreads throughout a cup of coffee. f) Water vapour can mix with air to make it humid. 2. How is the translational motion of particles different from the other two forms of motion? 3. Categorize the phases of matter according to the degree of disorder of the particles that compose the matter. Start with the phase with the least disorder. 4. What allows solids to have their own shape and to occupy a defined volume? 5. Why can we state that solids and liquids are virtually incompressible, while gases are compressible? 6. How does the disorder of a gas compare to that of a liquid or solid? Explain your answer. 7. Observe the following figures and for each describe the vibrational motion of the water molecule. a) b) c)

8. Why can we state that the mean kinetic energy of particles increases as the temperature rises?

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UNIT 1 Gases

9. Examine the following graph. Which curve shows the highest gas temperature? Explain your answer.

Number of particles

SECTION 2.1

3 2 1

Kinetic energy

10. According to the kinetic theory of gases, an ideal gas has all of the following characteristics but one. Which one? a) Particles whose size is nil or negligible. b) Particles that mutually attract each other. c) Particles that move randomly. d) Particles that are very far away from each other. 11. According to the kinetic theory of gases, explain each of the following observations: a) Gases are more compressible than liquids. b) The density of gases is less than that of solids. c) Gases do not have a fixed volume. d) A certain quantity of moles of water occupies much more space in gas form than in liquid form. e) At a given temperature, particles of nitrogen (N2) move, on average, more slowly than those of hydrogen (H2).

2.2 Behaviour of gases The behaviour of gases refers to the way in which gases react when some of their physical properties undergo change. This behaviour can be described qualitatively, based on observations, or quantitatively, based on laws. The kinetic theory can be applied to real gases under certain conditions. As long as the particles of gases are far enough away from each other that there is no interaction between them, real gas can be considered an ideal gas. Under such conditions of temperature and pressure, the kinetic theory can explain the behaviour and macroscopic properties of real gases, such as volume, shape and compressibility.

*

2.2.1

* Macroscopic Visible to the naked eye.

Compressibility

According to Hypothesis 1 of the kinetic theory of gases, the mean distance between particles is significantly greater than the size of the particles. This is why the particles of a gas move closer together and occupy a smaller volume when the gas is contained in a closed space and is subjected to the effect of a force. The capacity to reduce volume under the effect of a force is a property of gas called compressibility. For example, when you use a bicycle pump, the air inside it is compressed into a smaller volume (see Figure 14). The particles of air are forced closer together while they are inside the pump. However, even if the particles of air are compressed into a smaller space, they remain sufficiently distant from one another for the air to continue to behave like a gas. The compressibility of gases makes it possible to store large quantities of them in small spaces. For example, the tanks of compressed air used by divers usually contain 6 to 18 litres of compressed air at approximately 200 times normal atmospheric pressure (see Figure 15). By taking this quantity of air in such a small space under water, divers can spend close to one hour under approximately 10 metres of water.

2.2.2

Figure 14 Gases are compressible because there is a lot of space between particles. When a bicycle pump is pumped, the air is compressed before being released.

Expansion

According to Hypothesis 2 of the kinetic theory of gases, particles move continuously in a straight line in all directions. Since gases do not have shape or volume, they dilate indefinitely by filling any accessible space. The more a gas dilates, the greater the space between its particles. The phenomenon of gas dilatation is called expansion and varies according to atmospheric pressure.

Figure 15 The compressibility of gases allows this diver to dive with a tank of 18 litres of compressed air containing the equivalent of approximately 3600 litres of air at normal atmospheric pressure.

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63

For example, at ground level, the gas in a probe balloon occupies a relatively limited volume and the balloon is partially inflated. However, the gas dilates as the balloon gains altitude and the atmospheric pressure decreases. The expanding gas stretches the surface of the balloon and the gas occupies a greater volume (see Figure 16).

Figure 16 A probe balloon is only partially inflated at a low altitude. At a high altitude, the lower pressure allows the balloon to inflate.

2.2.3

Diffusion and effusion

Gas particles frequently collide with each other as well as with the walls of their container. This is why their trajectories resemble a series of randomly intersecting broken lines (see Figure 17).

Figure 17 The dotted line illustrates the possible trajectory of a particle of gas inside a volleyball. Each time a particle collides with other gas particles or with the wall of the ball, its trajectory changes direction.

This random dispersion of aimlessly colliding particles is called diffusion. Gas diffusion is a gradual process in which a gas mixes with other gases in a container through the movement of its particles. Gas diffusion allows the uniform distribution of the particles of all gases in a container over a period of time (see Figure 18).

Figure 18 After a period of time, diffusion distributes bromine (Br2) gas particles uniformly in the air of the container.

64

UNIT 1 Gases

Even though gas particles move very quickly, the diffusion process is relatively slow. This is because there are a large number of collisions among the particles, which cause a great many changes in trajectories. Moreover, not all gases diffuse at the same rate. The rate depends on the velocity of the particles. Consequently, at the same temperature, particles with a small mass diffuse more quickly than those with a greater mass because they move more quickly (see Figure 19). Number of particles of three gases based on their velocity at 25°C T  25 °C

Number of particles

Heavy gas (Cl2)

Medium gas (N2)

Light gas (He)

500

1000

1500

2000

2500

Particle velocity (m/s)

Figure 19 At the same temperature, the particles of a light gas, such as helium (He), move at a mean speed that is greater than that of the particles of heavy gases, such as nitrogen (N2) or chlorine (Cl2).

When a gas flows across a barrier through a small opening, the phenomenon is called effusion. An example of effusion is a helium (He) balloon that slowly deflates overnight (see Figure 20). In this case, the helium atoms effuse through the many pores in the rubber membrane. The duration of effusion can be considered the rate of diffusion, since they are proportional to each other.

Figure 20 A deflating helium (He) balloon is an example of effusion.

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65

According to Hypothesis 4 of the kinetic theory of gases, when two different gases are at the same temperature, the mean kinetic energy of their particles is identical. Consequently, the mean kinetic energy of a particle of two hypothetical gases called 1 and 2 is expressed as follows: Ek  Ek 1

2

1 1 m v 2  m2v 22 2 1 1 2

or

m1v 12  m2v 22 v1 2 m 2 v2 2  m 1

which is equivalent to:

v1 m2  v2 m1

Since it is more practical to consider the molar mass (M) rather than the mass (m) of an individual molecule, we obtain the following formula, known as Graham’s law: Graham’s law v1 M2  v2 M1

where v1  Rate of diffusion or effusion of gas 1, expressed in metres per second (m/s) v2  Rate of diffusion or effusion of gas 2, expressed in metres per second (m/s) M1  Molar mass of gas 1, expressed in grams per mole (g/mol) M2  Molar mass of gas 2, expressed in grams per mole (g/mol) This law states that under identical conditions of temperature and pressure the relative rates of diffusion and effusion of two gases are inversely proportional to the square roots of their molar masses. In other words, light gases diffuse or effuse more quickly than heavy gases. This can be observed when, for example, helium (He) and nitrogen (N2), enclosed in separate containers connected by a valve, under the same conditions of temperature and pressure, move toward each other (see Figure 21). Since helium is lighter than nitrogen, it diffuses more quickly. After a period of time, the two gases are uniformly distributed. N2

He

a) Closed valve

b) Recently opened valve

c) Valve opened for a period of time

Figure 21 Helium (He) and nitrogen (N2) gases mix when the valve separating them is opened. The helium diffuses more quickly since it is lighter than nitrogen. However, at some point, the two gases will be uniformly mixed in both tanks.

66

UNIT 1 Gases

The following examples show how we can use the equation of Graham’s law in diffusion and effusion: Example A At a given temperature, the rate of diffusion of nitrogen (N2) molecules is 0.098 m/s. Find the rate of diffusion of oxygen (O2) molecules at this temperature. Data: vN  0.098 m/s 2

MN  28.014 g/mol 2

vO  ? 2

MO  31.998 g/mol

Calculation: vO 2 MN2  MO2 vN 2 vO2  vN2 

2

MN2 MO2

 0.098 m/s 

28.014 g/mol  0.092 m/s 31.998 g/mol

Answer: The rate of diffusion of the oxygen molecules is 0.092 m/s. Example B Helium (He) and an unknown gas (X) effuse through a hole pierced in the side of a container. What is the molar mass of the unknown gas if its rate of effusion is 0.077 m/s and that of helium is 0.256 m/s? What gas could it be? Data: vX  0.077 m/s MX  ? vHe  0.256 m/s MHe  4.003 g/mol

Calculation: MX vHe  vX MHe MX  MHe 

vHe vX

 4.003 g/mol 

0.256 m/s 0.077 m/s

MX  4.003 g/mol  (3.32)2  44.12 g/mol Answer: The molar mass of the unknown gas is 44 g/mol. The gas could be carbon dioxide (CO2) since its molar mass is 44.01 g/mol.

Furthering

your understanding

Gas permeation Certain technical fabrics have the twofold advantage of being waterproof and being able to “breathe.” This characteristic is due to the phenomenon of gas permeation, which is very similar to gas effusion. Developed in 1969 by Bob Gore, this type of membrane is made in a laboratory using a polymer called polytetrafluoroethylene (a chain of C2F4), which is incorporated into the fabric. This breathing membrane is pierced with millions of pores measuring 0.2 micrometers in diameter, that is, approximately 20 000 times smaller than a drop of water. However, this diameter is 700 times larger than an individual molecule of water, such as those found in the gas phase when they are very distant from each other. Consequently, perspiration can easily move through the membrane by means of gas permeation, but rainwater cannot. This technology provides greater comfort compared to traditional waterproof wear, often coated with a substance that blocks the pores of the fabric, preventing perspiration from escaping.

Figure 22 The breathing membrane (in yellow) magnified 100 times by an electronic microscope

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67

SECTION 2.2

Behaviour of gases

1. A submarine can vary its position by pumping water into its air tanks, known as ballasts. The water that enters the ballasts occupies space without evacuating the air, which increases the submarine’s weight and makes it descend toward the ocean floor (a). The submarine remains suspended in the water when the water level in the ballasts corresponds to that of the desired depth (b). In order to surface, the pumps expel the water in the tanks and the air reoccupies the space freed by the water, which reduces the weight of the submarine and allows it to rise (c).

4. Why do two gases at the same temperature have different diffusion rates? 5. A meteorological balloon containing helium (He) is launched into the atmosphere. During its ascent, the atmospheric pressure and temperature decrease. Explain why the size of the balloon increases. 6. What is the difference between the phenomena of gas diffusion and gas effusion? Give an example. 7. Rank in decreasing order of rate of diffusion the following gases, which are all under the same conditions of temperature and pressure: CO, HF, HI, NH3, NO2.

Weight

(a)

Weight Movement

No movement

(b)

(c)

Which macroscopic property of gases is used by the submarine to adjust its depth? Explain your answer. 2. What safety measures and precautions must be taken with compressed gases? Use your knowledge of the movement of particles to explain these precautions. 3. Car manufacturers have equipped hatchbacks with gas springs to make it easier to open and close them. These springs support the weight of the hatchback thereby reducing the strength needed to open them, and making it possible to close them gently. What macroscopic property of gases is used in a gas spring of this type when it slows the closing of a hatchback to absorb the shock with the body of the car?

68

UNIT 1 Gases

8. The rate of effusion of an unknown gas is estimated at 43.0 mL/min. Under the same conditions, the rate of effusion of pure carbon dioxide (CO2) is 32.0 mL/min. What is the molar mass of the unknown gas? 9. An unknown gas takes 192 s to effuse through a porous barrier, while the same volume of nitrogen (N2) effuses across the same barrier in 84 s. The conditions of temperature and pressure are identical for the two gases. What is the molar mass of the unknown gas? 10. Examine the graph below that illustrates the distribution of the particles of three gases at the same temperature. Which curve represents the gas with the greatest mass? Explain your answer.

Number of particles

Movement

Weight

3 2 1

Particle rate

2.3 Pressure of gases The pressure of gases corresponds to the force they exert on a surface. Gases exert pressure on all of the surfaces they come into contact with, in all directions (see Figure 23). Pressure is calculated by dividing the value of force by the area upon which the force is exerted. The following formula is used:

Pressure P

F A

where P  Pressure, expressed in newtons per square metre (N/m2) or in pascals (Pa) F  Force, expressed in newtons (N) A  Area where the force is exerted, expressed in square metres (m2)

According to Hypotheses 2 and 4 of the kinetic theory of gases, the particles of a gas are in constant movement in all directions, and the greater their kinetic energy, the faster they move. Since particles move, they can collide with each other or with any object, such as the walls of their container. When the particles of a gas hit the walls of their container, they exert a force on them that pushes in all directions. This is the force that inflates a football, for example, and gives the ball its firmness (see Figure 24).

Figure 23 When a champagne bottle is shaken, the carbon dioxide (CO2) dissolved in the champagne is freed in the form of gas bubbles. The freed gas exerts pressure in all directions on the walls of the bottle, as well as on the cork. When the metal wire that holds the cork in place is removed, the pressure is strong enough to pop the cork.

Figure 24 As gas particles hit the walls of the ball, the particles that keep it inflated exert pressure in all directions.

The greater the number of collisions, the greater the force exerted on the container per unit of surface area and the greater the pressure exerted by the gas. Therefore, the pressure of a gas on an object depends on the sum of the forces exerted by the collisions of its particles on the surface of the object. According to Hypothesis 4 of the kinetic theory of gases, at a given temperature the mean kinetic energy of the particles of two gases is the same. However, since kinetic energy also depends on mass, a light gas will move more quickly than a heavy gas. Therefore, there will be more collisions between the particles of a light gas, but the force exerted by these collisions will be smaller. Under similar conditions, the particles of a light gas will collide with less force

CHAPTER 2 Physical Properties of Gases

69

and more often than the particles of a heavy gas while the sum of the forces exerted by the collisions of these two gases is the same. This is why two gases subjected to the same conditions exert the same pressure, regardless of the size of their particles and their molar mass (see Figure 25).

2.3.1

a) Light gas

b) Heavy gas

Figure 25 Under the same conditions, the particles of the heavy gas exert the same pressure as the light gas.

Atmospheric pressure

The atmosphere is the gaseous layer that envelops the Earth. Due to its gravitational force, the Earth retains the gases that compose the atmosphere around it. The gases with the highest mass are located close to the Earth’s surface, while the gases with lower mass are at a higher altitude. The gas particles present in the air exert a force on all objects with which they are in contact. The force exerted by the air is called atmospheric pressure. It is equivalent to the weight of the column of air located above the surface that is subject to the force. Since air is denser at ground level than at higher altitudes, collisions between air particles are more frequent. Consequently, the atmospheric pressure is higher at ground level and rapidly decreases at higher altitudes (see Figure 26). 3.5 kPa Low atmospheric pressure

25 km

51.6 kPa

5 km

88.5 kPa

1 km

Empty

Mercury

High atmospheric pressure 101.3 kPa

Sea level

Figure 26 At sea level, the air is dense and there are a large number of collisions between particles causing a higher pressure than at higher altitudes, where the air is less dense.

Patm

Figure 27 A barometer similar to the one invented by Torricelli

70

UNIT 1 Gases

In addition to varying with altitude, atmospheric pressure depends on temperature and meteorological conditions. Atmospheric pressure is measured using a barometer, an instrument invented by Evangelista Torricelli in 1643. The barometer built by Torricelli is composed of a long glass tube closed at one end and filled with mercury (Hg) that is inverted into an open container that also contains mercury (see Figure 27).

Gravitational force attracts the mercury (Hg) in the reservoir, while the atmospheric pressure exerted on the surface of the mercury in the reservoir causes it to rise in the tube. When the level of the mercury stabilizes, the two forces cancel each other out. When atmospheric pressure is measured at sea level, the height is 760 mm, on average. This pressure is approximately equal to normal pressure, which, according to the International System of Units (SI), is equivalent to 101.3 kilopascals (kPa) or 1 atmosphere (atm). Conversions of these units of measure are as follows:

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Conversion of units of measure of pressure 101.3 kPa  760 mm Hg  1 atm The following example shows how we can use conversions to transform units of measure for pressure. Example The lowest atmospheric pressure ever measured on Earth was 87.0 kPa. This pressure was measured at sea level in the eye of Typhoon Tip in October 1979. What is the value of this low pressure in millimetres of mercury (mm Hg)? Data: P  87.0 kPa P?

Calculation: 760 mm Hg ?  101.3 kPa 87.0 kPa 760 mm Hg ?  87 kPa   652.7 mm Hg 101.3 kPa

Answer: The value of this low pressure is 653 mm Hg.

2.3.2

Measuring the pressure of gases

The barometer is exclusively used to measure atmospheric pressure. To measure the pressure of gases in a container, a manometer or pressure gauge is used (see Figure 28). A manometer is generally used to measure the pressure of gases. There are two main types of manometers: dial manometers (see Figure 29) and U-tube manometers.

Figure 28 A pressure gauge is a simple instrument used to determine air pressure inside tires.

EVANGELISTA TORRICELLI Italian mathematician and physicist (1608–1647) With his invention of the first barometer composed of a glass tube filled with mercury (Hg), Evangelista Torricelli demonstrated that the atmosphere has a weight and exerts pressure. Galileo’s work on the water pump inspired Torricelli to invent an instrument that measured atmospheric pressure. Torricelli did not publish his findings at the time. A few years later, Blaise Pascal repeated Torricelli’s experiment and pursued his research on pressure. The torr is a unit of pressure named in honour of Torricelli that corresponds to the pressure of one millimetre of mercury (1 Torr  1 mm Hg).

Figure 29 A dial manometer measures the mechanical displacement caused by the force exerted by the pressure of the gas and transcribes it in the form of a digital display or a dial with a needle. CHAPTER 2 Physical Properties of Gases

71

In U-tube manometers, the gas pressure is indicated by the height of a column of liquid. This liquid moves freely in the U-tube connected to a container of gas. While mercury (Hg) is toxic, it is a liquid that is commonly used due to its high density, which makes it possible to build relatively small manometers that are easy to handle. There are two types of U-tube manometers: closed-end and open-end. The difference in level between the two columns of mercury in the U-tube, measured in millimetres of mercury, makes it possible to determine the value of the pressure of the gas enclosed in the container. In a closed-end manometer, the pressure exerted by the gas in the container is equal to the difference in height in mm Hg between the two columns of liquid in the U-tube (see Figure 30). Real pressure of a gas using a closed-end manometer Ph where Pgas  Pressure of the gas in the container, expressed in millimetres of mercury (mm Hg) h  Height of the column of mercury, expressed in millimetres of mercury (mm Hg) Gas

(cm)

The following example illustrates how to determine real gas pressure using a closed-end manometer.

90 80 70 Pgas

60 50 40 30 20 10

Figure 30 Closed-end manometer

h

Example What is the real pressure of the gas contained in the manometer in Figure 30? Data: Pgas  ? h ?

1. Calculation of the height of the column of mercury: h  88 cm Hg  32 cm Hg  56 cm Hg  560 mm Hg 2. Calculation of the real gas pressure: Pgas  h P  560 mm Hg

Answer: The pressure of the gas in the container of the manometer is 560 mm Hg. With the open-end manometer, atmospheric pressure must be taken into account to determine the pressure of a gas. When the gas pressure is greater than atmospheric pressure, the level of mercury in the part of the tube in contact with the gas is lower than that of the open part of the tube. The gas pressure is determined by recording the difference in height and adding it to the atmospheric pressure (see Figure 31 on the following page). Conversely, when the gas pressure is less than atmospheric pressure, the level of the mercury is higher than that of the open part of the tube. The gas pressure is determined by recording the difference in height and subtracting it from the atmospheric pressure (see Figure 32 on the following page).

72

UNIT 1 Gases

Real pressure of a gas using an open-end manometer If Pgas  Patm Then

Pgas  Patm  h

If

Pgas  Patm

Then

Pgas  Patm  h

where Pgas  Gas pressure in the container, expressed in millimetres of mercury (mm Hg) Patm  Atmospheric pressure, expressed in millimetres of mercury (mm Hg) h  Height of the column of mercury, expressed in millimetres of mercury (mm Hg)

The following examples illustrate how to determine real gas pressure using an open-end manometer:

Gas

(cm)

90

Example A What is the real pressure of the gas contained in the manometer in Figure 31? Data: Patm  760 mm Hg Pgas  ? h ?

Patm 760 mm Hg

80 70

1. Calculation of the height of the column of mercury: h  75 cm Hg  45 cm Hg  30 cm Hg  300 mm Hg 2. Calculation of the real gas pressure: Pgas  Patm  h  760 mm Hg  300 mm Hg  1060 mm Hg

Pgas

60

h

50 40 30 20 10

Figure 31 Open-end manometer, when gas pressure is greater than atmospheric pressure

Answer: The pressure of the gas in the manometer's container is 1060 mm Hg.

Example B What is the real pressure of the gas contained in the manometer in Figure 32? Data: Patm  102.6 kPa Pgas  ? h ?

1. Calculation of the height of the column of mercury: h  62 cm Hg  25 cm Hg  37 cm Hg  370 mm Hg 2. Calculation of atmospheric pressure: 760 mm Hg ?  101.3 kPa 102.6 kPa 760 mm Hg ?  102.6 kPa  101.3 kPa  769.8 mm Hg 3. Calculation of gas pressure: Pgas  Patm  h  769.8 mm Hg  370 mm Hg  399.8 mm Hg

Answer: The pressure of the gas in the manometer's container is 400 mm Hg.

Gas

(cm) Patm  102.6 kPa

90 80 70 Pgas

60 50 40

h

30 20 10

Figure 32 Open-end manometer, when gas pressure is less than atmospheric pressure

CHAPTER 2 Physical Properties of Gases

73

SECTION 2.3

Pressure of gases

1. Explain how a gas exerts pressure on the walls of its container. 2. The pressure inside a plane flying at a very high altitude is 688 mm Hg. What is this pressure in atmospheres? in kilopascals? 3. Determine the pressure measured by each of the following manometers based on the atmospheric pressures provided: a)

b) Gas

Gas (cm) 90 80 Pgas

(cm)

Patm  99.9 kPa

Pgas

90 80

70

70

60

60

50 40

50

h

40

30

30

20

20

10

10

c)

d) Gas

Gas (cm)

(cm) 90 80

Pgas

Patm  99.9 kPa

Pgas

90 80

70

70

60

60

50 40

h

50 40 h

30

30

20

20

10

10

4. A closed-end manometer is used to measure the pressure of a sample of a given gas. The level of mercury (Hg) is 18 mm higher in the part of the tube connected to the sample than in the other part. Determine the gas pressure in kilopascals, if the atmospheric pressure is 99.7 kPa.

74

Patm  99.9 kPa

UNIT 1 Gases

Patm  99.9 kPa

2.4 Simple gas laws The simple gas laws help to solve problems that establish a relationship between two of the four variables that describe gases, namely pressure (P), volume (V), absolute temperature (T) and quantity of gas (n) expressed in number of moles while the other two variables remain constant. Sharing ideas and observations plays an essential role in scientific discoveries and the formulation of laws. As in the case of the gas laws, scientists often discover laws after having pooled their work at the end of a very long process. Beginning in the 17th century, several scientists worked independently to learn more about the behaviour of gases. They conducted experiments to discover the relationships between the four variables used to describe the behaviour of gases: pressure, volume, absolute temperature and number of moles. However, to clearly establish a relationship between these four variables, scientists could not study all four simultaneously. Instead, they designed experiments that established a relationship between only two of the four variables at a time, while the other two were held constant. Then, based on their conclusions for each pair of variables, scientists formulated simple gas laws, which often carry their name. At the time these laws were formulated, the scientific community had not yet adopted the International System of Units. Moreover, the conditions for working with gases varied greatly from one laboratory to the next, given that temperature and air pressure varied through time and according to the altitude of the location on Earth where the measurements were taken (see Figure 33).

Figure 33 A replica of Antoine Laurent de Lavoisier’s laboratory, at the Deutsches Museum, in Munich

Since that time, scientists have agreed on the use of two sets of temperature and pressure conditions which make it possible to compare the behaviour of different gases. The two sets of conditions, which are standards, are as follows (see Table 2): standard temperature and pressure (STP) and standard ambient temperature and pressure (SATP). Table 2 Two standards for the temperature and pressure of gases Standard

Temperature

Pressure

Standard temperature and pressure (STP)

0°C

101.3 kPa

Standard ambient temperature and pressure (SATP)

25°C

101.3 kPa

Combining the simple laws proposed for each pair of variables led to the formulation of a more general law, called the ideal gas law, which led to another law, the general gas law. While they were formulated on the basis of the concept of ideal gas, these laws can be applied to real gases under certain conditions. In fact, many observations suggest that at relatively high temperatures and fairly low pressures, such as under STP and SATP conditions, most real gases behave almost like ideal gases. It is only when the temperature is very low (close to the condensation point of the gas) and the pressure is very high (on the order of 1 Pa) that the differences between real gases and an ideal gas become significant enough to compromise the use of these laws.

CHAPTER 2 Physical Properties of Gases

75

2.4.1

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Relationship between pressure and volume

Scientists Robert Boyle and Edme Mariotte discovered, within a number of years of each other, the relationship between the pressure of a gas sample and its volume. In fact, during the second half of the 17th century, both researchers worked, independently, with gas samples trapped in mercury (Hg) in a glass J-tube, which is closed and calibrated on the shorter side (see Figure 34). Added Hg

PHg

Added Hg

ROBERT BOYLE Irish chemist and physicist (1627–1691)

Gas (Constant quantity of n)

h h

PHg

PHg

a)

b)

c)

Figure 34 A J-tube similar to the one used by Robert Boyle and Edme Mariotte to compress a gas sample

EDME MARIOTTE French physicist (1620–1684) Robert Boyle is considered the father of modern chemistry and the scientific method. He was the first chemist to isolate and collect a gas. In 1661, Boyle formulated a law regarding the compressibility of gases. For his part, priest Edme Mariotte was interested in many subjects pertaining to plant function, the movement of fluids and eye anatomy. By dissecting a human eye, he dis covered that part of the retina lacks cells capable of capturing light and named it the blind spot. In 1676, Mariotte unknowingly repeated Boyle’s experiments and com pleted the law known today as Boyle’s law.

76

UNIT 1 Gases

The sample is held at a constant temperature, and initially, the level of mercury is equal in each part of the tube, which indicates that gas pressure equals atmospheric pressure. The pressure exerted on the gas can be increased or decreased by adding or removing mercury from the open portion of the J-tube. The weight of the additional mercury increases the pressure exerted on the trapped gas and the volume of the gas decreases. The volume of trapped gas is proportional to the height of the gas (h) measured in the apparatus. Consequently, by meticulously measuring the volume of the trapped gas and the pressure exerted by the mercury, the two scientists observed that, at a constant temperature, the greater the pressure exerted on a gas, the more the volume of the gas decreases. This relationship between the pressure and the volume of a gas is called Boyle’s law (also called Boyle-Mariotte’s law) and is formulated as follows:

Boyle’s law At a constant temperature, the volume occupied by a given quantity of gas is inversely proportional to the pressure of the gas.

In other words, if the pressure exerted on a quantity of gas is doubled by compressing it while holding its temperature constant, the volume of the gas will decrease by half. Inversely, if a gas is allowed to occupy a volume that is two times greater while holding its temperature constant, the pressure of this gas will decrease by half.

Boyle’s law can be illustrated using a bicycle pump connected to a manometer (see Figure 35). Volume of a gas as a function of pressure 4

3

Volume (L)

P1  100 kPa

P2  200 kPa

P3  400 kPa

2 A

A

1

B

C

B C 0

100

200

300

400

500

Pressure (kPa)

Figure 35 When the pressure exerted on the gas by the pump’s piston increases, the volume of the gas decreases proportionately.

The curve of the graph plotted using the results obtained in the three situations shows that at a given temperature, the volume of a gas is inversely proportional to its pressure. This relationship can be written in mathematical form using the proportionality symbol (). The expression means that volume is inversely proportional to pressure. Mathematically, the proportionality symbol can be removed by introducing a proportionality constant (ka).

V

1 P

becomes V  k a  PV  k a

Expressed in this way, Boyle’s law means that the product of pressure by volume is constant. The value of this constant differs as a function of the number of moles and the temperature. It can be determined by plotting the graph of pressure (P) as a function of the mathematical inverse of volume (1/V), since it is an inversely proportional relationship. For example, for the air contained in the bicycle pump in Figure 35, the graph plotted using the values P and 1/V is a straight line (see Figure 36). The slope (m) of the straight line obtained gives the value of the proportionality constant (ka).

Pressure of a gas as a function of the inverse of its volume 400

m

y x



400 kPa 2 L1

 200 kPaL

300 Pressure (kPa)

or

1 P

200

100

0

1

2

3

4

5

1/Volume (L−1)

Figure 36 The relationship between the pressure of a gas and the inverse of its volume is directly proportional. CHAPTER 2 Physical Properties of Gases

77

The ratio PV  ka can also be expressed by comparing two sets of measurements of pressure and volume for the same gas sample as long as the temperature does not change. Therefore, even if the initial volume (V1) and the initial pressure (P1) of a given quantity of gas take on new values (V2 and P2), called final volume and final pressure, their product is always equal to the constant ka. P1 V 1  k a P2 V 2  k a By combining these two equivalences, we obtain Boyle’s law. Boyle’s law P1V1  P2V2* where P1  Initial pressure, expressed in kilopascals (kPa) or millimetres of mercury (mm Hg) V1  Initial volume, expressed in millilitres (mL) or litres (L) P2  Final pressure, expressed in kilopascals (kPa) or millimetres of mercury (mm Hg) V2  Final volume, expressed in millilitres (mL) or litres (L) * On condition that the number of moles (n) of gas and temperature (T ) are constant.

This relationship between pressure and volume is often applied in everyday objects such as office chairs. Most of these chairs are equipped with a gas piston that makes it easy to adjust the chair’s height (see Figure 37).

Figure 37 The vertical rod on this chair is a piston that contains compressible gas.

Figure 38 A gas piston, like the ones used in office chairs, makes it easier to open and close a car hatchback.

78

UNIT 1 Gases

The seat is connected to the foot of the chair by a waterproof cylinder that contains nitrogen (N2) and a piston that can freely move up and down. When someone sits on the seat, the piston exerts pressure on the gas inside the cylinder and reduces its volume, which lowers the seat. When the person gets up, the pressure on the piston decreases and the volume of the gas increases, which allows the seat to rise. This type of piston can also be found on car hatchbacks to make them easier to open (see Figure 38).

According to Boyle’s law, at a constant temperature, the product of pressure by volume of a gas sample remains constant even when pressure and volume vary from one situation to the next. The following example shows how Boyle’s law can be used to compare pressures and volumes that vary for a given gas sample at a constant temperature:

Example A sample of helium (He) gas is collected at ambient temperature in a 2.5-L elastic rubber ball at normal atmospheric pressure. The ball is then immersed in a container of water, also at room temperature, such that the external pressure exerted on its walls increases to 110.6 kPa. What is the final volume of the ball? Data:

Calculation:

P1  101.3 kPa

P1V1  P2V2

V1  2.5 L

V2 

P2  110.6 kPa V2  ?



P1 V 1 P2 101.3 kPa  2.5 L 110.6 kPa

 2.3 L Answer: The final volume of the ball is 2.3 L.

Connection with the kinetic theory of gases By specifying that the particles of a gas are very far from each other, Hypothesis 1 of the kinetic theory of gases (see page 61) explains the relationship between the pressure and the volume of a gas. At a constant temperature, as the external pressure exerted on a gas increases, the volume of the gas decreases. Therefore, the gas particles become closer together and move over a shorter distance before colliding with each other and the walls of the container (see Figure 39 ). Consequently, the collisions are more frequent, which increases the pressure. Inversely, if the volume of the container is increased, the distance travelled by the particles is greater, the number of collisions per unit of time is smaller and the gas therefore exerts less pressure. This is an inversely proportional relationship.

Pext Pext

Pgaz

Pgas  Pext

The relationship between the pressure and the volume of gases is what allows air to enter and exit the lungs through the respiratory movements associated with ventilation (breathing). When inha ling, contractions coordinated by the intercostal muscles and the diaphragm cause an increase in the volume of the thoracic cage and, consequently, the lungs. This increase in the volume available for air reduces the air pressure in the lungs. Given that the external air is at a higher pressure, it moves into the lungs through the respiratory tracts. During exhalation, the muscles release causing the thoracic cage to move down and the diaphragm to move back up. The volume of the thoracic cage and the lungs decreases, which increases the air pressure in the lungs. Since this pressure is now higher than that of the external air, the air moves out of the lungs. Normal pressure

Elevated pressure

T and n are held constant d2

d1

Respiration

Pgaz

A higher Pext causes a reduction in V, which leads to more collisions until Pgas  Pext

Figure 39 At a given temperature, when the volume of gas is reduced, its particles move over a shorter distance (d2  d1) before colliding with each other and the walls.

Figure 40 During exhalation, air moves from the lungs where the pressure is higher, to outside the lungs, where the pressur is lower.

CHAPTER 2 Physical Properties of Gases

79

2.4.2

Relationship between volume and absolute temperature

Jacques Charles discovered the relationship between the volume of a gas sample and its temperature. His work focused on the variation in volume of several types of gases when they are heated at a constant pressure. Charles observed that, at a constant pressure, as temperature increases, volume increases and vice versa. This is the principle used in hot air balloons. Heating the air in the balloon increases its volume and, consequently, lowers its density sufficiently to lift the balloon into the air (see Figure 41).

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Figure 41 For a hot air balloon to lift off the ground, the air inside the balloon is heated using a propane burner. The volume of the air increases as it is heated, the balloon stretches and the resulting bubble of hot (and less dense) air lifts the balloon into the atmosphere.

JACQUES CHARLES French chemist and physicist (1746–1823) Jacques Charles was interested in the hot air balloons that the Montgolfier brothers were studying in France during this period. He introduced the innovation of using hydrogen (H) instead of hot air to fly a balloon he rode in. While he was the first to formulate the law of the dilatation of gases, he did not publish the results of his work. The fruit of his work remained unknown until some 15 years later when chemist Louis Joseph Gay-Lussac (1778– 1850) compiled Charles’ results and compared them to his own. This is why Charles’ law is sometimes called Gay-Lussac’s law.

80

UNIT 1 Gases

Charles observed that the volume of a gas measured at 0°C increased by 1/273 of its initial value for each degree Celsius increase in temperature. He deduced that, at a constant pressure, the volume of a gas sample initially at 0°C doubles when it is heated to 273°C. This simple linear relationship between volume and temperature expressed in degrees Celsius translates graphically into a straight line with the equation y  mx  b (see Figure 42). This means that the relationship between volume and temperature is not directly proportional. Volume of a gas as a function of its temperature Volume (L)

The volume of a gas doubles as a result of an increase of 273°C.

3

(273. 2) 2

B (50. 1.18)

A

B

1 A

(100. 1.37)

(0. 1)

(273. 0) 300

200

100

0

100

200

300

Temperature (°C)

Figure 42 The relationship between the volume of a gas and its temperature in degrees Celsius is not directly proportional because it does not intersect the origin (0. 0).

When the straight line is extended downward by extrapolation, it intersects the horizontal x-axis at -273°C.

When the same experiment was repeated with different quantities of gas and with samples of different gases, the linear relationship was reproduced each time, and it was possible to plot a new straight line with a different slope (see Figure 43). The discovery by Jacques Charles showed that, regardless of the gas studied, the x-intercept is always -273°C. This temperature, called absolute zero, is the lowest possible temperature. Volume of three gases as a function of their temperature

3

2

2

1

1

Volume (L)

3

300

200 0

0

100 100

200

100 300

300 (°C)

200 400

500

(K)

Temperature

Figure 43 The linear relationship between volume and temperature is reproduced each time the experiment is repeated, and a new straight line can be plotted that always has the same x-intercept but a different slope.

A few years after Charles’ discovery, Lord Kelvin defined absolute zero as the temperature at which the kinetic energy of all particles would be zero. At this temperature, the volume of a gas would also hypothetically be equal to zero, as long as the gas was considered an ideal gas and its particles were considered to occupy no volume. In reality, particles of real gases occupy a volume, and at low temperatures, all real gases condense and change phases. Nevertheless, Kelvin proposed a new temperature scale called the absolute temperature scale, or the Kelvin scale. The starting point of this scale is called absolute zero and corresponds to zero kelvin or 0 K. Note that the degree symbol (°) is not used in the Kelvin scale. The modern value of absolute zero, obtained using more sophisticated equipment than what was available to Charles, corresponds to -273.15°C. Therefore, the Kelvin scale does not have any negative values. The absolute temperature scale and the Celsius scale use the same equally spaced divisions (see Figure 44). An interval of 1°C is therefore equivalent to an interval of 1 K. Consequently, to convert degrees Celsius into kelvin, simply add 273, the rounded conversion factor, to the temperature in degrees Celsius. Conversion of degrees Celsius into kelvin T  °C  273

Kelvin

Celsius Boiling point of water

373 K

100°C

100 divisions

100 divisions

Freezing point of water 273 K

0°C

Absolute zero 0K

273°C

Figure 44 The zero on the absolute temperature scale and the zero on the Celsius scale are located at an interval of 273 units from each other. The Kelvin scale does not have any negative values.

CHAPTER 2 Physical Properties of Gases

81

Using the Kelvin scale rather than the Celsius scale makes it possible to observe that the relationship between volume and absolute temperature is directly proportional, which is not the case in Celsius. This means that at a constant pressure, as the absolute temperature of a gas increases, the volume of the gas increases by an equal factor and vice versa. This relationship between the volume and temperature of a gas is called Charles’ law and is formulated as follows: Charles’ law At a constant pressure, the volume occupied by a given quantity of gas is directly proportional to the absolute temperature of the gas.

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

In other words, if the temperature in kelvin of a quantity of gas is doubled while pressure is held constant, the volume of the gas doubles (see Figure 45). Inversely, if the volume is reduced by half while the pressure is held constant, the absolute temperature of the gas decreases by half. Volume of a gas sample as a function of its absolute temperature

4

WILLIAM THOMSON (LORD KELVIN) British mathematician, physicist and engineer (1824–1907) William Thomson, also known as Lord Kelvin, is especially known for his work in thermodynamics, which led him to introduce the concept of entropy. A brilliant thin ker, his long scientific career was marked by works on subjects as varied as the Earth’s age, the composition of matter and the electrical conductivity of underwater cables. He promoted the construction of the first transatlantic cable, which earned him his noble rank. In recognition of his contributions, the absolute temperature scale, one of his greatest inventions, was named after him whereby zero corresponds to the absolute absence of thermal agitation and pressure.

82

UNIT 1 Gases

Volume (L)

3

2

1

0

1

2

3

4

5

6

7

8

Temperature (K)

Figure 45 When a gas is heated at a constant pressure in a cylinder with a mobile piston, it dilates. If its absolute temperature is doubled, its volume is multiplied by two.

This relationship can be written in mathematical form using the proportionality symbol . The expressionV  T means that volume is directly proportional to absolute temperature. Mathematically, the proportionality symbol can be removed by introducing a proportionality constant (kb). VT becomes or

V  kb  T V  kb T

Expressed this way, Charles’ law means that the quotient of the two variables is constant. The value of this constant changes according to the number of moles and pressure. It can be determined by calculating the slope (m) of the straight line on the graph of the volume of a gas as a function of its absolute temperature (T).

The ratio V/T  kb can also be expressed by comparing two sets of measurements of volume and temperature for the same gas sample whose pressure does not change. Therefore, even if the initial volume (V1) and the initial temperature (T1) of a given quantity of gas take on new values (V2 and T2), called final values, their quotient is always equal to the constant kb. V1  kb T1 V2  kb T2 By combining these two equivalences, we obtain Charles’ law. Charles’ law

V1 V *  2 T1 T2

where V1  Initial volume, expressed in millilitres (mL) or litres (L) T1  Initial temperature, expressed in kelvin (K) V2  Final volume, expressed in millilitres (mL) or litres (L) T2  Final temperature, expressed in kelvin (K) * On condition that the number of moles (n ) of gas and the pressure (P ) are constant.

According to Charles’ law, at a constant pressure, the quotient of the volume and of the absolute temperature of a gas sample remain constant even if the volume and temperature vary from one situation to the next. The following example shows how to use Charles’ law to compare volumes and temperatures that vary for a given gas sample at constant pressure: Example A 28.7-mL sample of oxygen (O2) gas is collected in a glass syringe at SATP. The syringe is placed in an oven at 65°C until the gas attains the temperature of the oven. What volume will the oxygen occupy if the atmospheric pressure in the oven is the same as outside the oven? If the syringe can contain a volume of 50 mL, will the mobile piston be able to retain the gas inside? Data: V1  28.7 mL T1  25°C V2  ? T2  65°C

1. Conversion of the temperature into kelvin: T1  25°C  273  298 K T2  65°C  273  338 K 2. Calculation of volume: V1 V  2 T1 T2 V2  

V1  T2 T1 28.7 mL  338 K  32.6 mL 298 K

Answer: The volume of oxygen (O2) is 32.6 mL. The syringe could therefore contain the gas.

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Connection with the kinetic theory of gases By specifying that an increase in the temperature of a gas translates into a more rapid movement of particles, Hypothesis 4 of the kinetic theory of gases (see page 61 ) provides an explanation of the relationship between the volume and the temperature of a gas. As the temperature of a gas increases, its particles move more quickly and collide with each other and the walls of the container more often (see Figure 46). Consequently, the particles collide with the walls more often per unit of time, which increases the pressure exerted by the gas. This additional pressure increases the volume of the container until the pressure exerted by the gas equals the external pressure. That is why it can be said that the pressure remains constant. Pext Pext

T increases

Pext

V increases

Pgas

n is constant Pgas

a) Pgas  Pext

T1

Pgas

T2

b) A higher temperature T increases the frequency of collisions: Pgas Pext

T2

c) V increases until Pgas  Pext

Figure 46 When the burner heats the gas (b ), the particles move more quickly, which increases the pressure exerted by the gas on the piston and walls of the cylinder. This additional pressure increases the volume of the gas and moves the piston upward (c ). The piston stops when the pressure exerted by the gas equals the external pressure; hot gas therefore occupies a larger volume than cold gas.

Using a microwave without making a mess When putting a plastic container of food in a microwave oven, the cover must be left partially open to avoid making a mess. Within the container is a mixture of air and water vapour that is released from the food as it heats up. When the temperature of the gas mixture increases, its volume also increases and the cover gradually bulges. If the temperature continues to rise, the force that keeps the cover on the container becomes insufficient. The dilatation of the gas pops the cover off and the food inside the container is projected in all directions. Some containers are specially designed for microwave ovens. These containers have a valve on the cover that opens to allow the gas to escape as it dilates, and prevents this kind of mess from occurring.

84

UNIT 1 Gases

Figure 47 When the mixture of air and water vapour inside a hermetically sealed container is heated in a microwave oven, the volume of the trapped gas can increase to the point of forcing the cover off and creating a mess.

2.4.3

Relationship between pressure and temperature

The relationship between the pressure of a gas sample and its temperature is another form of Charles’ law. This law was first published by Louis Joseph GayLussac, whose work was similar to that of Jacques Charles. Gay-Lussac observed that, at a constant volume, as temperature increases, pressure increases and vice versa. As with Charles’ law, this relationship is directly proportional when the temperature is expressed in kelvin (see Figure 48). Pressure of a gas sample as a function of its temperature

Pressure (kPa)

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY 500

500

400

400

300

300

200

200

100

100

A

B B

A

300

200 0

0

100 100

200

100 300

200 400

300 (°C) 500

(K)

Temperature

Figure 48 The relationship between the pressure of a gas and its absolute temperature is directly proportional.

This means that at a constant volume, as the absolute temperature of a gas increases, the pressure of the gas increases by an equal factor and vice versa. This relationship between the pressure and the temperature of a gas is called Gay-Lussac’s law and is formulated as follows:

Gay-Lussac’s law At a constant volume, the pressure of a given quantity of gas is directly proportional to the absolute temperature of the gas.

In other words, if we double the temperature in kelvin of a quantity of gas while holding its volume constant, the pressure of the gas doubles. Inversely, if the pressure is reduced by half while holding the volume constant, the absolute temperature of the gas decreases by half.

LOUIS JOSEPH GAY-LUSSAC French chemist and physicist (1778–1850) In pursuing Jacques Charles’ work, Louis Joseph Gay-Lussac discovered what would become GayLussac’s law. According to this law, the pressure of a gas is proportional to its absolute temperature. Since Charles’ work served as the basis for Gay-Lussac’s experiments, Gay-Lussac’s law is sometimes referred to as Charles’ second law. A few years after this discovery, Gay-Lussac formulated the combined gas law, regarding the volume of gas combinations.

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This relationship can be written in mathematical form using the proportionality symbol . The expression P  T means that pressure is directly proportional to absolute temperature. Mathematically, the proportionality symbol can be removed by introducing a proportionality constant (kc).

PT becomes or

P  kc  T P  kc T

Expressed this way, Gay-Lussac’s law means that the quotient of the two variables is constant. The value of this constant can be determined by the slope (m) of the straight line on the graph of the pressure of the gas as a function of its absolute temperature. It is also possible to express the ratio P/T  kc if two sets of measurements of pressure and temperature are compared for the same gas sample whose volume does not change. Therefore, even if the initial pressure (P1) and the initial temperature (T1) of a given quantity of gas take on new values (P2 and T2), referred to as final values, their quotient is still equal to the constant kc. P1  kc T1 P2  kc T2

By combining these two equations, we obtain Gay-Lussac’s law. Gay-Lussac’s law P1 P *  2 T1 T2 where P1  Initial pressure, expressed in kilopascals (kPa) or millimetres of mercury (mm Hg) T1  Initial temperature, expressed in kelvin (K) P2  Final pressure, expressed in kilopascals (kPa) or millimetres of mercury (mm Hg) T2  Final temperature, expressed in kelvin (K) * On condition that the number of moles (n ) of gas and the volume (V ) are constant.

According to Gay-Lussac’s law, at a constant volume, the quotient of the pressure and the absolute temperature of a gas remain constant even if pressure and temperature vary from one situation to the next.

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UNIT 1 Gases

The following example shows how to use Gay-Lussac’s law to compare pressures and temperatures that vary for a given sample of gas, at a constant volume: Example At 14.0°C, helium (He) gas stored in a metal tank exerts a pressure of 507 kPa. What will the pressure be if the tank is stored in a warehouse where the temperature increases to 40.0°C? Data: T1  14.0°C

1. Conversion of the temperature into kelvin: T1  14.0°C  273  287 K

P1  507 kPa

T2  40.0°C  273  313 K

T2  40.0°C

2. Calculation of the pressure: P1 P  2 T1 T2

P2  ?

P2  

P1  T2 T1

Cooking under pressure

507 kPa  313 K  552.9 kPa 287 K

Answer: The final pressure of the helium (He) will be 553 kPa.

Connection with the kinetic theory of gases By specifying that an increase in the temperature of a gas translates into a more rapid movement of its particles, Hypothesis 4 of the kinetic theory of gases (see page 61 ) explains the relationship between the pressure and the temperature of a gas. As the temperature of a gas increases, its particles move more quickly and collide with each other and the walls of the container more frequently (see Figure 49 ). Consequently, the particles hit the walls of the container more often per unit of time, which increases the pressure exerted by the gas. Given that the total force exerted by the gas increases and that the volume remains constant because the container is rigid, the pressure increases.

P1  100 kPa

P2  200 kPa

T increases

T1

A pressure cooker makes it possible to cook food in a closed container at very high temperatures. Since the pot is hermetically sealed using a powerful clamp, the pressure from the vapour trapped inside increases, which, according to Gay-Lussac’s law, causes an increase in temperature. Since the vapour can only escape in small quantities through a valve that controls its release, the internal temperature is maintained at a level that can reach 125°C. Without this valve to release the vapour, there would be an uncontrolled increase in pressure inside the pot which could force the cover off. The increase in the pressure and the temperature of the vapour inside the pot considerably reduces the cooking time.

Greater frequency of collisions

T2

Figure 50 The valve on the Figure 49 When the temperature increases from T1 to T2, particles move more quickly with an increase in the violence and frequency of collisions. Given that the total force exerted by the gas increases and that the volume remains constant because the container is rigid, the pressure increases.

pressure cooker allows vapour to escape in order to keep the pressure at a safe level.

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87

2.4.4

Relationship between the volume and quantity of gas expressed in number of moles

Toward the beginning of the 19th century, Louis Joseph Gay-Lussac demonstrated that a chemical reaction between gases always occurs in a simple whole number ratio of volume as long as pressure and temperature are held constant. Consequently, his work on the reaction between different gases showed, in particular, that hydrogen (H2) gas and oxygen (O2) gas unite to form water vapour (H2O) in the following proportion: 2 volumes of hydrogen for 1 volume of oxygen produce two volumes of water vapour (see Figure 51). 2 volumes



1 volume

n

2 volumes

2 H2 (g)



O2 (g)

n

2 H2O (g)

Hydrogen (H) Oxygen (O)

Figure 51 Gay-Lussac observed that when hydrogen (H2) gas and oxygen (O2) gas react with each other to form

water vapour, the ratio of the combination of volumes of H2 and O2 is 2 : 1 , and that of H2 and O2 with H2O is respectively 2 : 2 and 1 : 2.

By using nitrogen (N2) to obtain ammonia (NH3), Gay-Lussac also obtained whole number ratios of volume (see Figure 52). 3 volumes



1 volume

n

2 volumes

3 H2 (g)



N2 (g)

n

2 NH3 (g)

Hydrogen (H) Nitrogen (N)

Figure 52 Gay-Lussac obtained whole number ratios when he used nitrogen (N2) to obtain ammonia.

In the theory he published following this research, Gay-Lussac formulated a fundamental law of chemistry: the combined gas law. This law states that when gases react, the volumes of the reactants and the products are always present in the form of simple whole number ratios if they are measured at constant temperatures and pressures. A few years later, Amedeo Avogadro interpreted Gay-Lussac’s results and formulated a hypothesis that was daring for the time and would lay the groundwork for the modern concept of the molecule. His hypothesis stated that, under the same conditions of temperature and pressure, equal volumes of different gases contain the same number of particles.

88

UNIT 1 Gases

With time, Avogadro’s hypothesis became Avogadro’s law since it was never refuted by the scientific community. Avogadro’s law Under the same temperature and pressure conditions, the volume of a gas is directly proportional to its quantity expressed in number of moles. This means that, if the number of moles of gas doubles, while holding temperature and pressure constant, the volume of gas doubles (see Figure 53). Inversely, to decrease the volume of a gas by half, while holding temperature and pressure constant, the number of moles of gas must be reduced by half.

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Volume of a gas as a function of its quantity

Volume (L)

B

A

A 1

AMEDEO AVOGADRO Italian chemist and physicist (1776–1856)

B

2

3

4

Quantity (mol)

Figure 53 The relationship between the volume of a gas and its quantity is directly proportional.

This relationship can be written in mathematical form by using the proportionality symbol . The expression V  n means that the volume is directly proportional to the number of moles of gas. Mathematically, the proportionality symbol can be removed by introducing a proportionality constant (kd). Vn becomes

V  kd  n

or

V  kd n

Expressed this way, Avogadro’s law means that the quotient of the two variables is constant. The value of this constant can be determined by calculating the slope (m) of the straight line on the graph of the volume of the gas as a function of its number of moles.

The son of a magistrate, Avogadro initially followed in his father’s footsteps. However, after five years of practising law, he abandoned this profession to embark on studies in physics. His career as a physicist was prolific. Among other things, Avogadro formulated the law that states that equal volumes of different gases, under the same conditions of temperature and pressure, contain the same number of particles. Avogadro was also the first to make a distinction between atoms and molecules. In honour of his discoveries, scientists named the number representing the number of units present in one mole: Avogadro’s number (NA).

CHAPTER 2 Physical Properties of Gases

89

The ratio represented by this constant can also be used to compare two sets of measurements of volume and of the number of moles of a gas whose temperature and pressure do not change. Therefore, even if the initial volume (V1) and the initial number of moles (n1) of a gas take on new values (V2 and n2), referred to as final values, their quotient always equals the constant kd. V1  kd n1 V2  kd n2 By combining these two equivalences Avogadro’s law is obtained. Avogadro’s law V1 V *  2 n1 n2 where V1  Initial volume, expressed in millilitres (mL) or litres (L) n1  Initial quantity of gas, expressed in moles (mol) V2  Final volume, expressed in millilitres (mL) or litres (L) n2  Final quantity of gas, expressed in moles (mol) * On condition that temperature (T ) and pressure (P ) are constant.

According to Avogadro’s law, at a constant temperature, the quotient of the volume and the number of moles of a gas remains constant even if the volume and the number of moles vary from one situation to the next. The following example shows how to use Avogadro’s law to compare volumes and numbers of moles that vary for a given gas sample at a constant temperature: Example A helium (He) balloon occupies a volume of 15 L and contains 0.50 mol of helium at SATP. What will the new volume of the balloon be if 0.20 mol of helium is added to the balloon under the same conditions? Data: V1  15 L

1. Calculation of the quantity of helium: n2  0.50 mol  0.20 mol  0.70 mol

n1  0.50 mol

2. Calculation of volume: V1 V  2 n1 n2

V2  ? n2  ?

V2  

V 1  n2 n1 15 L  0.70 mol  21 L 0.50 mol

Answer: The final volume of the helium (He) balloon is 21 L.

90

UNIT 1 Gases

Connection with the kinetic theory of gases Hypothesis 2 of the kinetic theory of gases (see page 61 ) explains the relationship between the volume and the number of moles of a gas. As the number of particles of a gas increases, they collide with each other and the walls of the container more frequently (see Figure 54 ). Consequently, the number of collisions per unit of surface area increases, which increases the pressure exerted by the gas. At a constant temperature, this additional pressure increases the volume of the container until the pressure exerted by the gas equals the external pressure.

Pext

Pext

n increases

Pext

V increases

Pgas

T is constant Pgas

a) Pgas  Pext

Pgas

b) A greater number of molecules increases the number of collisions Pgas > Pext

c) V increases until Pgas  Pext

Figure 54 When the number of moles of particles increases from n1 to n2 (b), there are more collisions between particles and the pressure exerted by the gas increases. This additional pressure increases the volume of the gas and moves the piston upward (c) until the pressure exerted by the gas equals the external pressure.

The inflatable life jacket Many people, when boating, do not bother wearing a life jacket, even though it is mandatory, because they find it cumbersome or uncomfortable. However, there are models of life jackets on the market that take up very little space and fill with gas when the person falls in the water. These jackets are equipped with a small cylinder of compressed carbon dioxide (CO2) that can be manually or automatically activated on contact with water. Once activated, the cylinder empties its contents into the life jacket. In a fraction of a second, the quantity of gas increases in the jacket and the volume of the jacket increases until it is completely inflated. The person wearing the inflated jacket floats, since the density of the life jacket is less than that of water.

Figure 55 Once the cylinder is activated, the volume of an inflatable life jacket increases instantly as a function of the quantity of gas in the cylinder.

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2.4.5

Molar volume of gases

The molar volume of gases is the volume occupied by one mole of gas, regardless of the gas, under fixed conditions of temperature and pressure. The molar volume (Vm) is expressed in litres per mole (L/mol). According to Avogadro’s law, under the same conditions of temperature and pressure, equal volumes of any gas contain the same number of particles. Given that the value of one mole corresponds to Avogadro’s number, whose value is 6.02  1023, the molar volume of a gas represents the volume occupied by 6.02  1023 particles of this gas, regardless of the gas. Therefore, the size of the gas particles, and consequently the molar mass of the gas (M), has no influence on the volume occupied by the gas.

Connection with the kinetic theory of gases By specifying that the size of the particles is negligible in relation to the volume the particles occupy, Hypothesis 1 of the kinetic theory of gases (see page 61 ) is in agreement with Avogadro’s hypothesis and the resulting concept of molar volume. In fact, gas particles are so small in relation to the volume of the container they occupy that, while they have their own mass, their size is negligible and does not contribute to the total volume occupied by the gas. This is why it is also possible to state that, under the same conditions of temperature and pressure, one litre of carbon dioxide (CO2) contains the same number of particles as one litre of helium (He), even if the particles of carbon dioxide are larger than those of helium.

Experimental measurements have made it possible to determine the molar volume of a gas. Consequently, the results show that at STP, one mole of gas occupies a volume of 22.4 L (see Figure 56).

22.4 L

22.4 L

22.4 L Characteristics of three gases at STP Helium

n  1 mol

n  1 mol

P  101.3 kPa

P  101.3 kPa

P  101.3 kPa

T  0°C (273 K)

T  0°C (273 K)

T  0°C (273 K)

V  22.4 L

V  22.4 L

V  22.4 L

Number of particles of gas  6.02  1023

Number of particles of gas  6.02  1023

Number of particles of gas  6.02  1023

Mass  4.003 g

Mass  28.014 g

Mass  31.998 g

  0.179 g/L

  1.251 g/L

  1.428 g/L

Nitrogen (N) Oxygen (O)

Figure 56 At STP, gases like helium (He), nitrogen (N2) and oxygen (O2) behave like ideal gases and have a molar volume of 22.4 L/mol.

UNIT 1 Gases

Oxygen

n  1 mol

Helium (He)

92

Nitrogen

At SATP, the value of the molar volume is 24.5 L/mol, which corresponds to the volume of air found in approximately 12 2-L soft drink bottles (see Figure 57).

SATP: 25°C, 101.3 kPa

Figure 57 At SATP, one mole of any gas, or a mixture of several gases, like air, occupies a volume of 24.5 L, which is the space occupied by approximately 12 2-L soft drink bottles.

The concept of molar volume is very useful when measuring precise quantities of gas. In fact, gases are substances that have low density () and are relatively difficult to weigh compared to solids, for example. Therefore, it is much easier to measure a large volume of gas than to measure a very small mass using a balance. Consequently, working with the molar volume of gases under fixed conditions of temperature and pressure based on STP or SATP standards allows greater precision when determining the quantity of gas contained in a given volume. The following example shows how to use molar volume to convert the volume of a given gas into number of moles:

Furthering your understanding Molar volume of real gases The molar volume of 22.4 L actually corresponds to that of an ideal gas. However, the values of molar volumes of certain real gases, such as helium (He), nitrogen (N2) and oxygen (O2), are very similar to that of ideal gases at STP, while others are significantly different. Table 3 Molar volumes of various gases at STP

Gas

Molar volume (L)

Example How many moles of gas are there in a container holding 69.2 L of methane (CH4) gas at STP?

Oxygen (O2)

22.397

Nitrogen (N2)

22.402

Data:

Hydrogen (H2)

22.433

Helium (He)

22.434

Argon (Ar)

22.397

Carbon dioxide (CO2)

22.260

Ammonia (NH3)

22.079

V  69.2 L n? Vm  22.4 L/mol (at STP)

Calculation: 22.4 L 69.2 L  1 mol ? 69.2 L  1 mol ? 22.4 L  3.10 mol

Answer: There are 3.10 mol of methane (CH4) gas in the container.

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2.4.6

Figure 58 The manometer connected to the tank tells the diver what the gas pressure is. The more the pressure decreases, the less gas there is in the tank.

Relationship between the pressure and the quantity of gas expressed in number of moles

When a metal tank is filled with air before a dive, the quantity of compressed gas inside increases as the tank fills, and its mass increases as the quantity of compressed air increases. Given that the tank is made out of rigid metal, its volume remains constant. Therefore, if the temperature is held constant, as the quantity of air increases during filling, the pressure indicated on the manometer connected to the tank increases (see Figure 58). Inversely, when the diver breathes air from the tank, the pressure decreases and the manometer tells the diver when to begin surfacing. This shows that the relationship between the pressure and the number of moles of a gas at a constant temperature and volume is directly proportional. This relationship between pressure and number of moles is formulated as follows:

Relationship between pressure and quantity of gas expressed in number of moles Under the same conditions of temperature and volume, the pressure of a gas is directly proportional to its number of moles.

In other words, if the number of moles is doubled while holding temperature and volume constant, the pressure of the gas doubles (see Figure 59). Inversely, to reduce the pressure of a gas by half while holding temperature and volume constant, the number of moles of gas is also reduced by half. Pressure of a gas as a function of its quantity (number of moles) B

400

Pressure (kPa)

200

P2  400 kPa

P1  200 kPa

300

A

100 B

A

0

1

2

3

4

Quantity (mol)

Figure 59 The relationship between the pressure of a gas and its quantity expressed in number of moles is directly proportional.

94

UNIT 1 Gases

The relationship between the pressure of a gas and its quantity can be written in mathematical form using the proportionality symbol . The expression P  n means that the pressure is directly proportional to the number of moles of gas. Mathematically, the proportionality symbol can be removed by introducing a proportionality constant ke. Pn becomes

P  ke  n

or

P  ke n

Expressed this way, the relationship means that the quotient of the two variables is constant. The value of this constant can be determined by calculating the slope of the straight line (m) on the graph of the pressure of gas as a function of its number of moles. The ratio P/n  ke can also be expressed by comparing two sets of measurements of pressure and number of moles of a gas whose temperature and volume do not change. Therefore, even if the initial pressure (P1) and the initial number of moles (n1) of a gas take on new values (P2 and n2), referred to as final values, their quotient always equals the constant ke. P1  ke n1 P2  ke n2

By combining these two equivalences, the formula for the relationship between pressure and number of moles is obtained. Relationship between pressure and number of moles P1 n1



P2 * n2

where P1  Initial pressure, expressed in kilopascals (kPa) or in millimetres of mercury (mm Hg) n1  Initial quantity of gas, expressed in moles (mol) P2  Final pressure, expressed in kilopascals (kPa) or in millimetres of mercury (mm Hg) n2  Final quantity of gas, expressed in moles (mol) * On condition that temperature (T ) and volume (V ) are constant.

According to this relationship, at constant temperature and volume, the quotient of pressure and number of moles of gas remains constant even if pressure and number of moles vary from one situation to the next.

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The following example shows how to use the relationship between the pressure and the number of moles of a gas to compare the pressures and quantities that vary for a given gas sample at a constant temperature and volume. Example A car tire that is suitably inflated contains approximately 5.0 mol of air at a pressure of 220 kPa. How many moles of air remain inside the tire when it has been partially deflated and the pressure indicated on a manometer is 185 kPa? Data: n1  5.0 mol P1  220 kPa n2  ? P2  185 kPa

Calculation: P1 P  2 n1 n2 n2  

P2  n1 P1 185 kPa  5.0 mol 220 kPa

 4.2 mol Answer: There are 4.2 mol of air remaining in the tire.

Connection with the kinetic theory of gases Hypothesis 2 of the kinetic theory of gases (see page 61 ) explains the relationship between the pressure and the number of moles of a gas. As the number of particles of a gas increases, they collide with each other and the walls of the container more frequently. Consequently, the number of collisions per unit of surface area increases, which increases the pressure exerted by the gas. At a constant temperature and volume, this additional pressure is added to the initial pressure and the pressure of the gas increases.

Filling a propane tank

Sllde

Beam

Propane (C3H8) in tanks used for barbecues is sold by quantity not volume. The tank must be placed on a balance to determine the number of kilograms injected into it. As soon as it is filled, the balance yields to the additional mass of the gas and the pressure increases. The tank is filled until it contains a mass of approximately 12 kg of propane. Adding more runs the risk of causing a pressure that is greater than what the tank can withstand.

Figure 60 Propane (C3H8) tanks for barbecues are

sold by mass not volume.

96

UNIT 1 Gases

SECTION 2.4

Simple gas laws

2.4.1 Relationship between pressure and volume 1. The cylinder in a bicycle pump contains 600 mL of air at 100 kPa. What is the volume of this air when the pressure reaches 250 kPa? 2. A gas occupies a volume of 50 mL under a pressure of 95.0 kPa at 50°C. What volume will it occupy if the pressure is reduced to 48.5 kPa without varying the temperature? 3. A gas occupies a volume of 68 mL under a pressure of 90 kPa at 25°C. What is the gas pressure if the volume is compressed to 45 mL? 4. A gas occupies a volume of 63.4 mL under a pressure of 82.3 kPa at 20°C. What is its pressure if the volume is increased to 78.5 mL? 5. A gas occupies a volume of 4.6 L under a pressure of 98 kPa at 15°C. What volume will it occupy if the pressure is doubled without changing the temperature? 6. At a pressure of 93 kPa, a gas occupies 0.25 L. What pressure must be exerted on the gas to reduce the volume to 0.20 L if the temperature is constant? 7. At 22°C, an inflatable ball is filled with 48 g of helium (He). The pressure indicated on the manometer is 87 kPa and the volume of the ball at this point is 303 L. What is the volume if the pressure is increased by 3 kPa without changing the temperature?

10. Hydrogen gas (H2) under pressure is used as a fuel in certain car prototypes. What volume does a sample of 28.8 L of hydrogen occupy when its pressure increases from 100 kPa to 350 kPa? 2.4.2 Relationship between volume and absolute temperature 11. Convert these temperatures into kelvin (K). a) 25°C b) 37°C c) 150°C 12. Convert these temperatures into degrees Celsius (°C). a) 373 K b) 98 K c) 425 K 13. A gas occupies a volume of 20.0 L at 10°C. What volume would it occupy at 200°C if the pressure is constant? 14. A 2.5-L rubber balloon is completely filled with helium (He) at a temperature of 24.2°C. The balloon it taken outdoors on a cold winter day (-17.5°C). What is the volume of the balloon if the pressure is constant? 15. At 20°C, 10.0 L of neon (Ne) is left to dilate until the gas attains a volume of 30.0 L. If the pressure is constant, what is the final temperature of the gas, in degrees Celsius (°C)?

8. A cylinder with a mobile piston contains 0.75 L oxygen (O2) gas at 101.3 kPa. The piston is moved, and the gas is compressed to a volume of 0.50 L. What is the final pressure exerted on the oxygen if the temperature is constant?

16. A syringe with a maximum volume of 60.0 cm3 takes in a 14.5-cm3 sample of oxygen (O 2) gas at 24.3°C. What maximum variation in temperature does the oxygen undergo before the piston is completely pushed out of the syringe?

9. A student produces 38.3 mL of oxygen (O2) gas in a burette. The following day, the burette contains 40.2 mL of gas at a pressure of 103 kPa. a) What was the pressure the day before? b) What is particular about the weather in this region?

17. The carbon dioxide (CO2) produced by yeast makes bread dough rise before it is put in the oven and dilates during baking. What is the final volume of 0.1 L of carbon dioxide contained in dough that, as it bakes, increases in temperature from 25°C to 98°C?

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18. A given mass of chlorine (Cl2) occupies a volume of 30 mL at 20°C. What is its volume at 45°C if the pressure is the same? 19. At STP, a gas occupies 0.5 L. If the volume of the gas is increased to 0.6 L, to what temperature must it be heated if the pressure is to remain the same? 20. To cool a computer microprocessor, nitrogen (N2) in its liquid state is sometimes used, since its temperature is -196°C. However, liquid nitrogen becomes gaseous fairly quickly and envelops the microprocessor. A sample of nitrogen gas enveloping a microprocessor occupies a volume of 300 mL at 17°C and 100 kPa. What volume will the nitrogen occupy at 100°C if the pressure is constant? 21. A ball is filled with a given mass of hydrogen (H2) at 22°C. The volume of the ball is 18.4 L at this point. The ball is then submerged in water and its volume increases by one quarter. What is the temperature of the water? 2.4.3 Relationship between pressure and temperature 22. In a metal ball at 30°C, the pressure of a gas is 90 kPa. What temperature will the thermometer display if the pressure increases to 110 kPa? 23. In a rigid ball at 30°C, the pressure of a gas is 90 kPa. What temperature will the thermometer display if the pressure decreases by 10 kPa? 24. At 25°C, a gas has a pressure of 700 mm Hg. If the container is heated to 50°C, what is the final pressure exerted by the gas? 25. At STP, a gas occupies 0.50 L in a rigid ball. What pressure will the manometer display if the ball is submerged in a container filled with water at 20°C? 26. A gas enclosed in copper tubes acts as a coolant in a small freezer. The gas exerts a pressure of 110 kPa at 45°C. The gas is left to dilate in an expansion chamber so that the pressure exerted decreases to 89 kPa. What is the temperature inside the freezer? 27. To what temperature must a gas in a rigid container be heated in order to double the pressure it exerts at 20°C? 28. A tank is made to contain a given volume of butane (C4H10) gas at 150 kPa and 35°C. For security purposes, in case the pressure increases to dangerous levels, a release valve is added that opens at 250 kPa. At what temperature, in degrees Celsius (°C), will this valve open?

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UNIT 1 Gases

29. An unknown gas is collected in a 250-mL container which is then sealed. Using electronic instruments, the gas inside the container is found to exert a pressure of 135.5 kPa at 15°C. What pressure would the gas exert if the temperature in kelvin (K) is doubled? 30. At 18°C, a sample of helium (He) gas stored in a metal cylinder exerts a pressure of 17.5 atm. What will the pressure be if the container is placed in a closed room where the temperature is 40°C? 2.4.4 Relationship between the volume and the quantity of a gas expressed in number of moles 31. Under certain conditions, 4 mol of a given gas can be put in a 70-L container. What must the volume of the container be in order to hold 6 mol of gas under the same temperature and pressure conditions? 32. A balloon is inflated by introducing 5.2 mol of helium (He) at 25°C. The volume of the balloon is 6.7 L at this point. If 2.2 mol of helium is added, what is the volume of the balloon if the temperature does not change? 33. At 20°C, a balloon is filled with 16 g of helium (He) under a pressure of 122 kPa. The volume of the balloon increases to 81 L. What is the volume if 16 g of oxygen (O2) is added and neither pressure nor temperature change? 34. A tire has an initial volume of 35 L and contains 3.3 mol of air. It is deflated and its final volume is 29 L. How many molecules of air escaped from the tire? 35. An elastic rubber balloon is inflated with 0.0784 mol of carbon dioxide (CO2) until it reaches a volume of 1.76 L at STP. How many moles of carbon dioxide must be added to the balloon for its volume to attain 1.98 L under the same conditions? Explain what would happen if, instead of adding carbon dioxide to the already inflated balloon, helium was added to increase the volume. 36. A fish’s swim bladder contains 1.32  10-4 mol of gas when it is close to the bottom of a river. To come up to the surface, the fish increases the volume of its swim bladder from 2  10-3 L to 3  10-3 L by diffusing oxygen (O2) from its circulatory system into its swim bladder. Assuming that the gas added to the bladder is pure oxygen, and that the pressure and temperature conditions are unchanged, how many grams of oxygen are there in the swim bladder when the fish reaches the surface?

2.4.5 Molar volume of gases 37. At STP, what is the volume of 3.45 mol of argon (Ar) gas? 38. Convert each of the following volumes of gas into moles: a) 5.1 L of carbon monoxide (CO) at SATP b) 20.7 mL of fluorine (F2) at STP c) 90 mL of nitrogen dioxide (NO2) at SATP 39. Weather balloons inflated with hydrogen (H2) can rise to almost 40 km altitude. On lift-off, at SATP, what volume does 7.5 mol of hydrogen gas occupy in a weather balloon? 40. Swamps, volcanoes, and crude oil and natural gas refineries release sulphur dioxide (SO2) gas. What quantity of this gas (in moles) is contained in a volume of 50 mL at SATP? 41. At low pressure, neon (Ne) gas emits red light that shines in illuminated signs. What volume does 2.25 mol of this gas at STP occupy before it is introduced into the tubes of a sign? 42. Hydrogen (H2) is the most abundant element in the universe. What volume does 500 mol of hydrogen occupy at SATP? 43. Hydrogen sulfide (H2S) is present in sour natural gas. Calculate the volume of 56 mol of hydrogen sulfide at SATP.

2.4.6 Relationship between pressure and quantity of gas expressed in number of moles 49. A tire containing 28 mol of air at a pressure of 150 kPa is inflated by pumping in an additional 7 mol of air. What is the new air pressure in the tire if temperature and pressure are held constant? 50. If 57 mol of gas enclosed in a rigid ball at a given temperature exerts a pressure of 235 mm Hg, how many moles of gas will exert a pressure of 100 mm Hg under the same conditions? 51. Using the kinetic theory of gases, explain why the pressure inside a propane tank increases during its filling while the pressure inside an elastic rubber balloon does not increase during its filling. 52. Calculate the value of the unknown variables in order to complete the following table. n1 (mol)

P2 (kPa)

100

0.003 50-

c)

85.3

200

74.5

a)

1.75

202

2.35

79.8

b)

145

6.72

300

25.8

P1 (kPa)

d)

n2 (mol) 0.002 50 e)

13.6

44. At SATP, what is the volume occupied by 2.02 g of hydrogen (H2)? 45. A balloon contains 2.00 L of helium (He) gas at STP. a) How many moles of helium are there? b) How many molecules of gas are there in 11.2 L at STP? 46. The bubble wrap that lines envelopes to protect their contents contains air at SATP. Each bubble contains 3.36  10-3 L of air. Assuming there is one air bubble per square centimetre, how many envelopes measuring 26 cm by 40 cm would have to be crushed for the air released from the bubbles to be equal to the molar volume of a gas at SATP? 47. Using the kinetic theory of gases, explain why the values of molar volume of a gas are different at STP and SATP. 48. Why does a hiker wishing to bring along one mole of water not require a bottle with a minimum capacity of 24.5 L?

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2.5 Ideal gas law The ideal gas law describes the interrelationships between the four variables that characterize a gas sample at a given time, namely, pressure (P), volume (V), absolute temperature (T) and quantity of gas (n) expressed in moles, and the gas constant. Simple gas laws, such as Boyle’s law, Charles’ law, Gay-Lussac’s law and Avogadro’s law, in theory apply only to “ideal” gases. An ideal gas does not exist in reality. An ideal gas is a hypothetical gas that, theoretically, obeys all of the gas laws, regardless of conditions. Moreover, the behaviour of an ideal gas corresponds to the hypotheses of the kinetic theory of gases (see the following page). For example, an ideal gas does not condense even when its temperature decreases to 0 K or it is subjected to very strong pressures. In contrast, when conditions of temperature and pressure conditions are very different from STP and SATP conditions, real gases stop resembling ideal gases and differences appear in their behaviour (see Table 4). Table 4 Ideal gases and real gases in light of a few hypotheses of the kinetic theory of gases Hypotheses of the kinetic theory of gases Hypothesis 1 The particles of a gas are infinitely small and the size of a particle is negligible compared to the volume of the container that holds the gas.

Hypothesis 2 The particles of a gas are in constant motion and move in a straight line in all directions.

Hypothesis 3 The particles of a gas do not exert any force of attraction or repulsion on each other.

Comparison with real gases at high and/or low temperatures When the pressure is high, the particles move much closer together and their size becomes important in relation to the volume they occupy. In other words, under such conditions of pressure, the size of the particles of real gases is not negligible.

When the particles of real gases collide, they lose a little energy. This means that the pressure of a real gas is in reality slightly weaker than that of an ideal gas.

As the temperature decreases, the particles slow down. At a given temperature, the particles attract each other and move considerably closer, and the gas becomes a liquid.

However, when real gases are studied under conditions similar to STP or SATP, they behave like ideal gases. This is why, under such conditions, their behaviour can be predicted using the gas laws, which make it possible to compare two of the four variables at a time. Nevertheless, to describe the interrelationships between the four variables at a given moment, it is more useful to combine the individual laws and create a single broader law.

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Boyle’s law According to Boyle’s law, volume is inversely proportional to pressure.

V

Charles’ law According to Charles’ law, volume is directly proportional to temperature.

VT

Gay-Lussac’s law According to Gay-Lussac’s law, pressure is directly proportional to temperature.

PT

Avogadro’s law According to Avogadro’s law, volume is directly proportional to the number of moles.

V n

Combination of the four laws The combination of the four laws indicates that volume is directly proportional to the product of the temperature and the number of moles by the inverse of pressure.

V 

1 P

1 Tn P

Mathematically, the proportionality symbol can be replaced by a proportionality constant (R) that includes the five constants of the simple gas laws: ka, kb, kc, kd and ke.

V R

1 P

 T  n or V 

nRT P

This reformulated expression becomes the ideal gas law. Ideal gas law PV  nRT where P  Pressure of the gas, expressed in kilopascals (kPa) V  Volume of the gas, expressed in litres (L) n  Initial quantity of the gas, expressed in moles (mol) R  Gas constant, expressed in (kPaL)/(molK) T  Temperature of the gas, expressed in kelvin (K) The value of the gas constant (R) can be determined based on the value of the molar volume at STP and by replacing the terms of the ideal gas equation with the appropriate values.

R

PV nT

R

101.3 kPa  22.4 L 1 mol  273 K

R  8.31 (kPaL)/(molK)

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The value of the gas constant is thus obtained. Gas constant R  8.31 (kPaL)/(molK) The ideal gas law is an equation that illustrates the interdependence between pressure, temperature, volume and number of moles of a gas at a given time. It is used to determine an unknown variable of a gas at a specific time when the value of the three other variables is known. When using the ideal gas law, it is essential to express the variables in the units of measure included in the unit of the constant R. The following example illustrates how the ideal gas law is used to obtain an unknown variable when the three others are known: Example When inflated to their maximum capacity, human lungs contain approximately 4.09 L of air at 37.0°C. How many moles of air do lungs contain if the air pressure is 100 kPa? Data: P  100 kPa

1. Conversion of the temperature into kelvin: T  37.0°C  273  310 K

V  4.09 L

2. Calculation of the number of moles: PV  nRT PV n RT 100 kPa  4.09 L  8.31 (kPaL)/(molK)  310 K

n? R  8.31 (kPaL) / (molK) T  37.0°C

 0.159 mol Answer: The lungs contain 0.159 mol of air.

The spirometer and pneumotachometer The spirometer is an instrument used to diagnose and monitor patients suffering from respiratory diseases like asthma. Tests conducted with a spirometer are used to evaluate pulmonary capacity, pulmonary volume and pulmonary flow. With nose blocked, the patient must blow into a tube connected to the spirometer. The instrument measures the volume of air exhaled and the speed of exhalation, which are then compared to normal values in order to make a diagnosis. To better plan their training, many elite athletes use a special type of spirometer, the pneumotachometer, to determine their maximum oxygen (O2) consumption, also called VO2 max. VO2 max is the maximum oxygen flow consumed during exertion and is expressed in millimetres per minute per kilogram (mL/(minkg)). This corresponds to the maximum volume of oxygen drawn from the lungs and used by the muscles during maximum aerobic exercise. The direct measurements are taken in a laboratory, while the person, connected to a pneumotachometer, performs intense aerobic exercise. The instrument measures the difference between the oxygen inhaled and the oxygen exhaled by the person. Among young, healthy individuals, the VO2 max level is around 45 mL/(min kg) for men, and 35 mL/(minkg) for women. Cross-country skiers tend to have the highest VO2 max value among athletes. For example, Norwegian Bjørn Daehlie, considered the greatest cross-country skier of all time, maintained, at his peak, a VO2 max of 90 mL/(minkg), which is double that of a normal individual.

102

UNIT 1 Gases

Figure 61 An athlete performs a VO2 max endurance test.

2.5.1

Molar mass of a gas

Under the same conditions of temperature and pressure conditions, the molar mass of a gas tells us about its density. For a given volume, some gases are denser and therefore heavier than others. For example, the molar mass of carbon dioxide (CO2) is approximately 1.5 times greater than that of air. This is why the vapour that escapes from a block of dry ice, composed of solid carbon dioxide, flows downward and along surfaces (see Figure 62). The molar mass (M) of an element or a gaseous compound can be determined the same way as for any element or compound: by adding the masses of the atoms. The molar mass of a gas can also be determined by dividing the mass of the gas (m) by its number of moles (n). This simple ratio indicates that the number of moles contained in a gas sample is the following: Figure 62 Dry ice is often used to create special effects because it moves to the ground rather than rising into the air, as most fumes do.

m M n m n M

It is also possible to determine the molar mass of a gas using the ideal gas law. By replacing the number of moles n by its value m/M in the ideal gas law equation, the following relationship is obtained:

PV  nRT PV 

mRT M

By rearranging the variables to isolate M, the formula for the molar mass of a gas is obtained.

Molar mass of a gas M

mRT PV

where M  Molar mass of the gas, expressed in grams per mole (g/mol) or in kilograms per mole (kg/mol) m  Mass of the gas sample, expressed in grams (g) or in kilograms (kg) R  Gas constant, expressed in (kPaL) / (molK) T  Absolute temperature, expressed in kelvin (K) P  Pressure of the gas, expressed in kilopascals (kPa) V  Volume of the gas, expressed in litres (L)

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The following example illustrates how to use this formula, derived from the ideal gas law, to determine the molar mass of a gas: Example What is the molar mass of a sample of an unknown gas if, at a temperature of 0°C and under a pressure of 102 kPa, a volume of 2.30 L of the gas weighs 4.23 g? Data: M? T  0°C P  102 kPa V  2.30 L m  4.23 g R  8.31 (kPaL)/(molK)

1. Conversion of the temperature into kelvin: T  0°C  273  273 K 2. Calculation of molar mass: mRT M PV 4.23 g  8.31 (kPaL) / (molK)  273 K   40.9 g/mol 102 kPa  2.30 L

Answer: The molar mass of the sample is 40.9 g/mol.

SECTION 2.5

Ideal gas law

1. Three 2-L containers hold, respectively, helium (He), oxygen (O2) and carbon dioxide (CO2) under a pressure of 104 kPa at 22°C. a) Which container contains the most molecules of gas? b) What is the mass of the gas contained in each of the containers? 2. What pressure will 32.0 g of methane (CH4) exert on the walls of a 5.00-L container at 20°C? 3. Calculate the number of moles of methane (CH4) gas contained in a sample with a volume of 500 mL at 35°C and 210 kPa? 4. At what temperature does 10.5 g of ammonia (NH3) exert a pressure of 85 kPa in a 30-L container? 5. What quantity of methane (CH4) gas is found in a sample with a volume of 500 mL at 35°C and 210 kPa? 6. Determine the pressure exerted in a 50-L compressed air cylinder if it contains 30 mol of air and is heated to 40°C. 7. What volume does 50 kg of oxygen (O2) gas occupy at a pressure of 150 kPa and a temperature of 125°C? 8. At what temperature does 10.5 g of ammonia (NH3) gas exert a pressure of 85 kPa in a 30-L container? 9. What is the molar mass of a gas if 2.6 L of this gas has a mass of 5.4 g at 100°C and 26.6 kPa?

104

UNIT 1 Gases

10. Calculate the mass of sulphur dioxide (SO2) contained in a 20-L cylinder exposed to the Sun at 40°C if the pressure exerted by the gas is 200 kPa. 11. In a 10-L container at a temperature of 249°C, an unknown gas exerts a pressure of 200 kPa. If the mass of the gas is 62 g, calculate its molar mass. 12. The following observations are used to identify a sample of pure gas: – Mass of the empty container  7.02 g – Mass of the container and the gas  9.31 g – Volume of the container  1.25 L – Temperature of the gas  23.4°C – Pressure of the gas  102.2 kPa a) Based on these observations, what is the molar mass of the gas in question? b) What gas could this be? 13. A sample of nitrogen (N2) gas occupies a volume of 11.2 L at 0°C at a pressure of 101.3 kPa. How many moles of nitrogen does this sample contain? 14. What is the volume of a weather balloon if it contains 10 mol of air at 75.5 kPa and 45°C? 15. In a room at 20°C, a 180-mL light bulb contains approximately 5.8  103 mol of argon (Ar). What is the pressure of the argon inside the bulb? 16. A 13.65-L container holds 0.75 mol of chlorine (Cl2) at 135 kPa. What is the temperature of the chlorine gas in degrees Celsius (°C)?

2.6 General gas law The general gas law establishes a relationship between the four variables that describe gases: pressure (P), volume (V), absolute temperature (T) and quantity of gas (n) expressed in moles. It can be used to predict the final conditions of a gas once its initial conditions have been modified. When the initial conditions of a gas change and take on new values, it is possible to compare the initial and final conditions using the ideal gas law. Therefore, even if all of the initial conditions of pressure, volume, absolute temperature and quantity of gas (P1, V1, n1 and T1) take on new values, called final values (P2, V2, n2 and T2), their ratio is always equal to the constant R. Using the ideal gas law, the following relationships are obtained: P1V1  n1RT1 and P2V2  n2RT2 P1V1 R n1T1

P2V2 R n2T2

By combining the two equivalences, the general gas law is obtained. General gas law P1V1 PV  2 2 n1T1 n2 T2 where P1  Initial pressure of gas, expressed in kilopascals (kPa) or millimetres of mercury (mm Hg) V1  Initial volume of gas, expressed in millilitres (mL) or litres (L) n1  Initial quantity of gas, expressed in moles (mol) T1  Initial temperature of gas, expressed in kelvin (K) P2  Final pressure of gas, expressed in kilopascals (kPa) or millimetres of mercury (mm Hg) V2  Final volume of gas, expressed in millilitres (mL) or litres (L) n2  Final quantity of gas, expressed in moles (mol) T2  Final temperature of gas, expressed in kelvin (K) The general gas law is an equation that allows two series of variables to be compared after a gas has undergone changes. It is also used to deduce all of the individual gas laws since it encompasses all of these laws. For example, using the ideal gas law, if the number of moles (n) and the pressure (P) are constant, they are eliminated from the general equation and Charles’ law is obtained. P 1 V1 P2 V 2  n1T1 n2 T2 V1 V  2 T1 T2

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The following example shows how the general gas law is used to determine the final conditions of a gas when the initial conditions change: Example At SATP, 0.150 mol of water vapour occupies a volume of 55.0 mL. What is the new temperature in degrees Celsius (°C) if 0.030 mol of water vapour is removed while increasing the pressure to 115.0 kPa and decreasing the volume to 40.0 mL? Data: P1  101.3 kPa

1. Conversion of the temperature into kelvin: T1  25°C  273  298 K

V1  55.0 mL

2. Calculation of the final number of moles: n2  n1  0.030 mol  0.150 mol  0.030 mol  0.120 mol

n1  0.150 mol T1  25°C P2  115.0 kPa

2. Calculation of the final temperature: P1V1 PV  2 2 n1T1 n2 T2

V2  40.0 mL n2  ? T2  ?

T2  

n1T1 P2V2  P1 V 1 n2 0.150 mol  298 K 115.0 kPa  40.0 mL  101.3 kPa  55.0 mL 0.120 mol

 307.5 K 3. Conversion of the final temperature into °C: T2  307.5 K  273  34.5°C Answer: The new temperature is 34.5°C.

Tire pressure A perfectly inflated tire on a parked car is an example of a gas sample that has the characteristics of constant pressure, temperature, quantity and volume. However, if the tire is flat and air is pumped into it, these parameters will change. By inflating it, the number of moles in the tire increases. The volume increases for a period and then stops increasing. As air continues to be injected into the tire, the pressure increases as the quantity of air increases. When the tire is set in motion at a pressure that is too low, the friction with the road is too great and generates more heat. This increases the temperature of the air inside the tire, which can make it explode since the volume of the tire is held relatively constant. This is why it is better to adjust the pressure, which, for most car tires, should be approximately 220 kPa, the equivalent of approximately 32 pounds of pressure on a pressure gauge. For a racing car competing in races such as Nascar, tires are inflated with nitrogen (N2) at a pressure of approximately 345 kPa.

106

UNIT 1 Gases

Figure 63 Adding air to a tire whose volume does not change increases the air pressure inside the tire.

SECTION 2.6

General gas law

1. At -200°C and 110 kPa, a given quantity of hydrogen (H2) occupies 400 mL. What temperature must the gas be brought to in order for the pressure to be 100 kPa and the volume 600 mL? 2. A gas occupies a volume of 100 L at 50°C and 75 kPa. What would be the volume of this gas at 17°C and 100 kPa? 3. A balloon filled with gas has a volume of 5.0 L at 20°C and 100 kPa. What would be its volume at 35°C and 90 kPa?

7. A car tire has a volume of 27 L at 225 kPa and 18°C. a) What is the volume of the tire when air leaks out and the pressure drops to 98 kPa? b) What volume would the remaining air occupy in the tire at SATP? 8. Methane (CH4) gas can be condensed by cooling it and increasing pressure. A 600-mL sample of methane gas at 25°C and 100 kPa is cooled to 20°C. The gas is compressed until its pressure has quadrupled. What is its final volume? 9. A gas sample has a volume of 150 mL at 260 K and 92.3 kPa. What is the new volume at 376 K and 123 kPa?

4. In a diesel engine cylinder, 500 mL of air at 40°C and 1 atm is highly compressed just before fuel injection, attaining a pressure of 35 atm. If the volume of air is reduced to 23 mL, what will the final temperature in the cylinder be?

10. In a large syringe, 48 mL of ammonia (NH3) gas is compressed at STP to obtain a volume of 24 mL at 110 kPa. What is the new temperature of the gas?

5. A 1.0-L helium (He) balloon floats in the air on a day when the atmospheric pressure is 102.5 kPa and the temperature is 25°C. Clouds suddenly appear. The pressure plummets to 98.6 kPa and the temperature drops to 20°C. What is the new volume of the balloon?

11. A 100-W bulb has a volume of 180 cm3 at STP. The bulb is lit and the heated glass dilates slightly, which increases the volume to 181.5 cm3. The internal pressure is 214.5 kPa. What is the temperature of the bulb in degrees Celsius (°C)?

6. Calculate the unknown quantities in the following table:

12. A gas sample is collected at 25°C. If the temperature of the gas triples and the pressure doubles, what fraction of the initial volume of gas will remain?

Tinitial (°C)

Tfinal (°C)

Pinitial (kPa)

Pfinal (kPa)

Vinitial (L)

Vfinal (L)

100

100

101.3

110

5

d)

0

0

1.5  102 c)

25

10

35

150

101.3

101.3

750

e)

85

b)

125.0

125

27

45.5

102.5

a)

85

99.5

35.5

25.5

86.7

1

f)

88.7

450

500

13. Sulphur hexafluoride (SF6) is used to perform magic tricks, like the one involving making an object float on invisible water. A 5.0-L sample of this gas is collected at 205°C and 350 kPa. What pressure must be applied to the gas sample to reduce its volume to 1.7 L if the temperature is 25°C?

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2.7 Stoichiometry of gases The stoichiometry of gases is a calculation method based on the ratios between the quantities of gas involved in a chemical reaction. This method is used to predict the quantity of a reactant or product involved in a chemical reaction in which at least one of the components is a gas. Gases take part in many chemical reactions. For instance, at home, the combustion of propane (C3H8) in a barbecue is a chemical reaction that allows heat to be generated in order to cook food. By studying other chemical reactions involving gases, Gay-Lussac formulated a fundamental law of chemistry: the combined gas law. This law states that when gases react, the volumes of the reactants and the products, measured at constant temperatures and pressures, are always present in the form of simple whole number ratios. Based on these observations, Avogadro put forward the hypothesis that equal volumes of different gases contain the same number of particles when they are exposed to the same conditions of temperature and pressure. Combining the ideas of Gay-Lussac and Avogadro solves certain problems related to the stoichiometry of gases. For example, if a reaction occurs between two volumes of the same gas and a volume of another gas at a given temperature and pressure, it can be stated that two particles of the first gas react with one particle of the second gas. The reaction between nitrogren (N2) and hydrogen (H2), which produces ammonia (NH3), is an example of this law (see Figure 64).

Nitrogen (N) Hydrogen (H)

Coefficients Molar ratio Volume ratio at STP Volume ratio at SATP Volume ratio

N2 (g)



3 H2 (g)

n

2 NH3 (g)

1 1 22.4 L 24.5 L 1

: : : :

3 3 67.2 L 73.5 L 3

: : : :

2 2 44.8 L 49.0 L 2

Figure 64 When a volume of nitrogen (N2) reacts with three volumes of hydrogen (H2) to produce two volumes of ammonia (NH3), this means that each time one molecule of nitrogen reacts with three molecules of hydrogen, two molecules of ammonia are produced.

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Stoichiometric calculations are used to determine the quantities of reactants needed in a reaction and to predict the quantity of the product obtained following a reaction. These calculations apply to all chemical reactions. They calculate the number of moles and the volume or mass of a reactant or product, regardless of the phase in which it is found. However, when stoichiometric calculations are applied to gases, the conditions of temperature and pressure must not change. Using molar mass values at STP and SATP can simplify calculations involving the quantities of gas consumed or produced. The following example provides a procedure for carrying out stoichiometric calculations: Example Propane (C3H8) burns in the air according to the following equation: C3H8 (g)  5 O2 (g) n 3 CO2 (g)  4 H2O (g) a) What volume of oxygen (O2) is needed for the combustion of 35.0 L of propane if the volumes are measured at the same conditions of temperature and pressure? b) What volume of carbon dioxide (CO2) will be produced if 155 g of propane react with enough oxygen at SATP? a) Data: VO  ? 2

VC H  35.0 L 3 8

Coefficient ratios between the reactants and products and reporting data at STP: C3H8 (g)  5 O2 (g) n 3 CO2 (g)  4 H2O (g) 1 5 35.0 L ? 5  35.0 L 1  175 L

VO  2

Answer: The volume of oxygen needed is 175 L. b) Data: VCO  ? 2

MCO  44.009 g/mol 2

mC H  155 g 3 8

nC H  ? 3 8

1. Calculation of the number of moles of propane: m  M 155 g  44.009 g/mol n  3.52 mol 2. Coefficient ratios between the reactants and the products and reporting data at SATP: C3H8 (g)  5 O2 (g) n 3 CO2 (g)  4 H2O (g) 1 3 3.52 mol ? VCO  3  3.52 mol  24.5 L/mol 2

 258.7 L Answer: The volume of carbon dioxide (CO2) produced is 259 L.

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SECTION 2.7

Stoichiometry of gases

1. The combustion of ammonia (NH3) by oxygen (O2) is described by the following unbalanced equation: NH3 (g)  O2 (g) n N2 (g)  3 H2O (g) a) Balance the equation. b) Determine the mass of oxygen needed to completely burn 16.0 mol of ammonia. 2. Determine the mass in grams of nitrogen (N2) produced when 2.72 mol of hydrazine (N2H4) are consumed after having balanced the equation of the following reaction: N2H4 (l)  N2O4 (l) n N2 (g)  H2O (g) 3. A barbecue burns 1.5 L of propane (C3H8) gas. This combustion is expressed by the following equation: C3H8 (g)  5 O2 (g) n 3 CO2 (g)  4 H2O (g) a) What is the volume of carbon dioxide (CO2) gas produced? b) What is the volume of oxygen (O2) used? 4. Engineers design an inflatable cushion that is deployed almost instantly on impact. In order for the device to work, the inflatable cushion must receive a large quantity of gas in a very short amount of time. Many car manufacturers use solid sodium azide (NaN3) with the appropriate catalysts to supply the gas needed to inflate the cushion. The following balanced equation represents this reaction: 2 NaN3 (g) n 2 Na (g)  3 N2 (g) a) What is the volume of nitrogen (N2) gas produced if 117 g of sodium azide is stored in the steering wheel at 20.2°C and 101.2 kPa? b) How many molecules of nitrogen are contained in this volume? 5. Use the following equation to answer the following questions: SO2 (g)  O2 (g) n SO3 (g)

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a) Balance the equation. b) At 100°C, 12 L of sulphur trioxide (SO3) is produced. What is the volume of oxygen (O2) used? 6. In a barbecue, 35 g of propane (C3H8) gas are burned according to the following equation: C3H8 (g)  5 O2 (g) n 3 CO2 (g)  4 H2O (g) All gases are measured at SATP. a) What is the volume of water vapour produced? b) What is the volume of oxygen (O2) used? 7. The following chemical equation describes what occurs when a match is struck on a rough surface to produce light and heat: P4S3 (s)  O2 (g) n P4O10 (g)  SO2 (g) a) Balance this chemical equation. b) How many litres of sulphur dioxide (SO2) are produced if 5.3 L of oxygen (O2) gas are consumed? 8. Ethanol vapour (C2H5OH) burns in air according to the following equation: 2 C2H5OH (g)  6 O2 (g) n 4 CO2 (g)  6 H2O (l) a) If 2.5 L of ethanol burn at STP, what volume of oxygen (O2) is required? b) What volume of carbon dioxide (CO2) will be produced? 9. Nitrogen monoxide (NO) is one of the gases that cause smog. It is produced in different ways, especially during the combustion of ammonia (NH3). 4 NH3 (g)  5 O2 (g) n 4 NO (g)  6 H2O (l) What mass of nitrogen monoxide is produced if 25.0 L of ammonia react with 27.5 L of oxygen (O2) at STP?

2.8 Dalton’s law According to Dalton’s law, also known as the law of partial pressures, at a given temperature, the total pressure of a mixture of gases equals the sum of the pressure of each of the gases. John Dalton studied weather conditions, in particular, the composition of the atmosphere. The many experiments he conducted on air allowed him to estimate its composition at approximately 79% nitrogen (N2) and 21% oxygen (O2). He observed that atmospheric pressure is the sum of the pressure exerted by nitrogen and oxygen. Dalton concluded that since the atmosphere is mostly composed of nitrogen and oxygen, on days when the atmospheric pressure is 100 kPa, for example, the pressure of nitrogen is 79%  100 kPa  79 kPa and that of oxygen is 21%  100 kPa  21 kPa.

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Connection with the kinetic theory of gases According to Hypothesis 3 of the kinetic theory of gases (see page 61 ), gas particles do not exert any force of attraction or repulsion on each other. By pursuing his research on other mixtures of gases, Dalton established that the pressure exerted by each gas in a mixture is equal to the pressure it would exert if it were present alone in the same volume. In other words, each of the gases in a container maintains its physical properties and behaves independently of the others. Consequently, each of the gases in a mixture contributes to the total pressure of the mixture based on its percentage in the mixture.

The individual pressure exerted by each of the gases in a container is called partial pressure. In effect, it represents only a portion of the total pressure exerted by the mixture of all of the gases in the container (see Figure 65).

Mixture of gas A and gas B Gas A

Gas B Lowered piston

Closed valve

PA  P T  50 kPa nA  0.30 mol

Open valve

PB  P T  100 kPa nB  0.60 mol

PT  P A  P B  150 kPa nT  0.90 mol

Figure 65 When gases A and B are mixed without changing their volume, the collisions of the particles of gas A, which cause partial pressure, are added to those of gas B.

JOHN DALTON British chemist and physicist (1766–1844) John Dalton was interested in meteorology before he put forth the first description of a visual abnormality from which he himself suffered and that was later named after him—daltonism (also referred to as colour-blindness). Later, his research on the physical properties of atmospheric air and other gases prompted him to formulate his atomic theory. At the time, this theory, which proposed an atomic model that resembled a complete sphere, dispelled the belief that it was possible to chemically transmute metals into gold. In particular, Dalton’s research on gases led him to formulate the law of partial pressures of gases in a mixture. This law was named after him and is called Dalton’s law.

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At first, when the valve is closed, gas A exerts a pressure of 50 kPa on the piston, and the pressure of gas B in the tank is 100 kPa. When the valve is opened and gas A and gas B are added by lowering the piston, the total pressure exerted by the mixture equals the sum of the partial pressures of the two gases, that is, 150 kPa, provided that the volume of the container is held constant. The total number of collisions of the two gases, therefore, produces a total pressure that equals the sum of the partial pressures of each of the gases in the mixture. Based on these observations, Dalton proposed the law that is known today as Dalton’s law, which is expressed as follows: Dalton’s law At a given temperature, the total pressure of a gas mixture equals the sum of the partial pressures of all of the gases in the mixture. This law can be applied to any gas mixture, regardless of the number of gases that constitute the mixture. This is why the following equation ends with an ellipsis: Dalton’s law PT  PA  P B  PC  … where PT  Total pressure of the mixture, expressed in kilopascals (kPa) or millimetres of mercury (mm Hg) PA  Partial pressure of gas A, expressed in kilopascals (kPa) or millimetres of mercury (mm Hg) PB  Partial pressure of gas B, expressed in kilopascals (kPa) or millimetres of mercury (mm Hg) PC  Partial pressure of gas C, expressed in kilopascals (kPa) or millimetres of mercury (mm Hg)

A manometer cannot measure the partial pressure of a gas in a mixture. It can only measure the total pressure of the mixture. To determine the partial pressure of one of the gases in a mixture, the molar proportion of each gas (which is equivalent to its percentage) is multiplied by the total pressure of the mixture. The molar proportion of a gas is calculated by establishing the ratio of its number of moles to the total number of moles of the mixture. Partial pressure of a gas PA 

nA P nT T

where PA  Partial pressure of gas A, expressed in kilopascals (kPa) or millimetres of mercury (mm Hg) nA  Quantity of gas A, expressed in moles (mol) nT  Total quantity of gas, expressed in moles (mol) PT  Total pressure of the mixture, expressed in kilopascals (kPa) or millimetres of mercury (mm Hg)

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The following example demonstrates how Dalton’s law and the formula for partial pressures are used to solve problems: Example At a given temperature, a mixture of gas contains 3.35 mol of neon (Ne), 0.64 mol of argon (Ar) and 2.19 mol of xenon (Xe). What is the partial pressure of xenon if the total pressure of the mixture is 200.0 kPa? Data: nNe  3.35 mol nAr  0.64 mol nXe  2.19 mol nT  ? PXe  ? PT  200 kPa

1. Calculation of the total number of moles: nT  3.35 mol  0.64 mol  2.19 mol  6.18 mol 2. Calculation of the partial pressure of xenon: n PXe  Xe  PT nT 2.19 mol  200 kPa 6.18 mol  70.9 kPa 

Answer: The partial pressure of xenon (Xe) is 70.9 kPa.

SECTION 2.8

Dalton’s law

1. To accelerate a reaction in a container under a pressure of 98 kPa, a chemist adds hydrogen (H2) gas at 202.65 kPa. What is the resulting pressure? 2. A mixture of neon (Ne) and argon (Ar) is collected at 102.7 kPa. If the partial pressure of the neon is 52.5 kPa, what is the partial pressure of the argon? 3. A gas mixture contains 12% neon (Ne), 23% helium (He) and 65% radon (Rn). If the total pressure is 116 kPa, what is the partial pressure of each of the gases? 4. The partial pressure of argon (Ar), which constitutes 40% of a mixture, is 325 mm Hg. What is the total pressure of the mixture in kilopascals (kPa)? 5. A mixture of gas in a cylinder contains 0.85 mol of methane (CH4), 0.55 mol of oxygen (O2), 1.25 mol of nitrogen (N2) and 0.27 mol of propane (C3H8). The manometer indicates a pressure of 2573 kPa. What pressure does each gas exert on the cylinder?

6. The pressure of a mixture of nitrogen (N2) and carbon dioxide (CO2) is 1 atm, and its temperature is 278 K. If 30% of the mixture is nitrogen, what is the partial pressure of the carbon dioxide? 7. A warehouse used to preserve fruit has a controlledatmosphere room with a total pressure of 102.6 kPa. The gases present in this room are carbon dioxide (CO2), which occupies 3.5% of the total volume of the room, oxygen (O2), which occupies 4.0 % of the total volume of the room and water vapour, which occupies 92.5% of the total volume of the room. Calculate the partial pressure of each of these gases. 8. A hermetically sealed container with a total pressure of 1000 mm Hg contains three gases: oxygen (O2), hydrogen (H2) and chlorine (Cl2). If the oxygen has a partial pressure of 125 mm Hg and the hydrogen has a partial pressure of 235 mm Hg, calculate the percentage of chlorine in the container.

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APPLICATIONS Gasoline engines and diesel engines Whether an engine runs on gasoline or diesel, it transforms the chemical energy of fuel into mechanical energy through combustion, a chemical reaction between the fuel and the oxygen (O2) in the air when exposed to heat. In a four-stroke gasoline engine, a gaseous mixture of air and fuel is drawn into the cylinder when the piston descends (see Figure 66). When the piston moves back up, it compresses the gaseous mixture, thereby increasing its temperature to 300°C. When the piston reaches the top and compression is at its maximum, a spark plug produces a spark that ignites the mixture of air and fuel. The combustion of gasoline produces gases such as water vapour and carbon dioxide (CO2) and a lot of heat. The pressure exerted by this large quantity of gas at very high temperature in a small volume drives the piston back down with tremendous force. As it moves downward, the piston causes a continuous rotation movement of the crankshaft, which is transmitted to the vehicle’s wheels by way of the transmission. The rotation of the crankshaft causes the piston to move upward, which 1. Intake

2. Compression

pushes the burned gases in the cylinder out through exhaust valves. When the piston moves back up, it can once again draw in a mixture of air and fuel. The basic principle of a diesel engine is slightly different since it is based on Gay-Lussac’s law, according to which the pressure and temperature of a gas are proportional. At first, it is only air, without fuel, that is drawn into the cylinder (see Figure 67). As in a gasoline engine, the piston compresses the air, but, in this case, compression is higher and increases the temperature of the air to over 800°C. Diesel is then injected into the cylinder. Since the temperature is higher than the ignition point of this fuel, the fuel ignites without requiring a spark plug. The combustion produces hot gases at high pressure that force the piston down. A diesel engine produces fewer greenhouse gas emissions than a gasoline engine of equal power. However, it produces more polluting particles, such as soot. A more ecological choice is to use biodiesel, which is a renewable fuel. 3. Combustion Spark plug

4. Exhaust

Air-gasoline mixture

Residue

Rotation movement

Figure 66 A gasoline engine requires a spark plug. 1. Intake

2. Compression Injector

Air

Rotation movement

Figure 67 A diesel engine does not require a spark plug.

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UNIT 1 Gases

3. Combustion

4. Exhaust

Injector

Residue

Dirigible balloons During the 19th and 20th centuries, air transportation experienced rapid growth as humans set out to conquer the sky. In order to achieve flight, pioneers like the Montgolfier brothers based their work on the phenomenon in which a gas with a lower density than air tends to rise above air. It is fairly easy to obtain a gas less dense than air by heating the gas in order to distance its particles from each other. This technique, employed in a hot air balloon, is a direct application of Charles’ law, according to which the volume of a gas is proportional to its temperature. However, since the hot air balloon can only be controlled vertically and its trajectory is determined by the winds, it is not a practical mode of transportation. Dirigible balloons, as their name implies, were easier to manoeuvre. In 1852, a dirigible successfully reached the sky thanks to a steam engine and its balloon filled with hydrogen (H2), an inexpensive gas. A few years later, several models of soft or semi-soft balloons using electric and internal combustion engines were introduced. Instead of using heat to reduce the density of air, these balloons used gases that have a lower density than air such as hydrogen or helium (He), two gases composed of particles lighter than those of air. In the early 20th century, German Count Ferdinand Graf von Zeppelin developed dirigibles made with rigid balloons in aerodynamic shapes, which were named after him (see Figure 68).

Figure 68 The LZ 127 Zeppelin in a hangar in New Jersey

Zeppelins were used as bomber planes during the First World War but without much military success. They then became widespread and were even used for transatlantic flights between the United States and Germany. This journey was a luxury at the time and many people imagined that the dirigible would become the mode of transportation of the future. During its construction in 1930, the Empire State Building (in New York) was intended to be designed with a dock for dirigibles (see FigFigure 69 A helium (He) ure 69). This project, howdirigible next to the Empire State ever, was abandoned be- Building in New York City in the cause of the risks involved. 1930s Hydrogen is a highly flammable gas and it caused many fires and explosions. Hydrogen was gradually replaced by helium in the 1930s. Since dirigibles are voluminous and slow-moving, they are not widely used today. They are mainly used for scientific research or advertising campaigns (see Figure 70).

Figure 70 View of a football stadium from a dirigible used in an advertising campaign

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CHAPTER

2

Physical Properties of Gases

2.1 Kinetic theory of gases • Particles of matter exhibit three types of movement: vibrational motion, rotational motion and translational motion. • The three types of particle movement are found in gases. • The kinetic energy (Ek) of a particle of gas depends on its mass (m) and velocity (v). 1 Ek  mv2 2 • Maxwell’s distribution curve demonstrates that at high temperatures, there is a greater probability of gas particles having a high velocity.

• The kinetic theory of gases is based on the following hypotheses: 1. The particles of a gas are infinitely small, and the size of a particle is negligible compared to the volume of the container that holds the gas. 2. The particles of a gas are in constant motion and move in a straight line in all directions.

vpp

Number of particles

• The higher the temperature of a gas, the faster its particles move. This is why the mean kinetic energy of particles increases as the temperature of the gas rises.

Number of particles of nitrogen gas as a function of their velocity at three different temperatures

Low temperature vpp Intermediate temperature vpp High temperature

500 1 000 Molecular velocity (m/s)

3. The particles of a gas do not exert any force of attraction or repulsion on each other. 4. The mean kinetic energy of the particles of a gas is directly proportional to the absolute temperature.

2.2 Behaviour of gases • Gases are compressible, which means that they can expand and disperse through diffusion and effusion. • According to Graham’s law, under identical conditions of temperature and pressure, light gases diffuse more quickly than heavy gases. • Graham’s law establishes a relationship between the rates of diffusion of two gases (v1 and v2) as a function of their molar mass (M1 and M2). v1  v2

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UNIT 1 Gases

M2 M1

1 500

2.3 Pressure of gases • Gases exert pressure on all of the surfaces they come into contact with, in all directions. • The pressure of a gas on an object therefore depends on the sum of the forces exerted by the collisions of its particles on the surface of the object. • Atmospheric pressure is the force exerted by the weight of air on objects that are in contact with the air. Normal atmospheric pressure can be represented by this equivalence: 101.3 kPa  760 mm Hg  1 atm. • Manometers are instruments used to measure pressure.

2.4 Simple gas laws • The simple gas laws help to solve problems that establish a relationship between two of the four variables that describe gases: pressure (P), volume (V), absolute temperature (T) and quantity of gas (n) expressed in number of moles while the other two variables remain constant. • To facilitate the comparison of gases, the following standards are used: – standard temperature and pressure (STP): 0°C and 101.3 kPa – standard ambient temperature and pressure (SATP): 25°C and 101.3 kPa • According to Boyle’s law, at constant temperature, the volume occupied by a given quantity of gas is inversely proportional to the pressure of the gas. P1V1  P2V2 • According to Charles’ law, at constant temperature, the volume occupied by a given quantity of gas is directly proportional to the absolute temperature of the gas. V1 V  2 T1 T2 • According to Gay-Lussac’s law, at constant volume, the pressure of a given quantity of gas is directly proportional to the absolute temperature of the gas. P1 P  2 T1 T2 • According to Avogadro’s law, under the same conditions of temperature and pressure, the volume of a gas is directly proportional to its number of moles. V1 V  2 n1 n2 • The molar volume of a gas is the volume occupied by one mole of any gas, under specific conditions of temperature and pressure. The molar volume of a gas is 22.4 L at STP and 24.5 L at SATP. • Under the same conditions of temperature and volume, the pressure of a gas is directly proportional to its number of moles. P1 P  2 n1 n2

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2.5 Ideal gas law • An ideal gas is a hypothetical gas that, in theory, obeys all of the gas laws regardless of conditions and whose behaviour corresponds to the hypotheses of the kinetic theory of gases. • The four variables that describe a gas sample at a given time are pressure (P), volume (V), absolute temperature (T) and quantity of gas (n) expressed in moles. The ideal gas law is used to predict the interrelationship between these four variables and the gas constant (R). PV  nRT where R  8.31 (kPaL) / (molK) • The molar mass of a gas can be determined using the following formula derived from the ideal gas law: mRT M PV

2.6 General gas law • The general gas law establishes a relationship between the four variables of a gas: namely, pressure (P), volume (V), absolute temperature (T) and quantity of gas (n) expressed in moles. This law is used to predict the final conditions of a gas after its initial conditions have been changed. P1V1 P V  2 2 n1T1 n2T2

2.7 Stoichiometry of gases • The stoichiometry of gases is used to determine the quantity of a reactant needed in a reaction and to predict the quantity of the product obtained following a reaction in which at least one of the components is gaseous.

2.8 Dalton’s law • Dalton’s law, also known as the law of partial pressures, is described as follows: at a given temperature, the total pressure of a gas mixture is equal to the sum of the partial pressure of each of the gases. PT  PA  PB  PC  … • The following equation can be used to determine the partial pressure of a gas in a mixture: n PA  A  PT nT

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UNIT 1 Gases

CHAPTER 2

Physical Properties of Gases

1. In a closed-end manometer, the difference in height of the columns of mercury equals 522 mm Hg. What is the pressure of the gas sample in kilopascals (kPa)? 2. An open-end manometer is connected to a container filled with carbon dioxide (CO2). The level of the mercury is higher by 22 mm on the open side. Determine the pressure exerted by the carbon dioxide if the atmospheric pressure equals 100.3 kPa. 3. When atmospheric pressure decreases, in what direction does the mercury move inside the open section of an open-end manometer? 4. In a closed-end manometer, it is never specified which of the two sections of the mercury tube is higher. Why? 5. In the manometer represented below, what is the difference in pressure, in kilopascals (kPa), between the atmosphere and the gas contained in the ball? Determine the pressure of this gas if the atmospheric pressure equals 102.3 kPa. What would happen if the gas pressure also equalled 102.3 kPa?

Gas

Patm

(cm)

80 70 60 50 40 30 20 10

7. Hydrogen (H2) gas occupies a volume of 500 dm3, at 125°C. If the pressure remains constant, what volume will the hydrogen occupy at 25°C? 8. A sample of natural gas occupies a volume of 350 mL at 20°C. The pressure remains constant and the temperature is increased until the final volume of the natural gas is 385 L. What is the final temperature of the gas in degrees Celsius (°C)? 9. The production of vapour during cooking explains in part why bread and cakes rise. What volume of water vapour is formed in a cake when 1.0 g of water vaporizes at 98°C and 103 kPa? 10. Large quantities of chlorine (Cl2) gas derived from salt are used to make bleach and to purify water. Calculate the volume of 26.5 mol of this gas at 400 kPa and 35°C. 11. Freon gas is a form of chlorofluorocarbon (CFC) formerly used as a coolant in air conditioners and refrigerators. If 500 mL of freon is compressed at 1.5 atm and 24°C in order to obtain 250 mL at 3.5 atm, what would the final temperature of this gas be?

90

Pgas

6. A sample of chlorine (Cl2) gas is used as a disinfectant in a pool. The sample occupies a volume of 500 mL in a cylinder equipped with a mobile piston. The piston injects the gas into the water as needed. The gas is trapped under a pressure of 25 atm, at normal temperature. If the temperature remains constant and the piston compresses the gas to 220 mL, what is the pressure inside the cylinder?

h

12. In a power plant, mechanical energy is obtained by using vapour under pressure to activate turbines. These rotating devices are connected to an electrical generator. The vapour penetrates a turbine at a high temperature and pressure, then exits, still in a gaseous state, at a lower temperature and pressure. Determine the final pressure of a sample of vapour of 10 kL, at 600 kPa and 150°C, that increases to 18 kL at 110°C.

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13. Convert the following temperatures into kelvin (K): a) 45°C b) 67°C c) 350°C

23. A sample of 75.3 mL of oxygen (O2) at 25.7°C is cooled until it occupies a final volume of 10 L. If pressure is constant, what is the final temperature in degrees Celsius (°C)?

14. Convert the following temperatures into degrees Celsius (°C): a) 473 K b) 108 K c) 225 K

24. A truck leaves Drummondville in early January when the temperature is -20°C. The truck’s tires are inflated to 210 kPa. Four days later the truck arrives in California where the temperature is 30°C. What is the air pressure in the tires when the truck reaches its destination?

15. Determine the volume occupied by 1.0 g of carbon dioxide (CO2) gas trapped in bread dough at SATP. 16. Gas chromatography is a technique used to separate the molecules in a gas mixture. Argon (Ar) is a noble carrier gas that displaces other gases in the mixture through a column specially designed for this purpose. What volume does 4.2 g of argon gas occupy at SATP? 17. Answer the following questions: a) A gas at 107 kPa and at 300 K is cooled to 145 K and maintains the same volume. What is the new pressure? b) A gas occupies a volume of 17 L. The temperature of this gas decreases from 300 K to 145 K, and pressure remains the same. What is the new volume of the gas? 18. Why is there a chance that a fully inflated balloon, which is floating above a hot stove, may pop? 19. The pressure exerted on 0.25 L of nitrogen (N2) is 120 kPa. What volume will this gas occupy at 60 kPa if the temperature and number of moles are constant? 20. In a large party balloon, 25 g of dry air at 20°C occupy a volume of 20 L. If the temperature increases to 40°C at a constant pressure, what volume will the balloon occupy? 21. In a house at 24.2°C, a 2.5-L elastic rubber balloon is completely filled with helium (He). The balloon is taken outside on a cold winter day at -17.5°C. What is the volume of the balloon if pressure is constant? 22. At 20°C, 10 L of neon is left to dilate until the gas occupies a volume of 30 L. If the pressure is constant, what is the final temperature in degrees Celsius (°C)?

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25. Sulphur hexafluoride (SF6) can be used as electric insulation in high-voltage instruments in electrical substations. A 5.0-L sample of this gas is collected at 205°C and 350 kPa. What pressure must be exerted on the gas sample to reduce its volume to 1.7 L if the temperature is 25°C? 26. A sample of nitrogen (N2) gas has a volume of 10 L at 101.3 kPa and 20°C. To preserve strawberries, the nitrogen is cooled to -190°C, which comes close to the temperature of liquid nitrogen. The gas pressure remains at 101.3 kPa. What volume will the nitrogen occupy at this new temperature? 27. A birthday balloon contains 2.0 L of air at STP. What will be its volume at SATP? 28. A 250-mL glass container is filled with krypton (Kr) gas at a pressure of 700 mm Hg at 25°C. If the container has been made to withstand a pressure of 2 atm, to what maximum temperature, in degrees Celsius (°C), can it be safely heated? 29. Certain electric light bulbs are filled with argon (Ar) gas because it is unreactive. At normal temperature and pressure, 650 cm3 of argon gas is heated in order to double its volume. The final pressure of the gas is 101.3 kPa. What is its final temperature in degrees Celsius (°C)? 30. Neon (Ne) is commonly used as a luminescent gas in illuminated signs. A neon sample has a volume of 5.5 L at 750 mm Hg and 10°C. If the gas is dilated to obtain a volume of 7.5 L at a pressure of 400 mm Hg, what is its final temperature in degrees Celsius (°C)? 31. Oxygen (O2) destined for a school laboratory is stored in a 2.0-L pressurized cylinder. The cylinder contains 25 g of oxygen at 20°C. What is the pressure of the oxygen in the cylinder?

32. Halogen light bulbs are generally filled with bromine (Br2) or iodine (I2) vapours at a pressure of 5 atm. When the light is turned on, the glass bulb can heat to more than 1150°C. If the ambient temperature is 20°C, what will the pressure in the bulb be when it attains 1150°C? 33. A diver swims 30 m below the surface of Lake Memphremagog. At this depth, the water pressure is 4 atm and the temperature is 8°C. A water bubble with a volume of 5.0 mL escapes from the diver’s mask. What is the volume of the bubble when it bursts at the water’s surface? Assume that atmospheric pressure is 101.3 kPa and the water temperature at the lake’s surface is 24°C. 34. A 1.56-L gas sample has a mass of 3.22 g at 100 kPa and 281 K. What is the molar mass of the gas? 35. Two litres of halothane have a mass of 14.1 g at 344 K and 1.01 atm. What is the molar mass of the halothane? 36. A gas has a mass of 0.548 g and a volume of 237 mL at 373 K and 755 mm Hg. What is the molar mass of this gas? 37. What volume does 2.0 mol of oxygen (O2) occupy at 30°C and 750 mm Hg? 38. What volume does 15 g of nitrogen (N2) gas occupy at 100°C and 5 atm? 39. Pressurized carbon dioxide (CO2) is used to manufacture soft drinks. a) How many grams of carbon dioxide are there in a 500-cm3 container at –50°C and 2 atm? b) How many grams of oxygen (O2) can the container hold at the same temperature and pressure? 40. A 60-g sample of nitrogen (N2) gas is stored in a 5-L tank at 10 atm. What is the temperature in degrees Celsius (°C)? 41. A sample of oxygen (O2) gas occupies a volume of 10 mL at 546 K. At what temperature, in degrees Celsius (°C), will the gas occupy a volume of 5.0 L?

42. The industrial revolution, which took place from the middle of the 18th century until the middle of the 19th century in Europe, resulted in technological progress, such as the development of an engine in which the pressure from water vapour could transform heat into mechanical energy. What is the increase in pressure, expressed as a percentage, in the boiler of a steam locomotive when the temperature increases from 100°C to 200°C? 43. During photosynthesis, plants release oxygen (O2) which is then taken in by plants and animals. How many moles of gas are there in 20 L of air at STP? Assume that air contains 20% oxygen per volume. 44. At the same temperature and pressure, nitrogen (N2) and an unknown gas escape from their respective containers through an equal number of tiny identical openings. This effusion takes 84 s for 1 L of the unknown gas and 32 s for an equal volume of nitrogen. Is the molar mass of the unknown gas greater than, equal to or less than that of nitrogen? Explain your answer. 45. Ammonia (NH3) gas is used in the production of fertilizer. A sample of ammonia gas at 55°C exerts a pressure of 7.5 atm. What pressure will the gas exert if it occupies one-fifth of its initial volume at 55°C? 46. A cylinder equipped with a mobile piston contains hydrogen (H2) gas collected at 30°C. The piston moves until the volume of hydrogen is reduced by half. The pressure in the cylinder at that point is 125 kPa at 30°C. What was the initial pressure inside the cylinder? 47. At STP, a container holds 14.01 g of nitrogen (N2) gas, 16.00 g of oxygen (O2), 66.00 g of carbon dioxide (CO2) and 17.04 g of ammonia (NH3). What is the volume of the container? 48. How many kilograms of chlorine (Cl2) gas are there in 87.6 m3 at 290 K and 2.40 atm, if 1 m3 = 1000 L? 49. Calculate the volume of 3.03 g of hydrogen gas (H2) if the pressure is 560 mm Hg and the temperature is 139 K.

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121

50. Helium (He) gas is stored in a 100-L steel cylinder at 20°C. The manometer attached to the cylinder indicates a pressure of 25 atm. The cylinder is used to inflate a weather balloon at 25°C. If the final pressure in the cylinder and in the balloon is 1.05 atm, what is the size of the balloon? 51. Given the following equation: H2 (g)  O2 (g) n H2O (g) a) Balance the equation. b) What mass of oxygen (O2) reacts to produce 0.62 L of water vapour at 100°C and at 101.3 kPa? 52. How many molecules are there in 0.250 m3 of oxygen (O2) at STP? 53. A 5.0-L metal container holds 9.0 g of an unknown gas at 0°C and 202 kPa. To determine the nature of the gas, a researcher decides to find its molar mass. a) What is the molar mass of the gas? b) What is this gas? 54. A 25-L container at -20°C contains 10 g of helium (He) gas and 10 g of nitrogen (N2) gas. a) What is the total number of moles of gas in the container? b) What is the total pressure, in kilopascals (kPa), in the container? c) What is the partial pressure of the helium in the container? 55. Carbon dioxide (CO2) is prepared using sodium carbonate (Na2CO3) and hydrochloric acid (HCl). Na2CO3 (s)  2HCl (aq) n 2NaCl (aq)  CO2 (g)  H2O (l) What quantity of sodium carbonate is required to react with hydrochloric acid to produce 1.0 L of carbon dioxide at 24°C and 760 mm Hg? 56. Answer the following questions: a) At a constant pressure, the temperature, in kelvin (K), is doubled. What influence will this variation have on a gas? Explain your answer. b) At a constant pressure, the temperature, in degrees Celsius (°C), is doubled. How is this situation different from the one in a)? How will the influence on a gas be different? Explain your answer.

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UNIT 1 Gases

c) At a constant temperature, the pressure on a gas is reduced by a factor of 5. What influence will this variation have on the volume of the gas? Why? 57. At a given temperature, the speed of oxygen (O2) molecules is 0.076 m/s. Find the speed of helium (He) atoms at this temperature. 58. Determine the ratio of the speed between helium (He) atoms and radon (Rn) atoms when these two gases are at the same temperature. 59. Carbon dioxide (CO2) has many applications, including making soft drinks fizzy. a) What does the volume of a 300-L sample of this gas become when its pressure is doubled? b) What does the volume of a 300-L sample of this gas become when the temperature is increased from 30°C to 60°C? c) What is the molar volume of carbon dioxide gas at 22°C and 84 kPa? d) Design an experiment to determine the volume of carbon dioxide dissolved in a soft drink. 60. In bread dough, yeast cells transform sugar (C6H12O6) into carbon dioxide (CO2) as well as into water or ethanol (C2H5OH), as illustrated in the following chemical equations: C6H12O6 (s)  6 O2 (g) n 6 CO2 (g)  6 H2O (l) C6H12O6 (s) n 2 CO2 (g)  2 C2H5OH (l) a) Predict the volume of carbon dioxide produced when 50 mL of oxygen (O2) gas reacts with glucose (C6H12O6). b) When the baking is finished, which of the two reactions will have contributed most to the dough rising? Explain your answer. 61. The space shuttle orbits the Earth at an altitude of approximately 300 km. Canadian astronauts worked outside the shuttle using the Canadian space arm to repair a satellite. They worked in what is referred to as a “vacuum.” However, precise measurements show that the atmospheric pressure at this altitude is 1.33  10−9 kPa. In sunshine, the mean temperature is 223°C. How many molecules of gas are contained in one litre of what is usually referred to as the vacuum in space?

62. Determine the ratio of the speed between hydrogen (H2) molecules and oxygen (O2) molecules when these two gases are at the same temperature.

 67.

63. Methane (CH4), a natural gas used as a fuel to heat a home, is stored in a 100-L tank at -10°C and a pressure of 125 atm. The central heating system consumes a mean of 500 L of methane per day. How much time will this reserve of methane last if the methane is burned at 450°C at a pressure of 102 kPa?

 64.

Methanol (CH3OH) can be used as a fuel. It burns in the presence of oxygen (O2) to produce carbon dioxide (CO2) and water.

2 NaHCO3 (s) n Na2CO3 (s)  H2O (g)  CO2 (g) Assuming that 5 mL (approximately 3.0 g) of bicarbonate is required to make a dozen muffins, what volume of carbon dioxide (CO2) will be generated at 195°C and 100 kPa to make the dough rise?

 68.

With each breath, we inhale approximately 0.50 L of air. How many molecules of each of the following gases do we inhale in one breath, at 22°C and 101.3 kPa? a) Nitrogen (N2) c) Argon (Ar) b) Oxygen (O2) d) Carbon dioxide (CO2)

 69.

After nitrogen (N2) and oxygen (O2), argon (Ar) is the most present gas in air. If, out of a total of 1000 mol, air comprises 0.934 mol of argon: a) calculate the partial pressure of argon contained in air under normal conditions of temperature and pressure b) calculate the mass of argon contained in 300 L of air at a temperature of 25°C and under a pressure of 101.3 kPa

 70.

A student wishes to identify a sample of pure gas. As she tries to determine its molar mass, she makes the following observations: – mass of the empty container: 8.04 g – mass of the container and the gas: 25.53 g – volume of the container: 3.25 L – temperature of the gas: 25.7°C – pressure of the gas: 90.2 kPa a) Based on these observations, what is the molar mass of the gas in question? b) What gas could it be?

CH3OH (l)  O2 (g) n CO2 (g)  2 H2O (g) a) Balance this equation. b) If 10 L of oxygen are used at STP, what will be the volume of carbon dioxide produced? c) What mass of methanol is used in this reaction?

 65.

Complex carbohydrates are starches. The body converts carbohydrates into glucose (C6H12O6), a type of sugar. Simple carbohydrates contain glucose which the body can use immediately. Reacting with air, glucose produces energy in the muscles of runners through the process of cellular respiration, which has the following unbalanced equation: C6H12O6 (aq)  O2 (g) n CO2 (g)  H2O (i) Just before running, Tala eats two oranges. Oranges provide her body with 25 g of glucose which allows it to produce energy. The external temperature is 27°C, and atmospheric pressure is 102.3 kPa. While 21% of the air Tala inhales is composed of (O2), she exhales approximately 16% of the oxygen. In other words, she only uses 5% of the oxygen she inhaled. a) How many litres of air does Tala inhale as she runs to burn the glucose consumed? b) How many litres of water vapour does she produce? c) How many litres of carbon dioxide gas does she produce?

 66.

Sodium bicarbonate (sodium hydrogen carbonate) can be used to make muffins. The bicarbonate is used as a leavening agent to make the dough rise during cooking. One of the reactions produced during cooking is the following:

Plants consume carbon dioxide (CO2) which allows them to produce sugar (C6H12O6) through photosynthesis. 6 CO2(g)  6 H2O n C6H12O6(s)  6 O2(g) To produce 50 g of sugar, how many litres of carbon dioxide at SATP must a sugar beet absorb? CHAPTER 2 Physical Properties of Gases

123

124

CONTENTS CHAPTER 3

Energy Transfer . . . . . . . . . . . . . . . . . . . 127 CHAPTER 4

Enthalpy Change . . . . . . . . . . . . . . . . . 147 CHAPTER 5

Graphical Representation of Enthalpy Change . . . . . . . . . . . . . . . 171 CHAPTER 6

Molar Heat of Reaction . . . . . . . . . . . . . . . . . . . . . . . 185 CHAPTER 7

Hess’s Law . . . . . . . . . . . . . . . . . . . . . . . 197

Energy is the source of all change. Energy also makes life possible, whether it involves the conversion of radiant energy from the Sun into chemical energy through photosynthesis, or the release of chemical energy contained in coal through a combustion reaction that makes an iron rod more malleable. Without energy, human beings could not perform many daily activities.

In this unit, you will learn how to calculate the quantity of energy transferred during physical or chemical changes using data gathered in the laboratory or elsewhere, and how to graphically represent the energy changes of these transformations. You will also learn how to predict the energy changes of a reaction without having to carry it out in a laboratory. This is an essential skill when studying dangerous reactions. 125

3.1 Difference between heat and temperature

CHAPTER

3

ENERGY TRANSFER

3.2 The law of conservation of energy 3.3 Relationship between thermal energy, specific heat capacity, mass and temperature change 3.4 Calculating energy transfer

4.1 Enthalpy and enthalpy change

CHAPTER

UNIT

2

ENERGY CHANGES IN REACTIONS

126

CHAPTER

4 5

ENTHALPY CHANGE

4.2 Endothermic and exothermic reactions 4.3 Energy balance 4.4 Calculating enthalpy change using stoichiometry

GRAPHICAL REPRESENTATION OF ENTHALPY CHANGE

CHAPTER

6

MOLAR HEAT OF REACTION

CHAPTER

7

HESS’S LAW

5.1 Activated complex, activation energy and the energy diagram 5.2 Observing the use of an energy diagram to plot the progress of a conversion

6.1 Molar heat of dissolution 6.2 Molar heat of neutralization

7.1 Reaction mechanism 7.2 Summation of enthalpies

Energy Transfer

E

nergy on Earth comes from the Sun, but also from the radioactive decay of certain atoms. This decay occurs in the Earth’s crust, in the mantle made up of magma that rises to the surface through volcanoes, and also in the inner core. This geothermal heat is a source of sustainable energy that produces little pollution.

All the types of energy we observe are changes and energy transfers. Sometimes, energy is transformed and seems to disappear, but it can be found microscopically at the molecular and atomic levels.

Review Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Potential energy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 The law of conservation of energy . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Relationship between thermal energy, specific heat capacity, mass and temperature change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

This chapter will help give you a better understanding of the nature of energy transfer in the forms of work and heat. It will also provide the tools you need to calculate and predict the amounts of energy involved in heat transfers between different types of systems.

3.1

Difference between heat and temperature . . . . . . . . . . . . . . . . . . . . . . . . 128

3.2 3.3

The law of conservation of energy . . . . . 131

3.4

Calculating energy transfer . . . . . . . . . . . . 138

Relationship between thermal energy, specific heat capacity, mass and temperature change . . . . . . . . . . . . . . . . . . . .134

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127

3.1 Difference between heat and temperature Heat is a transfer of thermal energy that occurs when two systems with different temperatures come into contact with each other.

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Temperature is a measurement of the agitation of the atoms and particles in a system. The transfer of kinetic energy between two systems in contact with each other can take place in two ways: by work or by heat. Work is an orderly motion of particles that all move together, while heat is a random motion of particles that move in all directions (see Figure 1).

Exterior surroundings

Exterior surroundings

JAMES PRESCOTT JOULE British physicist (1818–1889) James Prescott Joule’s work made a major contribution to the formulation of the law of conservation of energy. He also worked with William Thomson (Lord Kelvin) to establish a temperature scale beginning at absolute zero. Joule became increasingly interested in electricity, a field in which he formulated a law regarding the heat lost by an electrical current passing through a resistant substance, now called Joule’s law. Legend has it that Joule and his brother experimented with an electrical current by giving each other electrical shocks. It was in honour of Joule, who studied the nature of heat and its relationship to work, that a unit of energy in the International System of Units was named.

128

Energy

Energy

System

System

a) When energy is transferred in the form of work, particle motion is orderly.

b) When energy is transferred in the form of heat, particle motion is random.

Figure 1 Kinetic energy is transferred either by work or by heat.

The work done by the molecules of a gas when we inflate a balloon is a good example of energy transfer in the form of work. The kinetic energy of the gas particles is transferred, in the form of work, to the surface of the balloon, which begins to expand. The particles that make up the surface of the balloon move together in an orderly way in the same direction. In daily life, our skin often feels changes in kinetic energy in the form of heat. Skin can feel the heat that is transferred to it when it comes into contact with a hot object such as a cup of hot chocolate. At the microscopic level, the kinetic energy of the molecules in the drink is transmitted to the molecules of the cup, and then to those of the environment and your hand. The motion is random: the particles vibrate and move in all directions, and no overall movement is visible. Kinetic energy that can be transferred in the form of heat is also called thermal energy.

UNIT 2 Energy Changes in Reactions

The temperature of a system is a measurement of the thermal agitation of the particles that make it up. This agitation includes the displacement and rotation of molecules and the vibration of the chemical bonds within the molecules. The greater the agitation, the higher the temperature is (see Figure 2). Temperature is measured experimentally in degrees Celsius (°C), a scale based on the melting and boiling points of pure water. On the Celsius scale, at normal atmospheric pressure (101.3 kPa), 0°C is the temperature at which ice melts and 100°C is the temperature at which pure water boils. So, if we measure a temperature of 100°C at the centre of a roast that is cooking, it indicates that the molecules of the roast are as agitated as those of water at its boiling point. Conversely, a lower temperature indicates that the particles are less agitated. Therefore, there is a temperature at which particles exhibit no thermal agitation: absolute zero. Absolute zero is 273°C, or 0 K, the lowest point on the Kelvin scale. The kelvin is another unit of measurement of temperature often used in chemistry. When two bodies at different temperatures come into contact with each other, the agitation of the molecules is transmitted from one body to another. This transmission of thermal agitation between two systems constitutes heat, and the process can move in only one direction: from the hotter substance toward the cooler one. In physics and chemistry, it is recognized that spontaneous reactions occur in such a way as to cause and increase particle disorder. This disorder is also called entropy. Thus, when two bodies come into contact, the random thermal motion of the hotter body is transmitted to the cooler body. However, if it were cold that was transmitted from one body to another, that would mean that the slightly agitated particles of the cold body would create order among the particles in the hotter body and stop them from moving, which is contrary to the tendency to spread disorder. So, the sensation of cold that you feel if you put your hand into the snow does not come from cold transmitted from the snow but rather from the heat lost by your hand while it is in contact with the snow. Lava flowing from a volcano into an expanse of water is a spectacular example of heat transfer. The lava emerges from the volcano at about 1000°C. If it reaches the ocean quickly, it will heat the water by transmitting its own heat to the water and the lava will cool. After a while, the water and lava will be the same temperature, which will be between that of the lava and water (see Figure 3).

Figure 3 Lava from Kilauea Volcano, in Hawaii, flows right into the Pacific Ocean, where it heats the water

60 °C

20 °C

Figure 2 The thermal agitation of molecules is weak at lower temperatures and more intense at higher temperatures. See Relationship between volume and absolute temperature, p. 80.

The quest for absolute zero Absolute zero, the temperature at which particles stop moving, is theoretical and thus impossible to observe experimentally. Never theless, physicists have come increasingly close. The current record is 0.000 000 045 K, which was achieved with a technique of laser-cooling atoms. This technique consists of placing a small gaseous sample between two lasers facing each other. The lasers regularly emit luminous particles, called photons that strike the atoms symmetrically and stop them from moving forward or backward. Since the atoms are immobilized and demonstrate no thermal agitation, the measured temperature of the gas is extremely low.

Figure 4 A sample of calcium (Ca) atoms cooled by lasers

as it cools.

CHAPTER 3 Energy Transfer

129

T2  25°C

T1  5 °C a)

T2

T1

T2  15°C T1  15 °C b)

T1

Tf

T2

Figure 5 When two bodies at different temperatures come into contact, there is a transfer of heat from the hotter body to the cooler, until the temperatures of the two bodies are the same.

When two bodies at different temperatures come into contact, some heat passes from the hotter body, which has greater thermal energy, to the cooler body (see Figure 5). The transfer of thermal energy in the form of heat continues until the temperatures of the two bodies are the same, that is, until the agitation of the molecules reaches the same intensity. The temperature of the two bodies will then be between their initial temperatures. However, in reality, situations are often complex, and many systems may come into contact with each other. Returning to the example of the volcano, the flowing lava transfers thermal energy not only to the water but also to the surrounding air and to the earth that it flows over on its way to the water. The quantity of heat exchanged between two systems depends on the thermal energy of each of the systems. The thermal energy of a substance depends on the agitation of its molecules, and therefore its temperature, and also on how much of the substance there is. A 1-kg block of iron (Fe) and a 1-g bead of iron heated to glowing red (700°C) have the same temperature, thus the identical thermal agitation, but the block contains more agitated particles and so it is able to transfer much more heat than the bead.

SECTION 3.1

Difference between heat and temperature

1. What is involved in the following situations: heat alone; work alone; both heat and work; or neither one? If the situations involve heat, indicate in which direction the heat is being transferred. If they involve work, indicate what work is being done. a) An inflated party balloon is put into the freezer and becomes smaller. b) The mercury in a thermometer rises when the thermometer is placed in hot water. c) A weightlifter holds weights over his head and tries not to move. d) An electric light bulb is lit. 2. There are three cups on a table: – the blue cup contains 250 mL of water at 80°C – the red cup contains 200 mL of water at 80°C – the yellow cup contains 250 mL of water at 60°C a) If we move the blue and yellow cups together until they touch, in which direction will the heat be transferred? b) If we want to heat a 1-L bowl of cold water, which of the three cups should be poured into the bowl?

130

UNIT 2 Energy Changes in Reactions

3. Describe the heat exchanges in the following situations: In each case, indicate where the energy comes from and what type it is, the direction of the transfer and what temperature changes can be observed. a) A car whose motor has just been turned off in February. b) A parked car on a July morning. c) A water heater keeps a home’s hot water temperature at 55°C. d) A child uses a magnifying glass to burn paper with sunlight. e) The temperature rises during the day and falls during the night. 4. In studying the behaviour of gases, it has been demonstrated that increasing the temperature of a gas in a rigid container increases the pressure the gas exerts on the inner surface of the container. In the case of a non-rigid balloon, raising the temperature increases the volume of the balloon. Explain these two phenomena in terms of heat, kinetic energy and work. Indicate in what way (work or heat) energy is transferred to the exterior of the container.

3.2 The law of conservation of energy The law of conservation of energy states that energy can be transferred or transformed but that it cannot be created or destroyed. The famous statement “Nothing is lost, nothing is created, everything is transformed,” attributed to Lavoisier applies to matter as well as to energy. An object cannot begin to move on its own without any energy being applied to it. In the same way, energy does not cease to exist at the end of a process, even if it is sometimes difficult to detect. When a car’s engine has been running and is then turned off, the remaining thermal energy in the engine is gradually transferred in the form of heat to the molecules of the air surrounding it. Since energy can be neither created nor destroyed, it changes from one form to another or moves from one place to another. For example, when we light a light bulb, electrical energy is obviously transformed into light energy but also into thermal energy which then dissipates by spreading out into the room in the form of heat. In daily life, many processes involve changes from one type of energy to another. With fireworks, for example, there is, first of all, mechanical energy (friction) or electrical energy that produces a spark to light a fuse. Then there is a chemical reaction of combustion that generates enough energy to propel the rocket, and chemical energy is transformed into kinetic and thermal energy. When the rocket rises, its kinetic energy is transformed into potential energy. At a certain height, the combustion that was set off earlier reaches the core of the rocket, producing another explosion which emits the coloured sparks typical of fireworks. At that moment, chemical energy is transformed into light, sound and thermal energy (see Figure 6).

3.2.1

Figure 6 Fireworks are an example of an energy change.

Types of systems

*

The law of conservation of energy is the first principle of thermodynamics . When studying energy changes, the type of system and its interactions with its environment must be determined. The system is the location being observed. It could be a beaker in which a chemical reaction is taking place, an electrical appliance or one of the batteries that make it work, entire human beings or just one of their cells, the solar system or a single planet. Outside the system being studied, there may be one or more systems that are in contact with it. The surroundings, the system’s environment, should also be considered and the system should be viewed as being linked to other systems and constituting part of an environment. There is just one system that has the unique characteristic of containing all other systems: the universe. There are three types of systems: open, closed and isolated (see Figure 7). An open system is in contact with its surroundings and exchanges matter and energy with it. In chemistry, this type of system can be a beaker with a reaction going on inside it. Since it is possible to heat or cool a beaker, or to add or remove reactants, energy and matter transfers with the environment are also possible.

The study * Thermodynamics of energy changes. Surroundings Energy Matter a) An open system

Matter

Energy

b) A closed system Energy Matter c) An isolated system

Figure 7 The three types of systems in thermodynamics

CHAPTER 3 Energy Transfer

131

A closed system does not allow exchanges of matter but allows energy to be exchanged with the surroundings. For example, a balloon full of air is a closed system once its opening is tied. No more air can be put in or taken out. Yet if the balloon is heated, put in a freezer or compressed, there is an energy transfer between the exterior and interior of the balloon. When the balloon is put into an environment with a different temperature, energy is transferred in the form of heat. If the surface of the balloon is compressed, energy is transferred in the form of work.

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

An isolated system is completely closed to exchanges of matter and energy. An insulated bottle full of hot chocolate is a good example of an isolated system. This type of bottle has rigid walls; no type of work, and thus no change in volume, can occur. If we leave the insulated bottle in a car in winter, the chocolate will be lukewarm at the end of the day and cold after a few days, because the agitation of the molecules in the hot chocolate is transmitted to the surroundings through the stopper, by the gas molecules between the walls and by the walls themselves. However, it is possible to consider insulated containers as isolated systems for periods of a few minutes up to a few hours.

3.2.2 JULIUS ROBERT VON MAYER German doctor and physicist (1814–1878) Julius Robert von Mayer was the first person to declare the first law of thermodynamics, which he summarized as follows: “Energy can neither be created nor des troyed.” While travelling in Indonesia, he observed that the blood of people living in the tropical climate was darker because the Indonesians used less oxygen and therefore less energy to maintain a constant body temperature. Von Mayer was very interested in chemical energy used by living things and also demonstrated that plants use light as a source of energy in a process called photosynthesis. Von Mayer experimented with heat and work and obtained the same results as his colleague, Joule, sparking a controversy over who had been first to reach these conclusions. It was only many years later that von Mayer’s work was recognized.

132

Calorimetry and the calorimeter

Calorimetry is used to experimentally determine the quantities of heat involved in transformation. An instrument called a calorimeter measures heat transfers (See Figure 8). A calorimeter consists of a rigid barrel-like container placed in a container of water, all of which is separated from the surroundings by an insulating wall. The barrel constitutes a closed sub-system that allows energy, but not matter, to escape. Once the substances being studied have being placed in the calorimeter, it becomes an isolated system, because no exchange of matter or energy takes place with the surroundings during the observation period.

Calorimeter isolated from its surroundings

Stirring rod

Thermometer Reaction vessel

Water

Figure 8 A calorimeter is an isolated system. The reaction takes place in the reaction vessel and the quantity of heat transferred is determined by observing the change in the water temperature.

The chemical reaction or physical change takes place in the reaction vessel, whose rigid material prevents the expansion or contraction of its walls. This type of calorimeter does not allow energy transfer in the form of work between the reaction vessel and its surroundings.

UNIT 2 Energy Changes in Reactions

Once substances are added to the reaction vessel and it is hermetically sealed, it is considered to be a closed system because there can be no exchange of matter. However, the thermal agitation of the particles contained in the reaction vessel can spread to the water surrounding it. During the reaction, the water temperature is measured. If the chemical reaction releases heat, the water temperature rises. Conversely, if the reaction must have energy in order to take place, the water cools because it transmits some of its thermal energy to the reaction vessel in the form of heat.

A home calorimeter It is possible to make an inexpensive calorimeter with simple materials in order to study heat transfer in certain chemical or physical changes. Fit one polystyrene (Styrofoam) drinking cup upside down into another and put them both in a beaker, which will hold the cups steady and provide additional insulation. This will create a sufficiently isolated system to allow you to observe short-term thermal phenomena. Pour some water into the cup to create an environment where the reactions can take place. A thermometer pushed through a hole made in one of the cups will show the temperature changes to be studied. With this type of calorimeter, it is important to avoid very high temperatures so that the polystyrene does not melt. Also, since some solvents can dissolve polystyrene, this type of system is most appropriate for changes in aqueous mediums.

SECTION 3.2

The law of conservation of energy

1. In the following situations, indicate what constitutes the systems and whether they are open, closed or isolated: a) A beaker of sulphuric acid (H2SO4) diluted with water. b) Six bottles of soft drinks in a cardboard box. c) A hot water bottle full of hot water. d) Liquid nitrogen (N2) in a cylinder. e) A neon (Ne) tube. f) A calorimeter whose reaction vessel is full of a gaseous mixture. 2. What types of energy are involved in each of the following situations? a) An incandescent bulb is lit. b) A pencil falls on the ground. c) Bleach whitens clothes. d) The Sun warms the Earth.

Figure 9 An inexpensive calorimeter can be made quickly with two polystyrene cups, a thermometer and water.

3. Indicate whether the temperature measured by the calorimeter’s thermometer rises or falls after the following operations. The calorimeter begins at the ambient temperature. a) In a reaction chamber, water and metallic sodium (Na) react violently, sometimes bursting into flames. b) When dry ice, that is, solid carbon dioxide (CO2), is in a reaction chamber, it solidifies at -78.5°C. c) A calorimeter is taken outside in winter. d) A temperature change caused by a chemical reaction that releases heat is measured. Then the water from the outside of the reaction chamber is removed to make this part of the calorimeter a vacuum chamber (containing rarefied air).

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3.3 Relationship between thermal energy, specific heat capacity, mass and temperature change When a substance of a certain mass and specific heat capacity, also called specific heat, undergoes a change in temperature, thermal energy is released. This thermal energy is proportional to the substance’s mass as well as to the observed temperature difference and the nature of the substance. Thus it takes more heat to cause the same temperature increase in a saucepan full of water than in a cup of water. Also, more heat must be applied to achieve a greater temperature increase in a given quantity of water (see Figure 10).

20°C 1L

45°C

45°C

2L

1L

40°C (T  20°C)

30°C (T  10°C) 1L

a) When the water mass doubles, twice the quantity of heat must be applied to obtain the same temperature.

1L

b) When the quantity of heat doubles, the temperature difference doubles.

Figure 10 The heat that a substance must absorb so that its temperature rises is proportional to its mass as well as to the temperature change. Table 1 Specific heat capacity of several substances Substance

Specific heat capacity (J/(g  °C))

Water

4.184

Water vapour

1.41

Ice

2.05

Antifreeze, a compound of ethylene glycol (C2H6O2)

2.20

Aluminum (Al)

0.900

Lead (Pb)

0.16

Copper (Cu)

0.385

Iron (Fe)

0.444

Dry air

1.02

Concrete

2.10

Gypsum

1.09

Wood

1.76

Glass

0.84

Oil

2.00

134

UNIT 2 Energy Changes in Reactions

Thermal energy also depends on the nature of the substance, because each substance has its own specific heat capacity (see Table 1). The specific heat capacity corresponds to the quantity of energy required to raise the temperature of 1 g of the substance by 1°C. This means that the higher the heat capacity of a substance, the more it has to be heated to raise its temperature. In the same way, the substance will release more heat if its temperature falls. So, the specific heat capacity of a substance is proportional to the difficulty of raising or lowering its temperature. For example, during the day the Sun heats concrete, which is found in large quantities in major urban centres. These areas release a great deal of heat at night and are called islands of heat. Urban planners try to minimize this effect by preserving and creating green spaces and covering roofs with vegetation.

The specific heat capacity of water is 4.184 J/(g°C), which means that 4.184 J of energy must be applied to 1 g of water to raise its temperature by 1°C. In comparison, antifreeze, an ethylene glycol compound (C2H6O2), needs only 2.20 J to achieve the same result. This means that if there are equal masses of both, the water must be heated almost twice as much to achieve the same increase in temperature. Metals are good conductors of heat and thus have lower heat capacity. An empty aluminum saucepan heats quickly when placed on a stove element that is turned on, but it cools equally quickly, and it is possible to touch it a few minutes after it has been removed from the hot stove. The specific heat capacity of a substance depends on many factors: the nature of its atoms and their chemical bonds, the size, mass and structure of the molecules, as well as the potential attraction between the molecules.

Furthering

your understanding

Specific heat capacity of water The specific heat capacity of liquid water is very high in comparison to other liquids, most solids and even gases. Therefore, a lot of thermal energy must be applied to it to raise its temperature. At the microscopic level, this indicates that it is difficult to increase the thermal agitation of water molecules because of the attraction between them. In fact, there are intermolecular bonds in water, called hydrogen bonds, which involve an attraction between the hydrogen (H) atom in one molecule and the oxygen (O) atom in an adjacent molecule. So to raise the temperature of water, and its thermal agitation, a large quantity of energy has to be applied to break some of these intermolecular attractions.

Figure 11 It takes a great deal of energy to break the intermolecular bonds in water molecules because they are much stronger than the bonds in most compounds.

Water as a temperature regulator The climate in places that have the same latitude, and therefore the same amount of daylight, can vary enormously depending on whether a region is a desert or a seacoast. Because of the high specific heat capacity of water, the Sun’s heat does not raise the temperature of oceans very much. In a desert, the same quantity of energy from the Sun raises the temperature of sand significantly. When the Sun sets, the low specific heat capacity of the sand allows it to cool rapidly. Water, however, releases its heat very slowly. A desert region is thus very hot during the day and much colder at night. In the Sahara Desert in August, the temperature may reach 45°C during the day and drop to 18°C at night. Maritime regions are cooler during the day, and the temperature drops less at night because the water releases its heat during the night.

Figure 12 The Sahara Desert is one of the world’s hottest desert regions.

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We can calculate the energy that a certain mass of a substance releases into an environment or absorbs from it when its temperature changes (T), that is, when it changes from an initial temperature (Ti) to a final temperature (Tf). When a substance cools, the value of the temperature change obtained is negative, and the thermal energy (Q) calculated is negative. It is customary to consider negative energy values as released energy. Thermal energy Q  mcT where Q  Quantity of heat energy, expressed in joules (J) m  Mass of a substance, expressed in grams (g) c  Specific heat capacity of a substance, expressed in joules per gram degree Celsius (J/(g  °C)) T  Temperature change (Tf  Ti), expressed in degrees Celsius (°C) Below are examples of calculations using this mathematical relationship. Example A Calculate the quantity of thermal energy absorbed by a 5.00-kg block of concrete to raise its temperature from 17.1°C to 35.5°C. Data: m  5.00 kg  5000 g c  2.10 J/(g°C) Ti  17.1°C Tf  35.5°C T  ? Q?

1. Calculation of the temperature change: T  Tf  Ti  35.5°C  17.1°C  18.4°C 2. Calculation of thermal energy: Q  mcT  5000 g  2.10 J/g°C  18.4°C  193 200 J  193.2 kJ

Answer: The concrete block must absorb 193 kJ. Example B A 1.35-g pellet of aluminum foil is heated to 205°C, and then removed from the heat. After a few seconds, the pellet has released 176 J of heat. What is its final temperature? Data: m  1.35 g c  0.90 J/(g°C) Ti  205°C Tf  ?°C T  ? Q  176 J

1. Calculation of the temperature change: Q  mcT Q T  mc 176 J  1.35 g  0.90 J/(g°C)  144.9°C 2. Calculation of the final temperature: T  Tf  Ti Tf  T  Ti  144.9°C  205.0°C  60.1°C

Answer: The final temperature of the aluminum pellet is 60°C.

136

UNIT 2 Energy Changes in Reactions

SECTION 3.3

Relationship between thermal energy, specific heat capacity, mass and temperature change

1. In a grocery bag, a 500-g container of ice cream, which has a specific heat capacity of 0.74 J/(g°C), is at a temperature of -2°C. It is put into the freezer, where it reaches a temperature of -5°C. How much energy is transferred during this temperature change? What form of energy is transferred? In which direction is the energy transferred? 2. How much heat is released by a 2.35-g piece of copper whose temperature falls from 35°C to -10°C? 3. What mass of gypsum can have its temperature raised 10°C by 6000 J? 4. When a 1.35-g sample of an unknown liquid cools by 10°C, it releases 45 J of energy. What is its specific heat capacity? 5. A 2.5-kg plank of wood is initially 22.2°C and receives 9.69 kJ of energy from a radiator. What will its final temperature be? 6. What was the initial temperature of a 5.5-g silver ring, if it changed to 45°C when 50 J of thermal energy was applied? The specific heat capacity of silver is 0.24 J/ (g°C). 7. The temperature of a 30-g stone rises from 25°C to 60°C when 2.6 kJ of energy is applied. What is its specific heat capacity? 8. What mass of mercury (Hg) should be used if its temperature is to be raised from 0°C to 100°C with 0.02 J of heat? The specific heat capacity of mercury is 0.1395 J/(g°C). 9. Is it quicker to boil a potato or to deep-fry it? Compare the energy required to boil 100 g of water and to boil 100 g of oil, both starting from an ambient temperature of 20°C. The boiling point of oil is 210°C.

11. There are three pies: a 1.25-kg apple pie at 75°C; a 1.06-kg cherry pie at 75°C; and a 1.25-kg strawberry pie at 65°C. Assuming the three pies have the same specific heat capacity, the pie that releases the least heat will cool more rapidly than the others. Which one will reach ambient temperature (20°C) most quickly? 12. Compare the quantities of heat required to produce the same temperature change in equal masses of wood and glass. What mass requires more energy? How many times more energy does it require? 13. Answer the following questions: a) A silver spoon weighing 23.9 g is placed in a cup of hot chocolate. It takes 0.343 kJ to change the temperature of the spoon from 24.5°C to 85.0°C. What is the specific heat capacity of solid silver? b) The same quantity of heat (0.343 kJ) is gained by 23.9 g of liquid water. What is the temperature change of the water? 14. In a fireplace, the bricks absorb heat and then release it over a long period after the fire is extinguished. A student performs an experiment to find the specific heat capacity of the brick. The results show that a 938-g brick receives 16 kJ of energy as its temperature rises from 19.5°C to 35.0°C. Calculate the specific heat capacity of the brick. 15. What quantity of heat is needed to raise the temperature of 789 g of liquid ammonia (NH3) from 25.0°C to 82.7°C? 16. A solid substance has a mass of 250.00 g. It is cooled by 25.00°C and loses 4937.50 J of heat. What is its specific heat capacity? Using Table 1 on page 134, locate which substance this is.

10. In order to identify a liquid, a chemist decides to measure its specific heat capacity. The chemist takes a 12.5-g sample of the liquid and applies 92.3 J of energy. She observes that the liquid’s temperature rises from 20.0°C to 23.3°C. What is the specific heat capacity of the liquid? Of the following three liquids, what is the most probable nature of the chemist’s sample: water, oil or antifreeze?

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3.4 Calculating energy transfer Calculating energy transfer between two systems is done by using the mathematical relationship Q  mcT, assuming that the heat released by the first system is equal to the heat absorbed by the second. When two systems at different temperatures are in contact with each other, the thermal energy of the hotter system, system 1, is transferred to the cooler system, system 2. The quantity of heat (Q1) released by system 1 is equal to the quantity of heat (Q2) absorbed by system 2, if the two systems together are considered to be an isolated system, with no energy loss. since

Q1  Q2

therefore

m1c1T1  m2c2T2

This is how to obtain a formula to calculate the mass of a substance required to bring another substance to a certain temperature. For example, it is possible to calculate the mass of cold water needed to cool a substance to ambient temperature. Heat transfer between two systems m1c1T1  m2c2T2

where m1  Mass of the substance in system 1, expressed in grams (g) c1  Specific heat capacity of the substance in system 1, expressed in joules per gram degree Celsius (J/(g°C)) T1  Temperature change in system 1 (Tf – Ti), expressed in degrees Celsius (°C) m2  Mass of the substance in system 2, expressed in grams (g) c2  Specific heat capacity of the substance in system 2, expressed in joules per gram degree Celsius (J/(g°C)) T2  Temperature change in system 2 (Tf – Ti), expressed in degrees Celsius (°C)

Below is an example of how to calculate the mass of water required to cool a system. Example Calculate the mass of cold water at 10°C needed to cool to 30°C a 10-g piece of glass at 95°C. Data: m1  10 g c1  0.84 J/(g°C) Ti1  95°C Tf1  30°C m2  ? c2  4.184 J/(g°C)

1. Calculation of the temperature change in system 1: T1  Tf1  Ti1  30°C  95°C  65°C 2. Calculation of the temperature change in system 2: T2  Tf2  Ti2  30°C  10°C  20°C 3. Calculation of the mass of water: m c T m1c1T1  m2c2T2 ⇒ m2  1 1 1 c2T2 m2 

Answer: It would take 6.5 g of water.

138

UNIT 2 Energy Changes in Reactions

(10 g  0.84 J/(g°C)  65°C)  6.5 g 4.184 J/(g°C)  20°C

The final temperature can also be determined for two systems in contact with each other, when each had a different initial temperature. Since the two systems exchange heat until their temperatures equalize, the final temperature will be halfway between the initial temperatures of the two systems. For example, a café au lait will have a final temperature between the temperature of the brewed coffee and that of the milk added to it (see Figure 13). The quantity of heat exchanged between the two systems is proportional to the mass and the specific heat capacity of the two substances. It is calculated from the heat transferred by the two systems. m1c1T1  m2c2T2

Figure 13 The temperature of a café au lait is a value halfway between the temperature of the brewed coffee and that of the milk added to it.

m1c1(Tf  Ti1)  m2c2(Tf  Ti2) m1c1Tf  (m1c1Ti1)  m2c2Tf  m2c2Ti2 m1c1Tf  m1c1Ti1  m2c2Tf  m2c2Ti2 m1c1Tf  m2c2Tf  m2c2Ti2  m1c1Ti1 m1c1Tf  m2c2Tf  m2c2Ti2  m1c1Ti1 which equals

Tf (m1c1  m2c2)  m2c2Ti2  m1c1Ti1

We also obtain the following formula: Temperature of the two systems Tf 

m2c2Ti2  m1c1Ti1 m1c1  m2c2

where Tf  Final temperature of the two systems, expressed in degrees Celsius (°C) m1  Mass of the substance in system 1, expressed in grams (g) c1  Specific heat capacity of the substance in system 1, expressed in joules per gram degree Celsius (J/(g°C)) Ti1  Initial temperature of system 1, expressed in degrees Celsius (°C) m2  Mass of the substance in system 2, expressed in grams (g) c2  Specific heat capacity of the substance in system 2, expressed in joules per gram degree Celsius (J/(g°C)) Ti2  Initial temperature of system 2, expressed in degrees Celsius (°C)

Convection Conduction

Furthering

your understanding

How heat energy is transferred Heat is the transfer of the thermal agitation of particles at the microscopic level. At the macroscopic or human level, heat is transferred in three ways: by conduction, convection and radiation. Conduction is the transfer of thermal agitation through direct contact of molecules located close to each other. Convection is the movement of molecules in a fluid substance such as a liquid or a gas. Hotter particles move and can transfer their thermal agitation to particles farther away from them. Lastly, radiation is the transmission of radiant energy through space by light (visible or not).

Radiation Irradiating heat

Figure 14 The three means of transferring heat in boiling of water: conduction between the pot and the hand, convection when the cold water at the bottom of the pot rises to the surface and radiation from the heating element.

139

Below is an example of a calculation for this type of problem. Example When removed from the freezer, a 500-g package of frozen raspberries is at a temperature of -4.0°C. It is then thawed in an insulated container filled with 2 kg of tepid water initially at a temperature of 40.0°C. Assuming that the raspberries have a specific heat capacity of 3.50 J/(g°C), what is the final temperature of the water and raspberries? Data: m1  500 g c1  3.50 J/(g°C) Ti1  4.0°C Tf  ? m2  2 kg  2000 g

Calculation: m c T  m1c1Ti1 Tf  2 2 i2 m1c1  m2c2 

2000 g  4.184 J/(g°C)  40.0°C  500 g  3.50 J/(g°C)  4.0°C 500 g  3.50 J/(g°C)  2000 g  4.184 J/(g°C)

 32.4 °C

c2  4.19 J/(g°C) Ti2  40.0°C Answer: The temperature of the water and raspberries will be 32°C.

Walking on fire Photos of people walking on hot coals are spectacular, and it is generally thought that special abilities are needed to perform such a stunt. However, science shows that it is a fairly easy challenge to meet. No one thinks it is unusual to put your hand in a hot oven without feeling pain. The air inside an oven that is turned on, just like hot coals of many kinds of wood, has a low specific heat capacity and is not a good conductor of heat. So, if you walk quickly over 400°C embers, very little heat is transferred to your skin, and it does not cause pain. Still, it is not completely safe because not all types of wood have insulating properties. Also, if you walk too slowly or stand still on the embers, the heat transfer to your skin will continue and you will soon feel pain.

Figure 15 Chinese Malaysians walking on a bed of hot coals during a ceremony

140

UNIT 2 Energy Changes in Reactions

SECTION 3.4

Calculating energy transfer

1. What mass of antifreeze at 5°C must be used to cool to 15°C a 50-g piece of lead that was initially at 200°C? 2. What volume of water at 12°C would be needed to cool to 20°C a 1-tonne block of concrete at 35°C? The density of water is 1.00 g/mL. 3. A bathtub contains 100 kg of water. Since the water is too cold, 25 kg of water at 60°C is added. The final temperature of the water is 40°C. What was its temperature before the hot water was added? 4. A metal cup has an unknown specific heat capacity. The cup weighs 230 g and is initially at 20°C. If 100 g of water at 80°C is poured into the cup and a thermometer is placed into it, the water temperature falls to 75°C. Ignoring the heat loss to the surrounding air, calculate the specific heat capacity of the metal cup. 5. A mixture made up of 50% water and 50% antifreeze, a compound of ethylene glycol (C2H6O2), is at 5°C. The water was initially at 12°C. What was the initial temperature of the antifreeze? 6. 150 g of oil whose temperature is 21.5°C is taken out of a cupboard. Next, the oil is poured into a 2.5-kg castiron frying pan whose temperature is 140°C. The frying pan is not on a stove element. Cast-iron is an alloy of iron (Fe) and carbon (C) with a specific heat capacity of 0.46 J/(g°C). What is the final temperature of the oil?

9. In an isolated system, is the heat released by 100 g of water at 5°C enough to melt 100 g of ice initially at -5°C? 10. 250 mL of coffee at 80°C is poured into a 300-g glass cup at ambient temperature, which is 20°C. We add 10 g of cream, just out of the refrigerator is added, with a temperature of 4°C and a specific heat capacity of 3.77 J/(g°C). Assume the specific heat capacity of the coffee is the same as water, just like its density, which is 1.00 g/mL. System 1 will be the coffee, system 2 will be the cup and system 3 will be the cream. What is the final temperature of the mixture? (Here's a clue: the heat lost by the coffee is equal to the sum of the quantities of heat absorbed by the cup and the cream.) 11. Ice cubes are placed in a container of liquid nitrogen (N2) to determine whether the final temperature will be high enough to cause the nitrogen to boil. The boiling point of nitrogen is -195.79°C. From the laboratory’s freezer, which has a temperature of -10°C, 25 g of ice is taken, which has a specific heat capacity of 2.050 J/(g°C). The ice is added to 30 g of liquid nitrogen kept at -200°C. The specific heat capacity of liquid nitrogen is 2.042 J/(g°C). Will the nitrogen boil? Calculate the final temperature and the heat transfer between the two substances, and indicate the direction of the transfer.

7. A thermometer containing 2.3 g of mercury (Hg), initially at 80°C, is placed in a cup containing 10 g of antifreeze, a compound of ethylene glycol (C2H6O2), at -5 °C. The specific heat capacity of mercury is 0.1395 J/(g°C).What is the final temperature of the antifreeze and the mercury? 8. Water vapour at 125°C is mixed with gaseous carbon dioxide (CO2) whose specific heat capacity is 0.839 J/(g°C), at 30°C. What is the final temperature of the mixture, if it is made up of 2.5 mol of water vapour and 1.8 mol of CO2?

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141

APPLICATIONS The ground-source heat pump

The heat-conducting fluid circulates through a network of underground pipes at varying depths (see Figure 16). Because the soil temperature is higher than that of the liquid, the heat is transferred from the hot substance to the cold one. The heat that the fluid gains causes it to evaporate, which allows it to store thermal energy because of its enthalpy change.

Horizontal collector (1 m deep)

Low-pressure liquid at low temperature

Heat

Heat

Low-pressure gas at low temperature

High-pressure liquid at high temperature Expansion valve

Figure 17 The ground-source heat pump transfers heat from the soil to a heating system.

Some pumps can reverse the heating cycle so that the heat pump can be used to cool air rather than heat it. They do so by transferring heat from inside the building to the soil, acting as an air conditioner. A refrigerator operates on the same principle, except that the heat pump uses air rather than the soil to transfer its heat.

Ground-source heat pump

Vertical collector (between 30 and 100 m deep)

Figure 16 Pipes can be buried horizontally, requiring a longer network, or vertically, requiring a shorter network but deeper excavation.

142

Compressor High-pressure gas at high temperature

Evaporator

This thermal energy can be used to heat a building by means of a device called a ground-source heat pump. This device takes heat from the soil, concentrates it and transfers it to a heating system. To do this, the pump uses a heat-conducting fluid, usually a mixture of water and antifreeze such as ethylene glycol (C2H6O2).

The gas that results is first compressed in a compressor to raise its temperature (see Figure 17). It then moves to a condenser which forces it to return to the liquid state and thus release its stored energy in the form of heat. This heat can be used to heat a building. The heat-conducting fluid then returns to the pump system, where an expansion valve reduces its pressure and the evaporation-condensation cycle begins again.

Condenser

Geothermal energy is energy held in the soil in the form of heat. This heat comes from two sources: near the surface it is primarily generated by solar radiation, but at greater depths, it comes from the extremely high temperatures of the earth’s crust. A few metres below the surface, the soil’s temperature corresponds to average annual air temperatures in the region. It is estimated that this temperature rises linearly by approximately 2 or 3°C with every 100 metres of depth.

UNIT 2 Energy Changes in Reactions

Ground-source heat pumps reduce heating costs as well as the ecological footprint. Still, they are not considered a source of renewable energy because they simply transfer heat from one substance to another.

Calorimetry There was little progress in the field of calorimetry until the second half of the 18th century. This is because the very nature of heat was unknown. The first scientist to clearly define the difference between heat and temperature was chemist and physicist Joseph Black.

calorimeter that worked by measuring the water temperature change during reactions in aqueous media. However, these devices could only study reactions at constant pressure with solid or liquid reactants.

Inspired by Black’s work, Lavoisier and Laplace perfected the first ice calorimeter during the winter of 1783. This device measures the quantity of ice that a chemical reaction can melt before reaching 0°C. It is made up of three concentric containers (see Figure 18). The reactants for the chemical reaction are placed in the innermost container, which is inside a container full of ice. This second container is inserted into a third container which also contains ice in order to insulate the calorimeter from the ambient temperature. The heat that is released during the reaction melts the ice in the second container and the resulting water is collected in a graduated cylinder. However, little came of the work of Lavoisier and Laplace in the scientific community.

To solve this problem, Marcellin Berthelot, a French chemist, invented the bomb calorimeter in 1879. This device consists of a hermetically sealed container immersed in a water calorimeter. The sealed container holds the reactants and a large quantity of oxygen (O2). Two electrodes connected to the bomb ignite the oxygen, causing the sample to burn. The heat produced inside the bomb is transferred to the water in the calorimeter, which allows the thermal energy of the reaction at constant volume to be measured. The bomb calorimeter is still used today.

It was not until the latter half of the 19th century that real advances were made in calorimetry, in particular because of James Prescott Joule’s research, which led him to the finding that heat is a form of energy. The calorimeter developed into a water

Among the newest technologies, the differential scanning calorimeter, invented in 1960 by Watson and O’Neill is noteworthy (see Figure 19). This device can study the enthalpy of a reaction as well as a substance’s phase changes, by measuring its boiling and melting points. The differential scanning calorimeter works by comparing heat exchanges between the sample and a reference substance, usually alumina (Al2O3) or air. Calorimeters are used in many industries today, particularly in polystyrene quality control.

Inner container holding the reactants Second container holding ice Outer container holding ice

Pipe through which water flows out

Figure 18 The ice calorimeter invented by Antoine Lavoisier and Pierre-Simon Laplace

Figure 19 A differential scanning calorimeter

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143

CHAPTER

3

Energy Transfers

3.1 Difference between heat and temperature 60 °C

• Heat is the transfer of thermal energy between two systems at different temperatures that come into contact with each other. • Temperature is a measurement of the agitation of the particles in a system. The higher the temperature, the greater the agitation of the particles. • Heat is transferred from a system with a higher temperature to a system with a lower temperature.

20 °C

• Energy transfers between systems can be made by work, which is an orderly motion of particles, or by heat, which is a random motion of particles.

3.2 The law of conservation of energy

Energy Matter

• The law of conservation of energy states that energy can be transferred or transformed but that it cannot be created or destroyed. • An open system allows the exchange of matter and energy with its surroundings.

Matter

• A closed system allows the exchange of energy, but not matter, with its surroundings. • An isolated system allows no exchange of matter or energy with its surroundings. • A calorimeter is an instrument used to study the heat involved in certain chemical or physical changes. It constitutes an isolated system.

Energy Matter

3.3 Relationship between thermal energy, specific heat capacity, mass and temperature change • It is possible to calculate the quantity of heat (Q) involved when a substance’s temperature changes. It is the product of the mass (m), the specific heat capacity (c) and the temperature change (T). Q  mcT The specific heat capacity of a substance is proportional to the degree of difficulty of raising the substance’s temperature.

144

UNIT 2 Energy Changes in Reactions

Energy

3.4 Calculating energy transfer • The thermal energy (Q) transferred between two systems, 1 and 2, can be calculated by assuming that the heat released by one system is equal to the heat absorbed by the other. Q1  Q2 m1c1T1  m2c2T2 • The final temperature (Tf) of a mixture composed of two systems can be determined by using the following formula: Tf 

CHAPTER 3

m2c2Ti2  m1c1Ti1 m1c1  m2c2

Energy Transfer

1. For each of these situations, determine: • what the systems are composed of • whether the systems are open, closed or isolated • what the environment is considered to be • whether the energy transfer takes the form of heat or work • in which direction the energy is transferred a) A glass of water turns to ice in a closed freezer whose exterior surface is at ambient temperature. b) An astronaut enjoys some hot chocolate in a space station orbiting the Earth. c) A cook gets burned when hot oil splatters. 2. Calculate the quantity of heat absorbed by 222 g of oil when it is heated from 20°C to 120°C. 3. Is the heat released when a sample of ice is cooled by 10°C equal to the amount of heat released when a sample of water is cooled by 10°C? Is it the ice or the water that releases more heat? Calculate the heat released by a 1-g sample of each of these substances. The specific heat capacity of the ice is 2.05 J/(g°C).

4. A 9.4-g piece of lead (Pb) is placed in a calorimeter at 18.9°C. The calorimeter’s water reservoir has a capacity of 1 L. The density of water is 1.0 g/mL. What was the initial temperature of the lead if the temperature of the water was 19.1°C when the heat exchange was finished? 5. At what temperature should antifreeze be kept if 100 g of it is to be used to cool a 20-g piece of copper (Cu) from 300°C to 20°C? 6. 50 g of hexane (C6H14) at 20°C is mixed with 75 g of benzene (C6H6), a very toxic substance with a specific heat capacity of 1.72 J/(g°C), at 30 °C. The final temperature is 25.3°C. What is the specific heat capacity of hexane? 7. A piece of metal with a mass of 14.9 g to 98.0°C is heated. When the metal is placed in 75.0 g of water at 20.0°C, the water’s temperature rises by 28.5°C. What is the metal’s specific heat capacity? 8. There exists a piece of gold (Au) with a mass of 45.5 g and a temperature of 80.5°C. It is dropped into 192 g of water at 15.0°C. Using Table 8.12 of Appendix 8, find the final temperature of the system.

CHAPTER 3 Energy Transfer

145

9. The specific heat capacity of aluminum (Al) is 0.902 J/(g°C). The specific heat capacity of copper (Cu) is 0.389 J/(g°C). The same quantity of heat is applied to equal masses of aluminum (Al) and copper (Cu). Using Table 8.12 of Appendix 8, find which metal’s temperature will rise more. Explain your answer. 10. What is the final temperature of a mixture of 2.1 mol of water vapour initially at 103°C and 2.1 mol of oxygen (O2) initially at 24°C? The specific heat capacity of oxygen is 0.918 J/(g°C).

12. Compare the quantities of heat lost during the same cooling of equal masses of aluminum (Al) and copper (Cu). Which metal will release more energy? How many times more energy will this metal release?

16. 20 g of pure water and 20 g of seawater, initially at 22°C, are placed into a freezer. The sample of seawater has a freezing point of -2.60°C and a specific heat capacity of 3.99 J/(g°C). a) Compare the quantity of heat lost from each of the samples in reaching their freezing points. b) What mass of seawater would be needed so that the two liquids would solidify while releasing the same quantity of heat?

13. The water in a calorimeter is replaced by oil. The water (or oil) chamber of the calorimeter has a capacity of 2 L, and the calorimeter is placed in a room at 20°C. If the water temperature rises to 23°C during a certain chemical change, what would the temperature of the oil be after the same change? The density of water is 1 g/mL and that of oil is 0.9 g/mL.

17. The water in a calorimeter is replaced with 400 g of gaseous argon (Ar), with a specific heat capacity of 0.5203 J/(g°C). The calorimeter’s water chamber has a capacity of 1.2 L. The calorimeter is placed in a room at 22°C. If the water temperature rises to 21°C during a certain chemical change, what would the temperature of the argon be if it went through the same chemical change?

11. What mass of water at 100°C will freeze while releasing the same quantity of heat as contained in 300 g of water at 20°C?

14. A mixture whose mass is composed of 60% water and 40% ethanol (C2H5OH) is at 15°C. The water used was initially at 20°C, while the ethanol was at 2.13°C. Find the specific heat capacity of ethanol.

146

15. To conduct an experiment, a chemist wonders which of two solvents to use, methanol (CH3OH) or acetone (CH3COCH3). The chemist wants to use the solvent which, for an equal mass, can absorb the greatest possible quantity of heat before boiling. The boiling point of methanol is 65°C and that of acetone is 56.1°C. The specific heat capacity of methanol is 2.60 J/(g°C) and that of acetone is 2.15 J/(g°C). Which solvent should the chemist choose?

UNIT 2 Energy Changes in Reactions

18.

Calculate the final temperature of a mixture of two gases at standard pressure. The mixture is made from one 3-L bottle of helium (He) at 15°C and one 2-L bottle of hydrogen (H2) at 7.5 °C. The specific heat capacity of helium is 5.19 J/(g°C) and that of hydrogen is 14.30 J/(g°C).

19.

An electric kettle has a power of 1600 W. Power is energy per unit of time and is expressed in watts (1 W  1 J/s). This means that the kettle transfers energy at a rate of 1600 J per second. Since the power (P) is energy (E ) per unit of time (t ), it can be assumed that P  E/t. Since the form of energy in this case is heat, it can be said that P  Q/t. If the kettle is an isolated system, ignoring the heat loss to the environment, how long will it take 1.5 L of water at 19°C to boil in this kettle?

Enthalpy Change

C

ertain reactions have the property of releasing a large quantity of energy, such as the raging fire over this oil tanker heat. These changes are characterized by their enthalpy change, which is a measurement of this energy. In general, breaking the forces of attraction between atoms or molecules requires energy, while forming new bonds releases energy.

Review Lewis notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Enumeration of matter. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Phase changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Stoichiometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Synthesis, decomposition and precipitation . . . . . . . . . . . . . . . . . . . 25 Endothermic and exothermic reactions. . . . . . . . . . . . . . . . . . . . . . . 26 Oxidation and combustion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

This chapter will cover endothermic and exothermic changes, and how to perform an energy balance to determine the overall heat involved in a chemical reaction.

4.1 4.2

Enthalpy and enthalpy change . . . . . . . . 148

4.3 4.4

Energy balance . . . . . . . . . . . . . . . . . . . . . . . . 156

Endothermic and exothermic reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

Calculating enthalpy change using stoichiometry . . . . . . . . . . . . . . . . . . . . 161 CHAPTER 4 Enthalpy Change

147

4.1 Enthalpy and enthalpy change HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Enthalpy (H) is the total energy of a system, that is, the sum of all the potential and kinetic energies that the system contains at constant pressure. Enthalpy change (H) is the energy exchanged between a system and its environment during a physical change or a chemical reaction at constant pressure. It is also referred to as the heat of reaction. The sum of all of the types of energy found in a system at constant pressure constitutes the enthalpy of the system. A system contains energy in various forms. The kinetic energy of a system includes the movement of electrons around atomic nuclei, as well as the movement of molecules and atoms when they vibrate, rotate and move. The potential energy of a system is produced by the forces of attraction between nucleons, and between nucleons and electrons. This energy also comes from the chemical bonds between the atoms of molecules, as well as the interaction between molecules.

JOSIAH WILLARD GIBBS American physicist, chemist and mathematician (1839–1903) Josiah Willard Gibbs was one of the founders of chemical thermodynamics. He introduced the concept of the free energy of a system, defined as the energy available to do work, and associated it mathematically to the entropy (disorder) and enthalpy of the system. He was interested in the phase changes of matter and the equilibrium between different phases. He explored the concept of chemical potential, which can be seen as a force that pushes matter toward another phase or outside of the system, toward the attainment of equilibrium. While his work was fundamental, for years it remained beyond the grasp of specialists in experimental chemistry due to its great mathematical complexity.

148

Enthalpy, symbolized by the letter H, is a thermodynamic property of a system that encompasses all of the kinetic and potential energies. It also includes energy in the form of the work necessary to maintain constant pressure in the system. However, in practice, it is impossible to precisely measure the enthalpy (H) of a system, since too many factors that contribute to total energy must be taken into account. It is more useful, in chemistry, to consider enthalpy changes during the reactions that occur in the system. The total enthalpy of a system is never known, but the enthalpy change that occurs during a reaction is accessible. Enthalpy change, symbolized by ΔH, is the change in the total energy of the system during a chemical or physical change under constant pressure. During a physical change or chemical reaction, a quantity of energy is exchanged with the environment, primarily in the form of heat. Consequently, it is possible to evaluate the enthalpy change of a physical change or chemical reaction by measuring the heat it absorbs or releases if pressure is kept constant.

Enthalpy change H  Hp – Hr where H  Enthalpy change of the reaction, expressed in kilojoules (kJ) Hp  Product enthalpy, expressed in kilojoules (kJ) Hr  Reactant enthalpy, expressed in kilojoules (kJ)

During a physical change, like the fusion of water (H2O (s) n H2O (l)), the ice can be considered the reactant and the water the product in the formula for calculating enthalpy change.

UNIT 2 Energy Changes in Reactions

4.1.1

Standard molar enthalpy change

Standard molar enthalpy change (H°), sometimes simply called molar enthalpy, constitutes the enthalpy change that accompanies a reaction for one mole of substance at SATP. Its unit of measure is the kilojoule per mole (kJ/mol). Enthalpy changes can be measured using a calorimeter. This device evaluates the difference in temperature in the environment of a chemical reaction, that is, it determines the heat involved in the reaction. If the reaction occurs under constant pressure (which is often normal atmospheric pressure, that is, 101.3 kPa), the heat exchange with the environment corresponds to standard enthalpy change. A standard molar enthalpy change has been determined for a large number of physical and chemical changes. For example, the value of the standard molar enthalpy change (H°) of the vaporization of water is +40.66 kJ/mol at 100°C, which means 40.66 kJ must be supplied to one mole of water to vaporize it at normal atmospheric pressure. Meanwhile, the fusion of ice has a standard molar enthalpy change of 6.01 kJ/mol at 0°C. The positive sign of the enthalpy change means that for the reaction to occur, it must absorb heat. Standard enthalpy change varies with temperature. It is generally given for a temperature of 25°C, unless another temperature is provided. In chemical reactions, standard enthalpy change refers to the heat that must be supplied or that is released during a chemical reaction, which occurs at normal atmospheric pressure, from the reactants in their standard state until the formation of products in their standard state. The standard state of a substance is the state in which it is found at SATP.

See The calorimeter, p. 132.

APPENDIX 8 Table 8.4: Standard molar enthalpies of formation, p. 418.

See Simple gas laws, p. 75.

To simplify the vocabulary, the terms “enthalpy change” or “molar heat of reaction” are often used, without specifying if they are standard. The data used to calculate the enthalpy changes are standard since they were measured in standard conditions. However, in reality, fractions of a mole of a substance are often involved in a reaction under conditions that are not standard. The enthalpy changes obtained in these circumstances are not standard and are symbolized by H. For example, ethanol (CH3CH2OH), an alcohol derived from the fermentation of plants, is increasingly used as a fuel according to the following combustion reaction: CH3CH2OH (l)  3 O2 (g) n 2 CO2 (g)  3 H2O (g) The standard enthalpy change of the combustion of one mole of ethanol is sign of the enthalpy change means the reaction releases heat. Ethanol is increasingly used as a fuel, particularly in decorative fireplaces, since its combustion produces less residue in the air compared to the combustion of wood (see Figure 1).

1367 kJ/mol at 25°C. The negative

Figure 1 An ethanol fireplace

CHAPTER 4 Enthalpy Change

149

4.2 Endothermic and exothermic reactions Endothermic reactions absorb heat from the environment, while exothermic reactions release heat into the environment. Physical and chemical changes lead to an exchange of heat with the environment. Endothermic reactions require heat from the environment, while exothermic reactions release heat. In general, a process that requires the forces of attraction between particles to be broken is endothermic, while a process leading to the formation of interactions is exothermic.

4.2.1

See Particle behaviour in the various phases of matter, p. 54.

Endothermic and exothermic physical changes

Changes in the phase of matter causes forces of attraction between the particles that compose it to be broken or formed. For example, when a substance changes from the solid phase to the liquid phase, there is a greater separation of its particles. Consequently, fusion (melting), sublimation and vaporization are endothermic physical changes, since they require energy to break the forces of attraction between the particles. It is possible to feel the effect of an endothermic reaction by placing some ice cubes in your hands (see Figure 2). The fusion of ice requires heat to occur and absorbs heat from your hand. The cold sensation you feel on your skin is from this loss of heat.

Figure 2 The fusion of ice is an endothermic process.

Figure 3 The formation of snow crystals is an exothermic change.

150

Conversely, reactions in which the attractions between particles become greater are exothermic and release energy. Solidification, the condensation of a solid and the condensation of a liquid, are exothermic physical changes. An example of solid condensation is the formation of snow crystals in the upper atmosphere from water vapour (see Figure 3). This phase change involves the formation of intermolecular bonds between water molecules, which releases energy called latent heat into the surrounding air. This quantity of heat is difficult to measure on a small scale given the small mass of a few snowflakes. However, the latent heat released from the condensation of water provides the energy for various meteorological phenomena, including hurricanes. During a phase change, the substance absorbs or releases heat without a change in temperature. For example, if a sample of ice is gradually heated, its temperature increases regularly. The heat amplifies the agitation of the water molecules in the solid, which translates into a rise in temperature. However, once the ice reaches 0°C, the heat provided serves to trigger the phase change toward the liquid phase. The heat is then used to break the bonds between the water molecules rather than to increase the temperature of the sample. When there are no more bonds to be broken, the fusion is complete and all of the water is in the liquid phase. The heat added during this phase change corresponds to the enthalpy of the fusion of water. The same principle applies during the vaporization of water at 100°C. The heat used to break all of the bonds between the water molecules corresponds to the enthalpy change of vaporization.

UNIT 2 Energy Changes in Reactions

This phenomenon can be visualized on a graph which shows the temperature changes according to the heat absorbed or released by a sample (see Figure 4). Temperature varies as a function of heat for the three phases of matter. However, during phase changes, the temperature remains constant until the system has absorbed or released enough heat for the change to occur. This quantity of heat corresponds to the enthalpy change of the reaction. Figure 4a represents temperature change as a function of the heat absorbed for the endothermic phase changes of fusion and vaporization. Figure 4b, which resembles a mirror image of Figure 4a, illustrates the exothermic phase changes of condensation and solidification. The temperature of a sample as a function of the heat supplied for phase changes

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Gas

Temperature

Boiling point Liquid Fusion point

Hvaporization Solid Hfusion

Heat supplied a) Endothermic phase changes The temperature of a sample as a function of the heat released during phase changes

Gas

French engineer and physicist (1799–1864)

Temperature

Condensation point Liquid Solidification point

BENOÎT PAUL ÉMILE CLAPEYRON

Hcondensation Hsolidification

Solid

Heat released b) Exothermic phase changes

Figure 4 During phase changes, the temperature does not change until the phase change is complete. The heat involved while the temperature is stable corresponds to the enthalpy change of the reaction.

Note that the enthalpy change of a reaction is always equal, in absolute value, to that of the reverse reaction. It is only the sign that changes. If a reaction is endothermic, the reverse reaction is always exothermic. For example, the fusion of ice, which is endothermic, has a standard enthalpy change of 6.01 kJ/mol. The solidification of water, which is exothermic, has a standard enthalpy change of 6.01 kJ/mol.

Before turning his attention to thermodynamics, Benoît Paul Émile Clapeyron had a passion for trains and railways. As an engineer, he made a major contribution to the development of the French railway system. He then devoted himself to studying the phase changes of matter. He established a mathematical equation that linked enthalpy change, previously called latent heat, to changes in temperature, pressure and volume that occur during a phase change. As a tribute to him, his name was engraved in gold letters on the Eiffel tower.

CHAPTER 4 Enthalpy Change

151

4.2.2

Endothermic and exothermic chemical reactions

During a chemical reaction, the reactant molecules decompose and their atoms form new product molecules. To break the bonds that unite the atoms, energy must be supplied, while the formation of new bonds releases energy (see Figure 5). Bond breaking  ENERGY

Photosynthesis Photosynthesis is a process that occurs in plants and algae. It involves a great number of chemical reactions that convert carbon dioxide (CO2) from the atmosphere and water into more complex molecules, such as sugars, and release oxygen (O2) into the atmosphere. Plants use the sugars as nutrients to grow and reproduce. This is an endothermic process, since it requires light energy to occur. The light is absorbed by the chlorophyll molecules contained in chloroplasts, organs found in the leaves of plants. This is the site of a complex chemical reaction: the light energy is converted into chemical energy that can be used by the plants. Chlorophyll is the pigment responsible for the green colour of plants and algae.

Figure 6 The phenomenon of photosynthesis in plants and algae is an endothermic chemical reaction that uses light energy.

152

Molecule a) Endothermic process

Separated atoms

Bond formation  ENERGY

Molecule Separated atoms b) Exothermic process

Figure 5 Bond breaking is an endothermic process, whereas the formation of bonds between separate atoms is an exothermic process.

If more energy is required to break the reactant bonds than the energy released during the formation of the product bonds, the reaction is endothermic. Conversely, if the energy released during the formation of the products is greater than the energy absorbed to decompose the reactant molecules, the reaction is exothermic. For example, the decomposition of calcium carbonate (CaCO3), the primary compound in chalk, into calcium oxide (CaO) and carbon dioxide (CO2) is an endothermic reaction with a standard enthalpy change of 178 kJ/mol. This reaction can be expressed by a thermochemical equation in two different ways by including the enthalpy change value in the equation, or by excluding it. Thermochemical equations of an endothermic reaction 1)

CaCO3 (s) n CaO (s)  CO2 (g)

2)

CaCO3 (s)  178 kJ n CaO (s)  CO2 (g)

H  178 kJ/mol

In the first equation, the enthalpy change is written outside the equation and indicates the quantity of heat required by this reaction for one mole of decomposed reactant. It is indicated in kilojoules per mole. However, the enthalpy change may be indicated for more than one mole of substance: in this case, it is only indicated in kilojoules (kJ). The positive sign of the enthalpy change indicates that it’s an endothermic reaction. The positive sign of the enthalpy change indicates that heat was absorbed by the reaction, and that this reaction is endothermic.

UNIT 2 Energy Changes in Reactions

In the second equation, the energy absorbed is on the side of the reactants, since it must be supplied to the reactants in order for the reaction to occur. This means breaking the reactant bonds requires more energy than the energy released when the chemical bonds of the products are formed. When the enthalpy change is included in the equation, it is indicated in kilojoules (kJ). Inversely, in the case of an exothermic reaction, the enthalpy change is negative. The energy is placed on the side of the products in the second equation, since it is produced by the reaction. For example, the formation of iron oxide (Fe2O3), commonly called rust, from iron (Fe) is an exothermic reaction whose thermochemical equation can be expressed as follows. Thermochemical equations of an exothermic reaction 1)

4 Fe (s)  3 O2 (g) n 2 Fe2O3 (s)

2)

4 Fe (s)  3 O2 (g) n 2 Fe2O3 (s)  1648.4 kJ

H  824.2 kJ/mol

In this example, since iron oxide is being formed, the enthalpy change is expressed in reference to the iron. Thus, the first equation indicates that when one mole of iron oxide is produced, there is a release of 824.2 kJ of energy. When the value of the enthalpy change is integrated into the thermochemical equation, the stochiometric coefficient of the iron oxide must be considered, since it is the iron oxide that is formed. The value of the enthalpy change of the formation of iron oxide must be multiplied by its stoichiometric coefficient. The value of 1648.4 kJ represents the enthalpy change of this reaction for two moles of formed iron oxide, since 824.2  2  1648.4. The same applies to all types of reactions. To include the molar enthalpy change in the equation, it must be adapted to the number of moles of substance involved in the balanced equation. To determine what substance is involved, the type of reaction must be known. Consequently, in the case of a synthesis reaction, for example, the molar enthalpy change will be that of the substance formed. Therefore, it is a molar heat of formation. For a combustion reaction, the given molar enthalpy change is that of the substance undergoing combustion. In the case of a dissolution or neutralization reaction, the molar enthalpy change is given for one mole of the dissolved or neutralized substance.

4.2.3

Enthalpy diagram of endothermic and exothermic reactions

To visualize the progress of reactions from an energy point of view, it is possible to plot a simple diagram that shows the relative enthalpy of the reactants and products using horizontal lines at different levels. In the enthalpy diagram, the enthalpy change during the conversion of reactants into products is the difference in height between the lines, while the sign indicates if the reaction is endothermic or exothermic (see Figure 8 on page 154).

CHAPTER 4 Enthalpy Change

153

Enthalpy diagram of a reaction

Enthalpy diagram of a reaction

Hp

Positive H : endothermic reaction

Enthalpy

Enthalpy

Hr

Negative H : exothermic reaction

Hr

Hp

Progress of the reaction

Progress of the reaction

a) An endothermic reaction

b) An exothermic reaction

Figure 8 Enthalpy diagrams showing that the enthalpy change is positive for an endothermic reaction, and negative for an exothermic reaction, since H  Hp  Hr.

During an endothermic reaction (see Figure 8a), the enthalpy of the reactants is less than that of the products. This means the potential energy stored in the chemical bonds is greater in the products than in the reactants. The enthalpy change is therefore positive. Self-heating meals There are several types of selfheating meals on the market for outdoor activities or for use as emergency food for disaster victims. These meals are packaged in a special container with a flame-free heating element that functions according to the following exothermic reaction:

In an exothermic reaction (see Figure 8b), the enthalpy of the reactants is greater than that of the products. This means the potential energy stored in the chemical bonds is greater in the reactants than in the products. The enthalpy change is therefore negative. It is possible to create an enthalpy diagram if the enthalpy change of a chemical reaction is known, as illustrated in the following example:

Mg (s)  2 H2O (l) n Mg(OH)2 (aq)  H2 (g)

Plot the energy diagram of this reaction. Solution:

Enthalpy diagram of the combustion reaction of ethanol

CH3CH2OH (l)  3 O2 (g) Enthalpy (kJ/mol)

Water is added to the pieces of magnesium (Mg) to trigger the reaction and release heat. These meals are designed to be ready in about 15 minutes, heated at a temperature of 80˚C.

Example The combustion of ethanol occurs according to the following reaction: CH3CH2OH (l)  3 O2 (g) n 2 CO2 (g)  3 H2O (g)  1267 kJ

H  1267

2 CO2 (g)  3 H2O (g)

Figure 7 Components of a selfheating meal

154

UNIT 2 Energy Changes in Reactions

Progress of the reaction

SECTION 4.1 SECTION 4.2

Enthalpy and enthalpy change Endothermic and exothermic reactions

2. Are the following reactions endothermic or exothermic? a) 2 NO (g)  Cl2 (g)  77.4 kJ n 2 NOCl (g) b) N2 (g)  O2 (g) n 2 NO (g) H  180.4 kJ/mol 1 c) Li (s)  F2 (g) n LiF (s)  617 kJ 2 d) The manufacture of freon (CF2Cl2), a gas long used in refrigeration: CH4 (g)  2 Cl2 (g)  2 F2 (g) n CF2Cl2 (g)  2 HF (g)  2 HCl (g) H  1194 kJ/mol 3. Which of the following four diagrams best represents the sublimation of carbon dioxide (CO2), whose enthalpy change of sublimation is 571 kJ/mol ? a) T

b) T

5. According to the following diagram, a substance X, initially in the gas phase at high temperature, passes through the liquid and solid phases of matter as it cools.

Temperature (°C)

1. Indicate if the following physical changes are endothermic or exothermic: a) rubbing alcohol, which creates a cold sensation on the skin and evaporates quickly b) the formation of condensation on the bathroom mirror during a shower c) C4H10 (l) n C4H10 (s) d) ice cubes that shrink in volume after several days in the freezer e) the explosion of fireworks

40

Gas

30

Liquid

10

Solid 10

20 28 41 Heat released (kJ/mol)

a) What is the vaporization temperature of substance X? b) What is its solidification temperature? c) What is its enthalpy change of liquid condensation? d) What is its enthalpy change of fusion? 6. The following two enthalpy diagrams represent the progress of two reactions. For each diagram, provide the thermochemical equation of the reaction in two different ways. Then specify if the reaction is endothermic or exothermic. a) H(kJ) 2 Li (s)  2 HCl (g)

571 kJ/mol

Q

c) T

571 kJ/mol

562

Q

2 LiCl (s)  H2 (g)

d) T

Progress of the reaction

571 kJ/mol

Q

571 kJ/mol

Q

b) H(kJ) C2H6 (g)  2 NO2 (g)

Q: Heat absorbed or released (kJ/mol) T:Temperature of the substance (°C)

4. Plot the enthalpy diagram of each of the following reactions. Indicate on each diagram the value of the enthalpy change. a) Fe3O4 (s)  CO (g)  35.9 kJ n 3 FeO (s)  CO2 (g) b) 2 KClO3 (s) n 2 KCl (s)  3 O2 (g)  89.4 kJ

861

2 NH3 (g)  2 CO2 (g) Progress of the reaction

CHAPTER 4 Enthalpy Change

155

4.3 Energy balance The energy balance is the sum of the energy required to break the chemical bonds of the reactants and the energy released at the moment of the formation of the product bonds. Since it is impossible to know the initial enthalpy and the final enthalpy of a system in which a chemical reaction occurs, another method must be used to determine the total enthalpy change of the reaction. This method consists of examining and comparing the chemical bonds present in the molecules of the reactants and products. A chemical reaction involves the breaking of chemical bonds and the formation of new ones. Taking into account the energy values that exist during the breaking and formation of these bonds, it is possible to determine the enthalpy change of any reaction. APPENDIX 8 Table 8.5: Average bond energy, p. 419.

Table 1 Approximate mean values of the energies of some bonds

Bond

Energy (kJ/mol)

HOH

436

HOO

460

HOF

570

HOCl

432

COH

413

COC

347

CPC

607

COO

358

C N

891

OPO

498

ClOCl

243

N N

418

NPO

631

First, all of the bonds present in the reactants and products must be identified. Lewis diagrams are very helpful during this step, since they can be used to detect the presence of double or triple bonds. Each type of chemical bond possesses a characteristic bond energy that corresponds to the energy needed to break the bond or to energy released when the bond is formed. The energy of an AOB bond is expressed as follows: EAOB. It is given in kilojoules per mole (kJ/mol) of bonds. Bond strength depends on the atoms that come into play and the number of electrons shared between these atoms. A double bond, where four electrons are shared between the same atoms, is stronger than a single bond, where only two electrons are shared (see Table 1).

4.3.1

Performing an energy balance

To perform the energy balance of a chemical reaction, the type of bonds contained in the substances involved must first be determined. Then, the energy required to break all of the bonds of the reactants must be calculated by adding the energy values specific to each of the bonds that are present. Since bond breaking is an endothermic process, the value of the enthalpy change will be positive. Next, the enthalpy change associated with the formation of all the new chemical bonds of the products must be calculated. Since bond formation represents an exothermic process, the value obtained for this step will be negative. The energy balance allows us to determine the enthalpy change, which constitutes the addition of these two values. Energy balance H  Hbonds broken  Hbonds formed where Hr Hbonds broken Hbonds formed

 Enthalpy change of the reaction, expressed in kilojoules per mole (kJ/mol)*  Enthalpy change of the breaking of reactant bonds, expressed in kilojoules per mole (kJ/mol)*  Enthalpy change of the formation of product bonds, expressed in kilojoules per mole (kJ/mol)*

* Kilojoules can be used when an energy balance calculation is performed since bond energy values can be expressed for more than one mole.

156

UNIT 2 Energy Changes in Reactions

A graphical representation of the energy balance shows a three-level diagram (see Figure 9). The energy supplied to break all of the reactant bonds (Hbonds broken) creates a higher intermediate level, which is equivalent to the enthalpy of the atoms separated from each other. When the atoms group together to form the products, they release energy, which corresponds to Hbonds formed. An energy balance diagram of a reaction

An energy balance diagram of a reaction

Separated atoms

Separated atoms

Hbonds broken

Hp

Enthalpy

Enthalpy

Hbonds formed

Hbonds broken Hr

Hbonds formed

H negative

H positive Hr

Hp

Progress of the reaction a) An endothermic reaction

Progress of the reaction b) An exothermic reaction

Figure 9 The enthalpy change of the reaction corresponds to the balance between the enthalpy change to break the reactant bonds and the enthalpy change to form the product bonds.

If the reactant bonds are stronger than those of the products, the energy needed to break them is greater than the energy released by the formation of products. The value of Hbonds broken will be greater, in terms of absolute value, than the value of Hbonds formed. The addition of these two values produces a positive balance and indicates an endothermic reaction (see Figure 9a). In the case of an exothermic reaction (see Figure 9b), it is the value of Hbonds formed that is greater, but negative. The balance, therefore, provides a negative value of the enthalpy change of reaction. It is important to note that the enthalpy change of reaction calculated in this way is very often different from the experimental value. This is because the values of the bond energies used are mean values, and these energies vary slightly depending on the molecule being studied and its particular phase. In many chemical reactions, not all of the chemical bonds break when the reactants separate into atoms. For example, many reactions carried out with large reactant molecules involve only a small part of the molecule, while the rest remains intact. However, since these reactions sometimes involve the reorganization of atoms which makes it difficult to determine exactly which bonds are broken and formed, it is easier, when calculating enthalpy change, to consider that all of the reactant bonds are broken and new ones are formed. From an energy balance perspective, this produces the same result. In fact, if a bond that was broken is formed, the energy released during this formation cancels out the energy provided to break the bond, which respects the law of conservation of energy.

CHAPTER 4 Enthalpy Change

157

Example A Calculation of the enthalpy change of a reaction involving simple bonds The following is a reaction between hydrogen (H2) gas and chlorine (Cl2) gas: H2 (g)  Cl2 (g) n 2 HCl (g) Calculate the enthalpy change of the reaction by performing an energy balance, then indicate if the reaction is endothermic or exothermic. Then, create an energy balance diagram. Use the table of average bond energy (see Appendix 8 on page 419 ). Solution : 1. Determine the bonds broken in the reactants using Lewis diagrams:

ClOCl

HOH

The HOH bond contained in H2 and the ClOCl bond contained in Cl2. 2. Determine the bonds formed in the products using Lewis diagrams:

HOCl 2 HOCl bonds, since there are two molecules of HCl. 3. In the table of average bond energy (see Appendix 8 on page 419 ), find the values that correspond to the energy of these bonds: (EHOH): 436 kJ/mol (EClOCl): 243 kJ/mol (EHOCl): 432 kJ/mol 4. Calculate the total bond energy of the reactants and products, that is, the enthalpy of the bonds broken and the enthalpy of the bonds formed: Reactants: Hbonds broken  EHOH  EClOCl  436 kJ  243 kJ  679 kJ Products: Hbonds formed  (2  EHOCl)  (2  432 kJ)  864 kJ 5. Calculate the enthalpy change of the reaction by making an energy balance. H  Hbonds broken  Hbonds formed  679 kJ – 864 kJ  185 kJ 6. Create the energy balance diagram: Energy balance diagram of the reaction

Enthalpy (kJ/)

2 H  2 Cl

Hbonds broken  679

Hbonds formed  864

H2  Cl2 H  185

2 HCl

Progress of the reaction

Answer: The reaction is exothermic since its enthalpy change is hydrogen chloride (HCl), that is, -92.5 kJ/mol.

158

UNIT 2 Energy Changes in Reactions

185

kJ for two moles of

Example B Calculation of the enthalpy change of a reaction involving double bonds Below is the combustion of methane (CH4). CH4 (g)  2 O2 (g) n CO2 (g)  2 H2O (g) Calculate the enthalpy change of this reaction by performing an energy balance, then indicate if the reaction is endothermic or exothermic. Then, create an energy balance diagram. Use the table of average bond energy (see Appendix 8 on page 419 ).

H A O H COH A H

Solution : 1. Determine the bonds broken in the reactants using Lewis diagrams.

OPO

4 COH bonds contained in CH4 and 2 OPO bonds contained in 2 O2. 2. Determine the bonds formed in the products using Lewis diagrams.

OPCPO

HOOOH

2 CPO bonds contained in CO2 and 4 OOH bonds contained in 2 H2O. 3. Find the values corresponding to the energy of these bonds in the table of average bond energy (see Appendix 8 on page 419 ). (ECOH): 413 kJ/mol (ECPO ): 745 kJ/mol (EOPO): 498 kJ/mol (EOOH): 460 kJ/mol 4. Calculate the total bond energy of the reactants and products, that is, the enthalpy of the bonds broken and the enthalpy of the bonds formed. Reactants: Hbonds broken  4  ECOH  2  EOPO Products : Hbonds formed

 4  413 kJ  2  498 kJ 2648 kJ  (2  ECPO  4  EOOH)  (2  745 kJ  4  460 kJ)  3330 kJ

5. Calculate the enthalpy change of the reaction by making an energy balance. H  Hbonds broken  Hbonds formed  2648 kJ – 3330 kJ  682 kJ 6. Create the energy balance diagram. Energy balance diagram of the reaction

Enthalpy (kJ)

C4H4O Hbonds broken Hbonds formed  VENIR  3330 A2648 CH4  2 O2 Hr  682

CO2  2 H2O

Progress of the reaction

Answer : The reaction is exothermic since its enthalpy change is -682 kJ for one mole of methane CH4, that is, -682 kJ/mol. CHAPTER 4 Enthalpy Change

159

Furthering

your understanding

Different chemical formulas There are several ways to represent molecules and their structures. The empirical chemical formula of a substance like propane (C3H8) can be represented by a developed or semi-developed chemical formula. In this type of representation, a line is used to represent the bonds between the atoms. When there is only one bond, a single line is used, while double or triple lines indicate two or three bonds between two atoms. In the developed molecular formula, all of the bonds between the atoms are illustrated, while in the semi-developed molecular formula, the bonds between the carbon (C) and hydrogen (H) atoms are not illustrated. For example, propane can be represented by a developed molecular formula in which all of the bonds are visible, or by a semi-developed molecular formula that shows that a CH3 group is attached to a CH2 group, which is in turn attached to another CH3 group.

SECTION 4.3

1. Compare the energies required to break the following bonds: a) a single NON bond b) a double NPN bond c) a triple N N bond Which bond is the strongest? Which is the weakest? Explain your answers. 2. The following reaction is carried out in the lab and an enthalpy change of 1650 kJ/mol is obtained. 2 CO (g) 2 C (g)  O2 (g) Given this, determine if the bond between the carbon (C) and the oxygen (O) is single, double or triple in the carbon monoxide (CO) molecule. 3. Consider the following reaction: A2  B2 n 2 AB If the AOB bond is stronger than the AOA and BOB bonds, will the reaction be endothermic or exothermic?

UNIT 2 Energy Changes in Reactions

CH3OCH2OCH3 Figure 10 The developed molecular formula (above) and the semi-developed formula (below it) of propane, whose empirical molecular formula is C3H8.

Energy balance

To answer questions 1,4 and 6, use Table 8.5 on page 419 of the Appendices.

160

H H H A A A HOCOCOCOH A A A H H H

4. Calculate the enthalpy change of the following reactions and indicate if the reactions are endothermic or exothermic: a) H2 (g) n 2 H (g) b) H2 (g)  Br2 (g) n 2 HBr (g) c) BrF3 (g)  2 H2 (g) n HBr (g)  3 HF (g) 5. Consider the following reaction: X2  Y2 n 2 XY If the enthalpy change of reaction is 20 kJ/mol of XY, and the energy of the XOX bond is 100 kJ/mol and that of YOY is 50 kJ/mol, what is the energy of the XOY bond? 6. Calculate the enthalpy change of the following reactions. Here, the use of a Lewis diagram is recommended, since certain reactants and certain products contain double or triple bonds. Indicate if the reactions are endothermic or exothermic. a) 2 N (g) n N2 (g) b) 2 C4H10 (g)  13 O2 (g) n 8 CO2 (g)  10 H2O (g) c) CO2 (g) n C (s)  O2 (g) d) 2 HCCH (g)  5 O2 (g) n 4 CO2 (g)  2 H2O (g)

4.4 Calculating enthalpy change using stoichiometry Calculating enthalpy change using stoichiometry allows us to obtain the enthalpy change that accompanies a chemical reaction whose mass of reactants and products is known. It is possible to calculate the energy absorbed or released by a chemical reaction whose quantity of reactant or product is known by using stoichiometry. The molar enthalpy change of a reaction expressed in kilojoules per mole (kJ/mol) represents the energy involved in this reaction, based on the stoichiometric coefficients. The energy supplied or released during a chemical reaction is proportional to the quantity of reactants and products, in moles. For example, the synthesis of water occurs according to the following thermochemical equation:

H2 (g) 

1 O n H2O (g)  244 kJ 2 2 (g)

Therefore, this reaction releases 244 kJ, for the formation of one mole of water. If this reaction was carried out with two moles of water, the energy released would be twice as great, therefore, the enthalpy change would be 488 kJ. For a reaction carried out in a laboratory, it is pertinent to know the energy involved in the reaction and the quantities of reactants and products actually used. When the enthalpy change is known for the reaction, and the mass of a reactant is provided, it is possible to calculate the energy supplied or released for this given mass of reactant. The following example shows how to calculate the energy released or supplied by a reaction when the enthalpy of reaction or the reactant mass is known: Example A Consider the following reaction: H2 (g)  F2 (g) n 2 HF (g)  546.6 kJ Calculate the energy released during this reaction if a mass of 5.50 g of hydrogen (H2) is used with a sufficient amount of fluorine (F2). Data: m  5.50 g M  2.016 g/mol

Solution: 1. Calculation of the number of moles of H2: m m 5.50 g M n   2.73 mol n M 2.016 g/mol

Energy  ? 2. Calculation of the energy released for 2.73 mol of H2 : 546.6 kJ ?  2.73 mol 1 mol 546.6

kJ • 2.73 mol  1492 kJ 1 mol Answer: The reaction carried out with 5.50 g of hydrogen (H2) releases 1492 kJ of energy. CHAPTER 4 Enthalpy Change

161

Example B Ethane (C2H6) is one of the substances that makes up natural gas. Its combustion occurs according to the following equation: 2 C2H6 (g)  7 O2 (g) n 4 CO2 (g)  6 H2O (g)  2857 kJ Calculation of the energy needed to produce 10.0 g of water. Data:

Solution : m  10.0 g M  18.015 g/mol

Energy  ?

1.Calculation of the number of moles of H2O: m M n m 10.0 g n   0.555 mol M 18.015 g/mol 2.Calculation of the energy needed to produce 0.555 mol of H2O: 2857 kJ ?  6 mol 0.555 mol

2857 kJ • 0.555 mol  264.3 kJ 6 mol Answer : The reaction that produces 10.0 g of water releases 264 kJ of energy.

SECTION 4.4

Calculating enthalpy change using stoichiometry

1. Consider the synthesis reaction of: 3 A  2 B n 4 C. If the standard enthalpy change of this reaction is 500 kJ/mol, what energy is needed to make 2.4 moles of A react? 2. Consider this reaction: X  2 Y n 2 Z  200 kJ. What energy will be released if 1.5  103 moles of Y are used? 3. What quantity of energy is needed to melt 1.0 kg of ice, if the melting of ice has a enthalpy change of 6.01 kJ/mol at 0°C? 4. What quantity of energy is released by the condensation of 300.0 g of water vapour into liquid phase, if the vaporization of water has an enthalpy change of 40.66 kJ/mol? 5. The dissolution of hydrochloric acid (HCl) in water is an exothermic phenomenon with an enthalpy change of 18.0 kJ/mol. What quantity of heat will be released when preparing 200.0 mL of a solution with a concentration of 2.50 mol/L?

162

UNIT 2 Energy Changes in Reactions

6. The following synthesis reaction has an enthalpy change of 411 kJ/mol: 1 Na (s)  Cl2 (g) n NaCl (s) 2 What mass of sodium (Na) would be needed for the reaction to release 2000 kJ? 7. Below is how iron oxide (Fe2O3) is formed: 4 Fe (s)  3 O2 (g) n 2 Fe2O3 (s)  1648.4 kJ Using 100.0 g of iron and sufficient oxygen (O2), what energy is released by the reaction? 8. The neutralization of hydrochloric acid (HCl) by sodium hydroxide (NaOH), is the following: HCl (aq)  NaOH (aq) n NaCl (aq)  H2O (l) What is the enthalpy change of this reaction if 3.21 g of sodium chloride (NaCl) are produced while releasing 3.19 kJ?

9. Below is the XO synthesis reaction involving an unknown element, X. 2 X (s)  O2 (g) n 2 XO (s) Its enthalpy change is 602 kJ/mol. A mass of 18.23 g of X is used with the right quantity of oxygen (O2), and the reaction releases 451.5 kJ of heat. Find the molar mass of X and identify element X. 10. The formation of glucose (C6H12O6) by photosynthesis occurs according to the following reaction, using energy in light form: 6 CO2 (g)  6 H2O (l)  2803 kJ n C6H12O6 (s)  6 O2 (g) What mass of glucose will be produced if a lamp supplies 2000 kJ and there is a sufficient quantity of reactants? 11. The formation of ammonia (NH3) occurs according to the following reaction: N2 (g)  3 H2 (g) n 2 NH3 (g) a) Calculate the enthalpy change of this reaction and indicate if it is endothermic or exothermic. Use the Table of average bond energy (see Appendix 8 on page 419 ). b) Calculate the quantity of energy absorbed or released during the reaction of 10.0 g of nitrogen (N2) with 10.0 g of hydrogen (H2), if there is a surplus of one of these two reactants. 12. If the enthalpy change of combustion of ethane (C2H6) is 1.56 kJ/mol, calculate the quantity of heat transferred by the combustion of: a) 5.0 mol of ethane b) 40.0 g of ethane 13. Tetraphosphorus decaoxide (P4O10) is an acidic oxide. It reacts with water to produce phosphoric acid (H3PO4) according to the following reaction: P4O10 (s)  6 H2O (l) n 4 H3PO4 (l)

H  64.3 kJ/mol

a) Rewrite the thermochemical equation and include the enthalpy change. b) What quantity of energy is released from the reaction of 5.0 moles of tetraphosphorus decaoxide with surplus water? c) What quantity of energy is released from the formation of 235.0 g of phosphoric acid?

14. On sunny days, chemical substances can store solar energy which they release later on. Certain hydrated salts dissolve in water in an endothermic process when they are heated and release heat when they solidify. For example, Glauber’s salt (Na2SO4  10 H2O (s)) solidifies at 32°C and releases 78.0 kJ per mole of salt. What is the enthalpy change during the solidification of 454.0 g of Glauber’s salt, which is used to supply household energy, if we consider that the molar mass of this salt includes 10 molecules of water? 15. For each of the following balanced equations and enthalpy changes, write the symbol and calculate the enthalpy change of combustion for one mole of the substance that reacts with oxygen (O2). a) 2 H2 (g)  O2 (g) n 2 H2O (g)  488 kJ b) 4 NH3 (g)  7 O2 (g) n 4 NO2 (g)  6 H2O (g)  1272.1 kJ c) 2 N2 (g)  O2 (g)  163.2 kJ n 2 N2O (g) d) 3 Fe (s)  2 O2 (g) n Fe3O4 (s)  1118.4 kJ 16. Aluminum (Al) reacts quickly with chlorine (Cl2) gas to produce aluminum chloride (AlCl3). This reaction is very exothermic. 3 Al (s)  Cl2 (g) n 2 AlCl3 (s) H  704 kJ/mol 2 What is the enthalpy change when 25.0 g of aluminum react completely with the surplus chlorine? 17. Calculate the enthalpy change during the combustion of 100.0 g of methane (CH4) in a natural gas water heater, given that the enthalpy change of the combustion of methane is 802.0 kJ/mol. 18. Calculate the enthalpy change during the combustion of 10.0 g of butane (C4H10) in a camping stove, according to the following equation: 2 C4H10 (g)  13 O2 (g) n 8 CO2 (g)  10 H2O (g) The enthalpy change of combustion of butane is 1328.7 kJ/mol. 19. The combustion of ethanol (CH3CH2OH) occurs according to the following reaction: CH3CH2OH (l) 3 O2 (g) n 2 CO2 (g) 3 H2O (g) Calculate the energy released by the reaction if 25.0 g of water is produced. Use the table of average bond energy (see Appendix 8 on page 419).

CHAPTER 4 Enthalpy Change

163

APPLICATIONS Phase-changing materials In winter, thermal insulation considerably reduces heat loss. Meanwhile, in summer, air conditioners are widely used. In addition to their high cost and energy consumption, air conditioners produce greenhouse gases. An ecological alternative is to store energy using phase-changing materials. These materials change phase as a function of ambient temperature. Consequently, they are able to store or release heat. One of the first applications of phase-changing materials was in building insulation. Placed in the partitions of walls and ceilings, they melt during the day by absorbing excess thermal energy produced when the outside temperature exceeds their melting point. At night, when the temperature drops, they solidify by releasing the accumulated energy.

Consequently, it is important to find materials whose melting temperature and crystallization temperature are as close as possible to human body temperature. For textiles, as in the case of building insulation, paraffin is mainly used. Paraffin is a mixture of solid and liquid hydrocarbons which, once mixed, maintain a mean temperature of 30 to 34°C, which is very comfortable for the human body. In the case of intelligent textiles, paraffin is inserted in minuscule amalgamated impermeable bags to create a membrane. When the outside temperature exceeds the melting point of paraffin, the paraffin melts and absorbs energy in the form of heat. Inversely, when the ambient temperature cools, it solidifies and releases the stored heat (see Figure 12).

The functioning of these materials is based on micro encapsulation, a procedure in which a product is enclosed in microparticles (see Figure 11).

Exterior Microencapsulated paraffin membrane Body heat absorbed by

The heat stored

the membrane

is released to Body

the body when necessary

Figure 12 How phase-change materials function in regulating body heat

Figure 11 These paraffin beads are ready to be incorporated into construction material.

There are a growing number of applications for phase-changing materials, particularly intelligent textiles. Intelligent textiles are gaining ground in an increasing number of applications such as in clothing, shoes, bedding and sleeping bags. The goal is to passively regulate body temperature as a function of ambient temperature.

164

UNIT 2 Energy Changes in Reactions

Clothing made of this type of fabric is more comfortable because it reduces perspiration to a minimum. Increasingly widespread use of phasechanging materials in construction and in mass-market goods will reduce heating and air conditioning demands, thus helping to protect the environment.

The use of chemical energy Eating represents the most personal use of chemical energy. The human organism converts the chemical energy in foods into mechanical energy and body heat using cellular respiration, which is a form of combustion.

heating system that distributes steam to the community through underground piping (see Figure 14).

Today, humans rely on several energy sources such as wood, oil, nuclear, wind and hydroelectric energy, as well as that derived from the Sun and the Earth. However, only wood and oil are chemical forms of energy. As with all forms of energy, the process of extracting chemical energy usually requires heat. Over 450 000 years ago, human beings started burning wood as a source of heat and to feed themselves. In this case, fire transforms the chemical energy of combustion into thermal energy. During the Industrial Revolution of the 19th century, human beings burned large quantities of coal and oil in order to produce steam to power various types of engines (see Figure 13). Thereafter, heat became an intermediate product that powered manufacturing and contributed to economic development. At the same time, the use of gasoline to power automobile motors, which were revolutionizing transportation, rapidly increased. These types of engines use gasoline combustion to produce very hot gases that generate the high pressure needed to power pistons and wheels.

Figure 13 Steam engines required an enormous amount of work on the part of those who had to feed them coal.

Wood is still used today as a fuel in homes with wood-burning stoves as well as in certain community projects that use industrial waste from the logging industry. These projects recycle and burn large quantities of woodchips to power a central

Figure 14 All of the buildings in the Cree community of Oujé-Bougoumou, north of Chibougamau, are heated by a thermal power plant that burns woodchips.

The chemical energy in coal and other fossil fuels, such as natural gas, is still used to produce electricity. In a thermal power plant, the steam produced through heat provides the work needed to turn the turbines to produce electricity. However, this fuel emits large quantities of greenhouse gases. In homes, chemical energy also comes from the combustion of fossil fuels like propane or methane, the main component of natural gas, which assures the cooking of food and heating. Research into finding new, less polluting sources of energy is ongoing. The most notable avenues of the future are hydroelectricity, wind turbines and solar panels (see Figure 15).

Figure 15 Solar panels are increasingly used to heat homes.

CHAPTER 4 Enthalpy Change

165

CHAPTER

4

Enthalpy Change

4.1 Enthalpy and enthalpy change • Enthalpy (H) is the total energy of a system, that is, the sum of all the potential and kinetic energies that the system contains at constant pressure. • Enthalpy change (H) is the energy exchanged between a system and its environment during a physical change or a chemical reaction at constant pressure. • Enthalpy change of a reaction corresponds to: H  Hp – Hr

4.2 Endothermic and exothermic reactions Gas Boiling point Temperature

• Changes in phase are endothermic physical changes when they involve a decrease in the forces of attraction between the molecules in the following direction: solid n liquid n gas.

Liquid Melting point

Hvaporization Solid Hmelting

• Changes in phase are exothermic physical changes when they involve an increase in the forces of attraction between the molecules in the following direction: gas n liquid n solid.

Heat supplied

Gas

Temperature

Boiling point Liquid Hcondensation

Melting point

Hsolidification

• An endothermic chemical reaction must absorb heat in order to occur. Its enthalpy change (H) is positive, since the enthalpy of the products is greater than that of the reactants. The bonds are stronger in the reactants than in the products.

Heat released

Bond breaking  ENERGY

Molecule

• An exothermic chemical reaction releases heat when it occurs. Its enthalpy change (H) is negative since the enthalpy of the products is less than that of the reactants. The bonds are weaker in the reactants than in the products.

166

UNIT 2 Energy Changes in Reactions

Solid

Separated atoms

Bond formation  ENERGY

Molecule Separated atoms

• An endothermic reaction can be expressed by a thermochemical equation in two different ways: 1) CaCO3 (s)n CaO (s)  CO2 (g) H 178 kJ/mol 2) CaCO3

(s)

 178 kJ n CaO (s)  CO2 (g)

• An exothermic reaction can be expressed by a thermochemical equation in two different ways: 1) 4 Fe (s)  3 O2 (g) n 2 Fe2O3 (s) H 824.2 kJ/mol 2) 4 Fe (s)  3 O2 (g) n 2 Fe2O3 (s)  1648.4 kJ

4.3 Energy balance An endothermic process

An exothermic process

Separated atoms

Separated atoms

Hbonds broken

Hp

Enthalpy

Enthalpy

Hbonds formed

Hbonds broken Hr

Hbonds formed

H

H Hr

Hp

Progress of the reaction

Progress of the reaction

• Breaking chemical bonds is an endothermic process, while forming new ones is an exothermic process. • The greater the energy needed to break a bond, the stronger the bond. • To calculate the enthalpy change of a chemical reaction, the enthalpy change of the bonds broken (Hbonds broken) and the enthalpy change of bonds formed(Hbonds formed) must be added together. H  Hbonds broken  Hbonds formed

4.4 Calculating enthalpy change using stoichiometry • The energy absorbed or released by a reaction is proportional to the number of moles of reactants and products involved.

CHAPTER 4 Enthalpy Change

167

CHAPTER 4

Enthalpy Change

To answer questions 7, 9, 10, 20 and 21, use Table 8.5 on page 419 of the Appendices. 1. Which of the following four graphs best represents the solidification of mercury (Hg), whose enthalpy change of solidification is 2295 kJ/mol ? a) T

b) T

2295 kJ/mol

Q

c) T

2295 kJ/mol

Q

d) T

3. Draw the enthalpy diagram of each of the following reactions. On each graph, indicate the value of the enthalpy change of the reaction. a) NaOH (s)  CO2 (g) n NaCO3 (s)  127.5 kJ b) 2 C (s)  H2 (g)  227 kJ n C2H2 (g) 4. Create the diagram of phase change of acetone (CH3COCH3), from the solid phase to the gas phase, indicating its temperature as a function of the heat supplied. Using the following data note the intervals which correspond to the enthalpy changes. – Melting point: 94.6°C – Boiling point: 56.1°C – Enthalpy change of melting: 5.7 kJ/mol – Enthalpy change of vaporization: 31.3 kJ/mol 5. Examine the formation reaction of nitrogen monoxide (NO).

2295 kJ/mol

Q

2295 kJ/mol

Q

N2 (g)  O2 (g) n 2 NO (g)

Q : Heat absorbed or released (kJ/mol) T : Temperature of the substance (°C)

Temperature (oC)

2. The following graph indicates the phase changes of argon (Ar) according to temperature, as a function of the heat released. Argon, initially in the gas phase at high temperature, passes through the three phases of matter as it cools according to the following diagram:

Liquid

189.36

Solid 1.188 kJ/mol Heat released

a) b) c) d)

168

What is the vaporization temperature of argon? What is its solidification temperature? What is its enthalpy change of solidification? What is its enthalpy change of liquid condensation?

UNIT 2 Energy Changes in Reactions

a) Rewrite the thermochemical equation by integrating the enthalpy change into the equation. b) Use an enthalpy diagram to represent this reaction. c) What is the enthalpy change associated with the formation of one mole of nitrogen monoxide? d) What is the enthalpy change associated with the reaction of 2.5  102 g of nitrogen (N2) with a sufficient quantity of oxygen (O2) ? 6. The reaction of iron (Fe) with oxygen (O2) is very common. You can observe the resulting rust on buildings, vehicles and bridges.

Gas 185.85

6.447 kJ/mol

H  90.2 kJ/mol

4 Fe (s)  3 O2 (g) n 2 Fe2O3 (s)  1648.4 kJ a) What is the enthalpy change associated to this reaction? b) Use an enthalpy diagram to represent this reaction. c) What is the enthalpy change associated with the formation of 23.6 g of iron oxide (Fe2O3) ?

7. The Ostwald process is used to produce nitric acid (HNO3) from ammonia (NH3) for use in the manufacturing of nitrate-based fertilizers. Calculate the standard enthalpy change of the reactions in this process, as they appear below: a) 4 NH3 (g)  5 O2 (g) n 4 NO (g)  6 H2O (g) b) 2 NO (g)  O2 (g) n 2 NO2 (g) c) 3 NO2 (g)  H2O (l) n 2 HNO3 (l)  NO (g) 8. A chemist wants to determine the enthalpy change of the combustion of octane (C8H18). She places 5.0 g of octane in a bomb calorimeter immersed in 3.0 L of water. At the end of the combustion, which occurred at constant pressure, the temperature of the water in the calorimeter increased from 22.0°C to 40.91°C. What is the enthalpy change of this combustion? 9. Calculate the enthalpy for each of the reactions below and determine which of these molecules of fuel releases the most heat during combustion: a) Propane: C3H8 (g)  5 O2 (g) n 3 CO2 (g)  4 H2O (g) 25 n b) Octane: C8H18 (g)  O 2 2 (g) 8 CO2 (g)  9 H2O (g) c) Methanol: CH3OH (l)  2 O2 (g) n CO2 (g)  2 H2O (g) d) Ethylene: C2H4 (g)  3 O2 (g) n 2 CO2 (g)  2 H2O (g) e) Paraffin: C25H52 (s)  38 O2 (g) n 25 CO2 (g)  26 H2O (g) 10. Consider the following reaction: 2 H2S (s)  2 O2 (g) n 2 H2O (g)  2 SO2 (g) If the sulphur dioxide (SO2) is composed of an SPO double bond and an SOO single bond, and the enthalpy change of the reaction is 1036 kJ, determine the energy of the SPO bond. 11. Methanol (CH3OH) is a fuel used in the warming tray of fondue sets. In an experiment in which a jam jar filled with water is used as a calorimeter, the combustion of 2.9 g of methanol increases the temperature of 650 g of water by 20.9°C. Based on this observation, calculate the enthalpy change of the combustion of methanol.

12. Consider the following reaction: 2 A  3 B  1000 kJn 4 C where A, B and C are molecules. A mass of 35.5 g of A is used with the right quantity of B, and the mixture of reactants is heated with an energy of 750 kJ. Find the molar mass of A. 13. Decane (C10H22) is one of the hundreds of gasoline compounds. The enthalpy change of combustion of decane is 6.78 kJ/mol. What mass of decane must be burned to increase the temperature of 500 mL of water from 20.0°C to 55.0°C? 14. A lab experiment involves the following reaction: Ba(NO3)2 (s)  K2SO4 (aq) n BaSO4 (s)  2 KNO3 (aq) A researcher adds 261 g of barium nitrate (Ba(NO3)2) to 2.0 L of potassium sulfate (K2SO4) solution in a beaker. As the barium nitrate dissolves, a barium sulfate (BaSO4) precipitate is formed. The water in the beaker increases from 26.0°C to 29.1°C. Calculate the enthalpy change of the reaction of the barium nitrate. The reaction mixture has approximately the same density and the same specific heat capacity as water, that is, 1 g/mL and 4.184 J/(g°C). 15. Consider the reaction D2  E2 n 2 DE. The DOD bond has an energy of 400 kJ/mol, and the EOE bond has an energy of 500 kJ/mol. If the reaction is exothermic, what is the energy of the DOE bond? Select one of the following possibilities. Explain your answer. a) EDOE  900 kJ/mol e) EDOE  450 kJ/mol b) EDOE  900 kJ/mol f) EDOE  450 kJ/mol c) EDOE  900 kJ/mol g) EDOE  450 kJ/mol d) EDOE  450 kJ/mol 16. Examine the following thermochemical equation: H2 (g)  I2 (g)  53.0 kJ n 2 HI (g) a) What is the enthalpy change of reaction? b) What quantity of energy is required for 4.57  1024 molecules of iodine (I2) to react with surplus hydrogen (H2)? c) Draw the enthalpy diagram corresponding to this reaction.

CHAPTER 4 Enthalpy Change

169

17. The combustion of methane (CH4) occurs according to the following reaction and releases 802 kJ/mol: CH4 (g)  2 O2 (g) n CO2 (g)  2 H2O (g) Calculate the energy released during the complete combustion of a 3.5-L container of methane at SATP. 18. Consider the following reaction: B2H6 (g)  3 O2 (g) n B2O3 (s)  3 H2O (g) Calculate the enthalpy change of the reaction if it produces 2.3 L of water vapour at SATP and re leases 182.5 kJ. 19. A quantity of ice at 0.0°C is placed in a bomb calorimeter immersed in 2.0 L of water. Immediately upon completion of the melting, at constant normal pressure, while the temperature of the melted ice has not yet increased, the water in the calorimeter (2.0 L) decreased from 20.0°C to 19.9°C. What was the initial mass of the ice placed in the calorimeter, if the melting of ice has an enthalpy change of 6.01 kJ/mol at 0°C? 20. Calculate the enthalpy change of the following reaction by performing the energy balance. Indicate if they are endothermic or exothermic. Verify the presence of double or triple bonds. a) 2 Cs  4 H2 (g)  O2 (g) n 2 CH3OH (l) b) CH3COOH (l)  4 O2 (g) n 2 CO2 (g)  2 H2O (g) c) H2CPCH2 (g)  F2 (g) n CH2FCH2F (g) d) 2 CH2CHCH3 (g)  2 NH3 (g)  3 O2 (g) n 2 CH2CHCN (g)  6 H2O (g) e)

H A n O O O H2C CH2  HCN HOC C C A O H

H A N A H

21. Consider the following reaction: H2CCHCH3 (g)  H2 (g) n CH3CH2CH3 (g) Calculate the energy absorbed or released if 30 g of propylene (H2CCHCH3) reacts with 0.76 g of hydrogen (H2). Draw the Lewis diagrams of the reactants and products to verify the presence of double or triple bonds.

170

UNIT 2 Energy Changes in Reactions

22. When 50 mL of hydrochloric acid (HCl) at 1.0 mol/L reacts with 75 mL of sodium hydroxide (NaOH) at 1.0 mol/L in a beaker, the temperature of the solution increases from 20.2°C to 25.6°C. Determine the enthalpy change of this neutralization reaction. HCl (aq)  NaOH (aq) n NaCl (aq)  H2O (l) 23. A neutralization reaction between nitric acid (HNO3) and sodium hydroxide (NaOH) is carried out in a bomb calorimeter. HNO3 (aq) NaOH (aq) n NaNO3 (aq) H2O (l) A volume of 250 mL of a solution of HNO3 at 6.0 mol/L is placed in a bomb, and a sufficient quantity of NaOH is added to complete the reaction. The 1.0 litre of water in the calorimeter increases from 20.0°C to 40.6°C. What is the enthalpy change of this neutralization? The reaction mixture has approximately the same density and the same specific heat capacity as that of the water, that is, 1 g/mL and 4.184 J/(g°C). 24. Energy is converted during animal cellular respiration and plant photosynthesis. Cellular respiration involves a series of exothermic reactions where food substances (like glucose) undergo the following combustion inside animal cells. C6H12O6 (s)  6 O2 (g) n 6 CO2 (g)  6 H2O (g) H  2803 kJ/mol Photosynthesis involves a series of endothermic reactions in which green plant cells use light energy to produce glucose obtained from carbon dioxide (CO2) and water. This combustion reaction is the reverse of that of animal cellular respiration. a) Write the thermochemical equation of photosynthesis in two different ways. b) Draw the enthalpy diagram for the respiration and photosynthesis reactions.

Graphical Representation of Enthalpy Change

F

or many years, chlorofluorocarbons (CFCs) used in human activity have been slowly destroying the ozone layer of the Earth’s atmosphere. This situation raises environmental concerns. The ozone layer absorbs some of the Sun’s ultraviolet rays and prevents them from reaching us. Ozone molecules (O3) absorb the energy from these rays and are transformed into oxygen (O2) in an exothermic reaction. This decomposition reaction is reversible. It is therefore possible to trigger the reverse reaction, that is, to produce ozone using oxygen, but this requires a very large quantity of energy, similar to a major electrical discharge.

This chapter will explain how an energy diagram can be used to observe the change in the potential energy of chemical reactions. This type of diagram shows that a minimum quantity of energy is needed to trigger a reaction, whether endothermic or exothermic: this is called activation energy. Energy diagrams enable us to observe and evaluate the ease and reversibility of chemical reactions.

5.1

Activated complex, activation energy and the energy diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

5.2

Observing the use of an energy diagram to plot the progress of a conversion . . . . . . . . . . . . . . 176

Review Endothermic and exothermic reactions . . . . . . . . . . . . . . . . . . . . . . 26 Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Potential energy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 The law of conservation of energy . . . . . . . . . . . . . . . . . . . . . . . . . 32

CHAPTER 5 Graphical Representation of Enthalpy Change

171

5.1 Activated complex, activation energy and the energy diagram The activated complex is an unstable cluster of colliding reactant atoms that is formed during the partial conversion of reactants into products. Activation energy is a minimum quantity of energy needed for a chemical reaction to occur, whether endothermic or exothermic. The energy diagram of a reaction is a graph that is used to visualize the energy change of the substances involved in that reaction. The enthalpy of a system is the sum of all of the types of potential and kinetic energy that it contains. The energy contained in the chemical bonds between the atoms of molecules constitutes an important form of potential energy. During a chemical reaction, the reactant molecules collide with each other, causing bonds to be broken and new ones to be formed, which create different molecules. These new molecules do not contain exactly the same potential energy as the initial molecules because their bonds are different.

See Endothermic and exothermic chemical reactions, p. 152.

The change in kinetic energy of the molecules during the reaction is exchanged with the environment in the form of heat energy. Therefore, the kinetic energy of the products is the same as that of the reactants. This means that the difference in potential energy between the reactants and the products corresponds to the enthalpy change of the reaction (H). In an endothermic reaction, the potential energy of the reactants is less than that of the products, which produces a positive enthalpy change. When the products contain more potential energy than the reactants, energy, primarily in the form of heat, is absorbed during the reaction. The opposite occurs during an exothermic reaction, where the surplus energy of the reactants dissipates into the environment during the formation of the products, whose bonds contain less potential energy. However, whether the reaction is endothermic or exothermic, energy must be supplied to break the bonds of the reactants, while the formation of new bonds releases energy. It is the difference between these two types of energy that determines the enthalpy change of the reaction.

5.1.1

Activated complex

In all chemical reactions, there is an intermediary moment when the reactant molecules come into contact with each other, but have not yet separated to form the new compounds constituting the products. During this very brief moment, the molecules that have just collided begin to interact. Some of the bonds weaken while new bonds begin to form. This cluster of intermediary atoms between the reactants and the products is called the activated complex. It is an unstable cluster of atoms that lasts for such a brief time during the reaction that it cannot be detected. Its existence is, therefore, hypothetical, but all experiments seem to confirm it.

172

UNIT 2 Energy Changes in Reactions

The activated complex created during the formation reaction of hydrogen chloride (HCl) is a good example. H2 (g)  Cl2 (g) n 2 HCl (g) The activated complex corresponds to the arrangement of atoms grouped together during the collision between a hydrogen molecule (H2) and a chlorine molecule (Cl2) before they separate into two distinct molecules of hydrogen chloride (see Figure 1). Reactants

Activated complex

Products

Chlorine (Cl) Hydrogen (H)

Figure 1 The reaction of hydrogen (H2) and chlorine (Cl2) to form hydrogen chloride (HCl). For a short period of time, the reactant molecules are joined before separating in order to form products: this is the activated complex.

5.1.2

Activation energy

The level of energy of the activated complex is characterized by a very high potential energy, since it involves the reorganization of the forces of attraction between all of the atoms of the reactants. The activated complex therefore possesses more energy than the reactants and the products. In order for a chemical reaction to occur, the reactants must receive enough energy to form the activated complex. This explains why some reactions, although they are exothermic, do not occur spontaneously; a sufficient quantity of energy is needed to trigger them. For example, even though firewood is flammable, it can be safely stored in the house, since combustion does not occur spontaneously when firewood comes into contact with the oxygen (O2) in the air (see Figure 2). The firewood must be lit, providing energy in the form of heat, in order for the reaction to begin. The difference in energy between the activated complex and the reactants corresponds to the activation energy (Ea). This energy constitutes an energy barrier that must be crossed in order for the reaction to occur. To fully grasp this concept, use the analogy of a soccer player (reactants) who wants to get the ball over a hill and into the net (products) which is on the other side (see Figure 3). The player must transfer enough energy to the ball to get it over the top of the hill and into the net. When the ball is at the top of the hill, it is unstable and is not immobile; it can roll down either side of the hill.

Figure 2 Firewood can be stored without spontaneous combustion, an exothermic reaction, occurring spontaneously.

Figure 3 To score a goal, the player must transfer enough energy to the ball to get it over the hill.

CHAPTER 5 Graphical Representation of Enthalpy Change

173

Combustion reactions of hydrocarbons, such as gas, natural gas and propane (C3H8), are generally very exothermic. However, they do not occur spontaneously on contact with ambient oxygen (O2). A source of heat is needed to set off the reaction. The flash point of a liquid corresponds to the temperature at which its evaporation is great enough for the vapour to mix with the ambient air and for a single spark to trigger combustion. The flash point of natural gas is –40°C, which makes it an extremely flammable substance at ambient temperature.

5.1.3

Energy diagram

During a chemical reaction, the energy diagram shows the change in potential energy and allows for the visualization of the difference in potential energy between the reactants and the products (see Figure 4). The energy diagram also illustrates the increase in potential energy that occurs with the formation of the activated complex, which corresponds to the activation energy (Ea). Energy diagram of a reaction Activated complex

Potential energy

The flash point

This state corresponds to the activated complex, an unstable cluster of atoms with a very short lifespan, since it can easily change back into reactants if the energy supplied is insufficient, or convert themselves into products if the energy is sufficient.

Ea

Products

H Reactants

Progress of the reaction a) An endothermic chemical reaction Energy diagram of a reaction

Potential energy

Activated complex

Figure 5 A single spark can supply enough activation energy to trigger the combustion of natural gas in a lighter.

Ea Reactants

H Products

Progress of the reaction b) An exothermic chemical reaction

Figure 4 The energy diagram shows the change in potential energy over the course of a chemical reaction. See Energy balance, p. 156.

174

The three levels of the energy diagram, namely the reactants, the activated complex and the products, are similar to the energy balance diagram used to calculate the enthalpy change of a reaction. Despite the similar form of these two diagrams, they have important distinctions. To begin with, in the energy diagram, the y-axis represents the potential energy only, while in the energy balance diagram, the y-axis indicates enthalpy, which includes kinetic energy as well as potential energy.

UNIT 2 Energy Changes in Reactions

Then, when plotting an energy balance diagram, the enthalpy change values used are those that serve to completely break bonds (Hbonds broken) or to form bonds from free atoms (Hbonds formed). The highest enthalpy peak corresponds to free atoms. However, in an energy diagram, the highest peak corresponds to the activated complex. This is when the reactant bonds are partially broken, while the product bonds are partially formed. Therefore, the activation energy is not equivalent to the enthalpy change needed to break the bonds. Its value is determined by experimentation, not calculation. In reality, the atoms are never free; new bonds form at the same time as the old bonds break. In contrast, the highest peak of the energy balance does not correspond to a physical reality. Rather, the activated complex illustrated by the energy diagram, although hypothetical, reflects the real structure of molecules during the reaction. However, the energy balance diagram can be used to calculate the enthalpy change of the reaction. The results obtained are valid, since the reactant bonds are indeed broken and the product bonds are truly formed. If the enthalpy change of a reaction that takes place in a single step and its activation energy are known, its energy diagram can be plotted, as shown in the following example: Example Plot the energy diagram of a reaction between methane (CH 4) and sulphur (S2), which reacts according to the following equation: CH4 (g)  2 S2 (g) n CS2 (g)  2 H2S (g) The activation energy of this reaction is 140 kJ/mol its enthalpy change is 14 kJ/mol. Data: Ea  140 kJ/mol H  14 kJ/mol

Solution: 1. Since this is an endothermic reaction, the energy of the products will be 14 kJ/mol greater than that of the reactants. 2. The activation energy will be 140 kJ/mol greater than that of the reactants. Energy diagram of a methane sulphur reaction

Potential energy (kJ/mol)

Activated complex

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

JOHN CHARLES POLANYI Canadian chemist (1929– ) John C. Polanyi won the Nobel Prize in Chemistry in 1986. With his colleagues, he developed a type of laser that captures the very weak infrared light emitted by the newly formed molecules in a reaction. He also used spec troscopy to study the “molecular dance," when reactant atoms change partners to form products. In addition to his research in chemistry, Polanyi was the founding chairman of the Canadian Pugwash Group, which advocates for peace and disarmament. He was a proponent of pure scientific research, which he likened to artistic creativity.

Ea  140

CS2 (g)  2 H2S (g) CH4 (g)  2 S2 (g)

H  14

Progress of the reaction

CHAPTER 5 Graphical Representation of Enthalpy Change

175

5.2 Observing the use of an energy diagram to plot the progress of a conversion Observing the progress of a reaction using an energy diagram allows us to determine if the reaction is endothermic or exothermic, evaluate the quantity of activation energy, compare the direct and reverse reactions, and establish whether the reaction is reversible or irreversible. The diagram of the change in the potential energy of the reaction allows us to visualize the steps of a chemical reaction and provides a lot of information about the reaction. The shape of its curve helps us to determine if the reaction is endothermic or exothermic.

5.2.1

Height of the activation energy barrier and the spontaneous reaction

The activation energy value indicates the size of the energy barrier to cross in order for the reaction to occur. In the example of the soccer player, this represents the height of the hill and indicates to the player the quantity of energy she must transfer to the ball in order for it to get over the hill (see Figure 3 on page 173). The higher the hill, the more energy the player must transfer when kicking the ball. Therefore, the higher the activation energy, the more energy must be supplied to the reactants, whether the reaction is endothermic or exothermic. The reaction rate depends particularly on the activation energy, and the reaction tends to be slower when the activation energy is high.

See Types of collisions, p. 236.

The increase in the potential energy of the system until the activated complex is formed is derived from the conversion of kinetic energy into potential energy. If kinetic energy is transferred to the system, it can be converted into potential energy that will supply the activation energy. In order for this to occur, the substance can be heated, which increases the thermal agitation of the molecules, and supplies it with kinetic energy. Moreover, the agitation of reactant molecules favours the reaction by increasing the frequency of the collisions between these molecules. Frequent collisions between reactants increases the probability of the formation of the activated complex, which requires the molecules to come into contact with each other. In an endothermic reaction, more energy than the enthalpy change of the reaction must be supplied in order to cross the activation energy barrier. In an exothermic reaction, the necessary activation energy must first be supplied to trigger the reaction, which then releases heat. The activation energy is different for each reaction (see Figure 6 on the following page). Sometimes, it is very low. Some reactions have such a low activation energy value that the heat provided by the ambient air is enough to cross the energy barrier. When the reactant molecules have enough energy to overcome the activation energy barrier without requiring an additional input of energy, this is called a spontaneous reaction. The case of zero activation energy (Ea 0 kJ/mol) is theoretical, since the formation of an activated complex always involves an increase in potential energy.

176

UNIT 2 Energy Changes in Reactions

Energy diagram of different reactions

Potential energy

Activated complex

Ea

Ea

Ea  0

Products

H

Reactants

Progress of the reaction a) An endothermic reaction

Energy diagram of different reactions

Potential energy

Activated complex

Reactants

H

Ea

Ea

Ea  0 Products

Progress of the reaction b) An exothermic reaction

Figure 6 These energy diagrams show the potential energy change for different chemical reactions with activation energy values that are high (red), moderate (green) and zero (blue).

Furthering

your understanding

Reaction of alkali metals in water Placing a piece of an alkali metal in a container of water causes a violent reaction. The reaction is exothermic and produces hydrogen (H2), a flammable gas. The lower the alkali metal is on the periodic table, the more violent the reaction. The reaction of any alkali metal with water has an enthalpy change of approximately 200 kJ/mol. The major difference resides in the activation energy, which becomes smaller as you go down the elements and therefore results in increasingly faster spontaneous reactions. Hydrogen is released more rapidly, causing the water to spatter, and the heat from the reaction immediately ignites the potassium.

Figure 7 The violent reaction of potassium (K), an alkali metal, with water

CHAPTER 5 Graphical Representation of Enthalpy Change

177

5.2.2

Direct and reverse reactions

In daily life, most reactions are irreversible, that is, the reaction can only occur in one direction, from reactants to products. This is the case, for example, with a match that burns: it cannot be recreated from the products of the combustion. However, certain chemical reactions can occur in both directions: they are called reversible reactions. Reversible reactions can occur in their direct or reverse form. A direct reaction is a reaction that occurs when reactants become products, and a reverse reaction is a reaction that occurs when products once again become reactants. For example, hydrogen peroxide (H2O2) decomposes according to the following exothermic reaction:

H2O2 (l) n H2O (l) 

1 O  187.8 kJ 2 2 (g)

If this is the direct reaction, the reverse reaction is obtained by reversing the reactants and products. Moreover, the reverse reaction of an exothermic reaction will be endothermic, and vice versa. The formation reaction of hydrogen peroxide from water and oxygen (O2), which is the reverse reaction, is as follows: H2O (l) 

1 O  187.8kJ n H2O2 (l) 2 2 (g)

When a chemical reaction is reversed, the curve of its energy diagram is reversed, and the activation energy required is not the same in both directions (see Figure 8). However, the enthalpy change is the same in absolute value, since the enthalpies of the reactants and products are simply reversed. Energy diagram of a reaction Activated complex

Potential energy

Products Ea H Reactants

Progress of the reaction a) An endothermic chemical reaction

178

UNIT 2 Energy Changes in Reactions

Energy diagram of a reaction

Potential energy

Activated complex

Products

Ea

H

Reactants

Progress of the reaction b) An exothermic chemical reaction

Figure 8 The progress of a direct endothermic chemical reaction and of its reverse exothermic reaction.

The activation energy of the endothermic reaction is always greater than that of the exothermic reaction. To draw a parallel with the female soccer player (see Figure 3 on page 173), imagine a male player where the net is. The female player is on the side of the reactants and the male player is on the side of the products. The male player, who is positioned on the side with less energy, must kick the ball much more energetically than the female player to get the ball over to the other side of the hill. In the reverse chemical reaction of a reversible reaction, the change in activation energy must be considered. In the case of the decomposition of hydrogen peroxide (H2O2), the direct exothermic reaction has an enthalpy change (H 0) of 98 kJ per mole of H2O2 and an activation energy of 37.7 kJ per mole of H2O2 (see Figure 9). However, its reverse endothermic reaction has an activation energy of 135.7 kJ per mole of H2O2, that is, the sum of the enthalpy change (in absolute value) and the activation energy of the direct reaction. The reverse reaction requires more energy to occur and is unlikely. Therefore, this is an irreversible reaction. Energy diagram of the decomposition of hydrogen peroxide

Potential energy (kJ/mol)

Activated complex Ea  37.7

H 2O2

H  98

H 2O2 

1 2

O2

Progress of the reaction a) The direct exothermic reaction

CHAPTER 5 Graphical Representation of Enthalpy Change

179

Energy diagram of the reverse reaction of the decomposition of hydrogen peroxide

Potential energy (kJ/mol)

Activated complex H 2O2

Ea  135.7 H  98 H 2O2 

1 2

O2

Progress of the reaction b) The reverse endothermic reaction

Figure 9 The progress of the direct reaction of the decomposition of hydrogen peroxide (H2O2) and its reverse reaction show that their activation energies are different.

A reversible reaction is characterized by low enthalpy change, which means that the activation energy is almost the same for the direct reaction and the reverse reaction. For example, the formation reaction of hydrogen iodide (HI) is endothermic, and its activation energy is 90 kJ per mole of H. 1 1 H2 (g)  I2 (g) n HI (g)  26.5 kJ 2 2 Since the enthalpy change of this reaction is low, the reverse exothermic reaction requires an activation energy to occur, that is, 63.5 kJ/mol. Therefore, the direct and reverse reactions occur with equal ease (see Figure 10).

Energy diagram of the formation of hydrogen iodide

Potential energy (kJ/mol)

Activated complex

Ea  90

HI 1 2

H2 

1 2

H  26.5

I2

Progress of the reaction a) Direct endothermic reaction

180

UNIT 2 Energy Changes in Reactions

Energy diagram of the reversible reaction of the formation of hydrogen iodide

Potential energy (kJ/mol)

Activated complex

Ea  63.5 HI 1 H  26.5

2

H2 

1 2

I2

Progress of the reaction b) Reverse exothermic reaction

Figure 10 The energy diagrams of the reversible reaction of the formation of hydrogen iodide (HI) show that the activation energy is similar for the direct reaction and the reverse reaction.

However, it sometimes occurs that a reaction with low enthalpy change is experimentally irreversible. This is the case if the products are in the gas phase, especially if the reaction occurs in an open system. If the molecules of the gaseous products are released into the room and dissipate, it will be more difficult for them to collide again and trigger the reverse reaction. Therefore, for a reaction to be reversible, it must occur in a closed system so that no matter is lost. Moreover, the number of molecules formed must also be considered. For example, if a massive molecule decomposes into many small molecules, they will have difficulty reassembling to recreate the initial large molecule. To determine with certainty that a reaction is reversible, it is often preferable to conduct the experiment in the lab.

See Types of systems, p. 131.

Femtochemistry Femtochemistry studies chemical phenomena, as well as compounds like activated complexes, on timelines on the order of a femtosecond, that is, 1015 seconds. The principle is based on the use of a laser that emits pulses of light that lasts a femtosecond. The first pulse provides the activation energy of the reaction, then, shortly thereafter, a series of pulses make the molecules vibrate over very short intervals of time. A detector then captures the energy of these vibrations in order to obtain information on the structure of the compounds over the course of the reaction.

Figure 11 This optic installation redirects the beam of a laser emitting pulses of light for a femtosecond.

CHAPTER 5 Graphical Representation of Enthalpy Change

181

CHAPTER

5

Graphical Representation of Enthalpy Change

Energy diagram of a reaction

5.1 Activated complex, activation energy and the energy diagram

Potential energy

Activated complex

• The activated complex is an unstable cluster of atoms formed during the collision of reactant molecules before they separate to form the product molecules. The activated complex has high potential energy, since it involves a significant reorganization of the bonds between the atoms.

Energy diagram of a reaction

Potential energy

Activated complex

• The energy diagram of a chemical reaction is a graph that is used to visualize the potential energy change during a reaction.

• When a reaction is reversed, the reaction follows the curve of its energy diagram in the reverse direction. The enthalpy change is the same, in absolute value. However, the endothermic reaction has a higher activation energy than the exothermic reaction in the example shown here.

H Products

Energy diagram of the formation of hydrogen iodide Activated complex Potential energy (kJ/mol)

• A reversible reaction can occur in its direct or reverse form. A reaction with low enthalpy change is more likely to be reversible.

Ea Reactants

Progress of the reaction b) An exothermic chemical reaction

Ea  90 HI 1 2

H2 

1I 2 2

H  26.5

Progress of the reaction a) An endothermic chemical reaction Energy diagram of the reversible reaction of the formation of hydrogen iodide Activated complex Potential energy (kJ/mol)

• A reaction is considered spontaneous when the reactant molecules already have enough energy to cross the energy barrier without being given any additional energy.

H Reactants

Progress of the reaction a) An endothermic chemical reaction

• Activation energy is a quantity of energy needed for a chemical reaction to occur, whether endothermic or exothermic. Activation energy corresponds to the potential energy gain required to form the activated complex.

5.2 Observing the use of an energy diagram to plot the progress of a conversion

Products

Ea

Ea  63.5 HI

1 H  26.5

Progress of the reaction b) Reverse exothermic reaction

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UNIT 2 Energy Changes in Reactions

2

1 H2 

2

I2

Graphical Representation of Enthalpy Change

CHAPTER 5

Reactants

Products

1)

Potential energy

Products 100 kJ/mol

75 kJ/mol

Reactants

Progress of the reaction

Reactants

Products

Progress of the reaction

2. The energy diagram below represents the progress of a chemical reaction. a) Is the reaction endothermic or exothermic? b) Determine the enthalpy change and the activation energy of this reaction. c) Determine the enthalpy change of the reverse reaction (products n reactants). Does this reaction seem to be reversible? Explain your answer. 160 150

Potential energy

2)

Reactants 100 kJ/mol

75 kJ/mol Products

Progress of the reaction 3)

Potential energy

Potential energy

Progress of the reaction

Potential energy (kJ/mol)

3. Observe the diagrams of the four reactions below. a) Which diagrams represent an endothermic reaction? b) If diagram 1 represents a direct reaction, which diagram represents its reverse reaction? c) Which reaction or reactions are most likely to be reversible? Explain your answer. d) Which reaction or reactions are the fastest? Explain your answer. e) Determine the enthalpy change and activation energy for each of these four reactions.

100 kJ/mol Products Reactants

25 kJ/mol

Products Progress of the reaction

140 4)

130 120 110

Reactants

100

Potential energy

Potential energy

1. A chemist ponders the reversibility of two reactions he carried out in his laboratory. The energy diagrams of the two reactions are represented below. Which reaction is more likely to be reversible?

100 kJ/mol Reactants 25 kJ/mol Products

Progress of the reaction

Progress of the reaction

CHAPTER 5 Graphical Representation of Enthalpy Change

183

4. Reaction A  B n C  D has an activation energy of 600 kJ/mol while reaction C  D n A  B has an activation energy of 500 kJ/mol. a) Which of these two reactions is exothermic? b) What is the enthalpy change of each of these reactions? c) Which of these reactions will be faster? Explain your answer. 5. What is the activation energy of a reaction if its reverse reaction has an enthalpy change of 200 kJ/mol and an activation energy of 250 kJ/mol? Plot the energy diagram of this reaction. 6. Determine the enthalpy change of the decomposition reaction of acetaldehyde ethanal (CH3CHO), which is represented below using the Lewis notation. Then, using Table 8.5 in Appendix 8, plot the energy diagram of this reaction if its activation energy is 174 kJ/mol.

H O H A B A n O O O O H C C H H COH  C A A H H

O

CO (g)  Cl2 (g) n COCl2 (g) a) Calculate the enthalpy change of this reaction using Table 8.5 in Appendix 8. b) Plot the energy diagram of this reaction.

NO (g)  O3 (g) n NO2 (g) O2 (g) a) Using Table 8.5 in Appendix 8, calculate the enthalpy change of this reaction if the ozone molecule (O3) contains a single bond and a double bond. b) Plot the energy diagram of this reaction.

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UNIT 2 Energy Changes in Reactions

9. The reaction X  Y n Z has an activation energy of 120 kJ/mol of Z. The molar mass of Z is 50 g/mol. a) Calculate the enthalpy change of this reaction if, during the production of 37 g of Z in a calorimeter whose reaction vessel is hermetically sealed, the 2 L of water contained in the apparatus decreases from 20.0°C to 19.3°C. (Note:To trigger the reaction, a flame is used in the reaction vessel to cross the activation energy barrier. Assume that this flame provides the exact amount of required energy and does not have any other effects.) b) Plot the energy diagram of this reaction. c) Considering only the energy diagram and the apparatus in which the reaction occurred, can it be stated that this reaction is reversible? Explain your answer.

10.

The decomposition reaction of nitrogen pentoxide (N2O5) has an activation energy of 200 kJ/mol. 2 N2O5 (g) n 4 NO2 (g)  O2 (g)

7. The following is a reaction with an activation energy of 135 kJ/mol:

8. The following is a reaction with an activation energy of 63 kJ/mol:



a) Calculate the enthalpy change of this reaction if, when 150 g of reactant decomposes in a calorimeter whose reaction vessel is hermetically sealed, the 5.49 L of water contained in the apparatus decreases from 22.00°C to 12.75°C. (Note: To trigger the reaction, a flame is used in the reaction vessel to cross the activation energy barrier. Assume that this flame provides the exact amount of energy required and does not have any other effects.) b) Plot the energy diagram of this reaction. c) Considering only the energy diagram and the apparatus in which the reaction occurred, can it be stated that this reaction is reversible? Explain your answer.

Molar Heat of Reaction

W

hen a solution contains an electrolyte, ions are dissolved in the solvent. However, some solutions do not generate ions. For example, when we dissolve a sugar cube, its molecules remain intact in the form of dissolved solute.

In this chapter, you will study enthalpy change caused by reactions involving the transformation of a mole of substance. Enthalpy change, which can be measured experimentally in the laboratory, is also called the molar heat of reaction.

Dissolution and electrolytic dissociation are accompanied by enthalpy change, just as in acid-base neutralization, which occurs when the ions in acidic or basic solutions come into contact with each other.

Review Molar mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Dissolution and solubility. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Electrolytes and electrolysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Acid-base reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Oxidation and combustion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

6.1 6.2

Molar heat of dissolution . . . . . . . . . . . . . . 186 Molar heat of neutralization . . . . . . . . . . . . 192 CHAPTER 6 Molar Heat of Reaction

185

6.1 Molar heat of dissolution Molar heat of dissolution (Hd) is the quantity of energy absorbed or released in the dissolution of one mole of solute in a solvent. The enthalpy change of reaction, also called heat of reaction (H), is a general term for the energy changes caused by the transformation of reactants into products in a chemical system. The value of the heat of reaction is often expressed as a quantity of reactant transformed or product created. The quantity used to express this value is usually the mole. Enthalpy change is also called molar heat of reaction and is expressed in kilojoules per mole (kJ/mol) of reactant transformed or product created.

Figure 1 Gasoline with a higher octane content produces more energy during combustion.

APPENDIX 8 Table 8.3: Thermodynamic properties of certain elements, p. 417. Table 8.6: Molar heat of dissolution, p. 420.

Since there are many types of reactions, the molar heat of reaction (H) can be called by the particular name of the transformation, designated as either chemical or physical. For example, for a chemical change such as the combustion of octane (C8H18), the molar heat of reaction has the specific name of molar heat of combustion (Hc) of octane. It indicates the quantity of energy produced in the combustion of one mole of octane (see Figure 1). Similarly, in a physical change like the melting of ice, the molar heat of reaction has the specific name of molar heat of fusion (Hfusion). When a solute dissolves in a solvent, the molar heat of reaction is called molar heat of dissolution (Hd). It indicates the quantity of energy involved in the dissolution of one mole of solute in the solvent, which is usually water. It is expressed in kilojoules per mole (kJ/mol) of dissolved solute. Energy can be released or absorbed in the dissolution process. This is explained in part by the strength of the interactions in both the solute and the solvent, or between the two of them. Thus, when a solute dissolves in a solvent, the particles of solute disperse uniformly in the solvent. The solute particles fit themselves in the spaces in between the solvent particles. The ease with which the solute particles can enter the spaces between solvent particles depends on the relative strength of three types of interaction: solvent-solvent interaction, solute-solute interaction and solvent-solute interaction. Solvent-solvent interaction is the strength with which the solvent particles attract each other. In the case of water, which is a polar molecule, this force of attraction is exerted between the hydrogen (H) atoms in a water molecule and the oxygen atoms (O) in nearby molecules. This type of bond, called the hydrogen bond, is established between all the water molecules within the liquid and on its surface (see Figure 2).

 

Oxygen Hydrogen (H)

Figure 2 The hydrogen bond is an interaction between two atoms with opposite partial charge. In water, this bond occurs between a hydrogen atom (H) with a partial positive charge (), and an oxygen atom (O), with a partial negative charge ().

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UNIT 2 Energy Changes in Reactions

Thus it is the hydrogen bond between the water molecules that causes the surface tension of water, allowing some types of insects to move about on its surface (see Figure 3). The solute-solute interaction depends on what type of solid is dissolved. When the solid is a molecular compound, such as sucrose (see Figure 4a), commonly known as ‘‘sugar’’, the forces between the molecules of the solid are usually hydrogen bonds. The dissolution of this type of solid is a molecular dissolution. In this case, the molecules of solute do not dissociate and produce ions, but remain intact. They then form molecules of dissolved sucrose (see Figure 4b).

Figure 3 The surface tension of water, which allows this hemipteroid insect to slide on its surface, is caused by hydrogen bonds between the water molecules.

Water molecule: H2O (l)

Dissolved sucrose molecule: C12H22O11 (aq) Dissolved sucrose molecule: C12H22O11 (aq) Carbon (C) Oxygen (O) Hydrogen (H)

Solid sucrose molecules: C12H22O11 (s)

a) A sucrose molecule (C12H22O11)

b) The dissolution of sucrose

Figure 4 Sucrose molecules dissolved in water become aqueous molecules surrounded by water molecules. To simplify the representation, the sucrose molecule is shown as a solid sphere in Figure b).

When sucrose is dissolved in water, it forms an aqueous sucrose solution, which is represented by the following equation: C12H22O11 (s) n C12H22O11 (aq) On the other hand, when a dissolved solid is an ionic compound, such as silver chloride (AgCl), the forces between the particles of the crystalline solid before dissolution are ionic bonds. After dissolution, interactions are formed between the polarized water molecules and the silver (Ag) ions and aqueous chrlorine (Cl) (see Figure 5).

Chlorine (Cl) Silver (Ag) AgCl (s)

n

Ag (aq)



Cl (aq)

Oxygen (O) Hydrogen (H)

Figure 5 When silver chloride (AgCl) is solid, its particles are held together by ionic bonds (represented here by rods). Once it dissolves, interactions occur (represented here by the dotted lines) between the polar water molecules and the aqueous silver (Ag+) and chlorine (Cl-) ions.

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187

Contrary to molecular dissociation, where the molecules remain whole, the electrolytic dissociation of an ionic compound, or ionic dissociation, produces oppositely charged ions. For example, the ionic dissociation of silver chloride produces positive and negative ions according to the following equation: AgCl (s) n Ag (aq)  Cl (aq)

Solvent-solute interaction results in different types of interactions that occur between the solvent molecules and the solute particles. Taking into account these interactions, it is possible to observe dissolution, in terms of the energy involved, as a process with three distinct stages (see Figure 6).

Solvent

Stage 1

Stage 2

H1

H2

Stage 3

Solute

H3

Solution

Figure 6 At the particulate level, the dissolution process has three stages: the separation of solvent and solute (stages 1 and 2) and the mixing of the particles of solvent and solute (stage 3). Each stage may be endothermic or exothermic, depending on the substances involved.

In stage 1, the particles of solvent, which are attracted to each other through solvent-solvent interaction, have to separate. Likewise in stage 2, the particles of solute, which are attracted to each other through solute-solute interaction, have to separate. These two stages absorb the energy necessary to break the force of interaction between the particles. These, then, are stages whose enthalpy change is positive (H1  0 and H2  0). Stage 3 consists of the rearrangement of the particles of solvent and solute that now make up the solution. This stage is usually exothermic (H3  0), because new interactions form between the particles of solvent and solute. The molar heat of dissolution (Hd) is the energy balance between the energy absorbed and the energy released by the breaking and forming of interactions. It can be represented by the following expression:

Hd  H1  H2  H3

Thus, if the sum of H1 and H2 is greater than H3 in absolute value, the molar heat of dissolution is endothermic (Hd  0). Conversely, if the sum of H1 and H2 is less than H3 in absolute value, the molar heat of dissolution is exothermic (Hd  0).

188

UNIT 2 Energy Changes in Reactions

Most dissolutions take place in water, a solvent at ambient temperature, unless otherwise stated. At this temperature, a given mass of water contains a relatively large quantity of heat energy because of water’s high specific heat capacity (C), which is 4.184/(g°C). During stages 1 and 2 of the dissolution process, the energy absorbed to break the interactions between the particles of solvent and solute is taken from the water’s own heat energy. Nevertheless, at stage 3, when new interactions are formed between particles of solvent and solute, chemical energy is released in the form of heat. This heat is then absorbed by the water in the form of heat energy.

APPENDIX 8 Table 8.12: Specific heat capacity of various substances, p. 423.

Thus, if more heat is released at stage 3 than is absorbed in stages 1 and 2, the molar dissolution is exothermic and the water temperature rises. Conversely, if more heat is absorbed in stages 1 and 2 than is released in stage 3, the molar dissolution is endothermic and the water temperature falls. The molar heat of dissolution of a substance can be calculated from calorimetric experiments in which temperature measurements can be taken. The molar heat of dissolution can also be used to find the final temperature of a solution after dissolution of a solute.

See Relationship between thermal energy, specific heat capacity, mass and temperature change, p. 134.

The astonishing stickiness of a gecko’s toes Tree geckos can walk on almost any surface, even the smoothest, because of their adhesive toe pads. These pads are composed of millions of minuscule bristles, a thousand times finer than a human hair. The end of each bristle, which is made up of just a few molecules of keratin in width, can be seen only with an electron microscope. At the microscopic level, the end of each bristle comes in contact with only a few molecules of the surface the gecko is standing on, and an interaction is set up between these molecules. Scientists at first Figure 7 The gecko’s feet have believed that this interaction was the same as that bristles so fine that the animal can walk which causes the surface tension of water, that is, on any surface. the hydrogen bond. It was believed that this interaction occurred between the water molecules on the surface of objects and on the gecko’s bristles. However, further research showed that the gecko’s toe pads adhere to a surface because of electrical interactions between the atoms or molecules with opposite charges that make up the bristles and the object. These interactions are called van der Waals forces. The gecko’s adhesion mechanism is so effective that it can support its entire body weight with just one toe. Advances in nanotechnology allowed scientists to create a fabric of carbon nanofibre with properties like those of the gecko’s toe pads, so that humans or robots can walk on a variety of surfaces.

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The following examples demonstrate how to find a substance’s molar heat of dissolution using a calorimeter, and how to predict the final temperature of a solution from the value of its molar heat of dissolution. In these examples, the density of the water used is 1 g/mL, so 100 mL = 100 g. Example A Dissolve 6.69 g of lithium chloride (LiCl) in 100 mL of water at 24.2°C in a calorimeter. The final temperature of the water is 37.4°C. a) What is the molar heat of dissolution (Hd) of lithium chloride? b) Write the thermochemical equation for the dissolution of lithium chloride, incorporating the calculated value of Hd. a)

Data: m  100 g c  4.84 J/(g°C) Ti  24.2°C Tf  37.4°C t  13.2°C Q ?

1. Calculation of the quantity of heat transferred Q  mcT  100 g  4.84 J/(g°C)  13.2°C  5522.9 J of heat absorbed by the water 2. Determination of whether the value of the dissolution heat is negative or positive: Since the water absorbs heat, the dissolution of 6.36 g of LiCl released this heat (Hd). Hd  Q  5522.9 J 3. Calculation of the number of moles of LiCl: m M n m n M 6.69 g  42.39 g/mol  0.158 mol 4. Calculation of the molar heat of dissolution: 5522.9 J ?  1 mol 0.158 mol 1 mol  5522.9 J ? 0.158 mol  34 955.1 J  35.0 kJ

Answer: The molar heat of dissolution (Hd) of lithium chloride is 35.0 kJ/mol. b)

Solution: Since the final temperature is greater than the initial temperature, the dissolution reaction of lithium chloride is exothermic. This explains why the value of the molar heat of reaction is placed on the side of the products: LiCl (s) n Li (aq) + Cl (aq)  energy. Answer: The equation for the dissolution of lithium chloride is: LiCl (s) n Li (aq)  Cl (aq)  35 kJ

190

UNIT 2 Energy Changes in Reactions

Example B Dissolve 4.25 g of sodium nitrate (NaNO3) in 100 mL of water at 23.4°C. Given that the molar heat of dissolution of sodium nitrate is 21.0 kJ/mol, what will the final temperature of the water be? Data: m  100 g  0.100 kg c  4.84 kJ/(kg°C) Tf  ? Hd   21.0 kJ/mol

1. Calculation of the number of moles of NaNO3 : m M n m n M 4.25 g   0.0500 mol 84.96 g/mol 2. Calculation of the quantity of heat transferred: 21.0 kJ ?  1 mol 0.0500 mol 21.0 kJ  0.05 mol ?  1.05 kJ 1 mol 3. Determination of whether the value of the dissolution heat is negative or positive: Since the Hd of NaNO3 is positive, this heat is released by the water. Therefore, Qwater  Hd  1.05 kJ 4. Calculation of the temperature change: Q  mcT Q T  water mc 1.03 kJ   2.51°C 0.1 kg  4.184 kJ/(kg°C)

Intensity

5. Calculation of the final temperature of the solution: T  Tf  Ti Tf  T  Ti  2.51°C  23.4°C  20.89°C Answer: The final temperature of the solution (Tf) is 20.9°C.

The spicy heat of hot peppers When cooking, we sometimes say that hot peppers “burn” the tongue. This does not refer to a burn caused by energy in the form of heat that would be released from the dissolution of a chemical substance in the pepper. Rather, it refers to “spicy heat,” that is, a sensation of increased temperature in the mouth caused by the capsaicin molecule in the peppers that comes into contact with the nerve endings of the taste buds. This natural irritant is produced by different species of capsicum, a plant that uses this property to protect itself from animal predation. Since various varieties of peppers contain different amounts of capsaicin, the Scoville scale allows cooks to adjust the degree of spicy heat in the meals they prepare according to the capacity of their guests to tolerate spicy heat.

Pepper

Explosive

10

Habanero

Volcanic

9

Tabasco

Torrid

8

De Arbol

Burning

7

Cascabel

Fiery

6

Cayenne

Strong

5

Jalapeno

Hot

4

Espelette

Spicy

3

Ancho

Warm

2

Anaheim

Mild

1

Mild paprika

Neutral

0

Sweet pepper

Figure 8 Simplified Scoville scale of pseudo-heat. Capsaicin content is not specified but, for example, small habanero peppers contain approximately 40 times more capsaicin than jalapenos.

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6.2 Molar heat of neutralization The molar heat of neutralization (Hn) is the quantity of energy that is absorbed or released in the neutralization of one mole of an acid or base. Acid-base neutralization is a chemical reaction in which an acid reacts with a base to form a salt and water. This reaction is revealed by an indicator such as phenolphthalein, which turns pink in the presence of a base (see Figure 9). See Theories of acids and bases, p. 320.

Figure 9 An acid-base neutralization is complete when the indicator changes colour permanently.

According to Arrhenius theory, in this reaction the ions that react are hydrogen (H) ions from the acid in aqueous solution and hydroxide (OH ) ions from the base in aqueous solution. When these ions combine, they form water (H2O). The other ions are spectator ions which do not participate in the reaction. For example, we can simplify the net ionic equation (without the spectator ions) that represents the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) in aqueous solution in this way.

H(aq)  OH(aq) n H2O (l) The neutralization reaction is accompanied by a heat transfer that results from the interaction between the ions in the reaction. The heat involved at the moment of neutralization of one mole of a substance (acid or base) is called molar heat of neutralization (Hn). It is expressed in kilojoules per mole (kJ/mol) of neutralized acid or base. Given that the neutralization reaction usually takes place in an aqueous solution, the molar heat of neutralization obtained can also be expressed in terms of the number of moles of water formed, rather than in terms of the number of moles of neutralized acid or base. It is possible to study neutralization reactions between dilute solutions of a strong acid and a strong base in the laboratory using a calorimeter. To make it easier to measure, since the solutions are dilute, assume that their density and specific heat capacity are equal to those of water. In addition, since the reactions occur fairly rapidly, the heat is released in a relatively short time, minimizing heat loss to the environment. The following example shows how to calculate molar heat of neutralization in the laboratory. In this example, the density of the water used is 1 g/mL, that is, 100 mL = 100 g.

192

UNIT 2 Energy Changes in Reactions

Example In a calorimeter, completely neutralize 100 mL of an aqueous solution of sodium hydroxide (NaOH) at 0.5 mol/L by adding 100 mL of a hydrochloric acid (HCl) solution at 0.5 mol/L. The initial temperature of the solutions before being mixed is 22.5°C. The highest temperature obtained during the neutralization, after the two solutions are mixed, is 25.9 °C. Calculate the molar heat of neutralization of sodium hydroxide. Data: m total  100 g  100 g  200 g c  4.184 J/(g°C) Tf  25.9°C Ti  22.5°C T  3.4°C Q ?

1. Calculation of the quantity of heat transferred: Q  mcT  200 g  4.184 J/(g°C)  3.4°C  2845.1 J of heat absorbed by the water from the reaction mixture 2. Determination of whether the value of the neutralization heat is negative or positive: Since the water absorbs heat, the neutralization of 100 mL at 0.5 mol/L released this heat (Hn). Hence, Hn  Q  2845.1 J 3. Calculation of the number of moles of NaOH neutralized: n C V nCV  0.5 mol/L  0.1 L  0.05 mol of NaOH neutralized 4. Calculation of the molar heat of neutralization: 2845.1 J ?  1 mol 0.05 mol 1 mol  285.1 J ?  56 902.0 J 0.05 mol  56.9 kJ

Answer: The molar heat of neutralization (Hn) of sodium hydroxide is 57 kJ/mol.

Gastric acid The approximately 35 million gastric glands that line the internal surface of the stomach play an important role in digestion. Each day, they secrete close to three litres of gastric sugar into the stomach. Gastric sugar is composed in part of hydrochloric acid (HCl), which makes it very acidic. To protect itself from this strong acidity, the cells of the internal wall of the stomach secrete a mucous. The normal pH of the stomach, which is between 1.5 and 2.5, is also neutralized, in part by the food that is mixed in the stomach. However, the stomach can become acidic again over the course of a meal in order to allow the digestion process to run its course. Some people suffer from gastric hyperacidity, a disorder caused primarily by heredity, stress or poor eating habits. Medication like antacids can reduce gastric acidity. Some contain sodium bicarbonate (NaHCO3) which, when dissolved in water, acts as a base by neutralizing the acidity of the contents of the stomach. However, this reaction produces carbon dioxide (CO2) in the organism, which can cause discomfort for some people.

Figure 10 This antacid tablet contains sodium bicarbonate that produces carbon dioxide as it dissolves in water.

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Chewing gum Some people chew gum to freshen their breath. Freshness, in this case, is not exactly the result of endothermic dissolution, which would involve the absorption of a little heat inside the mouth. While the dissociation of a small quantity of sugar contained in chewing gum can have a small endothermic effect, the sensation of freshness is due instead to the dissolution of certain molecules, such as menthol, which interact with the sensory receptors lining the tongue and the inside of the mouth.

great deal of heat before it was soft enough to chew comfortably. Thomas Adam, who believed he had the solution to this problem, submitted a patent in 1871 for a machine that produced a small lozenge composed of chicle coated in sugar (see Figure 12). In 1890, William Wrigley Jr., a chemical yeast manufacturer, started producing chewing gum after noticing that the gum that he gave out for free upon purchase of his chemical yeast was more popular than the chemical yeast itself.

Due to the popularity of chewing gum over the years, certain gums now have a chemical composition that allows them to neutralize the acid produced by certain types of bacteria responsible for bad breath. However, chewing gum has existed for many years and over time many flavours and forms have emerged. The inhabitants of Ancient Greece were the first to chew gum with a strong odour which was derived from the sap of a tree, the lentisc pistachio tree, which they called the “mastica,” derived from the term “masticate” (see Figure 11). Across the world, the Mayans used the sap of another tree sap harvested from the lentisc pistachio tree species, the sapodilla, to make white latex called chicle, which was used to make chewing gum. Figure 11 Dried pellets of

Meanwhile, aboriginal peoples in North America used spruce sap as chewing gum and introduced it to the colonists who came to settle in North America. This spruce gum was later used as a basic ingredient in the first industrially produced chewing gums. The first chewing gum factory in North America was built in 1848 in the State of Maine, in the United States, and produced gum from spruce resin. In 1869, William Semple, a dentist from Ohio, obtained a patent to invent a paraffin-based chewing gum. Its major disadvantage, however, was that it required a

194

UNIT 2 Energy Changes in Reactions

Figure 12 Workers cutting gum in a factory, 1919

After the Second World War, with the development of petrochemicals, the main ingredient in chewing gum became synthetic. Various additives were included to enhance its flavour, colour, texture and certain properties properties. For example, today, latex-based chewing gum contains zinc, which inhibits the acid production of certain bacteria like streptococcus, which are often responsible for bad breath and proliferate in an acidic environment, like the mouth. Similarly, the evolution of chewing gum has allowed certain types of gum sweetened with xylitol to become an occasional substitute for tooth brushing in order to provide dental hygiene after a meal outside the home, and to reduce the risk of dental cavities in children.

CHAPTER

6

Molar Heat of Reaction

6.1 Molar heat of dissolution

Water molecule: H2O (l)

• Molar heat of dissolution (Hd ) is the quantity of energy absorbed or released in the dissolution of one mole of solute in a solvent. It is expressed in kilojoules per mole (kJ/mol) of dissolved solute.

Dissolved sucrose Dissolved molecule: C12H22O11 (aq) sucrose molecule: C12H22O11 (aq)

Carbon (C) Oxygen (O) Hydrogen (H)

Solid sucrose molecules: C12H22O11 (s)

a) A sucrose molecule (C12H22O11)

b) The dissolution of sucrose

6.2 Molar heat of neutralization • Molar heat of neutralization (Hn) is the quantity of energy absorbed or released in the neutralization of one mole of an acid or base. It is expressed in kilojoules per mole (kJ/mol) of neutralized acid or base.

CHAPTER 6

Molar Heat of Reaction

To answer questions 1, 5, 6, 8 and 9, refer to Table 8.6 on page 420 of the Appendices. 1. Write the equations for the dissolution of the following electrolytes and say whether the dissolution is endothermic or exothermic. a) HClO3 (s) b) HI (g) c) KNO3 (s) d) CuSO4 (s) e) Li2CO3 (s) 2. Calculate the molar heat of dissolution of sodium hydroxide (NaOH), given that 4 g of sodium hydroxide dissolved in 20 mL of water raises the water temperature by 50°C.

3. If 24.42 g of potassium chlorate (KClO3) are dissolved in 1.0 kg of water, the temperature of the water drops by 2°C. Calculate the molar heat of dissolution of potassium chlorate and write the dissolution equation, incorporating the value of the molar heat of dissolution. 4. The dissolution of potassium nitrate (KNO3) is endothermic, and that of sodium hydroxide (NaOH) is exothermic. Explain, in terms of particles interactions, the energy changes that accompany the transformations of the solute and solvent in each case.

CHAPTER 6 Molar Heat of Reaction

195

5. A student dissolves 1 mol of an unknown salt in 1000 mL of water and observes a drop in temperature of 5.5°C. Identify this salt. 6. A student dissolves 8 g of solid copper sulfate (CuSO4) in 100 mL of water at 20°C. What will the final temperature of the water be after dissolution? 7. This is the dissolution equation for solid lithium chloride (LiCl). LiCl (s) n Li (aq)  Cl (aq)

Hd  35.0 kJ/mol

If 8.42 g of lithium chloride are dissolved in 200 mL water at 22°C, what will the final temperature of the water be after dissolution? 8. How many grams of potassium hydroxide (KOH) must be dissolved in 100 mL water at 25°C to change the water temperature after dissolution to 80°C? 9. A student wants to find out about the nature of some de-icing salt which she has been told is harmless to vegetation and the concrete steps of her house. She decides to dissolve a sample of this salt to compare the value of its molar heat of dissolution with that of sodium chloride (NaCl), which is also used to de-ice roads and sidewalks in her municipality. Find out whether the de-icing salt that the student wants to use at home is the same kind as that used in the municipality, using the following results obtained in the laboratory: Mass of dissolved salt  10.00 g Volume of water  50.0 mL Initial temperature  22.3°C Final temperature  15.3°C 10. A laboratory technician prepares 2.0 L of sodium nitrate (NaNO3) solution at 0.1 mol/L. The initial temperature of the water used is 24°C. What is the final temperature of the solution? 11. In a neutralization 250 mL of potassium hydroxide (KOH) at 0.5 mol/L is reacted with 250 mL of nitric acid (KOH) at 0.5 mol/L. Calculate the molar neutralization heat if the temperature goes from 23.5°C to 27°C.

196

UNIT 2 Energy Changes in Reactions

12. Sink and bathtub drainage systems are often blocked by clumps of hair mixed with soap scum. To begin melting this blockage, how many grams of sodium hydroxide (NaOH) must be dissolved in 500 mL of hot water at 75°C? (Assume that the melting point of the soap scum is 120 °C.) 13. Calculate the molar neutralization heat of nitric acid (HNO3) using the results of an experiment compiled in the following table. The neutralization of the acid is complete. After neutralization

Before neutralization HNO3 (aq) (0.5 mol/L)

LiOH (aq) (1 mol/L)

Mixture

V (mL)

Ti (°C)

V (mL)

Ti (°C)

V (mL)

Tf (°C)

200

23.0

200

25.0

400

27.5

14.

After a natural disaster like a hurricane, self-heating meals are distributed to populations in need. The meal may consist of a small plastic pouch containing the food and a large pouch in which to mix water and two types of salt. The dissolution and neutralization that take place when they are mixed produces heat. To heat the meal, dissolve the salts in the water in the larger pouch, insert the hermetically sealed food pouch and close the large pouch. After a few minutes, the meal will be hot enough to eat (about 60°C). One could theoretically make a home version of this type of meal by dissolving sodium hydroxide (NaOH) in water, without neutralizing it. Calculate how many grams of sodium hydroxide are required to heat a 300-g meal of vegetable soup if the pouch in which the dissolution takes place can hold the meal and 500 mL of water. The soup and water used for the dissolution are at 40°C at the beginning, and the soup has about 90% of the heat capacity of the water.

Hess’s Law

C

ertain chemical reactions occur so naturally that they do not seem to be the result of a complex process. However, many reactions are extremely complex and require several successive steps to convert reactants into products. An example is photosynthesis, a process in which plants use light energy, atmosphe ric carbon dioxide (CO2) and water to produce organic molecules like sugars.

Review Balancing chemical equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Stoichiometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Potential energy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

This chapter will focus on reactions that occur in several steps and how to calculate the enthalpy changes of these reactions. You will also learn how to mathematically determine, without using a calorimeter, the molar heat of any chemical reaction, using Hess’s law.

7.1 7.2

Reaction mechanism . . . . . . . . . . . . . . . . . . . 198 Summation of enthalpies . . . . . . . . . . . . . . . 201 CHAPTER 7 Hess’s Law

197

7.1 Reaction mechanism A reaction mechanism is a series of simple reactions that convert reactants into products over the course of a complex reaction. A chemical reaction is represented in the form of the following balanced equation: x reactants(s) n y products(s) This representation symbolically describes the nature of the substances converted (the reactants) as well as that of the substances obtained (the products), following a chemical change. Moreover, the stoichiometric coefficients (x and y) that precede the chemical formulas of the reactants and products, indicate the proportions in which the reactants combine to form products. While this representation is simple and practical for conveying data regarding a chemical reaction, it does not tell us anything about how the reaction actually occurs. In most chemical reactions, the reactants are not directly changed into products. Consequently, the representation of a complex chemical reaction in the form of a balanced equation actually hides the sum of a series of simple reactions leading to the formation of intermediate substances. Each of these simple reactions, also called intermediate steps, constitutes a simple reaction, that is, a chemical reaction during which the reactants are directly converted into products.

Figure 1 Nitrogen dioxide (NO2) causes the brownish colour of fog that can be observed above large urban centres.

A complex chemical reaction can be broken down into a succession of several simple reactions and can be explained by means of a reaction mechanism. For example, the formation of nitrogen dioxide (NO2), one of the main components of smog, from nitrogen tetroxide (N2O4) is an example of a simple reaction (see Figure 1). It is described by the following balanced reaction: N2O4 (g) n 2 NO2 (g) However, nitrogen dioxide (NO2) can also be produced from nitrogen oxide (NO) and oxygen (O2) according to the following balanced equation: 2 NO (g)  O2 (g) n 2 NO2 (g) In this case, the nitrogen dioxide does not form directly from the nitrogen oxide and oxygen. During the reaction, it is possible to detect the presence of another compound: dinitrogen dioxide (N2O2). The presence of this compound indicates that this reaction is complex. In fact, the reaction mechanism of dinitrogen dioxide formation involves two intermediate steps, each one constituting a simple reaction.

198

UNIT 2 Energy Changes in Reactions

Firstly, two moles of nitrogen oxide are converted into one mole of dinitrogen dioxide. This first simple reaction is represented by the following balanced equation: 2 NO (g) n N2O2 (g)

1)

Then, one mole of dinitrogen dioxide combines with one mole of oxygen to produce two moles of nitrogen dioxide. This second simple reaction is presented by the following balanced equation: 2)

N2O2 (g)  O2 (g) n 2 NO2 (g)

The overall balanced equation of the reaction corresponds to the sum of the equations of the two simple reactions. 1)

2)

Balanced equation of the first simple equation: Balanced equation of the second simple equation:

Sum of the equations: Overall balanced equation:

2 NO (g) n N2O2 (g) N2O2 (g)  O2 (g) n 2 NO2 (g) 2 NO (g)  N2O2 (g)  O2 (g) n N2O2 (g)  2 NO2 (g) 2 NO (g)  O2 (g) n 2 NO2 (g)

Since there is one molecule of N2O2 (g) on either side of the equation, they cancel each other out and do not figure in the overall balanced equation. The overall balanced equation is an equation that summarizes the reaction mechanism of a complex reaction. During a complex reaction, reaction intermediates are formed during simple reactions that make up the reaction mechanism. In general, reaction intermediates constitute the products of one step in the reaction, and then become reactants in the following step. In the example of nitrogen dioxide (NO2) formation, the dinitrogen dioxide (N2O2) is a reaction intermediate. Reaction mechanisms cannot be directly observed. They are models that explain all of the observations made over the progress of a reaction. Scientists are continuously collecting new data regarding these mechanisms, which evolve over time as a function of this new data. Consequently, the reaction mechanisms used are generally the ones proposed by the scientists who study them.

Nitrogen gases Nitrogen gases are widespread in the atmosphere. Nitrogen tetroxide (N2O4) and nitrogen dioxide (NO 2) play a role in acid rain, the greenhouse effect and the destruction of the ozone layer. They are among many nitrogen oxides released during the combustion of fossil fuels, such as gas and coal, or during the incineration of organic waste. Transportation continues to be the main source of nitrogen oxides, despite the antipollution mechanisms now installed in cars. Figure 2 An organic waste incinerator

CHAPTER 7 Hess’s Law

199

7.1.1 See Activated complex, activation energy and the energy diagram, p. 172. See Observing the use of an energy diagram to plot the progress of a conversion, p. 176.

Graphical representation of a reaction mechanism

The progress of a complex reaction can be visualized using an energy diagram. The diagram can be interpreted to obtain useful data on the reaction mechanism. For example, the reaction mechanism of the hypothetical complex reaction of the formation of compound E from compound A involves four steps (see Figure 3) which are not apparent in the overall following equation. AnE

The energy diagram of a reaction

Potential energy

2nd activated complex

3rd activated complex

1st activated complex

Ea

Ea2 A

4th activated complex

Ea1 H1

B

Ea3

Ea4 D

H2

1st simple reaction

C

2nd simple reaction

H3

3rd simple reaction

H H4

E

4th simple reaction

Progress of the reaction

Figure 3 This energy diagram presents the reaction mechanism of an overall complex reaction between reactants A and products E, and is comprised of four simple reactions.

Each step is simple reaction in which the product of the first reaction becomes the reactant of the second, and so on, until the formation of the final products. These substances are the reaction intermediates (B, C and D). Each simple reaction possesses its own activation energy (Ea1, Ea2, Ea3 and Ea4) that leads to the formation of an activated complex. The energy diagram illustrates the activation energy of the overall reaction (Ea), which corresponds to the potential energy required to form the highest activated complex from the energy level of the initial substance. In the example of the preceding reaction, the activation energy of the overall reaction corresponds to the potential energy gain required between the energy level of substance A and that of the activated complex of the second step in the reaction. This energy diagram also provides information on the enthalpy change of the overall reaction (H) and each of the simple reactions (H1, H2, H3 and H4).

200

UNIT 2 Energy Changes in Reactions

7.2 Summation of enthalpies According to Hess’s law, also known as the law of constant heat summation, if a reaction can be broken down into several simple reactions, its enthalpy change is equal to the algebraic sum of the enthalpy changes of each of its simple reactions. For many chemical reactions, the enthalpy or heat of the reaction (H ) can be determined experimentally in the lab using a calorimeter. For example, calorimetric experiments can be used to determine the dissolution heat of a piece of sugar or the heat released through the neutralization of a basic aqueous solution by an acid. However, it is not always possible to study the heat released or absorbed by reactions using this type of experiment. Certain reactions occur very slowly, and the resulting temperature changes are not great enough to be measured with a calorimeter. This is the case in oxidation reactions that cause the aging of wine in wooden barrels (see Figure 4).

See Calorimetry and the calorimeter, p. 132. See Molar heat of dissolution, p. 186. See Molar heat of neutralization, p. 192.

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Figure 4 The oxidation reactions that cause wine to age in wooden barrels occurs over a period of several months or even years.

Wood is permeable to oxygen, which reacts with the wine over a long period of time. This is one of the many reactions wine undergoes over the course of ageing, until the desired taste is attained. The use of different species of wood, often oak, makes it possible to produce wine with a woody aroma. In contrast, other reactions are violent and release so much energy that it is impossible to observe them using a calorimeter. For example, the combustion of magnesium occurs too quickly to be measured (see Figure 5). Moreover, the light it emits is so bright that it can damage the eyes. The first camera flashes, invented in 1887, used magnesium powder.

Figure 5 The combustion of magnesium

In addition, the tremendous variety of chemical compounds and the complexity of many reactions leading to the formation of each of these compounds make it impossible to calculate the reaction enthalpies using a calorimeter. As a result, chemists have developed a method to predict the enthalpy change caused by chemical reactions. This algebraic method was derived from the summation of enthalpies, also called Hess’s law.

GERMAIN HENRI HESS Swiss chemist and physician (1802–1850) At a young age, Germain Henri Hess and his parents left their native Switzerland for Russia, where Hess earned a diploma in medicine. Around 1830, his interests turned to chemistry. He studied the heat produced by combustion reactions and formulated the law of constant heat summation, which was named after him. Recognized as the founder of thermochemistry, he also discovered a new mineral. Since he was the first to analyze this mineral, it was named hessite in his honour.

CHAPTER 7 Hess’s Law

201

7.2.1

Presentation of Hess’s law

According to Hess’s law, the enthalpy change of a chemical reaction depends solely on the reactants and products and is independent of the reaction mechanism and the number of intermediate steps necessary for the conversion of reactants into products. Hess’s law If a reaction can be broken down into several simple reactions, its enthalpy change is equal to the algebraic sum of the enthalpy changes of each of these simple reactions. This law can be expressed by the following general equation: Hess’s law H  H1  H2  H3  … where H  Enthalpy change of the overall reaction, expressed in kilojoules per mole (kJ/mol) H1, H2, H3  Enthalpy change of each simple reaction of the overall reaction, expressed in kilojoules per mole (kJ/mol) In other words, the enthalpy of a reaction remains the same, whether there is a direct conversion of reactants into products, or the reaction occurs over several intermediate steps. For example, the formation reaction of carbon dioxide (CO2) from oxygen (O2) and carbon (C) can occur according to two different mechanisms.

See Endothermic and exothermic chemical reactions, p. 152.

In the first mechanism, the carbon reacts with the oxygen and directly forms carbon dioxide. This reaction mechanism is represented by the following thermochemical equation. C (s)  O2 (g) n CO2 (g)

H  393.5 kJ/mol

The second mechanism occurs in two steps. First, the carbon combines with the oxygen and forms carbon monoxide (CO). Then, the carbon monoxide reacts with the oxygen to produce carbon dioxide. This reaction mechanism is represented by the following two thermochemical reactions:

1)

2)

Overall reaction

1 O n CO (g) 2 2 (g) 1 CO (g)  O2 (g) n CO2 (g) 2 C (s) 

C (s)  O2 (g) n CO2 (g)

H1  110.5 kJ/mol H2  283.0 kJ/mol H  393.5 kJ/mol

Regardless of the reaction mechanism, the final result is the production of one mole of carbon dioxide from one mole of carbon and one mole of oxygen. The enthalpy change of the first reaction mechanism is the same as that of the second mechanism.

202

UNIT 2 Energy Changes in Reactions

The sum of the enthalpy changes of the reaction mechanism occurring in two steps corresponds to the enthalpy change of the reaction mechanism involving only one step. The energy diagram will change according to the reaction mechanism (see Figure 6).

The energy diagram of a reaction

The energy diagram of a reaction

Reactants C (s)  O2 (g)

Reactants C (s) 

1 O2 (g) 2

H  395.5

Potential energy (kJ/mol)

Potential energy (kJ/mol)

H1  110.5

CO (s)  H  395.5

2

O2 (g) H2  283.0

Product CO2 (g)

Product CO2 (g)

Progress of the reaction a) A reaction mechanism occurring in one step

1

Progress of the reaction b) A reaction mechanism occurring in two steps

Figure 6 The enthalpy change of carbon dioxide (CO2) formation from carbon (C) and oxygen (O2) is the same, whether the reaction occurs in one step (a) or

two steps (b).

Hess’s law can be compared to the potential energy change caused by the change in altitude of two hikers who are descending a mountain. Regardless of the path each hiker takes to reach the base of the mountain, the total change in potential energy caused by the change in altitude is the same.

7.2.2

Applying Hess’s law

Hess’s law is a very useful tool for determining the enthalpy of complex reactions. By applying it, a complex reaction can be broken down into several steps or elementary reactions whose enthalpies are already known or have already been obtained using a calorimeter. The sum of these enthalpies provides the enthalpy of the complex reaction. To obtain the enthalpy of the complex reaction, the thermochemical equations of each of the intermediary reactions are added algebraically. The standard enthalpy change of formation (H f°) indicates the heat released or absorbed during the formation of a compound based on its elements in their standard state. Standard enthalpy formation is generally expressed in kilojoules per mole (kJ/mol) of product. By convention, the enthalpy formation of an element in its standard state is equal to zero. For example, given that the most stable form of the element nitrogen (N) is the diatomic form (N2), the standard enthalpy formation of nitrogen gas (N2 (g)) is 0 kJ/mol.

See Standard molar enthalpy change, p. 149.

CHAPTER 7 Hess’s Law

203

APPENDIX 8 Table 8.4: Standard molar enthalpies of formation, p. 418.

Generally, standard molar enthalpies of formation tables do not provide the complete formation equations. They must be reconstituted using the elements in their standard state. For example, water is 285.8 kJ/mol. The thermochemical equation of the formation of water based on its elements in their standard form is written as follows:

H2(g) 

1 O n H2 O (l) 2 2 (g)

H  285.8 kJ/mol

As illustrated by this equation, the most stable state of hydrogen and oxygen is their diatomic form, H2 and O2. Consequently, fractional coefficients must sometimes be used so that the balanced equation corresponds to the formation of one mole of product, since the values of standard enthalpies of formation are expressed in kJ/mol of product. Therefore, in the equation for the formation of water, the fractional coefficient 1/2 is written in front of the diatomic oxygen. To carry out the algebraic sum of thermochemical equations, one must respect the following rules:

• The identical terms situated on the same side of the equation are added. • The identical terms situated on either side of the equation are subtracted. • If an equation is reversed, the H sign must also be reversed. • If the coefficients of a chemical reaction are modified by multiplying or dividing them by a common factor, the value of H must also be multiplied or divided by this same common factor. The application of Hess’s law allows for the calculation of the enthalpy reaction associated with the combustion of butane (C4H10), which is translated by the following overall balanced equation:

C4H10 (g) 

13 O n 4 CO2 (g)  5 H2O (g) 2 2 (g)

H 2657.4 kJ/mol

To calculate the enthalpy reaction of the combustion of butane, the overall balanced equation must first be broken down into several steps or intermediate reactions. The values of standard molar enthalpies of formation are then used to write the thermochemical formation equations of the different terms in the overall balanced equation. To find the enthalpy reaction associated with the combustion of butane, the following three formation equations are needed: 1) 2) 3)

204

UNIT 2 Energy Changes in Reactions

4 C (s) 5 H2 (g) n C4H10 (g)

H1  125.6 kJ/mol

C (s)  O2 (g) n CO2 (g)

H2  393.5 kJ/mol

1 O n H2O (g) 2 2 (g)

H3  241.8 kJ/mol

H2 (g) 

By examining these equations and comparing them with the equation of the overall reaction, note that, in the first equation, the butane is not situated on the side of the reactants. Consequently, this equation must be reversed so that the butane is on the side of the reactants, as in the overall equation. The sign H1 must also be reversed. This ensures that the butane is on the side of the reactants when the sum of the reactions is carried out. 1)

C4H10 (g) n 4 C (s) 5 H2 (g)

H1  125.6 kJ/mol

Moreover, the number of moles of carbon (C) and water (H2O) vapour in the overall equation is greater than those in the second and third equation. The second equation must, therefore, be multiplied by four and the third equation by five to obtain a number of carbon and water vapour moles identical to those in the overall equation. Based on the rule, the values of H must also be multiplied. 2)

4 C (s)  4 O2 (g) n 4 CO2 (g)

H2  (393.5 kJ/mol  4 mol)  1574.0 kJ

3)

5 5 H2 (g)  O2 (g) n 5 H2O (g) 2

H3 (241.8 kJ/mol  5 mol)  1209.0 kJ

Since equations are manipulated this way and enthalpy change is no longer indicated for one mole of substance, it is simpler to carry out these kinds of calculations using values in kilojoules (kJ). The different equations can then be added by cancelling the identical substances present on both sides of the equation. Thus, we obtain the reaction enthalpy associated with the overall reaction of the combustion of butane. C4H10 (g) n 4 C (s)  5 H2 (g)

1) 2) 3)

4 C (s)  4 O2 (g) n 4 CO2 (g) 5 5 H2 (g)  O2 (g) n 5 H2O (g) 2 C4H10 (g) 

13 O n 4 CO2 (g)  5 H2O (g) 2 2 (g)

H1  125.6 kJ H2  1574.0 kJ H3  1209.0 kJ H  2657.4 kJ

The value of the enthalpy reaction of the combustion of butane can then be converted into molar heat of reaction (in kJ/mol). To do so, the enthalpy reaction obtained is simply divided by the number of moles of butane in the overall reaction. In this case the balanced reaction of combustion indicates one mole of butane, the molar heat of the combustion of butane is therefore 2657.4 kJ/mol. When adding equations, special attention must be paid to the different compounds involved in the reactions. For example, water in liquid state (H2O (l)) and water vapour (H2O (g)) must be considered as two different compounds since they have a different enthalpy change of formation (see Table 8.4, on page 418). Therefore, these compounds cannot be added or subtracted, nor can they cancel each other out.

APPENDIX 8 Table 8.4: Standard molar enthalpies of formation, p. 418.

The following example shows the method for determining the enthalpy change of a reaction using Hess’s law: CHAPTER 7 Hess’s Law

205

Example The combustion of methane (CH4) is described by the following balanced equation: CH4 (g)  2 O2 (g) n CO2 (g)  2 H2O (g) What is the molar heat of the combustion of methane? 1. Write the equations that, once combined, allow the overall equation to be obtained by consulting the standard enthalpy formation table (see page 418) : 1) C (s)  2 H2 (g) n CH4 (g) H1  74.6 kJ 2) C (s)  O2 (g) n CO2 (g) H2  393.5 kJ 3)

1 H2 (g)  O2 (g) n H2O (g) 2

H3  241.8 kJ

2. Manipulate the equations in order to obtain the overall equation by adding them at a later step:

Figure 7 Methane was used to feed the flame of the Olympic torch at the Turin Games in 2006.

The first equation is reversed so that the methane is on the side of the reactants and not the products. The third reaction is multiplied by 2 since there are two moles of water vapour in the products of the overall equation. 1) CH4 (g) n C (s)  2 H2 (g) H1  74.6 kJ 2) C (s)  O2 (g) n CO2 (g) H2  393.5 kJ 3) 2 H2 (g)  O2 (g) n 2 H2O (g) H3  483.6 kJ 3. Add the thermochemical equations: 1) CH4 (g) n C (s)  2 H2 (g) 2) C (s)  O2 (g) n CO2 (g) 3) 2 H2 (g)  O2 (g) n 2 H2O (g)

H1  74.6 kJ H2  393.5 kJ H3  483.6 kJ

CH4 (g)  2 O2 (g) n CO2 (g)  2 H2O (g)

H  802.5 kJ

4. Convert the enthalpy value of the overall reaction into molar heat: There are 802.5 kJ for one mole of methane, that is, 802.5 kJ/mol. Answer: The molar heat of the combustion of methane is 802.5 kJ/mol.

Furthering

your understanding

Spectroscopy

Spectroscopy is based on two principles: all compounds release or absorb radiant energy called spectra, and each compound has its own spectrum. The spectrum of a compound constitutes a kind of fingerprint that allows it to be identified. Measuring and interpreting the different spectra emitted during a chemical reaction makes it possible to identify the different reaction intermediates and reconstitute the reaction mechanism.

Chlorophyll a Chlorophyll b

Absorption

Since a reaction mechanism cannot be observed directly, chemists use spectroscopy to learn about the various steps in a complex reaction. This technique focuses on reaction intermediates, which often exist for only a few nanoseconds (109). Because the wavelengths are short, they are measured in nanometres (nm), that is to say in billionths of a metre.

Absorption spectrum of chlorophylls a and b

400

500 600 700 Wave lengths () in nm

Figure 8 Spectroscopy can be used to distinguish between chlorophylls a (C55H72O5N4Mg) and b (C55H70O6N4Mg), even though they have almost identical chemical formulas.

206

UNIT 2 Energy Changes in Reactions

CHAPTER

7

Hess’s Law

7.1 Reaction mechanism • A simple reaction is a chemical reaction during which the reactants directly become products, without any reaction intermediate on the molecular level. • A complex chemical reaction can be broken down into a succession of several simple reactions and explained by means of a reaction mechanism.

• A reaction intermediate is a compound that appears in the equations of the simple reactions of the reaction mechanism, but not in the overall reaction equation. • The progress of a complex reaction can be visualized using an energy diagram.

Potential energy (kJ/mol)

• A reaction mechanism is a series of simple reactions that converts reactants into products over the course of a complex The energy diagram of a reaction reaction. The reaction mechanism is summarized 2nd activated complex by the overall reaction equation. 1st activated complex A

3rd activated complex

Ea2

E a3

Ea1 H1

B D

H2

1st simple reaction

4th activated Ea complex Ea4

C

2nd simple reaction

H3

3 simple reaction

H H4

E

4th simple reaction

Reaction in progress

7.2 Summation of enthalpies • Hess’s law, or the “law of constant heat formation” can be used to algebraically determine the enthalpy of a reaction. • According to Hess’s law, if a reaction can be broken down into several simple reactions, its enthalpy change (H) is equal to the algebraic sum of the enthalpy changes of each simple reaction (H1, H2, H3, etc.). The enthalpy change of the overall reaction can be expressed by the following general equation: H  H1  H2  H3  … • To carry out the algebraic sum of the thermochemical equations, the following rules must be respected: – The identical terms situated on the same side of the equation are added. – The identical terms situation on either side of the equation are subtracted. – If an equation is reversed, the H sign must also be reversed. – If the coefficients of a chemical reaction are modified by multiplying or dividing them by a common factor, the value H must also be multiplied or divided by this same common factor.

CHAPTER 7 Hess’s Law

207

Hess’s Law

1. Determine the overall equation by adding each of the following simple reactions. Certain equations must be balanced. 1 a) NO2 (g) n N2 (g)  O2 (g) H  33.2 kJ/mol 2 1 1 N  O n NO (g) H  90.2 kJ/mol 2 2 (g) 2 2 (g) b) CO2 (g)  2 H2O (l) n CH4 (g)  2 O2 (g) H  890.4 kJ/mol C (s)  O2 (g) n CO2 (g) H  393.5 kJ/mol 2 H2 (g)  O2 (g) n 2 H2O (l) H  571.6 kJ/mol 11 c) C12H22O11 (s) n 12 C (s)  11 H2 (g)  O2 (g) 2 H  2225.5 kJ/mol 1 H2 (g)  O2 (g) n H2O (g) 2 H  241.8 kJ/mol n C (s)  O2 (g) CO2 (g) H  393.5 kJ/mol 1 d) H2SO4 (l) n S8 (s)  H2 (g)  2 O2 (g) 8 H  814.0 kJ/mol 1 H2 (g)  O2 (g) n H2O (g) H  241.8 kJ/mol 2 1 S  O2 (g) n SO2 (g) H  296.8 kJ/mol 8 8 (s) e) H2S (g) n H2 (g)  S (s) H  20.6 kJ/mol 2 H2 (g)  O2 (g) n 2 H2O (g) H  483.6 kJ/mol S (s)  O2 (g) n SO2 (g) H  296.8 kJ/mol 2. Ethanol (C2H5OH) is an alcohol that can be used as a biofuel. It is considered a renewable energy source since it can be synthesized from glucose (C6H12O6) produced through the fermentation of grain. This synthesis reaction is translated by the following overall equation: C6H12O6 (s) n 2 C2H5OH (l)  2 CO2 (g) Based on the two chemical reactions below, what is the molar heat of the formation of ethanol? C6H12O6 (s)  6 O2 (g) n 6 CO2 (g)  6 H2O (l) H  2803.1 kJ/mol C2H5OH (l)  3 O2 (g) n 2 CO2 (g)  3 H2O (l) H  1366.8 kJ/mol

208

UNIT 2 Energy Changes in Reactions

3. Below are two thermochemical reactions: 3 H  1675.7 kJ/mol 2 Al (s)  O2 (g) n Al2O3 (s) 2 3 Fe2O3 (s) n 2 Fe (s)  O2 (g) H  824.2 kJ/mol 2 Using these two reactions, determine the molar heat of the reaction below: Fe2O3 (s)  2 Al (s) n Al2O3 (s)  2 Fe (s) 4. The decomposition of diatomic hydrogen (H2) into atomic hydrogen (H) is translated by the following equation: H2 (g) n 2 H (g) The standard enthalpy change of this reaction is 434.6 kJ. Calculate the standard enthalpy of formation of atomic hydrogen. 5. Examine the graph below and answer the following questions:

Potential energy

CHAPTER 7

150 100 50

Reaction in progress

a) How many simple reactions are involved in this reaction? b) How many reaction intermediates are formed during this reaction? c) How many activated complexes are formed during this reaction? d) What is the activation energy of the second step? e) What is the enthalpy of the third activated complex? f) What is the enthalpy change of the first step? g) What is the enthalpy change between the second and fourth reaction? h) Is the first reaction exothermic or endothermic? i) What is the enthalpy change of the overall reaction? j) Is the overall reaction exothermic or endothermic?

6. Based on the standard molar enthalpy of formation table provided in the appendix (see page 418), what are the heats of the following reactions? 1 a) NO (g)  O2 (g) n NO2 (g) 2 1 b) CO (g)  O2 (g) n CO2 (g) 2

9. The combustion of propane (C3H8) is described by the following overall equation: C3H8 (g)  5 O2 (g) n 3 CO2 (g)  4 H2O (g) Determine the enthalpy change of this reaction by using the thermochemical reactions below: 1) CO (g)  H2 (g) n H2O (g)  C (s)  130 kJ 2) C (s)  O2 (g) n CO2 (g)  393.5 kJ 1 3) H2O (g)  241.8 kJ n H2 (g)  O2 (g) 2 4) 3 C (s)  4 H2 (g) n C3H8 (g)  104.7 kJ 1 5) CO (g)  O2 (g) n CO2 (g)  283 kJ 2

7. Acetylene (“ethene”) (C2H2) is an extremely flammable colourless gas. The flame produced by its combustion is very bright. The formation of acetylene using carbon (C) and hydrogen (H2) is translated by the following overall reaction: 2 C (s)  H2 (s) n C2H2 (g) Using the following equations, determine the molar heat of formation of acetylene. 5 C2H2 (g)  O2 (g) n 2 CO2 (g)  H2O (l) 2 H  1299.6 kJ/mol C (s)  O2 (g) n CO2 (g) 1 H2 (s)  O2 (g) n H2O (l) 2

H  393.5 kJ/mol

10. Photosynthesis can be summarized by the following equation: 6 CO2 (g)  6 H2O (l) n C6H12O6 (s)  6 O2 (g) How would you determine the enthalpy change of this reaction experimentally?

11.

H  285.8 kJ/mol

8. Consider the overall reaction described by the following equation: A  B n C  D  energy The steps of the reaction mechanism can be described by the following three equations: Step 1 : A  B  energy n X Step 2 : X  energy n Y Step 3 : Y n C  D  energy

Potential energy

Potential energy

Of the following four graphs, which one illustrates the reaction mechanism of the overall reaction? a) b)

Reaction in progress

Reaction in progress

Molar heat of the dissolution of NaOH(s)  53.4 kJ/mol Molar heat of the neutralization of a solution of NaOH by a solution of HCl = -54.0 kJ/mol What is the molar heat of the neutralization of solid sodium hydroxide by a solution of hydrochloric acid?

12.

Under normal conditions of temperature and pressure, certain elements exist in more than one form. This is the case for carbon (C), which, at normal temperature and pressure, exists in the form of both graphite and diamond. Graphite can be changed into diamond according to the following equation: Cgraphite n Cdiamond

Potential energy

d)

Potential energy

c)

Reaction in progress

A student wants to know the molar heat of the neutralization of solid sodium hydroxide (NaOH) by a solution of hydrochloric acid (HCl). She decides to determine in the lab, using a calorimeter, the molar heat of the dissolution of solid sodium hydroxide and the neutralization heat of a sodium hydroxide solution by a hydrochloric acid solution. She obtains the following results:

Consider the following reaction equations: Cgraphite  O2(g) n CO2(g) H  393.5 kJ/mol Cdiamond  O2(g) n CO2(g) H  393.4 kJ/mol Reaction in progress

Which form of carbon—graphite or diamond—is considered the standard state of carbon?

CHAPTER 7 Hess’s Law

209

210

CONTENTS CHAPTER 8

Measuring Reaction Rate . . . . . . . . 213 CHAPTER 9

Collision Theory . . . . . . . . . . . . . . . . . . 235 CHAPTER 10

Factors that Affect Reaction Rates . . . . . . . . . . . . . . . . . . . 245

The combustion of kerosene in a plane engine is a very rapid reaction that releases a large quantity of energy. However, fuel transported in the tank does not burn spontaneously when in contact with oxygen in the air, and the corrosion of metal that composes certain plane parts occurs very slowly. Why are some reactions so rapid and others so slow? How can we speed up reactions that are too slow or slow them down when they are too fast in order to fully benefit from their potential?

In this unit, you will learn how to express, calculate and compare chemical reaction rates. After studying the mechanism behind a chemical reaction from a particular perspective, you will be able to understand the factors that affect the rate of a chemical reaction.

211

8.1 Expressing reaction rate

CHAPTER

8

MEASURING REACTION RATE

8.2 Reaction rate in terms of the stoichiometric coefficients of a balanced chemical equation 8.3 Ways to measure reaction rate 8.4 Average reaction rate and instantaneous reaction rate

UNIT

3

REACTION RATE

CHAPTER

9

COLLISION THEORY

9.1 Types of collisions 9.2 Reaction mechanism explained by collision theory

10.1 Nature of reactants 10.2 Surface area of reactants CHAPTER

10

FACTORS THAT AFFECT REACTION RATES

10.3 Concentration of reactants and the rate law 10.4 Temperature of the reaction environment 10.5 Catalysts

212

Measuring Reaction Rate

C

hemical reactions do not all occur at the same rate. For example, the corrosion of a car’s body due to the oxidation of iron (Fe) happens very slowly. However, if a car is in a collision that causes its airbags to deploy, the chemical reaction that inflates the airbags occurs within a few dozen milliseconds. Studying the rate of a chemical reaction provides a better understanding of the factors that influence it with the goal of using the reaction for a particular purpose or to better control it.

Review The Mole concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 The law of conservation of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Balancing chemical equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Stoichiometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

This chapter will cover how the rate of a chemical reaction is expressed, in addition to explaining the variety of ways it can be measured.

8.1 8.2

Expressing reaction rate . . . . . . . . . . . . . . . . 214

8.3 8.4

Ways to measure reaction rate . . . . . . . . . 220

Reaction rate in terms of the stoichiometric coefficients of a balanced chemical equation . . . . . . 216

Average reaction rate and instantaneous reaction rate . . . . . . . . . . . . 226 CHAPTER 8 Measuring Reaction Rate

213

8.1 Expressing reaction rate The reaction rate is a positive value that expresses the change in the quantity of a reactant or a product as a function of time, during a chemical change. A chemical reaction can be represented by this general equation: reactant(s) → product(s) This equation indicates that during the reaction, the number of particles of reactant decreases because they are transformed, while the number of particles of product increases because they are formed. The elementary reaction in which reactants A and B are transformed into product C can be represented by the following equation: AB→C The reaction takes place gradually. In fact, at the beginning of the reaction, there are only particles of reactants A and B (see Figure 1). As time passes, particles of product C appear.

Figure 1 The progress of a reaction that transforms reactants into a product, at 10-second intervals. At first, there are only particles of the reactants. As time passes, particles of the product are formed.

Reactants Product

The graph of the progress of this reaction shows that at the beginning, particles of product C appear at a relatively rapid rate, as shown by the slope of the curve (see Figure 2 on the following page). After a while, the rate slows, as shown by the drop in the curve of product C. It is worth noting that this rate of appearance of particles of product C is tied to the rate of disappearance of the particles of reactants A and B. Consequently, in this simple reaction, the disappearance curve for reactants A and B is the inverse of that for the appearance of product C. The rate of change in the quantities of reactants A and B and product C during the reaction is called the reaction rate.

214

UNIT 3 Reaction Rate

Variation of number of particles as a function of reaction time

Number of particles

20 Reactant A Reactant B Reactant C

15

10

5

0

10

20

30 Time (s)

40

50

Reaction rate

60

Figure 2 During this reaction, the quantity of particles of reactants A and B decreases at the same rate as the particles of product C appear.

Reaction rate indicates a rapid or slow change in reactants and products. To calculate it, measure the speed at which reactants are changed and products are formed. As a rule, mathematical expressions of the reaction rate of reactants or products are expressed in this way:

In a few milliseconds, the tongue flick out at a speed of about 20 km/h. However, this reaction rate must not be confused with a chemical reaction rate, because during a chemical reaction the rate usually slows over time.

Reaction rate rreactant(s) rproduct(s)

Quantity

The term “reaction rate” belongs to the world of chemistry, but also to the world of physiology. For example, the chameleon must flick its tongue out very quickly to catch its prey: the time between the moment it sees its prey and the moment its tongue muscles respond is its reaction rate.

of reactant(s) t

Quantity of product(s) t

where rreactant(s)  Reaction rate of reactant(s) Quantity of reactant(s)  Change in quantity of reactant(s) (Quantityf of reactant)  Quantityi of reactant(s)) rproduct(s)  Reaction rate of product(s) Quantity of product(s)  Change in quantity of product(s) (Quantityf of product(s)  Quantityi of product(s)) t  Change in time (t f  t i)

Since the quantity of reactants decreases during a chemical reaction, the change that results is always negative. To make it positive, add a minus sign in front of the rate. This will make the value of the reaction rate of the reactants positive and it will conform to convention.

Figure 3 A chameleon can flick its tongue out of its mouth at a speed of about 20 km/h to capture its prey.

CHAPTER 8 Measuring Reaction Rate

215

8.2 Reaction rate in terms of the stoichiometric coefficients of a balanced chemical equation Reaction rate is proportional to the coefficient, the reactant or the product of the balanced chemical equation chosen to express it. Stoichiometry is the study of the quantitative relationships between the amounts of matter (reactants and products) that react in a chemical change. The coefficients placed before the chemical formulas of reactants and products in a balanced chemical equation indicate the proportions of the reactants that combine to form the products. These coefficients indicate both the number of molecules and the number of moles. It is important to pay particular attention to the coefficients of the reactants and the products used to express a more complex reaction rate. For example, in the decomposition reaction of nitrogen pentoxide (N 2O5), brown nitrogen dioxide (NO2) and colourless oxygen (O2) form (see Figure 4). The coefficients indicate that when two moles of nitrogen pentoxide decompose, they form four moles of nitrogen dioxide and one mole of oxygen. 2 N2O5 (g) → 4 NO2 (g)  O2 (g) Considering that this reaction takes place in a closed system at a constant volume, the coefficients can also represent concentrations expressed in moles per litre (mol/L).

Figure 4 Nitrogen pentoxide (N2O5) decomposes to yield brown nitrogen dioxide (NO2) and colourless oxygen (O2).

The rate of this reaction can be expressed in terms of the decomposition of the reactant, nitrogen pentoxide, or in terms of the formation of the products, nitrogen dioxide and oxygen. When the transformation rate of one of the reactants, or the formation rate of one of the products is known, it is possible to deduce the reaction rate of the other substances involved in the reaction from the stoichiometric coefficients of the balanced equation. In this case, for example, the rate of nitrogen dioxide production can be found if the rate of oxygen production is known. In fact, the stoichiometric coefficients indicate that the rate of nitrogen dioxide production is four times greater than that of oxygen. Consequently, the rate of oxygen production is equal to one quarter the rate of nitrogen dioxide production, and is expressed according to the following relationship: [O2] 1 [NO2]  t 4 t Similarly, when one mole of oxygen is produced, two moles of nitrogen pentoxide decompose.

216

UNIT 3 Reaction Rate

Consequently, the rate of oxygen production is equal to half the rate of nitrogen pentoxide decomposition, and is expressed according to the following relationship: [O2]  1 [N2O5]  t 2 t

8.2.1

General reaction rate

The general reaction rate is an expression of the change in the concentration of a given substance divided by its stoichiometric coefficient as a function of time. The mathematical relationship that describes the general rate of a chemical reaction is expressed in the following way: General reaction rate r where r

1 [A]  1 [B] 1 [C] 1 [D] *    a t b t c t d t

 General reaction rate, expressed in moles per litre-second (mol/(Ls) a, b, c, d  Coefficients of the substances involved in the reaction [A], [B], [C], [D]  Changes in concentration of the substances involved in the reaction, expressed in moles per litre (mol/L) t  Change in time, expressed in seconds (s) * For a chemical reaction of type a A  b B → c C  d D.

Since changes in concentration in relation to changes in time are in fact reaction rates whose value is positive according to convention, the general reaction rate can also be expressed in the following way: General reaction rate r where r a, b, c, d rA, rB, rC, rD

1 1 1 1 rA  rB  rC  rD a b c d

 General reaction rate, expressed in moles per litre-second (mol/(Ls))  Coefficients of the substances involved in the reaction  Reaction rates of the substances involved in the reaction, expressed in moles per litre-second (mol/(Ls))

These mathematical relationships can be used to find the general reaction rate and the reaction rate for each substance. The choice of one substance over another to express the reaction rate will be dictated by the indicators provided in the problem to be solved and how they relate to the reactants or the products.

CHAPTER 8 Measuring Reaction Rate

217

Below are some examples of how these mathematical relationships are used to solve problems. Example A Ammonia (NH3) is used in agriculture as a chemical fertilizer. It is produced by the reaction of nitrogen (N2) with hydrogen (H2) as shown in this equation: N2 (g)  3 H2 (g) → 2 NH3 (g) The rate of ammonia production is 5.0  106 mol/(Ls). a) What is the general reaction rate? b) What are the corresponding rates of nitrogen and hydrogen transformation? a) Data: rNH  3

5.0  106 mol/(Ls)

r ?

Calculation: 1 1 1 r  rA  rB  rC a b c  

1 1

rN  2

1 3

rH  2

1 2

rNH

3

1 1 rNH   5.0  106 mol/(Ls)  2.5  106 mol/(Ls) 3 2 2

Answer: The general reaction rate is 2.5  106 mol/(Ls). b) Data: rNH3  5.0  106 mol/(Ls) rN2  ? rH2  ?

Calculation: 1 1 1 r  rN  rH  rNH 2 2 3 1 3 2 ⇒ rN  2

1 1 rNH   5.0  106 mol/(Ls)  2.5  106 mol/(Ls) 3 2 2

1 1 r  3   5.0  106 mol/(Ls)  7.5  106 mol/(Ls) 2 NH3 2 6 Answer: The rate of nitrogen (N2) transformation is 2.5  10 mol/(Ls) and the rate of hydrogen (H2) transformation is 7.5  106 mol/(Ls). ⇒ rH  3  2

Example B Hydrogen cyanide (HCN) is a highly toxic gas produced from methane (CH4) and ammonia (NH3) as shown in this equation: CH4 (g)  NH3 (g) → HCN (g)  3 H2 (g) What is the rate of hydrogen (H2) formation if the concentration of ammonia goes from 0.20 mol/L to 0.060 mol/L in 90 s? Data: [NH3]i  0.20 mol/L [NH3]f  0.060 mol/L t  90 s r ? rH2  ?

1. Calculation of the general reaction rate: 1 [CH4]  1 [NH3] 1 [HCN] 1 [H2] r     1 t 1 t 1 t 3 t 

1 [NH3] ([NH3]f  [NH3]i)  1 t t

(0.060 mol/L  0.20 mol/L)  1.6  103 mol/(Ls) 90 s 2. Calculation of the reaction rate of hydrogen: 1 1 1 1 r  rCH4  rNH3  rHCN  rH2 1 1 1 3 ⇒ rH2  3r  3  1.6  103 mol/(Ls)  4.8  103 mol/(Ls) 

Answer: The rate of hydrogen (H2) formation is 4.8  103 mol/(Ls).

218

UNIT 3 Reaction Rate

SECTION 8.2

Reaction rate in terms of the stoichiometric coefficients of a balanced chemical equation

1. Carbon monoxide (CO) reacts with hydrogen (H2) to form methanol (CH3OH). CO (g)  2 H2 (g) → CH3OH (g) Which of the following expressions can be used to express the rate of methanol synthesis? a) Quantity of methanol produced/unit of time. b) Quantity of carbon monoxide produced/unit of time. c) Quantity of hydrogen consumed/unit of time. d) Quantity of methanol consumed/unit of time. e) Quantity of hydrogen produced/unit of time. 2. Write the mathematical expressions that designate the reaction rate of the reactants and the products for each of the following chemical equations: a) I(aq)  OCI(aq) → CI(aq)  OI(aq) b) 3 O2 (g) → 2 O3 (g) c) 4 NH3 (g)  5 O2 (g) → 4 NO(g)  6 H2O(g) 3. Ammonia (NH3) reacts with oxygen (O2) to produce nitrogen monoxide (NO) and water vapour as shown in the following equation: 4 NH3 (g)  5 O2 (g) → 4 NO(g)  6 H2O(g) At a specific moment in the reaction, the ammonia reacts at a rate of 0.068 mol/(L • s). What is the corresponding rate of water vapour production? 4. Gaseous hydrobromic acid (HBr) reacts with oxygen (O2) to produce bromine (Br2) and water vapour. 4 HBr(g)  O2 (g) → 2 Br2 (g)  2 H2O(g) How does the rate of hydrobromic acid decomposition, in moles per litre-second (mol/(Ls)), compare to the rate of bromine formation, also in moles per litresecond? Answer in the form of a mathematical relationship.

5. Metallic magnesium (Mg) reacts with hydrochloric acid (HCl) to produce magnesium chloride (MgCl2) and hydrogen (H2). Mg (s)  2 HCl (aq) → MgCl2 (aq)  H2 (g) In an interval of 1 s, the mass of magnesium changes by -0.011 g. a) What is the corresponding rate of hydrochloric acid consumption, in moles per second (mol/s)? b) Calculate the corresponding rate of hydrogen production, in litres per second (L/s), at 20°C and 101 kPa. 6. A hypothetical reaction takes place, as shown in this equation: A (s)  4 B (l) → 2 C (l)  3 D (g) After 45 s, the reaction is complete, and 6 mol of A have reacted. a) Calculate the transformation rate of A. b) Use a mathematical relationship to express the reaction rate of B and the reaction rate of D. c) Calculate the formation rate of 150 L of D at 25°C and 101.2 kPa. 7. You are notified that the electricity will be cut off for an unspecified length of time in the next few days. You prepare by calculating the combustion rate of a quantity of wax. To make the task interesting, you want to find out what quantity of carbon dioxide (CO2) and water will be formed and at what rate. You burn a candle for 45 min, and it loses 126 g. The reaction is represented by this equation: C25H52 (s)  38 O2(g) → 25 CO2 (g)  26 H2O(g)  energy a) Calculate the combustion rate of one mole of wax (C25H52). b) Calculate the formation rate of one mole of carbon dioxide. c) Calculate the production rate of water in millilitres per second (mL/s) at 25°C and 101.2 kPa. d) Express the concentration of carbon dioxide formed in parts per million per second (ppm/s).

CHAPTER 8 Measuring Reaction Rate

219

8.3 Ways to measure reaction rate The ways to measure reaction rate depend on the state of the reactant or product to be measured, the ease with which it can be measured in the laboratory and the way in which the results will be used. To determine reaction rates in the laboratory, measure the parameters of the reactants that are being transformed or the products that are being formed. The choice of the most appropriate measurement method depends on the state of the substance to be measured, and also on the ease with which this measurement can be made and the way in which the results will be used. For example, it is sometimes useful to take note of the data at regular time intervals if the results will be used to create a graph.

8.3.1

Measuring reaction rate in terms of the state of the reactant or the product and the ease of using the measurment method

There are a great many chemical reactions involving reactants and products of all kinds and in every physical state. These substances can be in a solid, liquid, gaseous or aqueous state, according to the type of reaction.

Figure 5 A solution of hydrochloric acid (HCl) reacts chemically on contact with metallic magnesium (Mg).

To measure the reaction rate experimentally, consider the state of at least one of the reactants or products in the reaction in order to choose the appropriate measurment method for that state. For example, the following equation illustrates the reaction between metallic magnesium (Mg) and a hydrochloric acid (HCl) solution of a known concentration (see Figure 5). Mg (s)  2 HCl (aq) → MgCl2 (aq)  H2 (g) To find the rate of this reaction, determine which parameter (mass, volume, pressure, number of particles, concentration, pH, etc.) of one of the reactants or products it would be most relevant to measure. The chosen parameter is dictated by the state of the substances and by the ease of measurement.

Figure 6 A pH meter is useful for measuring reaction rate if the concentration of hydronium (H3O) ions or hydroxyl (OH) ions changes during the reaction.

220

UNIT 3 Reaction Rate

Because magnesium is solid, its mass is relatively easy to measure with a balance. After weighing the magnesium, measure the time it takes for it to react completely in a sufficient quantity of a hydrochloric acid solution. The reaction rate is then expressed as a unit of mass as a function of time, that is, grams per second (g/s). Using molar mass, convert this measurement to moles per second (mol/s). The pH can also be used to measure the rate of this reaction. Since hydrochloric acid is consumed in the reaction, record the change in pH values in relation to time. A pH meter probe connected to a computer is very useful for this task (see Figure 6).

In this case, the reaction rate is expressed in pH units per second (pH/s). It is also possible to convert the pH units to concentration values of the hydro nium ([H3O]) ions, using the following relationship: Conversion of pH units into concentration of hydronium ions [H3O]  10pH This relationship can be used to express the reaction rate as a concentration of hydronium ions per second ([H3O]/s). Another way to measure the rate of this reaction is to time the production of gaseous hydrogen (H2) and to measure the total volume collected by displacement of water in a gas burette. The volume produced at regular time intervals can also be measured. In this case, the reaction rate is expressed as millilitres of hydrogen produced per second (mL/s). Chemical reactions that take place in the laboratory or elsewhere can proceed very rapidly, such as fireworks explosions (see Figure 7), or very slowly, such as the oxi dation of iron (Fe) that rusts nails. For rapid reactions, the rates have to be expressed in units of time appropriate to the duration of the reaction. So, the rate of the explosion of a firework can be expressed in milliseconds, while the reaction rate of the oxidation of a nail may take months or even years.

8.3.2

Measuring the rate in terms of the intended use of the results

Figure 7 Fireworks explosions are chemical reactions that occur very rapidly.

In a reaction between magnesium (Mg) and hydrochloric acid (HCl), the simplest way to measure the reaction rate is to weigh the magnesium and to record the length of time it takes for the reaction to be complete. However, this method provides only one datum, that is, the total mass of magnesium that reacted in the time measured, which cannot be used to draw a curve showing a change in rate. To draw a graph showing the change in the reaction over time, the most appropriate technique is to measure the volume by water displacement in a gas burette (see Figure 8).

APPENDIX 3 Determining concentration, p. 385.

Measurement of the volume at regular intervals can be used to draw a curve of the reaction rate expressed in terms of the production of gaseous hydrogen (H2) as a function of time. The choice of experimental technique to measure the parameters also depends on the intended use of the results. Figure 8 An apparatus for measuring gas volume by water displacement at regular time intervals

CHAPTER 8 Measuring Reaction Rate

221

APPENDIX 7 Common units of measure, p. 409.

The following table shows the units of measurement of reaction rate used most often in the laboratory, according to the parameter measured and the physical state of the substances, as well as the equation for calculating the rate.

Table 1 The main units of measurement of reaction rates, according to the parameter measured and the physical state of the substances Units of measurement of rate

Parameter measured

g/s

Mass

Solid, liquid, gas

mL/s

Volume

Liquid, gas

mol/(Ls)

Molar concentration

Gas, aqueous solution

Number of particles

Solid, liquid, gas

mol/s

Equation for calculating the rate

Physical state

État

r

m t

r

V t

r

[A] t

r

n t

where m  Change in mass, expressed in grams (g) t  Change in time, expressed in seconds (s) where V  Change in volume, expressed in millilitres (mL) t  Change in time, expressed in seconds (s) where [A]  Change in molar concentration, expressed in moles per litre (mol/L) t  Change in time, expressed in seconds (s) where n  Change in number of moles, expressed in moles (mol) t  Change in time, expressed in seconds (s)

The following examples show how to calculate a reaction rate using different measurement techniques, according to the physical state of the sub stances measured: Example A Calculating the reaction rate when the substance is in a solid state A 12.0-mg piece of magnesium (Mg) reacts in 100 mL of a solution of hydrochloric acid (HCl) at 1.00 mol/L, as shown in this equation: Mg (s)  2 HCl (aq) → MgCl2 (aq)  H2 (g) The piece of magnesium completely disappeared after 4 min 30 s. a) Calculate the rate of this reaction in grams per second (g/s). b) Calculate the rate of this reaction in moles per second (mol/s). a) Data: mi  12.0 mg  0.0120 g

Calculation: m r t

mf  0 g t f  4 min 30 s  270 s



ti  0 s

(0 g  0.0120 g)  4.44  105 g/s 270 s  0 s

r ? Answer: The reaction rate of the magnesium (Mg) is 4.44  105 g/s. b) Data: mi  12.0 mg  0.0120 g ti  0 s t f  4 min 30 s  270 s M  24.305 g/mol Mg

?

r ?

1. Calculation of the number of moles: m m 0.0120 g M  ⇒n   4.94  104 mol n M 24.305 g/mol

2. Calculation of the reaction rate: n  (0 mol  4.94  104 mol) r   1.83106 mol/s t 270 s  0 s

Answer: The reaction rate of the magnesium (Mg) is 1.83  106 mol/s.

222

UNIT 3 Reaction Rate

Example B Calculating the reaction rate when the substance is in a gaseous state Gaseous hydrogen (H2) escaping from the reaction site in Example A is collected and measured by water displacement every 30 seconds. The following table shows the results of the volume measurements: Volume of gaseous hydrogen (H2) produced as a function of time

Time (s)

Volume (mL)

Time (s)

Volume (mL)

0

0

150

28

30

7

180

32

60

13

210

35

90

18

240

38

120

23

270

40

a) Calculate the rate of gaseous hydrogen production in millilitres per second (mL/s). b) Create a graph showing the volume of gas produced as a function of time. b) Creation of the graph:

a) Data: Vi  0 mL Vf  40 mL ti  0 s tf  4 min 30 s  270 s r ?

Volume of gaseous hydrogen (H2) produced as a function of time 40 35

Calculation: V 40 mL  0 mL r   0.15 mL/s t 270 s  0 s

Volume (mL)

30 25 20 15 10 5

Answer: The rate of hydrogen (H2) production is 0.15 mL/s.

0 60

120 Time (s)

180

240

Example C Calculating the reaction rate when the substance is an aqueous solution The reaction between an aqueous solution of bromine (Br2) and an aqueous solution of methanoic (“formic”) acid (CH2O2) is shown by this equation: Br2 (aq)  CH2O2 (aq) → 2 Br(aq)  2 H(aq)  CO2 (g) At the beginning of the reaction, the solution is a dark brown colour and the concentration of bromine is 0.0120 mol/L. As the bromine is consumed, the colour of the solution becomes lighter. After 5 minutes, it is 0.0042 mol/L. Calculate the reaction rate of aqueous bromine in moles per litre-second (mol/(Ls)). Data: [Br2]i  0.0120 mol/L [Br2]f  0.0042 mol/L ti  0 s

Calculation: [Br2]  (0.0042 mol/L  0.0120 mol/L) r   t 300 s  0 s 5  2.6  10 mol/(Ls)

t f  5 min  300 s r ? Answer: The reaction rate for bromine (Br2) is 2.6  105 mol/(Ls). CHAPTER 8 Measuring Reaction Rate

223

APPLICATIONS Galvanization Steel is a highly valued material in the construction field. Known for hardness and strength, it is an alloy of iron (Fe) and carbon (C). The problem is that it oxidizes rapidly; in other words, it rusts (see Figure 9). The answer to this problem is hot galvanizing, a process that slows the reaction rate of oxidation and protects steel against corrosion. With this protection, steel can last for more than 25 years.

The galvanized product offers many advantages beyond its resistance to corrosion (see Figure 11). It withstands shocks and abrasion better because the iron-zinc alloy is even harder than steel alone. The extended service life reduces the cost of maintaining the steel.

There are many steps in the galvanization process. First, all dirt, such as rust, scale and oil, must be removed from the steel. The metal is dipped in a degreasing solution and then in a stripping solution. Fluxing, which prevents the steel from oxidizing again before the final phase of the process, completes the preparation of the surface to be galvanized. The steel is then plunged into an aqueous solution of ammonium chloride (NH4Cl) and zinc chloride (ZnCl2) at 60°C. Ashes formed by the process rise to the surface and are recovered. Next comes the crucial step. The steel is immersed in a bath of molten zinc (Zn) at 450°C. The duration of the dip varies according to the size and shape of the pieces. It can take between 3 and 15 minutes, depending on whether the piece is a bolt or a large piece of framework. Then there is a metallurgical reaction, which chemists call diffusion bonding, between the steel and the zinc. This means that when the steel is taken out of the bath it is not only covered in solidified zinc but its surface has formed an alloy with the zinc that is several layers deep (see Figure 10 and Table 2). The coating has a high concentration of zinc but the farther into the piece, the lower the zinc content. Between 60 and 70 kg of zinc are required to protect one tonne of steel.

Figure 9 A rusty adjustable wrench

224

UNIT 3 Reaction Rate

1

2 3 4 5

Figure 10 A microscopic view of layers of solidified zinc (Zn) coating

Table 2 The zinc (Zn) and iron (Fe) content of the layers of galvanized steel Layer

Percentage of zinc (Zn)

Percentage of iron (Fe)

1

/ 100

 0.03

2

94 to 95

5 to 6

3

88 to 93

7 to 12

4

72 to 79

21 to 28

5

0

/ 100

Figure 11 A galvanized open-end wrench

SECTION 8.3

Ways to measure reaction rate

1. The speed of a moving vehicle is expressed in kilometres per hour. What unit of speed would you use for each of the following? a) Pages printed by a printer. b) Downloading music from the Internet. c) A family’s consumption of soft drinks. d) Your Internet use. e) The time it takes you to read a book. 2. What parameters could be measured to find the speed of the following reactions? Explain your answers. a) The neutralization of hydrochloric acid (HCl) by sodium hydroxide (NaOH): HCl (aq)  NaOH (s) → NaCl (aq)  H2O (l) b) The rate of production of copper sulfate (CuSO4): Cu(s)  H2SO4 (aq) → CuSO4 (aq)  H2 (g) 3. A lighter contains about 4 mL of butane (C4H10). It takes about 25 min to burn the entire contents of a lighter. Since the density of liquid butane is 585 kg/m3 at 15°C and that of gaseous butane is 2.50 kg/m3 at 15°C, express the rate of this reaction using as many different units of measure as possible. 4. In an experiment, you observe that the concentration of a basic substance goes from 0.4760 mol/L to 0.0175 mol/L in 16 min. Calculate the transformation rate for this base in moles per litre-second (mol/(Ls)) and in moles per litre-hour (mol/(Lh)). 5. Under specific conditions, chlorine (Cl2), which is very toxic, can be bubbled through a solution of sodium iodide (NaI). Solid iodine (I2) is obtained, as shown in this reaction: Cl2 (g)  2 NaI (aq) → 2 NaCl (aq)  I2 (s) How would you go about measuring the rate of this reaction? 6. You create this reaction in the laboratory: CaCl2 (aq)  Na2CO3 (s) → CaCO3 (s)  2 NaCl (aq) How can you calculate the rate of formation of sodium chloride (NaCl) after 15 minutes?

7. Iron (Fe) is produced in a reaction combining iron oxide (Fe2O3) from ore with carbon monoxide (CO). This is the equation for the reaction: Fe2O3 (s)  3 CO (g) → 2 Fe (s)  3 CO2 (g) Since the production rate for iron is about one tonne per 30 minutes, calculate the mass of carbon dioxide produced and its volume at 25°C and 101.2 kPa. 8. Alkaline metals react violently with water. If 3 grams of sodium (Na) are deposited on the surface of some water, the reaction is violent and produces a great deal of heat. The sodium will be finished reacting after 1.5 s. This is the equation for the reaction: 2 Na (s)  2 H2O (l) → 2 NaOH (aq)  H2 (g) Calculate the rate of production of hydrogen (H2) in moles per litre-second (mol/(Ls)) at STP. 9. You neutralize 25 mL of sulphuric acid (H2SO4) at 0.5 mol/L with an excess of sodium hydroxide (NaOH). After three minutes, the pH-meter indicates 7.0. Calculate the mass of sodium sulfate (Na2SO4) produced and its rate of production in moles per litre-second (mol/(Ls)). 10. You put 0.95 g of magnesium (Mg) into a solution of hydrochloric acid (HCl) at 0.1 mol/L. At the end of 280 s, the reaction stops and you observe that there are still 0.5 g of magnesium left. a) Calculate the reaction rate in grams per second (g/s). b) Calculate the reaction rate in moles per second (mol/s). c) Calculate the reaction rate in grams per minute (g/min). 11. The gunpowder in some fireworks contains about 15% carbon (C), 10% sulphur (S) and 75% potassium nitrate (KNO3). If the duration of an explosion is 0.005 s, what is the combustion rate of potassium nitrate when an item containing 80 g of powder explodes? Express the rate using the appropriate time units for the reaction.

CHAPTER 8 Measuring Reaction Rate

225

8.4 Average reaction rate and instantaneous reaction rate The average reaction rate of a reaction is the change in the quantity of a reactant or a product over a given length of time. The instantaneous reaction rate of a reaction is the reaction rate at a specific time during the reaction. Not only do reaction rates vary according to the type of chemical reaction being studied, but the chemical reaction rate also varies during the progress of the reaction. In fact, the transformation rate of the reactants (or the formation rate of the products) is usually faster at the beginning of the reaction and tends to slow as the reaction continues. The rate calculated for a given time interval is the average reaction rate during that time interval (rx). The average reaction rate thus gives a general idea of the speed of the reaction. The instantaneous reaction rate indicates the reaction rate at a specific time during the course of the reaction.

A straight line intersecting * Secant a curve at two or more points. A straight line touching a * Tangent curve at just one point.

APPENDIX 4 Slope of a tangent to the curve, p. 390.

The rate calculations offered earlier (see examples A, B and C, on pages 222 and 223) are examples of calculations of average reaction rate, because they incorporate changes in quantities for a given time period. This period can be the total time for the reaction to be complete. It is also possible to calculate the average reaction rate for time intervals at any point in the progress of a reaction. Comparing average reaction rates at different times, at the beginning, in the middle or at the end of a reaction, for example, can provide more information about the change in the reaction rate. Calculating the average reaction rate involves graphing the value of the slope of the secant, which links the two points of the chosen time interval. By minimizing this time period, it is possible to graph the value of the reaction rate at a single point. This reaction rate is the instantaneous reaction rate. Because it represents the reaction rate at a precise point or “instant” on the curve, it is the calculation of the slope of the tangent at this precise point on the curve that will determine the value of the instantaneous reaction rate.

*

*

Average reaction rate

Figure 12 The speed of runners in a 100-metre race is not constant. That is why it is called average speed.

226

UNIT 3 Reaction Rate

Describing a chemical reaction rate or the speed of an athlete in a 100-metre race almost always refers to average rate/speed. But in both cases, it is not constant. When athletes leave the starting blocks at the beginning of a race, they enter an acceleration phase. Their strides gradually become longer and faster. After the first 60 metres, they reach their maximum speed of 39 km/h for women and 43 km/h for men. They maintain this velocity for about 30 metres. The last 10 metres are usually a deceleration phase. An athlete’s speed is not the same over the entire distance. This is also the case with a chemical reaction rate, which tends to slow over time. The rate for a given time interval is thus always the average .

Below is an example of how to calculate the average reaction rate at the beginning of a reaction and the instantaneous reaction rate at a specific moment. Example A hypothetical reaction: A (g) → B (g)  C (g) The data for the concentration of gas C is listed in the table below and plotted on the graph. In this graph, the slope of the secant determines the average reaction rate. The slope of the tangent on the curve of concentration as a function of time represents the instantaneous reaction rate. a) Calculate the average reaction rate during the first five seconds. b) Calculate the instantaneous reaction rate at exactly ten seconds from the beginning of the reaction (t  10 s). Concentration of C as a function of time 7 6

[C] (mol/L)

5

Time (s) 0.0

0.00

5.0

3.12  103

10.0

4.41  103

15.0

5.40  103

20.0

6.24  103

(mol/L  103)

Concentration of gas C

4

Tangent (instantaneous reaction rate)

3 2

Secant (average reaction rate)

1 0 5

a) Data: To graph the average reaction rate, we have to determine the slope of the secant (see graph above).

10 Time (s)

15

20

Calculation: [C]  t (rx)



3.1  103 mol/L  0.0 mol/L 5.0 s  0.0 s

 6.2  104 mol/(Ls) Answer: The average reaction rate during the first 5 seconds is 6.2  104 mol/(Ls) for the concentration of C. b) Data: To find the instantaneous reaction rate, we must first draw the tangent at point t  10 s on the curve. Then we must calculate the slope of this straight line by using x and y for two points on this straight line (see graph above).

Calculation: [C] rt  10 s  t 

5.4  103 mol/L  3.1  103 mol/L 15.0 s  5.0 s

 2.3  104 mol/(Ls)

Answer: The instantaneous speed at 10 s from the beginning of the reaction is 2.3  104 mol/(Ls).

CHAPTER 8 Measuring Reaction Rate

227

Average reaction rate and instantaneous reaction rate

SECTION 8.4

1. Give examples of average reaction rate and instantaneous reaction rate for each of the following situations: a) travelling in a car b) riding a bicycle c) walking d) jogging 2. Taking your laboratory experience into account, explain how a chemical reaction rate changes. 3. The results obtained from the decomposition of nitrogen pentoxide (N2O5) are compiled in the following table:

c) Calculate the instantaneous reaction rate at the following times: t = 10 s; t = 20 s; and t = 50 s. d) During which time interval is the reaction rate highest? 5. The concentration of carbon monoxide (CO), measured at regular intervals, reacts as shown in the following equation: CO (g)  NO2 (g) → CO2 (g)  NO (g) The data from this experiment have been compiled in a table and then graphed to express the change in the concentration of carbon monoxide over time. Concentration of carbon monoxide (CO) during a reaction

Concentration of nitrogen pentoxide (N2O5) during decomposition

Time (s)

[N2O5] (mol/L)

Time (s)

6.00  102

7.1  103

0

0.200

1.80  103

6.4  103

10

0.134

3.00  103

4.8  103

20

0.100

4.20  103

2.1  103

30

0.080

40

0.066

Graph the rate of decomposition of nitrogen pentoxide and calculate the instantaneous reaction rate of the reaction at the 2000th second.

Concentration (mol/L)

20 15 10

0.150

0.100

0.050

0 10

5 0

10

20

30 40 Time (s)

50

60

a) Is this substance a reactant or a product? Explain your answer. b) Calculate the average reaction rate between the 10th and 20th seconds.

228

0.200 Concentration of CO (mol/L)

4. The following graph illustrates the change in the concentration of a substance as a function of time during the course of a chemical reaction:

[CO] (mol/L)

UNIT 3 Reaction Rate

20 Time (s)

30

40

a) Calculate the average reaction rate during the first 10 seconds. b) Calculate the average reaction rate between the 20th and 30th seconds. c) Calculate the instantaneous reaction rate at the 10th second.

Food preservation Food preservation is not a recent invention. From the time when agriculture and animal breeding began in the Near East about 8000 BCE, preserving food was a problem. Until the 19th century, the typical method to preserve meat and fish was by salting (see Figure 13). This method was often accompanied by drying or smoking. Dehydration caused by these methods limited microbial activity, since micro-organisms, whether harmful or not, require a certain amount of water to survive. The fewer bacteria, yeasts and fungi there are on foods, the slower the chemical reactions that cause decomposition.

Figure 13 Salting was the first preservation method for fish and meat.

Although the first refrigerator was not manufactured until 1913 in Chicago, it was well known through the ages that lowering temperature preserved food. The Romans caught fish in the Rhine and wrapped them in ice and snow to transport them back to Rome. Low temperatures allow food to be kept longer by inhibiting microbial growth. When food is refrigerated (at about 5°C), the growth of micro-organisms is slowed (see Figure 14). When food is frozen (at about 10°C), bacterial colonization stops completely. When bacteria are inactive, they cannot cause food to decompose.

Figure 14 Preservation by refrigeration was not available until 1913.

Sterilization was developed in the late 18th century. Nicolas Appert invented the procedure, which consists of sterilizing foods in hermetically sealed containers through the use of heat (see Fig ure 15). This was the be ginning of food preservation. In 1865, Louis Pasteur developed pasteurization. This method of preservation involves heating the food to about 75°C for at least 15 seconds, which kills the majority of the bacteria and inactivates the enzymes that accelerate chem- Figure 15 Sterilization ical reactions. With fewer active allows for the consumption enzymes and micro-organisms, de- of garden vegetables terioration is slowed down. To fur- all year long. ther improve food preservation, pasteurized products are refrigerated (see Figure 16). Pasteurized milk reduced the infant mortality rate at the time of its introduction. Since then, food preservation has continued to develop. Irradiation of some foods is allowed in many countries, including Canada. Radiation kills bacteria by damaging their DNA. Nevertheless, the effectiveness of this method depends on a number of factors, such as the type and number of micro-organisms.

Figure 16 Pasteurized milk must be kept cool.

CHAPTER 8 Measuring Reaction Rate

229

CHAPTER

8

Measuring Reaction Rate

8.1 Expressing reaction rate • The reaction rate (r) is a positive value that expresses the change in the quantity of a reactant or a product as a function of time (t) during a chemical change. rreactant(s) 

 Quantity

of reactant(s) t

rproduct(s) 

Quantity of product(s) t

8.2 Reaction rate in terms of the stoichiometric coefficients of a balanced chemical equation • The reaction rate is proportional to the coefficient of the reactant or the product of the balanced chemical equation chosen to express it. • The general reaction rate of a chemical reaction (r) is the change in the concentration of a given substance ([A], [B], [C] and [D]) divided by its stoichiometric coefficient (a, b, c and d) as a function of time (t). • The mathematical relationships that allow us to find the general reaction rate and the reaction rate for each substance in a chemical reaction of type a A  b B → c C  d D are expressed this way: r 

1 [A]  1 [B] 1 [C] 1 [D]    a t b t c t d t

r 

1 1 1 1 r  r  rC  rD a A b B c d

8.3 Ways to measure reaction rate • The ways to measure the reaction rate depend on the state of the reactant or product to be measured and the ease with which it can be measured in the laboratory.

8.4 Average reaction rate and instantaneous reaction rate • The average reaction rate of a reaction is the average change in the quantity of a reactant or a product as a function of time for a given length of time. It is graphed by calculating the slope of the secant that links the two points of the chosen time interval. • The instantaneous reaction rate is the reaction rate at a specific time. It is graphed by calculating the slope of the tangent touching a point that corresponds to the precise moment in consideration.

230

UNIT UNIT3 3 Reaction ReactionRate Rate

CHAPTER 8

Measuring Reaction Rate

1. What are the advantages of measuring the reaction rate in the first moments of a reaction? 2. A piece of chalk (CaCO3) weighing 2.0 g reacts completely with hydrochloric acid (HCl) in 85 s according to the following equation: CaCO3 (s)  2 HCl (aq) → CaCl2 (aq)  CO2 (g)  H2O (l) a) Name two easy ways to measure the reaction rate experimentally. b) Calculate the reaction rate for the chalk in grams per second (g/s) and in moles per second (mol/s). 3. Name two very slow reactions (that take days or longer to complete) and two very rapid reactions (that are completely finished within a few minutes or seconds). 4. Propane (C3H8) is a gas that is commonly used as a source of chemical energy through combustion in internal combustion motors, barbecues and boilers, according to the following balanced equation: C3H8 (g)  5 O2 (g) → 3 CO2 (g)  4 H2O(g) a) Use a mathematical relationship to express the reaction rate for the concentration of each of these substances. b) Use a mathematical relationship to express the general reaction rate as a function of reaction rate for each substance involved in the reaction. 5. Explain why the expression of the rate of decrease of a reactant is always a negative value even though the convention is that reaction rates are always positive values. 6. Under what circumstances is the rate at which the concentration of a reactant decreases numerically equal to the rate at which the concentration of product increases? 7. Write mathematical expressions for the reaction rates of the reactants and products for each of the following chemical equations: a) H2 (g)  I2 (g) → 2 HI (g) b) 2 H2 (g)  O2 (g) → 2 H2O(g) c) 5 Br(aq)  BrO3(aq)  6 H(aq) → 3 Br2 (aq)  3 H2O (l)

8. In a laboratory experiment, a researcher calculated this stoichiometric equation: 2 B2O3 (s) → 4 B (s)  3 O2 (g) Which of the following statements is true? Explain your answer. a) The decomposition rate of boron oxide (B2O3) is equal to twice the production rate of boron (B). b) The decomposition rate of boron oxide is equal to half the production rate of boron. c) The production rate of oxygen (O2) is equal to 2/3 of the decomposition rate of boron oxide. d) The production rate of oxygen is equal to 4/3 of the production rate of boron.

9. Consider the following reaction: N2 (g)  3 H2 (g) → 2 NH3 (g) At a specific time during the course of the reaction, the hydrogen (H2) molecule reacts at a rate of 0.0074 mol/s. a) At what rate will ammonia (NH3) form? b) At what rate will nitrogen (N2) react?

10. A student measures the decomposition of metallic barium (Ba) in water, as shown in the following equation: Ba (s)  2 H2O (l) → Ba(OH)2 (aq)  H2 (g) The student observes that 3.2 g of barium decomposes completely in 1 min 35 s. What is the formation rate of hydrogen (H2) in moles per second (mol/s)?

11. The balanced equation for electrolysis of water is the following: 2 H2O (l) → 2 H2 (g)  O2 (g) a) How many times faster will hydrogen (H2) be produced than oxygen (O2)? Explain your answer. b) If 1 mole of hydrogen is produced in 1 minute, how many moles of oxygen will be produced in the same time? c) How many moles of oxygen will be produced in 1 hour? d) Write a mathematical relationship to express the general reaction rate for hydrogen and its coefficient in the balanced equation.

CHAPTER 8 Measuring Reaction Rate

231

12. In an experiment on the decomposition rate of a piece of chalk (CaCO3) in hydrochloric acid (HCl), a student observed these results: mass of decomposed chalk: 2.3 g; time of decomposition: 4 min 30 s; and [HCl]: 1 mol/L.

16. Sodium bicarbonate (NaHCO3) can extinguish small grease fires in a kitchen. When it is thrown on the flames, it absorbs the energy and decomposes according to the following reaction: 2 NaHCO3  E → Na2CO3  H2O  CO2

If the reaction begins when the chalk comes into contact with the acid, calculate the average reaction rate of decomposition of the chalk in grams per second (g/s) and in moles per second (mol/s).

a) Which of the gases in the reaction will be produced in the largest quantity? Explain your answer. b) Write the mathematical expressions for the general reaction rate of the reactants and the products. c) If at a given moment methanol reacts at a rate of 1 mol(Ls), what will the corresponding rate of consumption for oxygen (O2) be? d) What is the general reaction rate in moles per litre-second (mol/(Ls)) if 224 L of carbon dioxide (CO2) are produced in 1 min at STP?

14. In the following reaction, the production rate of sulphate (SO42) ions is 1.25  103 mol/(Ls): 2 HCrO4  3 HSO3  5 H → 2 Cr3  3 SO42  5 H2O a) What is the corresponding rate at which the concentration of hydrogen sulphite ions (HSO3) decreases in the same time interval? b) What is the corresponding rate at which the concentration of hydrogen chromate (HCrO4) ions decreases in the same time interval?

15. To provide a good illustration of the reaction between a solid and an aqueous solution, use solid copper (Cu) and silver nitrate (AgNO3) in a solution. At the beginning, the solution is colourless; at the end it is blue, and a silver substance has settled on the bottom of the beaker. This is the equation for the reaction: Cu (s)  2 AgNO3 (aq) → Cu(NO3)2 (aq)  2 Ag (s) If the mass of silver is 5 grams after 4 hours, calculate the transformation rate of solid copper into copper ions (Cu2) in grams per second (g/s).

232

UNIT 3 Reaction Rate

Decomposition of sodium bicarbonate (NaHCO3) 2.5

2.0 Quantity of NaHCO3 (102 mol)

13. The following balanced equation represents the combustion of methanol (CH3OH): 2 CH3OH (l)  3 O2 (g) → 2 CO2 (g)  4 H2O (g)

In a laboratory, an experiment is conducted in which the results presented in the following graph are obtained:

1.5

1.0

0.5

0

2

4

6 Time (min)

8

a) What is the reaction rate during the first 2 minutes? b) What is the average reaction rate of the reaction? c) What is the reaction rate during the interval from 6 to 8 minutes? d) What is the instantaneous reaction rate at 3 minutes? e) How do you explain the difference in rate between the beginning and end of the reaction?

17. For each of the following reactions, suggest one or more methods for monitoring the progress of the reaction. Explain your answers. a) C6H12 (l)  Br2 (aq) → C6CH12Br2 (l) (Hint: Bromine (Br2) is a brownish orange colour. The other compounds are colourless.) 1 b) H2O2 (aq) → H2O (l)  O2 (g) 2 c) CaCO3 (s)  H2SO4 (aq) → CaSO4 (aq)  CO2 (g)  H2O (l) d) N2 (g)  3 H2 (g) → 2 NH3 (g)

18. The neutralization of phosphoric acid (H3PO4) by magnesium hydroxide (Mg(OH)2) occurs according to the following balanced equation: 2 H3PO4 (aq)  3 Mg(OH)2 (aq) → Mg3(PO4)2 (aq)  6 H2O(l) Qualitative analysis of this neutralization showed that the phosphoric acid was consumed at a rate of 0.02 mol/(Ls). a) What is the corresponding consumption rate for magnesium hydroxide (Mg(OH)2) in moles per litre-second (mol/Ls))? b) Calculate the corresponding production rate for the salt (Mg3(PO4)2) in moles per litresecond (mol/(Ls)). c) If the total neutralization of the phosphoric acid occurred after 22 s, what was the initial concentration of the acid?

19 When heated to 300°C at standard atmospheric pressure, phosphorus pentachloride (PCl5) decomposes completely into phosphorus trichloride (PCl3) and chlorine (Cl2), as shown in this balanced equation: PCl5 (g) → PCl3 (g)  Cl2 (g) A quantity of phosphorus pentachloride is put in a 2-L flask and analysis shows that after 2.4 min 0.036 mol of chlorine has formed. What is the corresponding consumption rate of phosphorus pentachloride in grams per litre-second (g/(Ls))?

20. Inhaled oxygen (O2) allows human beings to produce energy through the oxidation of food substances, such as glucose (C6H12O6), according to the following equation: C6H12O6 (aq)  6 O2 (g) → 6 CO2 (g)  6 H2O (aq) At rest, human consumption of oxygen is assessed at about 3.5 (mL/kg)/min at SATP. Find the consumption rate of glucose in grams per hour (g/h) for a resting man whose mass is 75 kg. (Assume that all the oxygen consumed is used to produce energy through the oxidation of glucose).

21. Chemists are conducting an experiment to find the decomposition rate of nitrogen pentoxide (N2O5): 2 N2O5 (g) → 4 NO2 (g)  O2 (g) They collect data at a constant temperature presented in the following table: Time (s)

[O2] (mol/L)

0.00

0.0

6.00  102

2.1  103

1.20  103

3.6  103

1.80  103

4.8  103

2.40  103

5.6  103

3.00  103

6.4  103

3.60  103

6.7  103

4.20  103

7.1  103

4.80  103

7.5  103

5.40  103

7.7  103

6.00  103

7.8  103

a) Using graph paper, graph the production of oxygen (O2) as a function of time. b) Find the average reaction rate during the first 4800 seconds. c) Find the instantaneous reaction rates at 1200 s and 4800 s. d) Suggest a reason for the difference between the instantaneous reaction rates at 1200 s and at 4800 s. e) For a particular set of data, two students determined different average reaction rates. If neither of the students made a calculation error, explain the difference between the reaction rates. f) When comparing the reaction rates of reactions that occurred under different conditions, measurements are often taken at the beginning of the reactions. What is (are) the advantage(s) of this practice? (Hint: Think of slow reactions.)

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233

22. The decomposition of gaseous nitrogen pentoxide (N2O5) produces nitrogen dioxide (NO2) and oxygen (O2). Measurements taken during a laboratory experiment show the change in concentration of each of the substances in moles per litre (mol/L) as a function of time. This table shows the results of the measurements. Time (min)

NO2

O2

0

1.50  102

0.00

0.00

10

1.10  102

8.00  103

2.00  103

20

8.25  103

1.35  102

3.38  103

30

6.10  103

1.78  102

4.45  103

a) Write the balanced equation for this chemical change. b) Calculate the reaction rate for each of these substances between the 10th and 30th minutes. c) Why are these reaction rates, calculated for the same reaction under the same conditions, different?

23. The combustion of octane (C8H18) in a vehicle’s engine produces carbon dioxide (CO2) and water (H2O) vapour. a) What is the balanced equation for this chemical change? b) If the combustion rate of the octane is 6.80 mol/s, what is the formation rate of carbon dioxide in moles per second (mol/s)? c) What is the general reaction rate of this reaction? d) What are the mass, in grams (g), and the volume, in litres (L), of the carbon dioxide produced by a vehicle running for 15 min at 25°C and 101.2 kPa?

234

Suggest a simple device to monitor the decomposition rate of n mol of gaseous nitrogen pentoxide (N2O5) into nitrogen dioxide (NO2) and oxygen (O2) in a fixed volume at constant temperature.

UNIT 3 Reaction Rate

Two substances react to form a third, according to the following equation: AB→C Measurements of the concentration of substance C have been compiled in the following table: a) Graph the curve for the change in the concentration of substance A, if its initial concentration was 20.00 mol/L. b) Calculate the average reaction rate of this reaction for B. c) Calculate the instantaneous reaction rate at the halfway point of the reaction in the graph for substance A.

Concentrations (mol/L) N2O5

 24.

 25.

Time (s)

[C] (mol/L)

Time (s)

[C] (mol/L)

Time (s)

[C] (mol/L)

0

0.00

21

12.54

42

15.41

1

1.48

22

12.75

43

15.50

2

2.76

23

12.96

44

15.56

3

3.88

24

13.15

45

15.65

4

4.85

25

13.33

46

15.73

5

5.71

26

13.51

47

15.80

6

6.49

27

13.67

48

15.87

7

7.18

28

13.83

49

15.93

8

7.80

29

13.98

50

16.00

9

8.37

30

14.12

51

16.06

10

8.89

31

14.25

52

16.12

11

9.36

32

14.38

53

16.18

12

9.80

33

14.51

54

16.24

13

10.20

34

14.62

55

16.29

14

10.57

35

14.74

56

16.35

15

10.91

36

14.85

57

16.40

16

11.23

37

15.95

58

16.45

17

11.53

38

15.05

59

16.50

18

11.80

39

15.15

60

16.55

19

12.06

40

15.24

–

–

20

12.31

41

15.33

–

–

Collision Theory F

ootball is a sport of many collisions. Sometimes, these collisions lead to fumbles and the ball changes hands. Similarly, the changes that occur during chemical reactions result from collisions occurring between substances.

You will also discover that using collision theory to explain the reaction mechanism provides insight into the interactions between reactant particles, and the energy at every step of a reaction’s evolution.

In this chapter, you will learn about collision theory, which explains the reaction rates at the particle level.

Review Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 The law of conservation of energy . . . . . . . . . . . . . . . . . . . . . . . . . . 31

9.1 9.2

Types of collisions . . . . . . . . . . . . . . . . . . . . . . 236 Reaction mechanism explained by collision theory . . . . . . . . . . . 239

CHAPTER 9 Collision Theory

235

9.1 Types of collisions In an elastic collision, reactant particles hit each other and there is no chemical reaction. In an inelastic collision, reactant particles hit each other and set off a chemical reaction that transforms them into product particles. This strange dark stain in the sky is made up of tens of thousands of birds returning to Scotland to nest (see Figure 1). The birds creating this shape-shifting mass are European starlings. They fly astonishingly close to each other and then abruptly change direction, calling all the while to scare predators. Yet despite the fact that such a large number of birds are beating their wings frenetically in a relatively small space, few of the birds collide. Their excellent spatial orientation and rapid reaction time let them fly in all directions at speeds of over 30 km/h without ever touching each other. In the case of chemical reactions, the exact opposite must occur for reactants to transform into products. According to collision theory, particles must collide with each other and comply with certain conditions; if not, a reaction is not possible. Figure 1 Unlike these massed European starlings that almost never collide, reactant particles must collide to produce a chemical reaction.

APPENDIX 6 Scientific notation, p. 406.

Chemists developed collision theory to explain various macroscopic observations associated with the reaction rate. According to collision theory, for a chemical reaction to be produced, two reactant particles (atoms, molecules or ions) must collide with each other with a certain amount of energy and effect. The theory also stipulates that the reaction rate depends on the number of inelastic collisions between particles in a given period of time. Yet the relationship between the number of collisions and the reaction rate is more complex than it seems. Every collision does not automatically lead to a chemical reaction between the colliding particles. For example, in a 1-mL sample of gas at SATP, roughly 10 28 collisions occur every second. If every collision produced a reaction, the rate of all reactions between gaseous phase particles would be explosive. Yet some reactions between gases occur slowly. This is why some collisions are called ineffective, which have the effect of reducing the reaction rate. These ineffective collisions are called elastic collisions. A collision is elastic when the sum of the kinetic energies of the particles hitting each other remains unchanged. Following an elastic collision, the particles bounce and take off again with the same level of kinetic energy they had before impact. For a collision between reactant particles to be inelastic, it must meet the following two criteria: Inelastic collisions • Reactant particles must have the correct orientation. • Reactant particles must have a collision energy equal or superior to the reaction’s activation energy.

236

UNIT 3 Reaction Rate

9.1.1

Orientation of reactants

For a reaction to occur, it is essential that the reactant particles collide. However, the angle at which they hit each other plays an important role in the collision’s result. The colliding reactant particles must have just the right orientation to each other for a molecular rearrangement to lead to product formation. When the orientation is correct, it is called the appropriate collision geometry. For example, when nitrogen monoxide (NO) collides with nitrogen trioxide (NO3), nitrogen dioxide (NO2) can be formed if the collisions are inelastic, as shown by the following equation: NO (g)  NO3 (g) → 2 NO2 (g) There are several ways in which nitrogen monoxide and nitrogen trioxide can collide (see Figure 2). Only one of these, however, has the appropriate collision geometry for a reaction to occur. In the time before the collision, only a specific orientation of the nitrogen atom (N) in nitrogen dioxide and of one of the oxygen atoms (O) in nitrogen trioxide leads to the formation of two molecules of nitrogen dioxide because the collision is inelastic. Any other orientation produces elastic collisions.

Elastic

Elastic

Consequently, the relatively low number of properly oriented collisions between particles of nitrogen monoxide and particles of nitrogen trioxide leads to a reaction rate that, in this case, is slower.

9.1.2

Elastic

Activation energy

As well as hitting each other according to the appropriate collision geometry, the particles must collide with a minimum of collision energy to produce a reaction. To break the bonds in the reactants and begin forming bonds in the products, the reactants must collide with a certain amount of kinetic energy. In most reactions, only a small fraction of collisions occur with enough energy to produce a reaction. The activation energy (Ea) of a reaction is the minimum collision energy required for the reaction to occur.

Elastic

Inelastic

Figure 2 Only a specific orientation of the reactant particles produces an inelastic collision. All other orientations produce elastic collisions.

Collision energy depends on the kinetic energy of the particles hitting each other. Temperature is a measure of the average kinetic energy of a substance’s particles.

See Activation energy, p. 173. See Kinetic theory of gases, p. 54.

CHAPTER 9 Collision Theory

237

The Maxwell-Boltzmann distribution curve can be compared to a bicycle race, such as the Tour de France. Most racers are concentrated in the bunch (the central part of the curve) while a smaller group trails behind (the left side of the graph) and another group breaks off in the lead. It is this last group which has the energy to win the race. The risk of collision is greater within the bunch and, like two particles involved in a chemical reaction, the energy released when two cyclists collide de pends on their orientation and respective kinetic energies. Like the cyclists in the leading bunch who go faster than the rest, the particles with sufficient energy to participate in the chemical reaction are generally those that are the fastest.

Maxwell-Boltzmann distribution curve

Number of particles

Maxwell-Boltzmann distribution curve

The Maxwell-Boltzmann distribution curve can be used to show the distribution of the number of collisions between a substance’s particles at a given temperature as a function of each collision’s kinetic energy (see Figure 3). The dotted line on the graph shows the minimum level of activation energy the particles must have for there to be any reaction. The shaded part shows the number of particles with energy equal or superior to the activation energy.

Ea

Energy

Figure 3 The area below the Maxwell-Boltzmann distribution curve shows the distribution of particles’ kinetic energy at a constant temperature. The larger the hatched area, the faster the reaction is.

More generally, the rate of a reaction is dictated by the value of the activation energy. The higher the activation energy, the slower the reaction is because few molecules have the minimum energy to match the high level of activation energy required. Conversely, reactions with a low activation energy are faster reactions. It is the number of inelastic collisions per unit of time that determines the reaction rate. To accelerate a reaction, the number of inelastic collisions must be increased by changing the parameters affecting the orientation of reactants or activation energy.

Figure 5 Cyclists in the leading bunch in a race can be compared to particles with sufficient energy to participate in a chemical reaction.

For example, with a higher concentration of reactants, the total number of collisions increases because there are more particles per unit of volume. Because there are statistically more collisions whose orientation is likely to be correct, the reaction rate increases (see Figure 4). Similarly, if the particles’ kinetic energy level is higher (because of an increase in temperature, for instance), the reaction rate increases.

See Factors that affect reaction rate, p. 245.

238

UNIT 3 Reaction Rate

Figure 4 Metallic zinc (Zn) reacts faster in concentrated hydrochloric acid (HCl) (left) than in diluted hydrochloric acid (right).

Finally, the reaction rate increases if the reaction’s level of activation energy is reduced. In this case, a larger number of molecules ends up in the hatched part of the graph, and the number of inelastic collisions per second increases significantly. Any variation of one or more of the following factors, namely the nature of the reactants, their concentration, the presence of a contact surface and the addition of a catalyst, has the effect of changing the reaction rate.

9.2 Reaction mechanism explained by collision theory Collision theory explains the interactions between reactant particles and the energy present at each stage in the evolution of a reaction as illustrated by the reaction mechanism. When two cars collide, the kinetic energy (Ek) of the vehicles transforms into potential energy (Ep), which is used to perform the work that causes the cars to crash (see Figure 6).

Ep Maximum

The scale of the damage depends on the speed of the vehicles and therefore Ek Minimum on the total kinetic energy of the two cars. The sequence of energy changes occurring throughout the collision shows that as kinetic energy is transformed into potential energy, the potential energy increases proportionally to the decline of kinetic energy, Figure 6 Collisions between cars are not elastic collisions. according to the law of conservation of energy. At a fraction of a second when the two vehicles are immobile, the level of kinetic energy is nearly null and that of potential energy is nearly at its peak. See The law of conservation In a chemical reaction, the particles’ kinetic energy is also transformed into potential energy when the particles collide. The increase in potential energy is proportional to the decrease in the kinetic energy of the particles hitting each other. The reaction sequence can be represented graphically by an energy diagram. This diagram is a means of visualizing the energy variations occurring when reactant particles collide.

9.2.1

of energy, p. 131.

See Energy diagram, p. 174.

Simple reaction using collision theory

Collision theory provides a way of visualizing the changes occurring at a molecular scale throughout a chemical reaction. For example, the reaction between methyl bromide (CH3Br) and a hydroxide ion (OH) is expressed by the following equation: CH3Br  OH → CH3OH  Br The reaction can be illustrated by snapshots of different moments in the reaction (see Figure 7 on the following page).

CHAPTER 9 Collision Theory

239

Before the activated complex

Reactants

Activated complex

After the activated complex

Products

Energy

Ea

CH3Br  OH

Hreaction CH3OH  Br Progress of the reaction

Figure 7 As reactants collide, chemical bonds, shown here as sticks, break and form.

For a reaction to occur, the reactants’ orientation must be optimal and the collision energy at least equal or superior to the activation energy. When a collision is inelastic, the reaction follows its course and the activated complex breaks up to form product particles. Thus, when kinetic energy is sufficient, and the hydroxide ion (OH) is approaching the methyl bromide (CH3Br) from the side opposite to the bromine atom (Br), a reaction is possible. A partial bond is formed between the oxygen atom (O) in the hydroxide ion and the carbon atom (C) in the methyl bromide. Simultaneously, the bond C – Br weakens and the activated complex takes shape. When a collision is elastic, the activated complex decomposes to form reactants once more, and their kinetic energy remains unchanged.

9.2.2

Figure 8 Like most chemical reactions, assembling a car on a production line requires more than one step. See Reaction mechanism, p. 198.

240

UNIT 3 Reaction Rate

Complex reaction using collision theory

Most chemical reactions do not pass directly from reactant to product. In more complex reactions, which bring several reactant particles into play at a time, it is unlikely that all of the inelastic collisions will occur simultaneously. Similar to a production line in a car factory, where the manufacturing process is divided into several operations, complex chemical reactions occur in a series of intermediate steps, which constitute the reaction mechanism (see Figure 8). Each step is an elementary reaction that sets a single event at the molecular level into motion. At the end of a step, the products formed become the reactants of the next step, and so on, until the final product is formed.

Adding all of the elementary reactions together gives the overall reaction (see Figure 9). Complex chemical reaction

Ea

Potential energy (kJ/mol)

Ea

Ea

3

2

Ea

1

1st elementary reaction

2nd elementary reaction

3rd elementary reaction

4

4th elementary reaction

Progress of the reaction

Figure 9 An overall reaction can consist of several intermediate steps in which reactants transform into products at rates that depend on the activation energy value (Ea).

At each step in a complex reaction, particles form between the reactants at the beginning, and the products at the end, of the overall reaction. These particles are called reaction intermediates. A reaction intermediate is unstable and has a very short life span. However, when these particles react, they are subject to the same criteria for orientation and collision energy as those that apply to simple reactions. Consequently, each step has an activation energy all its own that leads to the formation of an activated complex when the energy barrier is crossed. The energy level of an activated complex determines the rate of each step in the overall reaction. The step in which the energy level of the activated complex is highest is the slowest step. This step is called the rate-determining step of the overall reaction. Like an automobile production line, whose speed is dictated by its longest operation, an overall reaction cannot occur at a rate faster than its slowest step. In Figure 9, step 3 determines which energy level of the activated complex is highest. The rate of the third elementary reaction therefore dictates the rate of the overall reaction.

Cooking or reacting? The secrets of cooking are not the prerogative of chefs alone; they can be scientifically explained by a set of complex chemical reactions. Throughout the cooking process, a number of essential factors regulate these chemical reactions, particularly time and duration. The two most frequent phenomena are the Maillard reaction and caramelization. In the first case, proteins, amino acids and sugars (present in meat, for instance) are heated at the same time. Food properties such as colour, aroma and nutritional value are therefore altered. Caramelization is a simple feat involving sugars and water. When sugar is heated to a temperature above its boiling point, the molecules decompose into simple sugars and then recompose into complex sugars of an entirely different kind, causing a browning reaction and altering the flavour. This recomposition creates caramel.

Figure 10 Caramel is the result of a chemical reaction.

CHAPTER 9 Collision Theory

241

CHAPTER

9

Collision Theory

9.1 Types of collisions • Collision theory stipulates that for a reaction to occur the reactant particles must hit each other in an effective manner.

Elastic

• An elastic collision occurs when reactant particles hit each other and there is no chemical reaction. The particles bounce off again with the same level of kinetic energy.

Elastic Elastic

• There is an inelastic collision when reactant particles hit each other and set off a reaction that transforms them into product particles.

Elastic

• To be inelastic, a collision must meet two criteria: – Reactant particles must have the correct orientation. – Reactant particles must have a collision energy equal or superior to the reaction’s activation energy.

Inelastic

• The reaction rate is determined by the number of inelastic collisions per unit of time.

9.2 Reaction mechanism explained by collision theory • When a collision between particles is inelastic, the activated complex forms and then breaks up, forming new product particles. • In a complex reaction, there are several steps known as elementary reactions. • In a complex reaction, the rate-determining step of the reaction mechanism is the one in which the energy level of the activated complex is highest. It dictates the rate of the overall reaction. Complex chemical reaction

Ea

Potential energy (kJ/mol)

Ea

Ea

3

2

Ea

1

1st elementary reaction

2nd elementary reaction

3rd elementary reaction

Progress of the reaction

242

UNIT 3 Reaction Rate

4

4th elementary reaction

CHAPTER 9

Collision Theory

2. A student council is preparing its program of school activities. It plans on offering eight activities to the student body. Three students must: – Put the eight pages describing the activities in order. – Present them neatly. – Staple them together. a) What is the rate-determining step of the operation? b) Suggest scenarios that might improve the speed of the operation.

4. For each of the reaction mechanisms represented in the graphs below, indicate: 1) The number of steps it comprises. 2) The activation energy (Ea) values for each step of the reaction mechanism. 3) The rate-determining step (justify your answer with numerical values). a) Potential energy (kJ/mol)

1. Explain the difference between an overall reaction and an elementary reaction. Indicate in your explanation what determines the rate of the overall reaction.

3. Below is the energy diagram of a chemical reaction.

600 500 400 300 200 100 0 100 200 300

Progress of the reaction b)

b) Potential energy (kJ/mol)

d)

e)

a)

Progress of the reaction

For each part indicated on the graph, state which corresponds to the following descriptions: 1) the number of collisions per second 2) the number of inelastic collisions per second 3) the activated complex 4) the enthalpy of reactants 5) the variation in enthalpy of the reaction 6) the number of elastic collisions per second 7) the enthalpy of the products 8) the activation energy of the reaction

Progress of the reaction

c)

100 Potential energy (kJ/mol)

Energy

c)

80 70 60 50 40 30 20 10 0 10

80 60 40 20 0 20 Progress of the reaction

CHAPTER 9 Collision Theory

243

6. A reaction occurs in three steps: Fast Slow Fast x →y y →z z → Final product a) Indicate the rate-determining step and explain the reasons for your choice. b) If you examined the flask’s contents in midreaction, what relative quantity of each substance would you find? 7. Draw a potential energy diagram as a function of the progression of the following reaction: – The reaction mechanism takes place over three elementary reactions. – The first elementary reaction is endothermic. – The second elementary reaction is exothermic. – The overall reaction is endothermic. In your diagram, identify the activation energies for each of the primary reactions. 8. Does an increase in temperature have an effect on activation energy? Explain your answer.

244

UNIT 3 Reaction Rate

9. Study the following graph and answer the questions below: Ea

Potential energy (kJ/mol)

5. Hydrogen gas (H2) and iodine gas (I2) are pumped into a closed flask at 25°C. What effect(s) will each of the following three changes have on the reaction rate? Explain your answer with arguments based on collision theory. a) The number of hydrogen gas (H2) particles is doubled. b) The pressure is doubled at a constant temperature. c) The temperature is decreased at a constant volume.

Ea Ea

3

2

1

1st elementary reaction

2nd elementary reaction

3rd elementary reaction

Progress of the reaction

Identify the elementary endothermic reaction(s). Identify the elementary exothermic reaction(s). Identify the rate-determining step. Is the overall reaction endothermic or exothermic? e) If you add a substance and the rate remains constant, on what primary reaction will the substance have an effect? f) What would you do to increase the reaction rate? a) b) c) d)

10. You know that methane (CH4) reacts with oxygen (O2) and produces a great deal of energy; this is an exothermic reaction. How can you explain the fact that methane and oxygen can be mixed and kept indefinitely without any apparent change?

Factors that Affect Reaction Rates

C

hemical reactions occur at extremely varied rates. The formation of stalactites through the precipitation of calcium carbonate (CaCO3) contained in runoff water takes thousands of years. This natural phenomenon is an example of a slow chemical reaction. Other reactions occur so quickly they appear instantly. This chapter explores and explains the various factors that affect reaction rates in light of collision theory.

Review Electrolytes and electrolysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Covalent bonding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Knowing the factors that affect chemical reaction rates allows everyone from chefs to chemical engineers to maximize the production of their goods.

10.1 10.2 10.3

Nature of reactants . . . . . . . . . . . . . . . . . . . 246

10.4

Temperature of the reaction environment . . . . . . . . . . . . . . . . . 260

10.5

Catalysts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262

Surface area of reactants . . . . . . . . . . . . . 251 Concentration of reactants and the rate law . . . . . . . . . . . . . . . . . . . . . . 253

CHAPTER 10 Factors that Affect Reaction Rates

245

10.1 Nature of reactants The nature of reactants, that is, the phase of the reactants, along with the number and strength of the bonds they contain, affects the reaction rate. See Collision Theory, p. 235.

According to collision theory, a chemical reaction is the result of the effective collisions that occur between the particles of a reactant. For the collisions to be effective, the reactant particles must have the correct orientation and the collisions must occur with the minimum energy required to break the bonds that join the atoms of the reactant. The greater the number of effective collisions per second, the faster the reaction is. Reactions that involve reactants that are all in the same phase (liquid, gas or aqueous solution) are called homogeneous. Reactions that occur, at least in part, at the interface of two phases (e.g. the surface of a chain that rusts in contact with air or water) are called heterogeneous reactions (see Figure 1).

Figure 1 Upon contact with oxygen (O2), iron (Fe) rusts and forms a layer of iron trioxide (Fe2O3).

In general, homogeneous reactions are faster than heterogeneous reactions. However, two homogeneous reactions that involve reactants composed of different particles do not necessarily occur at the same rate. This is the case, for example, with the reaction of nitrogen monoxide (NO) and carbon monoxide (CO) with oxygen (O2) in the air, which occurs entirely in the gas phase under SATP conditions (see Figure 2).

Fast reaction at SATP

a) 2 NO (g)



O2 (g)

2 NO2 (g)

Oxygen (O) Slow reaction at SATP b) 2 CO (g)



O2 (g)

Nitrogen (N) Carbon (C)

2 CO2 (g)

Figure 2 The difference in rate of these two reactions depends on the bonds of the atoms that form the molecules of the reactants.

Describes interactions * Intramolecular that occur between the atoms that compose a molecule.

246

UNIT 3 Reaction Rate

While the equations in Figure 2 are identical in form and both involve substances in the gas phase, they occur at different rates. This demonstrates that aside from the phase of the reactants, other factors can explain the variability of the rates of homogeneous reactions. The intramolecular properties of reactants are another factor that affect reaction rates.

*

Factors that affect reaction rates Reaction rates depend primarily on two factors related to the nature of the reactants: – phase of the reactants – number and strength of the bonds of the reactants to be broken

10.1.1

Phase of the reactants

According to the particle model of matter, the particles that compose matter are attracted to each other according to various forces of attraction. The closer the particles are to each other, the greater the forces of attraction. In a solid, particles are very close to each other and the bonds between particles are stronger than within a liquid or gas (see Figure 3).

a) In the particle arrangement of a solid, particles vibrate without changing position.

b) In the particle arrangement of a liquid, particles move slightly.

c) In the particle

arrangement of a gas, particles move quickly and in all directions.

Figure 3 Representation of the phases of matter according to the particle model

At equal temperature, particles move much more quickly in a gas than in a liquid (where movement is limited) or a solid (where movement is negligable). Consequently, reactions involving solid substances are slower than those involving substances in the liquid or gas state. In fact, for a reaction to occur with solids, the collision energy must be greater than that required for liquids or gases due to the greater number of bonds to be broken in solids. Moreover, since gas particles move very quickly and collide frequently, their very large number of collisions facilitates the formation of the activated complex. This explains why gas particles react very quickly. In reactions that occur in solutions containing aqueous ions with opposite charges, the reaction occurs even faster than for gases. For example, when sodium chloride (NaCl) and silver nitrate (AgNO3) come into contact, the reaction occurs instantly and a white precipitate of silver chloride (AgCl) appears (see Figure 4).

Figure 4 The white precipitate formed is silver chloride (AgCl).

CHAPTER 10 Factors that Affect Reaction Rates

247

The rate of this reaction is explained by the fact that the ions are uniformly dispersed in the solution and are surrounded by water molecules to which they are weakly bonded (see Figure 5).

Sodium (Na) Chlorine (Cl) Oxygen (O) Hydrogen (H) NaCl (s)

→

Na (aq)



Cl (aq)

Figure 5 During the electrolytic dissociation of sodium chloride (NaCl) into positive ions (Na) and negative ions (Cl) in water (H2O), the positive ions are surrounded by the negative ends of the polar water molecules while the negative ions are surrounded by the positive ends of the polar water molecules.

Consequently, following the electrolytic dissociation of sodium chloride (NaCl), the aqueous sodium ions (Na) and aqueous chloride ions (Cl) are free to circulate in the solution. They can collide with oppositely charged ions resulting from the electrolytic dissociation of silver nitrate (AgNO3) to form a precipitate. When this collision occurs, the reaction happens almost instan taneously, since the force of attraction between the ions that are colliding is greater than the bond strength that joins them to the water molecules (H2O). In summary, reaction rates can be classified from slowest to fastest as a function of the phase of the reactants, as follows: solid → liquid → gas → aqueous ions slow reaction fast reaction

10.1.2

See Enthalpy and enthalpy change, p. 148.

Number and strength of the bonds of the reactants to be broken

The activation energy (Ea) in a chemical reaction represents the energy barrier that must be overcome for a reaction to occur. The internal energy contained in the particles of the reactant, also called enthalpy, corresponds to the sum of all of the forces or movements, in one form or another, present in the reactants. These include intramolecular bond forces, which join the atoms to one another within the molecules of the reactant. The higher the energy of these intramolecular bond forces, the harder it is to break the molecules so that they can form the activated complex. A chemical reaction in which the reactant molecules contain a high level of bond energy, due to a large number of bonds or the strength of these bonds, is therefore slower than a reaction in which the reactants have a weak level of intramolecular bonds.

248

UNIT 3 Reaction Rate

This is the case, for example, with oleic acid molecules (C18H34O2), a component of olive oil, and stearic acid molecules (C18H36O2), a component of animal fat (see Figure 6). Their structures are very similar, but they react at different rates when they are put in contact with hydrogen peroxide (H2O2). The difference between the chemical formulas for oleic acid and stearic acid is summarized by two additional atoms of hydrogen (H) and a double bond between two atoms of carbon (C  C) in oleic acid (see Figure 7). Oleic acid reacts more slowly with hydrogen peroxide because it possesses a double bond between two atoms of carbon (C  C). Consequently more energy is required, namely 620 kJ/mol, to break this bond than to break the single bond between two atoms of carbon (C  C), which requires only 347 kJ/mol. Consequently, in the reaction involving oleic acid, the number of reactant particles that possess a sufficient level of activation energy (Ea) to produce an effective collision is less than the number of reactant particles in the reaction of stearic acid, which slows the reaction rate.

Figure 6 Oleic acid (C18H34O2) is a component of olive oil while stearic acid (C18H36O2) is contained in animal fat.

Reactions in organic chemistry In the lab, chemists often carry out reactions involving molecules with a large number of atoms, but only a small section of the molecule is involved in the reaction in an effort to modify it. Consequently, during these reactions, certain bonds remain intact while others are broken. This occurs naturally during the hydrolysis reaction of ethyl acetate (CH 3COOCH2CH3), a compound that is present in young wines and is responsible for their fruity taste. With time, the ethyl acetate reacts with water to form acetic acid (CH3COOH), a component of vinegar, and ethanol (CH3CH2OH), an alcohol. Consequently, wine loses its fruity taste over the years to develop a little more acidity and a greater alcohol content. During the reaction, the molecule is fragmented into two parts. Several C-H bonds as well as C-O, C=O and C-C bonds are not affected, while others are broken. The atoms of the water molecule combine with the fragments of the reactant molecules and several new bonds are formed.

O H H H B G DH D H C C H O G D D G D  C O C DA AG H H H H ethyl acetate

water

O B H C H G D D C O DA H H acetic acid



a) An oleic acid molecule

H H G D H C D G DH O C A GH H ethanol

Figure 8 The hydrolysis of ethyl acetate: certain bonds (in blue, green and red) are not broken during the reaction, while others are in order to form new bonds. b) A stearic acid molecule

Figure 7 Oleic acid (C18H34O2) and stearic acid (C18H36O2) have similar molecular structures. Oleic acid reacts more slowly with hydrogen peroxide (H2O2) than stearic acid due to its double bond between two atoms of carbon (C=C).

CHAPTER 10 Factors that Affect Reaction Rates

249

SECTION 10.1

Nature of reactants

1. Examine the following four reactions, then answer the following questions: 1) C2H5OH (g)  3 O2 (g) → 2 CO2 (g)  3 H2O (g)

a) C3H8 (g)  5 O2 (g) → 3 CO2 (g)  4 H2O (g)

2) H2SO4 (aq)  2 NaOH (aq) → 2 H2O (g)  Na2SO4 (aq)

b) 2 H2 (g)  O2 (g) → 2 H2O (I)

3) CaCO3 (s)  2 HCl (aq) → CaCl2 (aq)  CO2 (g)  H2O (g)

c) Li (aq)  OH (aq) → LiOH (s)

4) CH4 (g)  2 O2 (g) → CO2 (g)  2 H2O (g) a) Which reaction has the most atoms to be separated in the reactant’s molecules? b) Will this reaction be slower or faster than the others? c) Put the four equations in order of increasing reaction rates.

a)

b)

Number of molecules

Number of molecules

2. The graphs below represent, in random order, the distribution curve of the kinetic energies of the molecules of the reactants for each of the reactions in question 1. Match each graph with the reaction that best corresponds to it.

c)

d) Number of molecules

Kinetic energy

Number of molecules

Kinetic energy

Kinetic energy

250

3. Which of the three reactions below is the fastest? Explain your answer.

UNIT 3 Reaction Rate

4. Put the rates of the following four reactions in decreasing order, explaining your reasoning based on the nature of the reactants: a) Ca2 (aq)  SO42 (aq) → CaSO4 (s) 25 → 8 CO2 (g)  9 H2O (g) b) C8H18 (l)  O 2 2 (g) c) C3H8 (g)  5 O2 (g) → 3 CO2 (g)  4 H2O (g) d) 2 HI (g) → H2 (g)  I2 (g) 5. Examine the following two reactions. Using Table 8.5 in Appendix 8, answer the questions below: 1) H2 (g)  Cl2 (g) → 2 HCl (g) 1 2) H2 (g)  O2 (g) → H2O (g) 2 a) Calculate the enthalpy of the reactants in each of the two reactions. b) Indicate which of the two reactions is likely to occur most slowly given the energy level of the intramolecular bonds of the reactants. 6. Determine the reaction that will occur most quickly in the following pairs of reactions. Explain the reason for your choice. 73 → 24 CO2 (g)  25 H2O (g) a) 1) C24H50 (s)  O 3 2 (g) 73 → 24 CO2 (g)  25 H2O (g) 2) C24H50 (s)  O 3 2 (g) b) 1) C12H22 (l)  25 O2 (g) → 12 CO2 (g) + 26 H2O (g) 2) C12H26 (l)  25 O2 (g) → 12 CO2 (g) + 26 H2O (g)

Kinetic energy

7. At room temperature, vegetable oil is liquid. To convert it into solid matter, like margarine, it must be hydrogenated through a reaction with hydrogen (H). Why does olive oil oxidize faster than margarine manufactured from this same oil?

10.2 Surface area of reactants An increase in the surface area of the reactants generally increases the reaction rate. One of the factors that affect reaction rate is the surface area of the reactants in their solid state. For example, it is easier to light a campfire with wood split into small pieces than with whole logs. A greater surface area between the wood fragments and the oxygen (O2) in the air allows for faster combustion. Similarly, cutting vegetables into pieces creates a greater surface area in contact with heat so that they can be seared more quickly at a high temperature, thereby accelerating the chemical reaction that causes the characteristic browning of grilled food (see Figure 9). According to collision theory, chemical reactions will not occur without collisions between atoms, ions or molecules. In a heterogeneous reaction involving a solid and a gas, the particles can only collide with the exterior surface of the solid. If this surface is limited, the reaction rate is generally slower. When the solid is divided into finer particles (see Figure 13), the surface area is greater and the number of collisions increases, which makes the substance react more quickly (see Figure 10).

Figure 9 The increased surface area of the sliced vegetables makes it possible to cook them more quickly than vegetables that are kept whole.

m

1c

m

2 cm

1 cm

m 2c

m

1c

2c

a) Total surface area  24 cm2 Total volume  8 cm3

b) Total surface area  48 cm2 Total volume  8 cm3

Figure 10 When a cube measuring 24 cm2 is divided into eight cubes of equal dimensions, the surface area is doubled without increasing the volume of the substance.

Consequently, there is a higher risk of explosion when the dust of a combustible material, like sawdust in a sawmill or dust in a grain silo, combines with air and a spark is produced.

CHAPTER 10 Factors that Affect Reaction Rates

251

The very large surface area between the dust particles and the oxygen (O2) in the air allows for spontaneous combustion that sometimes results in an explosion (see Figure 11). Dissolving a solid in a solvent is the most effective method for increasing the surface area of a solid during a homogeneous reaction. The solid is reduced to its smallest element or compound particles (atoms, ions or separated mol ecules). If each of the reactants involved is already dissociated, the number of collisions becomes very high, which increases the reaction rate. This type of situation is only observed in a liquid or gas solution. In certain chemical reactions, it can be useful to stir the reaction environment to provoke greater contact between the reactants. This occurs in cheese production, for example, when rennet, a substance that provokes coagulation, is added at the curdling stage (see Figure 12). Stirring the reaction environ ment maximizes the contact between the rennet and the milk. This accelerates the precipitation of casein, a milk protein, into a granular mass that is then processed into cheese.

Figure 11 The port silo in Blaye, France, was the site of a dust explosion that destroyed it on August 20, 1997, killing 11 people.

Figure 12 Mixing milk after the addition of the rennet increases the contact between the substances that react during the production of cheese.

Explosion hexagon The explosion hexagon summarizes the six factors that must be taken into account in order to prevent the risk of explosion in industries that generate combustible dusts: oxidizer, fuel, con centration, confinement, energy source and dispersion of the fuel.

Oxidizer Example: oxygen (O2) in the air

Dispersion of the fuel

Fuel Examples: flammable dust or gas The explosion is the result of the presence of the six factors

Energy source

Concentration of the fuel Confinement

Figure 13 The six factors that must be present for combustible dust to explode

252

UNIT 3 Reaction Rate

10.3 Concentration of reactants and the rate law The concentration of the reactants has an effect on the reaction rate: in general, the higher the concentration of the reactants, the faster the reaction rate is. The rate law is a mathematical relationship that exists between the reaction rate and the concentration of the reactants. In the case of an elementary reaction, the reaction rate depends on the stoichiometric coefficients of the reactants in the balanced equation. One of the factors that affects reaction rate is the concentration of reactants. It is possible to quantitatively express this influence by a mathematical relationship called the rate law.

10.3.1

Concentration of reactants

The effect of concentration on the reaction rate can be studied through the hypothetical reaction between two gases, A and B, mixed in a 1-L volume maintained at a temperature of 25°C (see Figure 14).

Particle of gas A a)

Particle of gas B 1L

b)

1L

Effective collision

Figure 14 With higher concentrations of reactants A and B, there is a greater number of collisions per second, which, statistically, translates into a greater number of effective collisions.

At this temperature, the particles of gas A move at approximately 1500 km/h. If 0.1 mol of A and 0.1 mol of B are used, one molecule of gas A collides with one molecule of gas B approximately 109 times per second. Despite the greater number of collisions, only a few are effective and produce an activated complex, resulting in few particles of AB. Doubling the concentration of gas A will increase the number of collisions likewise, and if the concentration of gas B is also doubled, the number of collisions will quadruple. If there are four times more collisions, then, statistically, the number of effective collisions will also increase fourfold, thereby quadrupling the reaction rate. Most, but not all, reactions follow this rule. Data collected in the laboratory show that sometimes the rule does not apply. However, it is possible to affirm that, in general, an increase in the concentration of a reactant has the effect of increasing the reaction rate.

CHAPTER 10 Factors that Affect Reaction Rates

253

10.3.2

Rate law

The reaction between an aqueous solution of bromine (Br2) and an aqueous solution of methanoic (“formic”) acid (CH2O2) translates into the following equation. Br2 (aq)  CH2O2 (aq) → 2 Br (aq)  2 H (aq)  CO2 (g) Analyzing the results obtained during an experiment involving these reactants helps to better understand how the reaction rate evolves over the entire course of the reaction and to formulate the mathematical relationship of the rate law (see Table 1). Table 1 The change in concentration of aqueous bromine during its reaction with aqueous methanoic acid at 25°C Time (s)

[Br2] (mol/L)

0.0

0.0120

50.0

0.0101

100.0

0.0084

150.0

0.0071

200.0

0.0059

250.0

0.0050

300.0

0.0042

Mean rate ( 10 5 mol/(Ls)) ---------- 3.80 ---------- 3.40 ---------- 2.60 ---------- 2.40 ---------- 1.80 ---------- 1.60

For the first 50 seconds, the concentration of bromine is high and the mean rate is high, that is, 3.80  10 5 mol/(Ls). During the reaction, as the concentration of bromine decreases, the reaction rate also decreases until it nearly reaches zero when the bromine has completely reacted. The reaction rate also depends on the concentration of the methanoic acid, but its effect on the rate can become negligible if there is a surplus of acid, that is, if there are more moles of methanoic acid than of bromine. Studying the instantaneous rate at three points during the reaction also makes it possible to verify the slowing of the reaction as a function of the concentration of bromine (see Figure 15). The change in concentration of aqueous bromine as a function of time 0,012 0

Instantaneous mean at 100 s: 2.96  105 mol/(Ls)

[Br2] (mol/L)

0,010 0

Instantaneous mean at 200 s: 2.09  105 mol/(Ls)

0,008 0

Instantaneous mean at 300 s: 1.48  105 mol/(Ls)

0,006 0 0,004 0 0,002 0 0

50

100

150

200 250 Time (s)

300

350

400

Figure 15 The change in concentration of aqueous bromine (Br2) as a function of time. This explains why the instantaneous rate at the end of the reaction is weaker than at the beginning.

254

UNIT 3 Reaction Rate

The tangents drawn at 100, 200 and 300 seconds are straight lines that indicate that the reaction rate is directly proportional to the concentration: the greater the concentration of bromine (Br2), the faster the rate is. This relationship can be written in mathematical form using the proportionality symbol . The following expression means that the reaction rate is directly proportional to the concentration of the bromine:

APPENDIX 4 Slope of a tangent to the curve, p. 390.

r  [Br2] Mathematically, the proportionality symbol can be removed by introducing a proportionality constant (k). r  k [Br2] The constant k is a proportionality constant between the reaction rate and the concentration of the reactants, called the rate constant. There is a particular rate constant for each reaction, and it depends on the temperature at which the reaction occurs. To determine the value of the rate constant as a function of [Br2], it must be isolated in the equation below: k

r [Br2]

The value of the constant k is not affected by the concentration of the bromine. In fact, when the rate is high, the concentration is great, whereas when the rate is low, the concentration is lower, such that the value k remains constant if the temperature is stable (see Table 2). Table 2 The ratio k 

r for the reaction between aqueous bromine and aqueous methanoic acid at 25°C [Br2] r

k

Time (s)

[Br2] (mol/L)

Rate ( 105 mol/(Ls))

[Br2] ( 103 s1)

0.0

0.0120

4.20

3.50

50.0

0.0101

3.52

3.49

100.0

0.0084

2.96

3.50

150.0

0.0071

2.49

3.51

200.0

0.0059

2.09

3.51

250.0

0.0050

1.75

3.50

300.0

0.0042

1.48

3.52

CHAPTER 10 Factors that Affect Reaction Rates

255

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

The proportional relationship between the reaction rate and the concentration of the reactants is called the rate law. It can be expressed more generally, at a given temperature, as follows: Rate law r  k  [A]x  [B]y where r  k  [A], [B]  x, y 

PETER WAAGE Norwegian chemist (1833–1900) CATO MAXIMILIAN GULDBERG Norwegian chemist and mathematician (1836–1902) Peter Waage and Cato Maximilian Guldberg, united by ties of friendship and family, sealed their partnership with their discovery of the law of mass action, which made it possible to define the conditions of equilibrium of a chemical reaction system. This law stipulates that the rate of a chemical reaction is proportional to the concen trations of the reactants. First presented in 1864, no one took note of it for more than a decade. Other scientists later validated it, without knowledge of the earlier publications by these two chemists. Waage subsequently devoted himself to research with practical applications, inventing, among other things, new methods for preserving milk. Guldberg also pursued research, and, in 1890, discovered a law regarding the boiling points of liquids.

256

UNIT 3 Reaction Rate

Reaction rate, expressed in moles per litre-second (mol/(Ls)) Rate constant Concentrations of reactants, expressed in moles per litre (mol/L) Stiochiometric coefficients of the reactants in a balanced equation of the elementary reaction

The rate law applies only to elementary reactions. In the case of most other more complex reactions, there is no direct relationship between the stoichiometric coefficients of the balanced equation of the reaction and the rate law of the reaction. In this case, the values of exponents x and y must be determined experimentally. However, regardless of whether the reaction is elementary or complex, the rate law takes into account only the concentration of the reactants, not that of the products. Moreover, since the concentration of pure substances in their solid and liquid states do not vary, only the concentrations of reactants in the gas or aqueous state are taken into consideration in writing the algebraic expression of this law. For example, the reaction for the synthesis of ammonia (NH3), where all of the reactants are gaseous, is translated by the following reaction: N2 (g)  3 H2 (g) → 2 NH3 (g) The algebraic expression for the reaction rate of the reaction is as follows: r  k  [N2]  [H2]3 In the case where at least one of the reactants is in the solid or liquid state, as in the reaction between solid calcium carbonate (CaCO3) and aqueous hydro chloric acid (HCl), the concentration of the calcium carbonate is not considered because it is a solid. CaCO3 (s)  2 HCl (aq) → CaCl2 (aq)  CO2 (g)  H2O (g) Consequently, the algebraic expression for the reaction rate is as follows: r  k  [HCl]2

The following examples show how to use the rate law to compare the rates of elementary reactions when the concentrations of the reactants are changed. Example A At a given temperature, two aqueous ions A and B combine with a reaction rate (r1) as follows:

Furthering your understanding

r1  k  [A]2  [B]3

Exponents of the concentrations of reactants

How does the rate change if the concentration of the two reactants is doubled without changing the temperature? What is the new rate (r2) as a function of the initial rate? Data: [A]1  x [B]1  y r2  ?

1. Calculation of the initial reaction rate: r1  k [A]2  [B]3 r1  k  x 2  y 3 2. Calculation of the new reaction rate: [A]2  2x [B]2  2y r2  k  [A]2  [B]3  k  (2x)2  (2y)3  k  4x2  8y 3  k  32x2  y 3  32kx2y3 r1 r2  2 3 x y 32 x 2y 3

When a reaction is complex, the exponents that are given to the values of the concentrations of the reactants must be determined through experimentation. In this case, they do not correspond in an absolute manner to the stoichiometric coefficients of their reactants as in the case of elementary reactions. Their value is generally 1 or 2 but can also be 0. 3 or even a real number. The rate law is expressed as follows for complex reactions: r  k  [A]m  [B]n

32 x 2y 3 r1 x 2y 3  32 r1

r2 

Answer: The new reaction rate (r2), is 32 times greater than the initial rate (r1). Example B Two gases C and D combine with a reaction rate of 1.8 mol/(Ls) in a 1-L container maintained at a constant temperature according to the following equation:

The values of exponents m and n are determined by chan ging the initial concentration of each of the reactants, one after the other, while maintaining those of the other reactants constant.

2CD→ E If the concentration of C is 0.50 mol/L and that of D is 1.0 mol/L, what is the value of the rate constant at this temperature? Data:

Calculation:

r  1.8 mol/(Ls) [C]  0.50 mol/L [D]  1.0 mol/L k ?

r  k [C]2  [D] r k 2 [C]  [D] 1.8 mol/(Ls)  (0.50 mol/L)2  (1.0 mol/L)  7.2 L2/(mol2s)

Answer: The value of the rate constant is 7.2 L2/(mol2  s).

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Example C Consider the two gases in example B. What will happen to the reaction rate if the volume of the container in which these two gases were combined is decreased by half? Data:

Calculation:

r1  1.8 mol/(Ls) [C]1  0.50 mol/L [D]1  1.0 mol/L r2  ?

If the volume is decreased by half, the concentrations double: [C]2  2  0.50 mol/L  1.0 mol/L [D]2  2  1.0 mol/L  2.0 mol/L And then: r2  k  ([C]2)2  [D]2  7.2 L2/(mol2 s)  (1.0 mol/L)2  (2.0 mol/L)  14.4 mol/(Ls)

Answer: When the volume of the container is decreased by half, the new reaction rate (r2) is 14 mol/(Ls), that is 8r1.

SECTION 10.3

Concentration of reactants and the rate law

1. How does collision theory explain that higher concentrations lead to faster reaction rates? 2. The rate law indicates that the rate is proportional to the molar concentration of the reactants with, as exponent, the stoichiometric coefficients of the balanced equation. For each of the following reactions, express the mathematical relationship that makes it possible to calculate the rate: a) H2SO4 (aq)  Ca(OH)3 (aq)) → CaSO4 (s)  2 H2O (l) b) 5 I(aq)  6 H(aq)  3 IO3(aq) → 4 I2 (s)  3 H2O (l) c) CaCO3 (s)  2 HCl (aq) → CaCl2 (aq)  CO2 (g)  H2O (l) d) N2O4 (g)  H2O (g) → HNO2 (g)  HNO3 (g) 3. To dye hair, the hair pigment must first be decoloured using hydrogen peroxide (H2O2).The products used in hair salons often contain 6% hydrogen peroxide and generally require a waiting time of approximately 30 minutes. If a different product containing 9% hydrogen peroxide is used, how will this change the necessary waiting time? Explain why.

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4. The decomposition of hydrogen peroxide (H2O2) occurs according to the following reaction: 2 H2O2 (aq) → 2 H2O (l)  2 O2 (g) The rate constant of this reaction at 243°C is 5.32  107 L/(mols). If 22 g of hydrogen peroxide are put in a 500-mL container, what will the reaction rate be? 5. Nitrogen monoxide (NO) reacts in an aqueous solution with molecular bromine (Br2) according to the following reaction: 2 NO (aq)  Br2 (aq) → 2 NOBr (aq) If the initial reaction rate is 1.6 mol/(Ls), the concentration of nitrogen monoxide is 1 mol/L and the concentration of bromine is 2 mol/L. Calculate the rate constant. 6. When the initial concentrations of A and B are, respectively, 0.31 mol/L and 0.75 mol/L, the rate of formation of RS is 7.3  103 mol/(Ls). The equation is of the type A  B → RS. Calculate the rate constant k. 7. Two gases C and D combine according to the following reaction rate: r  k  [C]2  [D]3. How does the rate change if the volume of the system in which the reaction is occurring is decreased by half?

8. If k  3.14  102 s1 for an equation of the type A  B → C, calculate the reaction rate for the following two situations:

13. Using the following graph, explain how to change the reaction rate between time zero and 10 seconds.

b) [A]  3.2  103 mol/L and [B]  5.6  101 mol/L 9. Carbon (C) reacts with iron trioxide (Fe2O3) to form carbon dioxide (CO2) and iron (Fe) according to the following equation: C (s)  Fe2O3 (aq) → Fe (s) + CO2 (g) a) Balance the equation. b) What is the algebraic expression of the rate of this reaction? 10. Sulphur dioxide (SO2) gas forms according to the following equation: S8 (s)  8 O2 (g) → 8 SO2 (g) a) Express the rate law of this reaction. b) Eliminating all of the oxygen (O2) present in the container will take 12 minutes. If the initial concentration of the oxygen is 2.3 mol/L, what is the rate constant of this reaction? c) What is the reaction rate if the quantity of solid sulphur dioxide is doubled? d) What is the reaction rate if the concentration of oxygen is tripled? 11. A chemical reaction between reactants A and B forms product F according to the following reaction: 2 A (g)  4 B (g) → F (g) a) If the rate of this reaction is 5.25 → 107 mol/(Ls), the concentration of A is 3.0  104 mol/L and that of B is 5.2  102 mol/L, what is the rate constant of this reaction? b) If the pressure is doubled, what is the new reaction rate? 12. Nitrogen monoxide (NO) reacts on contact with bromine (Br2) as follows: 2 NO (g)  Br2 (g) → 2 NOBr (g) What is the effect on the reaction rate: a) if the pressure of the bromine is tripled by decreasing its volume? b) if the volume of the bromine is doubled?

Concentration (mol/L)

a) [A]  1.1 mol/L and [B]  0.83 mol/L

9 8 7 6 5 4 3 2 1 0 0

2

4

6 Time (s)

8

10

12

14. The combustion of ethane (C2H6) occurs according to the following equation: C2H6 (g)  O2 (g) → CO2 (g)  H2O (g) a) Balance the equation. b) If the combustion rate of the ethane is 123 mol/(Ls) and the volume in which the reaction occurs is doubled, what is the combustion rate? 15. Consider the following reaction: S2O82 (aq)  I (aq) → SO42 (aq)  I3 (aq) a) Balance the equation. b) What is the rate law of this reaction? c) If 0.45 mol/L of iodine (I) disappears each minute in the solution, and the concentration of the reactants at a given point in time is [S2O82]  0.20 mol/L and [I] 1.20 mol/L, what is the rate constant? 16. You are using aluminum trichloride (AlCl3) to form sodium chloride (NaCl) according to the following reaction: AICI3 (aq) 3 NaOH (aq) AI(OH)3 (s)  3 NaCl (aq) The reaction occurs at a rate of 0.050 mol/(Ls) as a function of the aluminum trichloride. a) If the initial concentration of the aluminum trichloride is 3.50 mol/L at 25°C, what is the concentration of aluminum trichloride and sodium chloride after 20 seconds? b) How would you increase the rate of the reaction? c) If you double the concentration of aluminum trichloride and you decrease by one third the concentration of the sodium hydroxide (NaOH), what is the new reaction rate?

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10.4 Temperature of the reaction environment An increase in the temperature of the reaction environment generally increases the reaction rate. The chemical reactions responsible for the decomposition of food can be slowed down or accelerated by a change in temperature. For example, putting bottles containing frozen juice (to serve as a coolant) in a lunch box is a trick for slowing down the action of bacteria and preserving the food so that it is still edible when the time comes for it to be consumed. Conversely, forgetting a lunch box in a car exposed to the Sun, with the windows closed, causes the temperature to rise, and if the food is left in these conditions for too long, it may no longer be edible. These examples demonstrate that temperature has an effect on reaction rate. See Reaction mechanism explained by collision theory, p. 239

The temperature of the reaction environment is another factor that affects the reaction rate, which can be explained by collision theory. Collision energy depends on the kinetic energy of particles that collide with each other. Maxwell-Boltzmann’s distribution makes it possible to represent the dis tribution of kinetic energy in a gas sample that reacts at two different tem peratures, T1 and T2, with T2  T1 (see Figure 16).

Number of particles

The kinetic energy of the particles of a gas sample at two temperatures

Greater number of particles at T2 with enough energy to react T1 T2 T1

Kinetic energy

Figure 16 At higher temperatures, a larger number of particles collide with each other with enough kinetic energy to react. In many reactions, the rate more or less doubles with each temperature increase of 10°C.

The activation energy of the reaction is indicated by the dotted vertical line. At temperature T1, few particles have more energy than the activation energy (hatched area in blue). Therefore, very few collisions have enough energy to produce an effective collision leading to a chemical reaction. The shape of the distribution curve changes for temperature T2. The curve flattens and extends,

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which indicates that the mean velocity of the particles, and consequently their kinetic energy, is greater at higher temperatures. As a result, there are a larger number of particles with kinetic energy that is above the threshold of activation energy at T2 than at T1 (hatched area in red). A larger number of particles are therefore able to produce effective collisions, and the reaction rate at T2 is much greater than at T1. This assertion does not only apply to the reaction environment of gases, but the reaction rate that occurs in an aqueous environment is also generally greater at higher temperatures. Chemists use the fact that reaction rates are dependent on temperature by causing chemical reactions to occur at high temperatures in order to accelerate them and maximize the industrial production of certain substances. Similarly, at home, food storage is based on the dependence of chemical reactions on temperature. Storing food in the refrigerator instead of at room temperature, for example, helps to keep it fresh and edible for a longer period of time. The cold air in the refrigerator does not stop food from spoiling; it merely slows down the rate of the reactions that cause food to spoil. Conversely, to cook food faster, the stove should be set to higher temperatures (see Figure 17). This increases the reaction rate that occurs while food is being cooked.

Figure 17 A pizza placed in the oven at a high temperature cooks faster than at a low temperature.

The thermometer cricket The hotter it gets, the more it sings. The stridulation (chirping) of the thermometer cricket, commonly found through out Canada, increases as the temperature rises (see Fi gure 17). The sounds it emits are, in fact, directly linked to its metabolism, which increases and decreases as a function of the outdoor temp era ture. Therefore, the temperature in degrees Celsius is roughly equal to the number of stridulations emitted in eight seconds, to which five se conds is added. A song of 120 stridulations per minute therefore corresponds to a temperature of 21°C, that is: (120  60  8)  5  21°C. These changes in metabolism, controlled by a set of internal enzyme reactions, are a characteristic of poikilothermal animals, commonly known as cold-blooded animals.

Figure 18 The thermometer cricket emits an increasing number of stridulations as the temperature rises.

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10.5 Catalysts Catalysts are substances that increase the reaction rate without changing the result of the conversion and without being consumed by the reaction. Without the use of a catalyst, many chemical reactions carried out in industries would not be profitable due to their slow reaction rate. Other reactions that occur in nature, such as the digestion of insects by certain carnivorous plants, would be impossible without the presence of natural catalysts called digestive enzymes (see Figure 19). Catalysis is a process that accelerates a reaction. It occurs due to the presence of a substance, a catalyst, that remains unchanged chemically and remains intact at the end of the process. Following catalysis, the products formed are identical to the ones that would have formed if the reaction had not been catalyzed. There are two types of catalysts: homogeneous catalysts and heterogeneous catalysts. While they have their own particular modes of action, they function according to the same principle. Figure 19 Certain carnivorous plants have adapted to poor soil by trapping and digesting insects using digestive enzymes contained in a digestive fluid at the bottom of their pitcher-shaped leaves.

10.5.1

How a catalyst functions

The role of a catalyst is to lower the activation energy of a reaction so that a greater number of reactant particles have enough energy to react (see Figure 20). The area below the curve, which represents the particles with sufficient energy to react, increases significantly when the activation energy barrier is lowered by the catalyst. This produces a larger number of effective collisions and, therefore, increases the reaction rate. The kinetic energy of reactant particles of a chemical reaction occurring with and without a catalyst Catalyzed Ea

Non-catalyzed Ea

Particles with sufficient energy to react in the non-catalyzed reaction

Number of particles

Particles with sufficient energy to react in the catalyzed reaction

Kinetic energy

Figure 20 A catalyst lowers the activation energy, which allows a greater number of particles to have sufficient kinetic energy to react.

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This change in the activation energy level does not, however, affect the energy level of the reactants and products, which means that the enthalpy change (H) of the reaction remains unchanged (see Figure 21).

See Enthalpy and enthalpy change, p. 148.

The energy diagram of a reaction occurring with and without a catalyst Non-catalyzed reaction

Potential energy

Non-catalyzed Ea

Catalyzed reaction

Catalyzed Ea Reactants

H

Products

Progress of the reaction

See Activated complex, activation

Figure 21 The use of a catalyst makes it possible to lower the activation energy without changing the level of energy of the reactants or products or the value of the enthalpy change (H).

energy and the energy diagram, p. 172.

By lowering the level of the activation energy of the direct reaction, the catalyst also accelerates the rate of the inverse reaction, when the latter is present. In fact, by reading the graph of the course of the reaction from right to left, it is possible to observe that the activation energy level of the inverse reaction is also lowered on the curve of the catalyzed reaction. Inhibitors are substances that, unlike catalysts, reduce the reaction rate. They act by increasing the activation energy of chemical reactions, thus reducing the number of particles with sufficient energy to react. Inhibitors are used primarily in the agri-food industry to slow down chemical reactions that cause food to spoil, or in the health sector, in the form of medications to slow down or stop certain biological processes (see Figure 22).

Figure 22 Among the treatments to fight HIV infection, zidovudine (AZT) is a molecule that inhibits the action of an enzyme, reverse transcriptase, used by HIV to create a DNA strand.

Bread-making Bread-making is an example of a process in which the reaction rate is accelerated, primarily by way of a catalyst. The holes in the soft part of bread are gas bubbles produced by the action of the yeast added to the mixture at the beginning of the process. Yeast is a unicellular fungus that produces enzymes which catalyze several chemical reactions that produce gases. One of these reactions occurs in bread dough and produces carbon dioxide (CO 2) gas, which remains trapped in the dough, causing it to rise and become softer.

Figure 23 Yeast used in bread-making produces gases that make the dough rise.

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10.5.2

Homogeneous catalysts

A homogeneous catalyst is a substance that is found in the same phase as the reactants. Generally, homogeneous catalysts catalyze reactions that involve substances in the gaseous or aqueous phase. Homogeneous catalysts play an active role in the reaction by producing one or several reaction intermediates that require less activation energy than in a non-catalyzed reaction. By lowering the activation energy of the reaction, the homogeneous catalyst provides a new reaction mechanism that facilitates the course of the reaction without changing the nature of the products formed or the enthalpy change (H). For example, the simple hypothetical reaction between A and B to form AB occurs in a single step, when it is non-catalyzed, according to the following equation: A  B → AB A homogeneous catalyst will increase the rate of this reaction by providing a new mechanism whose activation energy will be lower. This other mechanism for the catalyzed reaction can be represented by the following steps which, when added together, are identical to the overall reaction: Step 1 Step 2 Overall reaction

A  catalyst → A  catalyst A  catalyst  B  AB  catalyst A  B → AB

In the proposed mechanism, the two steps involved in producing two reaction intermediates at the vertex of the curve of the catalyzed reaction are faster than the original non-catalyzed reaction (see Figure 24). Consequently, the reactants and products of the overall reaction are identical in the catalyzed and non-catalyzed reactions, but the catalyzed reaction occurs faster. The energy diagram of a non-catalyzed and catalyzed reaction by a homogeneous catalyst Non-catalyzed reaction

Non-catalyzed Ea

Catalyzed reaction

Potential energy

Reaction intermediate

AB

Catalyzed Ea

H AB Step 1

Step 2 Progress of the reaction

Figure 24 The use of a catalyst triggers a new reaction mechanism, sometimes involving two or more intermediate steps, which decreases the activation energy while maintaining the same enthalpy change (H).

The substance A–catalyst is a reaction intermediate that is produced during step 1, but consumed in step 2. In contrast, the catalyst is regenerated and unchanged at the end of the overall reaction without becoming a reactant or

264

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product. It can therefore be used again in the catalysis process and form new reaction intermediates by combining with reactant particles that have not yet reacted. The phenomenon of catalyst regeneration can be observed in the reaction between sodium potassium tartrate (NaKC4H4O6) and hydrogen peroxide (H2O2), catalyzed by cobalt chloride (CoCl2) at 70°C (see Figure 25). Cobalt chloride is pink in solution. Initially, when the cobalt chloride is poured into the beaker, the catalyst is intact and the reaction system is pink in colour. The solution then quickly changes to dark green, which indicates the formation of a reaction intermediate. At the end of the reaction, the solution returns to its original pink colour, indicating that the cobalt chloride catalyst has been regenerated.

a) The reactants and the catalyst before the mixture

Biological catalysts: enzymes Human beings depend on a variety of chemical reactions for their survival. Most of the chemical reactions in the body would not occur without the presence of biological catalysts called enzymes. Indeed, enzymes enable reactions in the human body by allowing them to occur at the body’s normal temperature of 37.5°C, when they would normally only occur at higher temperatures. Enzymes are enormous protein molecules present in living cells. The action of enzymes is very specific and limited to a few molecules, called substrates, upon which enzymes can act. This explains why some cells contain up to 3000 different enzymes, each one catalyzing a particular reaction in which a substrate is converted into a product needed for a given biological process.

b) When the catalyst is added, the solution turns pink.

For example, amylase is an enzyme found in saliva and pancreatic juices that enables the digestion of starch by breaking down this macromolecule into tiny glucose molecules. Since enzymes are in the same state as the substrates upon which they act, they are considered homogeneous catalysts. Only a small part of the enzyme, called the active site, takes part in the reaction. The substrate molecule, which is essentially a reactant, binds to the active site of the enzyme during an enzyme reaction. In so doing, the enzyme enables the formation of the activated complex by guiding the reactants in such a way that they are converted into products.

c) The solution becomes dark green after approximately 20 seconds.

A few models can be used to explain the interaction between the enzyme and its substrate. The lock-and-key model suggests that an enzyme is like a lock, and the substrate, the key that fits into it perfectly (see Figure 26).

Substrate

Products Active site Fast

Enzyme

Figure 26 The lock-and-key model

d) The solution returns to its pink colour after approximately 2 minutes.

Slow Enzyme-substrate complex

Enzyme

Figure 25 The reaction of sodium potassium tartrate (NaKC4H4O6) with hydrogen peroxide (H2O2) is catalyzed by cobalt chloride (CoCl2) at 70 °C.

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HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

The second model, called the induced-fit model, suggests that the active site of the enzyme changes its shape so that the substrate can fit into it, like a piece of modelling clay into which an object is pressed (see Figure 27). Substrate

Products Active site

Fast Enzyme

MAUD LEONORA MENTEN Canadian biochemist (1879–1960) In 1912, women were still not authorized to conduct medical research in Canada. Consequently, Maud Leonora Menten chose exile in Berlin in order to pursue her work with biochemist Leonor Michaelis. One year later, their collaboration bore fruit: they developed the Michaelis-Menten equation, which made it possible to estimate the rate of a chemical reaction produced by an enzyme. This equation applies to many enzymes, although it is not very effective for complex behaviour. Menten was also interested in hemoglobin and conducted many experiments that contributed significantly to histochemistry, the study of tissues and cells.

Slow Enzyme-substrate complex

Enzyme

Figure 27 The induced-fit model

10.5.3

Heterogeneous catalysts

A heterogeneous catalyst is a substance in a phase that is different from the reactants in the reaction that it catalyzes. Generally, heterogeneous catalysts are found in the solid phase, while the reactants are usually in the gaseous or aqueous phase. In heterogeneous catalysis, the reaction usually occurs on the surface of the solid catalyst, which is often a metal. Heterogeneous catalysts are often used in industries, since they enable various industrial processes, from the manufacture of margarine and nitric acid (HNO3) to the synthesis of ammonia (NH3), which would otherwise be too slow or produce an insufficient yield. The Haber process, a chemical reaction that produces ammonia needed to manufacture chemical fertilizers and explosives, uses iron (Fe) as a heterogeneous catalyst (see Figure 28). Iron (Fe) Nitrogen (N) Hydrogen (H)

a) Hydrogen (H2) and nitrogen (N2) molecules attach to the surface of the catalyst.

b) This interaction weakens the hydrogen and nitrogen bonds, and the molecules dissociate.

c) Hydrogen (H) and nitrogen (N) atoms combine to form molecules of ammonia (NH3), which detach from the surface of the catalyst.

Figure 28 The Haber Process uses iron (Fe) as a heterogeneous catalyst.

Without a catalyst, this reaction would not produce sufficient yield for commercial purposes. However, when the reaction is catalyzed by iron, the reaction rate increases significantly. Nitrogen (N 2) and hydrogen (H 2) molecules create bonds with the surface of the metal. This phenomenon weakens the covalent bonds of the nitrogen and hydrogen, which dissociate into nitrogen (N) atoms and hydrogen (H) atoms. The nitrogen atoms and the hydrogen atoms combine to form molecules of ammonia in a sufficient yield from an economic perspective.

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UNIT 3 Reaction Rate

SECTION 10.4 SECTION 10.5

Temperature of the reaction environment Catalysts

1. Why can food be preserved in a freezer for months without spoiling? Explain your answer. 2. In the following graph, which curve represents the highest temperature of the reaction?

Number of particles

1

2 3

5. Examine the following three reactions: 1) C3H8 (g)  5 O2 (g) → 3 CO2 (g)  4 H2O (g) 7 2) C2H6 (g)  O2 (g) → 2 CO2 (g)  3 H2O (g) 2 25 3) C8H18 (g)  O → 8 CO2 (g)  9 H2O (g) 2 2 (g) At room temperature: a) Which is the fastest reaction? Explain your answer. b) Which is the slowest reaction? Explain your answer. 6. What effect would an inhibitor have on the following graph? Explain your answer.

Kinetic energy

Step 2

Enthalpy H (kJ)

Step 1

Noncatalyzed Ea

Number of particles

3. Examine the following graph, then answer the questions below:

Particles with sufficient energy to react in a noncatalyzed reaction

Kinetic energy

a) If a catalyst were used in step 1, what would be its effect on the overall reaction rate and what would the graph show? b) If a catalyst were used in step 2, what would be its effect on the overall reaction rate and what would the graph show? c) If an inhibitor were used in step 1, what would be its effect on the overall reaction rate and what would the graph show? d) If an inhibitor were used in step 2, what would be its effect on the overall reaction rate and what would the graph show? 4. Can a catalyst increase the final concentration of a product in a reaction? Why or why not?

1

Potential energy

Progress of the reaction

7. The following graph represents the progression of a chemical reaction subjected to three different conditions: • without the addition of a catalyst or inhibitor • with the addition of a catalyst • with the addition of an inhibitor

2 3

Progress of the reaction

a) Match each of the curves with the corresponding condition. b) Under which condition (1, 2 or 3) is the reverse reaction the fastest? Explain your answer. CHAPTER 10 Factors that Affect Reaction Rates

267

APPLICATIONS What is hydrogenation? At room temperature, vegetable oil is a liquid. In order to convert it into a solid fat like margarine, or a semisolid (fat) like the oils we find at the grocery store, it must be hydrogenated, that is, hydrogen (H) is added to react with the oil. Vegetable oil contains a mixture of fatty acids composed of carbon (C) chains bonded to hydrogen atoms. In a saturated fatty acid, all of the bonds between the carbon atoms are single. This means that each carbon atom has the maximum hydrogen atoms possible bonded to it. Unsaturated fatty acids contain one or several C  C double bonds. The hydrogenation of unsaturated fatty acids converts the double bond into a single bond by adding a hydrogen atom to each carbon atom involved in the bond. The oil becomes saturated in hydrogen. Often, the process is not fully completed; this is when vegetable oil is referred to as being partially hydrogenated (see Figure 29).

Hydrogenation was widely used in the 1950s. One method consisted of heating oil to 200°C, under a pressure of at least three atmospheres, that is, three times the normal pressure. To trigger the reaction, nickel (Ni), in the form of a fine grey powder, was often used as a catalyst. At the end of the process, the consistency of the oil went from liquid to solid or semi-solid (fat). In this new state, the oil is more stable and breaks down more slowly when used in cooking (see Figure 30).

Figure 30 Solid hydrogenated cooking oil

Products that contain hydrogenated oils (e.g. pastries, soups, canned foods) can also be preserved longer. The same is true of oils that are of a lesser quality, such as fish oils, which are hydrogenated in order to be used to make soap and wax (see Figure 31).

Figure 29 Liquid hydrogenated cooking oil

The problem with trans fats arises in this last process. When carbon bonds are destroyed, certain fatty acids are rearranged, at the molecular level, and assume a "trans" configuration that allows them to join together more easily. This increases the lifespan of the oil, but also increases the risk of cardiovascular problems in the human body. Total hydrogenation does not produce any trans fats.

268

UNIT 3 Reaction Rate

Figure 31 Soap made with hydrogenated oils

Catalysis Chemists have long been intrigued by the action of catalysts. They have had difficulty conceiving of a substance that could take part in a chemical reaction without undergoing any transformation itself. Almost all effective catalysts have been discovered by purely empirical methods, that is, by repeated trial and error in order to determine what works. In the 19th century, Swedish chemist Jöns Jacob Berzelius was the first to use the term catalysis for experiments carried out by other chemists before him. But in the early 20th century, German chemist Wilhelm Normann successfully carried out the hydrogenation of liquid oleic acid (C18H34O2) into solid stearic acid (C118H36O2) on nickel (Ni) broken into small pieces. This process is still used today in the food, pharmaceutical and perfume industries. Gradually, chemists realized that metals prepared in such a way that they could present a large surface area (in powder or shavings) could be used as catalysts in

Figure 32 The decomposition reaction of hydrogen peroxide (H2O2) in the presence of platinum (Pt) allows contact lenses to be cleaned.

many reactions. This led to many industrial and daily applications. In fact, people who wear contact lenses are very familiar with the decomposition reaction of hydrogen peroxide (H2O2) in the presence of platinum (Pt) since they use it daily to clean their lenses (see Figure 32). In industrial applications, the platinum, palladium (Pd) and rhodium (Rh) used as catalytic converters for car exhaust systems are an example of the everyday use of catalysis. These catalysts process exhaust fumes and transform them into gases that are less toxic for the environment (see Figure 33). Catalysts play a vital role in today’s chemical technology and industry, since they make it possible to use lower temperatures in various industrial processes. Not only do they reduce energy consumption, but they also prevent the decomposition of reactants and products while reducing the risks of unwanted secondary reactions. The effectiveness of many chemical reactions is thus enhanced, as is their economic profitability.

Figure 33 A catalytic converter reduces the toxicity of a car’s exhaust fumes. It converts carbon monoxide (CO), unburned hydrocarbons and nitrogen oxides (NOx ) into less toxic substances: water and carbon dioxide (CO2).

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CHAPTER

10

Factors that Affect Reaction Rates

10.1 Nature of reactants • The nature of reactants, specifically the state (or phase) in which reactants are found, and the number and strength of the bonds they contain, affect the reaction rate. • In general, reaction rates can be classified from the slowest to the fastest as a function of the state of the reactants, as shown below. solid → liquid → gas → aqueous ions • In general, a chemical reaction in which reactant molecules contain high total bond strength, due to their number or strength, is slower than a reaction whose reactants have weak intramolecular bonds.

10.2 Surface area of reactants • In general, increasing the surface area of the reactant increases the reaction rate.

10.3 Concentration of reactants and the rate law • The concentration of the reactants has an effect on the reaction rate; in general, the higher the concentration of the reactants, the faster the reaction rate. • The rate law is a mathematical relationship that exists between the reaction rate and the concentration of the reactants, and it depends on the stoichiometric coefficients of the reactants that appear in the balanced equation. • The reaction rate law can be expressed more generally, at a given temperature, as follows: r  k  [A]x  [B]y

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UNIT 3 Reaction Rates

The kinetic energy of the particles of a gas sample at two temperatures

• In general, increasing the temperature of the reaction environment increases the reaction rate.

Number of particles

10.4 Temperature of the reaction environment

Largest number of particles at T2 with sufficient energy to react T1 T2 T1

Kinetic energy

10.5 Catalysts • Catalysts are substances that increase the reaction rate without changing the result of the conversion and without being consumed by the reaction. • The use of a catalyst reduces the activation energy, which allows a greater number of particles to possess sufficient kinetic energy to react. • An inhibitor is a substance that reduces the reaction rate. It acts by increasing the activation energy of the chemical reaction. • A homogeneous catalyst is a substance whose state (or phase) is the same as that of the reactants. • Human beings depend on chemical reactions catalyzed by enzymes, which are homogeneous biological catalysts. • A heterogeneous catalyst is a substance whose state (or phase) is different than that of the reactants of the reaction it is catalyzing. The kinetic energy of reactant particles of a chemical reaction occurring with and without a catalyst Catalyzed Ea

Non-catalyzed Ea

Particles with sufficient energy to react in a noncatalyzed reaction

Number of particles

Particles with sufficient energy to react in a catalyzed reaction

Kinetic energy

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Factors that Affect Reaction Rates

1. Examine the graph below: a) Where are the particles that possess sufficient energy to produce effective collisions situated? Use the activation energy line as a reference. Number of particles

3) 2 CH3COOH (aq)  Ba(OH)2 (aq) → Ba(CH3COO)2 (aq)  2 H2O (l) 4) CH4 (g)  2 O2 (g) → CO2 (g)  2 H2O (g)

According to the rate law, what will be the effect of each of the following four changes? Explain your answer for each situation. a) Pieces of magnesium are added. b) The same concentration of the sulphuric acid is added. c) Magnesium powder is added. d) The concentration of sulphuric acid is increased.

Kinetic energy

Ea Mean energy

3. In the lab, magnesium (Mg) reacts with sulphuric acid (H2SO4) concentrated at 1.0 mol/L according to the following equation: Mg (s)  H2SO4 (aq) → MgSO4 (aq)  H2 (g)

Ea

Number of molecules (mol)

2. Using only one factor, name an action that increases the reaction rate and another that reduces it. Explain the changes in each of the following four situations: a) cooking potatoes b) preserving a carton of milk that was just bought c) browning of pears (oxidation) d) preserving fish that was just caught

5. Graph a) below represents the kinetic energy of the reactant particles and the activation energy of a reaction at a given temperature. Graph b) represents the same reaction subjected to different conditions. What change(s) in conditions explain(s) such a change in graph b) ? Explain your answer. a) b)

Mean energy

b) For each of the following four situations, explain the change that the curve will undergo: 1) The temperature is increased. 2) The temperature is decreased. 3) A catalyst is added. 4) Reactant particles are added without changing the volume.

a) Which is the fastest reaction at room temperature? Explain your answer. b) Which is the slowest reaction at room temperature? Explain your answer. c) Which of the four reactions in an aqueous solution is the fastest? Explain your answer.

Number of molecules (mol)

B

Kinetic energy

UNIT 3 Reaction Rate

2) KCl (aq)  AgNO3 (aq) → KNO3 (aq)  AgCl (s)

Ea

A

272

4. Examine the following four reactions, then answer the questions below: 1) 2 Li (s)  2 H2O (l) → 2 LiOH (aq)  H2 (g)

Kinetic energy

6. The salt used to de-ice the entrances to houses contains calcium chloride (CaCl2). If table salt (NaCl) is used instead, the same reaction occurs, but is much slower at similar temperatures. What factor is responsible for this difference? 7. A half-filled grain silo containing flour from wheat harvested from the fields is on fire. The fire department quickly determines that the fire was caused by a spark and that the flour dust was set ablaze almost immediately. How do you explain this when it is common knowledge that a bag of flour is very difficult to ignite?

8. In the lab, it is possible to separate the essential oil in a clove from the rest of its solid components. You have a choice between a whole clove and a ground clove. Which one will you choose? Explain your choice.

11. During an experiment, you use a powder that reacts with a substance in aqueous solution. You realize that you cannot record any data from your experiment because it occurs too quickly. Explain what you would do to reduce the reaction rate. 12. Why is smoking forbidden while putting gas in your car?

9. Examine the following graph, then answer the questions below: Step 1

Step 2

Step 3

 13.

A

Enthalpy H (kJ)

2.50 Concentration (mol/L)

B

Progress of the reaction

a) b) c) d)

Examine the following graph:

Which is the rate-determining step? Which is the fastest step? Which is the exothermic step? Which step must be modified in order to change the overall reaction rate?

10. Examine the following table, which represents a study of the decomposition rate of nitrogen pentoxide (N2O5) into nitrogen oxide (NO2) and oxygen (O2), then answer the questions below. Test

[N2O5] (mol/L)

Reaction rate (mol/(Ls))

1

0.1500

2.44  105

2

0.3000

1.03  104

3

0.6000

2.13  104

a) Determine the equation for the rate of this reaction. b) Determine the values of the rate constants in test 1 and test 3. c) How do you explain the difference between the rate constants in test 1 and test 3?

2.00 1.50 1.00 0.50 0.00

0

10

20 Time (minutes)

30

To what chemical equation does the graph correspond? Explain your choice. a) A → B c) 3A → B e) B → 2A b) 2A → B d) B → A f) B → 3A

 14.

A chemist wants to evaluate the effect of temperature on the combustion rate of propane (C3H8) which occurs according to the following equation: C3H8 (g)  5 O2 (g) → 3 CO2 (g)  4 H2O (g) She conducts the experiment at 20°C and collects the carbon dioxide (CO2) in a balloon during the reaction. The balloon inflates to 75.5 mL at a pressure of 100.6 kPa in 4 min 33 s. a) What is the reaction rate in moles per second (mol/s) as a function of the carbon dioxide (CO2)? b) What is the reaction rate as a function of the oxygen (O2)? c) What is the equation that expresses the reaction rate?

CHAPTER 10 Factors that Affect Reaction Rates

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274

CONTENTS CHAPTER 11

The Qualitative Aspect of Chemical Equilibrium . . . . . . . . . 277 CHAPTER 12

The Quantitative Aspect of Chemical Equilibrium . . . . . . . . . 307

Most of the reactions observed in daily life, such as the combustion of wood, are complete reactions that cease when there are no more reactants. However, there are reactions in which the reactants are not completely consumed. In these reactions, the reactants and products co-exist in a system where equilibrium is attained between direct and reverse reactions. In a system at equilibrium, nothing appears to be happening. Yet, like the tightrope walker whose muscles are constantly working in order to stay balanced on the

metal wire, the system is dynamic at the particle level, so that it can restore equilibrium when it is disturbed by external changes. In this unit, you will study both qualitative and quantitative aspects of chemical equilibrium. The quantitative aspect of equilibrium constants of acids and bases will allow you to determine, for example, whether an acid or a base is weak or strong.

275

11.1 Static equilibrium and dynamic equilibrium 11.2 Irreversible and reversible reactions CHAPTER

UNIT

11

THE QUALITATIVE ASPECT OF CHEMICAL EQULIBRIUM

11.3 Necessary conditions for attaining equilibrium 11.4 Le Chatelier’s principle 11.5 Factors that affect the state of equilibrium

4

11.6 Chemical equilibrium in everyday life

CHEMICAL EQUILIBRIUM CHAPTER

276

12

THE QUANTITATIVE ASPECT OF EQUILIBRIUM

12.1 Equilibrium constant 12.2 Ionic equilibrium in solutions

The Qualitative Aspect of Chemical Equilibrium

T

his volcano appears to be extinct; it is in a state of equilibrium. Yet, many reactions are occurring. For example, molten rock produces magma containing volcanic gases that are released from the magma when there are changes in temperature or pressure. These changes can disrupt the equilibrium, setting off a volcanic eruption.

Therefore, hidden beneath the apparent calm of these systems, reactions are occurring between the particles. This calm often results from the establishment of a state of equilibrium between these reactions. In this chapter, you will learn about the conditions necessary to achieve a state of equilibrium and the factors

Review Phase changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Dissolution and solubility. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Concentration and dilution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Endothermic and exothermic reactions. . . . . . . . . . . . . . . . . . . . . . 26

that affect this state. You will also learn how to make predictions about how an equilibrium system will react to disturbances.

11.1

Static equilibrium and dynamic equilibrium . . . . . . . . . . . . . . . . . 278

11.2 11.3

Irreversible and reversible reactions . 280

11.4 11.5

Le Chatelier’s principle . . . . . . . . . . . . . .288

11.6

Chemical equilibrium in everyday life . . . . . . . . . . . . . . . . . . . . . . .299

Necessary conditions for attaining equilibrium . . . . . . . . . . . . .283

Factors that affect the state of equilibrium . . . . . . . . . . . . . . . . . . . . . . . . .289

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11.1 Static equilibrium and dynamic equilibrium Static equilibrium exists when a system remains at a given point or is kept immobile without active dynamic processes being involved. Dynamic equilibrium is the result of two opposing processes occurring at the same rate so that no visible change takes place in a reaction system. There are three types of dynamic equilibrium: phase or “physical” equilibrium; solubility equilibrium; and chemical equilibrium. In everyday life, the term equilibrium is often used to designate the state of immobility or a lack of motion. For example, a structure made of stones piled on top of one another so each one is immobile is considered a structure at equilibrium (see Figure 1). This equilibrium is called static equilibrium because, in this case, it describes the immobile state of the structure. Static equilibrium exists when a system remains at a given point or is kept immobile without active dynamic processes being involved. Paradoxically, in chemistry, the term equilibrium is used to describe a state that is far from immobile. On the contrary, the various Figure 1 This structure of stones particles in a system that is in a state of piled on top of one another is an example equilibrium are constantly moving and change, of static equilibrium. even though there is no apparent change. This is why this type of equilibrium is described as dynamic. Dynamic equilibrium is the result of two opposing processes occurring at the same rate so that no visible change takes place in the reaction system. Dynamic equilibrium can be compared to the comings and goings of hockey players during a game. Players on the same team take turns on the ice and on the bench, but throughout the game, the number of players on the ice and the number of players on the bench never changes. In chemistry, there are generally three types of dynamic equilibrium: phase or physical equilibrium, solubility equilibrium and chemical equilibrium.

11.1.1

Phase or “physical” equilibrium

Dynamic equilibrium is described as phase equilibrium or physical equilibrium when a single substance is found in several phases within a system as a result of a physical change. Figure 2 Although nothing seems to be occurring in this bottle of water, the water molecules are constantly moving from the liquid phase to gas phase, and vice versa, thus creating a state of phase equilibrium.

278

UNIT 4 Chemical Equilibrium

The evaporation of water in a closed bottle at a constant temperature is an example of phase equilibrium (see Figure 2). Water poured into a bottle slowly evaporates. The increase in the number of water molecules in the gas phase gradually increases the partial pressure of the water vapour in the bottle until it reaches its maximum value and becomes constant.

At this point, the number of water molecules that evaporate is equal to the number of water molecules that condense. The water-water vapour system has attained equilibrium. In the bottle, the rate of evaporation is the same as the rate of condensation of water. Despite this constant movement of water molecules between the liquid phase and the gas phase, no change can be perceived by observing the water in the bottle with the naked eye. This is phase equilibrium since it involves a change in a single substance as it moves from one phase to another.

11.1.2

Solubility equilibrium

Solubility equilibrium is a state in which a solute is dissolved in a solvent or a solution, and an excess of the solute is in contact with the saturated solution. A cup of tea containing a deposit of sugar (C12H22O11), for example, is a solution in which the solvent is the water used to make the tea and the solute is the sugar. This solution appears immobile to the naked eye, and it is easy to assume that nothing is happening since the sugar cannot dissolve anymore in the water (see Figure 3). The reality is quite different: the excess solid sugar that is deposited at the bottom of the cup is continually dissolving in the water and then returning to a state of aqueous sugar. The dissolution of sugar occurs at the same rate as the rate at which dissolved sugar returns to its solid state. These two opposing processes occur at the same rate and at the same time, which creates a state of equilibrium.

11.1.3

Figure 3 In this cup of tea, there is solubility equilibrium between the molecules of solid sugar and the molecules of dissolved sugar.

Chemical equilibrium

Chemical equilibrium is dynamic equilibrium that results from two opposing chemical reactions that occur at the same rate, leaving the composition of the reaction system unchanged. Chemical equilibrium is more complex than phase or solubility equilibrium since it involves the presence of more than one substance and means that the opposing changes that come into play are chemical reactions. For dynamic equilibrium to be characterized as chemical equilibrium, it must be the result of a chemical change between at least two different substances: a reactant and a product. The conversion of nitrogen tetroxide (N2O4) into nitrogen dioxide (NO2) is an example of chemical equilibrium (see Figure 4). In this reaction, the reactant changes into a product and the product changes into a reactant simultaneously and at the same rate. Just like phase equilibrium or solubility equilibrium, the conversion of reactants into products and of products into reactants is not visible to the naked eye and everything appears to be immobile.

Figure 4 In this test tube, gaseous nitrogen tetroxide (N2O4) is constantly being converted into reddish-brown gaseous nitrogen dioxide (NO2) and vice versa. However, these changes are not visible.

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11.2 Irreversible and reversible reactions An irreversible reaction is a reaction that can only occur in one direction, from reactants to products. A reversible reaction is a reaction that can occur in both directions, from reactants to products and from products to reactants. Like phase and solubility equilibria, chemical equilibrium involves opposing changes. However, in chemical equilibrium, these changes must be chemical reactions. These reactions are generally described as irreversible or reversible.

11.2.1 Figure 5 The combustion of wood is an irreversible chemical reaction.

Irreversible chemical reactions

Some reactions occur so rarely in the opposite direction that it is virtually impossible for products to return to reactants. For example, in the combustion of a log, the wood reacts with oxygen in the air to produce ash and smoke (see Figure 5). This chemical reaction is irreversible: it is impossible to convert ash and smoke back into wood and oxygen. The decomposition of food is another example of an irreversible reaction (see Figure 6). The compounds found in food are converted into nitrogen compounds and carbon dioxide through the action of micro-organisms. A reaction is irreversible when it is complete and only ceases when one or all of the reactants have been completely consumed (see Figure 7).

Figure 6 The decomposition of foods is an irreversible chemical reaction.

Number of particles

A change in the number of particles of reactants and products in an irreversible reaction over time

Products Reactants

Time

Figure 7 In an irreversible reaction, reactants are completely used up as the reaction favours the production of products.

The combustion of methane is an example of an irreversible reaction. The formation of carbon dioxide (CO2) and water vapour from methane (CH4) and oxygen (O2) is represented by the following equation:

280

UNIT 4 Chemical Equilibrium

CH4 (g)  2 O2 (g) n CO2 (g)  2 H2O (g) The one-way arrow from left to right indicates that the reactants become products and that the reaction is irreversible.

11.2.2

Reversible chemical reactions

It may seem that most chemical reactions are irreversible and occur only in one direction. Reactants are converted into products until there aren’t any more, at which point the chemical reaction stops. Yet, in theory, all chemical reactions are reversible: if chemical bonds can be broken, they can also be reformed. For example, the synthesis of hydrogen iodide (HI) gas using hydrogen (H2) gas and iodine (I2) vapour is represented by the following equation: Direct reaction H2 (g)  I2 (g) n 2 HI (g) The arrow that goes from left to right indicates that the reactants become products. This is called a direct reaction. However, this equation does not fully represent what actually occurs between the molecules of hydrogen iodine vapour and the hydrogen iodide gas molecules. In reality, the bond that joins the hydrogen atom with the iodine atom to form a molecule of hydrogen iodide gas can be easily broken to produce, once again, hydrogen gas and iodine vapour. The decomposition of hydrogen iodide gas that produces hydrogen and iodine vapour is, in fact, the reverse of the synthesis reaction of hydrogen iodide gas. This reverse reaction is represented by the following equation: Reverse reaction H2 (g)  I2 (g) m 2 HI (g) The arrow that goes from right to left indicates that the products change back to reactants. This is called a reverse reaction. The combination of these two equations describes the actual reaction that occurs between the molecules of hydrogen and iodine vapour and the hydrogen iodide gas molecules. This is a reversible reaction and is represented by the following equation: Direct reaction H2 (g)  I2 (g) 2 HI (g) Reverse reaction

CHAPTER 11 The Qualitative Aspect of Chemical Equilibrium

281

The double arrow indicates that the reaction can occur in both directions. A reversible reaction is, therefore, a reaction that can occur in either direction. Reactants are converted into products, and these products are, in turn, converted into reactants. Since there is a constant to-and-fro motion between the molecules of the reactants and the molecules of the products, the reactants are never completely converted into products. Reversible reactions tend to attain a state of equilibrium in which the direct reaction and the reverse reaction occur at the same rate. State of equilibrium where rdir  Direct reaction rate rrev  Reverse reaction rate

r dir  r rev

In a reversible reaction equlibrium, the quantity of products and reactants remains constant and no change is apparent, since the number of particles of reactants and products remains constant (see Figure 8). Change in the number of particles of reactants and products in a reversible reaction over time

Number of particles

State of equilibrium attained

Products Reactants

Time

Figure 8 During a reversible reaction, the reactants are not completely converted into products.

Therefore, a reversible chemical reaction occurs in both directions: the reactants are converted into products, and the products are, in turn, converted into reactants. When the rates of these two opposing reactions are equal, chemical equilibrium is attained. Just as in chemical reactions, the conversions involved in phase equilibrium and solubility equilibrium must also be reversible.

282

UNIT 4 Chemical Equilibrium

11.3 Necessary conditions for attaining equilibrium Three conditions are necessary for attaining equilibrium: a reversible change must occur in a closed system whose macroscopic properties remain constant. Phase equilibrium, solubility equilibrium and chemical equilibrium are dynamic processes. However, movement between particles is not visible to the naked eye. This is why it is sometimes difficult to determine if a reaction system is at equilibrium. It is therefore important to verify if the three conditions necessary to attain equilibrium have been respected.

11.3.1

Change is reversible

For a reaction system to attain a state of equilibrium, regardless of the type of equilibrium, the changes that take place must occur in both directions. In other words, equilibrium can only be attained when the changes are reversible. Although all reversible changes can attain equilibrium, this state can only be observed under certain experimental conditions. For example, the quantity of solute and solvent in a reaction system can either enable or prevent it from attaining equilibrium. The dissolution of sodium chloride (NaCl) in water in the form of sodium ions (Na) and chloride ions (Cl) is a reversible change represented by the following equation: z Na (aq)  Cl (aq) NaCl (s) y

z Na (aq)  Cl (aq) a) NaCl (s) y

Depending on various experimental conditions, this change may or may not attain equilibrium. If only a portion of the sodium chloride is converted into sodium and chloride ions, a residue of sodium chloride forms at the bottom of the beaker, which indicates that the solution is saturated with salt (see Figure 9a). The simultaneous presence of a solvent and a solute in the beaker allows the reaction to attain solubility equilibrium. At this point, particles contained in sodium chloride are converted into sodium and chloride ions, while other sodium and chloride ions are converted into sodium chloride. When the rates of these two changes are equal, equilibrium has been attained. When all of the sodium chloride is converted into sodium and chloride ions, no precipitate is observed at the bottom of the beaker (see Figure 9b). This indicates that the solution is not saturated with sodium chloride and that all of the particles of the solute have been dissolved in the solvent. Since the solute is completely dissolved, this change is considered irreversible or complete under these experimental conditions, and equilibrium is not possible.

b) NaCl (s) n Na (aq)  Cl (aq)

Figure 9 The state of solubility equilibrium is only attained in Beaker A because the solution is saturated with salt. In Beaker B, the salt is completely dissolved and no change occurs.

For chemical equilibrium to be attained, the change must be a reversible reaction. For example, the synthesis reaction of calcium sulfate (CaSO4) and sodium chloride (NaCl) from sodium sulfate (Na2SO4) and calcium chloride (CaCl2) is a reversible reaction that can be represented by the following equation: Na2SO4 (s)  CaCl2 (aq)

CaSO4 (s)  NaCl (aq)

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When products and reactants are present simultaneously, the reversible reaction can attain chemical equilibrium. At this point, the reactants change into products while the products change into reactants. When the rates of these two reactions are the same, chemical equilibrium is attained.

11.3.2

See Types of systems, p. 131.

Change occurs in a closed system

When one or several products produced in a chemical reaction dissipate into the surrounding environment, the reaction cannot attain chemical equilibrium. This type of reaction system is called an open system, since there is an exchange of matter between the system and the surrounding environment. An open bottle of carbonated water is an example of an open system. Molecules of carbon dioxide (CO2) gas are released from the container into the surrounding environment (see Figure 10a). Inversely, a closed system is a reaction system that does not allow the transfer of energy or the exchange of matter with the surrounding environment. Since matter cannot enter or leave the system, the quantity of matter involved remains constant. A closed bottle of carbonated water is an example of a closed system (see Figure 10b).

a)

b)

Figure 10 Two bottles of carbonated water. The one on the left is open and the one on the right is closed. Since carbon dioxide (CO2) can be released from the bottle on the left, it represents an open system. In the bottle on the right, the carbon dioxide cannot be released because of the cap. This bottle is a closed system.

Closed systems enable reversible changes to attain equilibrium. When carbonated water is poured into a bottle that is immediately closed, a certain quantity of carbon dioxide (CO2) contained in the water is released and occupies the remaining space between the liquid and the cap (see Figure 11). The pressure from the carbon dioxide inside the bottle increases until it attains its maximum value. After some time, the pressure from the carbon dioxide becomes constant and solubility equilibrium is attained. At this point, the dissolution rate of the gaseous carbon dioxide in the water is equal to the rate of the gas molecules being released from the water. The equilibrium between the carbon dioxide dissolved in the water and the carbon dioxide gas is represented by the following equation: z CO2 (g) CO2 (aq) y

CO2 (aq) n CO2 (g)

z CO2 (g) CO2 (aq) y

CO2 (aq) z y CO2 (g) Solubility equilibrium attained

Oxygen (O) Carbon (C)

Figure 11 This closed system is composed of a closed bottle filled with carbonated water. In this system, the reversible change of the carbon dioxide (CO2) dissolving in the water finally reaches a state of equilibrium when the rate of dissolution of the carbon dioxide molecules in the water is equal to the rate of the carbon dioxide molecules released from the water.

284

UNIT 4 Chemical Equilibrium

A system that is not physically closed is not necessarily an open system. In some cases, the container in which the reaction occurs can be open without any exchange of matter between the system and the surrounding environment. This type of system is also considered closed. One example is a saturated solution of sodium chloride (NaCl) in a beaker (see Figure 12). This equilibrium is represented by the following equation: z Na (aq)  Cl (aq) NaCl (s) y Since the evaporation of water is negligible over a short period of time and dissolved sodium chloride and solid sodium chloride cannot leave the container, this system is closed, even though the beaker is open.

11.3.3

Figure 12 Even though it is open, this beaker constitutes a closed system where the reversible change of the dissolution of sodium chloride (NaCl) in water occurs.

Macroscopic properties are constant

The state of equilibrium is a phenomenon that is visible to the naked eye. In a system, when equilibrium is attained, there are no additional visible changes and everything seems immobile, as demonstrated by the macroscopic proper ties of the substances. These properties include colour, volume, pH, temperature and pressure, all of which can be observed and measured. These properties can be perceived with the five senses or can be detected using instruments that are extensions of the senses, such as a manometer, thermometer or pH meter. A system that has attained equilibrium is characterized by constant macroscopic properties. For instance, a quantity of water that does not change in a closed bottle is an example of a constant macroscopic property in a system at phase equilibrium.

a)

The colour purple, characteristic of a system at a chemical equilibrium between hydrogen iodide (HI), hydrogen (H2) and iodine (I2), is another example of a constant macroscopic property (see Figure 13). Hydrogen iodide and hydrogen are colourless gases, while iodine vapour is dark purple. This reaction is represented by the following equation: z 2 HI (g) H2 (g)  I2 (g) y colourless purple colourless

b)

When hydrogen and iodine are put inside a round bottom flask, the inside of the flask turns dark purple due to the presence of the iodine vapour (see Figure 13a). The reaction between the hydrogen and the iodine vapour thus begins, and hydrogen iodide is formed. Gradually, the quantity of iodine decreases and the colour of the contents of the flask begins to fade (see Figure 13b). However, even after a long lapse of time, the contents of the flask never become colourless. To the contrary, the colour remains constant, which indicates that the concentrations of the reactants and products are no longer varying. The contents of the flask remain pale purple, which shows that the state of equilibrium has been attained and there is a simultaneous presence of hydrogen, iodine and hydrogen iodide (see Figure 13c).

c)

Iodine (I) Hydrogen (H)

Figure 13 When equilibrium is attained between hydrogen (H2), iodine (I2) and hydrogen iodide (HI), the purple colouration of the iodine vapour in Flask C is faded but remains constant.

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Steady state Observing constant macroscopic properties is not in and of itself proof of a system’s state of equilibrium. In fact, this condition is also present in systems in a steady state. Contrary to balanced systems, systems in a steady state are often open, and the quantity of energy or matter released from the system is equivalent to the quantity introduced. This is why, from a macroscopic point of view, the properties of the system remain constant. For example, a pool filled with water whose temperature is held constant by a water heater constitutes a steady system. Energy, in the form of heat, is constantly transmitted by the water heater to the water in the pool. The quantity of energy that the water stores is equal to the heat it loses at the surface and is dissipated into the surrounding environment, which means that the pool maintains a constant temperature. However, since the system is in a steady state, equilibrium is not established.

Microscopic properties Certain properties cannot be seen with the naked eye or even with measuring instruments. These are called microscopic properties because they describe matter at the microscopic level of atoms and molecules. To explain these properties, abstractions and mathematical models are often used. The strength of the bonds that join two atoms and the collisions between the different atoms in a solution mixture are examples of microscopic properties. If it were possible to observe the microscopic properties of a system at equilibrium, the constant conversion between the reactants and products would reveal its dynamic state. In reality, on a microscopic scale, there is a constant movement between the particles involved in reaction mixtures which are at physical, solubility or chemical equilibrium.

Furthering

your understanding

Vital disequilibrium Nerve impulses travel in the form of an electrical signal created by a flow of ions that pass through the membrane that envelopes neurons. When a neuron is at rest, it is not in a state of equilibrium with the external environment, since the interior of the neuron is negatively charged while the exterior is positively charged. This disequilibrium is caused primarily by the presence of many negatively charged proteins inside the neuron. Neurons must maintain this state of disequilibrium in order to function. An enzyme called a sodium-potassium pump works to prevent the natural concentration equilibrium that tends to occur on either side of the membrane. By using the energy stored inside the neuron, this pump sends three sodium (Na) ions to the exterior of the neuron and brings in two potassium (K) ions. Without this pump, equilibrium would soon be attained and nerve impulses would stop flowing.

Na Na Na

Exterior of the neuron Cell membrane

Pump Na/ K

Interior of the neuron K K Proteins

Proteins

Proteins Proteins

Figure 14 Sodium-potassium pump action between the interior and exterior of a neuron

286

UNIT 4 Chemical Equilibrium

SECTION 11.2 SECTION 11.3

Irreversible and reversible reactions Necessary conditions for attaining equilibrium

1. Determine if the following conversions are reversible or irreversible. Explain your answer. a) Cooking an egg. b) The complete dissolution of a piece of metal in acid. c) A test tube containing an aqueous salt solution with a salt residue at the bottom. d) The dissolution of carbon dioxide (CO2) gas in a bottle of champagne. 2. Determine if the following systems are open or closed. a) a recycling bin b) soup in an isothermal container c) a propane tank when the valve is closed d) a bottle of perfume e) a lit candle f) an erupting volcano 3. The combustion of ethanol (C2H5OH) in open air is described by the following equation: C2H6O (g)  3 O2 (g) n 2 CO2 (g)  3 H2O (g)  1409 kJ Observe this equation and explain why the combustions that occur in open air are irreversible reactions. 4. Provide three examples of macroscopic properties other than colour, volume, pH, temperature and pressure. 5. Determine if the following systems are in a state of equilibrium. If the system is at equilibrium, specify what type of equilibrium is involved. If the system is not at equilibrium, determine which of the three conditions necessary for attaining a state of equilibrium are not met. a) A burner in which the height and colour of the flame are constant. b) A closed isothermal mug containing coffee with a sugar deposit at the bottom. c) The water in a hydroelectric dam maintained at a constant level for one week. The temperature and pressure of the system are constant. d) A closed can of gas. e) A closed round bottom flask containing a mixture of nitrogen dioxide (NO2), nitrogen monoxide (NO) and oxygen (O2). The pressure, temperature and colour of the mixture are constant. 6. Give two examples from daily life where a system at equilibrium can be observed and provide the conditions that prove that the system is at equilibrium.

7. Copper sulfate (CuSO4), a solid blue compound, dissolves into copper ions (Cu2) and sulphate ions (SO42) in water according to the following equation: z Cu2 (aq)  SO42 (aq) CuSO4 (s) y a) A certain quantity of copper sulfate is added to 1 L of water in a round bottom flask which is then closed with a stopper. After a certain period of time, the copper sulfate dissolves completely and the solution in the round bottom flask becomes light blue. The colouration of the solution, its temperature and its volume are constant. Explain why, despite its constant macroscopic properties, this system is not at equilibrium. b) How could you modify the experimental conditions to obtain a system that is able to attain a state of equilibrium? 8. Vinegar used in cooking is an acid solution formed by the dissociation of acetic ("ethanoic") acid (CH3COO) and hydrogen ions (H) in water. This reaction is represented by the following equation: z CH3COO (aq)  H (aq) CH3COOH (l) y The bottle on the right contains a vinegar solution. Can this solution have attained a state of phase equilibrium, despite the fact that there is no deposit at the bottom of the container? Explain your answer.

9. A hypothetical compound used for industrial purposes, called compound C, is synthesized from compounds A and B, according to the following equation: zC ABy When designing the industrial process to produce compound C, should the engineers ensure that this reaction attains a state of equilibrium? Explain your answer.

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11.4 Le Chatelier’s principle Le Chatelier’s principle makes it possible to qualitatively predict the direction of the reaction (direct or reverse) that will be favoured when the conditions of a balanced system are changed. The equilibrium of a system is fragile. Any change in the experimental conditions in which a chemical reaction occurs can disrupt the state of equilibrium and lead the chemical reaction to a new state of equilibrium. For example, if factors such as the concentration, temperature or pressure of the substances in a reaction system change, this can have an effect on its equilibrium.

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

There is a general statement that makes it possible to predict the direction in which a system at chemical equilibrium will move if its experimental conditions are changed. This statement is called Le Chatelier’s principle. Le Chatelier’s principle Le Chatelier’s principle states that if the conditions of a system in a state of equilibrium change, the system will react to partially oppose this change until it attains a new state of equilibrium. This principle makes it possible to qualitatively predict the manner in which a system at equilibrium will react to changes in experimental conditions.

HENRY LE CHATELIER French chemist and metallurgist (1850–1936) Henry Le Chatelier is best known for the principle he formulated, which carries his name, regarding the stability of physico-chemical equilibriums. However, Le Chatelier also saved many lives by creating safer explosives for coal mines. He also studied the structure of metals and alloys. In his lifetime, he published more than 500 articles on his work in chemistry, as well as on the scientific management of industries. This eminent scientist also founded the Revue de métallurgie in 1904, and published two books on his experiments.

288

UNIT 4 Chemical Equilibrium

Since a state of equilibrium is defined as the result of two opposing reactions occurring at the same rate, any change to the rate of the direct or reverse reaction will disrupt the equilibrium. If a change to the experimental conditions favours an increase in the rate of the direct reaction and this rate becomes greater than that of the reverse reaction, the equilibrium will be disrupted. Since the direct reaction is temporarily favoured compared to the initial state, the equilibrium will shift toward the formation of products. Reactants z y Products The longer single-headed arrow indicates the reaction that is favoured. Above, it is the direct reaction, that being the one pointing toward the products. The equilibrium will also be disrupted if a change to the experimental conditions favours an increase in the rate of the reverse reaction and this rate becomes greater than that of the direct reaction. Since the reverse reaction is favoured, the equilibrium will shift toward the formation of reactants.

z Products Reactants y The longer single-headed arrow indicates the reaction that is favoured. Above, it is the reverse reaction, that being the one pointing toward the reactants.

11.5 Factors that affect the state of equilibrium Three factors that affect the state of chemical equilibrium of a system are: the concentration of the reactants or products; temperature; and pressure. Adding a catalyst makes it possible for a system to attain a state of equilibrium more quickly, but does not have an effect on this equilibrium.

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

By applying Le Chatelier’s principle, it is possible to predict the consequences of a change in the concentration, temperature and pressure on a system at equilibrium and to determine the direction of the movement of the reaction. Depending on the direction of this movement, the reactants or the products will be favoured. In industry, when chemical engineers manufacture a chemical compound, they are usually able to obtain the greatest possible quantity of this compound by changing the factors that affect the state of equilibrium. In 1912, German scientist Fritz Haber determined the best conditions for obtaining a large quantity of ammonia (NH3) from nitrogen (N2) and hydrogen (H2) according to the following equation: FRITZ HABER

N2 (g)  3 H2 (g) z y 2 NH3 (g)

German chemist (1868–1934) Haber’s work involved calculating the quantity of ammonia formed under different experimental conditions. His experiments made it possible to determine the optimal temperature and pressure conditions for producing ammonia. His industrial process for manufacturing ammonia, called the Haber Process, is still used today to manufacture chemical fertilizers and many nitrogen-based compounds.

11.5.1

Changes in concentration

It is possible to predict the effect that a change in the concentration of reactants or products will have on the state of equilibrium of a chemical system using Le Chatelier’s principle as follows: Le Chatelier’s principle applied to a change in concentration Any increase in the concentration of a substance of a reaction system at equilibrium shifts this equilibrium in favour of the reaction that decreases, in part, the concentration of this substance. Any decrease in the concentration of a substance of a reaction system at equilibrium shifts this equilibrium in favour of the reaction that increases, in part, the concentration of this substance.

The effect of a change in the concentration of reactants or products on the chemical equilibrium of a reaction can be studied by observing the formation of ammonia (NH3) from nitrogen (N2) and hydrogen (H) (see Figure 15 on page 290).

Fritz Haber is known for the development of a process for syn the sizing ammonia directly from atmospheric nitrogen and hydrogen. This process was used to make up for the shortages of nitrogen compounds used in fer tilizers and explosives experienced by certain countries in the early 20th century. Haber won the Nobel Prize in Chemistry in 1918. He conducted important work on combustion reactions and in electrochemistry and attempted, in vain, to extract gold from seawater. He was also interested in pesticides. However, his contribution to the development of Zyklon B, a toxic gas that was used years later in the gas chambers of Nazi concentration camps, greatly tarnished his reputation.

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Addition of N2 Hydrogen (H)

z 2 NH N2  3 H2 y 3

b) Disequilibrium

c) New state of equilibrium

z 2 NH3 N2  3 H2 y

N2

a) Equilibrium

Nitrogen (N)

NH 3 z 2 H2 y 3 

Figure 15 The state of equilibrium of the reaction mixture (a) is disrupted by the increase in concentration of nitrogen (b). To oppose this change, the system reacts by favouring the formation of ammonia until a new state of equilibrium is attained (c).

The state of equilibrium (see Figure 15a) is disrupted by the addition of nitrogen (see Figure 15b). Le Chatelier’s principle states that the system reacts to oppose the change imposed on it by eliminating part of the added substance. To do so, the direct reaction is favoured (see Figure 15b) until a new state of equilibrium is attained (see Figure 15c).

See Factors that Affect Reaction Rates, p. 245. See Collision Theory, p. 235.

The shift of equilibrium is explained by the reaction rate and collision theory. In fact, when nitrogen is added, the number of collisions between the nitrogen molecules and the hydrogen molecules increases. Consequently, the number of effective collisions also increases, which makes the rate of the direct reaction increase. Therefore, the formation of ammonia is favoured and its concentration increases. At the same time, the concentration of nitrogen and hydrogen decreases. However, the greater quantity of ammonia also causes an increase in the rate of the reverse reaction, thus causing the disappearance of oxygen and hydrogen to slow down. When the rates of the direct and reverse reactions become equal again, a new state of equilibrium is attained (see Figure 16). Change in the concentration of the substances of a system as a function of time

Addition of N2 Change

Concentration

Initial equilibrium

New equilibrium

NH3 N2 H2

Time

Figure 16 When nitrogen is added to the system at equilibrium, part of the nitrogen reacts with the hydrogen to form ammonia until a new equilibrium has been attained.

290

UNIT 4 Chemical Equilibrium

When the new equilibrium is attained, the final concentration of ammonia is higher than it was in the initial state of equilibrium since the system was conducive to the direct reaction. That is why the concentration of the reactants decreased. However, the final concentration of nitrogen is greater than its initial concentration. In fact, as Le Chatelier’s principle states, the system partially opposes change, so that the added nitrogen is not completely consumed. Therefore, a new state of equilibrium is created, with a higher concentration of ammonia and nitrogen, and a lower concentration of hydrogen.

CaCO3

CaO

On the other hand, in some cases in which the system at equilibrium contains a sub stance in a solid or liquid phase, the addition or removal of some of this substance does not change the equilibrium. For example, the reaction of the decomposition of calcium carbonate (CaCO3) into calcium oxide (CaO) and carbon dioxide (CO2) is a reversible reaction represented by the following equation: z CaO (s)  CO2 (g) CaCO3 (s) y CaCO3

In this case, the addition or removal of some of the calcium carbonate or the calcium oxide will not change the rate of the direct and reverse reactions (see Figure 17). In fact, the concentration of carbon dioxide remains stable even if the quantity of the solid changes. The example on page 292 shows how to predict the direction of movement of a chemical reaction to a state of equilibrium following a change in concentration by applying Le Chatelier’s principle.

CaO

Oxygen (O) Carbon (C)

Figure 17 At a constant temperature, regardless of the quantity of the two solids present, the concentration of carbon dioxide (CO2) at equilibrium remains stable.

Changes in concentration Most chemical reactions tend to attain a state of equilibrium. However, when a reaction mixture is kept far from its state of equilibrium, it is possible that the concentration of the substances is not the same everywhere in the reaction mixture. The markings on the coat of a mammal, such as a leopard or zebra, are perfect examples of reactions that do not attain a state of equilibrium. Among mammals, specialized skin cells, called melanocytes, determine the colour of fur because of melanin, a pigment they produce. Two distinct types of melanin can be produced: one that produces black or brown fur and one that produces golden or red fur. During the embryonic development of an animal, the two types of melanin react with each other and travel toward the fur at different rates. When the two pigments travel at equal rates, the state of equilibrium is rapidly attained and the concentration of each of the pigments is stable; the animal’s fur will therefore be a uniform colour. If the rates of travel are different, however, and a state of equilibrium cannot be attained, the concentration of the two types of pigments vary from area to area, and patterns or markings emerge on the animal’s coat.

Figure 18 The leopard’s spotted coat is explained by changes in the concentration of pigments.

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Example Methanol (CH3OH) is an industrially produced alcohol made from carbon monoxide (CO) and hydrogen (H2). The state of equilibrium of this reaction is represented by the following equation: z CH3OH (g) CO (g)  2 H2 (g) y What effect will the addition of carbon monoxide have on the concentration of each substance involved in the reaction? 1

h

To predict the effects of this disequilibrium on concentration:

CO (g)

z CH3OH (g) y n

1 Determine if the substance added is a reactant or a product.

CO (g) 2 H2 (g) 2 3

2 Apply Le Chatelier’s principle to find the direction of the reaction that will be favoured.

direct reaction

g

g

3

h

Determine the effects of the shift in equilibrium on the concentration of each substance involved in the reaction.

Answer: Following the addition of CO (g), carbon monoxide (CO) and hydrogen (H2) decreases while that of methanol CH3OH increases. By proceeding this way, it is possible to determine the influence of the change in concentration of the other substances in this system at chemical equilibrium.

Effect of the change in the concentration of the substances in the system

Changes to the system at equilibrium

[CO]

[H2]

Direction of the shift

[CH3OH]

Reaction favoured

Addition of H2

h

g

g

n

h

Direct

Addition of CH3OH

h

h

h

m

g

Reverse

Removal of CO

g

h

h

m

g

Reverse

Removal of H2

g

h

h

m

g

Reverse

Removal of CH3OH

g

g

g

n

h

Direct

11.5.2

Changes in temperature

The study of the effect of temperature on equilibrium is done taking into account the energy involved over the course of the reaction. The effect will be different depending on whether the reaction is endothermic or exothermic. See Endothermic and exothermic reactions, p. 150.

In fact, based on this characteristic, it is possible to identify the consequences of changes in temperature according to Le Chatelier’s principle. To determine the reaction that will be favoured if the system is heated or cooled, energy can be considered a reactant or a product. Therefore, it is possible to predict the effect that a change in temperature will have on the state of equilibrium of a reaction system using Le Chatelier’s principle as follows: Le Chatelier’s principle applied to a change in temperature Any increase in the temperature of a reaction system at equilibrium shifts the equilibrium in favour of an endothermic reaction. Any decrease in the temperature of a reaction system at equilibrium shifts the equilibrium in favour of an exothermic reaction.

292

UNIT 4 Chemical Equilibrium

The effect of temperature on the state of equilibrium can be studied by heating a reaction system where a reversible reaction is occurring in which nitrogen dioxide (NO2) is transformed into nitrogen tetroxide (N2O4) (see Figure 19). 2 NO

2

z N2O4  energy 2 NO2 y

a) Equilibrium

yz N

2 O4

e

z N2O4  energy 2 NO2 y

nerg y

b) Disequilibrium

c) New state of equilibrium

Figure 19 The state of equilibrium of the reaction mixture (a) is disrupted by the increase in temperature (b). To oppose this change, the system favours an endothermic reaction until a new state of equilibrium is attained (c ).

At equilibrium, the reaction is exothermic and the energy is considered a product (see Figure 19a). Since the increase in temperature is equivalent to adding energy in the form of heat to the system (see Figure 19b), according to Le Chatelier’s principle, the reaction system reacts to partially oppose this change. Consequently, the equilibrium shifts toward the reaction that allows it to partially absorb this surplus energy, that is, the reverse reaction which is endothermic. The formation of reactants is, therefore, favoured until a new state of equilibrium is attained (see Figure 19c). Consequently, Le Chatelier’s principle makes it possible to predict that a decrease in the temperature of this system will shift the equilibrium toward the formation of products, namely, nitrogen tetroxide. The exothermic reaction, which in this case is the direct reaction, will be favoured until a new state of equilibrium is attained. The effect of temperature on the state of equilibrium is easily observed by the change in colour of the reaction system of a system when nitrogen dioxide (NO2) is converted into nitrogen tetroxide (N2O4). When the system is heated, the reverse reaction is favoured and the nitrogen tetroxide, a colourless gas, is converted into nitrogen dioxide (NO2), a brown gas (see Figure 20a). However, when the system cools, the direct reaction is favoured and some of the nitrogen dioxide is converted into nitrogen tetroxide (see Figure 20b). The contents of the round bottom flask become pale brown because of the reduced quantity of nitrogen dioxide.

a)

b)

Figure 20 An increase in temperature increases the rate of conversion of nitrogen tetroxide (N2O4), which leads to a change in the colour of the contents of the round bottom flask. It becomes dark brown because it contains more nitrogen dioxide (NO2).

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The following example shows how to predict the direction of the shift in equilibrium of a system following a change in temperature by applying Le Chatelier’s principle: Example Phosphorus pentachloride (PCl5) is used in the synthesis of many chlorine compounds. In a gaseous phase, it decomposes into chlorine (Cl2) and phosphorus trichloride (PCl3). At equilibrium, the system is described by the following equation: z Cl2 (g)  PCl3 (g) H  156.5 kJ PCl5 (g) y What effect will the increase in temperature have on the concentration of each substance involved in the reaction? z Cl2 (g)  PCl3 (g) 2 PCl5 (g)  156.5 kJ y

hT

direct reaction

3 4

To predict the effects of the increased temperature on concentration: 1. Determine if the reaction is endothermic or exothermic. Here, the change in enthalpy is positive, therefore the reaction is endothermic. 2 Integrate the energy of the reaction into the equation. If the reaction is endothermic, energy is added to the reactant side; if the reaction is exothermic, energy is added to the product side. 3 Apply Le Chatelier’s principle to determine the direction of the reaction that will be favoured. 4 Determine the effects of the shift in equilibrium on the concentration of each substance involved in the reaction. Answer: Following an increase in temperature, the concentration of phosphorus pentachloride (PCl5) decreases and that of the chlorine (Cl2) and phosphorus trichloride (PCl3) increases. By proceeding this way, it is possible to determine the effect of the change in temperature on the concentration of all of the substances in this system at chemical equilibrium. Effect of the change in temperature on the equilibrium in the system

Changes to the system at equilibrium

294

UNIT 4 Chemical Equilibrium

[PCl5]

Energy

Direction of the shift

[Cl2]

[PCl3]

Reaction favoured

Increase in temperature

h

g

h

n

h

h

Direct

Decrease in temperature

g

h

g

m

g

g

Reverse

11.5.3

Changes in pressure

It is possible to predict the effect that a change in pressure will have on the state of equilibrium of a chemical system using Le Chatelier’s principle as follows: Le Chatelier’s principle applied to a change in pressure In a reaction system at equilibrium, any increase in pressure shifts the equilibrium in favour of a reaction producing the least number of gas molecules. In a reaction system at equilibrium, any decrease in temperature shifts the equilibrium in favour of a reaction producing the greatest number of gas molecules. The effect of a change in pressure on a system at equilibrium is closely linked to the general gas law. This law states that pressure is directly proportional to temperature and the number of moles of gaseous particles, but inversely proportional to the volume occupied by the gas. In a system at equilibrium, the effect of a change in pressure is related to the fact that it causes a change in the concentration of all of the substances that are in a gas state in the system. Consequently, if the volume of a container holding gaseous substances is decreased by half, the concentration of these gases will double (see Figure 21).

See General gas law, p. 105.

Oxygen (O) Nitrogen (N)

1L

0.5 L

Figure 21 The increase in pressure caused by the reduced volume of this system compresses the gas molecules of nitrogen tetraoxide (N2O4) and nitrogen dioxide (NO2), and increases their concentration.

In a system at equilibrium, a change in pressure has no effect unless the system contains at least one substance in a gaseous phase. Since liquids and solids are virtually incompressible, a change in the pressure of the system will not have an effect on their concentration. To study the change in pressure caused by a change in volume, it is necessary to first consider the total number of moles of the gas reactants and the total number of moles of the gas products. One mole of nitrogen tetroxide (N2O4), which produces two moles of nitrogen dioxide (NO2), is a reversible reaction. This reaction is described by the following equation: z 2 NO2 (g) N2O4 (g) y

The effect of the change in pressure on this system at equilibrium can be seen by observing the changes in colour in the container holding nitrogen tetroxide, a colourless gas, and nitrogen dioxide, a brown-coloured gas (see Figure 22 on page 296). CHAPTER 11 The Qualitative Aspect of Chemical Equilibrium

295

Nitrogen (N) a)

b)

c)

z 2 NO2 (g) N2O4 (g) y

z 2 NO2 (g) N2O4 (g) y

z 2 NO2 (g) N2O4 (g) y

Oxygen (N)

Figure 22 The lightening of the brown colour of the mixture of nitrogen tetroxide (N2O4) and nitrogen dioxide (NO2) makes it possible to see the shift in equilibrium toward the reactants following an increase in the pressure of the system.

At equilibrium, the presence of nitrogen dioxide gives the system a light brown colour (see Figure 22a). When, in the system, the volume decreases and the pressure increases in the system, the concentration of nitrogen dioxide and nitrogen tetroxide increases, and the mixture appears dark brown due to the increase in the concentration of the nitrogen dioxide (see Figure 22b). Le Chatelier’s principle states that the reaction system reacts to the increase in pressure so as to oppose this change. Consequently, the equilibrium shifts toward the reaction that allows it to reduce the number of gas molecules. Therefore, according to the stoichiometric coefficients of the equation, the system favours the reverse reaction by forming nitrogen tetroxide since this reaction lowers the number of molecules in the system. In fact, by favouring the reverse reaction, each time two nitrogen dioxide molecules unite to form a nitrogen tetroxide molecule, the number of molecules decreases by half, which causes the pressure to decrease. The formation of reactants is favoured until a new state of equilibrium, determined by the change in pressure, has been established. The changes cause a gradual lightening of the colour of the mixture. When a new state of equilibrium is attained, the colour of the mixture is paler (see Figure 22c) than when the equilibrium was disrupted (see Figure 22b). Inversely, Le Chatelier’s principle makes it possible to predict that a decrease in the pressure of the system will shift the equilibrium toward the formation of nitrogen dioxide. Therefore, the direct reaction will be favoured until a new equilibrium has been attained. However, a change in pressure in a system does not have any effect on a reaction that contains an equal number of molecules of reactants and products in a gaseous phase. For example, hydrogen chloride (HCl) is formed from hydrogen (H2) and chlorine (Cl2) according to the following balanced equation: z 2 HCl (g) H2 (g)  Cl2 (g) y Since there are two moles of gas on the reactant side and two moles of gas on the product side, a change in pressure of the system will not have any effect on its state of equilibrium.

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UNIT 4 Chemical Equilibrium

Example Sulfuryl chloride (SO2Cl2) is synthesized from sulphur dioxide (SO2) and chlorine (Cl2) according to the following equation at equilibrium: z SO2Cl2 (g) SO2 (g)  Cl2 (g) y What effect will an increase in pressure have on the concentration of each substance involved in the reaction? hT

z SO2Cl2 (g) SO2 (g)  Cl2 (g) y 1 2 moles of gas direct reaction 2

1 1 mole of gas

3 To predict the effects of an increase in pressure on concentration: 1 Determine the number of moles of gas on the reactant side and product side. 2 Apply Le Chatelier’s principle to establish the direction of the reaction favoured. 3 Determine the effects of the shift in equilibrium on the concentration of each substance involved in the reaction. Answer: The concentration of the sulphur dioxide (SO2) and the chlorine (Cl2) decreases and that of the sulfuryl chloride (SO2Cl2) increases. By proceeding this way, it is possible to determine the effect of the change in pressure on all of the substances in a gaseous phase in this system at chemical equilibrium. Effect of pressure changes on the equilibrium of the system

Changes to the system at equilibrium

[SO2]

[Cl2 ]

Direction of the shift

[SO2Cl2]

Reaction favoured

Increase in pressure

h

g

g

n

h

Direct

Decrease in pressure

g

h

h

m

g

Reverse

11.5.4

Adding a catalyst

Catalysts are substances that increase the reaction rate without changing the result of the conversion. They reduce the activation energy by allowing a greater number of particles to have the kinetic energy required to react.

See Catalysts, p. 262.

In a reversible reaction, a catalyst increases the rate of the direct reaction and that of the reverse reaction. Consequently, it does not change the equilibrium of the system, but allows equilibrium to be attained more quickly by reducing the activation energy of the direct reaction and the reverse reaction of a system at equilibrium. When equilibrium is attained, the concentrations of the reactants and the products are identical to those obtained without a catalyst. For example, most hydrogen (H2) produced industrially is synthesized from methane (CH4) and water (H2O) vapour. This reaction is reversible and attains the state of equilibrium represented by the following equation: z CO (g)  3 H2 (g) CH4 (g)  H2O (g) y

In industry, nickel (Ni) is also added to the reaction mixture. Nickel acts as a catalyst and speeds up the attainment of equilibrium, which maximizes the yield of the reaction. The hydrogen produced is used in many industrial processes including the manufacture of chemical fertilizers and the production of rocket fuel (see Figure 23).

Figure 23 A rocket lift-off which uses liquid hydrogen and oxygen (O2) as its fuels

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Le Chatelier’s principle Factors that affect the state of equilibrium

SECTION 11.4 SECTION 11.5

1. Observe each of the reactions in a state of equilibrium below and answer the questions that follow: z CO (g)  H2O (g)  energy A CO2 (g)  H2 (g) y z 2 CO2 (g)  energy B 2 CO (g)  O2 (g) y z NO2 (g)  CO (g)  energy C NO (g)  CO2 (g) y z CaO (s)  CO2 (g) D CaCO3 (s)  energy y a) For each of the equations, what reaction will be favoured by adding carbon dioxide (CO2) in the system? b) For each of the equations, what reaction will be favoured by an increase in the pressure of the system? c) For each of the equations, what reaction will be favoured by a decrease in the temperature of the system? 2. Sulphur trioxide (SO3) is synthesized from sulphur dioxide (SO2) and oxygen (O2) according to the following equation:

3. The industrial production of ammonia (NH3) from atmospheric nitrogen (N2) and hydrogen (H2) is expressed by the following balanced equation: z 2 NH3 (g) N2 (g)  3 H2 (g) y Explain how the production of ammonia in this system can be favoured without adding nitrogen or hydrogen and without changing the temperature or pressure. 4. The state of equilibrium between the precipitation and dissolution of silver chloride (AgCl) is represented by the following equation: z AgCl (s) Ag (aq)  Cl (aq) y Will the addition of aqueous hydrogen chloride (HCl) favour the formation of silver chloride if the temperature and pressure of the system remain constant? Explain your answer. UNIT 4 Chemical Equilibrium

Illustrate the effect of a decrease in the quantity of hydrogen by drawing a graph showing the changes in concentrations of carbon monoxide, hydrogen and methanol as a function of time. Indicate the areas on the graph that correspond to the initial equilibrium, the change in concentration of the components of the system and the new state of equilibrium. 6. The following is a balanced reaction: z C (g) A (g)  B (g) y The graph below represents the change in reaction rate as a function of time of the conversion of compounds A and B into compound C.

H  98.9 kJ

What reaction will be favoured if: a) the temperature of the system is increased? b) the pressure of the system is decreased by decreasing its volume? c) oxygen (O2) is added? d) a catalyst is added?

298

z CH3OH (g) CO (g)  2 H2 (g) y

Reaction rate

z 2 SO3 (g) 2 SO2 (g)  O2 (g) y

5. Methanol (CH3OH) is synthesized from carbon monoxide (CO) and hydrogen (H2). The state of equilibrium of this reaction is represented by the following equation:

A (g)  B (g) n C (g)

C (g) n A (g)  B (g)

Time

What would this graph look like if the reaction were catalyzed? Illustrate your answer by copying this graph and adding the curves corresponding to the catalyzed direct and reverse reactions. 7. The process developed by Henry Deacon in 1874 helped to industrially manufacture chlorine (Cl2) from hydrogen chloride (HCl) and oxygen (O2). At equilibrium, this reaction is represented by the following equation: z 2 Cl2 (g)  2 H2O (g) 4 HCl (g)  O2 (g)  177 kJ y Using Le Chatelier’s principle, describe all of the changes that can be made to this system to produce the most chlorine possible.

11.6 Chemical equilibrium in everyday life The quality of life on Earth is very closely tied to the dynamic equilibrium of the biosphere. This equilibrium is established thanks to the biogeochemical cycles of elements essential to life such as carbon (C), nitrogen (N) and phosphorus (P). In fact, the Earth can be considered a closed reaction system, and the various factors that disrupt the biogeochemical cycles have repercussions on the planet’s equilibrium. For example, the increase in the greenhouse effect and global warming is caused in great part by disturbances to the carbon cycle related to human activity. Coal and oil have been present in the ground for hundreds of millions of years, and humans have harnessed these resources as fuel. These fossil fuels are carbon compounds derived from plant decomposition. Their combustion releases carbon in the form of carbon dioxide (CO2) gas into the Earth’s atmosphere. According to Le Chatelier’s principle, an increase in the concentration of carbon dioxide favours the photosynthesis reaction. However, the system can only partially oppose the increase in atmospheric carbon dioxide.

Figure 24 The scientific community is concerned about climate change, which seems to be causing an increasing number of natural disasters, such as flooding.

Since the beginning of the industrial era, human activity has produced a surplus of carbon, which the oceans and forests cannot sufficiently absorb. Moreover, intensive livestock production to meet human food needs has led to a considerable increase in emissions of methane (CH4), another greenhouse gas. The increase in the level of carbon dioxide in the atmosphere, as well as other greenhouse gases, is threatening the fragile equilibrium of the planet and is leading to worrisome weather disturbances. Consequently, it is important to develop new sources of renewable non-polluting energy as soon as possible. The intensification of human activity also has an impact on the biogeochemical cycles of phosphorus (P) and nitrogen (N). The disequilibrium of these cycles is largely caused by the massive use of fertilizers on agricultural land, which contributes to the eutrophication of watercourses (see Figure 25).

*

A gradual process * Eutrophication of enrichment of an aquatic environment with nutrients like phosphorus (P) and nitrogen (N).

Figure 25 Eutrophication turns the water in this lake green because it contains an abnormally high quantity of algae.

The leaching of agricultural fertilizers also changes the chemical equilibrium of the water and may be the cause of the coral bleaching phenomenon (see Figure 26). With coral providing an environment for over 4000 species of fish, this phenomenon is threatening the survival of several species, particularly those which use coral as a site for hatching and protecting their offspring.

Figure 26 The phenomenon of bleaching can kill immense coral reefs.

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APPLICATIONS Halogen light bulbs Most of the light bulbs that illuminate our homes are incandescent bulbs, which means that they light up when the metallic filament inside them is heated. Incandescent halogen bulbs, which are used for automobile headlights, for example, last two to four times longer than conventional incandescent bulbs. An incandescent bulb consists of a glass globe and a filament made of tungsten (W), a metal that can withstand very high temperatures without melting (see Figure 27). The melting point of tungsten is 3422°C, the highest of all the metals. The air inside a conventional incandescent bulb is replaced with an inert gas such as krypton (Kr) or argon (Ar). It is vital to replace the air with one of these gases, because the filament would burn if it came into contact with oxygen in the air. Glass bulb

Tungsten filament

Gas

Lead wires

Support wires

Figure 27 The components of an incandescent light bulb

When the bulb is turned on, the friction between electrons from the lead wires and atoms in the filament heats the filament to almost 2500°C. The filament then glows with a strong light. However, heating the filament to this high temperature causes its sublimation. This means that every time the filament is heated, gaseous tungsten escapes from it. When the bulb cools, the gaseous tungsten returns to a solid state and is deposited at the bottom of the globe. This physical transformation of tungsten between the solid and gaseous phases is a reversible trans-

300

UNIT 4 Chemical Equilibrium

formation that occurs at equilibrium. The result is that the tungsten filament gets a little thinner every time the bulb is used. When the filament breaks, the light bulb no longer lights. This problem is offset with incandescent halogen light bulbs. The inert gas in these bulbs is replaced with a halogen gas, usually iodine (I), bromine (Br) or fluorine (F). When the incandescent halogen light bulb is turned on, the gaseous tungsten lost from the filament combines with the halogen gas in the globe in a chemical reaction. This reaction is also reversible therefore a state of equilibrium is quickly reached inside the globe. Since the direct reaction is exothermic, the very high temperature in the light bulb causes a shift in equilibrium toward the reaction that can absorb the excess energy, that is, the reverse reaction. This causes gaseous tungsten to form, and some of it is deposited on the filament instead of falling to the bottom of the globe. This is why incandescent halogen bulbs last longer than conventional incandescent bulbs. However, the tungsten atoms are deposited randomly on the filament, which becomes thinner in some places; these fragile areas eventually give way, and then the bulb stops working.

Figure 28 An incandescent halogen light bulb

Furthermore, using a halogen gas allows the filament to work at a higher temperature than in a conventional incandescent bulb, as high as 2900°C. This makes the light whiter and brighter. It is more similar to the light from the Sun, which makes it more suitable for human vision. That is why incandescent halogen bulbs are used in reading lamps and for work requiring a high degree of precision.

The Nobel Prize Alfred Nobel (1833–1896) was a Swedish chemist and engineer who invented dynamite, an explosive based on nitroglycerine, in 1867 (see Figure 29). Nobel amassed a substantial fortune selling explosives and producing oil. One day, a French newspaper erroneously published the news of his death and condemned his invention of dynamite. When Nobel read Figure 29 Alfred Nobel that “the merchant of death is dead,” he decided to leave a better legacy. In his will, he made provisions for an institution to be set up after his death to award prizes of about $1.4 million. The Nobel Foundation was established in 1900.

Since the beginning of the last century, three Nobel chemistry prizes have been awarded to scientists who studied chemical equilibrium. In 1909, Friedrich Wilhelm Ostwald was rewarded for his work on catalysis and his research on the fundamental principles of chemical equilibrium and reaction speed. Fritz Haber won the Nobel Prize in 1918 for his work leading to the development of the Haber Process, based on the principles governing chemical equilibrium. The process is used to extract nitrogen gas (N2) from the atmosphere in the form of liquid ammonia (NH3), which can then be used to make a variety of substances, including agricultural fertilizers (see the History highlights box on page 289).

Every year the Nobel Foundation rewards those who have benefited humankind by contributing to significant advancement in a variety of fields. The prizes are awarded in five areas of knowledge and culture: chemistry, physics, physiology or medicine, literature and diplomacy or peace. The Nobel Prize in Chemistry is awarded by the Royal Swedish Academy of Sciences, which receives thousands of applications every year. Many eminent scientists die before being recognized for an important discovery, since a minimum of 20 years must elapse between the time of the discovery and the awarding of the prize. Nobel had stipulated in his will that candidates and their discoveries had to have withstood the test of time. Only a few women have been awarded the Nobel Prize in Chemistry. In 1911, Marie Curie was the first woman to receive the prize for her discovery of radium and polonium. In 1935, Irène Joliot-Curie, Marie Curie's daughter, received this prestigious prize, along with her husband, Frédéric Joliot, for their discovery of artificial radioactivity (see Figure 30). In 1965, Dorothy Crowfoot Hodgkin was the third woman to receive the Nobel Prize in Chemistry for her identification of the structure of important biological substances including cholesterol and penicillin (see Figure 31).

Figure 30 Irène Joliot-Curie

Figure 31 Dorothy Crowfoot Hodgkin

Finally, in 1931 Carl Bosch shared the Nobel Prize in Chemistry with Friedrich Bergius for the development of high-pressure chemistry. Bosch made it possible for chemical reactions to occur at pressures of over 101.3 kPa and went on to supervise the first industrial application of the Haber Process, which then became known as the Haber-Bosch Process. With these technological advances, the industrial production of fertilizer helped feed the world more efficiently (see figures 31 and 32).

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CHAPTER

11

The Qualitative Aspect of Chemical Equilibrium

11.1 Static equilibrium and dynamic equilibrium • Static equilibrium is the state of that which remains at a given point or which is kept immobile. • Dynamic equilibrium is the result of two opposing processes occurring at the same rate so that no visible change takes place in the reaction system. • Phase equilibrium (or physical equilibrium) is a dynamic equilibrium in which a single substance is found in several phases within a system following a physical conversion. • Solubility equilibrium is a state in which a solute is dissolved in a solvent or solution, and in which an excess of this solute is in contact with the saturated solution. • Chemical equilibrium is a dynamic equilibrium resulting from two opposed chemical reactions that occur at the same rate, which leave the composition of the reaction system unchanged.

11.2 Irreversible and reversible reactions • An irreversible reaction is a reaction that can occur only in one direction. These reactions are characterized by a complete transformation into products of at least one of the reactants in the system. CH4 (g)  2 O2 (g) n CO2 (g)  2 H2O (g) Irreversible reaction • A reversible reaction is a chemical reaction that can occur only in one direction (direct reaction) or the other (reverse reaction). In these reactions, the reactants are transformed into products and the products are transformed into reactants. Reversible reactions are recognizable by the simultaneous presence of reactants and products in the reaction system. z 2 HI (g) H2 (g)  I2 (g) y Reversible reaction

11.3 Necessary conditions for attaining equilibrium • Three conditions must be met for a system to attain equilibrium: – the reaction must be reversible – the reaction must take place in a closed system – the macroscopic properties of the system must remain constant • In the case of chemical equilibrium, the conversion must be a reversible chemical reaction.

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UNIT 4 Chemical Equilibrium

11.4 Le Chatelier’s principle • According to Le Chatelier’s principle, if the conditions of a system at equilibrium are changed, the system will react to partially oppose this change until a new state of equilibrium has been attained. • When the conditions of a system are changed, Le Chatelier’s principle makes it possible to qualitatively predict the direction of the shift in equilibrium.

11.5 Factors that affect the state of equilibrium • Three factors affect the state of chemical equilibrium of a system: the concentration of the reactants or products, temperature and pressure. • According to Le Chatelier’s principle, in a system at equilibrium: − an increase in the concentration of a substance shifts the equilibrium in favour of the reaction that will allow a decrease in the concentration of this substance − a decrease in the concentration of a substance shifts the equilibrium in favour of the reaction that will allow an increase in the concentration of this substance − any increase in temperature shifts the equilibrium in favour of an endothermic reaction − any decrease in temperature shifts the equilibrium in favour of an exothermic reaction − any increase in pressure due to a decrease in volume shifts the equilibrium in favour of the reaction that produces the smallest number of molecules of gas − any decrease in pressure due to an increase in volume shifts the equilibrium in favour of the reaction that produces the greatest number of molecules of gas − a change in pressure has an influence only if the system contains at least one substance in the gaseous phase • Adding a catalyst makes it possible for a system to attain a state of equilibrium more quickly, but does not have an effect on this equilibrium.

a) z 2 NO2 (g) N2O4 (g) y

b) N2O4 (g) z y 2 NO2 (g)

c) N2O4 (g) z y 2 NO2 (g) Nitrogen (N) Oxygen (O)

11.6 Chemical equilibrium in everyday life • The Earth can be considered a closed system in a state of dynamic equilibrium. • The equilibrium of biogeochemical cycles is fragile; it is threatened, among other things, by the increase in the production of greenhouse gases and global warming.

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The Qualitative Aspect of Chemical Equilibrium

1. The following reaction is in a state of equilibrium: z2 B2 C 4 Ay Observe the following three graphs below and answer the questions that follow: Compound A Compound B Compound C A

Initial equilibrium

Change

New equilibrium

a) Which graph represents the change in the concentrations of compounds A, B and C following an addition of compound A? b) Which graph represents the change in the concentrations of compounds A, B and C following a removal of compound B? c) Which graph represents the change in the concentrations of compounds A, B, and C following an addition of compound C? 2. Observe the reaction at equilibrium below and answer the questions that follow:

Concentration

z C (s)  2 H2 (g) CH4 (g) y

Time Initial equilibrium

Change

New equilibrium

Time Initial equilibrium

Change

z H2S (g)  energy H2 (g)  S (s) y Name the different factors that have an effect on the state of equilibrium of this system. Time

304

New equilibrium

4. When food decomposes, the different proteins in the food also decompose. During this process, the proteins that contain sulphur produce hydrogen sulfide (H2S). This compound produces most of the unpleasant odour released by decomposing food. The formation of hydrogen sulfide from its element components is described by the following reversible reaction:

Concentration

C

What reaction will be favoured following each of the changes below? a) A decrease in the concentration of methane (CH4). b) The removal of a small quantity of carbon (C). c) A decrease in temperature. d) A decrease in volume. e) An increase in hydrogen (H2). f) The addition of a catalyst. g) A decrease in pressure. h) An increase in temperature. i) The addition of methane (CH4). 3. Phase equilibrium and chemical equilibrium are phenomena that are both commonly observed in chemistry. a) Explain what these two phenomena have in common. b) Explain what makes them different.

Concentration

B

H  74.4 kJ/mol

UNIT 4 Chemical Equilibrium

5. Are the following statements true or false? Explain your answer for each statement. a) In a system in a state of equilibrium, the rate of formation of products is nil. b) A system whose total mass of reactants is equal to the total mass of products is a system in a state of equilibrium. c) A closed beaker containing oxygen (O2) gas cannot attain a state of equilibrium. d) It is impossible for a reaction occurring in an open container to attain a state of equilibrium. e) All reversible reactions occurring in an aqueous solution attain a state of equilibrium. f) A system whose temperature is constant has attained a state of equilibrium. 6. Among many reptiles, the sex of the embryo does not depend on chromosomes, but rather on ambient temperature. For example, among alligators, sex is determined by the temperature of the eggs during the second and third week of incubation. A temperature of 30°C or less during this period results in alligators of the female sex, while eggs incubated at over 34°C result in males. The mother alligator’s choice of nesting areas ensures an equilibrium between male and female newborns. What do you think the long-term consequences of global warming could be for the equilibrium of the ratio of male to female alligators? 7. A chemistry Internet site presents a new process for increasing the industrial production of ammonia. The production of ammonia (NH3) from nitrogen (N2) and hydrogen (H2) is described by the following equation: z 2 NH3 (g) N2 (g)  3 H2 (g) y The site mentions that this process is based on the use of a new catalyst that makes it possible to shift the equilibrium toward the production of ammonia. What do you think of the new process presented on the Internet site?

9. In a laboratory, a reaction is carried out between compound A and compound B with the goal of producing compound C. By observing the reaction, it is noted that the colour of the reaction mixture changes from pale blue to dark blue in a few minutes. After 10 minutes, the reaction mixture is still dark blue and it can be concluded that the state of equilibrium has been attained. The reaction of the formation of compound C, from compounds A and B, is expressed in a state of equilibrium by the following equation: A(g)  B(g) pale blue

colourless

z C(g) y dark blue

a) What reaction will occur by the addition of compound C in the reaction mixture? What will the colour of this mixture be when a new state of equilibrium is attained? b) A mystery compound X is added to the reaction mixture. After a few minutes, the colour of the mixture has changed from pale blue to dark blue. If compound X is neither compound A, nor compound B, nor compound C, and the temperature and pressure of the system have not changed, how can you explain the change in the colour of the mixture from pale blue to dark blue? Explain your answer using Le Chatelier’s principle. 10. The dissolution of cobalt chloride (CoCl2) in hydrochloric acid (HCl) attains a state of equilibrium represented by the following equation: z CoCl42 (aq)  6 H2O (l) Co(H2O)62 (aq)  4 Cl (aq) y pink

blue

Since Co(H2O)62 (aq) is pink and CoCl42(aq) is blue, the shift in equilibrium is easily observable by the change in colour of the solution. Design an experiment to determine if the direct reaction is endothermic or exothermic.

8. In ecosystems, the interaction between carnivorous and herbivorous animals is often a predator-prey relationship in which the carnivores are the predators and the herbivores are the prey. However, in most ecosystems, predators are in equilibrium with prey. Is this a dynamic equilibrium? Explain your answer.

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 12.

The shells of chicken eggs are largely composed of calcium carbonate (CaCO3). The formation of calcium carbonate from calcium ions (Ca2) and carbonate ions (CO32) is described by the following equation: z CaCO3 (s) Ca2 (aq)  CO32 (aq) y Carbonate ions are supplied through the carbon dioxide (CO2) produced by the chicken’s cellular metabolism. The carbon dioxide is converted into carbonic acid (H2CO3) through an enzyme called carbonic anhydrase, which catalyzes the following reaction: z H2CO3 (aq) CO2 (g)  H2O (l) y The carbonic acid is then converted into carbonate ions according to the following equation: z 2 H(aq)  z CO32 (aq) H2CO3 (aq) y a) Why are the shells of chicken eggs thinner in the summer, when chickens must pant to cool off? Use Le Chatelier’s principle to explain your answer. b) Suggest a way to resolve the problem of thinning eggshells.

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UNIT 4 Chemical Equilibrium

13.

Carbon dioxide (CO2) is formed in the air when carbon monoxide (CO) combines with oxygen (O2). At equilibrium, this reaction is represented by the following equation: z 2 CO2 (g)  energy 2 CO (g)  O2 (g) y The graph below represents the changes in concentration of the different compounds involved in the synthesis of carbon dioxide as a function of time:

Concentration

11. The production of sulphur trioxide (SO3) is an important process for industrial purposes. Sulphur trioxide can be converted to obtain sulphuric acid (H2SO4), a compound used for, among other things, the manufacturing of fertilizers and oil refinement. Sulphur trioxide gas is formed from sulphur dioxide (SO2) gas and oxygen (O2) gas. This reaction occurs at 400°C and is exothermic. In an industrial context, a catalyst is added to increase the yield of sulphur trioxide. a) Write the balanced equation that illustrates this reaction in a state of equilibrium and indicate the energy produced. b) What reaction will be favoured following an increase in the temperature of the system? c) What reaction will be favoured following a decrease in the pressure of the system? d) What effect will removing the catalyst have on the shift in equilibrium? e) What reaction will be favoured following the addition of sulphur dioxide to the system? f) What reaction will be favoured following the removal of sulphur dioxide from the system?

CO2 (g) CO (g) O2 (g)

A

B

C

D

Time

What are the changes imposed on the system at the four times indicated by the letters A, B, C and D?

14.

In the human body, the transportation of oxygen (O2) in the blood is ensured by a protein called hemoglobin (Hb). This protein makes it possible for red blood cells to transport oxygen from the lungs to the cells that compose the various tissues in the body. The bond between the oxygen and the hemo globin and the dissociation of the oxygen from the hemoglobin are represented by the following equation: Direct reaction toward the tissues

z Hb (aq)  O2 (g) HbO2 (aq) y Reverse reaction toward the lungs

A number of factors, including temperature, influence oxygen’s capacity to bond with hemoglobin. The active tissue in the body is usually hotter than the inactive tissue, and active tissue has a greater need for oxygen than does inactive tissue. Using this information, answer the following questions: a) Is the direct reaction exothermic or endothermic? Explain your answer. b) Body temperature decreases slightly during the night. What effect will this decrease have on oxygen’s capacity to bond with hemoglobin? Does this effect seem logical to you? Why or why not?

The Quantitative Aspect of Chemical Equilibrium

T

he equilibrium of the dissolution of carbon dioxide (CO2) in sea water is essential for many marine organisms. For example, corals must synthesize calcium carbonate (CaCO3), also known as limestone, in order to form their skeletons. The increased concentration of atmospheric carbon dioxide due to human activity could disrupt the equilibrium needed for coral skeletal formation, since carbon dioxide acidifies sea water as it dissolves in it. The acidification of our oceans reduces the quantity of calcium carbonate available, which threatens this species.

This chapter will focus on the state of equilibrium from a quantitative perspective, by establishing a relationship that can be used to predict the concentrations of substances at equilibrium. This relationship, called the equilibrium constant, will also be studied in particular cases involving water and the electrolytic dissociation of acids, bases and ionic compounds in solutions.

Review Representations of atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Electrolytes and electrolysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Measuring a physical change using the pH scale. . . . . . . . . . . . . . 20 Acid-base reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Synthesis, decomposition and precipitation. . . . . . . . . . . . . . . . . . 25

12.1 12.2

Equilibrium constant . . . . . . . . . . . . . . . . . .308 Ionic equilibrium in solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .320

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12.1 Equilibrium constant The equilibrium constant (Kc) is also called the equilibrium law. It is a relationship that shows that at a given temperature, in any simple chemical reaction at equilibrium, there is a constant ratio between the product concentration and the reactant concentration, each concentration being raised to a power corresponding to the stoichiometric coefficient. See Static equilibrium and dynamic equilibrium, p. 278.

Chemical equilibrium is a dynamic equilibrium resulting from two opposing chemical reactions that occur at the same rate, thereby leaving the composition of the reaction system unchanged. This equilibrium takes a certain amount of time to be established. In some instances, only the reactants may be present in the reaction system initially. Their concentration is therefore at their maximum, while product concen tration is non-existent, since products have not yet been formed. The direct reaction then begins and the reactants are converted into products according to the stoichiometric ratios of the balanced equation. As soon as the products are formed, the reverse reaction begins and they convert back into the initial reactants. However, as time passes, the concentration of the reactants decreases while that of the products increases, until equilibrium has been attained. At this point, the concentrations of the reactants and the products are constant and the direct reaction rate is equal to that of the reverse reaction (see Figure 1). Variation in a system's direct reaction rate and its reverse reaction rate

Reaction rate

A (g)  B (g) n C (g)

C (g) n A (g)  B (g)

Time

Figure 1 In a system that is at chemical equilibrium, the direct reaction rate is equal to that of the reverse reaction.

If the expressions of the direct and reverse reaction rates are known, it is possible to deduce the mathematical expression of the equilibrium constant. At the same rate, the system converts a few of the reactant particles into products and a few of the product particles into reactants. However, the concentrations of the reactants and the products are not necessarily equal.

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UNIT 4 Chemical Equilibrium

12.1.1

Expressing the equilibrium constant

According to the rate law, the ratio of the reaction rate to the concentration of reactants, raised to a power corresponding to the stoichiometric coefficient of the balanced equation, is always constant at a given temperature.

See Measuring Reaction Rate, p. 213.

This law can be applied to the reactants in direct and reverse reactions in a system at equilibrium. For example, the synthesis of hydrogen iodide (HI) gas from hydrogen (H2) gas and iodine (I2) vapour is a simple reaction which, at equilibrium, is represented by the following equation: x 2 HI (g) H2 (g)  I2 (g) w In this system at equilibrium, the reactants in the direct reaction are hydrogen and iodine. The reactant of the reverse reaction is hydrogen iodide. Since these are simple reactions, the direct reaction rate (rdir) and reverse reaction rate (rrev) can be described according to the rate law by using the direct rate constant (kdir) and the reverse rate constant (krev) as follows: rrev  krev [HI]2

rdir  kdir  [H2]  [ I2]

Given that at equilibrium these two rates are equal, the following relationship is obtained: rdir  rrev kdir  [H2]  [ I2]  krev  [HI]2 By rearranging this relationship, a ratio is obtained between the two rate constants that is translated by the following equation: kdir [HI]2  krev [H2]  [I2] The ratio between the two rate constants is also a constant. Consequently, this ratio can be replaced by the equilibrium constant (Kc).

Kc 

[HI]2 [H2]  [I2]

This value is called the equilibrium constant of the reaction system. The subscript “c” rather than “eq” is used to designate this since it is molar concentrations that are indicated inside the square brackets of the equation. To simplify this, the constant is expressed as a numerical value without units.

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Since each system at equilibrium involves a different chemical reaction, it is possible to formulate the equilibrium law so that it applies to all systems. Equilibrium constant For a reversible simple reaction at equilibrium, at a given temperature, the value of the equilibrium constant is equal to the product of the molar concentrations of the products divided by the product of the molar concentrations of the reactants, with each concentration of substance being attributed an exponent equal to its whole number stoichiometric coefficient in the balanced equation of this reaction. The expression of the equilibrium constant or the equilibrium law for a z c C  d D is as follows: hypothetical simple reaction of the type aA  b B y

Equilibrium constant Kc 

[C]c  [D]d [A]a  [B]b

where Kc  Equilibrium constant as a function of concentrations [C], [D]  Product concentrations at equilibrium, expressed in moles per litre (mol/L) [A], [B]  Reactant concentrations at equilibrium, expressed in moles per litre (mol/L) c, d  Whole number stoichiometric coefficients of the products in the balanced chemical equation a, b  Whole number stoichiometric coefficients of the reactants in the balanced chemical reaction

The value of the equilibrium constant is calculated by using only the concentrations of gaseous substances or substances in solutions. The concentration of pure substances in solid or liquid states remains constant as the reaction takes its course. Since these concentrations have fixed values, they do not modify the value of the equilibrium constant, which is only sensitive to changes in concentrations. Consequently, in a balanced chemical equation, the concentrations of substances designated by the subscripts “solid” and “liquid” are not taken into account in the expression of the equilibrium constant. By expressing the equilibrium constant using product concentrations as the numerator and reactant concentrations as the denominator, the value of the equilibrium constant of the direct reaction (Kcdir) is obtained.

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UNIT 4 Chemical Equilibrium

To determine the value of the equilibrium constant of the reverse reaction (Kcrev), we calculate the mathematical reverse of the equilibrium constant of the direct reaction (Kcdir). Equilibrium constant of the reverse reaction 1 Kcrev  Kcdir where Kcrev  Equilibrium constant of the reverse reaction Kcdir  Equilibrium constant of the direct reaction

The following examples show how to determine the value of the equilibrium constant as well as the product or reactant concentrations at equilibrium: Example A During the synthesis of ammonia (NH3) in the gaseous phase at 472°C, 1.207 moles of hydrogen (H2), 0.4020 moles of nitrogen (N2) and 0.0272 moles of ammonia at equilibrium are measured in a 10.0-L container. Calculate the equilibrium constant of the synthesis of ammonia at this temperature and the equilibrium constant of the decomposition of ammonia at this temperature, if the balanced equation of this reaction at equilibrium is as follows: x 2 NH3 (g) N2 (g)  3 H2 (g) w Data: nH2  1.2070 mol nN2  0.4020 mol nNH3  0.0272 mol V  10.0 L kcsyn  ? kcdec  ?

1. Calculation of concentration at equilibrium: n C V 1.2070 mol  0.120 70 mol/L 10 L 0.4020 mol [N2]   0.040 20 mol/L 10 L 0.0272 mol [NH3]   0.002 72 mol/L 10 L [H2] 

2. Calculation of the equilibrium constant of the synthesis of ammonia: [C]c  [D]d [A]a  [B]b [NH3]2 (0.002 72)2 Kcsyn    0.1031 [N2]  [H2]3 0.040 20  (0.120 70)3 Kc 

3. Calculation of the equilibrium constant of the decomposition of ammonia: 1 Kcdir 1 1 Kcdec    9.699 Kcsyn 0.1031 Kcrev 

Answer: The equilibrium constant of the synthesis of ammonia (NH3) is 0.103 and that of the decomposition of ammonia is 9.7.

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Example B Iodine (I2) and bromine (Br2) react to form iodine bromide (IBr) according to the following equation: x 2 IBr (g) I2 (g)  Br2 (g) w At 250°C, a mixture at equilibrium in a 2.0-L round bottom flask contains 0.024 moles of iodine and 0.050 moles of bromine. What is the concentration of iodine bromide if the value of the equilibrium constant is 120.33? Data: nI2  0.024 mol nBr2  0.050 mol V  2.0 L Kc  120.33 [I2]  ? [Br2]  ? [IBr]  ?

1. Calculation of the concentrations of iodine and bromine at equilibrium: n C V 0.024 mol  0.012 mol/L [I2]  2.0 L 0.050 mol [Br2]   0.025 mol/L 2.0 L 2. Calculation of the concentration of iodine bromide: [C]c  [D]d [A]a  [B]b [IBr]2 Kcsyn  [I2]  [Br2] Kc 

[IBr]2  Kcsyn  [I2]  [Br2] [IBr]  120.33  0.012 mol/L  0.025 mol/L  0.190 mol/L Answer: The concentration of iodine bromide (IBr) is 0.19 mol/L.

12.1.2

Interpreting the value of the equilibrium constant

The value of the equilibrium constant provides a better understanding of the course of a reaction at equilibrium. Given that the product concentration is divided by the reactant concentration, knowing the value of the constant makes it possible to deduce the extent of the reaction (see Figure 2).

Reactants

Products

Reactants Products

Products

Reactants

Reactants Products Kc  1 [Reactants]  [Products] a) The direct reaction is favoured.

Kc  1 [Reactants]  [Products] b) No reaction is favoured.

Kc  1 [Reactants]  [Products] c) The reverse reaction is favoured.

Figure 2 The value of the equilibrium constant Kc indicates the extent of a chemical reaction. A high value of the equilibrium constant indicates that the system favours the direct reaction, while a low value of the equilibrium constant indicates that the system favours the reverse reaction.

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Generally, if the value of the equilibrium constant is high (Kc  1), it means that the products of the direct reaction are primarily found at equilibrium. In this case, the position of equilibrium is on the right side of the equation, that is, the reaction favours the formation of products. Therefore, the greater the value of the equilibrium constant, the more the system will tend to favour the direct reaction, and the more the products are favoured at equilibrium. Similarly, if the value of the equilibrium constant is low (Kc  1), it means that the reactant concentration is higher than the product concentration. The position of equilibrium is therefore on the left side of the equation and the reaction favours the reactants. Therefore, the lower the value of the equilibrium constant, the more the system tends to favour the reverse reaction, and the more the reactants are favoured at equilibrium. A value of the equilibrium constant that is approximately equal to 1 (Kc  1), means that the concentrations of reactants and products at equilibrium are almost equal. The system favours neither the direct reaction nor the reverse reaction. However, the value of the equilibrium constant does not provide any indication of the time it takes for a reaction to attain equilibrium. The time it takes for a reaction to attain equilibrium depends on the reaction rate. This rate is determined by certain parameters, such as the activation energy value of the reaction and temperature. The equilibrium constant provides only a measurement of the position of equilibrium of a reaction.

See Activation energy, p. 173.

Generally, reactions in which the value of the equilibrium constant is greater than 1010 are considered complete reactions. Reactions in which the equilibrium constant is less than 1010 are generally considered reactions that have not occurred.

12.1.3

The effect of temperature on the value of the equilibrium constant

According to Le Chatelier’s principle, the change in concentration of a substance involved in a reaction at equilibrium temporarily disrupts the equilibrium and increases the reaction rate of this substance. However, this rate decreases as the concentration of the added substance decreases. Ultimately, the equilibrium is restored with new concentrations of reactants and products. Therefore, changes in concentrations, pressure or volume have no effect on the value of the equilibrium constant. Furthermore, experiments show that, in a given system at equilibrium, only temperature can change the equilibrium constant. This is why the temperature of a system must always be specified when providing its equilibrium constant. For example, the oxidation of carbon monoxide (CO) is an exothermic change (see Figure 3), translated by the following equation: x 2 CO2 (g)  566 kJ 2 CO (g)  O2 (g) w

See Le Chatelier’s principle, p. 288.

Figure 3 The oxidation of carbon monoxide (CO), an exothermic reaction, occurs in the catalytic converter of a scooter. This reduces polluting emissions.

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The expression of the equilibrium constant of this reaction is the following:

Kc 

[C02]2 [CO]2  [O2]

When the system is at equilibrium, at a given temperature, the concentrations of reactants and products are stable, and the constant takes on a given value. However, when the temperature of the system changes, the system reacts according to Le Chatelier’s principle to oppose the change, and the equilibrium shifts in order to minimize the effect of the change. When a system is cooled, it reacts by favouring an exothermic reaction, that is, the direct reaction. The reaction therefore produces heat to compensate for the heat lost at the time of cooling. Equilibrium shifts to the right, and the reactant concentration decreases while the product concentration increases. A new equilibrium is established according to proportions that are different from those of the initial equilibrium. This new concentration ratio changes the value of the equilibrium constant of the system. In the oxidation of carbon monoxide (CO), when the temperature is lowered, the numerator increases and the denominator decreases, and the value of the equilibrium constant increases. Generally, a decrease in the temperature of a system causes the equilibrium constant of an exothermic reaction to increase. Conversely, the same temperature decrease reduces the equilibrium constant of the system if the reaction is endothermic (see Table 1). Table 1 Effect of new temperature conditions on the value of the equilibrium constant (Kc) Type of reaction Exothermic (H  0) Reactants n products  energy Endothermic (H  0) Reactants  energy n products

12.1.4

Temperature change

Favoured reaction

Change in Kc

Increase

Reverse (m)

Decrease

Decrease

Direct (n)

Increase

Increase

Direct (n)

Increase

Decrease

Reverse (m)

Decrease

Calculating concentrations at equilibrium

The equilibrium constant (Kc) can be calculated by replacing the terms of the mathematical expression of the constant with the values of the concentrations at equilibrium of the reactants and the products. To do so, the reaction system must be left to react until it attains equilibrium. Then, certain properties of the reaction system are measured in order to determine the concentration of all of the substances present. In the case of gaseous substances, the measured properties can be pressure, colour or pH. The measured values can then be converted for the various substances of the system.

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UNIT 4 Chemical Equilibrium

Sometimes, it is impossible to experimentally determine all of the concentrations of the substances present at equilibrium. However, in some cases, unknown concentrations can be predicted algebraically if the initial concentration of the reactants and the final concentration of at least one of the products are known. It is also possible to algebraically predict the concentrations if the initial concentrations of the reactants and the value of the equilibrium constant are known. The study of a simple reaction, such as the synthesis of hydrogen iodide (HI), enables an understanding of the method used to make these kinds of predictions. When hydrogen (H2) is mixed with iodine (I2), the reaction begins quickly and equilibrium is established according to the following equation: x 2 HI (g) H2 (g)  I2 (g) w The initial and final concentrations of the substances were measured in a laboratory (see Table 2). Table 2 Concentration of the substances involved in the synthesis of hydrogen iodide (HI) at 448°C

Substance

Initial concentration (mol/L)

Final concentration (mol/L)

H2 (g)

1.00

0.22

I2 (g)

1.00

0.22

HI (g)

0.00

1.56

Using an Initial-Change-Equilibrium (ICE) table is an easy way to track changes in concentration that occur during a reaction and to calculate concentrations at equilibrium. The data in Table 2 is recorded in an ICE table, and then the unknown data is deduced (see Table 3). Table 3 Data in Table 2 recorded in an ICE table

w x 2 HI (g)

Concentration (mol/L)

H2 (g)

Initial

1.00

1.00

0.00

Change

?

?

?

Equilibrium

0.22

0.22

1.56



I2 (g)

The study of Table 3 allows us to assert that if 0.22 mol/L of hydrogen remains at equilibrium, the change that occurred during the reaction was 0.78 mol/L, that is, the concentration at equilibrium minus the initial concentration: 0.22 mol/L  1.00 mol/L  0.78 mol/L. Since the concentration changes respect the stoichiometric coefficients of the balanced equation, the change is also 0.78 mol/L for the iodine and 2  0.78 mol/L  1.56 mol/L for the hydrogen iodide. Given that during the course of the reaction, the reactants are consumed and the products are formed, the change in concentration of the reactants is negative and that of the products is positive. The unknown values on the Change line of the ICE table are therefore, respectively, 0.78 mol/L, 0.78 mol/L and 1.56 mol/L.

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Therefore, if the initial concentrations and at least one other indicator are known, it is possible to use this method, involving an ICE table, to deduce the concentrations at equilibrium. The data is then used to calculate the equilibrium constant. The following examples show how to calculate final concentrations in various situations: Example A Calculation of the equilibrium constant using initial concentrations and the concen tration of a substance at equilibrium At a given temperature, 10 moles of nitrogen oxide (NO) and 8 moles of oxygen (O2) are placed in a 2-L container. After a given period of time, the following equilibrium is obtained: x 2 NO2 (g) 2 NO (g)  O2 (g) w Once equilibrium is attained, only 4 moles of oxygen remain. Calculate the equilibrium constant. Data: nNOi  10 mol nO2 i  8 mol nO2 eq  4 mol V2L Kc  ?

1. Calculation of concentrations at equilibrium: n C V 10 mol [NO]i   5 mol/L 2L 8 mol [O2]i   4 mol/L 2L 4 mol [O2]eq   2 mol/L 2L 2. Recording of data and use of the ICE table: Concentration (mol/L)

2 NO (g) 

Initial (Ci) Change (C) Equilibrium (Ceq)

A

O2 (g)

5

B

4

w x 2 NO 2 (g)

4

A

C 1

2

2

0

B

4

C 4

First, the change in concentration of the oxygen is calculated. [O2]  [O2]eq  [O2]i  2 mol/L  4 mol/L  2 mol/L Given that oxygen is a reactant, its change is negative (2 mol/L).

B The changes of the other substances are deduced based on the stoichiometric ratios, which are 2:1:2, therefore 4, 2, 4. Given that nitrogen oxide is a reactant, its change is negative (4) while the change in the nitrogen oxide, a product, is positive (4). C The concentrations at equilibrium of each substance are calculated by adding the initial concentration and the change in the initial concentration, while taking into account the sign of the value of the concentration change. Ceq  Ci  C [NO]eq  [NO]i  [NO]  5  4  1 mol/L [NO2]eq  [NO2]i  [NO2]  4  2  2 mol/L 3. Calculation of the equilibrium constant: [NO2]2 (4)2 Kc   8 [NO]2  [O2] (1)2  2 Answer: The value of the equilibrium constant at this temperature is 8.

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UNIT 4 Chemical Equilibrium

Example B Calculation of concentrations at equilibrium using initial concentrations, which are different for each substance, and the equilibrium constant. At 1100 K, the equilibrium constant of the following reaction is 25. x 2 HI (g) H2 (g)  I2 (g) w 2 moles of hydrogen (H2) and 3 moles of iodine (I2) are placed in a 1-L container. What is the concentration of each substance when the reaction attains equilibrium at 1100 K? Data: nH2 i  2 mol nI2 i  3 mol V1L

1. Calculation of initial concentrations: n 2 mol  2 mol/L C [H2]i  V 1L 2. Recording of data and use of the ICE table: Concentration (mol/L)

H2 (g)

[H2]eq  ?

Initial (Ci)

2

[I2]eq  ?

Change (C)

[HI]eq  ?

Equilibrium (Ceq)

Kc  25.0

A

x

B 2x

I2 (g)



x

3x

3 mol  3 mol/L 1L

w x 2 HI (g)

3 A

[I2]i 

0 A

2x

B 2x

A The change in the unknown concentration is written as x for the reactants and 2x for the products according to the stoichiometric ratios, which are 1:1:2. B The concentrations at equilibrium are determined as a function of the change x. Ceq  Ci  C [H2]eq  [H2]i  [H2]  2  x [I2]eq  [I2]i  [I2]  3  x 3. Calculation of concentrations at equilibrium for each substance: [HI]2 Kc  [H2]  [I2]

25 

[HI]eq  [HI]i  [HI]  0  2x  2x

(2x)2 (2  x)  (3  x)

This second-degree equation is of the type ax 2  bx  c  0. Therefore, it must be rewritten as a quadratic equation. 25(2  x)  (3  x)  (2x)2 25(6  5x  x 2)  4x 2 150  125x  25x 2  4x 2 21x 2  125x  150  0 The possible values of x, can be found using: x x

b 

b 2  4ac 2a

(125) 

(125)  (125)2  4(21  150) (125)2  4(21  150)  1.71  4.29 or x  2(21) 2(21)

The value x  4.29 is impossible because it would provide a negative concentration of the products at equilibrium. It is therefore rejected. The value x  1.71 is used to calculate concentrations at equilibrium. 4. Calculation of the concentrations of each of the substances at equilibrium: Ceq  Ci  C [H2]eq  [H2]i  [H2]  2  x  2.00  1.67  0.33 mol/L [I2]eq  [I2]i  [I2]  3  x  3.00  1.67  1.33 mol/L [HI]eq  [HI]i  [HI]  0  2x  2x  2(1.67)  3.38 mol/L Answer: At equilibrium, the concentration of hydrogen (H2) is 0.33 mol/L, that of iodine (I2) is 1.33 mol/L and that of hydrogen iodide (HI) is 3.38 mol/L.

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SECTION 12.1

Equilibrium constant

1. Write the expression of the equilibrium constant of the reactions described by the following equations: x CO (g)  H2O (g) a) CO2 (g)  H2 (g) w xSn  2 CO b) SnO  2 CO w 2 (s)

(g)

(s)

2 (g)

x NO2 (g)  O2 (g) c) NO (g)  O3 (g) w d) 2 H2O (g) w x 2 H2O (l)  O2 (g) x 2 Hg  O e) 2 HgO w (s)

(l)

2 (g)

2. The following reaction occurs in a sealed round bottom flask at 250°C: x PCl3 (g)  Cl2 (g) PCl5 (g) w At equilibrium, the gases contained in the round bottom flask have the following concentrations: – [PCl5]  1.2  102 mol/L – [PCl3]  1.5  102 mol/L – [Cl2]  1.5  102 mol/L Calculate the value of the equilibrium constant at 250°C. 3. Hydrogen sulfide (H2S) is an acrid and toxic gas. At 1400 K, in a mixture at equilibrium, note the presence of 0.013 mol/L of hydrogen (H2), 0.046 mol/L of sulphur (S2) and 0.18 mol/L of hydrogen sulfide according to the following reaction: x 2 H2 (g)  S2 (g) 2 H2S (g) w Calculate the value of the equilibrium constant of the reaction at 1400 K. 4. At equilibrium, a hermetically sealed 1-L container holds 4 moles of hydrogen (H2), 6 moles of fluorine (F2) and 3 moles of hydrogen fluoride (HF), all in a gaseous state. This reaction is represented by the following equation: x 2 HF (g) H2 (g)  F2 (g) w Calculate the value of the equilibrium constant of the reaction. 5. The value of the equilibrium constant for the synthesis of nitrogen monoxide (NO) is 1  1030 at 25°C according to the following reaction: x 2 NO (g) N2 (g)  O2 (g) w Calculate the value of the equilibrium constant for the decomposition reaction of NO.

318

UNIT 4 Chemical Equilibrium

6. At 427°C, a 1-L round bottom flask contains, at equilibrium, 20 moles of hydrogen (H2), 18 moles of carbon dioxide (CO2), 12 moles of water vapour (H2O) and 5.9 moles of carbon monoxide (CO) according to the following reaction: x CO (g)  H2O (g) CO2 (g)  H2 (g) w Calculate the value of the equilibrium constant of the reaction. 7. A 1-L container holds a system at equilibrium at 55°C formed of nitrogen dioxide (NO2) and nitrogen tetroxide (N2O4) according to the following reaction: x N2O4 (g) 2 NO2 (g) w If the equilibrium constant is 1.15 and the concentration of the nitrogen oxide at equilibrium is 0.05 mol/L, calculate the concentration of nitrogen tetroxide at equilibrium. 8. Place the following reactions in increasing order of their tendency to form products. x 2 NO (g) a) N2 (g)  O2 (g) w Kc  4.7  1031 w 2 NO b) 2 NO  O x K  1.8  106 (g)

2 (g)

2 (g)

x 2 NO2 (g) c) N2O4 (g) w

c

Kc  0.025

9. Determine whether each of the following reactions occurs completely or not at all. x2 NCl3 (g) a) N2 (g)  3 Cl2 (g) w Kc 3.0  1011 xC2H6 (g) H2 (g) b) 2 CH4 (g)w Kc 9.5  1013 x N2 (g) 2 CO2 (g) c) 2 NO (g)  2 CO (g) w 59 Kc  2.2  10 10. Examine the following reaction: x 2 HCl (g) H2 (g)  Cl2 (g) w Hydrogen chloride (HCl) gas is placed in a reaction jar. The equilibrium constant is 2.4  1033 at 25°C. To what extent do you expect the mixture at equilibrium to dissociate into hydrogen (H2) and chlorine (Cl2)? Explain your answer.

11. In each of the following reversible reactions, determine if the direct reaction is favoured by high temperatures or low temperatures: x 2 NO2 (g) a) N2O4 (g) w H   59 kJ x I2 (g)  Cl2 (g) H  35 kJ b) 2 ICl (g) w w 2 CO (g)  O2 (g) c) 2 CO2 (g)  566 kJ x x H2 (g)  F2 (g) d) 2 HF (g) w H  536 kJ 12. The following is an exothermic reaction: x PCl5 (g) PCl3 (g)  Cl2 (g) w To favour the production of phosphorus pentachloride (PCl5), should the temperature be increased or decreased? 13. The following is an endothermic reaction: x NH3 (g)  H2S (g) NH4SH (s) w In which direction will the equilibrium shift if the following changes are made? a) An increase in the concentration of ammonia (NH3). b) A decrease in the concentration of hydrogen sulfide (H2S). c) A decrease in temperature. d) A decrease in the volume of the container. 14. At 25°C, the value of the equilibrium constant of the following equation is 82: x 2 ICl (g) I2 (g)  Cl2 (g) w If 0.83 mol of iodine (I2) and 0.83 mol of chlorine (Cl2) are placed in a 10-L container at 25°C, what is the concentration of each of the three gases at equilibrium? 15. At a given temperature, the value of the equilibrium constant of the following reaction is 4: x H2 (g)  F2 (g) 2 HF (g) w A 1-L reaction jar contains 0.045 moles of fluorine (F2) at equilibrium. What was the initial quantity of hydrogen fluoride (HF) in the reaction jar? 16. A chemist studies the following reaction: x NO (g)  SO3 (g) SO2 (g)  NO2 (g) w In a 1-L container, he adds 1.7  101 moles of sulphur dioxide (SO2) and 1.1  101 moles of nitrogen dioxide (NO2). The value of the equilibrium constant of this reaction at a given temperature is 4.8. What is the concentration at equilibrium of sulphur trioxide (SO3) at this temperature?

17. Phosgene (COCl2) is a very toxic gas. It was used during the Second World War, and is still used today to make pesticides, pharmaceutical products, dyes and polymers. It is prepared by mixing carbon monoxide (CO) and chlorine (Cl2) in a gaseous state according to the following equation: x COCl2 (g) CO (g)  Cl2 (g) w A 5-L container holds 0.055 moles of carbon monoxide and 0.072 moles of chlorine. At 870 K, the value of the equilibrium constant is 0.2. What is the concentration of the components of the mixture at equilibrium at 870 K? 18. At 700 K, hydrogen bromide (HBr) decomposes according to the following equation, and the value of its equilibrium constant is 4.2 109 : x H2 (g)  Br2 (g) 2 HBr (g) w A 2-L reaction jar contains 0.09 moles of hydrogen bromide which is heated to 700 K. What is the concentration at equilibrium of each of the gases? 19. The following is a general reaction: x E (g)  2 F (g) C (g)  D (g) w A 2-L container holds 1.0 moles of C and 1.0 moles of D. When equilibrium is attained, the container holds 0.1 moles of E. What is the value of the equilibrium constant? 20. A 5-L container, at 448°C, holds a mixture of 5.00  103 moles of hydrogen (H2) and 1.00  102 moles of iodine (I2). The reaction is described by the following equation: x 2 HI (g) H2 (g)  I2 (g) w Once equilibrium has been attained, the concentration of hydrogen iodide (HI) is 1.87  103 mol/L. What is the equilibrium constant? 21. A 2-L round bottom flask is filled with 4.6 moles of each of the following reactants in gaseous states: nitrogen monoxide (NO) and ozone (O3). The mixture is heated and, at equilibrium, 2.4 moles of nitrogen dioxide (NO2) gas is present. Under these conditions, the following reaction occurs: x NO2 (g)  O2 (g) NO (g)  O3 (g) w What is the value of the equilibrium constant?

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12.2 Ionic equilibrium in solutions Ionic equilibrium in solutions is a state of equilibrium established between the concentrations of the various ions following the dissociation of a chemical compound in a solution. In Senegal, Lac Retba, also called Lac Rose, contains extremely salty water. Its pink colour is due to the presence of microscopic algae, a cyanobacterium, that produces a red pigment that allows it to resist the high concentration of salt. The salt concentration is so great that a deposit is formed at the bottom of the lake where solubility equilibrium is established (see Figure 4). In the case of the salt contained in Lac Rose, which is a solid ionic compound, an equilibrium constant of the dissolution can be expressed. This constant is called the solubility product constant (Ksp).

Figure 4 The salinity of Lac Rose is so high that the salt deposited at the bottom of the lake in the form of a crust is harvested directly from the lake after being broken up with picks.

However, in aqueous solutions that are different from Lac Rose, the dissolved ions give the solution a particular character. It can, for instance, be neutral, acidic or basic. Nevertheless, in all of these cases, a given equilibrium will be attained and an equilibrium constant can be calculated. In the case of pure water, this constant is called the ionization constant of water (Kwater). In the case of acidic substances, the equilibrium constant is called the acidity constant (Ka) and in the case of basic substances, it is called the basic constant (Kb). Before studying these various equilibrium constants, it is essential to acquire a better understanding of the properties of acids and bases.

12.2.1

Theories of acids and bases

Acids and bases can be distinguished in various ways. One way is to describe their characteristic macroscopic properties and their molecular structure. However, these methods are limited and do not allow an adequate explanation or prediction of the behaviour of acids and bases. To remedy this situation, various theories have been developed over the years to better define these substances.

Arrhenius theory of acids and bases In 1887, Svante Arrhenius developed the first useful theoretical definition of acids and bases. This theory is still used today as a simplified explanation of the properties of acids and bases in solution. Arrhenius put forward his definition after having observed the tendency of acids and bases to form ions. He found that hydrochloric acid (HCl) dissolved in water dissociates into hydrogen ions (H) and chloride ions (Cl) according to the following equation: HCl (aq) n H (aq)  Cl (aq)

320

UNIT 4 Chemical Equilibrium

Arrhenius also noted that sodium hydroxide (NaOH), a base, dissolved in water dissociates to form sodium ions (Na) and hydroxide ions (OH) according to the following equation: NaOH (aq) n Na (aq)  OH (aq)

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

By observing a series of reactions involving the electrolytic dissociation of other acids and bases (see Table 4), Arrhenius noticed a tendency specific to each type of substance: in water, acids produce hydrogen ions (H), while bases produce hydroxide ions (OH). Table 4 Dissociation of some acids and bases and the ions produced according to the observations of Svante Arrhenius Acids that dissociate in water and the resulting ions

Bases that dissociate in water and the resulting ions

HBr (aq) n H (aq)  Br (aq)

LiOH (aq) n Li (aq)  OH (aq)

H2SO4 (aq) n H (aq)  HSO4 (aq)

KOH (aq) n K (aq)  OH (aq)

HClO4 (aq) n H (aq)  ClO4 (aq)

Ba(OH)2 (aq) n Ba2 (aq)  2 OH (aq)

Arrhenius published a theory that explains the nature of acids and bases in an aqueous solution. This theory was named after him. Arrhenius theory of acids and bases An acid is a substance that dissociates in water to produce hydrogen ions (H). A base is a substance that dissociates in water to produce hydroxide ions (OH). According to this theory, acids increase the concentration of hydrogen ions (H) in aqueous solutions. Consequently, the molecular formula of an Arrhenius acid must contain at least one hydrogen atom as a source of hydrogen ions. In contrast, bases increase the concentration of hydroxide ions (OH) in aqueous solutions. The molecular structure of an Arrhenius base must therefore contain at least one oxygen atom (O) and one hydrogen atom (H) that combine to form hydroxide ions. The Arrhenius theory is helpful in explaining the production of ions during the electrolytic dissociation of acids and bases. It also explains what occurs during the course of an acid-base neutralization reaction. During this reaction, an acid and a base react to form a salt and water. For example, hydrogen chloride, also called hydrochloric acid (HCl), and sodium hydroxide (NaOH) react as follows: HCl (aq)  NaOH (aq) n NaCl (aq)  H2O (l)

SVANTE ARRHENIUS Swedish chemist (1859–1927) In 1884, Arrhenius shook the world of chemistry with his doctoral thesis which proposed a theory of the dissociation of electrolytes in solution. His famous dissertation, entitled Investigations on the galvanic conductivity of electrolytes, just barely earned him a PhD, since his professors were not at all impressed by his theories. And yet, it was on the basis of this same publication that he won the Nobel Prize in Chemistry in 1903. It was also this dissertation that led to a theory that provided the first theoretical definitions of acids and bases, which he published in 1887. Arrhenius pursued his re search, which, among other things, paved the way to his theoretical concept of the activation energy of a chemical reaction.

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During this neutralization, the ions that react are hydrogen ions derived from the acid and hydroxide ions derived from the base. The other ions, which do not react, are called spectator ions. Thus, by omitting the spectator ions, the net ionic equation of this reaction indicates the main ions that take part in the reaction according to the Arrhenius theory. Net ionic equation of acid-base neutralization H (aq)  OH (aq) n H2O (l)

However, the Arrhenius theory has its limitations, particularly in the case of the hydrogen ion that causes acidity. According to this theory, the electrolytic dissociation of hydrogen chloride in an aqueous solution is translated by the following equation, in which the aqueous hydrogen ion is found: Extremity with partially positive charges +

+

H

H

O

–

Positive dipole

HCl (aq) n H (aq)  Cl (aq)

Chemists deduced that it is very unlikely that the aqueous hydrogen ion exists in water, as stated in the Arrhenius theory. In the preceding equation, which describes the reaction according to Arrhenius, water is not written as a component of the reaction even though the reaction occurs in an aqueous environment. Furthermore, when water is added to the equation, it remains Negative unchanged during the course of the reaction. dipole

Extremity with partially negative charges

Figure 5 Water is a V-shaped polar molecule. The extremity with partially negative charges (–) is found on the side of the oxygen atom (O), and the extremities with the partially positive charges (+)are situated on the side of the hydrogen atoms (H).

HCl (aq)  H2O (l) n H (aq)  Cl (aq)  H2O (l)

Water molecules are polar, which means that in each molecule, electrons are distributed irregularly, thus forming two dipoles: a negative dipole on the side of the oxygen atom and a positive dipole on the side of the hydrogen atoms (see Figure 5). Consequently, there is an interaction in the solution between the water, hydrogen ions (H) and chloride ions (Cl). The hydrogen ion, due to its simple particle composition, is the equivalent of a proton, which carries a positive charge. This charge is attracted by the negative bipole of the water molecule. As a result, the hydrogen ions (protons) do not remain in free state in an aqueous solution. Instead, when this type of proton gets close to polar water molecules, it binds itself to one or more of these molecules and becomes hydrated (see Figure 6). A hydrated proton is called a hydronium ion (H3O).

Oxygen (O) Hydrogen (H)

Figure 6 A hydronium ion (H3O) is pyramid-shaped. The summit of the pyramid is an oxygen atom (O) while the base is composed of three hydrogen atoms (H).

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UNIT 4 Chemical Equilibrium

The Arrhenius theory also has its limitations when attempting to explain certain chemical reactions. For example, aqueous solutions composed of ammonia (NH3) are basic. While ammonia does not contain oxygen or hydrogen atoms as a source of hydroxide ions (OH), its molecules nevertheless react with acids. Therefore, the Arrhenius theory does not explain the basic properties of ammonia. Nor can it explain why many aqueous salt solutions that do not contain oxygen or hydrogen atoms, such as calcium carbonate (CaCO3), also present basic properties.

The Arrhenius theory also has another serious drawback. It does not explain the reactions between acids and bases that occur in non-aqueous environments. For example, ammonium chloride (NH4Cl) can form during the course of a reaction between a basic gas, ammonia, and an acidic gas, hydrogen chloride, according to the following reaction: NH3 (g)  HCl (g) n NH4Cl (s) This reaction produces a white powder composed of tiny crystals of ammonium chloride in the region where the gases come into contact with each other (see Figure 7).

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

JOHANNES NICOLAUS BRØNSTED Danish chemist (1879–1947) Figure 7 The Arrhenius theory does not explain why ammonia (NH3) and hydrogen chloride (HCl) gases that are released from these open containers react by forming a white cloud composed of tiny crystals of solid ammonium chloride (NH4Cl) without the presence of water.

Brønsted-Lowry theory of acids and bases In 1923, two chemists independently proposed a more general theory of acids and bases, which expanded on the Arrhenius theory. Johannes Brønsted and Thomas Lowry developed a theory that emphasized the role of acids and bases in a chemical reaction, rather than the acidic or basic properties of their aqueous solutions, which was the focus of Arrhenius’s work. This theory, which solved the problems raised by the Arrhenius theory, is called the Brønsted-Lowry theory of acids and bases. Brønsted-Lowry theory of acids and bases An acid is a substance in which a proton (hydrogen ion, H) can be removed. An acid is seen as a proton donor. A base is a substance that can remove a proton from an acid. A base is seen as a proton acceptor. According to this theory, a Brønsted-Lowry acid must include a hydrogen (H) atom in its chemical formula, just like an Arrhenius acid.

Johannes Nicolaus Brønsted was a chemistry professor and specialist in chemical reactions. He also studied electrochemistry, amphoteric electrolytes, pH measurement and pH indicators. He is primarily known for his work on acids and bases, which in 1923 led him to publish a theory bearing his name. According to the theory, an acid is defined as a proton donor, while a base is a proton accepter. At the same time, Thomas Lowry independently published a similar theory on acids and bases. This is why the theory is now called the Brønsted-Lowry theory. Thomas Lowry studied aqueous solutions, the nature of ions, and the deflection of light by organic compounds.

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However, all negative ions, not just the hydroxide ion (OH), can constitute a Brønsted-Lowry base. Moreover, it is not necessary to use water as the reactant. In other words, water is not the only solvent that can be used. A single condition is required for a reaction to occur between a Brønsted-Lowry acid and base: one of the substances must supply a proton and the other must receive this proton. Based on this theory, acid-base neutralization leads to the transfer of a proton from an acid to a base. Proton transfer is an important concept for fully understanding the nature of acids and bases. According to the Brønsted-Lowry theory, any substance can behave like an acid as long as another substance behaves like a base at the same time. Similarly, any substance can behave like a base if, at the same time, another substance behaves like an acid. For example, in a reaction between hydrochloric acid (HCl) and water, one molecule of hydrochloric acid donates a proton to a molecule of water according to the following equation (see Figure 8): Transfer of H 



Chlorine (Cl) Oxygen (O) Hydrogen (H)

Conjugate acid-base pair HCl (aq) Acid



H2O (l) Base

n

H3O (aq) Conjugate acid



Cl (aq) Conjugate base

Conjugate acid-base pair

Figure 8 Reaction between hydrochloric acid (HCl) and water, according to the Brønsted-Lowry theory

In this reaction, hydrogen chloride plays the role of a Brønsted-Lowry acid because it supplies a proton, that is, a hydrogen ion (H), to the water. The water molecule receives the proton. Therefore, the water plays the role of a BrønstedLowry base in this reaction. When it receives the proton, it becomes a hydronium ion (H3O), located on the right side of the equation. The conjugate base of an acid is the particle that remains when the proton has been lost from the acid. The conjugate acid of a base is the particle that remains when the proton has been added to the base. In this case the chloride ion (Cl) is the conjugate base of the acid (the hydrogen chloride) and the hydronium ion constitutes the conjugate acid of the base (water). A pair of substances that differ by only one proton is called a conjugate acidbase pair. In this case, the hydrogen chloride molecule and the chloride ion form such a pair. Similarly, the second conjugate acid-base pair of this reaction is the water and the hydronium ion. A substance can be classified as a Brønsted-Lowry acid or base only for a given reaction. Sometimes, a substance loses a proton to a second substance, but in the presence of a third substance, it gains one. Consequently, this substance can play the role of an acid and then the role of a base. This is referred to as an amphoteric substance. Water is an example of an amphoteric substance.

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UNIT 4 Chemical Equilibrium

For example, in the preceding reaction of hydrogen chloride and water, the water plays the role of the base. However, in the reaction of ammonia (NH3) and water, the water molecule plays the role of the acid.

Transfer of H 

 Nitrogen (N) Oxygen (O) Hydrogen (H)

Conjugate acid-base pair NH3 (aq) Base



H2O (l)

n

Acid

NH4 (aq) Conjugate acid



OH (aq) Conjugate base

Conjugate acid-base pair

Figure 9 Reaction between ammonia (NH3) and water, according to the Brønsted-Lowry theory

The Brønsted-Lowry theory does not explain why substances donate or accept protons. Nevertheless, it makes it possible to define acids and bases as a function of chemical reactions rather than simply as substances that form solutions with acidic or basic properties. Due to the definitions proposed by this theory, it is possible to describe, explain and predict a large number of reactions involving substances in their pure state, in an aqueous or non-aqueous solution. Arrhenius’s definitions of an acid and a base can nevertheless be applied to acids and bases in aqueous solutions, and particularly to acid-base neutralization.

12.2.2

Ionization constant of water

The electrolytic dissociation of an acidic or basic compound in an aqueous solution produces ions that interact with the water. Equilibrium is established between the aqueous ions and the water molecules. The pH of an aqueous solution is a measure of the acidity of a solution and is determined by the position of equilibrium established between the aqueous ions and the water molecules. This equilibrium can also be quantified using an equilibrium constant, the ionization constant of water (Kwater). This constant makes it possible to understand the interdependence between the molar concentrations of hydronium ions (H3O) and hydroxide ions (OH). Before examining the equilibrium constant of water in greater depth, it is helpful to expand on the concept of pH and pOH.

Furthering your understanding Amino acids Amino acids are molecules that serve as the building blocks for proteins in living cells. Most amino acids are soluble in a polar solvent, such as water or alcohol. In solution, amino acids are amphoteric substances. This means that they can behave as acids or bases when in an aqueous solution. Their ionization varies depending on the pH of the solution. In an acidic environment, certain amino acids can behave like a base and capture a proton. In a basic environment, certain others can behave like an acid and release a proton. This allows the proteins to be positively charged when they are in an acidic environment and to behave like a base, and vice versa. The charge carried by the proteins is what allows researchers to separate a protein mixture using the technique of electrophoresis. A variant of this technique involves using a gel plate with a pH that varies from one extreme to the other. The proteins move along the length of the plate, stopping at different points depending on the pH of the gel and their acidic or basic character. Among other things, this technique is used to detect genetic diseases such as sickle-cell anemia, a very widespread hereditary disease characterized by an alteration of the red blood cells.

Calculating pH and pOH The degree of acidity of an aqueous solution can be expressed quantitatively by providing the hydronium ion concentration, symbolized by the expression [H3O]. However, given that this concentration is often very weak and requires the use of expressions with scientific notation, it is not very practical.

a)

10

10

10

9

9

9

8

8

8

7

7

7

6

6

6

5

5

5

4

4

4

3

3 b)

3 c)

Figure 10 The protein mixture is placed in the gel (a ), and the proteins move along the length of the pH gradient (b ) positioning themselves according to their pH (c ).

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HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

In 1909, Søren Sørensen designed the pH scale. According to this scale, aqueous solutions with a pH of less than 7 are acidic, those with a pH equal to 7 are neutral, and those with a pH greater than 7 are basic. This logarithmic scale is based on the value of 10. The pH of a solution corresponds to the opposite of the logarithm of the hydronium ion concentration in moles per litre (mol/L). pH is expressed as a numerical value without units. Expression of pH pH  log [H3O] [H3O]  10pH

SØREN PETER LAURITZ SØRENSEN Danish chemist (1868–1939) At the outset of his career, Søren Sørensen carried out work on the synthesis of amino acids. Shortly thereafter, his interest in analyt ical chemistry prompted him to study the action of enzymes. In 1909, Sørensen published a thesis on enzymes in which he developed a simple and practical method for measuring the acidity of a substance, namely, pH. The pH scale he invented made it possible to express different hydronium ion (H3O) concentrations, which are often very low, using the numbers 0 to 14.

where pH  pH value [H3O]  Hydronium ion concentration at equilibrium, expressed in moles per litre (mol/L) The first expression above indicates that pH is equal to the negative logarithm in base 10 of the hydronium ion concentration. The second expression, which is the reciprocal function of the first, makes it possible to determine the hydronium ion concentration when the pH is known. It is also possible to calculate the pOH of a solution, that is, the potential hydroxide ion (OH) concentration, using a similar expression. Expression of pOH pOH  log [OH] [OH]  10pOH where pOH  pOH value [OH]  Hydroxide ion concentration at equilibrium, expressed in moles per litre (mol/L) The following examples show how to use these equations: Example A Express, in the form of pH, the hydronium ion (H3O) concentration of 4.7  1011 mol/L in an aqueous solution. Is this solution acidic, neutral or basic? Data: [H3O]  4.7  1011 mol/L pH  ?

Calculation: pH  log [H3O]  log [4.7  1011]  10.33

Answer: The pH of the solution is 10.33. The solution is basic, because its pH is greater than 7. Example B Express a pOH of 3.60 in the form of ion hydroxide (OH) concentration. Data: pOH  3.60 [OH]  ?

Calculation: [OH]  10pOH  103.6  2.5  104

Answer: The ion hydroxide (OH) concentration is 2.5  104 mol/L.

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UNIT 4 Chemical Equilibrium

Calculating the ionization constant of water A very sensitive multimeter can detect slight electrical conductivity in very pure water (see Figure 11). According to the Arrhenius theory, this electrical conductibility is due to the presence of ions. In the case of pure water, these ions originate from the ionization of a few water molecules (see Figure 12). The ionization of water molecules is very rare. It has been observed that at 25°C, only two water molecules in a billion dissociate this way. Pure water is therefore a poor electrical conductor, and a very sensitive device would be required to detect this conductivity.

Figure 11 A very sensitive multimeter can detect that pure water has weak electrical conductivity.

H2O (l)



z y

H2O (l)

H3O (aq)



OH (aq)

Oxygen (O) Hydrogen (H)

Figure 12 The ionization of water occurs when molecules collide, resulting in the formation of hydronium ions (H3O) and hydroxide ions (OH).

At 25°C, pure water is neutral, which means that its pH is 7. The hydronium ion (H3O) concentration is therefore 10-pH, that is, 1  107 mol/L. Since each water molecule that ionizes produces as many hydroxide ions (OH) as hydronium ions, it is possible to state that their concentrations are equal. [H3O]  [OH]  1  107 mol/L The ionization of water is a reversible process that can be expressed by the following simplified equation: x H3O (aq)  OH (aq) 2 H2 O (l) w Once equilibrium has been attained, the equilibrium constant for the ionization of water (Kwater) is expressed as follows: Ionization constant of water Kw  [H3O]  [OH] where Kwater  Ionization constant of water [H3O]  Hydronium ion concentration at equilibrium, expressed in moles per litre (mol/L)  [OH ]  Hydroxide ion concentration at equilibrium, expressed in moles per litre (mol/L)

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The ionization constant of water is a concentration equilibrium constant. This is why the concentration of the water on the reactant side, where water is in the liquid phase, is not taken into account. To determine the numeric value of this constant at 25°C, insert the values of the concentration of the ions in the expression for the ionization constant of water. Kw  [H3O]  [OH] Kw  (1  107 mol/L)  (1  107) Kw  1  1014 Table 5 Ionization constant of water at different temperatures

Temperature (°C)

Kwater

0

1.14  1015

10

2.92  1015

20

6.82  1015

25

1.00  1014

30

1.47 

1014

40

2.92  1014

50

5.48 

1014

60

9.55  1014

70

1.64  1013

100

6.31  1013

As is the case for all equilibrium constants, the value of the ionization constant of water varies as a function of temperature (see Table 5). Unless another value is provided for a different temperature, this is the value used to carry out the calculations that establish a relationship between the pH and the molar concentration of the hydronium and hydroxide ions. This expression also makes it possible to establish a simple ratio between the values of pH and pOH in a solution at 25°C. According to the ionization constant of water, the product of the concentrations of hydronium and hydroxide ions is expressed as follows: [H3O]  [OH]  1  1014 at 25°C By carrying out the logarithmic inverse of each side of the equation, the following equivalences are obtained: (log

[H3O]  log [OH])  log (1  1014) log [H O]  log [OH]  14 3 pH  pOH  14

Using this equality, it is possible to quickly convert a pH value into a pOH value, and vice versa.

12.2.3

Relationship between the pH and the molar concentration of hydronium and hydroxide ions

The ionization constant of water is particularly important when studying the behaviour of acids and bases in aqueous solutions. According to the Arrhenius theory, an acid is a substance that produces H+ ions when it reacts with water. These ions bond with one or several water molecules to form hydronium ions (H3O). Therefore, the acid supplies additional hydronium ions, which increase the initial concentration of the ions while decreasing the pH. For its part, the base supplies hydroxide ions (OH), which increase the initial concentration of these ions while increasing the pH. Since the ionization constant of water applies to all aqueous solutions, it can be used to determine the relationship between the molar concentrations of hydronium and hydroxide ions in such solutions. Even if the concentration of hydronium and hydroxide ions varies, the ionization constant of water does not change. Consequently, it can be used to calculate either the concentration of hydronium ions or hydroxide ions, as long as one of these two concentrations or the pH of the solution is known.

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UNIT 4 Chemical Equilibrium

The following example shows how to use the ionization constant of water to determine the molar concentration of ions present in a solution when the pH is known: Example At 25°C, a hydrochloric acid (HCl) solution has a pH of 3.2. What is the concentration of each of the ions in this solution? Data: pH  3.2 [H3O]  ? [OH]  ?

1. Calculation of hydronium ions: [H3O]  10pH  103.2  6.3  104 2. Calculation of hydroxide ions: Kw  [H3O]  [OH]  1  1014 [OH]  

1  1014 [H3O] 1  1014  1.58  1011 6.3  104

Answer: The concentration of hydronium ions (H3O) is 6.3  104 mol/L and that of hydroxide ions (OH) is 1.6  1011 mol/L.

12.2.4

Acidity constant and basicity constant

The strength of electrolytes corresponds to the rate of electrolytic dissociation of the solute ions. The stronger the electrolyte, the higher its dissociation and the greater its production of ions in solution. Acids and bases are electrolytes that undergo electrolytic dissociation in aqueous solutions to varying degrees. This explains the strength of acids and bases.

Strength of acids A strong acid is an acid that completely dissociates into ions in water (see Table 6). For example, hydrogen chloride (HCl) is a strong acid, since it completely ionizes in water. All its molecules dissociate into hydronium ions (H3O) and chloride ions (Cl) in an aqueous solution (see Figure 13). At this point, the solution contains the same percentage of hydronium ions and chloride ions, that is, 100%.

Name

Chemical formula

Hydrochloric acid

HCl

Hydrobromic acid

HBr

Hydriotic acid

HI

Nitric acid

HNO3

Sulphuric acid

H2SO4

Perchloric acid

HClO4

After dissociation Percentage of molecules

Percentage of molecules

Before dissociation

Table 6 Some strong acids

100%

100%

100%

H3O

Cl

O% HCl

HCl (aq)

HCl



H2O (l)

n

H3O (aq)



Cl (aq)

Chlorine (Cl) Oxygen (O) Hydrogen (H)

Figure 13 The molecules of hydrogen chloride (HCl), a strong acid, dissociate at a rate of 100% when they are in an aqueous solution.

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The hydronium ion (H3O) concentration in a solution diluted with a strong acid is equal to the concentration of this acid. Consequently, a solution of hydrogen chloride at 1 mol/L of hydronium ions contains 1 mol/L of hydronium ions and 1 mol/L of chloride ions. A weak acid is an acid that dissociates very little in an aqueous solution. For example, acetic (ethanoic) acid (CH3COOH) is a weak acid. Most of its molecules remain whole after their dissolution. Only a very small percentage of its molecules separate into hydronium ions (H3O) and acetate (ethanoate) ions (CH3COO). This percentage of dissociation is called the ionization percentage. It is calculated as follows for a given acid (generally symbolized by HA):

Ionization percentage Ionization percentage 

[H3O]  100 [HA]

where [H3O]  Hydronium ion concentration at equilibrium, expressed in moles per litre (mol/L) [HA]  Concentration of an acid, expressed in moles per litre (mol/L) On average, approximately 1% of acetic acid molecules dissociate in a solution at 0.1 mol/L. In fact, the concentration of hydronium ions in a weak acid is always less than the concentration of the dissolved acid (see Figure 14).

After dissociation Percentage of molecules

Percentage of molecules

Before dissociation 100%

99%

Approximately Approximately

1% CH3COOH

CH3COOH (aq)  H2O (l)

CH3COOH

z y

H3O

1% CH3COO

H3O (aq)  CH3COO (aq)

Carbon (C) Oxygen (O) Hydrogen (H)

Figure 14 The molecules of acetic acid (CH3COOH), a weak acid, dissociate very little in an aqueous solution.

The concentration at equilibrium of the hydronium ions of this reaction can be calculated based on the ionization percentage and the concentration of the acetic acid. In this acetic acid solution at 0.1 mol/L, only 1% of the molecules dissociate to form hydronium ions.

[H3O] 

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UNIT 4 Chemical Equilibrium

1  0.1 mol/L  1  103 mol/L 100

When the concentrations at equilibrium of the various particles present in the solution are known, the value of the equilibrium constant of the reaction can be calculated.

Calculating the acidity constant Given that acetic acid (CH3COOH) is a weak acid, there are always acetic acid molecules in the solution after its dissociation and equilibrium has been attained. At equilibrium, the molecules of acetic acid dissociate at the same rate that the dissociated ions recombine to produce these same molecules. Most acids that attain such an equilibrium in an aqueous solution, are weak acids. The general formula used to describe the ionization of a weak acid is as follows: x H3O (aq)  A (aq) HA (aq)  H2O (l) w The strength of a weak acid can be expressed using an equilibrium constant. In general, the expression of the equilibrium constant for such a reaction is called an acidity constant (Ka) and is expressed as follows: Acidity constant Ka 

[H3O]  [A] [HA]

where Ka  Acidity constant [H3O]  Hydronium ion concentration at equilibrium, expressed in moles per litre (mol/L) [A]  Conjugate base concentration at equilibrium, expressed in moles per litre (mol/L) [HA]  Non-dissociated acid concentration at equilibrium, expressed in moles per litre (mol/L) The acidity constant is a variant of the concentration equilibrium con stant (Kc). This explains why the concentration of water, which is in liquid phase, does not appear. Moreover, since strong acids dissociate completely, there is no real equilibrium attained since, given that the reaction is complete, there are no more reactant molecules. This explains why the acidity constant of strong acids in an aqueous solution can be described using a number. The acidity constant makes it possible to classify acids according to their strength. The weaker the acid, the smaller the value of its acidity constant.

APPENDIX 8 Table 8.7: Ionization constants of acids, p. 421.

To obtain the value of the equilibrium constant of an acid in a given situation, the concentration of the different substances in the reaction at equilibrium must be known. The acidity constant can be used to predict the concentrations of the hydronium ions (H3O) at the equilibrium of an acid at various concentrations, as well as to calculate the pH or the percentage of dissociation of various acids. The following examples show a variety of problems that can be solved using the value of the acidity constant:

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Example A At 25°C, an acetic (ethanoic) acid solution (CH3COOH) at 1 mol/L dissociates according to the following equation: x H3O (aq)  CH3COO (aq) CH3COOH (aq)  H2O (l) w At equilibrium, the pH is 2.38. What is the value of the acidity constant of this acid? What is the dissociation percentage of the acid? Data: 1. Conversion of pH to molar concentration: [CH3COOH]  1 mol/L [H3O]  10pH pH  2.38  102.38  4.2  103 mol/L Ka  ? % dissociation  ? 2. Recording of data and use of the ICE table: Concentration (mol/L)

CH3COOH (aq)

Initial (Ci)

w x

H3O (aq)

1

Change (C)

B

Equilibrium (Ceq)

C

4.2

CH3COO (aq)



0

 103

A

0.9958 or 1

0

 103

B

4.2  103

C

4.2

4.2

 103

4.2  103

A Start with the following calculation: [H3O]  [H3O]eq  [H3O]i  4.2  103 mol/L  0 mol/L  4.2  103 mol/L Given that H3O is a product, its change is positive (4.2  103). B The changes in the other substances are deduced according to the stoichiometric ratios, which are 1:1:1. Given that CH3COOH is a reactant, its change is negative (4.2  103). Conversely, given that CH3COO is a product, its change is positive (4.2  103). C The concentrations at equilibrium of each substance are calculated by adding the initial concentrations and the changes in concentration and taking into account the sign of the value of the change in concentration. Ceq  Ci  C [CH3COOH]eq  [CH3COOH]i  [CH3COOH]  1  4.2  103  1 mol/L [CH3COO]eq  [CH3COO]i  [CH3COO]  0  4.2  103  4.2  103 mol/L 3. Calculation of the acidity constant: [H O]  [A] Ka  3 [HA] 

[H3O]  [CH3COO] 4.2  103  4.2  103   1.7  105 [CH3COOH] 1

4. Calculation of the dissociation constant: [H O] Ionization percentage  3 eq  100 [HA] 

[H3O]eq 4.2  103  100   100  0.42% [CH3COOH] 1

Answer: The value of the acidity constant of the acetic acid (CH3COOH) is 1.7  105 and the dissociation percentage is 0.42%.

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UNIT 4 Chemical Equilibrium

Example B Calculate the pH of an aqueous solution of formic (methanoic) acid (HCOOH) at 0.20 mol/L if the value of its acidity constant is 1.8  104. The equation at equilibrium of this reaction is as follows: x H3O (aq)  HCOO (aq) HCOOH (aq)  H2O (l) w Data: [HCOOH]  0.20 mol/L Ka  1.8  104 pH  ?

1. Recording of data and use of the ICE table: Concentration (mol/L)

x H3O (aq) HCOOH (aq) w

Initial (Ci)

0.2

A

Change (C) Equilibrium (Ceq)

 HCOO (aq)

0

0

x

A

x

A

x

B 0.20  x

B

x

B

x

A The change in the unknown concentration is written as equal to x for the reactants and x for the products according to the stoichiometric ratios, which are 1:1:1. B The concentrations at equilibrium are written as a function of the change x. Ceq  Ci  C [HCOOH]eq  [HCOOH]i  [HCOOH]  0.2   x [H3O]eq  [H3O]i  [H3O]  0  x  x [HCOO]eq  [HCOO]i  [HCOO]  0  x  x 2. Calculation of the acidity constant: Ka  1.8  104 

[H3O]  [A] [H3O]  [HCOO]  [HA] [HCOOH] xx x2  0.2  x 0.2  x

This second-degree equation is of the type ax 2  bx  c  0. Therefore, it must be rewritten as a quadratic equation. 1.8  104 (0.2  x)  x 2 3.6  105  1.8  104x  x 2 2 x  1.8  104x  3.6  105  0 The possible values of x can be found using: x x

(1.8  104) 

b 

(1.8  104)2  4(1  3.6  105) 2(1)

b 2  4ac 2a

or x 

(1.8  104) 

(1.8  104)2  4(1  3.6  105) 2(1)

x  6.09  103

x  5.9  103

The value x  6.09  103 is impossible because it would provide a negative concentration of H3O and HCOO at equilibrium. It is therefore rejected. The value x  5.9  103 is used to calculate concentrations at equilibrium. [HCOOH]eq  0.2  x  0.20  0.0059  0.194 mol/L [H3O]eq  [HCOO]eq  x  0.0059 mol/L 3. Calculation of pH: pH  log [H3O]  log (0.0059)  2.23 Answer: The pH of the aqueous solution of formic acid (HCOOH) is 2.23.

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Calculating the basicity constant Table 7 Some strong bases Chemical formula

Name Sodium hydroxide

NaOH

Potassium hydroxide

KOH

Calcium hydroxide

Ca(OH)2

Strontium hydroxide

Sr(OH)2

Barium hydroxide

Ba(OH)2

Like acids, bases can be classified as a function of their degree of dissociation. Strong bases completely dissociate into ions in water (see Table 7). For example, sodium hydroxide (NaOH) is a strong base, since it completely ionizes in water. All its molecules dissociate into sodium ions (Na) and hydroxide ions (OH) in an aqueous solution. The concentration of hydroxide ions in a solution diluted with a strong base is equal to the concentration of this base. Consequently, a hydroxide solution of sodium at 1 mol/L contains 1 mol/L of sodium ions and 1 mol/L of hydroxide ions. In reality, most bases are weak. A weak base dissociates very little in an aqueous solution. For example, a weak base, represented by the symbol B, reacts with skin to form an ionic solution in which equilibrium is attained according to the following equation: x HB (aq)  OH (aq) B (aq)  H2O (l) w One method for expressing the strength of a weak base is to use an equilibrium constant. In general, the expression of the equilibrium constant of such a reaction is called the basicity constant (Kb) and is expressed as follows: Basicity constant Kb 

[HB]  [OH] [B]

where Kb  Basicity constant  [HB ]  Concentration of the conjugate acid at equilibrium, expressed in moles per litre (mol/L) [OH]  Concentration of hydroxide ions at equilibrium, expressed in moles per litre (mol/L) [B]  Concentration of the non-dissociated base at equilibrium, expressed in moles per litre (mol/L) This equilibrium constant is a variant of the concentration equilibrium constant (Kc). This explains why the concentration of water, which is in liquid phase, does not appear. As with strong acids, the numerical value of the constant of strong bases in an aqueous environment cannot be established since, given that the reaction is complete, there is no equilibrium. The basicity constant makes it possible to classify bases according to their strength. The weaker the base, the smaller its constant.

APPENDIX 8 Table 8.8: Ionization constants of nitrogen bases, p. 421.

To obtain the value of the equilibrium constant of a base in a given situation, the concentration of the different substances in the reaction at equilibrium must be known. The basicity constant can be used to predict the hydroxide ion concentrations of a base at equilibrium at various concentrations, as well as to solve problems similar to those of weak acids. The following example shows how to solve a problem using the basicity constant:

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UNIT 4 Chemical Equilibrium

Example Ammonia (NH3) dissolves in water to form ammonium hydroxide (NH4OH) according to the following equation: x NH4 (aq)  OH (aq) NH3 (aq)  H2O (l) w The value of the basicity constant of ammonia at 25°C is 1.8  105. What is the pH of an ammonium hydroxide solution at 0.40 mol/L? Data: [NH3]  0.40 mol/L

1. Recording of data and use of the ICE table: Concentration (mol/L) Initial (Ci)

A

Change (C) Equilibrium(Ceq)

NH3 (aq)

w x NH4 (aq) 

OH (aq)

0.40

0

0

x

A

x

A

x

B 0.40  x

B

x

B

x

A The change in the unknown concentration is written as x for the reactants and x for the products according to the stoichiometric rations, which are 1:1:1. B The concentrations at equilibrium are written as a function of the change x. Ceq  Ci  C [NH3]eq  [NH3]i  [NH3]  0.40  x [NH4]eq  [NH4]i  [NH4]  0  x  x [OH]eq  [OH]i  [OH]  0  x  x 2. Calculation of the basicity constant: Kb  1.8  105 

[HB]  [OH] [B]

[NH4]  [OH] xx x2   [NH3] 0.40  x 0.40  x

This second-degree equation is of the type ax 2  bx  c  0. Therefore, it must be rewritten as a quadratic equation. 1.8  105 (0.40  x)  x 2 7.2  106  1.8  105x  x 2 x 2  1.8  105x  7.2  106  0 The possible values of x can be found using: x x

(1.8  105) 

x  2.7  103

b

 b 2  4ac 2a

(1.8  105)  (1.8  105)2  4(1  7.2  106) (1.8  105)2  4(1  7.2  106) or x  2(1) 2(1) 3  x  5.4  10

The value x  5.4  103 is impossible because it would provide a negative concentration of NH4 and OH at equilibrium. It is therefore rejected. The value x  2.7  103 is used to calculate concentrations at equilibrium. [NH3]eq  0.40  x  0.40  0.002 7  0.397 mol/L [NH4]eq  [OH]eq  x  0.002 7 mol/L 3. Calculation of pH: p0H  log [OH]  log (0.002 7)  2.57

pH  pOH  14 pH  14  pOH  14  2.57  11.43

Answer: The pH of the aqueous solution of ammonium hydroxide (NH4OH) is 11.4.

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12.2.5

Solubility product constant

A saturated solution that contains a non-dissolved solute deposited at the bottom of a container is an example of a system at equilibrium. In this case, solubility equilibrium has been attained. To determine the quantity of the solute present in the saturated solution, it is necessary to examine the solubility of the solute that is dissolved. The solubility of a substance corresponds to the maximum quantity of the substance that dissolves in a given volume of solvent at a given temperature. Generally, solubility is expressed in grams of solute per 100 millilitres of solvent (g/100 mL). The molar solubility of a substance can also be expressed in moles of solute per litre of solvent (mol/L). This is referred to as molar solubility. Barium sulfate (BaSO4) is an ionic compound that is not soluble (see Figure 15). It is used in radiology to take X-rays of the large intestine (see Figure 16). Figure 15 Barium sulfate (BaSO4)

precipitate

When crystals of solid barium sulfate are added to water, the barium ions (Ba2) and sulphate ions (SO42) leave the surface of the solid and enter the solution. Initially, the concentration of these ions is very weak. The direct reaction, dissolution, occurs at a greater rate than that of the reverse reaction, recrystallization. However, as the ions enter the solution, the recrystallization rate increases and eventually equals the dissolution rate. At this point, the system is at solubility equilibrium and can be represented by the following reaction: x Ba2 (aq)  SO42 (aq) BaSO4 (s) w Given the simultaneous presence, at equilibrium, of solid barium sulfate and its dissolved ions, it is possible to write the expression of an equilibrium constant as a function of the concentration of particles. Since the barium sulfate is in the solid phase, its concentration is not included in the expression of the equilibrium constant obtained, which is then called the solubility product constant (Ksp). Ksp  [Ba2]  [ SO42] The following general formula is used to describe the solubility equilibrium obtained following the partial dissolution of an ionic compound: x nX (aq)  mY (aq) XnYm (s) w

Figure 16 Barium sulfate (BaSO4), an ionic compound that is not

very soluble, is used to increase the clarity of X-rays, since it is opaque in X-rays and therefore creates a contrast.

336

UNIT 4 Chemical Equilibrium

The solubility product constant can be described even more generally by the following expression: Solubility product constant Ksp  [X]n  [Y]m where Ksp  Solubility product constant [X], [Y]  Concentration of ions at equilibrium, expressed in moles per litre (mol/L) n, m  Coefficients of each of the ions in the balanced equation

The solubility product constant is a variant of the concentration equilibrium constant (Kc). This explains why the concentration of the solid XnYm, which is constant, does not appear. Like all equilibrium constants, it varies with temperature. Consequently, experiments must be conducted to determine the value of the solubility product constant at various temperatures (see Table 8). Table 8 Solubility product constants at 25°C for some ionic compounds Compound

Solubility product constant

Magnesium sulfate (MgSO4)

5.9  103

Lead chloride (PbCl2)

1.7  105

Barium fluoride (BaF2)

1.84  107

Cadmium carbonate (CdCO3)

1.8  1014

Copper hydroxide (Cu(OH)2)

2.2  1020

Silver sulfide (Ag2S)

8.0  1048

The solubility product constant makes it possible to classify compounds according to their solubility. Consequently, when compounds containing ions in the same proportions are compared, such as magnesium sulfate (MgSO4) and cadmium carbonate (CdCO3), note that the greater the solubility of a compound in water, the greater the value of the solubility product constant. Table 8 lists the value of the solubility product constant of certain ionic compounds with low solubility. Compounds like sodium chloride (NaCl) and potassium nitrate (KNO3), which are strong electrolytes, are not on this list because the value of their constant is relatively high compared to the others. In order to obtain the value of the solubility product constant in a given situation, the concentrations of the different substances in the reaction at equilibrium must be known. The solubility product constant of a compound can also be used to predict the concentration of its ions at equilibrium in a saturated solution or to determine the molar solubility of an ionic compound.

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The following example shows how to solve a problem using the expression of the solubility product constant: Example The solubility of silver carbonate (Ag2CO3) is 3.6  103 g/100 mL of solvent at 25 °C. Calculate the value of the solubility product constant of silver carbonate. Data: SolubilityAg2CO3  3.6  Ksp  ?

103

g/100 mL

1. Conversion into units of molar solubility: m M n n

m 3.6  103 g   1.3  105 mol for 100 mL (or 0.1 L) M 275.8 g/mol

Molar solubility 

1.3  105 mol  1.3  104 mol/L 0.1 L  10

2. Calculation of the solubility product constant: x 2 Ag (aq)  CO32 (aq) Ag2CO3 (s) w Ksp  [X]n  [Y]m  [Ag]2  [CO32] [Ag]  2  [Ag2CO3]  2  1.3  104 mol/L  2.6  104 mol/L [CO32]  [Ag2CO3]  1.3  104 mol/L Ksp  [X]n  [Y]m  [Ag]2  [CO32]  (2.6  104 mol/L)2  1.3  104 mol/L  8.8  1012 Answer: The value of the solubility product constant of silver carbonate (Ag2CO3) at 25°C is 8.8  1012.

SECTION 12.2

Ionic equilibrium in solutions

1. Name and write the formula for the conjugate base of each of the following molecules or ions. a) HCl c) H2SO4 b) HCO3 d) N2H5 2. Name and write the formula for the conjugate acid of each of the following molecules or ions: a) NO3 c) H2O b) OH d) HCO3 3. A sample of household ammonium hydroxide (NH4OH) has a pH of 11.9. What is the pOH and the concentration of the hydroxide ions (OH) of the sample?

338

UNIT 4 Chemical Equilibrium

4. Calculate the hydronium ion (H3O) concentration in each of the following solutions: a) 4.5 mol/L of aqueous hydrochloric acid (HCl). b) 30.0 mL of aqueous hydrogen bromide (HBr) at 4.50 mol/L diluted to 100.0 mL. c) 18.6 mL of aqueous perchloric acid (HClO4) at 2.60 mol/L added to 24.8 mL of aqueous sodium hydroxide (NaOH) at 1.92 mol/L. d) 17.9 mL of aqueous nitric acid (HNO3) at 0.175 mol/L added to 35.4 mL of aqueous calcium hydroxide (Ca(OH)2) at 0.0160 mol/L.

5. In a sample of milk, the hydronium ion (H3O) concentration is 3.98  107 mol/L. Is the milk acidic, neutral or basic? Calculate the pH and the hydroxide ion (OH) concentration of the sample. 6. Identify the conjugate acid-base pairs in each of the following reactions: x H2S (aq)  OH (aq) a) HS (aq)  H2O (l) w w x 2 OH b) O  H O 2

(aq)

2

(l)

c) H2S (aq)  NH3 (aq) d) H2SO4 (aq)  H2O (l)

(aq)

w x NH4 (aq)  HS (aq) n H3O (aq)  HSO4 (aq)

7. The pH of an aqueous solution of phenol (C6H6O) is measured at 4.72. Is the phenol acidic, neutral or basic? Calculate the hydronium ion (H3O) concentration, the hydroxide ion (OH) concentration and the pOH of the solution. 8. At normal body temperature, that is, 37°C, the value of the equilibrium constant of water is 2.5  1014. Calculate the hydronium ion (H3O) concentration and the hydroxide ion (OH) concentration at this temperature. At 37°C, is pure water acidic, neutral or basic? 9. A sample of sodium bicarbonate (NaHCO3) is dissolved in water. The pOH of the solution is 5.81 at 25°C. Is the solution acidic, neutral or basic? Calculate the pH, the hydronium ion (H3O) concentration and hydroxide ion (OH) concentration.

13. 2.75 g of solid magnesium oxide (MgO) is added to 70 mL of aqueous nitric acid (HNO3) at 2.4 mol/L. Is the solution obtained following this reaction acidic or basic? What is the ion concentration that determines the type of solution? 14. Determine if the reaction of each of the following pairs of reactants produces an acidic solution or a basic solution. Then, calculate the ion concentration that makes the solution acidic or basic. (Assume that the volumes in a) are additive and the volumes in b) remain the same.) a) 31.9 mL of aqueous hydrochloric acid (HCl) at 2.75 mol/L added to 125 mL of aqueous magnesium hydroxide (Mg(OH)2) at 0.0500 mol/L. b) 4.87 g of solid sodium hydroxide (NaOH) added to 80 mL of aqueous hydrogen bromide (HBr) at 3.5 mol/L. 15. Find the concentration of hydronium ions (H3O) and the concentration of hydroxide ions (OH) in each of the following solutions. a) Hydrochloric acid (HCl) at 0.45 mol/L. b) Sodium hydroxide (NaOH) at 1.1 mol/L. 16. Find the concentration of hydronium ions (H3O) and the concentration of hydroxide ions (OH) in each of the following solutions. a) Hydrogen bromide (HBr) at 0.95 mol/L. b) Calcium hydroxide (Ca(OH)2) at 0.012 mol/L.

10. Calculate the hydroxide ion (OH) concentration in each solution. a) 3.1 mol/L of aqueous potassium hydroxide (KOH). b) 21 mL of potassium hydroxide (KOH) at 3.1 mol/L diluted to 75 mL. c) 23.2 mL of aqueous hydrochloric acid (HCl) at 1.58 mol/L added to 18.9 mL of aqueous sodium hydroxide (NaOH) at 3.50 mol/L. d) 16.5 mL of aqueous sulphuric acid (H2SO4) at 1.5 mol/L added to 12.7 mL of sodium hydroxide (NaOH) at 5.50 mol/L.

17. At low dosages, barbiturates act as sedatives. Barbiturates are manufactured using barbituric acid (C4H4N2O3), a weak acid that was first prepared by German chemist Adolph von Baeyer in 1864. A chemist prepares a solution of barbituric acid at 0.1 mol/L. He finds a pH of 2.50 for the solution. What is the acidity constant of the barbituric acid? What percentage of its molecules dissociate?

11. A chemist dissolves a few tablets of aspirin in water. She then measures the pH of the solution and obtains a value of 2.73 at 25°C. What is the hydronium ion (H3O) concentration and hydroxide ion (OH) concentration in the solution?

19. The concentration of hydronium ions (H3O) in a calcium hydroxide (Ca(OH)2) solution is 1.7  1014. What is the molar concentration of the calcium hydroxide?

12. If the hydronium ion (H3O) concentration is 2.9  104 mol/L in a glass of orange juice, calculate the pH of this juice. Is this solution acidic or basic?

18. The concentration of hydroxide ions (OH) in a solution of hydrochloric acid (HCl) is 5.6  1014 mol/L. What is the molar concentration of the hydrochloric acid?

20. Butanoic acid (C4H8O2), also called butyric acid, gives fresh parmesan its characteristic smell. Calculate the pH of a solution at 1.0  102 mol/L of butanoic acid if the value of the acidity constant is 1.51  105.

CHAPTERCHAPTER 12 The Quantitative 12 The Quantitative Aspect of Aspect Chemical of Equilibrium

339

21. Complete the following table by calculating the unknown values and indicating if the solutions are acidic or basic: [H3O] (mol/L) a) b) c)

pH

[OH] (mol/L)

pOH

Acidic or basic

3.7  105 10.41 7.0  102 8.90

d)

22. Calculate the pH of a sample of vinegar that contains acetic acid (CH3COOH) at 0.83 mol/L. What is the ionization percentage of the vinegar? 23. A solution of hydrofluoric acid (HF) has a molar concentration of 0.0100 mol/L. What is the pH of this solution? 24. Hypochlorous acid (HOCl) is used as a bleaching agent and a germicide. A chemist finds that 0.027% of hypochlorous acid molecules are dissociated in a solution at 0.40 mol/L of this acid. What is the value of the acidity constant of hypochlorous acid? 25. Caproic acid (C5H11COOH) is present in its natural state in palm and coconut oils. It is a weak acid, with an acidity constant of 1.3  105. A given aqueous solution of caproic acid shows a pH of 2.94. What quantity of acid is dissolved to prepare 100 mL of this solution? 26. Calculate the pH of the following solutions: a) The hydronium ion (H3O) concentration in a solution diluted with nitric acid (HNO3) is 6.33  103 mol/L. b) The hydronium ion (H3O) concentration in a sodium hydroxide (NaOH) solution is 6.59  1010 mol/L. 27. 5% of a 0.10 mol/L solution of a weak acid dissociates. Calculate the value of the acidity constant. 28. Morphine (C17H19NO3) is a base of natural origin used to control pain. A solution of 4.5  103 mol/L has a pH of 9.93. Calculate the basicity constant of the morphine. 29. An aqueous solution of ammonium hydroxide (NH4OH) has a pH of 10.85. What is the concentration of the solution?

340

UNIT 4 Chemical Equilibrium

30. What are the concentrations of hydronium ions (H3O+) and hydroxyde ions (OH−), and the pH and pOH of a 0.048 mol/L solution of benzoic acid (C6H5COOOH)? 31. At 25°C, vinegar (CH3COOH) has a 1.3 mol/L concentration of hydronium ions (H3O+). Calculate the concentration of hydroxide ions (OH−) in this substance. 32. At 25°C, the concentration of hydroxide ions (OH−) in human blood is normally 2.20 107 mol/L. Calculate the concentration of hydronium ions (H3O+) and the pH of the substance. 33. An aqueous solution of household ammonium hydroxide (NH4OH) has a molar concentration of 0.105 mol/L. Calculate the pH of the solution. 34. For each of the following weak bases, write the equation of chemical equilibrium and the equation of the basicity equilibrium constant: a) Aqueous cyanide (CN) b) Aqueous sulphate (SO42) 35. The concentration of hydroxide ions (OH ) in a solu tion of aqueous sodium propionate (NaC2H5COO) at 0.157 mol/L is 1.1  105 mol/L. Calculate the basicity constant of the propionate ion. 36. Write the balanced chemical equation representing the dissociation of each of the following compounds in water, and the expression of the corresponding solubility product: a) Copper chloride (CuCl2) b) Barium chloride (BaCl2) c) Silver sulfide (Ag2SO4) d) Silver carbonate (Ag2CO3) 37. The maximum solubility of silver cyanide (AgCN) is 1.5  108 mol/L at 25°C. Calculate the value of the solubility product constant of silver cyanide. 38. A saturated solution of calcium fluoride (CaF2) contains 1.2  1020 molecules of calcium fluoride per litre of solution. Calculate the value of the solubility product constant of the calcium fluoride. 39. Based on their degree of dissociation, what is the difference between a strong acid and a weak acid, and between a strong base and a weak base?

APPLICATIONS Saltwater pools The effect of Sun and wind-borne microbes, and animal and human contact can deteriorate the quality of swimming pool water. Potentially pathogenic algae and mould can appear and pose a threat to swimmers’ health and hygiene (see Figure 17).

Because salt molecules contain chlorine, the electrolyzer separates the saltwater’s constituents with an electrical current forming aqueous ions (see Figure 18). These ions then recombine to form sodium hypochlorite (NaClO), which is then transformed into hypochlorous acid (HClO). Release of H2

NaCI

NaCIO solution

NaClO NaOH

H2O

CI2

H(aq)

H2O

Ha(aq)

NaCI

OH(aq) CI(aq)





Cathode

Anode

Figure 17 Pool water purity is essential to swimmers’ health and hygiene.

There are a number of ways to treat and disinfect pool water and maintain its quality. First and foremost is mechanical filtration. By eliminating most water-borne particles, mechanical filtration solves up to 80% of water treatment problems. However, because this physical treatment does not kill bacteria, the water must be disinfected by chemical or electrolytic treatment. The most common treatment is chemical disinfection. The most popular chemical disinfectants for pool water, such as the bleach ingredient sodium hypochlorite (NaClO), contain chlorine-based compounds (Cl). When dissolved in water, they are transformed into a disinfectant active ingredient, hypochlorous acid (HClO). However, handling these highly concentrated chlorinated compounds is hazardous, and their secondary effects, such as reddened eyes, are causing increasing numbers of people to opt for electrolytic disinfection. Salt electrolysis is a treatment that requires no chemical products toxic to humans or the environment. The simple addition of a small quantity of salt (NaCl) and the use of an apparatus called an electrolyzer are enough to disinfect the water in a swimming pool.

Figure 18 Diagram of a salt-water electrolyzer

The process begins with the production of sodium hydroxide (NaOH), chlorine gas (Cl2) and hydrogen gas (H2) according to the following chemical reaction, which occurs at the apparatus’s electrodes: 2 NaCl (aq) + 2 H2O (l)  2 NaOH (aq) + Cl2 (aq) + H2 (g) The sodium hydroxide and the chlorine gas then react to form sodium hypochlorite according to the following equilibrium reaction: 2 NaOH (aq) + Cl2 (aq) z w NaCl (aq) + NaClO (aq) + H2O (l) The sodium hypochlorite is then converted into hypochlorous acid in the pool water according to the following equilibrium reaction: NaClO (aq) + H2O (l) z w NaOH (aq)+ HClO (aq) After destroying the bacteria and micro-organisms in the water, the hypochlorous acid converts naturally into salt under the action of the Sun’s UV rays, making further additions of salt unnecessary. The chemical equilibrium between these reversible reactions continuously disinfects the water in the pool, with no need for chemical agents.

CHAPTER 12 The Quantitative Aspect of Chemical Equilibrium

341

Sunglasses The retina is a membrane of the eye that is sensitive to invisible solar radiation, especially the type-A ultraviolet rays that are particularly dangerous in case of overexposure. Around the year 1300, Chinese judges began wearing lenses made of coloured quartz to conceal their expressions when they presided in court. The ancestors of Nunavik’s inhabitants fashioned iggaaks, protective eyewear made of caribou antlers or whale bone. Still used today, the long slits of the iggaaks effectively limit entry into the retina of rays reflected by snow and ice (see Figure 19).

With the advent of photochromic lenses in the 1960s, sunglasses became increasingly more sophisticated. Photochromic lenses have the ability to darken in sunlight, then to gradually lighten in shade or filtered light. This type of lens owes its effectiveness to a reversible chemical reaction that oc curs within the glass reaching equilibrium. The Figure 20 General Douglas basic principle is based on MacArthur, photographed in 1950 the fact that when it is wearing aviator glasses manufactured, the melted lens glass is supplemented with a silver chloride solution (AgCl) and traces of copper chloride (CuCl). After absorbing UV light, each copper (Cu+) and silver (Ag+) ion donates an electron to an ion (see Figure 21).

UV light

Figure 19 An Inuit person wearing wooden iggaaks

Ag CI

Cu CI

AgCI

Cu CI e Cu2

Ag (s)

In 1752, James Ayscough accidentally invented the first pair of sunglasses by manufacturing tinted glass to correct certain anomalies of the eye. However, sunglasses would not become popular until 1929, when Sam Foster opened an eyewear store on the boardwalk in Atlantic City, in the U.S.

Figure 21 The glass-darkening process. When the light intensity diminishes, the opposite process begins.

Sunglasses continued to evolve throughout the 1930s when US army scientists perfected a dark green tint specially designed for airplane pilots, who were often blinded by high-altitude glare. Improved glass and the addition of a polarizing filter developed by Edwins H. Land led to the design and commercialization of aviator glasses in 1937 (see Figure 20).

Silver atoms form and group together to form clusters of atoms large enough to block light and darken the lenses. Copper chloride makes the process reversible and restores the initial equilibrium, which corresponds to the time the lens is lightened. The glass of the lens returns to its transparent state, and equilibrium is maintained until the light intensity changes.

342

UNIT 4 Chemical Equilibrium

Ag (s) clusters

CHAPTER

12

The Quantitative Aspect of Chemical Equilibrium

12.1 Equilibrium constant • The equilibrium constant (Kc), also called the equilibrium law, is a relationship that shows that at a given temperature, in any simple chemical reaction at equilibrium, there is a constant ratio between the concentration of the products (C, D) and the concentration of the reactants (A, B), each concentration being raised to a power corresponding to the stoichiometric coefficient. [C]c  [D]d * Kc  [A]a  [B]b x cC  dD * For a hypothetical simple reaction of the type aA  bB w • The value of the equilibrium constant is calculated by using only the concentrations of the substances in the gas phase or in solution. • The value of the equilibrium constant indicates the extent of a chemical reaction. A high value of the equilibrium constant indicates that the system favours the direct reaction, while a low value of the equilibrium constant indicates that the system favours the reverse reaction. • For a given system at equilibrium, only temperature can change the equilibrium constant. Temperature change

Type of reaction

Reaction favoured

Change in Kc

Exothermic (H  0) Reactants n products  energy

Increase

Reverse (m)

Decrease

Decrease

Direct (n)

Increase

Endothermic (H  0) Reactants  energy n products

Increase

Direct (n)

Increase

Decrease

Reverse (m)

Decrease

12.2 Ionic equilibrium in solutions • Over the years, various theories have been formulated to better define acids and bases. Comparison of the Arrhenius and Brønsted-Lowry theories Theory

Arrhenius

Brønsted-Lowry

Acid

Any substance that dissociates in water to produce hydrogen ions (H)

Any substance from which a proton (hydrogen ion, H) can be removed

Base

Any substance that dissociates in water to produce hydroxide ions (OH-)

Any substance that can remove a proton from an acid

CHAPTER 12 The Quantitative Aspect of Chemical Equilibrium

343

• Ionic equilibrium in solutions is a state of equilibrium that is established between the concentrations of the various ions following the dissociation of a chemical compound in a solution. • The pH of a solution corresponds to the opposite of the logarithm of the hydronium ion concentration (H3O) in mol/L. pH is expressed as a numerical value without units. pH  log [H3O] [H3O]  10pH • The pOH of a solution, that is, the potential of hydroxide ion (OH) concentration, is expressed as follows: pOH  log [OH] [OH]  10pOH • The equilibrium constant (Kwater) for the ionization of water is expressed as follows: Kwater  [H3O]  [OH] • The ionization constant of water makes it possible to deduce the molar concentrations of hydronium ions (H3O) and hydroxide ions (OH). The numerical value of this constant is 1  1014 at 25°C. Consequently, pH  pOH  14. • The strength of a weak acid can be expressed using an acidity constant (Ka). The higher the acidity constant, the stronger the acid. For a weak acid, the acidity constant is expressed as follows: [H O]  [A] * Ka  3 [HA] * For the following reaction: HA

(aq)

 H2O

(l)

w x H3O (aq)  A (aq)

• The strength of a weak base can be expressed using a basicity constant (Kb). The lower the basicity constant, the weaker the base. For a weak base, the basicity constant is expressed as follows: [HB]  [OH] * Kb  [B] * For the following reaction: B

(aq)

 H2O

(l)

w x HB (aq)  OH (aq)

• The solubility of a substance corresponds to the maximum quantity of substance that dissolves in a given volume of solvent at a given temperature. Generally, solubility is expressed in grams of solute per 100 millilitres of solvent (g/100 mL). Solubility can also be expressed in moles of solute per litre of solvent (mol/L). This is referred to as molar solubility. • The solubility equilibrium attained after the partial dissolution of an ionic compound is described by the following equation: w nX x X Y  mY n m (s)

(aq)

(aq)

The resulting solubility product constant (Ksp) is represented by the following expression: Ksp  [X]n  [Y]m

344

UNIT 4 Chemical Equilibrium

CHAPTER 12

The Quantitative Aspect of Chemical Equilibrium

1. a) What is the conjugate acid of a base? Give an example. b) What is the conjugate base of an acid? Give an example. 2. Write the formula for the conjugate acid of the following substances: a) Hydroxide ion (OH) b) Carbonate ion (CO32) 3. Among the following compounds, which is an acid according to the Arrhenius theory? a) Water (H2O) b) Calcium hydroxide (Ca(OH)2) c) Phosphoric acid (H3PO4) d) Hydrogen fluoride (HF) 4. Among the following compounds, two are Arrhenius bases. Which ones? a) Potassium hydroxide (KOH) b) Barium hydroxide (Ba(OH)2) c) Hypochlorous acid (HOCl) d) Phosphoric acid (H3PO4) 5. What is the pH of a 100-mL sample of sulphuric acid (H2SO4) at 0.002 mol/L? 6. A 1-L solution contains 1.04 g of potassium hydroxide (KOH). What is the pH of this solution? 7. Write the expression of the equilibrium for each of the following homogeneous reactions: x SbCl3 (g)  Cl2 (g) a) SbCl5 (g) w xN b) 2 H  2 NO w 2H O 2 (g)

(g)

2 (g)

2

(g)

x 4 H2 (g)  CS2 (g) c) 2 H2S (g)  CH4 (g) w 8. When 1 mole of ammonia (NH3) gas is injected into a 0.5-L round bottom flask, the following reaction occurs until it attains equilibrium: x N2 (g) 3 H2 (g) 2 NH3 (g) w At equilibrium, the presence of 0.3 moles of hydrogen (H2) gas is noted. a) Calculate the concentration of nitrogen (N2) gas and ammonia at equilibrium. b) What is the value of the equilibrium constant of this reaction?

9. At a given temperature, the value of the equilibrium constant of the following reaction of sulphur dioxide (SO2) and nitrogen dioxide (NO2) is 4.8: x NO (g)  SO3 (g) SO2 (g)  NO2 (g) w The reactants have the same initial concentration of 0.36 mol/L. What quantity of sulphur trioxide (SO3) gas is present at equilibrium in a 5-L container? 10. The following is an expression of the equilibrium constant: Kc 

[NO]4  [H2O]6 [NH3]4  [O2]5

a) Write the chemical equation of this reversible reaction. b) If, at a given temperature, the concentrations of nitrogen monoxide (NO) and ammonia (NH3) are equal, and the concentration of the water vapour is 2.0 mol/L and that of the oxygen (O2) is 3 mol/L, what is the value of the equilibrium constant at this temperature? 11. In a sealed container, nitrogen dioxide (NO2) is at equilibrium with nitrogen tetroxide (N2O4), with an equilibrium constant of 1.15 at 55°C, according to the following equation: x N2O4 (g) 2 NO2 (g) w a) What is the equation of the equilibrium constant of this chemical system? b) If the concentration at equilibrium of nitrogen dioxide is 0.05 mol/L, what is the concentration of nitrogen tetroxide? c) What shift in equilibrium will lead to an increase in the concentration of nitrogen dioxide? 12. Categorize the reactants of the following equations based on whether they are an acid or a base according to the Brønsted-Lowry theory: x F (aq)  HSO3 (aq) a) HF (aq)  SO32(aq) w x b) CO32 (aq)  CH3COOH (aq) w CH3COO (aq)  HCO3 (aq) x H2PO4 (aq)  HOCl (aq) c) H3PO4 (aq)  OCl (aq) w x SO42 (aq)  H2CO3 (aq) d) HCO3 (aq)  HSO4 (aq) w CHAPTER 12 The Quantitative Aspect of Chemical Equilibrium

345

13. A hydrated proton is called a: a) hydronium ion b) hydroxide ion c) hydroxyl group d) Brønsted-Lowry base

20. In which of the following reactions does the water behave as an acid? x OH (aq)  NH4 (aq) a) H2O (l)  NH3 (aq) w

14. The hydroxide ion (OH) concentration in a window cleaning solution is 2.1 mol/L. The hydronium ion (H3O) concentration of this solution is: a) 2.1  1015 mol/L b) 4.8  1015 mol/L c) 2.1  1012 mol/L d) 4.8  1012 mol/L 15. A pH metre indicates that the pH of a soft drink is 3.46. The pOH of the soft drink is: a) 10.46 c) 14.46 b) 10.54 d) 14.54 16. The recipe for preparing a cleaning solution requires 5 g of sodium hydroxide (NaOH) be dissolved in 4 L of water. The pH of the solution obtained is: a) 1.51 c) 13.1 b) 12.49 d) 13.9 17. 5% of a 0.10 mol/L solution of a weak acid dissociates. Calculate the acidity constant. 18. Put the following aqueous solutions in increasing order of pH: a) Perchloric acid (HClO4) at 2 mol/L b) Sodium chloride (NaCl) at 2 mol/L c) Acetic acid (CH3COOH) at 0.2 mol/L d) Hydrochloric acid (HCl) at 0.02 mol/L 19. The following equilibrium has an equilibrium constant of 4.8  103 : x 2 NO2 (g) N2O4 (g) w Which of the following sets of concentrations represents the conditions at equilibrium?

346

Concentration of nitrogen tetroxide (N2O4)

Concentration of nitrogen dioxide (NO2)

a)

4.8  101

1.0  104

b)

1.0  101

4.8  104

c)

1.0  101

2.2  102

d)

2.2  102

1.0  101

e)

5.0  102

1.1  102

UNIT 4 Chemical Equilibrium

x H3O (aq)  H2PO4 (aq) b) H2O (l)  H3PO4 (aq) w 1 x H2 (g)  c) H2O (l) w O 2 2 (aq) x BaCl2 (aq)  2 H2O (l) d) 2 H2O (l)  BaCl2 (s) w x 4 NaOH (aq)  O2 (aq) e) 2 Na2O2 (s)  2 H2O (l) w 21. A saturated solution of silver acetate (ethanoate) (CH3COOAg) contains 2  103 moles of silver ions (Ag) per litre of solution. What is the value of the solubility product constant for the silver acetate? a) 2  103 d) 2  106 6 b) 4  10 e) 4  103 c) 1  103 22. Calculate the pH of each of the following solutions given the hydronium ion (H3O) concentration: a) [H3O]  0.0027 mol/L b) [H3O]  7.28  108 mol/L c) [H3O]  9.7  105 mol/L d) [H3O]  8.27  1012 mol/L 23. If the hydronium ion (H3O) concentration is approximately 5.0  103 mol/L in a cola, calculate the pH of this soft drink. Is this solution acidic or basic? 24. Put the following foods in increasing order of acidity: a) Beets, pH  5 b) Camembert, pH  7.4 c) Egg white, pH  8 d) Sauerkraut, pH  3.5 e) Yogurt, pH  4.5 25. Calculate the pH of each of the following body fluids given the hydronium ion (H3O) concentration in each: a) Tears, [H3O]  4  108 mol/L b) Gastric acid, [H3O]  4  102 mol/L 26. The following equation represents the dissociation of hydrogen iodide (HI) in the gaseous state: xH 2 HI w I (g)

2 (g)

2 (g)

At 430°C, the value of the equilibrium constant is 0.20. Hydrogen iodide (HI) is placed in a closed container at 430°C. The analysis of the equilibrium indicates that the concentration of iodine (I2) is 5.6  104 mol/L. What are the concentrations of hydrogen (H2) and hydrogen iodide (HI) at equilibrium?

27. A saturated solution of an ionic compound with low solubility does not contain a solid solute. Is this system at equilibrium? Briefly explain your answer. 28. Codeine (C18H21NO3), one of the ingredients in migraine relief tablets, has a basicity constant of 1.73  106. Calculate the pH of a codeine solution at 0.02 mol/L. 29. For the equilibrium below, the value of the equilibrium constant is 6.00  102 : x 2 NH3 (g) N2 (g)  3 H2 (g) w Explain why the value of the equilibrium constant does not apply when the equation is written as follows: 1 3 x NH3 (g) N  H w 2 2 (g) 2 2 (g) 30. In the following equilibrium, the value of the equilibrium constant is 1.0  1015 at 25°C and 0.05 at 2200°C: x 2 NO (g) N2 (g)  O2 (g) w Based on this information, is the reaction exothermic or endothermic? Briefly explain your answer. 31. 8.5 g of sodium hydroxide (NaOH) is dissolved to obtain 500 mL of a cleaning solution. Determine the pOH of this solution. 32. Acetic (ethanoic) acid (CH3COOH) is the weak acid most commonly used in industry. Determine the pH and pOH of 1.25 kL of a solution prepared by dissolving 60 kg of this pure acid in liquid form. 33. Determine the mass of sodium hydroxide (NaOH) that must be dissolved to obtain 2 L of a solution with a pH of 10.35. 34. Examine an equilibrium in which oxygen (O2) reacts with hydrogen chloride (HCl) gas to form water vapour and chorine (Cl2) gas. At equilibrium, these gases have the following concentrations: – [O2]  8.6  102 mol/L – [HCl]  2.7  102 mol/L – [H2O]  7.8  103 mol/L – [Cl2]  3.6  103 mol/L a) Write the balanced chemical equation of this reaction. b) Calculate the value of the equilibrium constant.

35. Sulphur trioxide (SO3) gas reacts with hydrogen fluoride (HF) gas to produce sulphur hexafluoride (SF6) gas and water vapour. The value of the equilibrium constant is 6.3  103. a) Write the balanced chemical equation of this reaction. b) 2.9 moles of sulphur trioxide is mixed with 9.1 moles of hydrogen fluoride in a 4.7-litre round bottom flask. Write an equation to determine the concentration of the sulphur hexafluoride at equilibrium. c) Explain why this equation cannot be solved. 36. Answer the following questions: a) Write a definition of an acid and a base according to the Brønsted-Lowry theory. b) What points do the Brønsted-Lowry theory and the Arrhenius theory have in common? How are these theories different? 37. During various reactions in an aqueous solution, the bicarbonate ion (HCO3) can behave as an acid or a base. a) Write the chemical formula of the conjugate acid and the conjugate base of this ion. b) Complete the following equations and indicate if the bicarbonate ion is a Brønsted-Lowry acid or base. – HCO3 (aq)  H3O (aq) n ? – HCO3 (aq)  OH (aq) n ? 38. The maximum solubility of barium fluoride (BaF2) at 25°C is 1.3 g/L. a) Calculate the value of the solubility product constant of barium fluoride at 25°C. b) Calculate the solubility of barium fluoride in molecules of barium fluoride per litre (L). 39. Why do tables of values of the solubility product constant of compounds not give the value of soluble compounds? 40. Values of the solubility product constant are provided for a given temperature. How does the value of the solubility product constant of most salts vary if the temperature of the solution increases? 41. For a solubility system to attain equilibrium, a quantity of undissolved solid must be present. Explain why.

CHAPTER 12 The Quantitative Aspect of Chemical Equilibrium

347

42. Lead iodide (PbI2) and barium sulfate (BaSO4) have almost the same solubility product constant: – Lead iodide: Ksp  9.8  109 – Barium sulfate: Ksp  1.1  1010 What is the ratio of the number of lead ions (Pb2) to barium ions (Ba2) in solutions saturated with these salts? 43. The concentration of iodine (I) in a saturated solution at 25°C is 1.5  104 ppm. a) Calculate the value of the solubility product constant for mercury iodide. The solubility equilibrium is written as follows:

47. A saturated solution of copper phosphate (Cu3(PO4)2) has a concentration of 6.1  107 g of copper phosphate per 1.00  102 mL of solution at 25°C. What is the value of the solubility product constant for copper phosphate at 25°C? 48. A student dissolves 5 g of vitamin C in 250 mL of water. The value of the solubility product constant of the ascorbic acid (C6H8O6) is 8  105. Calculate the pH of the solution.

49.

x H2O (g)  CO (g) CO2 (g)  H2 (g) w At 2000 K, the concentrations of the compounds in the system are: – Concentration of carbon dioxide (CO2) gas: 0.30 mol/L – Concentration of hydrogen (H2) gas: 0.2 mol/L – Concentration of water vapour and carbon monoxide (CO) gas: 0.55 mol/L a) What is the value of the equilibrium constant? b) When the temperature is decreased, 20% of the carbon monoxide is reconverted into carbon dioxide. Calculate the equilibrium constant at this lower temperature. c) Rewrite the equilibrium equation and indicate on which side of the equation the heat term should be placed.

x Hg2 2 (aq)  2 I (aq) Hg2I2 (s) w b) Write down the hypotheses you formulated, converting parts per million (ppm) into moles per litre (mol/L). 44. An 11.5-g sample of diatomic iodine (I2) is sealed in a 250-mL flask. At equilibrium, it is established that this molecular form of diatomic iodine dissociates into atoms of iodine (I), as shown by the following equation: x 2 I (g) I2 (g) w The value of the equilibrium constant of this reaction is 3.8  105. Calculate the concentration at equilibrium of these two forms of iodine. 45. Salicylic acid (C6H4OHCOOH) is one of the active ingredients in cleansing solutions used to treat acne. Since the value of its acidity constant is not provided in any reference book, a student decides to determine it experimentally. If he finds that a saturated solution of this acid (1 g/460 mL) has a pH of 2.4 at 25°C, calculate the ionization constant of salicylic acid. 46. Does a greater value of the solubility product constant always indicate greater solubility? To illustrate your answer, calculate and compare the molar solubility of silver chloride (AgCl), whose solubility product constant is 1.8  1010, and that of silver chromate (Ag2CrO4), whose solubility product constant is 2.6  1012.

348

UNIT 4 Chemical Equilibrium

Consider the following equilibrium:

50.

Observations suggest that the hydrogenosulfide ion (HSO3) is an amphoteric substance. A solution of sodium hydrogenosulfide (NaHSO3) can partially neutralize a spill of sodium hydroxide (NaOH) or hydrochloric acid (HCl). a) Write the net ionic equation of the reaction between hydrogenosulfide ions in an aqueous solution and hydroxide ions (OH) in solution. Indicate if the reactants are acids or bases. b) Write the net ionic equation of the reaction of hydrogenosulfide ions and hydronium ions (H3O) produced by a hydrochloric acid solution. Indicate if the reactants are acids or bases.

Supplement

CONTENTS 1

Oxidation and reduction. . . . . . . . . . . . . . . . . . . . 350 1.1 General equation for redox reactions . . . . . . . . 351

2

Oxidation number . . . . . . . . . . . . . . . . . . . . . . . . . 353 2.1 Oxidation number of ionic compounds . . . . . . . 353 2.2 Oxidation number of covalent compounds . . . . 353

3

Reducing power of metals. . . . . . . . . . . . . . . . . . 357 3.1 Spontaneous reactions . . . . . . . . . . . . . . . . . . . 357

Review Chemical formulas and ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Representation of atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Nature of chemical bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4

Electrochemical cell. . . . . . . . . . . . . . . . . . . . . . . 358

5

Reduction and oxidation potential. . . . . . . . . . . 360 5.1 Standard electrode potential . . . . . . . . . . . . . . 360 5.2 Cell potential . . . . . . . . . . . . . . . . . . . . . . . . . . . 363

SUPPLEMENT Redox Reactions

349

1 Oxidation and reduction A redox (oxidation-reduction) reaction involves two simultaneous half-reactions: oxidation and reduction. Oxidation is a chemical reaction in which atoms lose electrons. Reduction is a reaction in which atoms gain electrons.

Figure 1 Rust is the result of a redox reaction. Oxidized iron becomes iron oxide (Fe2O3).

A redox reaction consists of an electron exchange between reactants. Despite what the name would seem to indicate, oxygen is not necessary for oxidation. Every chemical reaction involving a loss of electrons is an oxidation reaction, while every reaction involving a gain of electrons is a reduction reaction. Since half of a redox reaction consists of an oxidation reaction and half consists of a reduction reaction, we describe these events as oxidation half-reactions and reduction halfreactions. As indicated by their position in the periodic table, some atoms tend to accept electrons to become anions (negative ions). The opposite is also possible: some atoms tend to lose electrons to become more stable cations (positive ions). When electrons leave an atom, they typically join another atom. The presence of free electrons in the environment is highly unusual. That is why oxidation and reduction reactions always occur simultaneously in an oxidation-reduction reaction. Furthermore, redox reactions take place most often in an aqueous solution, which allows the electrons to move from one atom to another. Each half-reaction can be represented by an equation like the one below for the oxidation half-reaction of iron: Oxidation half-reaction

state An atom is in ground * Ground state when it is electrically neutral, with an equal number of electrons and protons.

Fe (s)

n

Fe2 (aq)



2 e

Fe

n

Fe2



e

e

In this example, an atom of iron in solid state loses two electrons (2 e) to form the ferrous ion (Fe2) that has become aqueous. The iron atom’s charge has gone from 0 to 2 because it is now missing two electrons in comparison to its ground state . Since the iron has lost electrons, it has been oxidized (see Figure 1).

*

Conversely, a reduction half-reaction occurs when an element gains one or more electrons, as in the following equation: Reduction half-reaction Au3 (aq)



3 e

n

Au (s)

Au3



e e e

n

Au

In this reaction, the gold ion (Au3) is reduced to an atom of solid gold by gaining three electrons. After the electron transfer, the ion’s charge increases by three to return to zero, which is the atom’s charge in ground state. Since the gold has gained electrons, it has been reduced.

350

SUPPLEMENT Redox Reactions

A reactant that gains electrons in a redox reaction is an oxidizing agent because it causes the oxidation of another substance. It is reduced in the reduction halfreaction. Conversely, a reactant that loses electrons is a reducing agent because it causes the reduction of another substance. By giving up electrons, the reducing agent is oxidized in the oxidation half-reaction (see Table 1). Table 1 Characteristics of reducing agents and oxidizing agents Agent

Electron exchange

Half-reaction

The atom is

Reducing agent

It loses electrons

Oxidation

Oxidized

Oxidizing agent

It gains electrons

Reduction

Reduced

It is important to note that in a redox reaction the oxidizing agent and the reducing agent are a pair, since there is always an exchange of electrons between two substances. Since the oxidizing agent gains electrons, it is the one that causes the oxidation of the reducing agent. In return, the reducing agent causes the reduction of the oxidizing agent by giving it electrons (see Figure 2).

1.1

e Reducing agent

Oxidizing agent

Figure 2 In a redox reaction, electrons are transferred from a reducing agent to an oxidizing agent.

General equation for redox reactions

The general equation for a redox reaction can be found by simply adding the two half-reactions that make it up, if the number of electrons involved is the same in each half-reaction. For example, the redox reaction of solid zinc and copper ions in an aqueous solution is shown in the following equation: Zn (s) n Zn2 (aq)  2 e

Oxidation half-reaction Reduction half-reaction

Cu2 (aq)  2 e n Cu (s)

Overall reaction

Zn (s)  Cu2 (aq) n Zn2 (aq)  Cu (s) Zn

Cu2

Zn2

Cu

Reducing agent

Oxidizing agent

Oxidized

Reduced

The overall reaction can be found by eliminating the identical terms on either side of the arrow, such as the two electrons in this example. These two electrons were transferred from the zinc atom to the copper ion, which became a groundstate copper atom. At the same time, the zinc atom became a positive ion. Still, depending on what substances are involved, the number of electrons in one half-reaction can be different from the number in the other half-reaction. In this case, to balance the equations, we multiply one or both of the half-reactions by an integer so that the number of electrons in each half-reaction is the same.

SUPPLEMENT Redox Reactions

351

For example, the equation for the reduction half-reaction of silver must be multiplied by two so that the number of exchanged electrons is equal in the two half-reactions.

Figure 3 A copper wire is immersed in a silver nitrate (AgNO3) solution. During the reaction, the solution turns blue, indicating the presence of aqueous Cu2 ions. The deposit on the copper wire is solid silver.

Oxidation of an apple Oxygen (O 2) is the oxidizing agent in many redox reactions, whether they are mineral, like the corrosion of iron, or organic, like the oxidation of an apple. The flesh of an apple oxidizes, that is, it changes from white to brown, when it is exposed to oxygen in the ambient air. The oxygen atom acts as the oxidizing agent, and the reduced substances are organic carbon-based molecules, such as quinone (C6H4O2). Once these substances are oxidized, they form complex molecules with a brownish colour and a different texture from the unoxidized flesh.

Figure 4 Oxidation of an apple in the presence of oxygen (O2) in the air. The left side of the apple has just been cut and shows no signs of oxidation, unlike the right half.

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SUPPLEMENT Redox Reactions

Reduction half-reaction

2 Ag(aq)  2 e

n

2 Ag (s)

Oxidation half-reaction

Cu (s)

n

Cu2 (aq)  2 e

Overall reaction

2 Ag (aq)  Cu (s)

n

Cu2 (aq)  2 Ag (s)

In this reaction, a copper wire (Cu), formed of solid copper atoms, is immersed in a solution of silver nitrate (AgNO3) containing aqueous silver ions (Ag) (see Figure 3). When two silver ions come into contact with the surface of the copper, two electrons are exchanged between a copper atom and two aqueous silver ions. The silver ions gain electrons and are reduced. By doing so, they become solid and are deposited onto the surface of the copper. At the same instant, the copper loses two electrons, oxidizes and moves into the aqueous environment in the form of an aqueous copper ion (Cu2). The diameter of the copper wire decreases as the solid copper is transformed into aqueous copper ions. Over time, the solution turns blue, a sign of the appearance of the copper ions in solution. The following example demonstrates how to recognize the half-reactions involved in a redox reaction. It also shows how to identify the oxidizing agent and the reducing agent: Example The reaction of a piece of magnesium (Mg) in hydrochloric acid (HCl) results in the formation of magnesium dichloride (MgCl2). The release of hydrogen (H2) can also be observed. During this reaction, metallic magnesium is oxidized into aqueous Mg2 ions, while the aqueous H ions of the acidic solution are reduced to hydrogen gas. a) What are the half-reactions in this reaction? b) What is the general equation for oxidation-reduction? c) Find the oxidizing agent and the reducing agent. Method a) Since the magnesium is oxidized, it loses electrons. The Mg2 aqueous ion indicates that two electrons were lost in oxidation. This shows that the equation for the oxidation half-reaction is the following: Answer: Oxidation equation: Mg (s) n Mg2(aq)  2 e The H ions were reduced, so, each H ion gained 1 electron during its reduction, which gives the following reduction equation: 1 Answer: Reduction equation: H(aq)  1 e n H 2 2 (g) b) To add the half-reactions, multiply the reduction equation by an integer so that the number of electrons is the same for both oxidation and reduction. n Mg2 (aq)  2 e Oxidation: Mg (s) n Mg2 (aq)  2 e Mg (s) 1 Reduction: 2 (H (aq)  1 e n H2 (g)) 2 H (aq)  2 e n H2 (g) 2 Answer: Overall reaction: 2 H (aq)  Mg (s) n Mg2 (aq)  H2 (g) c) Answer: The oxidizing agent is the aqueous Hion, since it gains electrons. The reducing agent is solid magnesium (Mg) because it loses electrons.

2 Oxidation number The oxidation number, also called the oxidation state, indicates the number of electrons an element has lost or gained, in relation to its ground state, during a redox reaction. According to convention, all elements in ground state or in a molecule of an element have an oxidation number of 0. These elements are considered to be atoms that have not lost or gained any electrons. When atoms are involved in chemical reactions, their oxidation numbers vary. The oxidation number increases in oxidation because atoms lose electrons. Conversely, the oxidation number decreases during reduction because atoms gain electrons. The sum of the oxidation numbers of the atoms in a neutral molecule is equal to zero. To find the oxidation number of an atom, we must first determine whether it is part of an ionic or covalent compound.

2.1

Oxidation number of ionic compounds

The atoms in ionic compounds are ionized because electrons have been exchanged between the metal and the non-metal when the compound was formed. The metal then forms a positive ion, and the non-metal forms a negative ion. When an element is ionized, its oxidation number is equal to its charge. Calcium chloride (CaCl2), for example, is a salt used to de-ice roads when winter driving conditions are hazardous (see Figure 5). The CaCl2 molecule consists of three ions: one Ca2 ion and two Cl ions. To distinguish between an ion’s charge and its oxidation number, their negative and positive values are expressed differently. According to convention, the charges of the Ca2 and Cl ions are 2 and 1, respectively. Conversely, the oxidation numbers of these two ions are 2 and 1. Table 2 shows the charges and oxidation numbers of a variety of categories of substances.

Figure 5 Calcium chloride (CaCl2) is an ionic compound that is spread on icy roads.

Table 2 Charges and oxidation numbers of some substances Substances

Charge

Oxidation number

Elements in ground state (Li, Mg, Al, Fe, etc.)

0

0

Molecules of elements (H2, O2, Cl2, N2, S8, etc.)

0

0

Ions of alkali metals (Li, Na, K, etc.)

1

1

Ions of alkaline earth metals (Ca2, Mg2, Be2, etc.)

2

2

2.2

Oxidation number of covalent compounds

When an atom is part of a molecule or a polyatomic ion, convention determines its oxidation number by assigning each pair to the more electronegative atom in the bond, that is, the atom that is more likely to attract electrons to fill its outermost shell. To easily determine the oxidation number of a molecule, the molecule can be represented with Lewis notation.

SUPPLEMENT Redox Reactions

353

H

O

H

Figure 6 Lewis notation permits finding the oxidation number of the oxygen atom (2) and the hydrogen atoms (1) in the water molecule.

For example, in the water molecule, the pair in each covalent bond between oxygen and hydrogen is assigned to oxygen because it is more electronegative (see Figure 6). Since oxygen usually has six electrons in ground state, this gives it two additional electrons. So, the oxidation number of oxygen is -2. Since each hydrogen atom has lost an electron, the oxidation number of hydrogen is +1. Still, in this type of covalent compound, the oxidation number is not the same as the atom’s charge. For example, a water molecule contains no O2 or H ions. Assigning the electrons to the oxygen atom is just a means of calculating the oxidation number. In reality, the electrons are actually shared between the hydrogen and oxygen atoms, creating a covalent bond between them. The following example shows how to calculate the oxidation number of atoms in a molecule with Lewis notation: Example A What are the oxidation numbers of the atoms in the ammonia (NH3) molecule? Method 1. Use Lewis notation to illustrate the molecule. Since the NH3 molecule contains no metallic atoms, it is formed of covalent bonds.

H N H H

2. Use the periodic table to determine which atom is more electronegative and find its oxidation number. In this case, it is nitrogen. The pairs in the bonds must therefore be assigned to nitrogen, bringing the number of valence electrons to 8, or 3 more than nitrogen has in ground state. The oxidation number of nitrogen is thus 3. 3. Find the oxidation number of hydrogen atoms. Since the electron in each hydrogen atom has been assigned to the nitrogen, each hydrogen atom is missing an electron, which gives it an oxidation number of 1. Answer: The oxidation number of nitrogen is 3, while that of hydrogen is 1. Table 3 Oxidation numbers of some elements in compounds

Substances

Oxidation number (ON)

Hydrogen bonded to a non-metal

1

Alkali metals

1

Alkaline earth metals

2

Hydrogen bonded to a metal

1

Oxygen

2

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Some atoms can exist in the same state with a variable oxidation number, that is, they can have a different oxidation number according to the compound they are part of. This is true of chromium, which can be found in variable oxidation states. Chromium can exist in the Cr3 state in chromium oxide (Cr2O3), or in the Cr6 state in potassium dichromate (K2Cr2O7). To find the oxidation number of an atom in such a compound, Lewis notation can be used, and the oxidation number of the other atoms in the molecule can then be deduced. In fact, some elements have a stable oxidation number (See Table 3). These include alkali metals, which usually have an oxidation number of 1 when they are in an ionic compound. It is the same for alkaline earth metals, which always have an oxidation number of 2 when they are joined to other atoms by ionic bonds.

There are some non-metals that usually have the same oxidation number. Oxygen is one of them. It most often has an oxidation number of 2, except when it is in a peroxide compound. Then it has an oxidation number of 1. Similarly, hydrogen almost always has an oxidation number of 1, but it can sometimes have an oxidation number of 1 when it is in metallic compounds such as lithium hydride (LiH), sodium hydride (NaH) or calcium hydride (CaH2). In these compounds, hydrogen is more electronegative than the metallic atoms, and the pair in the bond is assigned to it. In cases of complex ions which would be difficult to represent with Lewis notation, an atom’s oxidation number can be found by using the following method: Example B What is the oxidation number of the manganese atom in the permanganate ion (MnO4), a very strong oxidizing agent? The permanganate ion’s negative charge indicates that the net charge is 1. Furthermore, since the oxidation number of oxygen is 2 and there are four oxygen atoms in this molecule, use the following mathematical equation: Mn  4 O  1 X  4 (2)  1 X  8  1 X7 Answer: The oxidation number of the manganese atom is thus 7.

In this example, find the oxidation number of each atom: Example C What are the oxidation numbers of all of the atoms in the following redox reaction? Fe2O3 (s)  2 Al (s) n Al2O3 (s)  2 Fe (l) Method 1. According to the information in Table 3 (see previous page), each oxygen atom in the Fe2O3 molecule has an oxidation number of 2, for a total charge of 6. Since the molecule’s overall charge is 0, each iron atom has a charge of 3 . 2. Since aluminum is in ground state, its oxidation number is 0. 3. The Al2O3 molecule is similar to iron oxide. Each oxygen atom in the molecule has an oxidation number of 2, for a total charge of 6. Since the molecule’s overall charge is 0, each aluminum atom has a charge of 3. 4. Iron has an oxidation number of 0, since it is also in ground state. Fe2O3 (s)  Oxidation number:

3

2

2 Al (s) n Al2O3 (s)  0

3 2

2 Fe (l) 0

Answer: In Fe2O3, the oxidation number of Fe is 3 and that of oxygen is 2. The oxidation number of atomic Al is 0. In Al2O3, the oxidation number of aluminum is 3 and that of oxygen is 2. The oxidation number of atomic Fe is 0.

SUPPLEMENT Redox Reactions

355

SECTION 1 SECTION 2

Oxidation and reduction Oxidation number

1. Determine whether the following half-reactions involve oxidation or reduction: a) Na (s) n Na (aq)  1 e 1 b) O  2 e n O2 (aq) 2 2 (aq) c) Pb2 (aq)  2 e n Pb (s) d) Zn (s) n Zn2 (aq)  2 e 2. For each of the following redox reactions, indicate which substance loses electrons and which gains electrons: a) 2 H (aq)  Pb (s) n Pb2 (aq)  H2 (g) b) Sn4 (aq)  Ni (s) n Ni2 (aq)  Sn2 (aq) 3. Which of the following definitions is incorrect? a) Oxidation is the process in which a substance loses one or more electrons. b) The terms oxidation and oxidizing agent are synonymous. c) A reducing agent is a substance that oxidizes. d) An anion is a negative ion. 4. Choose the right answer. The aqueous copper ion Cu2 comes from: a) a copper ion that has lost two electrons b) a copper atom that has gained two electrons c) a copper atom that has lost three electrons d) a copper atom that has lost two electrons 5. Find the general redox reaction equation from the following half-reactions: a) Fe (s) n Fe3 (aq)  3 e Pb2 (aq)  2 e n Pb (s) b) In3 (aq)  3 e n In (s) Co (s) n Co2 (aq)  2 e 6. For each of the following reactions, write the corresponding pair of half-reactions and find the reducing agent and the oxidizing agent: a) Pb (s)  Cu2 (aq) n Pb2 (aq)  Cu (s) b) Cl2 (aq)  2 Br (aq) n 2 Cl (aq)  Br2 (l) 7. Find the oxidizing agent and the reducing agent in the following reactions: a) PbO2 (s)  C (s) n Pb (s)  CO2 (g) b) Sn (s)  Br2 (l) n SnBr2 (s) c) NiO (s)  H2 (g) n Ni (s)  H2O (l)

356

SUPPLEMENT Redox Reactions

8. Find the oxidation number of the atoms that make up the following substances: a) Cr2O72 f) CO2  b) MnO4 g) H2O2 c) H2O h) H2SO4 d) HClO4 i) HBr e) PbO2 j) NO2 9. In which of the following substances does chlorine have the lowest oxidation number? a) HClO4 b) HClO3 c) HClO d) HCl e) Cl2 10. For each of the following situations, write the equation for the half-reaction, indicate whether it is oxidation or reduction and explain why: a) Under the action of certain bacteria in the soil, ammonia (NH3) changes into nitrite ions (NO2) in the presence of acid. b) During the bleaching of paper pulp, hydrogen peroxide (H2O2) changes into water in the presence of acid. 11. Indicate which of the following reactions are redox reactions: a) Cu (s)  Cl2 (g) n CuCl2 (s) b) 2 C2H6 (g)  7 O2 (g) n 4 CO2 (g)  6 H2O (g) c) S2O32(aq)  I2 (aq) n S4O6 (aq)  2 I (aq) d) BF3 (g)  H2O (l) n HF (aq)  H3BO3 (l) e) HCl (aq)  NaOH (aq) n NaCl (aq)  H2O (l) 12. In the following reactions, find the oxidizing substances and the reducing agent: a) Zn (s)  2 HCl (aq) n ZnCl2 (aq)  H2 (g) b) Cu2S (s)  O2 (g) n 2 Cu (s)  SO2 (g) d)

SiH4 (g)  2 O2 (g) n SiO2 (s)  2 H2O (l) Fe2 O3 (s)  3 CO (g) n 2 Fe (s)  3 CO2 (g)

e)

3 NO2 (g)  H2O (l) n 2 HNO3 (aq)  NO (g)

c)

3 Reducing power of metals When immersing a metal rod in a solution of metallic ions, a deposit sometimes forms spontaneously on the rod. This deposit comes from the solution’s ions, which are reduced by the metal rod. On the other hand, when immersing a piece of silver in a copper solution, no deposit is formed. This phenomenon can be explained by comparing the reducing power of the substances involved. The reducing power of a metal is its likelihood of giving up its electrons, that is, of oxidizing. This tendency to give up electrons varies from one metal to another. As a rule, the alkali metals and the alkaline earth metals are viewed as good electron donors (see Figure 7). By giving up one or two valence electrons, respectively, these metals acquire the most stable configuration, that of noble gases. Metals can be classified according to their reducing power (see Figure 8).

Au

Pt

Ag

Hg

Cu

Pb

Sn

Ni

Cd

Fe

Cr

Zn

Al

Mg

Figure 7 Lithium is an alkali metal that oxidizes on contact with air. It is an excellent reducing agent because it gives up its electrons easily.

Na

Weak reducing agent

Ca

Ba

K

Li

Strong reducing agent

Figure 8 Classification of metals according to their reducing power

3.1

Spontaneous reactions

For a spontaneous redox reaction to occur between two metals, the stronger reducing agent must be in a solid state and the weaker reducing agent must be in an aqueous state, in the form of ions (see Table 4). In fact, for the metal that is the stronger reducing agent to give up its electrons, the other metal must be in an ionized state to be able to accept the electrons and thus be reduced. When an aluminum (Al) rod is immersed in a solution of Cu2, a copper (Cu) deposit forms spontaneously (see Figure 9). The explanation for this is that aluminum is a stronger reducing agent than copper. Since the copper ions are able to accept electrons, the reaction is spontaneous. Conversely, when immersing gold (Au) in a solution of Cu2(aq), there is no reaction. In fact, since copper is a stronger reducing agent than gold, it should give up its electrons to the gold. But the copper has already lost its electrons since it is in the aqueous Cu2 state. Also, the gold is unable to accept electrons because it is in a solid state. It is therefore impossible to form gold ions with a surplus of electrons in aqueous conditions.

Figure 9 Copper deposit on an aluminum rod immersed in a Cu2 solution

Table 4 Necessary conditions for a spontaneous reaction Reaction

Stronger reducing agent

Weaker reducing agent

Spontaneous

Solid

Aqueous solution

None

Aqueous solution

Solid

SUPPLEMENT Redox Reactions

357

4 Electrochemical cell An electrochemical cell is a device that can spontaneously generate an electrical current. An electrochemical cell is composed of two electrodes, also called half-cells because a reduction half-reaction and an oxidation half-reaction takes place on them. Each electrode is made up of a piece of metal immersed in an ionic solution. The two electrodes are joined by a wire, and the two solutions are connected by a salt bridge that consists of a tube with a porous membrane at each end. The tube is filled with a solution of inert ions, which do not take part in the reaction. An aqueous solution of potassium chloride (KCl) is usually used. A zinc-copper battery is a classic example of an electrochemical cell. It is made of a piece of zinc (Zn) immersed in a solution of zinc sulfate (ZnSO4) and a piece of copper (Cu) in a solution of copper sulfate (CuSO4) (see Figure 10). Since zinc is a better reducing agent than copper (see Figure 8 on the previous page), it is the zinc that gives up its electrons to the copper; this means that the zinc is oxidized. According to convention, the electrode on which the oxidation takes place is called the anode and it accepts anions; it is the positive terminal of the cell. The electrode on which reduction takes place is called the cathode and it accepts cations; it is the negative terminal of the cell. Redox reaction Zn (s)  Cu2 (aq) n Zn2 (aq)  Cu (s)

Anode

Cathode

K Cl Salt bridge Zinc

Copper

Zn2

ZnSO4 solution

SO42

Voltmeter

CuSO4 solution

Cu2 SO42

e

e

Oxidation

2 e

Cu2 Zn2

2 e Reduction Cu

Zn

Zn (s) is oxidized into Zn2 (aq) Zn (s) n Zn2 (aq)  2 e

Cu2 (aq) is reduced into (s) Cu2 (aq)  2 e n Cu (s)

Figure 10 A zinc-copper battery. A piece of zinc and a piece of copper are immersed in solutions of their respective ions. A salt bridge containing potassium chloride (KCl) connects the two containers.

358

SUPPLEMENT Redox Reactions

When the cell is working, the concentration of each half-cell’s ions varies. On the anode side, the concentration of Zn2 ions increases, while on the cathode side, the concentration of Cu2 ions decreases. Since the quantities of negative and positive charges in a solution are always equal, the function of the salt bridge between the two half-cells is to balance the ions in the two solutions. That is why the K ions in the salt bridge move from the right side to replace the missing copper ions, thereby balancing the charges in the solution. The same principle applies to the Cl ions that move toward the zinc solution as it becomes enriched with Zn2 cations.

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

The Cu2 ions in the cathode’s solution attract the electrons released by the anode’s zinc atoms. Therefore, the electrons lost by the zinc atoms do not go into the solution. They are held in the metal piece and move toward the cathode through the electrical wire that joins the two electrodes. The electrical current that runs from the anode to the cathode is generated by this movement of electrons. When an electrochemical cell is running, some characteristic phenomena taking place on the electrodes can be observed. During an oxidation reaction, which takes place on the anode, more and more zinc atoms are oxidized and changed into Zn2 ions, and then dissolved in the solution. Thus, the anode disintegrates and loses mass during the course of the reaction. Conversely, the copper ions are reduced to copper atoms and are deposited as solid copper on the surface of the cathode. This is why the mass of the cathode increases during the redox reaction process. The same phenomenon occurs in all electrochemical cells: the anode loses atoms while the cathode gains them, in sufficient quantities to change the mass of the electrodes.

Furthering F th i g Furt Furth Furthe urthering thering thering

your understanding d t di g

Cathodic protection Iron must be prevented from oxidizing in order to protect it from corrosion. When iron is involved in a redox reaction, for example, after contact with moist air, it is possible to prevent iron from losing electrons to the anode by making it the cathode. To reverse the redox reaction process, a sacrificial anode can be used. In this method, the piece of iron to be protected is placed in contact with a metal that oxidizes more easily than iron. Zinc and magnesium are the metals usually used to protect service station gasoline storage tanks or ships’ hulls from corrosion. The metal that is used supplies the iron with electrons as it gradually oxidizes so that it will become solid again. After a while, the piece of metal must be replaced because it has been reduced.

Underground steel storage tank (cathode)

Copper wire

Electron flow Block of magnesium (sacrificial anode)

Figure 11 A cathodic protection system prevents corrosion of a gasoline storage tank.

ALESSANDRO VOLTA Italian physicist (1745–1827) Most of Volta’s work centred on electricity. During his career, he designed many electrical machines and demonstrated that electricity could be generated by contact between metals. In 1800, he published an article on a cell that he had invented, the first of its kind, the voltaic cell which was named after him. This type of battery is made up of layers of copper and zinc, separated by paper soaked in a saline solution. Volta con nec ted a piece of metal at the bottom of the battery with one at the top and observed that an electrical current ran between them. The difference in potential between the metal pieces and the ions in the saline solution is what causes the electrons to move and generates the current.

SUPPLEMENT Redox Reactions

359

5 Reduction and oxidation potential During a redox reaction, one of the metals has greater potential energy than the other because it has a greater reducing power. Without this potential difference between two metals of different types, no chemical reaction would take place. It is possible to calculate the potential difference between two metals used in a battery. Since this difference depends on the types of metal used in each electrode, there must be a reference electrode to index the potentials of different metals. The reference electrode is the standard against which all the others are measured. The commonly used reference electrode in electrochemistry is the hydrogen electrode (see Figure 12). This electrode consists of a small platinum grid immersed in an acidic solution, usually hydrochloric acid (HCl), with a hydronium (H3O) ion concentration of 1 mol/L.

H2 at 101.3 kPa

HCl (aq)

Platinum electrode

Figure 12 The hydrogen electrode is a standard against which to measure the potential of different metals.

Hydrogen gas is introduced at a pressure of 101.3 kPa into the solution containing the platinum electrode so that the following reaction reaches equilibrium: z H2 (g) 2 H (aq)  2 e y

E 0  0.00 V

To compare the electrodes, assign a value of 0.00 V to the standard potential (E0) of the reference electrode. By comparing the electrodes with the reference electrode, the value of their potential can be found and they can be classified.

5.1

Standard electrode potential

The redox potential (E0) of an electrode is the potential difference between this electrode and a reference electrode at STP. To measure redox potential, calculate the difference in potential between this electrode and the hydrogen electrode. To do this, simply make a battery and connect a voltmeter between the hydrogen electrode and the electrode whose redox potential needs to be measured.

360

SUPPLEMENT Redox Reactions

For example, it is possible to measure experimentally the redox potential of a copper electrode by immersing a rod of copper in a solution containing copper ions (Cu2) (see Figure 13). A salt bridge is used to connect the copper electrode to a hydrogen electrode. This will permit the calculation of the difference in potential of the two electrodes.

Anode

Cathode

H2 (g) Salt bridge

25°C 100 kPa

Cu (s) 0.34 V

HCl (aq)

Cu2 (aq)

Oxidation

Reduction

Figure 13 A standard copper-hydrogen battery

The reaction is spontaneous and a value of 0.34 V can be seen on the voltmeter. A copper deposit can be observed on the electrode. This confirms that reduction took place on the surface of the copper electrode and that oxidation took place on the surface of the reference electrode. Since the redox potential of the hydrogen electrode is always zero, the positive value of 0.34 shown on the voltmeter indicates that the standard electrode potential of reduced copper is 0.34 V. The redox potential of a half-cell can be expressed like this:

E 0Cu 2/ Cu  0.34 V

The positive redox potential of an electrode connected to a hydrogen electrode indicates that the latter is acting as the cathode and that it is the location of the reduction half-reaction. Conversely, a negative redox potential indicates that the electrode connected to a hydrogen electrode is acting as an anode and that it is the location of the oxidation half-reaction. By calculating the potential difference between the electrodes and a reference electrode, the redox potential of all known electrodes can be found (see Table 5 on the following page). The electrodes in this table are ranked in descending order of standard electrode potential. So, if two electrodes of different metals on this table are used to make a fuel cell, the one with the lowest redox potential will cause the oxidation half-reaction. This ranking permits the comparison of the oxidizing or reducing power of substances. For example, chlorine gas (Cl2) is a better oxidizing agent than bromine (Br2), and potassium (K) is a better reducing agent than silver (Ag).

Hydrogen fuel cell The hydrogen fuel cell was developed by NASA, which is still using it today. The cell runs on hydrogen gas and oxygen from the ambient air circulating in the space capsule. The reaction produces water, which is recovered from the shuttle for the use of the astronauts.

Figure 14 Astronaut Leland Melvin, on board the shuttle Atlantis in 2009, watches a drop of water coming out of a bottle.

SUPPLEMENT Redox Reactions

361

Table 5 Redox potentials for reduction at 25°C Reduction half-reaction

Hydrogen fuel cell Silver (Ag) is a metal that can oxidize with sulphur (S) in the air in the same way that iron forms rust. A very tarnished silver spoon is covered with a layer of silver sulfide (Ag2S). To remove this undesirable coating, the silver ion Ag in the silver sulfide molecule must be reduced. To do this, first cover the bottom of a Pyrex dish with aluminum foil, since the reaction will occur between the aluminum and the silver sulfide (see Figure 15). Then fill the dish with an aqueous solution of equal quantities of table salt (NaCl) and sodium bicarbonate (NaHCO3). Lastly, put the silver into the solution. The silver is reduced to its metallic form and the sulphur forms aluminum sulfide on the spoon; this will be rinsed off. This process has caused a redox reaction to occur. 3 Ag2S  2 Al n 6 Ag  Al2S3

Figure 15 Silver can be cleaned with aluminum foil and an aqueous solution of sodium chloride (NaCl) and sodium bicarbonate (NaHCO3).

362

SUPPLEMENT Redox Reactions

Best oxidizing agents z 2 F (aq) F2 (g)  2 e y z Ag (aq) Ag2 (aq)  e y 3  z Co2 (aq) Co (aq)  e y   z 2 H2O (l) H2O2 (aq)  2 H (aq)  2 e y z MnO2 (aq)  2 H2O (l) MnO4 (aq)  4 H (aq)  3 e y    z IO3 (aq)  H2O (l) 2 H (aq)  IO4 (aq)  2 e  2 y    z Mn2 (aq)  4 H2O (l) MnO4 (aq)  8 H (aq)  5 e y z Au (s) Au3 (aq)  3 e y  z 2 Cl (aq) Cl2 (g)  2 e y z 2 H2O (l) O2 (g)  4 H (aq)  4 e y MnO2 (s)  4

H

e

(aq)  2 Br2 (l)  2 e NO3 (aq)  4 H (aq)  3 e ClO2 (aq)  e 2 2 Hg (aq)  2 e Ag (aq)  e 2 Hg2 (aq)  2 e Fe3 (aq)  e O2 (g)  2 H (aq)  2 e MnO4 (aq)  e I2 (s)  2 e Cu (aq)  e O2 (g)  2 H2O (l)  4 e Cu2 (aq)  2 e AgCl (s)  e Cu2 (aq)  e

z y z y z y z y z y z y z y z y z y z y z y z y z y z y z y z y

z y z y z y z y z y z y z y z y z y z y z y z y z y z y z y z y z y

2.87 1.99 1.92 1.78 1.68 1.60 1.51 1.50 1.36

(aq)  2 H2O (l) 2 Br (aq) NO (g)  2 H2O (l) ClO2 (aq) Hg22 (aq) Ag (s) 2 Hg (l) Fe2 (aq) H2O2 (aq) MnO4 2 (aq) 2 I (aq) Cu (s) 4 OH (aq) Cu (s) Ag (s)  Cl (aq)

1.23 1.22 1.07 0.96 0.95 0.92 0.80 0.80 0.77 0.70 0.56 0.54 0.52 0.40 0.34 0.22

Cu (aq)

0.15

Mn2

z H2 (g) 2 H (aq)  2 e y Fe3 (aq)  3 e Pb2 (aq)  2 e Ni2 (aq)  2 e Cd2 (aq)  2 e Cr3 (aq)  e Fe2 (aq)  2 e Cr3 (aq)  3 e Zn2 (aq)  2 e 2 H2O (l)  2 e Mn2 (aq)  2 e Al3 (aq)  3 e Mg2 (aq)  2 e Na (aq)  e Ca2 (aq)  2 e Ba2 (aq)  2 e K (aq)  e Li (aq)  e

E° (V)

Fe (s) Pb (s) Ni (s) Cd (s) Cr2 (aq) Fe (s) Cr (s) Zn (s) H2 (g)  2 OH (aq) Mn (s) Al (s) Mg (s) Na (s) Ca (s) Ba (s) K (s) Li (s)

Best reducing agents

0.00 0.04 0.13 0.26 0.40 0.41 0.45 0.74 0.76 0.83 1.18 1.66 2.37 2.71 2.89 2.91 2.93 3.04

5.2

Cell potential

The cell potential (E0cell), is equal to the sum of the oxidation potential and reduction potential of the cell. Calculating the potential of an electrochemical cell made up of two different electrodes can be done without having to use a reference electrode Take the example of an electrochemical cell with a silver (Ag) electrode and a magnesium (Mg) electrode. Below are the reduction equations for these two metals (see Table 5 on the previous page): Ag (aq)  e n Ag (s) Mg2 (aq)  2 e n Mg (s)

E 0  0.80 V E 0  2.37 V

The first step is determining the stronger reducing agent. In this example, magnesium is the stronger reducing agent, since it is found below silver in the table of redox potentials (see Table 5). So it is the electrode that will lose its electrons and undergo oxidation. To find the equation for the oxidation halfreaction, we simply reverse the reduction equation. Consequently, the potential of this electrode will become positive. Reduction

2 Ag (aq)  2 e n 2 Ag (s)

E 0 = 0.80 V

Oxidation

Mg (aq) n Mg2 (s)  2 e

E 0 = 2.37 V

Multiplying the reduction equation by two has no effect on the standard electrode potential. In fact, whether one or two electrons migrate, the potential difference is the same. To calculate cell potential, add the potential of the oxidation half-reaction and the potential of the reduction half-reaction, as in the following equation:

Cell potential E 0cell  E 0oxidation  E 0reduction

where E 0cell  Difference in cell potential, expressed in volts (V) E 0oxidation  Oxydation potential, expressed in volts (V) E 0reduction  Reduction potential, expressed in volts (V)

Thus, the potential value of the cell made up of the silver and magnesium electrodes is 3.17 V, that is, the sum of its oxidation potential (2.37 V) and its reduction potential (0.80 V). This positive value indicates that the reaction will be spontaneous. In fact, when the potential difference of a fuel cell is greater than zero, the reaction is spontaneous. Conversely, when the potential difference of a fuel cell is less than zero, the reaction is not spontaneous.

Spontaneous reaction if E 0cell  0 Non-spontaneous reaction if E 0cell is  0 SUPPLEMENT Redox Reactions

363

A fuel cell can be represented in the following simplified way: Simplified representation of a fuel cell Oxidation Reduction Mg  Mg2  Ag  Ag

According to convention, the oxidation reaction is shown on the left and the reduction reaction is shown on the right. An electrode and its solution are separated by a single stroke (), and a double stroke () represents the salt bridge. Below is an example that shows how to calculate the potential of a fuel cell: Example An electrochemical cell is made of a copper electrode immersed in a solution of Cu(NO3)2 at 1 mol/L and an electrode of zinc in a solution of Zn(NO3)2 at 1 mol/L. Calculate the potential of the fuel cell at 25°C. Solution : 1. Find the half-reactions that take place and determine the potentials of each electrode, using Table 5. Find the stronger reducing agent and reverse its reduction reaction to change it to oxidation, by changing the E0 sign. Zn2 (aq)  2 e n Zn (s)

E 0  0.76 V

Cu2 (aq)  2 e n Cu (s)

E 0  0.34 V

2. Balance the number of electrons. Since the zinc electrode has a more negative reduction potential than the copper electrode, the zinc will give up its electrons and will be oxidized. Oxidation Zn (s) n Zn2 (aq)  2 e E 0  0.76 V Reduction

Cu2 (aq)  2 e n Cu (s)

E 0  0.34 V

3. Calculate the battery's potential by adding E 0oxidation and E 0reduction. E 0cell  E 0oxidation  E 0reduction  0.76 V  0.34 V  1.10 V Answer: The cell potential of the zinc-copper fuel cell is 1.10 V.

A reaction that melts iron The redox reaction between iron oxide (Fe2O3) and aluminum (Al) is represented by the following equation: Fe2O3 (s)  2 Al (s) n 2 Fe (l)  Al2O3 (x) + energy Since this reaction is strongly exothermic, it produces liquid iron. In fact, the temperature of this reaction is over 1535 °C, the melting point of iron at standard temperature and pressure (STP). This is called a thermite reaction. At the time when the railways were built, this reaction was often used to solder rails. The low cost of the reactants and the small amount of equipment required meant that rails could be soldered efficiently in places where electricity was not yet available.

364

SUPPLEMENT Redox Reactions

Figure 16 A railway employee is creating a thermite weld between two rails.The liquid iron produced by this reaction can be seen.

SECTION 3 SECTION 4 SECTION 5

Reducing power of metals Electrochemical cell Redox potential

1. Immerse a piece of copper (Cu) in a solution of silver nitrate (AgNO3). Observe the formation of silver crystals on the surface of the piece of copper. What is the oxidation equation for this redox reaction? 2. In your opinion, if a solution of lead ions (Pb2) is mixed with a solution of silver ions (Ag), can a redox reaction take place between these two metals? Explain your answer. 3. Choose the right answer. Substances that have a higher standard electrode potential than hydrogen are considered to be: a) strong oxidizing agents b) strong reducing agents c) weak reducing agents 4. Choose the right answer. Substances that have a lower standard electrode potential than hydrogen are considered to be: a) strong oxidizing agents b) strong reducing agents c) weak reducing agents 5. Which of the following solutions will react if a strip of nickel is immersed in it? a) Sn(NO3)2 b) Zn(NO3)2 c) Al(NO3)3 d) Ni(NO3)2

9. Which chemical group do many strong reducing agents belong to? Explain your answer. 10. In your opinion, does cesium (Cs) have a negative or positive redox potential? Why? 11. Choose the right answer. In an electrochemical cell, reduction is the partial reaction during which this occurs: a) loss of electrons b) loss of mass c) loss of electrons and gain of mass d) gain of electrons 12. The following processes take place in an electrochemical cell. For each one, indicate whether it occurs on the anode or the cathode: a) reduction half-reaction b) oxidation half-reaction c) the reaction of the stronger oxidizing agent d) the reaction of the stronger reducing agent 13. This is an electrochemical cell: Cl− Cd (s)

K

K Cl− K

Cl−

Cd2 (aq)

Cu (s)

Cu2 (aq)

6. Arrange the following elements from the strongest oxidizing agent to the weakest: Mg, Al, Br2, Li and Au 7. Study these two reactions: Mg (s) n Mg2(aq)  2 e Co (s) n Co2(aq)  2 e Using the information from these reactions, say which of the following substances can most easily be reduced: a) Co2 (aq) b) Co (s) c) Mg2(aq) d) Mg (s) 8. Which element is the stronger oxidizing agent: nickel or mercury?

a) b) c) d)

In which direction do the electrons move? Toward which container do the Cl ions move? Which is the anode in this cell? On which electrode does the reduction take place?

14. Calculate the potential of the following cells: a) Zn (s)  Zn(NO3) (aq)  Cu(NO3)2 (aq)  Cu (s) b) Pb (s)  Pb2(aq)  Au3(aq)  Au (s) c) Ag (s)  Ag(aq)  Cd2(aq)  Cd (s) 15. In your opinion, which pair of metals will provide the greatest cell potential?

SUPPLEMENT Redox Reactions

365

APPLICATIONS Domestic batteries Batteries are portable devices that transform chemical energy into electrical energy to power portable electrical appliances. The main types of battery are primary cells, secondary cells and fuel cells. Most of the batteries used today work because of the redox reaction. Primary cells cannot be recharged and must be discarded after a while, when the chemical products inside them are spent. The redox reaction that occurs inside the primary cells is irreversible, but they have the advantage of being simple, reliable and relatively inexpensive. This is particularly true of the zinccarbon (Zn-C) and alkaline batteries often used in toys, clocks and flashlights. The electrodes in alkaline batteries are immersed in an alkaline electrolyte of potassium hydroxide (KOH), a porous paste damp enough to allow the ions to circulate (see Figure 17).

There is a wide range of secondary batteries, including nickel-metal hydride (Ni-MH) and nickelcadmium (Ni-Cd) batteries. They are practical for appliances that use a lot of energy in a short time and must be recharged frequently: hand vacuum cleaners, power tools, cordless telephones. Lithium-ion batteries are lighter and hold a charge longer, but are more expensive. They are often used in cell phones, laptop computers and cameras. Lead batteries, which provide energy to start conventional motors, are made up of six secondary cells. (see Figure 18). Anode Partition Positive plate: lead grid filled with PbO2 (s)

Cathode Negative plate: lead grid filled with solid lead

A cell H2SO4 (aq) electrolyte in each of the battery’s cells

Anode: Zn and KOH mixture Electron collector Separator

Cathode: MnO2 and KOH mixture Sealing agent Direction in which the electrons move

Figure 17 Inside an alkaline battery

It is also possible to buy very strong miniature alkaline batteries that last a long time. They are used in many smaller devices, such as watches and hearing aids. Secondary batteries are also called rechargeable or storage batteries. When secondary batteries’ chemical energy is used up, they can be recharged by applying an electrical current to reverse the oxidationreduction reaction and reconstitute the original reactants, restoring chemical energy. A charger pumps the electrons from the positive pole to the negative pole, a process that can be repeated more than 1000 times, in some cases.

366

SUPPLEMENT Redox Reactions

Figure 18 A car battery is a storage battery made up of six cells, usually of 2 V each, separated by partitions.

A fuel cell produces electricity by causing a reaction between a fuel such as hydrogen (H2) and an oxidant such as oxygen (O2), as with an H2-O2 fuel cell, for example. In this type of cell, the two gases flow continuously toward porous electrodes on which the redox reaction occurs between the ions from the gases. This reaction produces water, allowing electricity to be generated without directly producing greenhouse gas emissions. Still, in order to have no environmental impact at all, the hydrogen used should not be derived from oil, since making it produces a considerable quantity of greenhouse gas emissions. A great deal of research is currently being conducted in this area to make hydrogen use more economical and ecological, particularly for cars.

Supplement CHAPITRE 1 Redox Reactions

Les propriétés chimiques des gaz 1 Oxidation and reduction • A redox reaction comprises two half-reactions, oxidation and reduction, which occur simultaneously. • Oxidation is a reaction in which an atom loses one or more electrons. This reducing atom is thus oxidized since it reduces the oxidizing agent. • Reduction is a reaction in which an atom gains one or more electrons. This oxidizing atom is thus reduced since it oxidizes the reducing agent.

2 Oxidation number • The oxidation number of an atom in ground state is 0. When an element is ionized, its oxidation number is equal to its charge. • The oxidation number of a substance increases during oxidation and decreases during reduction.

3 Reducing power of metals • Metals are good electron donors; some metals are better donors than others.

4 Electrochemical cell • An electrochemical cell generates an electrical current spontaneously. • In an electrochemical cell, oxidation takes place on the anode and reduction takes place on the cathode.

5 Reduction and oxidation potential • Measure redox potential against the hydrogen reference electrode which, according to convention, has a redox potential of 0 V. • A fuel’s cell potential corresponds to the measured potential difference between its two electrodes. This is how it is calculated: E 0cell  E 0oxidation  E 0reduction • The reaction is spontaneous if E 0cell  0. The reaction is not spontaneous if E 0cell  0.

SUPPLEMENT Redox Reactions

367

Redox Reactions 1. Find the half-reaction for the following redox pairs: a) Al3  Al b) MnO4  Mn2 c) NO  NO3 2. Which of the following statements about redox reactions is true? a) The substance that accepts electrons is the reducing agent in the system. b) An element is reduced when it gains electrons. c) Oxidation is a partial reaction in which the oxidation number decreases. d) An oxidizing agent is a substance that is oxidized during a reaction. 3. Identify the oxidizing agent and reducing agent in the following redox reactions: a) 2 Al3  3 Mg n 2 Al  3 Mg2 b) Cu2  Na n Na  Cu 4. What is the reaction that represents the oxidation of the chrome (Cr2) ion? 5. Find the oxidation number of sulphur in the following compounds: a) HSO3 b) SF6 c) H2S 6. This is an electrochemical cell: Cl− Fe (s)

K Cl−

Fe2 (aq)

K Cl− K

Mg (s)

Mg2 (aq)

a) Which is the anode? b) Toward which electrode do the cations in the salt bridge move? c) Which electrode’s mass will decrease? d) On which electrode does oxidation occur?

368

SUPPLEMENT Redox Reactions

7. Of the following cells, which one has a cell potential of 0.50 volts ? a) Zn-Pb b) Cu-Ag c) Zn-Ni d) Ni-Ag 8. In this reaction, Zn  H2SO4 n ZnSO4  H2, a) which is the reducing agent? b) which is the oxidizing agent? 9. Immerse a solid aluminum (Al) rod in a copper sulfate (CuSO4) solution. Observe the formation of solid copper on the rod. Analysis shows the presence of aqueous Al3 ions in the solution. Give the general redox reaction equation that represents this phenomenon. 10. Aqueous silver ions (Ag) react with lead (Pb) to form a deposit of metallic silver and lead ions Pb2, as shown in this reaction: 2 Ag(aq)  Pb (s) n 2 Ag (s)  Pb2(aq) a) What changes do the reactants undergo? b) Identify the oxidizing agent and reducing agent that react with each other. c) What half-reactions occur? 11. An electrochemical cell has a zinc (Zn) electrode in a Zn(NO3)2 solution and a silver (Ag) electrode in an AgNO3 solution. Which of the following statements is false? a) Oxidation occurs on the zinc electrode. b) The mass of the silver electrode increases while the cell is running. c) The electrons move from the silver electrode toward the zinc electrode. d) The zinc electrode is the anode. 12. An electrochemical cell is made up of these halfcells: Ag  Ag and Ni  Ni2. Write the equation for the reaction that occurs under standard conditions and calculate the cell’s cell potential.

SUPPLEMENT Redox Reactions

13. In the laboratory, strips of different metals are placed in beakers containing different solutions. Notice that there is a reaction in beakers 1 and 3. Arrange the Cu2, Pb2, Zn2 and Mg2 ions from weakest reducing agent to strongest. Zn

Cu2

Cu

Pb2

Beaker 1

18. Hydrogen peroxide (H2O2) is mixed with an acidic solution of potassium iodide (KI). The mixture is colourless at first but gradually becomes brownish. The half-reactions are H2O2  H2O and I2  I. a) Find the equations for the half-reactions. b) Write the general equation for the reaction. c) What substance is responsible for the colour?

19.

Beaker 2 Mg

17. There is an electrochemical cell, with one lead (Pb) electrode and one copper electrode (Cu). a) What salts should be dissolved in each half-cell? b) Which metal will provide the electrons? c) What is the name of this electrode?

Pb

Drivers suspected of drunk driving have to blow into a device that measures the amount of alcohol in their blood. This is the reaction upon which the breath test is based: 3 C2H6O  2 K2Cr2O7  8 H2SO4 n 3 C2H4O2  2 Cr2(SO4)3  2 K2SO4  11 H2O

Zn2

Beaker 3

Determine which substance is reduced and which is oxidized, as well as the reducing agent and the oxidizing agent in this reaction.

Mg2

Beaker 4

14. Why is a salt bridge necessary between the two electrodes in an electrochemical cell? 15. Do the following reactions occur spontaneously? Explain your answer. a) a bar of zinc (Zn) in a solution of lithium chloride (LiCl) b) solid nickel (Ni) in a solution of copper sulfate (CuSO4) c) a cadmium (Cd) rod in a solution of silver nitrate (AgNO3) d) a solution of sodium chloride (NaCl) on a chromium (Cr) plate

20.

An electrochemical cell with a copper (Cu) electrode is immersed in a solution of copper nitrate (CuNO3) and an aluminum (Al) electrode immersed in a solution of aluminum nitrate (AlNO3). Both solutions have an initial concentration of 1 mol/L. a) Which electrode will be the anode and which will be the cathode? b) What are the two half-reactions in this cell? c) Give the equation for the overall reaction. d) What is the potential of this cell?

16. Will the following reactions occur spontaneously, given that all of the solutions have an initial concentration of 1 mol /L? a) Ba (s)  Cd2 (aq) n Ba2 (aq)  Cd (s) b) 2 Ag (s)  Co2 (aq) n 2 Ag (aq)  Co (s) c) Cu (aq)  Fe3 (aq) n Cu2 (aq)  Fe2 (aq)

SUPPLEMENT Redox Reactions

369

370

CONTENTS 1

Laboratory Safety . . . . . . . . . . . . . . . . . . . . . . . . . . 372

5.4 Significant figures in results of mathematical operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396

1.1 Danger symbols . . . . . . . . . . . . . . . . . . . . . . . . 372

5.5 How to round off a number . . . . . . . . . . . . . . . 397

1.2 Safety symbols used in the Quantum collection . . . . . . . . . . . . . . . . . . . . . 375 1.3 Safety rules . . . . . . . . . . . . . . . . . . . . . . . . . . . 375

2

6

Mathematics in Science . . . . . . . . . . . . . . . . . . . . 398 6.1 Review of some mathematical equations . . . . 398

Methods Used in Chemistry . . . . . . . . . . . . . . . . . 377

6.2 Transformation of algebraic expressions . . . . . 404

2.1 Summary table of methods . . . . . . . . . . . . . . . 377

6.3 Scientific notation . . . . . . . . . . . . . . . . . . . . . . 406

2.2 The four steps of problem-solving . . . . . . . . . . 378 2.3 Method of unit analysis problem-solving . . . . 379

3

7

Units of Measure in Chemistry . . . . . . . . . . . . . . . 408

2.4 Experimental method . . . . . . . . . . . . . . . . . . . . 381

7.1 Common SI prefixes . . . . . . . . . . . . . . . . . . . . . 408

2.5 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382

7.2 Common units of measure . . . . . . . . . . . . . . . . 409

Laboratory Instruments and Techniques . . . . . . 383 3.1 How to prepare a solution . . . . . . . . . . . . . . . . 383 3.2 How to collect samples . . . . . . . . . . . . . . . . . . 387

8

Reference Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 8.1 Alphabetical list of elements and their atomic mass . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 8.2 Everyday chemical products . . . . . . . . . . . . . . 412

4

Presenting Scientific Results . . . . . . . . . . . . . . . . 388 4.1 Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388 4.2 Broken-line graph . . . . . . . . . . . . . . . . . . . . . . . 389 4.3 Line of best fit . . . . . . . . . . . . . . . . . . . . . . . . . 389 4.4 Slope of a tangent to the curve . . . . . . . . . . . . 390 4.5 Bar graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390 4.6 Histogram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391 4.7 Circle graph . . . . . . . . . . . . . . . . . . . . . . . . . . . 391 4.8 Laboratory report . . . . . . . . . . . . . . . . . . . . . . . 392

5

Interpreting Measurement Results . . . . . . . . . . . 394 5.1 Uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394 5.2 Significant figures . . . . . . . . . . . . . . . . . . . . . . 395

8.3 Thermodynamic properties of certain elements . . . . . . . . . . . . . . . . . . . . . . . 417 8.4 Standard molar enthalpies of formation . . . . . 418 8.5 Average bond energy . . . . . . . . . . . . . . . . . . . . 419 8.6 Molar heat of dissolution . . . . . . . . . . . . . . . . 420 8.7 Ionization constants of acids . . . . . . . . . . . . . . 421 8.8 Ionization constants of nitrogen bases . . . . . . 421 8.9 Solubility-product constants in water at 25°C for various compounds . . . . . . . . . . . . . . 422 8.10 Common polyatomic ions . . . . . . . . . . . . . . . . . 422 8.11 Solubility (in water) of common ionic compounds . . . . . . . . . . . . . . . . . . . . . . . 423 8.12 Specific heat capacities of various substances . . . . . . . . . . . . . . . . . . . . . 423

5.3 Measurements with unknown uncertainty . . . 396

371

1 Laboratory Safety

APPENDIX

1

It is extremely important to pay attention to health and safety in the laboratory. You must thoroughly understand the meaning of all danger symbols. You must also be familiar with safety rules and follow them to avoid accidents.

1.1

Danger symbols

Various danger symbols and safety symbols are commonly used in the laboratory. These symbols always provide a warning. You will therefore need to take special precautions whenever you handle substances or materials labelled with any of these symbols. Anyone handling dangerous substances must know what these symbols mean and which precautions to take when using these substances.

Danger symbols on household products Many products found in homes, at school and in the workplace contain dangerous substances. These products must be used with caution. Manufacturers warn of the dangers associated with their products by labelling the containers with symbols. Each symbol found on a household product has a specific meaning (see Table 1). Table 1 Meanings of danger symbols found on household products Symbol

Meaning This product may burn the skin or eyes. If swallowed, it causes injury to the throat and stomach.

Figure 1 Oven cleaner is a

Corrosive

corrosive product.

This product may explode if its container is heated or punctured.

Explosive

This product and the fumes it releases are highly flammable if they are placed near a heat source, flames or sparks. Flammable

This product is toxic and may cause serious health complications. It may be fatal if swallowed and, in certain cases, if inhaled. Poison

372

APPENDIX 1 Laboratory Safety

WHMIS danger symbols

APPENDIX

1

The Workplace Hazardous Material Information System (WHMIS) is the Canadian standard for all information and communication concerning dangerous materials. The WHMIS danger symbols identify substances that pose health risks. These substances include household products and solvents found in schools and homes. Each WHMIS symbol indicates which precautions should be taken (see Table 2). Table 2 Safety precautions for products identified with WHMIS danger symbols Symbol

Precaution Handle this product with care; keep it away from heat and flames.

Compressed gas Keep this product away from heat, flames and sparks.

Flammable and combustible material Keep this product away from heat, flames, sparks and any combustible materials. Oxidizing material Avoid all contact with this product.

Material causing immediate and toxic effects Avoid all contact with this product.

Material causing other toxic effects Avoid all contact with this product.

Infectious material Avoid all contact with skin and eyes. Wear safety glasses, gloves, a lab coat or an apron. Corrosive material Use only in a well-ventilated area.

Dangerously reactive material

APPENDIX 1 Laboratory Safety

373

GHS danger symbols

APPENDIX

1

In 2003, the member countries of the United Nations (UN) adopted the Globally Harmonized System of Classification and Labelling of Chemicals (GHS). The GHS danger symbols are gradually replacing the various systems, such as the WHMIS, used in different countries. The new system was designed to increase human and environmental safety. Each GHS danger symbol has a specific meaning. The symbols also indicate which precautions should be taken when handling certain products (see Table 3).

Table 3 Risks and precautions identified by GHS danger symbols Symbol

374

Risks

Precautions

Unstable, explosive material or object

Risk of mass explosion, blast, projection or fire

Handle the container with caution; keep away from heat or flames; do not puncture.

Flammable gas, liquid or solid

Risk of fire or explosion

Keep the container and its contents away from heat, flames and sparks.

Oxidizing gas, liquid or solid

Risk of fire, intensification of fire or explosion

Keep the container and its contents away from heat, flames, sparks and any other combustible material, such as wood, gas or solvents.

Gas under pressure

Risk of explosion

Handle the container with caution; keep away from heat and flames; do not puncture.

Corrosive material

Risk of burns, irritation of the skin or eyes, or eye injury

Avoid all contact with skin and eyes; in case of contact, rinse with water. Wear safety glasses, gloves and a lab coat or apron when handling this product.

Highly toxic material

Risk of poison if swallowed, if in contact with skin or if inhaled

Avoid all contact with this product. Let an experienced person handle it for you.

Harmful material

Risk of skin allergies, irritation of the skin, eyes and respiratory tracts, dizziness or drowsiness

Avoid all contact with this product. Let an experienced person handle it for you.

Material that presents various hazards

Risk of allergic reactions, asthma or breathing difficulties Avoid all contact with this product. Let an experienced if inhaled. Risk of genetic defects, cancer or organ damage. person handle it for you. Risk of infertility or harm to fetus. Risk of harm to breastfed babies.

Material that is hazardous to the aquatic environment

Risk of harmful long-term effects

APPENDIX 1 Laboratory Safety

Avoid all contact with this product. Let an experienced person handle it for you.

Safety symbols used in the Quantum collection

1.2

APPENDIX

1

It is extremely important that you understand safety symbols when working in the laboratory. Each safety symbol used in the Quantum collection has a specific meaning and indicates the precautions that must be taken (see Table 4). Table 4 Safety symbols used in the Quantum collection Symbol

Precautions

Protect eyes

Wear safety glasses.

Protect hair

Tie hair back.

Protect skin

Wear gloves.

Protect clothing

Wear a lab coat or an apron.

Caution: may cause burns

Exercise caution when using hot substances or objects.

Caution: hot

Wear heat-resistant mitts.

Caution: dangerous fumes

Work under a fume hood or in a well-ventilated area.

Caution: sharp

Exercise caution when using sharp edges.

Safety rules

1.3

Safety depends upon responsible behaviour and attitude. This includes obeying certain rules. It is important to be thoroughly familiar with these safety rules and, above all, to follow them at all times. Laboratory safety rules 1

Inform the person in charge about any health problems such as an allergy or illness that might interfere with the activity.

2

Identify the closest first-aid kit, fire extinguisher, fire blanket, emergency shower, laboratory eyewash and fire alarm. Learn how to adequately use each of these.

APPENDIX 1 Laboratory Safety

375

APPENDIX

1

Figure 2 In the laboratory, tie back your hair, and wear safety glasses, gloves and a lab coat.

376

APPENDIX 1 Laboratory Safety

3

Handle the equipment with care, act calmly and pay attention to what you are doing.

4

Wear safety glasses. Wear gloves and a lab coat or an apron when handling materials that are dirty or corrosive.

5

Do not wear jewellery, clothing that might get in the way, or untied shoes. Tie back your hair.

6

Never eat or drink during an activity.

7

Keep your work surface clean and neat.

8

Follow directions concerning the use of dangerous products.

9

Before beginning an experiment, make sure that you fully understand every aspect of the procedure, such as how to handle the equipment and materials, waste disposal and safety symbols (see Table 4 on page 375 ).

10

Obtain the approval of the person in charge before beginning the experiment if you have developed your own procedure or if you have modified an approved procedure.

11

Never leave an experiment unattended.

12

Never touch or taste a substance. Never smell a substance directly.

13

Wash your hands after each experiment.

14

Immediately inform the person in charge of any accident, even if it does not appear serious. Examples include injury, spills and equipment breakage.

Handling electrical devices and heating appliances 1

Before using an electrical device or heating appliance, make sure that it works properly. Inform the person in charge of any problems.

2

Never distract someone who is working with an electrical device or heating appliance.

3

Never use a burner to heat a flammable substance; use a hot plate instead.

4

Never leave an electrical device or heating appliance that is in use unattended.

5

Never touch an electrical device or electrical outlet with wet hands.

6

Never touch an electrical cord that appears to be damaged.

7

Never leave electrical cords out on the floor.

8

When you are finished with an electrical device, unplug it by pulling on the plug (not the cord) and then put it away.

2 Methods Used in Chemistry

APPENDIX

There are different ways of finding an effective solution to a problem. Each method involves a logical process of trial and error.

2.1

2

Summary table of methods

To solve a problem in chemistry, you may use a general problem-solving method or more specific methods depending on the nature of the problem. The choice of a specific method is based on the problem to be solved. Two or more methods may be combined to arrive at a result. Five of these methods are explored in the third year of Secondary Cycle Two (see Table 5). Table 5 Methods used in chemistry Method Modelling Modelling provides a concrete representation of something that is difficult to imagine. This representation can take on several forms, such as a text, a drawing, a mathematical formula or a chemical equation. Over time, the model becomes more refined and complex. It may even be modified or rejected. A model must take into account the characteristics of the phenomenon or process that is being studied. It must also help people understand a given reality, explain certain properties of that reality and predict new observable phenomena. Observation method This method consists of observing a phenomenon and interpreting facts while taking into account various criteria. Through observation, you can learn new facts that result in a different understanding of the phenomenon. Analysis This method involves analyzing an object or a system with the aim of identifying the components from which it is constructed as well as the interactions between these componenets. This analysis helps in understanding the purpose of the object or system, as well as how it works and how it is made. In some cases, this method involves using a broader understanding of a system to determine the function of its parts and the relationships between them (see Section 2.5 on page 382 ).This aspect of the analytical method is particularly useful in studying phenomena and applications. Experimental method This method seeks an answer to a problem through an experiment. Students can begin looking for an answer and defining the framework of the experiment. The person conducting the experiment must then develop an experimental procedure in order to identify a certain number of variables to be manipulated. The aim of the procedure is to identify and compare observable or quantifiable phenomena and check them against the initial hypotheses. Analysis at the end of the experiment raises new questions, helps to formulate new hypotheses, adjust the experimental procedure and take the limitations of the experiment into account (see Section 2.4 on page 381). Empirical method This method involves field research without any manipulation of variables. This makes it possible to explore the elements of a problem or to obtain a new representation of the problem. This method is often based on intuition. It can lead to the development of hypotheses and possible avenues for other research projects. Surveys are an example of the empirical method.

APPENDIX 2 Methods Used in Chemistry

377

APPENDIX

2

2.2

The four steps of problem-solving

Regardless of the method used to solve a problem, the same four steps are always followed. Figure 3 illustrates the four steps used to find a solution to a scientific problem. You can always return to an earlier step if you need to modify a hypothesis, action plan or procedure. If this is the case, continue with the procedure from the point at which the modification was made. Adjustments will probably have to be made to the rest of the procedure. Define the problem Begin by describing the problem. Next, determine your goal, taking into account the context of the problem. Then, you can formulate your working hypothesis by referring to the relevant scientific concepts.

Develop an action plan You must explore all problem-solving methods to determine which one is the most appropriate to the situation. Then, develop an action plan that can work with the available resources and applicable constraints. You must also plan out each step of the action plan.

Carry out an action plan Follow the steps of your action plan. Make a note of any information that may be useful during the problem-solving process.

Analyze the results You must establish connections between the information you obtained and scientific concepts. You may also draw conclusions and provide an explanation or a solution. Figure 3 The four steps of problem-solving

Attention! You may, at any time, return to an earlier step to make modifications. Continue with the procedure from this step. It is very important to keep a record of all modifications.

378

APPENDIX 2 Methods Used in Chemistry

2.3

Method of unit analysis problem-solving

APPENDIX

When solving problems, it is important to use a method that involves steps leading to the discovery of the answer. One way to do this is through unit analysis problem-solving. This method consists of analyzing the units and determining conversion factors. It involves matching the units and arranging them so that they cancel out and you will be left only with the unit required in the answer. Finally, multiply and divide the numbers corresponding to these units.

2

The following examples show how to apply this method to solve numerical problems: Example A What is the final volume of a weather balloon that contains 200.0 L of Helium (He) at STP at a pressure of 50.0 kPa and a temperature of 20.0°C at its maximum altitude? Step 1 Write down the data without rounding off the numbers, and determine what you are looking for. Data: V1  200.0 L T1  25.0°C P1  101.3 kPa

V2  ? T2  20.0°C P2  50.0 kPa

Step 2 Select the appropriate formula for the data, and isolate the unknown variable by eliminating any non-essential variables. P1V1 PV  2 2 n1T1 n2 T2 V2 

P1 V 1 n2 T2 PVnT PVT   1 1 2 2 1 1 2 n1 T1 P2 n1 T1 P2 T1P2

Step 3 Convert the data as required so that they are expressed in the appropriate units of measure. Replace the variables with their values, perform the calculations, and carry forward the units of measure. Then, cancel out the units and check your calculations. 1. Conversion of temperature into kelvin: T1  25.0°C  273  298 K T2  20.0°C  273  253 K 2. Calculation of the final volume: PVT V2  1 1 2 T1 P2 

101.3 kPa  200.0 L  253 K  344.01 L 298 K  50.0 kPa

Step 4 Check whether your answer is reasonable, and express it appropriately. Answer: The final volume of the weather balloon is 344 L.

APPENDIX 2 Methods Used in Chemistry

379

APPENDIX

2

Example B In the past, pharmacists used a unit called the grain (gr) to measure the active ingredients in many medications. One grain is equal to 64.8 mg. If a pain reliever for headaches contains 5.0 gr. of active acetylsalicylic acid (ASA), how many grams of ASA would there be in two tablets? Step 1 Write down the data without rounding off the numbers, and determine what you are looking for. Data: ? g of ASA in 2 tablets 1 tablet  5.0 gr Step 2 Determine which conversion factor to use. A conversion factor is usually a proportion between two numbers and includes the appropriate units (for example, 1000 g/1 kg). Multiply the data provided by the conversion factor to obtain the units needed in the answer. 64.8 mg /1 gr 1 g / 1000 mg Step 3 Arrange the data and conversion factors in order to eliminate unnecessary units, and perform the required multiplications. Calculation: 5.0 gr 64.8 mg 1g ?g    2 tablets 1 tablet 1 gr 1000 mg 

5.0 gr  64.8 mg  1 g  2 1 gr  1000 mg

 0.648 g Step 4 Check whether your answer is reasonable, and express it appropriately. Answer: There are 0.65 g of ASA in two headache tablets.

380

APPENDIX 2 Methods Used in Chemistry

2.4

Experimental method

APPENDIX

The problem-solving process applied to the experimental method includes four steps (see Table 6).

2

Table 6 The four steps of problem-solving applied to the experimental method Step

Example: Determining the melting point of ice

Define the problem a) Define the problem to be solved.

a) Ice begins to melt at a certain temperature. I must accurately determine this temperature.

b) Identify the goal.

b) Determine the melting point of ice.

c) Use relevant scientific concepts (formulas, theories) for solving the problem, if necessary.

c) Every pure substance possesses a characteristic melting point.

d) Formulate a hypothesis.

d) I believe that the melting point of ice is 1°C.

Develop an action plan a) Specify the variables you need to observe and measure.

a) Variables: state of water based on temperature, and water temperature (as a function of time)

b) Make a list of the equipment required and make a drawing of the set-up, if necessary.

b) Equipment and materials: 250 mL of crushed ice, a 500 mL beaker, a thermometer, a stopwatch, etc.

c) Write out the steps of the procedure.

c) Procedure 1. Place the ice in the beaker. 2. Place the thermometer in the beaker. Etc.

Carry out an action plan a) Perform the experiment in a safe manner.

a) I follow safety rules while conducting the experiment.

b) Gather the data and make a note of any useful observations.

b) I record the temperature every 30 seconds and when the ice begins to melt.

c) Process the data (by making diagrams, calculations, etc.) gathered from observations made during the experiment.

c) I write this information in a table. I prepare a broken-line graph representing water temperature as a function of time. I use a colour code to indicate the state of the water at each temperature.

Analyze the results a) Analyze your data and observations.

a) I observe that ice begins to melt at 0°C and that this temperature remains stable until all of the ice has melted.

b) Draw conclusions that either confirm or disprove the initial hypothesis.

b) I conclude that the melting point of ice is 0°C. This conclusion does not correspond to my initial hypothesis.

c) Formulate explanations and, if necessary, suggest new hypotheses or improvements that could be made to the experiment.

c) The period during which the temperature remains steady is a plateau. The experiment could be repeated to determine whether a plateau occurs when water transforms to vapour.

APPENDIX 2 Methods Used in Chemistry

381

2.5

APPENDIX

2

Vacuum

Mercury

Analysis

In chemistry, you can use analysis to recognize interactions among the various elements that make up a system, a phenomenon or an object. For example, by analyzing an object such as the Torricelli mercury barometer (see Figure 4), we can identify its structural components (mercury-filled glass tube inverted into a container also containing mercury), and determine the connections among these components. Determining the function of each part and the relationship among these parts helps to illustrate the dynamics of the system, which, in the case of this type of barometer, measures atmospheric pressure.

Glass tube

Patm

Container

Figure 4 An analysis of the Torricelli mercury barometer provides an understanding of how it works.

It is possible to analyze a system either quantitatively or qualitatively. An example of qualitative analysis is the study of a system at equilibrium using Le Chatelier’s principle. In this case, studying the system after adding or removing some of its components helps to explain the dynamics of the equilibrium established (see Figure 5). This type of analysis allows us to predict how the system will behave and then use it in various applications, for example, in the industrial production of substances such as ammonia, which is used to manufacture chemical fertilizers. Addition of N2 Hydrogen (H) Nitrogen (N)

N2  3 H2 z y 2 NH3 N2

a) Equilibrium

H2 3

zy

2 NH 3

b) Disequilibrium

N2  3 H2 z y 2 NH3

c) New state of equilibrium

Figure 5 A qualitative analysis of a system at equilibrium using Le Chatelier’s principle helps to predict how it will behave. For example, an increase in the nitrogen (N2) concentration in b ) causes the system to react by the formation of ammonia (NH3) until a new equilibrium is reached in c ).

382

APPENDIX 2 Methods Used in Chemistry

3 Laboratory Instruments and Techniques

APPENDIX

3

Conducting experiments in the laboratory is essential to applying the concepts you have learned in class. You can use observation instruments, prepare various solutions or measure the melting point and boiling point of a substance in the laboratory. It is essential that you work with the appropriate material and follow specific techniques while observing safety rules at all times.

How to prepare a solution

3.1

In the laboratory, you will often need chemical solutions whose concentration is determined according to the requirements of an experiment or of a specific analysis. These solutions are prepared from a solid compound or from a liquid concentrate.

Dissolution You can prepare a solution of a determined concentration by dissolving a solid solute. Follow this procedure when preparing an aqueous solution: 1

Determine the required volume of solution, and select a volumetric flask of an appropriate volume.

2

Rinse the volumetric flask thoroughly with distilled water.

3

Determine the mass of the solute required to obtain the desired concentration by performing the appropriate calculation: • If the required concentration is expressed in grams per litre (g/L), use the following equation to perform the calculation: C(solution) 

m (solute) V(solution)

• If the required concentration is expressed in moles per litre (mol/L), use the following equation to perform the calculation: C

n V

4

Place an empty dish on a balance and “tare” the scale, adjusting it so that the mass of the weighing dish is not included in the reading. By preloading the scale, you can subtract the mass of the weighing dish at the beginning of the experiment.

5

Place the required quantity of dry powdered solute in the dish.

6

Using a funnel, pour the solute into the volumetric flask.

7

Rinse the dish and funnel with distilled water from a wash bottle and pour the rinse water into the volumetric flask.

APPENDIX 3 Laboratory Instruments and Techniques

383

APPENDIX

3

8

Fill the volumetric flask half-way with distilled water.

9

Stopper the volumetric flask and swirl it gently using a circular motion until the solute is completely dissolved (see Figure 6 ).

10 Add enough distilled water to the volumetric flask so that the meniscus of the liquid reaches the yellow line. 11 Stopper the volumetric flask and swirl it gently to homogenize the mixture.

If the experimental procedure requires a lesser degree of accuracy, you may replace the volumetric flask with an Erlenmeyer flask. In this case, measure the volume of the distilled water using a graduated cylinder of the appropriate volume. In step 8, pour half of the distilled water into the Erlenmeyer flask. In step 10, add the remaining distilled water to obtain the total volume of solution.

Dilution Figure 6 Swirling the solution accelerates the particles’ motion, which facilitates dissolution.

Dilution consists of preparing a solution with a lower concentration than the initial solution, which has a higher concentration. Follow these steps to dilute an aqueous solution: 1

Determine the required volume of solution, and select a volumetric flask of the same volume.

2

Rinse the volumetric flask thoroughly with distilled water.

3

Determine the volume of the initial solution (V1) needed to obtain the required concentration using the following equation: C1V1  C2V2

Figure 7 The initial solution is a concentrated solution.

4

Determine the volume of the initial solution (V1) by measuring it accurately using a graduated cylinder, and then pour it into the volumetric flask (see Figure 7 ).

5

Fill the volumetric flask with distilled water about half-way.

6

Stopper the volumetric flask and swirl it gently using a circular motion to homogenize the mixture (see Figure 6 ).

7

Add enough distilled water to the volumetric flask so that the meniscus of the liquid reaches the yellow line (see Figure 8 ).

8

Stopper the volumetric flask and swirl it gently to homogenize the mixture.

If the experimental procedure requires a lesser degree of accuracy, you may replace the volumetric flask by an Erlenmeyer flask. In this case, measure the volume of the distilled water using a graduated cylinder of the appropriate volume. In step 5, pour half of the distilled water in the Erlenmeyer flask. In step 7, add the remaining distilled water to obtain the total volume of the solution.

Figure 8 The final solution is a diluted solution.

384

APPENDIX 3 Laboratory Instruments and Techniques

Determining concentration

APPENDIX

Titration is a method used to determine the concentration of a solution by causing it to react with another solution with a known concentration. The solution with the known concentration is called the standard solution. Titration is often used to determine the concentration of a base by causing it to react with an acid in an acid-base neutralization, or vice versa. Often, this technique requires an indicator in order for the change in colour of the basic solution to be observed, which indicates that the neutralization reaction is complete.

3

The purpose of acid-base titration is to determine the exact volume of the standard solution that caused the complete neutralization of the ions present in the solution of unknown concentration. At that point, all moles of OH ions present in the initial volume of the basic solution of unknown concentration are neutralized. Since we know the initial volume of the solution of unknown concentration, we can calculate this concentration. For your titration to be successful, you need to: • gradually add known quantities of the standard solution • be able to determine when the chemical reaction is complete A burette is used for gradually pouring a solution and for determining the volume of the solution you have poured. Usually, you can tell that the neutralization reaction is complete when a colour change occurs because an indicator is present. Follow these steps to determine the concentration of a base through acidbase titration:

Filling the burette 1

Assemble the set-up shown in Figure 9 (see the following page).

2

Fill the burette to the top with the acidic standard solution.

3

Slowly open the tap to fill the part under the tap, and adjust the zero mark. Collect the acidic solution as required, using a beaker, and discard it.

Preparing the basic solution to be titrated 4

Using a graduated cylinder and dropper bottle, carefully measure a known volume of the basic solution to be titrated (VB) and pour it into a clean Erlenmeyer flask.

5

Add the required number of drops of the indicator to the basic solution to be titrated (see Figure 9 on the following page).

The titration step 6

Open the burette’s tap and let the standard acidic solution flow slowly into the base being titrated, gently swirling the Erlenmeyer to mix well (see Figure 10 on the previous page).

APPENDIX 3 Laboratory Instruments and Techniques

385

APPENDIX

3

7

When the solution in the Erlenmeyer flask changes colour consistently where the flow of acid contacts the base being titrated, close the tap partially to reduce the amount of acid flowing into the burette, and add one drop at a time, swirling constantly.

8

Continue to add acid drop by drop until the colour change is permanent in the entire solution being titrated. When this occurs, close the tap. Neutralization is complete.

9

On the burette, take a reading to determine the total volume of the standard acid solution poured into the beaker (VA), and record it in a table.

10 Calculate the number of moles of acid using the following formula: NA  CA  VA. 11 Use the chemical equation from the acid-base neutralization reaction to determine, by stoichiometry, the number of moles of the base (NB) that reacted. Calculate the concentration of the base (CB) using the formula CB  NB , where VB is the initial volume of the basic solution to be titrated in the Erlenmeyer. VB

Figure 9 Titration set-up

Figure 10 Close the tap when the indicator changes colour permanently.

Approximate concentrations may be obtained using other methods such as pH paper for the H3O ion.

386

APPENDIX 3 Laboratory Instruments and Techniques

How to collect samples

3.2

APPENDIX

Samples are collected to, among other things, monitor water quality, for environmental studies and environmental protection, and to monitor certain industrial processes. Sample collection requires much attention and thoroughness because these samples serve as the basis for analyzing the real world in the laboratory.

3

Follow these steps when collecting samples: 1

Plan your sample collection. For example, ask yourself questions such as: “What will the samples be used for?”, “What will the samples consist of?”, “When and how will the sample collection take place?” and “Who will collect the samples?”

2

Prepare a sufficient number of appropriate containers (flasks, bottles, hermetically sealed pouches, etc.).

3

Avoid contaminating the samples: use gloves and a mask if necessary. If you need to use a tool such as a spatula or pipette to collect the sample, prepare a separate tool for each sample, or clean the tool between samples. For example, do not use the same spatula to collect different soil samples because this might lead to inaccurate results due to contamination.

4

Label each sample with a different identification number.

5

During sample collection, use a notebook to record detailed information about each sample, such as the location of the collection, the method used, the conditions, etc.

6

To collect a liquid, fill the container to the top so that it does not contain any air. To collect a gas, it is preferable to use the water displacement method (see Figure 11).

Lastly, it is important to analyze the samples as soon as possible to avoid changes to sample characteristics due to external factors.

Figure 11 Collecting a gas using the water displacement method. The gas produced by the chemical reaction will gradually displace the water in the test tube. When it is full, it must be hermetically sealed before removing it from the water.

APPENDIX 3 Laboratory Instruments and Techniques

387

4 Presenting Scientific Results

APPENDIX

4

In chemistry, data is often presented in a table or diagram, which helps in understanding and interpreting the data. The laboratory report is another essential tool for presenting results obtained from an experimental method.

4.1

Table

Tables organize data (number or words) into columns and rows, which makes it easier to analyze information. Tables are also used to record data that will be presented subsequently in the form of a diagram (see sections 4.2 to 4.7). The units of measurement should be provided in parentheses in the table under the column title; this avoids having to repeat them on each line. Data may be recorded for an independent variable and a single dependent variable.

Table title

Table 7 Average monthly precipitation for Montréal

Independent variable

Value of the independent variable

Month

Precipitation (mm)

January

87

February

66

March

91

April

81

May

91

June

96

July

97

August

100

September

97

October

91

November

98

December

93

Dependent variable Unit of measure

Value of dependent variable

We can also record data for several dependent and independent variables in tables with double entries. Table 8 Temperature difference between two pieces of cardboard observed after 30 minutes

Independent variables

388

APPENDIX 4 Presenting Scientific Results

Under white cardboard

Under black cardboard

Initial temperature (°C)

22

22

Final temperature (°C)

27

33

Variation (°C)

+5

+ 11

Dependent variables

Calculations

4.2

Broken-line graph

APPENDIX

A broken-line graph is used to illustrate continuous data. It provides a graphical representation of the relationships between the independent variable and the dependent variable. When preparing a broken-line graph, begin by drawing the axes. Make them long enough so that you can clearly indicate all values of the variables. Represent each value with a point. Connect each point to the next. This creates the broken line (see Figure 12, which was created based on Table 7 on page 388).

Average monthly precipitation for Montréal

4

Title

Unit of measure Dependent variable

Precipitation (mm)

Axis of ordinates 105 100 95 90 85 80 75 70 65 60 55 50

Point representing a value Broken line connecting each point

0

J

Independent variable

F

M

A

M

J

J

A

S

O

Month

N

D Axis of abscissas

Figure 12 Example of a broken-line graph

4.3

Line of best fit

In a diagram, you may draw a broken-line graph or a line of best fit, depending on the situation. If the variation in the dependent variable is discontinuous, as in the previous example, draw a broken-line graph. However, if the variation of the dependent variable is continuous, draw a line of best fit (see Figure 13). This may be either a straight line or a curve.

Cooling time of 250 mL of hot water

Unit of measure Dependent variable

Temperature (°C)

Axis of ordinates

60 55 50 45 40 35 30 25 20 15

Title

Point representing a value Curve of best fit in a scatter diagram Axis of abscissas

0

10 20 30 40 50 60 70 80 90 100 Time (min) Unit of measure Independent variable

Figure 13 Example of a line of best fit

APPENDIX 4 Presenting Scientific Results

389

4.4

APPENDIX

4

Slope of a tangent to the curve

In a graph illustrating position as a function of time, rate is determined by calculating the slope of a tangent to the curve drawn at a specific point. To draw the tangent of the curve at point A, you must draw a line: • touching the curve only at point A • with the same slope as the curve at this point This line is a tangent to the curve. The tangent is properly oriented if the points on the curve move away from the tangent more or less symmetrically on each side of point A (see Figure 14). To calculate the slope, choose two points on the tangent and calculate m  y . x

y (m)

A

y

x 0

x (s)

Figure 14 A tangent drawn to the curve at point A

4.5

Bar graph

A bar graph illustrates discontinuous data. It provides a graphical representation of the relationship between an independent variable and a dependent variable. One variable is represented by numbers, the other, with words. Bar graphs allow you to quickly compare the number of elements belonging to different set categories. The bars can be horizontal or vertical (see Figure 15).

Type of snack chosen by students Axis of ordinates

70

Number of students

Dependent variable

Title

Bar indicating value of variable

60 50 40 30 20 10 0

Banana

Granola bar Chocolate bar

Trail mix

Orange

Apple

Axis of abscissas Independent variable

Snack

Figure 15 Example of a vertical bar graph

390

APPENDIX 4 Presenting Scientific Results

4.6

Histogram

APPENDIX

4

Histograms illustrate continuous data grouped into categories. They show the distribution of a continuous variable along an axis, which provides an overview of the data distribution. Once a unit is assigned to an axis, the upper and lower limits of the categories are recorded on this axis. Then, a series of bars is drawn. The base of each bar represents a category interval; the height of each bar is proportional to the category frequency (see Figure 16).

Age of people who own an mp3 player Axis of ordinates

Title

30 28 26

Dependent variable

Frequency (number of people)

24

Bar indicating value of variable

22 20 18 16 14 12 10 8 6 4 2

Axis of abscissas Independent variable

0-9

10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 Age category

Figure 16 Example of a histogram

4.7

Circle graph

Circle graphs can be used to represent the percentages of a range of categories within a set, based on a given criterion. Each percentage corresponds to a section of the diagram. These sections must be proportional to the angle of the sector they represent (see Figure 17). Students who take part in team sports

Title Section of the diagram corresponding to the percentage associated with one category

To draw the sections of a circle graph, you must calculate the angle that corresponds to the percentage of a particular category. Angle  Percentage  360° Example of the calculation of the section representing the volleyball category:

Basketball (50%) Volleyball (30%)

Angle 

Badminton (10%) Other (10%)

Figure 17 Example of a circle graph

Legend

30  360°  108° 100

The angle of the section representing the volleyball category is 108°.

APPENDIX 4 Presenting Scientific Results

391

APPENDIX

4

4.8

Laboratory report

A laboratory report is a summary of the experimental method used. It describes each step of the method and presents the results. A laboratory report must always have a title page. The following example shows the different parts of a laboratory report:

Goal of the experiment The goal of the experiment is explained at the beginning of the laboratory report. The goal is to answer one of the questions regarding a phenomenon that has been observed. For example, the question might be, “How does this phenomenon occur?” or “What conditions must be present in order for the phenomenon to occur?” The goal is expressed in a statement using scientific words or expressions. This statement must begin with an action verb in the infinitive, such as "measure" or "determine."

Hypothesis The hypothesis is a proposition that must be validated through experimentation. It must be based on knowledge or observations. The results of the experiment will either confirm or disprove the hypothesis. The hypothesis is a statement that always begins with “It is suggested that …” or “It is believed that …” The wording used should, if possible, identify two variables: the independent variable and the dependent variable.

Experimental procedure The experimental procedure is a plan that describes exactly how the experiment will be conducted. It includes clear, concise descriptions of each step. The procedure should be easily understood by anyone else who wishes to repeat the experiment.

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An experimental procedure usually includes these elements: • a list of equipment and materials with specified quantities • a diagram of the set-up, such as a plan, drawing, photo or flow chart • the steps to follow when conducting the experiment (they must be numbered and described in short sentences)

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4

Results The experiment results are an essential part of the laboratory report. The data obtained during the experiment may be presented in a table or diagram (see sections 4.1 to 4.7). Whatever form you choose, it is essential to indicate the units of measure used in the experiment. The observations recorded during the experiment must also be mentioned in this part of the report. It is important to provide an example of each type of calculation performed with the data. For each calculation, show the data, equation, calculations and answer.

Analysis of the results In this part of the report, the meaning of the results obtained is discussed. You must also answer any questions your teacher asked at the beginning of the experiment. If the results are very different from what was expected, you must indicate the source of any possible errors. For example, you could question whether the measuring instruments were accurate enough or whether handling or calculation errors occurred.

Conclusion In the last part of the report, you must state whether or not the hypothesis was supported and summarize what you observed by conducting the experiment. For example, you should use the results of the experiment to explain the relationship between the independent variable and the dependent variable. Improvements that could be made to the experiment are also discussed in this section. Finally, suggest possible avenues for new experiments that could lead to a better understanding of the phenomenon studied. You may also suggest experiments to explore similar phenomena.

APPENDIX 4 Presenting Scientific Results

393

5 Interpreting Measurement Results

APPENDIX

5

5.1

Uncertainty

All measurements taken with instruments or devices involve uncertainty. One source of uncertainty is the measuring instrument itself. Another is the skill of the person using the instrument at taking and interpreting the readings. No measurement can be taken with absolute certainty. Errors relating to measuring instruments depend on three factors: repeatability, sensitivity and accuracy.

Absolute uncertainty and relative uncertainty The absolute uncertainty of a reading associated with a measuring instrument is generally equal to half of the instrument’s smallest graduation. In the case of an electronic device, absolute uncertainty is the smallest unit of the graduation displayed. Absolute uncertainty is usually recorded in a table of results with a  symbol in front of the uncertainty. For example, the absolute uncertainty of a measurement taken with a 50 mL graduated cylinder, whose smallest graduation is 1 mL, is 0.5 mL (see Figure 18). This is written as follows. V  (30.0  0.5) mL This means that the volume of the measured water is not exactly equal to 30.0 mL, but that it is between 29.5 mL and 30.5 mL. The last digit (the one on the far right) is always an estimated value. V  (30.0  0.5) mL € 29.5 mL  V  30.5 mL

The initial volume: 30 mL.

Figure 18 The volume of the measured water is 30.0 mL  0.5 mL.

To know how accurate a measurement is, express the uncertainty in the form of a percentage. This is referred to as relative uncertainty. Relative uncertainty is the ratio between the absolute uncertainty and the measured value and is calculated as follows: Relative uncertainty 

Absolute uncertainty  100% Value of measurement

The following example shows how to determine the relative uncertainty of the volume of a graduated cylinder: Example What is the relative uncertainty of the volume of the solution in the cylinder shown in Figure 18? Data:

Calculation: V  30.0 mL

Absolute uncertainly  0.5 mL Relative uncertainly  ?

Relative uncertainly 

Value of measurement 0.5 mL   100% 30.0 mL  1.7%

Answer: The volume of the water measured is 30.0 mL  2%.

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Absolute uncertainty

 100%

Repeatability, sensitivity and accuracy The repeatability of a measuring instrument is its ability to repeat the same result for the same measurement taken under the same conditions. To determine the repeatability of an instrument or device, each measurement must be taken more than once.

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5

The sensitivity of a device refers to its ability to detect small variations in measurement. If several measuring instruments or several calibrations of the same instruments are available, it is important to choose the one that offers the greatest sensitivity. The accuracy of a measuring instrument refers to its ability to take measurements with very few errors. Calibrating a device before using it improves its accuracy. For example, before a balance is used, we must adjust the needle to zero using the thumbwheel.

5.2

Significant figures

The digits recorded when taking a measurement are called significant figures. Significant figures include one digit that is certain and a final, uncertain digit that is estimated when the measurement is taken. The degree of certainty of a measurement is the number of significant figures it contains. Use the following rules to determine the number of significant figures of a measurement: Rule 1 All digits other than zero are significant. • 7.886 has four significant figures. • 19.4 has three significant figures. • 527.266 992 has nine significant figures. Rule 2 All zeros located between figures other than zero are significant. • 408 has three significant figures. • 25 074 has five significant figures. Rule 3 Zeros located at the beginning of a number are not significant. • 0.0927 has three significant figures: 9, 2 and 7. Rule 4 Zeros located at the end of a number are significant. • 22 700 has five significant figures. • 0.00210 has three significant figures: 2, 1 and the 0 on the far right. Rule 5 When counting the number of significant figures, do not take into account decimals, multiples, submultiples or powers of 10. • Ek  45 786 J; Ek  45.786 kJ; and Ek  4.5786  104 J all have five significant figures (4, 5, 7, 8 and 6). In scientific notation, all significant figures are shown. For example, the numbers shown in Rule 4 would be written as follows: • 22 700  2.2700  104

• 0.002 10  2.10  103 APPENDIX 5 Interpreting Measurement Results

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5.3

Measurements with unknown uncertainty

When it is not specifically indicated, uncertainty relates to the last significant figure. By convention, the uncertainty is equal to one unit of the least significant figure. • P  0.22 W means P  (0.22  0,01) W • T  1000°C means T  (1000  1)°C • F  100.2 N means F  (100.2  0.1) N • V  0.078 mL means V  (0.078  0.001) mL It is important to maintain the same number of significant figures when transforming units or when converting from units to their multiples or submultiples. • l  22.4 km  22.4  103 m (three significant figures in both notations), and not l  22 400 m (five significant figures) • V  5.75   106 mL (three significant figures in both notations), and not V  5750 mL (four significant figures)

5.4

Significant figures in results of mathematical operations

Significant figures are mainly used to determine the degree of certainty of a result obtained from the calculation of several measurements. For example, on a calculator, the result of the following calculation does not consist solely of significant figures. 0.024 89 mol  6.94 g/mol  0.172 736 6 g The following rules help to express the results of calculations with the correct number of significant figures: Rule 1 Addition and subtraction In a calculation, the value with the smallest number of decimal places determines the number of decimal places in the answer. • 1.2 g  1.22 g  1.222 g  3.6 g, because 1.2 only has one decimal place. Rule 2 Multiplication and division In a calculation, the value with the smallest number of significant figures determines the number of significant figures in the answer. • 1.2 m  1.33 m  1.6 m2, because 1.2 only has two significant figures.

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Rule 3 Complex calculations When calculations include additions (and subtractions) as well as multiplications (and divisions), these calculations must be performed separately. Data: F  23.55 N s  12.5 m

Calculation P2  P1 

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5

(F  s) t (23.55 N  12.5 m) 0.021 s

t  0.021 s

 5.23  104 W 

P1  5.23  104 W

 5.23  104 W  1.4  104 W  6.6  104 W, because 1.4  104 W has only two significant figures

Note : When successive calculations are required, it is generally advised to keep the maximum number of significant figures on the calculator during the intermediate calculations. Round off the final result only according to the rules previously explained.

5.5

How to round off a number

There are several methods for rounding off a number while taking into account the rules mentioned above. The method used in this textbook is called arithmetic rounding and is explained as follows: Rule 1 Select the last digit you need to retain as well as the following one. Consider the number 5.074 68. To round off to the nearest hundredth, truncate the number at 5.074 (last digit to retain  7; the following digit  4). Rule 2 If the digit that follows that last digit to be retained is less than 5, do not change the last digit (rounding off). In the truncated number 5.074, the digit 4, which follows the last digit to be retained (7), is less than 5. Rounded off to the nearest hundredth, the result is 5.07. Rule 3 If the digit that follows the last digit is 5 or more, increase the last digit to be retained by one unit (rounding up). Consider the number 5.074 68. To round off to the nearest thousandth, truncate the number at 5.0746 (last number to retain  4; the following digit  6). In the truncated number 5.0746, the digit 6, which follows the last digit to be retained (4), is more than 5. Rounded off to the nearest thousandth, the result is 5.075. Note : This method consists of separating the 10 decimal digits into two groups: the first group is 0, 1, 2, 3, 4: rounded off the second group is 5, 6, 7, 8, 9: rounded up

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6 Mathematics in Science

APPENDIX

6

Mathematics provides chemistry with the tools for better representing and better understanding the Universe. It also helps to solve scientific problems using algebraic expressions.

6.1

Review of some mathematical equations

Appendix 6.1 includes all of the formulas used in the Quantum collection. Acidity constant [H O]  [A] Ka  3 [HA]

Avogadro’s law V1 V2 *  n 1 n2

where

where

Ka  Acidity constant [H3O]  Hydronium ion concentration at equilibrium, expressed in moles per litre (mol/L) [A]  Conjugate base concentration at equilibrium, expressed in moles per litre (mol/L) [HA]  Non-dissociated acid concentration at equilibrium, expressed in moles per litre (mol/L)

V1  Initial volume, expressed in millilitres (mL) or in litres (L) n1  Initial quantity of gas, expressed in moles (mol) V2  Final volume, expressed in millilitres (mL) or in litres (L) n2  Final quantity of gas, expressed in moles (mol)

* On condition that temperature (T ) and pressure (P ) are constant.

Basicity constant [HB]  [OH] Kb  [B]

Boyle’s law P1V1  P2V2*

where

where

 Basicity constant  Concentration of the conjugate acid at equilibrium, expressed in moles per litre (mol/L) [OH]  Concentration of hydroxide ions at equilibrium, expressed in moles per litre (mol/L) [B]  Concentration of the non-dissociated base at equilibrium, expressed in moles per litre (mol/L) Kb  [HB ]

P1  Initial pressure, expressed in kilopascals (kPa) or millimetres of mercury (mm Hg) V1  Initial volume, expressed in millilitres (mL) or litres (L) P2  Final pressure, expressed in kilopascals (kPa) or millimetres of mercury (mm Hg) V2  Final volume, expressed in millilitres (mL) or litres (L)

* On condition that the number of moles (n) of gas and temperature (T) are constant.

Charles’ law V1 V2 *  T 1 T2

where

V1  Initial volume, expressed in millilitres (mL) or litres (L) T1  Initial temperature, expressed in kelvins (K ) V2  Final volume, expressed in millilitres (mL) or litres (L) T2  Final temperature, expressed in kelvins (K)

* On condition that the number of moles (n) of gas and the pressure (P ) are constant.

Concentration of solutions C1V1  C2V2 where

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C1  Concentration of the initial solution V1  Volume of the initial solution C2  Concentration of the initial solution V2  Volume of the initial solution

Dalton’s law PT  PA  PB  P C  …

where

PT  Total pressure of the mixture, expressed in kilopascals (kPa) or in millimetres of mercury (mm Hg) PA  Partial pressure of gas A, expressed in kilopascals (kPa) or in millimetres of mercury (mm Hg) PB  Partial pressure of gas B, expressed in kilopascals (kPa) or in millimetres of mercury (mm Hg) PC  Partial pressure of gas C, expressed in kilopascals (kPa) or in millimetres of mercury (mm Hg)

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6

Energy balance Hreaction  Hbonds broken  Hbonds formed where Hreaction  Enthalpy change of the chemical reaction, expressed in kilojoules per mole (kJ/mol) Hbonds broken  Enthalpy change when the reactant bonds are broken, expressed in kilojoules per mole (kJ/mol) Hbonds formed  Enthalpy change of the formation of product bonds, expressed in kilojoules per mole (kJ/mol) Enthalpy change H  Hp  Hr

where H  Enthalpy change of the reaction, expressed in joules (J) or kilojoules (kJ) Hp  Product enthalpy, expressed in joules (J) or kilojoules (kJ) Hr  Reactant enthalpy, expressed in joules (J) or kilojoules (kJ)

Equilibrium constant [C]c  [D]d Kc  where [A]a  [B]b

Kc  Equilibrium constant as a function of concentrations [C], [D]  Product concentrations at equilibrium, expressed in moles per litre (mol/L) [A], [B]  Reactant concentrations at equilibrium, expressed in moles per litre (mol/L) c, d  Whole number stoichiometric coefficients of the products in the balanced chemical equation a, b  Whole number stoichiometric coefficients in the balanced chemical reaction

Equilibrium constant of the reverse reaction 1 Kcrev  where Kcrev  Equilibrium constant of the reverse reaction Kcdir Kcdir  Equilibrium constant of the direct reaction Expression of pH pH  log [H3O] [H3O]  10pH where pH  pH value [H3O]  Hydronium ion concentration at equilibrium, expressed in moles per litre (mol/L) Expression of pOH pOH  log [OH] [OH]  10pOH where pOH  Value of pOH [OH]  Hydroxide ion concentration at equilibrium, expressed in moles per litre (mol/L)

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Gay-Lussac’s law P 1 P2 *  where T 1 T2

P1  Initial pressure, expressed in kilopascals (kPa) or in millimetres of mercury (mm Hg) T1  Initial absolute temperature, expressed in kelvin (K) P2  Final pressure, expressed in kilopascals (kPa) or in millimetres of mercury (mm Hg) T2  Final absolute temperature, expressed in kelvin (K)

* On condition that the number of moles (n) of gas and the volume (V ) are constant.

General gas law P1V1 P2V2  where n1T1 n2T2

P1  Initial pressure of gas, expressed in kilopascals (kPa) or in millimetres of mercury (mm Hg) V1  Initial volume of gas, expressed in millilitres (mL) or in litres (L) n1  Initial quantity of gas, expressed in moles (mol) T1  Initial absolute temperature of gas, expressed in kelvin (K) P2  Final pressure of gas, expressed in kilopascals (kPa) or in millimetres of mercury (mm Hg) V2  Final volume of gas, expressed in millilitres (mL) or in litres (L) n2  Final quantity of gas, expressed in moles (mol) T2  Final absolute temperature of gas, expressed in kelvin (K)

General reaction rate 1 [A] 1 [B] 1 [C] 1 [D] r    where a t b t c t d t r  General reaction rate, expressed in moles per litre-second (mol/(Ls) a, b, c, d  Whole stoichiometric coefficients of the substances involved in the chemical equation in the balanced reaction reaction [A], [B], [C], [D]  Changes in concentration of the substances involved in the reaction, expressed in moles per litre (mol/L) t  Change in time (tf  ti), expressed in seconds (s) Graham’s law r1 M2  r2 M1

where

r1  Rate of diffusion or effusion of gas 1, expressed in metres per second (m/s) r2  Rate of diffusion or effusion of gas 2, expressed in metres per second (m/s) M1  Molar mass of gas 1, expressed in grams per mole (g/mol) M2  Molar mass of gas 2, expressed in grams per mole (g/mol)

Gravitational potential energy Ep mgh where

Ep  Gravitational potential energy, expressed in joules (J) m  Mass of an object, expressed in kilograms (kg) g  Gravitational acceleration, which is 9.8 m/s2 on Earth h  Height of an object in relation to a reference point, expressed in metres (m)

Heat transfer between two systems 1c1T1  m2c2T2 where m1  Mass of the substance in system 1, expressed in grams (g) c1  Specific heat capacity of the substance in system 1, expressed in joules per gramdegree Celsius (J/(g°C)) T1  Temperature change in system 1 (Tf  Ti), expressed in degrees Celsius (°C) m2  Mass of the substance in system 2, expressed in grams (g) c2  Specific heat capacity of the substance in system 2, expressed in joules per gram degree Celsius (J/(g°C)) T2  Temperature change in system 2 (Tf  Ti), expressed in degrees Celsius (°C)

m

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Hess’s law H  H1  H2  H3  …

Ideal gas law PV  nRT

where

where

Mechanical energy Em  Ep  E c

6

Kwater  Ionization constant of water [H3O]  Hydronium ion concentration at equilibrium, expressed in moles per litre (mol/L) [OH]  Hydroxide ion concentration at equilibrium, expressed in moles per litre (mol/L)

where

Ek  Kinetic energy, expressed in joules (J) m  Mass of an object, expressed in kilograms (kg) v  Velocity of an object, expressed in metres per second (m/s)

where

Em  Mechanical energy, expressed in joules (J) Ep  Potential energy, expressed in joules (J) Ek  Kinetic energy, expressed in joules (J)

Molar concentration n C where V

Molar mass of a gas mRT M where PV

APPENDIX

P  Pressure of the gas, expressed in kilopascals (kPa) or in millimetres of mercury (mm Hg) V  Volume of the gas, expressed in litres (L) n  Quantity of gas, expressed in moles (mol) R  Gas constant, expressed in (kPaL)/(molK) T  Temperature of the gas, expressed in kelvin (K)

Ionization constant of water Kwater  [H3O]  [OH] where

Kinetic energy 1 Ek  mv 2 2

H  Enthalpy change of the overall reaction H1, H2, H3  Enthalpy change of each intermediate step or elementary reaction of the overall reaction

C  Molar concentration, expressed in moles per litre (mol/L) n  Amount of solute, expressed in moles (mol) V  Volume of solution, expressed in litres (L)

M  Molar mass of the gas, expressed in grams per mole (g/mol) or in kilograms per mole (kg/mol) m  Mass of the gas sample, expressed in grams (g) or in kilograms (kg) R  Gas constant, expressed in (kPaL)/(molK) T  Absolute temperature, expressed in kelvin (K) P  Pressure of the gas, expressed in kilopascals (kPa) V  Volume of the gas, expressed in litres (L)

Partial pressure of a gas n PA  A  PT where nT

PA  Partial pressure of gas A, expressed in kilopascals (kPa) or in millimetres of mercury (mm Hg) nA  Quantity of gas A, expressed in moles (mol) nT  Total quantity of gas, expressed in moles (mol) PT  Total pressure of the mixture, expressed in kilopascals (kPa) or in millimetres of mercury (mm Hg)

APPENDIX 6 Mathematics in Science

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Pressure F P A

Rate law r  k  [A]x  [B]y

where

P  Pressure, expressed in newtons per square meter (N/m2) or in pascals (Pa) F  Force, expressed in newtons (N) A  Area, expressed in square meters (m2)

where

r  Reaction rate, expressed in moles per litre-second (mol/(Ls)) k  Rate constant [A], [B]  Concentration of reactants, expressed in mol/L x, y  Coefficients of the reactants in a balanced equation of the elementary reaction

Reaction rate Quantity of reactant(s) rreactant(s)  where rreactant(s)  Reaction rate of reactant(s) t Quantity of reactant(s)  Change in quantity of reactant(s) Quantity of product(s) Quantityf of reactant(s)  Quantityi of reactant(s)) rproduct(s)  t r  Reaction rate of product(s) product(s)

Quantity of product(s)  Change in quantity of product(s) (Quantityf of product(s)  Quantityi of product(s)) t  Change in time (tf  ti)

Real pressure of a gas using a closed-end manometer Ph where Pgas  Pressure of the gas in the container, expressed in millimetres of mercury (mm Hg) h  Height of the column of mercury, expressed in millimetres of mercury (mm Hg) Real pressure of a gas using an open-end manometer If Pgas  Patm Then Pgas  Patm  h If Then

Pgas  Patm Pgas  Patm  h

where

Pgaz  Gas pressure in the container, expressed in millimetres of mercury (mm Hg) Patm  Atmospheric pressure, expressed in millimetres of mercury (mm Hg) h  Height of the column of mercury, expressed in millimetres of mercury (mm Hg) Relationship between pressure and number of moles P1 P2 *  where P1  Initial pressure, expressed in kilopascals (kPa) or in millimetres of mercury (mm Hg) n 1 n2 n1  Initial quantity of gas, expressed in moles (mol) P2  Final pressure, expressed in kilopascals (kPa) or in millimetres of mercury (mm Hg) n2  Final quantity of gas, expressed in moles (mol) * On condition that temperature (T ) and volume (V ) are constant.

State of equilibrium rdir  rrev

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where

rdir  Direct reaction rate rrev  Reverse reaction rate

Solubility product constant Kps  [X]n  [Y]m where

[X],

Kps  Solubility product constant [Y]  Concentration of ions at equilibrium, expressed in moles per litre (mol/L) n, m  Coefficients of each of the ions in the balanced equation

APPENDIX

6

Temperature of two systems m c T  m1c1Ti1 where Tf  Final temperature of the two systems, expressed in degrees Celsius (°C) Tf  2 2 i2 m1c1  m2c2 m1  Mass of the substance in system 1, expressed in grams (g) c1  Specific heat capacity of the substance in system 1, expressed in joules per gram degree Celsius (J/(g°C)) Ti1  Initial temperature of system 1, expressed in degrees Celsius (°C) m2  Mass of the substance in system 2, expressed in grams (g) c2  Specific heat capacity of the substance in system 2, expressed in joules per gram degree Celsius (J/(g°C) Ti2  Initial temperature of system 2, expressed in degrees Celsius (°C) Thermal energy Q  mcT

where

Q  Amount of heat, expressed in joules (J) m  Mass of a substance, expressed in grams (g) c  Specific heat capacity of a substance, expressed in joules per gram per degree Celsius (J/(g°C)) T  Variation in temperature (Tf – Ti), expressed in degrees Celsius (°C)

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6.2

Transformation of algebraic expressions

Many scientific phenomena are described by algebraic expressions. These expressions translate the relationship among different physical variables involved in a phenomenon. It is not always possible to use an algebraic expression “as is” to solve a specific problem. For example, the expression PV = nRT does not allow us to immediately determine the temperature (T) of a gas when its pressure (P), volume (V), number of moles (n) and the gas constant (R) are known. The expression must first be transformed to T  PV . nR The purpose of transforming an algebraic expression is to express a variable of the problem as a function of the others, that is, to isolate the variable. To do this you must: • determine where the variable you want to isolate is located in the algebraic expression • gradually isolate this variable by performing the necessary inverse mathematical operations on each side of the equation Example Isolate T in the ideal gas law PV  nRT. Solution : 1. T is on the right side of the equation. PV  nRT

2. Isolate T by dividing the terms on each side of the equation by nR. PV nRT  nR nR 3. Write out the expression in the usual manner, placing the isolated variable on the left side of the equation. PV T nR

Sometimes, the variable you need to isolate is in a second-order equation of the type y  ax2  bx  c. In this case, follow the steps for solving a secondorder equation and determine the values of x using the following expression:

x

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b

 b 2  4ac 2a

Example Determine the value of x in the following expression:

APPENDIX

6

6x 2  4  8x  3x  4x 2  7 Solution : 1. Gather all of the terms on the same side of the equation by performing the appropriate mathematical operations among similar terms, and rearrange them so that they are in the form ax 2  bx  c  0. 6x 2  4  8x  3x  4x 2  7 6x 2  4x 2  4  7  8x  3x  0 2x 2  5x  3  0

2. Determine the values of a, b and c. a2 b5 c  3

3. Calculate the possible values of x. x 

b  5 

b 2  4ac 2a 5  49 52  4(2  3) 5  25  24   2(2) 4 4

 0.5 or x 

b  5 

b 2  4ac 2a 5  49 52  4(2  3) 5  25  24   2(2) 4 4

 3

4. Determine which of the values of x makes more sense based on the nature of the problem. Only keep the value that is relevant. For example, if the value you are looking for is molar concentration, reject the second value obtained for x because a negative molar concentration is impossible. Answer: The value of x is 0.5 or -3.

APPENDIX 6 Mathematics in Science

405

APPENDIX

6

6.3

Scientific notation

The order of magnitude of numbers used in science is often very small or very large. This type of number is difficult to handle when performing calculations. Scientific notation helps to simplify these numbers. The following two examples show how to write numbers using scientific notation. This notation is the product of a number called the mantissa whose absolute value is between 1 and 10 (excluding 10) and a power of 10. Example A The number of particles contained in a mole of a substance (Avogadro’s number) is usually rounded off to 602 000 000 000 000 000 000 000. To simplify this number, it is expressed in the form of a number between 1 and 10 that is multiplied by a power of 10. To do this, move the decimal to the non-zero number on the far left, and count how many places you have moved the decimal. This number becomes the exponent of the base 10 (see Figure 19 ).

6.02 000 000 000 000 000 000 000. 23 21

18

15

12

9

6

3

6.02  10

23

Figure 19 To express a large number using a power of 10, move the decimal to the left.

Example B The mass of a molecule of water is 0.000 000 000 000 000 000 000 029 9 g. To express this number using scientific notation, that is, in the form of a number between 1 and 10 that is multiplied by a power of 10, move the decimal toward the right until the first non-zero number you meet, and count how many places you have moved the decimal. This number becomes a negative exponent. (see Figure 20 ).

0.000 000 000 000 000 000 000 02.9 9 3 2.99  10

6

9

12

15

18

21

23

23

Figure 20 To express a small number using a power of 10, move the decimal to the right.

406

APPENDIX 6 Mathematics in Science

Calculation rules for scientific notation

APPENDIX

Every measurement of a quantity must include the correct number of significant figures. This number can be provided using scientific notation. Use the following rules when performing calculations with numbers expressed in scientific notation:

6

Rule 1 To multiply two numbers expressed in scientific notation, multiply the mantissas. The result is then multiplied by a power of 10 to which is attributed the sum of the exponents of the initial powers of 10. (a  10m) (b  10n)  (a  b)  10(m  n) (7.32  103)  (8.91  102)  (7.32  8.91)  10(3  2)  65.221 2  105  6.52  105 Rule 2 To divide two numbers expressed in scientific notation, first divide the mantissas. The result is then multiplied by a power of 10 to which is attributed the difference between the exponents of the initial powers of 10. (a  10m) a   10(m  n) n (b  10 ) b (1.842  106) 1.842   10(6  2) (1.078 7  102) 1.078 7  1.707 611  104  1.708  104 Rule 3 To add or subtract numbers in scientific notation, first convert the numbers so that they have the same exponents. Each number must have the same exponent as the number with the greatest exponent of 10. Once all the numbers are expressed with the same exponent of 10, the mantissas are added or subtracted and the exponent of 10 remains as is. (3.42  106)  (8.53  103)  (3.42  106)  (0.008 53  106)  (3.42  0.008 53)  106  3.428 53  106  3.43  106 (9.93  101)  (7.86  101)  (9.93  101)  (0.0786  101)  (9.93  0.078 6)  101  9.851 4  101  9.85  101

APPENDIX 6 Mathematics in Science

407

7 Units of Measure in Chemistry

APPENDIX

7

This appendix presents the units of measure most often used in chemistry to express quantities. These units of measure are part of the International System of Units (SI). This system is recognized internationally because it facilitates exchanges in the fields of science, technology and education.

7.1

Common SI prefixes

In addition to defining units of measure, the SI proposes certain prefixes for multiples and submultiples of units of measure. The prefix of the multiple and the unit of measure form a new unit. The symbol for this new unit is made up of two symbols: the symbol for the multiple and the symbol for the unit of measure. Some SI prefixes are quite common (see Table 9).

Table 9 The most common SI symbols with examples Multiple

408

Prefix

Symbol

Example

109  1 000 000 000

giga-

G

gigametre (Gm)

106  1 000 000

mega-

M

megametre (Mm)

103  1000

kilo-

k

kilometre (km)

102  100

hecto-

h

hectometre (hm)

101  10

deca-

da

decametre (dam)

101  0.1

deci-

d

decimetre (dm)

102  0.01

centi-

c

centimetre (cm)

103  0.001

milli-

m

millimetre (mm)

106  0.000 001

micro-

μ

micrometre (μm)

109  0.000 000 001

nano-

n

nanometre (nm)

APPENDIX 7 Units of Measure in Chemistry

7.2

Common units of measure

APPENDIX

Some units of measure are commonly used in chemistry (see Table 10). The table on the next page lists each quantity measured and its symbol, as well as the name of the corresponding unit of measure and its symbol.

7

Table 10 The most common units of measure used in chemistry* Name and symbol of quantity measured

Name and symbol of unit of measure

Amount of heat energy (Q )

• joule (J)

Amount of matter (n)

• mole (mol)

Area (A)

• square centimetre (cm2) • square kilometre (km2)

• square metre (m2)

Concentration of a solution (C )

• milligram per litre (mg/L)

• gram per litre (g/L)

• gram per millilitre (g/mL)

• gram per 100 mL (g/100 mL)

• mole per litre (mol/L)

• parts per million (ppm)

• volume/volume percent (V/V%) • mass/volume percent (m/V%) • mass/mass percent (m/m%) Density (ρ)

• gram per cubic centimetre (g/cm3)

Energy (E )

• joule (J)

Enthalpy (H )

• joule (J)

Force (F )

• newton (N)

Kinetic energy (Ek)

• kilojoule (kJ)

Mass (m)

• gram (g)

• kilogram (kg)

• tonne (t)* (1 t  1000 kg)

• megaton (Mt)*

• gram per millilitre (g/mL)

• kilojoule (kJ)

Mechanical energy (Em )

• kilowatt-hour (kWh) (1 kWh 3 600 000 J)

Molar mass (M )

• gram per mole (g/mol)

Potential energy (Ep )

• megawatt-hour (MWh) (1 MWh  3 600 000 000 J)

Pressure (P )

•newton per square metre (N/m2)

• pascal (Pa)

Solubility of a substance

• gram per litre (g/L)

• gram per 100 millilitres (g/100 mL)

Specific heat capacity (c )

• joule per gram per degree Celsius (J/(g°C)) • kilojoule per kilogram per degree Celsius (kJ/(kg°C))

Speed (v )

• metre per second (m/s)

• kilometre per hour (km/h)

• kilometre per second (km/s) Temperature (T )

• degree Celsius (°C)

• degree kelvin (K) (0 K  273°C)

Time (t )

• second (s)

• minute (min)

• hour (h)

• day (d)

• year (y) Volume (V )

Weight (Fg )

• cubic centimetre (cm3)

• millilitre (mL) (1 mL  1 cm3)

• cubic metre (m3)

• litre (L) (1 L  1 dm3)

• newton (N)

*Some units of measure commonly used by scientists are not SI units.

APPENDIX 7 Units of Measure in Chemistry

409

8 Reference Tables

APPENDIX

8

Table 8.1

Alphabetical list of elements and their atomic mass

Element

Symbol

Atomic number Atomic mass*

Actinium

Ac

89

[227]

Aluminum

Al

13

Americium

Am

Antimony

Element

Symbol

Atomic number

Erbium

Er

68

167.259(3)

26.981 538 6(8)

Europium

Eu

63

151.964(1)

95

[243]

Fermium

Fm

100

[257]

Sb

51

121.760(1)

Fluorine

F

9

18.998 403 2(5)

Argon

Ar

18

39.948(1)

Francium

Fr

87

[223]

Arsenic

As

33

74.921 60(2)

Gadolinium

Gd

64

157.25(3)

Astatine

At

85

[210]

Gallium

Ga

31

69.723(1)

Barium

Ba

56

137.327(7)

Germanium

Ge

32

72.64(1)

Berkelium

Bk

97

[247]

Gold

Au

79

196.966 569(4)

Beryllium

Be

4

9.012 182(3)

Hafnium

Hf

72

178.49(2)

Bismuth

Bi

83

208.980 40(1)

Hassium

Hs

108

[270]

Bohrium

Bh

107

[272]

Helium

He

2

4.002 602(2)

Boron

B

5

10.811(7)

Holmium

Ho

67

164.930 32(2)

Bromine

Br

35

79.904(1)

Hydrogen

H

1

1.007 94(7)

Cadmium

Cd

48

112.411(8)

Indium

In

49

114.818(3)

Calcium

Ca

20

40.078(4)

Iodine

I

53

126.904 47(3)

Californium

Cf

98

[251]

Iridium

Ir

77

192.217(3)

Carbon

C

6

12.0107(8)

Iron

Fe

26

55.845(2)

Cerium

Ce

58

140.116(1)

Krypton

Kr

36

83.798(2)

Cesium

Cs

55

132.905 451 9(2)

Lanthanum

La

57

138.905 47(7)

Chlorine

Cl

17

35.453(2)

Lawrencium

Lr

103

[262]

Chromium

Cr

24

51.9961(6)

Lead

Pb

82

207.2(1)

Cobalt

Co

27

58.933 195(5)

Lithium

Li

3

[6.941(2)]

Copernicium

Cn

112

[285]

Lutetium

Lu

71

174.966 8(1)

Copper

Cu

29

63.546(3)

Magnesium

Mg

12

24.305 0(6)

Curium

Cm

96

[247]

Manganese

Mn

25

54.938 045(5)

Darmstadtium

Ds

110

[281]

Meitnerium

Mt

109

[276]

Dubnium

Db

105

[268]

Mendelevium

Md

101

[258]

Dysprosium

Dy

66

162.500(1)

Mercury

Hg

80

200.59(2)

Einsteinium

Es

99

[252]

Molybdenum

Mo

42

95.96(2)

* The number in parentheses indicates uncertainty in the last digit of the atomic mass. The number in square brackets indicates the mass of the isotope with the longest expected life.

410

APPENDIX 8 Reference Tables

Atomic mass*

APPENDIX

8

Table 8.1

Alphabetical list of elements and their atomic mass (cont.)

Element

Symbol

Atomic number Atomic mass*

Element

Symbol

Atomic number Atomic mass*

Neodymium

Nd

60

144.242(3)

Selenium

Se

34

78.96(3)

Neon

Ne

10

20.179 7(6)

Silicon

Si

14

28.08 55(3)

Neptunium

Np

93

[237]

Silver

Ag

47

107.868 2(2)

Nickel

Ni

28

58.693 4(4)

Sodium

Na

11

22.989 769 28(2)

Niobium

Nb

41

92.906 38(2)

Strontium

Sr

38

87.62(1)

Nitrogen

N

7

14.0067(2)

Sulphur

S

16

32.065(5)

Nobelium

No

102

[259]

Tantalum

Ta

73

180.947 88(2)

Osmium

Os

76

190.23(3)

Technetium

Tc

43

[98]

Oxygen

O

8

15.999 4(3)

Tellurium

Te

52

127.60(3)

Palladium

Pd

46

106.42(1)

Terbium

Tb

65

158.925 35(2)

Phosphorus

P

15

30.973 762(2)

Thallium

Tl

81

204.383 3(2)

Platinum

Pt

78

195.084(9)

Thorium

Th

90

232.038 06(2)

Plutonium

Pu

94

[244]

Thulium

Tm

69

168.934 21(2)

Polonium

Po

84

[209]

Tin

Sn

50

118.710(7)

Potassium

K

19

39.098 3(1)

Titanium

Ti

22

47.867(1)

Praseodymium

Pr

59

140.907 65(2)

Tungsten

W

74

183.84(1)

Promethium

Pm

61

[145]

Ununhexium

Uuh

116

[293]

Protactinium

Pa

91

231.035 88(2)

Ununoctium

Uuo

118

[294]

Radium

Ra

88

[226]

Ununpentium

Uup

115

[288]

Radon

Rn

86

[222]

Ununquadium

Uuq

114

[289]

Rhenium

Re

75

186.207(1)

Ununtrium

Uut

113

[284]

Rhodium

Rh

45

102.905 50(2)

Uranium

U

92

238.028 91(3)

Roentgenium

Rg

111

[280]

Vanadium

V

23

50.9415(1)

Rubidium

Rb

37

85.467 8(3)

Xenon

Xe

54

131.293(6)

Ruthenium

Ru

44

101.07(2)

Ytterbium

Yb

70

173.054(5)

Rutherfordium

Rf

104

[265]

Yttrium

Y

39

88.905 85(2)

Samarium

Sm

62

150.36(2)

Zinc

Zn

30

65.38(2)

Scandium

Sc

21

44.955 912(6)

Zirconium

Zr

40

91.224(2)

Seaborgium

Sg

106

[271]

* The number in parentheses indicates uncertainty in the last digit of the atomic mass. The number in square brackets indicates the mass of the isotope with the longest expected life. Source: International Union of Pure and Applied Chemistry (IUPAC), 2010.

APPENDIX 8 Reference Tables

411

APPENDIX

8

412

Table 8.2

Everyday chemical products

Common name

Chemical formula and other names

Physical properties

Potential dangers

Comments

Acetone

CH3COCH3 2-propanone

Clear, evaporates easily

Flammable, toxic if swallowed or inhaled

Solvent, present in some dissolving agents (nail polish)

Acetylene

C2H2 Ethene

Pleasant odour

Highly explosive

In oxyacetylene soldering torches, it burns when it comes into contact with oxygen and releases large amounts of heat; used in the production of a broad range of synthetic products

ASA

O-acetoxybenzoic acid (acetylsalicylic acid)

White crystals, slightly bitter taste

Excessive use can cause stomach irritation as well as hearing loss or Reye's syndrome, especially in young people

Used in some medications to relieve pain, fever and inflammation

Battery acid

H2SO4 Sulphuric acid

Clear and odourless

Corrosive

Used in lead-acid batteries (car batteries)

Bleach

NaClO (aq) Sodium hypochlorite in solution

Yellowish solution with an odour of chlorine

Toxic, strong oxidizing agent

Used to whiten laundry and in cleaning

Blue vitriol

CuSO45H2O Copper(II) sulfate pentahydrate (cupric sulfate pentahydrate)

Blue crystals or blue crystalline granules

Toxic if swallowed; strong irritant

Used in agriculture and in certain industries as a germicidal agent and to protect wood

Borax

Na2B4O710H2O

White crystals

None

Obtained primarily from mines; used in the glass and ceramic industries; used in manufacturing soap and fertilizer

Carborundum

SiC Silicon carbide

Hard black solid

None

Used as an abrasive

CFC

CCl2F2, CCl3F, CClF3 Chlorofluorocarbons (freon, freon 12)

Colourless and odourless gases

Prohibited by the Montreal Protocol

In the past was used as refrigerating agents and in aerosols

Citric acid

(HOOCCH2)2 C(OH) (COOH) Translucent crystals None 2-hydroxypropane-1,2,3- with a strong acidic tricarboxylic acid taste

Used in foods and soft drinks as an acidifying agent and as an antioxidant

Sodium bicarbonate

NaHCO3 Bicarbonate of soda

Tiny white crystals

None

Used in cooking and cleaning, as an antacid, as a mouthwash and in fire extinguishers

Wood charcoal / graphite

C(s) Pure carbon in a less structured form than diamond

Soft grey or black solid that readily marks other substances

None

Used in pencils and charcoal pencils, as a decolourant and filtration agent, as gunpowder and in barbecue briquettes

APPENDIX 8 Reference Tables

Table 8.2

Everyday chemical products (cont.)

Common name

Chemical formula and other names

Physical properties

Potential dangers

APPENDIX Comments

Cream of tartar

C4H5KO6 Potassium bitartrate

White crystalline solid

None

Used as leaven in chemical leavening agents

Dry ice

CO2 Solid carbon dioxide

Cold, white solid, capable of sublimation

Prolonged exposure causes damage to skin and tissues

Used as a refrigerating substance in laboratories to produce cold temperatures (up to -79°C)

Epsom salts

MgSO47H2O Magnesium sulfate heptahydrate

Colourless crystals

May cause abdominal cramps and diarrhoea

Used as bath salts, in beauty products, in food supplements and in various industries

Ethylene

C2H4 Ethene

Colourless gas with a pleasant odour and taste

Flammable

Used to accelerate the ripening of fruits and to synthesize polymers such as polystyrene; naturally occurring in plants

Ethylene glycol

CH2OHCH2OH Glycol

Clear, colourless, syrupy liquid

Toxic if swallowed or inhaled

Used in antifreeze, in beauty products and as a de-icing fluid on runways

Glucose

C6H12O6 Dextrose, grape sugar, corn sugar

White crystals, sweet taste

None

Used as an energy source by most organisms

Grain alcohol

C2H5OH Ethanol (ethyl alcohol)

Clear, volatile liquid with a distinctive odour

Flammable

Used as an alcoholic beverage, antiseptic, laboratory and industry solvent; produced by the fermentation of cereals or fruits

Gypsum rock

CaSO42H2O Gypsum

Hard, beige mineral

None

Used in moulding plaster as a main ingredient in dry-type construction

Hydrogen peroxide

H2O2

Clear, colourless liquid

Causes skin damage in high concentrations

Sold in 3% solutions in pharmacies; non-chlorinated bleaching agents often contain 6% hydrogen peroxide

Ibuprofen

C13H18O2 P-isobutyl-hydratropic acid

White crystals

Incompatible with other medications

Used as an ingredient in overthe-counter pain relievers

Laughing gas

N2O Nitrous oxide, dinitrogen oxide

Soluble, colourless, essentially odourless gas

May cause brain damage and lead to infertility after prolonged exposure

Used as an anesthetic in dentistry, as a propellant in aerosols and to increase fuel performance in race cars

APPENDIX 8 Reference Tables

8

413

APPENDIX

8

414

Table 8.2

Everyday chemical products (cont.)

Common name

Chemical formula and other names

Physical properties

Potential dangers

Comments

Lime

CaO Calcium oxide (hydrated lime, hydraulic lime, quicklime)

White powder

Reacts with water to form caustic calcium hydroxide, Ca(OH)2, releasing heat

Used in manufacturing cement and to clean and neutralize livestock odours

Limestone

CaCO3 Calcium carbonate

Soft white mineral

None

Used to manufacture lime; used in construction and in various industries

Lye

NaOH Sodium hydroxide (caustic soda)

White solid, usually in the form of small balls or tablets; quickly absorbs water and CO2 from the air

Corrosive and strong irritant

Produced by the electrolysis of salt water or through the reaction of calcium hydroxide and sodium carbonate; multiple uses in the laboratory and in various industries; used in manufacturing chemical products and soap

Malachite

CuCO3Cu(OH)2 Copper(II) carbonate

Hard, bright-green mineral

None

Decorative and ornamental stone; a copper ore

Milk of magnesium

Mg(OH)2 Magnesium hydroxide

White powder

No danger in small quantities

Used as an antacid and laxative

Monosodium glutamate

COOH(CH2)2CH(NH2)COON Sodium monoglutamate

White crystalline powder

Can cause headaches in some people

Used to enhance food flavours at concentrations of approximately 0.3%

Mothballs

C10H8 Naphthalene

Volatile white solid with an unpleasant odour

Toxic if swallowed or inhaled

Used to ward off insects in houses and gardens and in manufacturing synthetic resins; made from crude oil

Muriatic acid

HCl(aq) Hydrochloric acid

Colourless or slightly yellow aqueous solution

Toxic if swallowed or inhaled; strong irritant

Has numerous uses in laboratories and in certain industries; used in food processing, cleaning and pickling

Natural gas

Consisting of approximately 85% methane, CH4), 10% ethane, C2H6, as well as propane, C3H8, butane, C4H10, and pentane, C5H12

Odourless and colourless gas

Flammable and explosive; as a precaution, a distinctive odour is added to the gas for domestic use

Used in heating, cooking and as an energy source: approximately 3% is used as a basic product in the chemical industry

APPENDIX 8 Reference Tables

Table 8.2

Everyday chemical products (cont.)

Common name

Chemical formula and other names

Physical properties

Potential dangers

APPENDIX Comments

Oxalic acid

HO2CCO2H Ethanedioic (“oxalic”) acid

Strong-tasting acid; white crystals

Toxic if swallowed or inhaled; strong irritant in high concentrations

Naturally occurring in rhubarb, mountain woodsorrel and spinach; used as a whitening agent for wood and textiles, to remove rust and to clean platforms; many uses in laboratories and in certain industries

PCB

Polychlorinated biphenyls: family of compounds consisting of two benzene rings and two or more substituted chlorine atoms

Colourless liquids

Highly toxic, unreactive and persistent; harmful to the environment

In the past was used as cooling agents in electrical transformers

Potash

K2CO3 Potassium carbonate

Granular and translucent white powder

Solution causes irritation to tissues

Various applications in the laboratory and in certain industries; used in manufacturing certain types of glass, soap and as a dehydrating agent

PVC

(C2H3Cl)n Polyvinyl chloride, polychloroethene

Unreactive hard, white solid

None

Commonly used as a construction material

Road salt

CaCl2 Calcium chloride

White crystalline compound

None

By-product of the Solvay process

Rotten egg gas

H2S Hydrogen sulfide

Colourless gas with an unpleasant odour

Very flammable, causing increased risk of fire; explosive; toxic if inhaled, strong irritant to eyes and mucous membranes

Obtained from sulphurous gas during the production of natural gas

Rubbing alcohol

(CH3)2CHOH Isopropyl alcohol

Colourless liquid with a pleasant odour

Very flammable, causing increased risk of fire; explosive; toxic if swallowed or inhaled

Used for industrial and medical purposes

Salicylic acid

HOC6H4COOH 2-hydroxybenzoic acid

White crystalline solid

Harmful to skin in high concentrations

Used in various quantities in foods and dyes as well as in the treatment of warts

APPENDIX 8 Reference Tables

8

415

APPENDIX

8

416

Table 8.2

Everyday chemical products (cont.)

Common name

Chemical formula and other names

Physical properties

Potential dangers

Comments

Sand

SiO2 Silica

Large cubic crystals resembling glass

Toxic if inhaled; chronic exposure to silica dust may cause silicosis

Present throughout nature in the form of sand, quartz, flint and diatomite

Slaked lime

Ca(OH)2 Calcium hydroxide

White powder; insoluble in water

None

Used to neutralize soil acidity and to manufacture whitewash, decolourant and glass

Soda ash

Na2CO3 Sodium carbonate

Powdery white crystals

None

Used to manufacture glass, soap and detergent

Soda crystals

Na2CO3H2O Sodium carbonate monohydrate

White powdered crystals

May irritate skin

Used in cleaning, in photography and as a food additive; many industrial and laboratory applications

Sugar

C12H22O11 Sucrose (cane sugar or beet sugar)

Cubic white crystals

None

Used as a sweetener in foods; source of metabolic energy

Table salt

NaCl Sodium chloride (rock salt, halite)

Cubic white crystals

None

Produced through the evaporation of naturally salted water and by the solar evaporation of sea water, or extracted from underground mines; used in foods and to keep roads free of ice

TSP

Na3PO4 Sodium phosphate

White crystals

Toxic if swallowed; strong irritant to tissues

Used as a water softener and in cleaning (for example, metals and walls before painting); used for many industrial purposes

Vinegar

5% acetic acid, CH3COOH, in water

Colourless solution with a distinctive odour

None

Used in cooking and in household cleaning

Vitamin C

C6H8O6 Ascorbic acid

White crystals or powder with harsh acidic taste

None

Important in nutrition to prevent scurvy; present in citrus fruits, tomatoes, potatoes and green vegetables

Wood alcohol

CH3OH Methanol (methyl alcohol)

Clear, colourless liquid with a faint scent of alcohol

Flammable; toxic if swallowed, inhaled or absorbed by the skin; may cause blindness and death

Many industrial and household applications; used in fuel line antifreeze and to dilute lacquer and paint; mixed with vegetable oil and lye to obtain diesel

APPENDIX 8 Reference Tables

Table 8.3

Thermodynamic properties of certain elements* Formula

Hfus (kJ/mol)

Hvap (kJ/mol)

c (J/(g°C))

Aluminum

Al

10.79

294

0.897

Argon

Ar

1.18

Beryllium

Be

7.90

Name

Boron Bromine

6.43 297

1.825

B

50.2 10.57

29.96

0.474

117

—

0.709

20.41

0.479

C

Chlorine

Cl2

6.40

1.026

Chromium

Cr

21.0

339.5

0.449

Cobalt

Co

16.06

377

0.421

Copper

Cu

12.93

300.4

0.385

Fluorine

F2

0.51

6.62

0.824

Gallium

Ga

5.58

254

0.371

Germanium

Ge

36.94

334

0.320

Gold

Au

12.72

324

0.129

Helium

He

Hydrogen

H2

Iodine

I2

Iron

Fe

13.81

Krypton

Kr

1.64

0.014

0.08

5.193

0.12

0.90

14.304

15.52

41.57

0.214

340 9.08

0.449 0.248

Lead

Pb

4.78

128

0.248

Magnesium

Mg

8.48

128

1.023

Manganese

Mn

12.91

221

0.479

Mercury

Hg

2.29

Neon

Ne

0.33

Nickel

Ni

17.04

Nitrogen

N2

0.71

5.57

1.040

Oxygen

O2

0.44

6.82

0.918

Phosphorous

P4

0.66

Platinum

Pt

22.17

Radon

Rn

3.25

Scandium

Sc

Selenium

Se

6.69

Silicon

Si

50.21

359

0.705

Silver

Ag

11.28

258

0.235

Sulphur

S8

1.72

45

0.710

Tin

Sn

7.17

296.1

0.228

Titanium

Ti

14.15

425

0.523

Tungsten

W

52.31

806.7

0.132

Uranium

U

9.14

417.1

0.116

14.1

Vanadium

V

Xenon

Xe

21.5 2.27

Zinc

Zn

7.07

59.1 1.71 377.5

12.4 469 18.10 332.7 95.48

459 12.57 123.6

8

0.520

Br2

Carbon (graphite)

480

APPENDIX

0.140 1.030 0.444

0.769 0.133 0.094 0.568 0.321

0.489 0.158 0.388

* Molar enthalpies at 101.325 kPa (1 atm) and specific heat capacities under standard conditions at STP.

APPENDIX 8 Reference Tables

417

APPENDIX

8

Table 8.4

Standard molar enthalpies of formation

Chemical name

Formula

H°f (kJ/mol)

Chemical name

Formula

H°f (kJ/mol)

Chemical name Phenylethene (styrene)

C6H5CHCH2 (l)

103.8

Phosphorus pentachloride

PCl5 (g)

443.5

Phosphorus trichloride (liquid)

PCl3 (l)

319.7

H°f (kJ/mol)

1.2-dichloroethane

C2H4Cl2 (l)

126.9

Hydrogen bromide

HBr (g)

36.3

Acetone

(CH3)2CO (l)

248.1

Hydrogen chloride

HCl (g)

92.3

Aluminum oxide

Al2O3 (s)

1675.7

Hydrogen fluoride

HF (g)

273.3

Hydrogen iodide NH3 (g)

HI (g)

26.5

Ammonia

Hydrogen peroxide

187.8

Ammonium chloride

NH4Cl (s)

H2O2 (l)

Hydrogen sulfide

H2S (g)

20.6

287.0

Iodine (vapour)

I2 (g)

Phosphorous trichloride (vapour)

PCl3 (g)

62.4

Iron (II, III) oxide

Fe3O4 (s)

1118.4

Potassium chlorate

KClO3 (s)

397.7

Iron (III) oxide

Fe2O3 (s)

824.2

Potassium chloride

KCl (s)

436.7

Isobutane

C4H10 (g)

134.2

Potassium

KOH (s)

424.8

Lead(II) oxide

PbO (s)

219.0

hydroxide

Lead(IV) oxide

PbO2 (s)

277.4

Propane

C3H8 (g)

104.7

641.3

Silicon dioxide

SiO2 (s)

910.7

1095.8

Silver bromide

AgBr (s)

100.4

Silver chloride

AgCl (s)

127.0

Silver iodide

AgI (s)

45.9 314.4

Ammonium nitrate

NH4NO3 (s)

Barium carbonate

BaCO3 (s)

Barium hydroxide

Ba(OH)2 (s)

944.7

Barium oxide

BaO (s)

944.7

Barium sulfate

BaSO4(s)

553.5

Benzene

C6H6 (l)

49.0

Bromine (vapour)

Br2 (g)

30.9

Butane

C4H10 (g)

125.6

Calcium carbonate

CaCO3 (s)

1206.9

Calcium hydroxide

Ca(OH)2 (s)

986.1

Calcium oxide

CaO (s)

634.9

Carbon dioxide

CO2 (g)

393.5

Carbon disulfide

CS2 (l)

89.0

Carbon monoxide

CO (g)

110.5

Chloroethene

C2H3Cl (g)

365.6 1216.3

37.3

Magnesium chloride MgCl2 (s) Magnesium carbonate

MgCO3 (s)

Magnesium

Mg(OH)2 (s)

924.5

Sodium bromide

NaBr (s)

361.1

MgO (s)

601.6

Sodium chloride

NaCl (s)

411.2

Manganese(II) oxide MnO (s)

385.2

Sodium hydroxide

NaOH (s)

425.6

Manganese(IV) oxide MnO2 (s)

520.0

hydroxide Magnesium oxide

Sodium iodide

NaI (s)

90.8

Sucrose

C12H22O11 (s)

Mercury(II) sulphide HgS (s)

58.2

Sulphur dioxide

SO2 (g)

296.8

Sulphur trioxide (liquid)

SO3 (l)

441.0

Sulphur trioxide (vapour)

SO3 (g)

395.7

Sulphuric acid

H2SO4 (l)

814.0

Tin(II) oxide

SnO (s)

280.7

Tin(IV) oxide

SnO2 (s)

577.6

Trimethyl-2,2,4 pentane

C8H18 (l)

259.2

33.2 90.2

Urea

CO(NH2)2 (s)

333.5

1139.7

Copper(I) oxide

Cu2O (s)

168.6

Methanal (formaldehyde)

CH2O (g)

Copper(II) oxide

CuO (s)

157.3

Methane

CH4 (g)

Copper(I) sulphide

Cu2S (s)

79.5

HCOOH (l)

425.1

Copper(II) sulphide

CuS (s)

53.1

Methanoic acid (formic acid)

C2H6 (g)

83.8

Methanol

CH3OH (l)

239.1

Nickel(II) oxide

NiO (s)

239.7

Nitric acid

HNO3 (l)

174.1

Nitrogen dioxide

NO2 (g)

C2H4(OH)2 (l)

454.8

Ethanoic acid (acetic acid)

CH3COOH (l)

432.8

Ethanol

C2H5OH (l)

235.2

Ethene (ethylene)

C2H4 (g)

52.5

Ethyne (acetylene)

C2H2 (g)

Glucose

C6H12O6 (s)

Hexane

C6H14 (l)

Ethane-1,2-diol

Nitrogen monoxide NO (g)

108.6

74.4

287.8 2225.5

Nitromethane

CH3NO2 (l)

113.1

Water (liquid)

H2O (l)

285.8

226.8

Octane

C8H18 (l)

250.1

Water (vapour)

H2O (g)

241.8

1273.1

Ozone

O3 (g)

142.7

Zinc oxide

ZnO (s)

350.5

Pentane

C5H12 (l)

173.5

Zinc sulphide

ZnS (s)

206.0

198.7

Standard molar enthalpies of formation are measured at STP (25°C and 100 kPa). By definition, the molar enthalpy of an element in its standard state is zero.

APPENDIX 8 Reference Tables

61.8

HgO (s)

Mercury(II) oxide

Chromium(III) oxide Cr2O3 (s)

Ethane

418

Formula

Table 8.5 Bond

Average bond energy Energy (kJ/mol)

Energy (kJ/mol)

Bond

Hydrogen

Bond

Carbon

Energy (kJ/mol)

Nitrogen

Bond

Energy (kJ/mol)

436

COC

347

NON

160

POP

210

HON

339

CON

305

NOO

201

POS

444

HOO

460

COO

358

NOF

272

POF

490

HOF

570

COF

552

NOSi

330

POCl

331

HOSi

299

COSi

305

NOP

209

POBr

272

HOP

297

COP

264

NOS

464

POI

184

HOS

344

COS

259

NOCl

200

SOS

266

HOCl

432

COCl

397

NOBr

276

SOF

343

HOBr

366

COBr

280

NOI

159

SOCl

277

HOI

298

COI

209

SOBr

218

HOMg

126

COH

413

SOI

170

Silicon

8

Phosphorous and sulphur

HOH

Oxygen

APPENDIX

Halogens

Multiple bonds

OOO

204

SiOSi

226

FOCl

256

CPC

607

OOF

222

SiOP

364

FOBr

280

CPN

615

OOSi

368

SiOS

226

FOI

272

CPO

745

OOP

351

SiOF

553

ClOBr

217

NPN

418

OOS

265

SiOCl

381

ClOI

211

NPO

631

OOCl

269

SiOBr

368

BrOI

179

OPO

498

OOBr

235

SiOI

293

FOF

159

C

C

839

OOI

249

SiPO

640

ClOCl

243

C

N

891

BrOBr

193

C

O

1077

IOI

151

N

N

945

The values in this table represent mean values of bond dissociation between the pairs of atoms listed. The real values can vary for different molecules.

APPENDIX 8 Reference Tables

419

APPENDIX

Table 8.6

8

Molar heat of dissolution

Substance AgNO3 (s)

Hd (kJ/mol) absorbed 23.0

CO2 (g)

20.0

CuSO4 (s)

68.0

CUSO45H2O (s)

12.0

HCl (g)

74.0

Hl (g)

30.0

H2SO4 (l)

74.0

HCl (s)

18.0

HClO3 (s)

42.0

KClO3 (s)

42.0

KI (s)

21.0

KNO3 (s)

36.0

KOH (s)

55.0

LiCl (s)

35.0

Li2CO3 (s)

13.0

MgSO47H2O (s) NaCl (s) NaNO3 (s)

16.0 4.3 21.0

NaOH (s) Na2SO410H2O (s)

42.0 79.0

NH3 (g)

420

APPENDIX 8 Reference Tables

released

35.0

NH4Cl (s)

16.0

NH4NO3 (s)

26.0

Table 8.7

Ionization constants of acids

Base

APPENDIX

Formula

Conjugate acid

Ka

Acetic acid

CH3COOH

CH3COO

1.8  105

Benzoic acid

C6H5COOH

C6H5COO

6.3  05

Chlorous acid

HClO2

ClO2

1.1  102

Cyanic acid

HOCN HCHO2

OCN CHO2

3.5  104 1.8  104

Hydrobromic acid

HBr

Br

1.0  109

Hydrochloric acid

HCl

Cl

1.3  106

Hydrocyanic acid

HCN

CN

6.2  1010

Hydrofluoric acid

HF

F

6.3  104

Hydrogen oxide

H2O

OH

1.0  1014

Hypobromous acid

HBrO

BrO

2.8  109

Table 8.8

8

Ionization constants of nitrogen bases

Base

Formula

Conjugate acid

Kb

1,2-diaminoethane (ethylenediamine)

NH2CH2CH2NH2

NH2CH2CH2NH3

8.4  105

Dimethylamine (N-methylmethanamine)

(CH3)2NH

(CH3)2NH2

5.4  104

Ethanamine

C2H5NH2

C2H5NH3

4.5  104

Methanamine

CH3NH2

CH3NH3

4.6  104

Trimethylamine (N,N-dimethylmethanamine)

(CH3)3N

(CH3)3NH

6.4  105

Ammonia

NH3

NH4

1.8  105

Hydrazine

N2H4

N2H5

1.3  106

Hydroxylamine

NH2OH

NH3OH

8.8  109

Pyridine

C5H5N

C5H5NH

1.7  109

Aniline

C6H5NH2

C6H5NH3

7.5  1010

Urea

NH2CONH2

NH2CONH3

1.3  1014

APPENDIX 8 Reference Tables

421

8

Table 8.10

PO33

Bromates

AgBrO3 TlBrO3

5.38  105 1.10  104

Bromides

AgBr CuBr PbBr2

5.35  1013 6.27  109 6.60  106

Carbonates

Ag2CO3 BaCO3 CaCO3 MgCO3 PbCO3

8.46  1012 2.58  109 3.36  109 6.82  106 7.40  1014

Chlorides

AgCl CuCl

1.77  1010 1.72  109

Chromates

Ag2CrO4 BaCrO4 PbCrO4

1.12  1012 1.12  1010 02.3  1013

Cyanide

AgCN CuCN

5.97  1017 3.47  1020

Fluoride

BaF2 CdF2 CaF2 FeF2

1.84  107 6.44  103 3.45  1011 2.36  106

Hydroxides

Be(OH)2 Cd(OH)2 Ca(OH)2 Co(OH)2 Eu(OH)3 Fe(OH)2 Fe(OH)3 Pb(OH)2 Mg(OH)2 Ni(OH)2 Sn(OH)2 Zn(OH)2

6.92  1022 07.2  1015 5.02  106 5.92  1015 9.38  1027 4.87  1017 2.79  1039 1.43  1020 5.61  1012 5.48  1016 5.45  1027 3  1017

Iodates

Ba(IO3)2 Ca(IO3)2 Sr(IO3)2 Y(IO3)3

4.01  109 6.47  106 1.14  107 1.12  1010

Iodides

CuI PbI2 AgI

1.27  1012 09.8  109 8.52  1017

Phosphates

AlPO4 Ca3(PO4)2 Co3(PO4)2 Cu3(PO4)2 Ni3(PO4)2

9.84  1021 2.07  1033 2.05  1035 1.40  1037 4.74  1032

Sulphates

BaSO4 CaSO4 Hg2SO4

1.08  1010 4.93  105 06.5  107

Thiocyanate

CuSCN Pd(SCN)2

1.08  1013 4.39  1023

Common polyatomic ions

Formula 3

Solubility-product constants in water at 25°C for various compounds

Table 8.9

APPENDIX

Name

Formula

Name

Formula

Name

Chlorite

SiO32

Silicate

Hypochlorite

NH4

Ammonium

Sulphate

CrO42

Chromate

HPO42

Hydrogenophosphate or biphosphate

Sulfite

Cr2O72

Bichromate

H2PO4

Dihydrogenophosphate

Carbonate

C2H3O2

Acetate (also called ethanoate)

HPO32

Hydrogenophosphite

Nitrate

CN

Cyanide

H2PO3

Dihydrogenophosphite

NO2

Nitrite

OH

Hydroxide

HSO4

Hydrogenosulfate or bisulfate

ClO4

Perchlorate

MnO4

Permanganate

HSO3

Hydrogenosulphite or bisulphite

Chlorate

C2O42

Oxalate

HCO3

Hydrogenocarbonate or bicarbonate

PO4

SO4

2

SO32 CO3

2

NO3

ClO3

422



Phosphate

ClO2

Phosphite

ClO

APPENDIX 8 Reference Tables

Table 8.11

Solubility (in water) of common ionic compounds

APPENDIX

8

Positive ions (cations) Li

Transition elements and

Na K Negative ions (anions)

Rb

Ga3

Ge

Cs

Bi3

As3

Fr

As5

In3

Ag

H

Sn2

Sn4

Cu

2 NH4 Be 2 Ca2 Mg

CH3COO –

NO3–

Sr2

Ba2 Ra2

AI3

TI

Pb2

Hg22

CIO3–

SO4 2– SO32–

PO4 3–

CO32–

S2– OH– CI–

Br–

I– Ag

CrO4 2– Low solubility at 25°C: formation of a precipitate

Table 8.12

Specific heat capacity of various substances Substance

Heat capacity (J/(g°C)) at 25 °C Element

Aluminum

0.900

Carbon (graphite)

0.711

Copper

0.385

Gold

0.129

Hydrogen

14.267

Iron

0.444 Compound

Ammonium hydroxide (liquid)

4.70

Ethanol

2.46

Water (solid)

2.05

Water (liquid)

4.184

Water (gas)

1.41 Other materials

Air

1.02

Cement

0.88

Glass

0.84

Granite

0.79

Wood

1.76 APPENDIX 8 Reference Tables

423

GLOSSARY A Acidity constant (Ka) An equilibrium constant that describes the relative strength of partially ionized acids in aqueous solution. The stronger the acid, the higher its ionization and the greater the value of Ka. Activated complex An unstable cluster of colliding reactant atoms that is formed during the partial conversion of reactants into products. Activation energy The minimum quantity of energy needed for an endothermic or exothermic chemical reaction to occur. Arrhenius acid A substance that dissociates in water to produce hydrogen ions (H+). Arrhenius base A substance that dissociates in water to produce hydroxide ions (OH–). Average reaction rate The change in the quantity of a reactant or product as a function of a given interval of time. It is determined by calculating the slope of the secant that intersects the two points that delineate the time interval. Avogadro’s hypothesis A hypothesis that states that, under the same conditions of temperature and pressure, equal volumes of different gases contain the same number of particles. Avogadro’s law A law that states that under the same conditions of temperature and pressure, the volume of a gas is directly proportional to its quantity expressed in number of moles. The mathematical expression of this law is written as follows:

V1 V  2. n1 n2

B Basicity constant (Kb) An equilibrium constant that describes the relative strength of partially ionized bases in aqueous solution. The stronger the base, the higher its ionization and the greater the value of Kb. Behaviour of gases The way in which gases react when some of their physical properties change. This behaviour can be described qualitatively, based on observations, or quantitatively, based on laws. Boyle’s law A law that states that at a constant temperature, the volume occupied by a given quantity of gas is inversely proportional to the pressure of this gas. The mathematical expression of this law is written as follows: P1V1  P2V2.

424

GLOSSARY

Brønsted-Lowry acid A substance that can donate a proton (hydrogen ion, H). An acid is considered a proton donor. Brønsted-Lowry base A substance that can accept a proton (hydrogen ion, H). A base is considered a proton acceptor.

C Calorimetry An experimental method used to determine quantities of heat involved in certain reactions. Catalyst A substance that increases the reaction rate without changing the result of the conversion and without being consumed by the reaction. A catalyst reduces the activation energy by allowing a greater number of particles to possess sufficient kinetic energy to react. Charles’ law A law that states that at a constant pressure, the volume occupied by a given quantity of gas is directly proportional to the absolute temperature of this gas. The mathematical expression of this law is written as follows:

V1 V2  . T1 T2 Chemical equilibrium A dynamic equilibrium that results from two opposing chemical reactions that occur at the same rate, leaving the composition of the reaction system unchanged. Chemical reactivity The tendency of a gas to undergo a chemical change due to various factors such as heat, light or contact with other substances. Closed system A system that allows the exchange of energy, but not matter, with the external environment. Collision theory A theory that explains the interactions between reactant particles and the energy present at each stage in the evolution of a reaction illustrated by the reaction mechanism. Concentration The ratio of the amount of dissolved solute to the total quantity of solution. Conditions for attaining equilibrium The three conditions that must be met for a chemical reaction to attain equilibrium, namely, a reversible reaction occurring in a closed system whose macroscopic properties are constant.

D Dalton’s law A law that states that at a given temperature, the total pressure of a mixture of gases is equal to the sum of the partial pressures of all of the gases in the mixture. Also known as the law of partial pressures. Direct reaction A reaction that occurs when the reactants become products. Dynamic equilibrium The result of two opposing processes occurring at the same rate so that no visible change takes place in the reaction system. There are three types of dynamic equilibrium: phase equilibrium or physical equilibrium, solubility equilibrium, and chemical equilibrium.

F Factors that affect reaction rate The five factors affect reaction rate, namely, the nature of the reactants, the surface area of the reactants, the concentration of the reactants, the temperature of the reaction environment, and the catalysts. Factors that affect the state of equilibrium The three factors that affect the state of chemical equilibrium of a system, namely, the concentration of the reactants or products, the temperature, and the pressure.

G E Elastic collision A collision in which reactant particles collide with each other without leading to a chemical reaction. In an elastic collision, the sum of the kinetic energies of the particles remains unchanged. Endothermic reaction A reaction that absorbs heat from the environment. Energy balance The sum of the energy required to break the chemical bonds of the reactants and the energy released during the formation of the product bonds. Used to determine the total enthalpy change of a chemical reaction. Energy diagram A graph that illustrates the energy change of the substances involved in a reaction. Equilibrium constant (Kc) The relationship that shows that at a given temperature, the ratio of product concentration to reactant concentration is constant in all elementary chemical reactions at equilibrium. Also called the equilibrium law.

Gas A substance whose particles are distant from each other, is compressible, occupies the entire volume of a container, is undefined in form, and diffuses quickly. Gay-Lussac’s law A law that states that at a constant volume, the pressure of a given quantity of gas is directly proportional to the absolute temperature of this gas. The mathematical expression of this law is written as follows:

P1 P2  . T1 T2 General gas law A law that establishes a relationship between the four variables that characterize gases, namely, pressure (P ), volume (V ), absolute temperature (T ) and quantity of gas (n) expressed in moles. It can be used to predict the final conditions of a gas once its initial conditions have been modified.

H

Enthalpy (H) The total energy of a system, that is, the sum of all the potential and kinetic energies of a system at constant pressure.

Heat The transfer of thermal energy that occurs when two systems with different temperatures come into contact with each other.

Enthalpy change (ΔH) The energy exchanged between a system and its environment during a physical change or a chemical reaction at constant pressure. Also called the heat of reaction.

Hess’s law A law that states that if a reaction can be broken down into several elementary reactions, its enthalpy change is equal to the algebraic sum of the enthalpy changes of each elementary reaction. Also called the law of constant heat summation.

Exothermic reaction A reaction that releases heat into the environment. Potential energy A form of energy stored in a substance that depends on the relative position of its particles.

Hydronium ion A hydrated ion, that is, a proton attached to a water molecule. Its chemical formula is H3O.

Thermochemical equation A chemical equation that includes information on the quantity of energy involved in the reaction.

GLOSSARY

425

I

M

Ideal gas law A law that establishes a relationship between the four variables that characterize a gas sample at a given time, namely, pressure (P ), volume (V ), absolute temperature (T ) and quantity of gas (n) expressed in moles, as well as the gas constant. Inelastic collision A collision in which reactant particles collide with each other and set off a chemical reaction that converts them into product particles. Instantaneous reaction rate The reaction rate at a specific time during the reaction. It is determined by calculating the slope of the tangent that intersects a point corresponding to the time in question. Ionic equilibrium in solutions A state of equilibrium established between the concentrations of the various ions following the dissociation of a chemical compound in solution. Ionization constant of water (Kw) An equilibrium constant that describes the interdependence between the molar concentrations of hydronium ions (H3O) and hydroxide ions (OH). Irreversible reaction A reaction that can only occur in one direction, from reactants to products. Isolated system A system that allows no exchange of matter or energy with the external environment.

Molar heat of neutralization (ΔHn) The quantity of energy absorbed or released during the neutralization of one mole of an acid or a base. Molar heat of reaction (ΔH) The quantity of energy absorbed or released during the reaction of one mole of substance. Molar volume of a gas The volume occupied by one mole of any gas, under specific conditions of temperature and pressure, that is, at STP, standard temperature and pressure (0°C and 101.3 kPa)) and SATP, standard ambient temperature and pressure (25°C and 101.3 kPa). The molar volume at STP  22.4 L; the molar volume at SATP  24.5 L.

N Nature of reactants The phase in which reactants are found. The nature of reactants as well as the number and strength of the bonds they contain affect reaction rate.

0 Open system A system that allows the exchange of matter and energy with the external environment.

K Kinetic energy A form of energy related to the movement of particles. Kinetic theory of gases A form of energy related to the movement of particles.

L Law of conservation of energy A law that states that energy can be transferred or converted but cannot be created or destroyed. Le Chatelier’s principle A principle that makes it possible to qualitatively predict the direction (direct or reverse) that will be favoured when the conditions of a system at equilibrium are changed.

426

Molar heat of dissolution (ΔHd) The quantity of energy absorbed or released during the dissolution of one mole of solute in a solvent.

GLOSSARY

P Phase equilibrium A state of dynamic equilibrium in which a single substance is found in several phases within a system following a physical change. Pressure The measurement of a force exerted on a surface. Product A substance that forms during a chemical reaction. Its chemical formula is located to the right of the arrow in the chemical equation of the reaction. Product solubility constant (Kps) An equilibrium constant that describes the solubility of a partially ionized solute in aqueous solution. In compounds with the same proportions of ions, the more soluble the solid, the higher its ionization and the greater the value of Kps.

R Rate law A law that states the mathematical relationship between the reaction rate and reactant concentration. In an elementary reaction, this law depends on the stoichiometric coefficients of the reactants present in the balanced equation. Reactant A substance that decomposes or combines with one or more other substances during a chemical reaction. Its chemical formula is located on the left of the arrow in the chemical equation of the reaction. Reactant surface area A factor that influences the reaction rate: in general, an increase in the surface area of the reactant increases the reaction rate. Reaction mechanism A series of elementary reactions that converts reactants into products over the course of a complex reaction. Reaction rate A positive value that expresses the change in the quantity of a reactant or a product as a function of time during a chemical reaction. The reaction rate is proportional to the coefficient, of the reactant or the product, of the balanced chemical equation chosen to express it.

S Solubility equilibrium A state in which a solute is dissolved in a solvent or a solution, and a surplus of this solute is in contact with the saturated solution. Static equilibrium A state in which a system remains at a given point or is kept immobile. Stoichiometry of gases A calculation method based on the ratios of the quantities of gas involved in a chemical reaction. This method is used to predict the quantity of a reactant or product involved in a chemical reaction in which at least one of the components is a gas. System A particular place where a group of substances are undergoing a conversion.

T Temperature The measurement of the agitation of the atoms and particles in a system.

Reverse reaction A reaction that occurs when the products convert back into reactants. Reversible reaction A reaction that can occur in both directions, from reactants to products and from products to reactants.

GLOSSARY

427

INDEX A Absolute zero, 81, 128-129 Acid(s), 9, 19-20, 25 amino, 241, 325, 326 Arrhenius, 192, 320-323, 328 Brønsted-Lowry, 323-325 conjugate, 324-325, 334 determine the concentration of an, 385-386 fatty, 268 ionization constants of, 421 rain, 199 strength, 329-331 strong, 192, 329-330, 331, 334 weak, 330-331, 334 Acid-base indicator(s), 44, 192, 323, 385-386 neutralization, 20, 25, 192-193, 321-322, 324-325, 385-386 Acid-base indicator, 385-386 Acid-base neutralization, 20, 25, 192-193, 321-322, 324-325, 385-386 Acidity constant, 329, 331-333, 398, 421 measuring the acidity of a substance, 20, 325-326 Activated complex, 172-173, 174-175, 176-177, 200, 240-241, 247, 248, 253, 265 Aerosol, 43 Alkaline, 9, 177, 353-354, 357 Alkaline earth, 9, 353-354, 357 Alphabetical list of elements and their atomic mass, 410-411 Amphoteric, 323, 324-325 Anesthesia (a brief history of…), 49 Anion, 4, 11, 28, 350, 353, 358, 423 Anode, 341, 358-359, 361, 366 sacrificial, 359 Arrhenius Svante, 320, 321, 323, 325 theory of, 192, 320-323, 327-328 Atmosphere, 27, 38-41, 42, 47, 48, 70, 71, 111, 150, 152, 199, 299 Atom(s), 4, 21 collisions of, 236-240, 246, 248-249, 251 enumeration, 13-14 interactions between atoms, 28-29, 135, 152, 156-157, 172-175, 186, 189, 321-322 motion, 54-58, 129, 148 representations of, 5-7 Atomic nuclei, 5-6, 28, 31, 148 Atomic number, 5-7, 8, 10 Avogadro Amedeo, 88, 89, 108 Avogadro’s hypothesis, 89-91, 92, 100-101 Avogadro’s law, 88-90, 92, 100-101, 400 Avogadro’s number, 13, 92, 406 constant, 13, 92, 406

B Balancing chemical equations, 22, 23-24 Barometer, 70-71, 382 Base(s), 19-20, 25 Arrhenius, 192, 320-323, 328 Brønsted-Lowry, 323-325 conjugate, 324-325, 331 determine the concentration of a, 385-386 strength, 329, 334 strong base (s), 192, 334 weak, 334

428

INDEX

Basicity basicity constant, 320, 329, 334-335, 398, 421 measuring the basicity of a substance, 20, 325-326 Behaviour of gases, 63 Berzelius, Jöns Jacob, 269 Biogas, 39 Bomb calorimeter, 132-133 Bond chemical, 4, 5, 28-29, 129, 135, 152-154, 249, 281 covalent, 29, 46, 266, 353-354 double, 156, 159, 160, 249, 268 energy of, 31, 148, 154, 156-157, 172, 248, 419 enthalpy of, 156-157, 172, 175, 237, 248 force of a, 58, 156, 246-247, 248-249 hydrogen, 186-188, 189 intermolecular, 56, 135, 150 ionic, 12, 28-29, 187, 354 simple, 158, 160, 268 triple, 46, 156, 160 Boyle, Robert, 76 Boyle’s law, 76-79, 100-101, 400 Bread-making, 263 Brønsted, Johannes Nicolaus, 323 Brønsted-Lowry theory of acids and bases, 323-325 Buoyancy of ice, 57 of fish, 41

C Calculating energy transfer, 138-140 concentrations at equilibrium, 314-317 enthalpy change using stoichiometry, 161-162 oxidation number, 354-355 Calorimeter, 132-133, 149, 190, 192, 201 bomb calorimeter, 132-133 home calorimeter, 133 Calorimetry, 132-133, 201, 203 Caramelization, 241 Catalysis (a brief history of…), 269 Catalyst(s), 262-266, 268, 269 adding a catalyst, 289, 297 biological, 265-266 function(s) of a catalyst, 262-263 heterogeneous, 262, 266 homogeneous, 262, 264-266 Cathode, 341, 358-359, 361, 363, 366 Cation, 4, 11, 28, 350, 351, 353, 358-359, 423 Cell domestic battery/ies, 366 fuel cell, 366 electrochemical cell, 358-359, 361, 363-364 hydrogen fuel cell, 361 simplified representation of a fuel cell, 363-364 voltaic cell, 359 Change energy, 26, 186, 239 enthalpy, 147-165, 172-181, 186, 188, 200-205, 263, 264, 403 in concentration, 217, 220, 222, 254, 289-292, 294-297, 313 in pressure, 294-297 in temperature, 32, 134-136, 138-141, 150-151, 292-294 Clapeyron, Benoît Paul Émile, 151 Charles Charles’ law, 82-84, 85, 100-101, 105, 115, 400 Jacques, 80-81, 85

Chemical formula(s), 4, 11-12, 17, 20, 22-23, 198, 216, 323, 329, 334 developed, 160 everyday chemical products, 412-416 or semi-developed, 160 Claude, Georges, 41 Climate, 41, 135 global warming, 39, 299 Coefficients fractional, 204 stoichiometric, 23, 108, 198, 216-217, 253, 256, 257, 296, 308-310, 315 Collision(s), 57, 59, 64-65, 69-70, 79, 84, 87, 91, 96, 111-112, 172-173, 176, 181 collision geometry, 237 collision theory, 236-241, 246, 247-248, 251-252, 253, 260-261, 262, 290 elastic, 236-238, 239, 240 inelastic, 236-238, 240, 246, 253, 260-261, 262, 290 perfectly elastic, 61 types of collisions, 236-238 Combustible(s), 39, 48, 149, 252, 373 clean fuel, 47 fossil fuels, 199, 299 fuel cell, 366 gas, 42, 44, 46-47 Combustion, 27, 39, 42, 46-48, 108, 114, 131, 149, 153, 174, 186, 199, 201, 204-205, 251-252, 280-281, 299 Compounds carbon, 299 chlorinated, 48, 314 covalent, 29, 353-355 halogen, 49 ionic, 28, 187, 320, 336-337, 353, 354 molecular, 187 nitrogen, 38, 280, 289 organic, 46 solubility-product constants in water at 25°C for various, 422 Compressed air, 33, 42, 43, 45, 63, 91, 94, 373 Compressibility, 33, 43, 58, 63, 76, 78, 114 Concentration, 16-18 change in concentration, 217, 220, 222, 254, 289-292, 294-297 313 concentration of solutions, 18, 397 determine the, 385 equilibrium, 286, 288, 289-292, 308-311, 313-317, 330-331, 334, 337 exponents, 257, 308-309 factors that affect the state of equilibrium, 289-297 molar, 17 predict the, 18, 314-317 rate law of the reaction, 254-257 reactant, 238, 253, 254-258, 309-312 relationship between the pH and the molar concentration of the hydronium and hydroxide ions, 325-326, 328-329 Conductivity electrical, 8-9, 82, 300, 327 thermal, 8-9, 135, 140 Configuration electrons, 4-6, 45, 357 trans, 268 Conjugate acid-base pair, 324-325 Conservation food preservation, 229, 261 of energy, 31-32, 128, 131-133, 157, 239 of mass, 21-22 Constant(s) acidity, 320, 329, 331-333 basicity, 320, 329, 334-335 direct rate constant, 255-256, 309, 310-311

equilibrium, 308-317, 320, 328, 331, 337 gas laws, 100-102 ionization constant of water, 320, 325, 327-329 ionization constants of acids, 421 ionization constants of nitrogen bases, 421 reverse rate constant, 309, 311 solubility product constant, 320, 336-338, 422 Contact surface, 238, 251-252 Cooking or reacting, 241 under pressure, 87 Corrosion, 27, 224, 352, 359

D Dalton Dalton’s law, 111-113, 400 John, 111-112 Decomposition, 25-26, 152, 216-217, 269, 280, 281, 291, 299 Diagram, bar graph, 390 broken-line graph, 389 circle graph, 391 energy balance diagram, 157-159, 174-175 energy diagram, 172-173, 174-181, 200, 239, 263-264 enthalpy, 153-154 Lewis notation, 5, 7, 29, 156, 353-355 observing the progress of a reaction using an energy diagram, 176181 Diffusion, 64-67 Dilution, 16, 18 steps to, 384 Dipole(s), 322 Dirigible balloons, 115 Disinfectants for pool water, 341 Dissolution, 15, 19, 153, 188-190, 201, 252, 279, 283, 284, 320, 330, 336 molar heat of dissolution, 186-187, 189, 190 ionic (electrolysis), 19-20, 188, 248, 320, 321-322, 325, 329, 330331, 334 procedure when preparing for, 383-384 molar, 189

E Effusion, 64-65, 66-67 Electrical current, 19, 31, 47-48, 128, 341, 358-359, 366 Electricity, 128, 359-366 conductors of, 8-9, 82, 300, 327 Electrode, 19, 358-361, 363-364, 366 hydrogen electrode, 360-361 reference electrode, 360-361, 363 Electrolysis, 47, 341 Electrolyte(s), 19-20, 321, 329, 337, 366 Electrolytic dissociation, 19-20, 188, 248, 320, 321-322, 325, 329, 330-331, 334 Electron(s), 4-6, 9, 12, 28, 40, 148, 300, 322, 350-352, 353-355, 357, 358-359, 363 electron pairs, 7, 353-354 electron transfer, 28, 342 350-351, 353 free electrons, 350 odd electrons, 7 sharing of electrons, 29, 156, 354 valence electrons, 5, 7, 9, 28, 45, 354, 357 Electronegativity, 8, 28-29, 353-354 Electron shell, 5-6, 10 outermost, 5, 45, 353 Electrophoresis, 325

INDEX

429

Electrophoresis, 325 Elements alphabetical list of, 410-411 and their atomic mass, 410-411 thermodynamic properties of certain, 417 Endothermic reaction, 26, 152-153, 154, 156-157, 172, 174, 176-177, 178-181, 188-189, 292-294, 314 transformation, 150-151 Energy balance, 46, 156-160, 174-175, 188, 398 graphical representation of, 157 performing an, 156-159 Energy diagram, 172-173, 174-181, 200, 239, 263-264 activation energy, 172, 173-175, 176-177, 178- 181, 200, 237-238, 240-241, 248, 260-261, 262-264, 297, 313, 321 average bond, 156, 419 bond, 31, 46, 148, 154, 156-157, 248 change of energy, 21, 26, 186, 239 chemical energy, 27, 31, 114, 131-132, 152, 189, 366 collisions, 236-238, 240-241, 247, 249, 260 conservation of energy, 31-32, 131-133, 128, 157, 239 elastic, 31 elastic potential energy, 31 electric, 31, 131, 366 electric potential energy, 31 energy transfer, 127-143, 284 exchange of energy, 46, 128, 148, 151, 161, 186 forms of energy, 30-32 gravitational potential energy, 30, 31, 399 kinetic, 30-32, 54, 58-60, 61, 66, 69, 81, 128, 148, 172, 176, 236, 238-240, 260-261, 297, 399, 409 luminous, 131, 152 mechanical, 32, 114, 131, 399, 409 nuclear, 31 pneumatic, 43 potential, 30-32, 131, 148, 154, 172-174, 176-177, 200, 203, 239, 360, 409 radiant energy, 31, 39, 206 renewable sources of energy, 47, 299 the law of conservation of energy, 31-32, 128, 131-133, 157, 239 thermal energy, 31-32, 128-130, 131, 132-133, 134-136, 138, 189, 286, 293, 399 transferred, 127-143 wind energy, 31, 47 Energy transfer, 127-209 Engine diesel, 114, 366 gasoline, 42, 114, 366 Enthalpy, 148, 248, 409 diagram, 153-154 enthalpy change, 147-165, 172-181, 186, 188, 200-205, 263, 264, 403 enthalpy in chemical reactions, 248 formation reaction, 149, 156, 203-205, 418 standard molar enthalpy change, 149, 156, 203-205, 418 summation of enthalpies, 201-206 Entropy, 82, 129, 148 Enumeration of matter, 13-14 Enzymes, 229, 262, 263, 265-266, 286, 326 Equation(s) balancing of an equation, 22, chemical equation, 21, 22, 27, 153, 204, 216, 310, 386 general equation for redox reactions, 351-352 mathematics, 398-403 review of some mathematical equations, 398-403 skeleton equation, 22 thermochemical equations, 152-153, 161, 202-204 Equilibrium calculation of concentrations at equilibrium, 314-317 chemical, 277-342

430

INDEX

conditions to attain equilibrium, 256, 283-286 equilibrium constant, 308-317, 320, 328, 331, 337 dynamic equilibrium, 278-279, 280-282, 299, 308 ionic equilibrium in solutions, 320-338 phase equilibrium, 278-279, 280, 282, 283, 285, 286 physical equilibrium, 278-279, 280, 282, 283, 285, 286 solubility equilibrium, 279, 280, 282-284, 286, 320, 336-338 state of equilibrium, 278-282, 283-286, 288-297A, 300, 313, 320, 399 static equilibrium, 278 the qualitative aspect of equilibrium , 277-301 the quantitative aspect of equilibrium, 307-342 Eutrophication, 299 Exothermic change, 150-151 exothermic reaction, 26, 150, 152-154, 156-157, 172-173, 176-177, 178-180, 188, 292-294, 300, 313-314, 364 Expansion, 63-64, 128, 132 Explosion hexagon, 252

F Factors that affect reaction rates, 246-269 that affect the state of equlibrium, 289-297 Fermentation, 39, 149 Femtochemistry, 181 Fire walking on fire, 140 fire triangle, 46 Fire triangle, 46 Fluids compressible fluids, 33 incompressible fluids, 33 Force gravitational force, 57, 70-71 strength of acids, 329-331 strength of bases, 334 van der Waals forces, 189

G Galvanization, 224 Gas anesthetizing gases, 49 behaviour of gases, 54, 63-67 chemical properties of gases, 38-49 chemical reactivity of gases, 44-48 combustible gases, 42, 46-47 compressed gases, 42, 43, 45, 63, 91, 94, 373 daily use of gases, 38-43 gas constant, 100-102, 105 general gas law, 75, 105-106 greenhouse gas(es), 39, 46-47, 114, 199, 299, 366 ideal gas, 54, 61, 63, 75, 81, 93, 100-101, 103-104 inert gases, 9, 12, 39, 42, 41, 44-45, 54, 300, 357 kinetic theory of gases, 54-61, 63, 66, 69, 79, 84, 87, 91, 92, 96, 100, 111 natural phenomena, 38-41 nitrogen gases, 199 noble gas, 9, 12, 39, 41, 42, 44-45, 54, 300, 357 oxidizing gases, 48 physical properties of gases, 53-115 pressure of a gas, 69-73, 77, 82, 85, 94-95, 100 rare gases, 9, 12, 39, 41, 42, 44-45, 54, 300, 357 real gases, 61, 63, 75, 81, 93, 100 simple gas laws, 75-96, 100-102 technological applications of gases, 38, 41-43

Gas permeation, 67 Gay-Lussac Gay-Lussac’s law, 85-87, 100-101, 114, 400 Louis Joseph, 80, 85, 88, 108 Gibbs, Josiah Willard, 148 Graphical representation of an energy diagram, 174-175 of a reaction mechanism, 200 Ground-source heat pump, 142 Groups, 8, 9, 12 Guldberg, Cato Maximilian, 256

polyatomic, 4, 11, 353, 422 positive, 4, 11, 28, 350, 351, 353, 358-359, 423 spectator, 192, 322, 358 Ionization constant of water, 320, 325, 327-329 percentage, 330-333 Ionization percentage, 330-331, 401 Isotope, 6-7

J Joule, James Prescott, 128, 132

H Haber Fritz, 289, 301 process, 266, 289, 301 Halogen, 9, 45, 48, 419 bulbs, 48, 300 Heat, 128-130 combustion, 46, 108, 114 heat exchanges, 128-130, 131, 132, 134-135, 138-140, 149, 150151, 153, 172-173, 176-177, 187-188-189, 192, 203, 293, 313314 latent, 150, 151 measurement, 132-133 molar heat of combustion, 149, 186, 205-206 molar heat of dissolution, 186-191, 201 molar heat of formation, 153 molar heat of fusion, 186 molar heat of neutralization, 192-193, 201 molar heat of reaction, 149, 186-193, 205 reaction, 147-165, 172, 175, 176, 178, 180-181, 186, 188, 201-203, 205 specific, 32, 134-136, 138, 139, 140, 189, 192 Hess Germain Henri, 201 Hess’s law, 202-206, 401 Hot air balloon, 80, 115 How to round off a number, 397 How to prepare a solution, 383-386 How to collect samples, 387 Hydrocarbons, 46, 174, 269 Hydrogenation, 268, 269 Hydronium, 220-221, 322, 324-326, 327-328, 329-331, 360 Hypothesis Avogadro’s hypothesis, 89-91, 92, 100-101 hypothesis of the kinetic theory of gases, 54, 61, 63, 66, 69, 79, 84, 87, 91, 92, 96, 100, 111

I Inhibitor, 263 Interaction solute-solute, 186, 187-188 solvent-solute, 186, 188 solvent-solvent, 186-187, 188 International System of Units (SI) common prefixes for, 408 Interpreting measurement results significant figures, 395 uncertainty, 394 Ion(s), 4, 19 chemical formulas and, 4 hydrogen, 320-324 hydronium, 220-221, 322, 324-326, 327-328, 329-331, 360, 386 hydroxide, 192, 220, 239-240, 321-322, 324-326, 327-328, 334 negative, 4, 11, 28, 350, 353, 358, 423

K Kelvin, Lord, 81-82, 128

L Laboratory instruments and techniques, 383-387 Laboratory safety, 372-376 Laboratory techniques, 383-387 Lavoisier, Antoine Laurent de, 75, 131 Law(s) Avogadro’s law, 89-91, 92, 100-101, 400 Boyle-Mariotte’s law, 76-79, 100-101, 400 Charles’s law, 82-84, 85, 100-101, 105, 115, 400 combined gas law, 85, 88, 108 Dalton’s law, 111-113, 400 equilibrium law, 308-317, 320, 328, 331, 337 Gay-Lussac’s law, 85-87, 100-101, 114, 400 general law, 75, 105-106 Graham’s law, 66-67, 401 Hess’s law, 201-206, 401 ideal gas law, 75, 100-104, 105, 401 Joule’s law, 128 law of conservation of energy, 31-32, 128, 131-133, 157, 239 law of conservation of mass, 21-22 law of constant heat summation, 201, 202-206 law of mass action, 256 law of the dilatation of gases, 80 law of partial pressures, 111-113 rate law, 253-258, 309, 401 simple gas laws, 75-96, 100-102 Le Chatelier Henry, 288 Le Chatelier’s principle, 288, 289-297, 299, 313-314 Life jacket, 91 Lowry, Thomas Martin, 323

M Manometer, 71-73, 112, 285 dial, 71 U-tube, 71-73 Mariotte, Edme, 76 Mass atomic, 8, 13, 409, 410-411 of electrodes, 359 molar, 8, 13-14, 66, 70, 92, 103-104, 220, 401, 409 number, 6-7 density, 8, 47, 57, 72, 80, 91, 93, 115, 192-193, 409 Mathematics in science, 398-407 scientific notation, 406-407 review of some mathematical equations, 398-403 transformation of algebraic expressions, 404-405

INDEX

431

Matter enumeration of, 13-14 organization of, 4 phases of, 15, 54-58, 148, 150-151, 246, 247-248, 264, 266, 278-279 Maxwell Boltzmann Maxwell’s distribution curve, 59-60, 238, 260 James Clerk, 59-60 Menten, Maud Leonora, 266 Metal, 8-9, 20, 28, 39, 135, 177, 224, 266, 269, 353-354, 359, 360, 361, 363 reducing power of, 357 Metalloids, 8-9 Method(s) analysis, 377, 382 observation method, 377 modelling, 377 empirical method, 377 experimental method, 377, 381 summary table of, 377 Microwave, 84 Model Dalton’s atomic model, 111 induced fit model, 266 lock-and-key model, 265 particle model of matter, 54-58, 247 Rutherford-Bohr’s atomic model, 5 simplified atomic model and the neutron, 5-7 Mole, 13-14, 17, 23, 75, 88-91, 92-93, 94-96, 103, 112, 149, 161, 186, 216, 295, 336 Molecule(s), 4, 55-58, 88-89, 135, 152, 157, 172-173, 175, 181, 236238, 248-249, 265-266, 282, 286, 295-296, 329-330, 334, 353-355 agitation of, 128-130, 132, 148, 150, 176 nearby, 186-187, 189, 248, 322 ways to represent, 160 Multimeter, 327

N Nature of the reactants, 238, 246-249, 360 Neutron, 5-7 Neon (Ne) signs, 41, 42 Nobel Alfred, 301 Nobel Prize, 175, 289, 301, 321 Non-metal, 8-9, 20, 28-29, 353, 354 Notation Lewis, 5, 7, 29, 156, 353-355 scientific, 325, 396, 406-407 Number Avogadro’s number, 13, 92, 406 oxidation, 353-355 mass, 6-7, 411

O Organic chemistry 249 Oxidation, 27, 39, 46, 201, 221, 224, 313-314, 350-352, 358-359, 360-361, 363-364 of an apple, 352 number, 353-355 Oxidation state, 353-355 Oxidizer(s), 27, 39, 46, 252, 373, 374 gas 42, 48 Oxidizing, 44, 351-352, 361-362

432

INDEX

P Particle behaviour in gases, 57-58 in liquids, 56-57, 58 in solids, 56, 58 Pascal, Blaise, 71 Pasteurization, 229 Period, 5, 8, 10 Periodic classification, 8-10 periodic table, C1-C2 pH calculating, 325-326 expression of, 326, 400 litmus paper, 20, 44, pH paper, 386 scale, 20, 325-326, 328-329, 385-386 Phase(s), 54-58 changes, 15, 54, 150-151 macroscopic properties in various, 58 reactants and reaction rates, 246, 247-248 Photosynthesis, 27, 38, 39, 132, 152, 299 pH scale, 20, 325-326, 328-329, 385-386 Plasma, 54 pOH, 325-326, 328, 400 Point boiling, 8, 81, 129, 151, 241 flash point, 174 fusion, 8, 151, 300, 364 ignition, 27, 46, 114 Polanyi, John Charles, 175 Potential difference in, 359, 360, 361, 363-364 hydrogen (pH), 20, 25, 325-326, 327, 328-329, 331 hydroxide (pOH), 325-326, 328 standard electrode, 360-362, 363 Precipitation, 25-26, 252 Precipitate, 21, 26, 247-248, 283 Presenting scientific results line of best fit, 389 bar graph, 390 broken-line graph, 389 circle graph, 391 histograms, 391 slope of a tangent to the curve, 390 laboratory report, 392-393 table, 388 Pressure, 33 atmospheric, 33, 63, 70-71, 72, 73, 76, 111, 129, 149 of gases, 69-73 inside tires, 71, 106 quantity of gas, 94-96 and temperature, 85-87 and volume, 76-79 measuring the, 71-73 partial, 111-113 changes in, 294-297 Pressure cooker, 87 Pressure gauge, 71, 106 Product(s) everyday chemical products, 412-416 Molar proportion, 112

Property/ies characteristic chemical, 44 chemical properties of gases, 38-49 intramolecular properties of reactants, 247 macroscopic, 58, 285-286 microscopic, 286 physical property characteristic of the substance, 32 physical properties of gases, 53-115 thermodynamic properties of certain elements, 417 Proton(s), 5-6, 28, 322, 350 acceptor, 323-325 donor, 323-325 hydrated, 322, 324-326, 327-328, 329-331, 360, 386 transfer of, 324-325

Q Quantity of heat, 32, 130, 134, 138-139, 150-151, 153 of gas, 41, 75-76, 82, 85, 88-91, 93, 94-96, 100, 105

R Ratio mole, 23, 108 Reactant(s), 46 concentration of, 253, 254-258 nature of, 246-249, 264 number and strength of the bonds of the reactants to be broken, 248-249 orientation of, 237 phase of, 247-248 surface area of, 251-252 Reaction complete, 313 complex, 198-206, 240-241, 256-257 direct, 176, 178-181, 263, 281-282, 288, 290-291, 293, 296, 297, 300, 308- 311, 312-314, 336 half-, 350-352, 358, 361, 363 endothermic, 26, 152-153, 154, 156-157, 172, 174, 176-177, 178181, 188-189, 292-294, 314 exothermic, 26, 150, 152-154, 156-157, 172-173, 176-177, 178-180, 188, 292-294, 300, 313-314, 364 heterogeneous, 246 homogeneous, 246, 252 irreversible, 176, 178-181, 280-281, 283 Maillard, 241 of alkaline metals in water, 177 in organic chemistry, 249 progress of a, 176-181, 199-200, 214 rate, 176, 214-227, 236-238, 278-279, 282-284, 288, 290-291, 297, 301, 308, 309, 313, 403 rate law, 253-258, 309, 401 reverse, 47, 151, 176, 178-181, 280-282, 288, 290, 293, 296, 300, 308-309, 311-314, 336, 343 reversible, 176, 178-181, 280-282, 283-284, 291, 293, 297, 300, 310, 341, 342 simple, 198-202, 214, 240-241, 253, 256-257, 310 spontaneous, 176-177, 357, 361, 363 thermite, 364 types of, 25-27 Reaction intermediate, 199-200, 206, 241, 264-265 Reaction mechanism, 198-200, 202, 206, 239-241, 264 graphical representation of a, 200

Reaction rate, 176, 214-227, 236-238, 278-279, 282-284, 288, 290291, 297, 301, 308, 309, 313, 403 average, 226-227, 254, 261 factors that affect, 245-269 general, 217-218, 403 instantaneous, 226-227, 254 measuring, 213-229 rate-determining step, 241 rate laws, 253-258, 309, 401 ways to measure, 220-223 Reactivity causes of the chemical reactivity of gases, 45-46 chemical reactivity of gases, 44-48 Redox (oxidation-reduction), 350-366 general equation, 351-352 Reducing agent, 351-52, 357, 362, 363 Reduction, 27, 350-352, 353, 358, 360, 363-364 Relationship between the pH and the molar concentration of hydronium and hydroxide ions, 328-329 Repeatability of a measuring instrument, 394, 395 Respiration, 27, 38, 79 cellular, 48 Review of some mathematical equations, 398-403 Rounding off, 397 Rules calculation rules for scientific notation, 407 notation, 11-12 of nomenclature, 11 of safety, 372-375, 375-376 to determine the number of significant figures of a measurement, 395

S Salt, 19-20, 25, 192, 321-322 to disinfect the water in a swimming pool, 341 SATP, 75, 93, 100, 149, 236, 246 Secant, 226-227 Sensitivity of a measuring instrument, 394, 395 Significant figures in results of mathematical operations, 396 rules to determine the number of significant figures of a measurement, 395 Solar wind, 40 Solubility, 15 product constant, 320, 336-338, 422 of common ionic compounds, 423 equilibrium, 279 molar, 336-337 Solute, 15-17, 19, 186-189, 279, 336 Solution, 15-16, 18 how to prepare a, 383-386 in an electrochemical cell, 358-359 ionic equilibrium, 320-338 saturated, 15, 279, 283, 336-337 standard, 385 Solvent, 15, 19, 186, 188-189, 252, 279, 283, 336 polar, 325 Solving problems method of unit analysis, 379-380 the four steps of, 378 Sørensen, Søren Peter Lauritz, 326 Specific heat capacity, 32, 134-136, 138, 139, 140, 189, 192, 409 of various substances, 423 Spectroscopy, 206 Standard, 149, 151-152

INDEX

433

State ground state, 350-351, 353-355 state of equilibrium, 278-286, 288-297, 300, 313, 320, 360, 399 steady state, 286 Sterilization, 229 Stoichiometry, 23-24, 216 calculating enthalpy change using, 161-162 of gases, 108-109 STP, 75, 92-93, 100, 364 Substance amphoteric, 323 Summation of heat of the reaction, 201-206 Sunglasses (a brief history of…), 342 Symbols danger, 372-374 safety, 375 Synthesis, 25, 27, 153 of ammonia, 256, 266 System(s) at equilibrium, 283-286, 288-297, 309-310, 336 closed, 131-132, 133, 181, 284-285 isolated, 32, 131-132 open, 131, 181, 284-285 types of, 131-133 System at equilibrium, 256, 283-286

T Table(s) groups of, 8, 9, 12 periodic, C1-C2, 5, 8-10, 44, 350 periods of, 5, 8, 10 reference, 410-423 Tangent, 226-227, 254-255, 390 Temperature, 128-130, 132-133, 149, 189-190, 238, 241, 313-314, 328,364 absolute, 61, 75, 80-82, 85, 86, 100, 105 and kinetic energy, 54, 58-60, 65-66 and pressure, 70, 85-87 and volume, 80-84 changes of, 32, 134-136, 138-141, 150-151, 292-294 difference between heat and, 128-130 effect on reaction rate, 255, 260-261 effect on the state of equilibrium, 292-294 of the reaction environment, 260-261 of two systems, 138-140, 403 scale of, 81, 129 Theory/ies acids and bases, 320-325 Arrhenius theory, 192, 320-323, 327-328 Brønsted-Lowry theory of acids and bases, 323-325 collision theory, 236-241, 246, 247-248, 251-252, 253, 260-261, 262, 290 kinetic theory of gases, 54-61, 63, 66, 69, 79, 84, 87, 91, 92, 96, 100, 111 Thermal agitation, 31, 32, 58, 61, 82, 128-130, 132-133, 135, 150, 176 Thermochemistry, 201 Thermodynamic, 82, 131-132 first principle of, 31-32, 128, 131-133, 157, 239 thermodynamic properties of certain elements, 417 Thermometer, 133, 285 cricket, 261

434

INDEX

Thomson, William, 81-82, 128 Titration, 385 steps to set up, 385-386 Torricelli, Evangelista, 70-71 Transfer energy, 127-143, 284 heat transfer, 128-130, 131, 132, 134-135, 138-140, 149, 150-151, 153, 172-173, 176-177, 187-188-189, 192, 203, 293, 313-314, 403 Transformation(s) chemical, 21-27 endothermic, 150-151 energy balance of, 156-160, 174-175, 188, 398 [This exact term not on Eng/Fre pdf p.46] energy changes in, 127-209 exothermic, 150-151 of algebraic expressions, 404-405 physical, 15-20, 54, 148, 150-151, 278-279, 300

U Uncertainty absolute, 394 relative, 394 unknown, 396 Units common units of measure, 409 of measure in chemistry, 408-409 Using an energy diagram to observe the progress of a conversion, 176-181

V Vital disequilibrium, 286 Volta, Alessandro, 359 Volume, 75 and pressure, 76-79 and quantity of gas, 88-91 and temperature, 80-84 molar volume of gases, 92-93 Von Mayer, Julius Robert, 132

W Waage, Peter, 256 Water as a temperature regulator, 135 ionization constant of water, 320, 325, 327-329 Ways to measure reaction rate, 220-223 Work, 128, 132, 148, 239

Z Zeppelin, 115

SOURCES Legend t: top b: bottom c: centre l: left r: right

PHOTOS COVER R.T. Wohlstadter/Shutterstock (fractal), Curtis Kautzer/Shutterstock (car), Ramunas Bruzas/Shutterstock (stalactites), Artmann Witte/Shutterstock (natural gas), Rich Carey/Shutterstock (coral), Oneo/Shutterstock (water drop on a leaf)

REVIEW p. 2br: Pedro Salaverría/Shutterstock • p. 2tr: Tischenko Irina/Shutterstock • p. 2bl: DAVID TAYLOR/SCIENCE PHOTO LIBRARY • p. 2tl: Pawel Gaul/Istockphoto • p. 2c: Sebastian Kaulitzki/Istockphoto • p. 2cr: Dirk Rietschel/ Istockphoto • p. 24: © NASA • p. 26: DAVID TAYLOR/SCIENCE PHOTO LIBRARY • p. 31 from top to bottom: Dariusz Gora/Istockphoto, Ian Poole/Istockphoto, Fesus Robert/Istockphoto, Photos.com, PATRICK LANDMANN/SCIENCE PHOTO LIBRARY, sint/Shutterstock, Paul Seheult; Eye Ubiquitous/CORBIS, David Arky/Corbis

UNIT 1 p. 34-35: Bertrand Rieger/Hemis/Corbis CHAPTER 1 CHEMICAL PROPERTIES OF GASES p. 37: Pakhnyushcha/Shutterstock • p. 38t: Jakub Cejpek/Shutterstock • p. 38b: Scimat/Photo Researchers, Inc • p. 39: AP Photo/Rick Bowmer • p. 40t: HEALTH PROTECTION AGENCY/SCIENCE PHOTO LIBRARY • p. 40 c: © Pablo Corral Vega/CORBIS • p. 40b: Alex0001/Shutterstock • p. 41cr: Getty Images • p. 41c: Roger Viollet/Getty Images • p. 42 from top to bottom: HomeStudio/ Shutterstock, wando studios/iStockphoto, DJ Mattaar/Shutterstock, Charles D. Winters/Photo Researchers, Inc., iStock, Andrew Lambert

Photography/Science Photo Library • p. 43b: webphotographeer/iStockphot • p. 43t: Christina Richards/Shutterstock • p. 44t: Dudarev Mikhail/ Shutterstock • p. 44b: Danny E Hooks /Shutterstock • p. 45t: Stefan Klein/iStockphoto, Shane White /Shutterstock • p. 46b: Arthur Jan Fijałkowski /Wikipedia Commons • p. 48c: © Wu Hong/epa/ Corbis • p. 48b: STEVE HORRELL/SCIENCE PHOTO LIBRARY • p. 49 SSPL via Getty Images • p. 49br: beerkoff/Shutterstock • p. 50t: HomeStudio/Shutterstock • p. 50b: Dudarev Mikhail/Shutterstock • p. 51: Istockphoto • p. 52l: Dane Wirtzfeld/Istockphoto • p. 52tr: Wikipedia Commons • p. 52br: Maciej Korzekwa/Istockphoto

CHAPTER 2 PHYSICAL PROPERTIES OF GASES p. 53: Simon Askham/Istockphoto • p. 54: Sinisa Botas/Shutterstock • p. 56: Cezar Serbanescu/ iStockphoto • p. 57: Charles D. Winters/Photo Researchers, Inc. • p. 59: SEGRE COLLECTION/ AMERICAN INSTITUTE OF PHYSICS/SCIENCE PHOTO LIBRARY • p. 63: Ocean Image Photography/ Shutterstock • p. 64tl: © Mike Smith • p. 64tr: © Mike Smith • p. 64b: Charles D. Winters/Photo Researchers, Inc • p. 65b: nito/Shutterstock • p. 67: DR. JEREMY BURGESS/SCIENCE PHOTO LIBRARY • p. 68: Sergiy Zavgorodny/Shutterstock • p. 69: Mikael Damkier/Shutterstock • p. 71bl: Brad Killer/iStockphoto • p. 71br: Janicke Morrissette/ Le Bureau Officiel • p. 71t: © Bettmann/Corbis • p. 75: © Deutsche Museum • p. 76t: Wikipedia Commons • p. 76c: Roger-Viollet/The Image Works • p. 78: Émilie Carrier • p. 80l: LIBRARY OF CONGRESS/SCIENCE PHOTO LIBRARY • p. 80t: Daniel Williams/Istockphoto • p. 82: ROYAL ASTRONOMICAL SOCIETY/SCIENCE PHOTO LIBRARY • p. 84: Photo Antoine Couture • p. 85: SHEILA TERRY/SCIENCE PHOTO LIBRARY • p. 87: MARTYN F. CHILLMAID/SCIENCE SOURCES

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CHAPTER 5 GRAPHIC REPRESENTATION OF ENTHALPY CHANGE

UNIT 2

p. 185: © 1986 Richard Megna, Fundamental Photographs, NYC • p. 186: Stéphanie Colvey • p. 187: optimarc/Shutterstock • p. 189c: Bjørn Christian Tørrissen/bjornfree.com • p. 189r: Eric Isselée/ Shutterstock • p. 192: Science Source/Photo Researchers • p. 193: Charles D. Winters/Photo Researchers, Inc. • p. 194l: Wikipedia Commons • p. 194r: Corbis

p. 124-125: Lo Mak/Redlink/Corbis CHAPTER 3 ENERGY TRANSFER p. 127: Photo by Woods Hole Oceanographic Institution/Advanced Imaging and Visualization Laboratory/NOAA/NSF • p. 128: Wikipedia Commons • p. 129bl: Bruce Omori/epa/Corbis • p. 129br: © Erling Riis • p. 131: Charles Knox/Istockphoto • p. 132: SCIENCE PHOTO LIBRARY • p. 135: Graeme Purdy /Istockphoto • p. 139: Graham Clarke/Istockphoto • p. 140: Bazuki Muhammad/Reuters/Corbis • 141: Charles D. Winters/Photo Researchers, Inc. • p. 143g: SSPL via Getty Images • p. 143r: ANDREW BROOKES, NATIONAL PHYSICAL LABORATORY/SCIENCE PHOTO LIBRARY • p. 146: Charles D. Winters/Photo Researchers, Inc.

p. 171: EUROPEAN SPACE AGENCY/SCIENCE PHOTO LIBRARY • p. 173: Istockphoto • p. 174: Robert Kyllo/Istockphoto • p. 175: Reuters/Corbis • p. 177: Charles D. Winters/Photo Researchers, Inc. • p. 181: CNRS Photothèque/ENSTA/LOCHEGNIES José CHAPTER 6 MOLAR HEAT OF REACTION

CHAPTER 7 HESS’S LAW p. 197: oneo/Shutterstock • p. 198: Getty Images • p. 199: The Canadian Press Images/Francis Vachon • p. 201c: robert paul van beets/Shutterstock • p. 201B: ANDREW LAMBERT PHOTOGRAPHY/ SCIENCE PHOTO LIBRARY • p. 201r: Wikipedia Commons • p. 206: Getty Images

UNIT 3 CHAPTER 4 ENTHALPY CHANGE

p. 210-211: Check Six/Getty Images

p. 147: AP Photo/Gloria Calvi • p. 148: SCIENCE, INDUSTRY AND BUSINESS LIBRARY/NEW YORK PUBLIC LIBRARY/SCIENCE PHOTO LIBRARY • p. 149: L-shape, an ethanol (or bio) fireplace by Planika • p. 150t: Foodfolio/maXx images • p. 150b: Kenneth Libbrecht/Visuals Unlimited, Inc. • p. 151: Bibliothèque MINES ParisTech • p. 152: Severe /Shutterstock • p. 154: Photo by Mark Wilson, Gizmodo.com • p. 164: press photo BASF

CHAPTER 8 MEASURING REACTION RATE

436

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p. 213: Curtis Kautzer/Shutterstock • p. 215: DAVID AUBREY/SCIENCE PHOTO LIBRARY • p. 216: © 2008 Richard Megna, Fundamental Photographs, NYC • p. 220t: MARTYN F. CHILLMAID/SCIENCE PHOTO LIBRARY • p. 220b: Martin Shields/Photo Researchers, Inc. • p. 221t: Galyna Andrushko /Shutterstock • p. 221b: Photo Diane Grenier

• p. 224bl: Luis Pedrosa/Istockphoto • p. 224t: Courtesy of American Galvanizers Association • p. 224br: Istockphoto • p. 226: Pete Niesen /Shutterstock • p. 229tl: Ekaterina Starshaya /Istockphoto • p. 229bl: Maurice van der Velden /Istockphoto • p. 229tr: Serguei Kovalev/Istockphoto • p. 229br: Tom Hahn/Istockphoto CHAPTER 9 COLLISION THEORY p. 235: Herbert Kratky/Shutterstock • p. 236: Andrew Parkinson/Corbis • p. 238bl: Richard Thornton /Shutterstock • p. 238br: © The McGraw-Hill Companies, Inc./Photo by Stephen Frisch • p. 239: AFP/ Getty Images • p. 240: Ricardo Azoury /Istockphoto • p. 241: Ryerson Clark/Istockphoto • p. 244: ROBERT BROOK/SCIENCE PHOTO LIBRARY CHAPTER 10 FACTORS THAT AFFECT REACTION RATES p. 245: Ramunas Bruzas/Shutterstock • p. 246: Joel Blit/Istockphoto • p. 247: Ian Crysler Photographer • p. 249: Ules Barnwell/Istockphoto, Leslie Banks/Istockphoto • p. 251: Sergey Kashkin /Istockphoto • p. 254l: AP Photo/Bob Edme • p.252r: Greenfire/Shutterstock • p. 256: Wikipedia Commons • p. 261t: Istockphoto • p. 261b: Thomas J. Walker • p. 262: Chai Kian Shin/Shutterstock • p. 263cr: CPimages.ca/Phototake • p. 263br: Istockphoto • p. 265: Taken from Chimie 12 © 2003 Les Éditions de la Chenelière • p. 266: University Archives, University of Pittsburgh • p. 268cl: Steve Lovegrove /Shutterstock • p. 268cr: CC STUDIO/SCIENCE PHOTO LIBRARY • p. 268br: Valeriy Velikov /Istockphoto • p. 269l: Melissa King/Shutterstock • p. 269r: GENERAL MOTORS DOCUMENT/REUTER R/CORBIS SYGMA • p. 273l: istockphoto • p. 273r: Richard Walters/Istockphoto, Stefan Klein/Istockphoto

UNIT 4 p. 274-275: Yoshio Shinkai/Getty Images CHAPTER 11 THE QUALITATIVE ASPECT OF CHEMICAL EQUILIBRIUM p. 277: Yoshio Shinkai/Getty Images • p. 278t: Leigh Prather/Shutterstock • p. 278b: Roman Sigaev/Shutterstock • p. 279r: Olga Sapegina/ Shutterstock • p. 279b: © 1999 Richard Megna, FUNDAMENTAL PHOTOGRAPHS, NEW YORK • 280t: Péter Gudella/Shutterstock • p. 280b: Colour/Shutterstock • p. 283: © 2001 Richard Megna, Fundamental Photographs, NYC • p. 285: © 2001 Richard Megna, Fundamental Photographs, NYC • p. 287: Nikola Bilic/Shutterstock p. 288: Academie des Sciences, Paris, France/Archives Charmet/The Bridgeman Art Library • p. 289: SSPL via Getty Images • p. 291: Ewan Chesser/Shutterstock • p. 297: NASA • p. 299l: DR. JEREMY BURGESS/SCIENCE PHOTO LIBRARY • p. 299br: GEORGETTE DOUWMA/SCIENCE PHOTO LIBRARY • p. 300l: Kuzma/Shutterstock • p. 300r: Duncan Walker/Istockphoto • p. 301l: Gösta Florman (1831–1900)/The Royal Library • p. 301c: Hulton-Deutsch Collection/Corbis • p. 301r: Bettmann/CORBIS

CHAPTER 12 THE QUANTITATIVE ASPECT OF CHEMICAL EQUILIBRIUM p. 307: John Anderson/Istockphoto • p. 313: Jacques Nadeau • 320: Pierre Holtz/epa/Corbis • p. 321: SCIENCE PHOTO LIBRARY • p. 323c: sciencephotos/Alamy • p. 323r: Bibliotheque Nationale, Paris, France/Archives Charmet/The Bridgeman Art Library • p. 326: SCIENCE PHOTO LIBRARY • p. 327: Courtesy of Techne USA • p. 336t: Charles D. Winters/ Photo Researchers, Inc. • p. 336b: MEDIMAGE/ SCIENCE PHOTO LIBRARY • p. 341: Wolfgang Steiner/Istockphoto • p. 342l: Getty Images p. 342r: epa/Corbis

SOURCES

437

SUPPLEMENT – REDOX REACTIONS

APPENDICES

p. 349: Copestello/Shutterstock • p. 350: MARTYN F. CHILLMAID/SCIENCE PHOTO LIBRARY • p. 352t: © 1986 Peticolas/Megna, Fundamental Photographs, NYC • p. 352b: Photo Antoine FafardCouture • p. 353: Y. DEROME/PUBLIPHOTO DIFFUSION/SCIENCE PHOTO LIBRARY • p. 357t: Theodore Gray/Visuals Unlimited, Inc. • p. 357b: © 2001 Richard Megna, Fundamental Photographs, NYC • p. 359: SCIENCE PHOTO LIBRARY • p. 361: NASA/Reuters/Corbis • p. 362: Dave Starrett Photographer • p. 364: Courtesy of the government of the Northern Territory, Australia

p. 370tl: Olivier Le Queinec/Shutterstock • p. 370tc: Sebastian Duda /Shutterstock • p. 370tr: Alexey Stiop/Shutterstock • p. 370bl: oriontrail/Shutterstock • p. 370br: Ariadna de Raadt/Istockphoto • p. 342l: Jason Smith/ Shutterstock • p. 376: Image Source/Corbis • p. 384: Janicke Morrissette/Le bureau officiel • p. 386: Janicke Morrissette/Le bureau officiel • p. 387: Janicke Morrissette/Le bureau officiel • p. 392: Janicke Morrissette/Le bureau officiel • p. 394: Janicke Morrissette/Le bureau officiel

ILLUSTRATIONS All illustrations are by Michel Rouleau, except: p. 372, 373, 374, 375 (Arto Dokouzian) p. 5, 10, 13, 19, 20, 21, 28, 29 (Late Night Studio)

438

SOURCES

CHEMISTRY The QUANTUM collection offers a flexible and concrete approach to the Chemistry Program for the Third Year of Secondary Cycle Two. It is a major asset that helps students develop their competencies, build their understanding of the scientific concepts in the Program and have a better grasp of the issues pertaining to their daily environment. The QUANTUM collection is: • an encyclopaedic textbook that offers substantial treatment of all concepts in the Program • an impressive number of exercises with varied levels of difficulty • a wide variety of laboratories, some previously tested and others new and original • a clear presentation of scientific formulas, mathematical equations, procedures and examples of calculations • numerous highly precise diagrams and illustrations to help students understand the concepts • a Review section for Chemistry concepts addressed in the first and second years of Secondary Cycle Two • appendices that address certain scientific procedures, strategies and techniques and that delve deeper into various aspects of problem-solving

• indispensable reference tables • animation clips that take into account different learning styles and facilitate the comprehension of certain phenomena • a choice of new learning and evaluation situations related to students’ interests and designed to help develop competencies • suggestions for the use of ICTs in the classroom • varied evaluation tools adapted to the reality of the classroom • a clear structure that encourages the development of student autonomy • flexible lesson plans

QUANTUM SUBSTANTIAL, SOLID AND RIGOROUS CONTENT!

The components of the QUANTUM collection – CHEMISTRY, Third Year of Secondary Cycle Two

• Student Textbook

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ISBN 978-2-7652-1438-0