AceGamsat - The General Chemistry Bible

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The General Chemistry Bible Everything you need to know for GAMSAT General Chemistry www.AceGAMSAT.com The information contained in this guide is for informational purposes only. The publication of such Third Party Materials does not constitute a guarantee of any information, instruction, opinion, products or services contained within the Third Party Material. Publication of such Third Party Material is simply a recommendation and an expression of our own opinion of that material. No part of this publication shall be reproduced, transmitted, or sold in whole or in part in any form, without the prior written consent of the author. All trademarks and registered trademarks appearing in this guide are the property of their respective owners. Users of this guide are advised to do their own due diligence when it comes to making decisions and all information, products, services that have been provided should be independently verified by your own qualified professionals. By utilising this guide, you agree that the company AceGAMSAT is not responsible for the success or failure relating to any information presented in this guide. ©2018 AceGAMSAT. All Rights Reserved. AceGAMSAT is not affiliated with ACER in any way.

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About GAMSAT The GAMSAT (Graduate Australia Medical School Admissions Test) has been developed by the Australian Council for Educational Research (ACER). This test selects for students that have the greatest capacity to advance their studies in medicine, dentistry, and pharmacy. The GAMSAT was developed for Australian universities and its component of use for admissions has spread to institutions in the UK and Ireland. GAMSAT evaluates the nature and extent of abilities and skills gained through prior experience and learning, including the mastery and use of concepts in basic science as well as the acquisition of more general skills in problem solving, critical thinking and writing. Candidates whose first degree is in a non-scientific field of study can still sit GAMSAT and succeed in an application for admission to one of the graduate-entry programs. A science degree is not always a prerequisite and institutions encourage applications from candidates who have achieved academic excellence in the humanities and social sciences. However, it must be stressed that success in GAMSAT is unlikely without knowledge and ability in the biological and physical sciences.

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Table of Contents ATOMS AND MOLECULES

1 1 2 7 17 21 23 24 24 26 29

ATOMS ELEMENTS THE PERIODIC TABLE IONS MOLECULES AND INTERACTIONS NAMING INORGANIC COMPOUNDS CHEMICAL REACTIONS AND EQUATIONS TYPES OF CHEMICAL REACTIONS OXIDATION NUMBERS AND REDOX REACTIONS OXIDISING VS REDUCING AGENTS

GASES

30 30 31

GASES KINETIC MOLECULAR THEORY

KINETICS AND CHEMICAL EQUILIBRIUM

36 36 37 40 41 42 43 45 47 48

THE COLLISION THEORY EQUATIONS FOR REACTION RATES DETERMINING THE RATE LAW BY EXPERIMENT REACTION ORDERS RATES OF REVERSIBLE REACTIONS CATALYSIS EQUILIBRIUM THE REACTION QUOTIENT LE CHATELIER’S PRINCIPLE

THERMODYNAMICS

52 53 54 55 56 56 57

THE FIRST LAW OF THERMODYNAMICS THE SECOND LAW OF THERMODYNAMICS THERMODYNAMIC FUNCTIONS INTERNAL ENERGY TEMPERATURE PRESSURE

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57 58 59

ENTHALPY ENTROPY GIBBS FREE ENERGY

SOLUTIONS

61

PHASE DIAGRAMS

62 63 65 67

ACIDS AND BASES

69

UNITS OF CONCENTRATION VAPOR PRESSURE SOLUBILITY

DEFINITIONS EQUILIBRIUM CONSTANTS FOR ACID-BASE REACTIONS LOGARITHMS TITRATIONS BUFFERS

69 73 75 77 84

ELECTROCHEMISTRY

85

OXIDATION-REDUCTION

85 86 87 88

POTENTIALS GALVANIC CELLS ELECTROLYTIC CELL

APPENDIX

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Atoms and Molecules Video: Elements and Atoms Video: Atomic number, Mass number, and Isotopes Video: The mole and Avogadro's number Video: Empirical, molecular, and structural formulas Video: Ionic, Covalent, and Metallic bonds Video: Naming Ionic and Covalent Molecular Compounds Video: Chemical Reactions Introduction

Atoms

Atoms are the basic building blocks of all matter. Each atom is composed of a nucleus. The nucleus contains protons and neutrons and is surrounded by one or more electrons.

Fig. 1 - The characteristics of an atom.

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Protons have a positive (+) charge. Electrons have a negative (-) charge. Neutrons are electrically neutral.

Fig. 2 - Properties of protons, neutrons, and electrons.

Electrons and protons have opposite charges of equal magnitude. The size of a proton and neutron are approximately equal; however, an electron is much smaller.

Elements

An element is a chemical substance containing atoms that have the same number of protons in their atomic nuclei. Elements are the building blocks of all compounds. Elements are displayed as follows:

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Fig. 3 - The display of an element.

A is the mass number of the element. This is the number of protons plus neutrons in the nucleus. Z is the atomic number. This is the number of protons in the nucleus.

Example:

Fig. 4 - The atomic and mass number of sodium.

An element is characterised by its number of protons. Any element usually has a select number of protons. However, the number of neutrons or electrons can change, yet the element will still retain its identity.

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If there are two atoms of the same element that have a different mass number (different number of neutrons), these are said to be isotopes. Take carbon for example. Carbon has three isotopes: 12C, 13C, and 14C.

Fig. 5 - Three isotopes of carbon.

Each isotope of carbon contains six protons (given the element identified is carbon). 12C contains six neutrons, 13C contains seven neutrons, and 14C contains eight neutrons. The atomic weight or molar mass (M) of an atom is given in atomic mass units (amu). The amu is a ratio and is defined by carbon-12.

Fig. 6 - Carbon-12, the standard for atomic weight measurement.

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One atom of 12C has an atomic weight or molar mass of 12 amu. All other atomic weights are measured against this standard. 12C

of

is also used to define a mole. A mole is the number of carbon atoms in 12 grams

12C.

Avogadro’s number also defines the number of carbon atoms in 12 grams of carbon as 6.022 x 1023. So, there are 6.022 x 1023 carbon atoms in 12 grams of 12C. The formula that surrounds these concepts is as follows:

n = moles m = mass in grams M = molar mass (found on the periodic table for individual atoms)

Examples 1. mass = moles x molar mass

Calculate the mass of 0.5 moles of water a. Write the equation: m(in grams) = n x M

b. Extract the data from the question: n = 0.5 mol c. Calculate the molar mass of the substance using a periodic table: M(H2O) = (2 x 1.008) + 16.00 = 18.016 g mol-1

d. Substitute the values into the equation and solve: mass = 0.5 x 18.016 = 9.008 g

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2. moles = mass ÷ molar mass

Calculate the moles in 50 g of oxygen gas. a. Write the equation: n=m/M

b. Extract the data from the question: m = 50 g

c. Calculate the molar mass of the substance using a periodic table: M(O2) = 2 x 16.00 = 32.00 g mol-1

d. Substitute the values into the equation and solve: n = 50 / 32.00 = 1.56 mol

3. molar mass = mass ÷ moles

Calculate the molar mass of a pure substance if 2.51 moles of the substance has a mass of 31.113 g a. Write the equation: M=m/n

b. Extract the data from the question: m = 31.113 n = 2.51 mol c. Substitute the values into the equation and solve: M = 31.113 ÷ 2.51 = 12.396 g mol-1

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The Periodic Table All elements are listed on the periodic table. Elements are listed based on their atomic number from left to right. The vertical column of the periodic table is called a group or family and the horizontal row is a period or row. Groups are numbered from 1 to 18, and periods are numbered from 1 to 7.

Fig. 7 - The periodic table.

The periodic table above divides the elements into several different groups. These groups are represented by different colours in the periodic table above. Note that nonmetals generally have lower melting points than metals and form negative ions.

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Table 1 - Common ions and their charges.

Molecular substances are usually made up from only nonmetals. For example, H2O is only made up of nonmetals (oxygen and hydrogen). Metals tend to lose electrons to form positive ions. All metals exist as solids at room temperature except for mercury, which stays a liquid.

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Group 1 is known as the alkali metals and form 1+ cations.

Fig 8. - Characteristics of group 1 – The Alkali Metals.

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Group 2 is known as the alkaline earth metals and form 2+ cations.

Fig. 9 - Characteristics of group 2 – The Alkaline Earth Metals.

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Groups 3-12 are known as the transition metals.

Fig. 10 - Characteristics of the transition metals.

Sometimes the transition metals are not counted in the group numbering. Group 1 and 2 remain the same. Group 13 is termed group 3; group 14 is termed group 4; all the way up to group 18, which is termed group 8.

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Group 14 elements (also termed group 4) can all form 4 covalent bonds with nonmetals.

Fig. 11 - Characteristics of group 4 (or group 14) – The Crystallogens. www.acegamsat.com

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Group 15 elements can form 3 covalent bonds.

Fig. 12 - Characteristics of group 5 (or group 15) – the Pnictogens. www.acegamsat.com

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Group 16 contains oxygen (important).

Fig 13. - Characteristics of group 6 (or group 16) – The Chalcogens. www.acegamsat.com

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Group 17 is known as the halogens and are very reactive, forming only one bond with other elements.

Fig. 14 - Characteristics of group 7 (or group 17) – The Halogens. www.acegamsat.com

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Group 18 is known as the noble gases and are nonreactive.

Fig. 15 - Characteristics of group 8 (or group 18) – The Noble Gases. www.acegamsat.com

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Ions An element becomes an ion when it has more or fewer electrons than protons. Negative ions are known as anions and positive ions are known as cations.

Fig. 16 - Anions and cations.

General rule: Metals form cations and non-metals form anions.

Table 2 - Common simple cations and anions www.acegamsat.com

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General predictions about elements can be made based on their position in the periodic table.

Fig. 17 - Trends of the periodic table.

Atomic size: Atomic size (radius) increases from right to left and from top to bottom. This is because the effective nuclear charge increases (more protons) from left to right, and each additional electron is pulled more strongly toward the nucleus. Atomic radius increases from top to bottom because the atom becomes larger with each added shell.

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Fig. 18 - Trends in atomic size (radius).

Note: When moving down the periodic table, each underlying period will have elements with an extra shell of electrons. For example, helium has one shell and lithium has two shells.

Ionisation energy

Ionisation energy is defined as the energy required to remove an electron from an atom. If an electron is more strongly attached to the nucleus, it requires more energy to be removed. Ionisation energy increases from left to right and from bottom to top. Again, elements have a greater nuclear charge (more protons) from left to right. As such, there are more protons to create a stronger attraction with the surrounding electrons and, thus, more ionisation energy is required to remove the electron from the atom. When moving down the periodic table, the distance of the electron from the nucleus increases because there are more electron shells when going down the periodic table. This increased distance creates a decrease in electric field strength and thus less energy is required to remove an electron. This explains why the ionisation energy increases when moving up the periodic table.

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Fig. 19 - Trends in ionisation energy.

Electronegativity Electronegativity is the tendency of an atom to attract an electron in a bond that it shares with another atom. Electronegativity tends to increase from left to right and from bottom to top.

Fig. 20 - Trends in electronegativity.

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Molecules and Interactions

In molecules, atoms are held together by bonds. The interactions between atoms within the molecule are known as intramolecular bonds. The interactions between atoms in different molecules are known as intermolecular bonds. Watch this video to clearly understand the difference between intermolecular and intramolecular bonds in greater details:

Video: Intermolecular and Intramolecular forces

Covalent bond: A covalent bond occurs when two electrons are shared by two nuclei. The electrons (negatively charged) are pulled toward both positively charged nuclei via electrostatic forces. Hydrogen bond: A hydrogen bond is a special case of dipole forces. A hydrogen bond is the attractive force between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule. Usually the electronegative atom is nitrogen, fluorine, or oxygen, which has a partial negative charge. The hydrogen then has the partial positive charge. Ionic bond: This is a type of chemical bond that involves the electrostatic attraction between oppositely charged ions. It is the primary interaction in ionic compounds. The ions are atoms that have gained one or more electrons (anions – negatively charged) and atoms that have lost one or more electrons (cations - positively charged). Van der Waals interactions: This is the interaction of electron clouds between molecules - intermolecular interaction. Van der Waals interactions are the weakest of all intermolecular attractions between molecules. Hydrophobic interactions: These interactions are intermolecular and occur between nonpolar substances.

A compound is formed when a substance contains one or more elements in a definite ratio. For example, water is a compound that contains oxygen and hydrogen in a definite ratio.

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Fig. 21 - Formation of a compound – water.

The empirical formula of a pure compound is the simplest whole number ratio between the number of atoms of the different elements in the compound. For example, the empirical formula for glucose is CH2O.

Table 3 - Examples of molecular and empirical formulas.

The molecular formula of a molecule states the exact number of the different atoms that make up the molecule. The molecular formula for glucose is C6H12O6.

Video: Empirical formula from mass composition

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Naming inorganic compounds

In the GAMSAT, it is uncommon that ACER will directly test the ability to name inorganic compounds. However, it is important that candidates can identify these compounds. Ionic compounds are named after their cation and anion. When naming an ionic compound, the cation name is placed in front of the anion name. For example, the cation Na+ and anion Cl- or NaCl (when combined) is called sodium chloride. Monoatomic ions and polyatomic ions are given the suffix –ide. For example, H- is a hydride ion and OH- is a hydroxide ion. The following figure details a range of common polyatomic ions.

Table 4 - Common polyatomic ions. www.acegamsat.com

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Chemical Reactions and Equations Compounds that initially react in a chemical reaction are termed reactants. Reactants are always written on the left-hand side of a chemical equation. The compounds that are produced in the reaction are termed the products of the chemical reaction. Products are always written on the right-hand side of a chemical equation. As illustrated:

2 HCl + 2 Na → 2 NaCl + H2 Reactants

Products

Fig. 22 - Coefficients of chemical reactions.

The coefficients in the above equation represent the relative number of moles of reactants that combine to form the relative number of moles of products. The law of conservation of mass states that the number of atoms of a given element remains constant during the process of a chemical reaction.

Types of Chemical Reactions

Students should be familiar with the four fundamental types of reactions:

Combination: C + D  E Decomposition: E  C + D Single Displacement: C + DE  D + CE Double Displacement: CD + EF  CF + DE

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The letters C, D, E and F represent hypothetical elements and/or molecules.

Table 5 - Types of reactions in general chemistry.

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Oxidation Numbers and Redox reactions

Some questions in the GAMSAT require students to choose the reaction from a range of reactions that shows a change in oxidation state. This change could either be an increase or decrease in oxidation state.

Fig. 23 - Change in oxidation state after a reaction.

An element is said to be oxidised in a reaction if its oxidation number increases, and reduced if its oxidation number decreases. A general set of rules are used to assign oxidation numbers to different elements in different compounds. These rules will be provided in the GAMSAT, but read these rules beforehand and go through the following practice examples to become familiar with assigning oxidation numbers. This will improve speed and confidence when taking the GAMSAT.

1. The oxidation number for an atom in its elemental form is always zero. o

o

A substance is elemental if both following statements are true: 

Only one kind of atom is present.



If the charge is 0.

Examples: 

S8: The oxidation number of S = 0



Fe: The oxidation number of Fe = 0

2. The oxidation number of a monoatomic ion is the charge of the monatomic ion. o

Examples:  

Oxidation number of S2- is -2.

Oxidation number of Al3+ is +3.

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3. The oxidation number of all Group 1 metals = +1 (unless elemental). 4. The oxidation number of all Group 2 metals = +2 (unless elemental). 5. Hydrogen (H) has two possible oxidation numbers: o

+1 when bonded to a nonmetal.

o

-1 when bonded to a metal.

6. Oxygen (O) has two possible oxidation numbers: o

-1 in peroxides (O22-)—uncommon.

o

-2 in all other compounds—most common.

7. The oxidation number of fluorine (F) is always -1. 8. The sum of the oxidation numbers of all atoms (or ions) in a neutral compound = 0. 9. The sum of the oxidation numbers of all atoms in a polyatomic ion = charge on the polyatomic ion.

Assigning oxidation states – quick reference guide: When assigning oxidation numbers to the elements in a substance, take a systematic approach with the following questions: 1. Is the substance elemental? 2. Is the substance ionic? 3. If the substance is ionic, are there any monoatomic ions present? 4. Which elements have specific rules and which do not? Use rule 8 or 9 from above to calculate these.

Example 1: Determine the oxidation number of each element in Na2SO4. 1. Is the substance elemental? o

No. There are 3 elements present.

2. Is the substance ionic? o

Yes, there is a metal and non-metal (ionic).

3. If the substance is ionic, are there any monoatomic ions present? o

Yes, the sodium ion (Na+) is monoatomic. www.acegamsat.com

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o

Thus, the oxidation number of Na = +1.

4. Which elements have specific rules and which do not? o o o o

Oxygen has a rule (-2 in most compounds). Oxidation number of O = -2. S does not have a rule.

Use rule 8 (algebraic manipulations) to find the oxidation number of S. 

Let S = oxidation number of sulfur. According to rule 8:



(# Na) (Oxid. # of Na) + (# S) (Oxid. # S) + (# Oxygens) (Oxid. # of O) = 0



So: 2(+1) + S + 4(-2) = 0



Solve for S using algebra.



S = +6.

Therefore, the oxidation number of S = +6. Example 2: Determine the oxidation number of each element in K2C2O4. 1. Is the substance elemental? o

No. There are 3 elements present.

2. Is the substance ionic? o

Yes, there is a metal and non-metal (ionic).

3. If the substance is ionic, are there any monoatomic ions present? 1. Yes, the potassium ion (K+) is monoatomic. 2. Therefore, the oxidation number of K = +1.

4. Which elements have specific rules?

3. Oxygen has a rule (-2 in most compounds). 4. Oxidation number of O = -2.

5. Which element(s) do(es) not have rules? 5. C does not have a rule. 6. Use rule 8 (and a little simple algebra) to find the oxidation number of C.

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

Let C = oxidation number of carbon.



According to rule 8:



(# K) (Oxid. # of K) + (# C) (Oxid. # C) + (# Oxygens) (Oxid. # of O) = 0



So: 2(+1) + 2C + 4(-2) = 0



Solve for C using algebra.



C = +3.

Therefore, the oxidation number of C = +3

Oxidising Versus Reducing agents

The following must be memorised for the GAMSAT: An oxidising agent will cause oxidation to occur, whilst the agent itself will be reduced. A reducing agent will cause reduction to occur, whilst the agent itself will be oxidised.

If something is reduced, it gains electrons. If something is oxidised, it loses electrons.

Cu2+ + Mg(s) Cu(s) + Mg2+ Oxidation Half Reaction: Mg(s) Mg2+ + 2e- – Mg solid is oxidised to Mg2+ Reducing Agent: Mg(s) Reduction Half Reaction: Cu2+ + 2e- Cu(s) – Cu2+ is reduced to Cu solid Oxidising Agent: Cu2+

Remember: OIL RIG - Oxidation Is Loss; Reduction Is Gain.

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Gases Video: Ideal Gas Equation Video: Ideal gas equation example 1 Video: Ideal gas equation example 2 Video: Ideal gas equation example 3 Video: Ideal gas equation example 4 Video: Partial pressure example Video: Vapor pressure example

Gases

A gas is a loose connection of weakly attracted atoms or molecules that move in random directions. Gases are less dense than liquids.

Fig. 24 - Movement of gas molecules in random directions (carbon dioxide).

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Zero degrees Celsius (0°C) and 1 atm are called the standard temperature and pressure (STP). All gases are miscible with each other, whereas liquids are not.

Kinetic Molecular Theory A model of an ideal gas known as the kinetic molecular theory was created to better understand the behaviour of gases. The ideal gas assumes some characteristics that are not experienced by real gases. These characteristics are: - Gas molecules make completely elastic collisions. - Gas molecules have zero volume. - The average kinetic energy of gas molecules is directly proportional to the temperature of the gas. - Gas molecules only exhibit forces due to collisions. An ideal gas will thus obey the following ideal gas law:

PV = nRT P – Pressure (in atm) V – Volume (in liters) n – Number of moles of gas R – Universal or ideal gas constant (0.08206 L·atm·mol−1·K−1 or 8.314 J·K−1·mol−1)

T – Temperature (in Kelvin)

There are special cases of the ideal gas law that should be understood for the GAMSAT. These include:

Charle’s Law: The volume of a gas (V) is directly proportional to the temperature (in Kelvin), when pressure (P) and amount of gas (n) is kept constant.

V1/T1 = V2/T2

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Fig. 25 - Representation of Charle’s law.

Boyle’s Law: When the volume (V) of an ideal gas is kept at a fixed temperature (T), the pressure and volume are inversely proportional.

P1V1=P2V2 By looking at the graph below it can be seen that as volume increases, pressure decreases.

Fig. 26 - Representation of Boyle’s Law.

For the GAMSAT, students must understand that the ideal gas law is the same for all “ideal” gases. Therefore, all gases will have the same volume, if they have the same number of molecules, temperature, and pressure.

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The kinetic molecular theory can be applied to pure gases and mixtures of gases. Each gas in a mixture in a container behaves as if it were the only one in the container. The amount of pressure contributed by any gas in a mixture is called the partial pressure of that particular gas. Therefore, the partial pressure of a particular gas is the total pressure of the gaseous mixture multiplied by the mole fraction of the particular gas.

P(B) = XB . P(total) P(B) – partial pressure of gas B XB – mole fraction of gas B P(total) – total gaseous pressure

Example: A synthetic atmosphere is created by blending 2 mol percent CO2, 20 mol percent O2 and 78 mol percent N2. If the total pressure is 0.987 atm, calculate the partial pressure of the oxygen component.

Mole fraction of oxygen is (20/100) = 0.2

P(B) = XB . P(total) Therefore, partial pressure of oxygen = (0.2)(0.987) = 0.1974 atm.

Dalton’s Law: The total pressure exerted by a gaseous mixture is the sum of the partial pressures of each of its component gases. This law shows that each gas behaves like it is in the container by itself, so that all the partial pressures of the individual gases add together to equal the total pressure.

P(total) = P1 + P2 + P3…

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Fig. 27 - Demonstration of Dalton’s law.

Students should be familiar with the two types of gaseous spreading for the GAMSAT— effusion and diffusion.

Fig. 28 - Diffusion and effusion.

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Diffusion: the spreading of one gas into another gas or into an empty area. Effusion: the spreading of a gas from a high pressure to a very low pressure through an opening that is smaller than the average distance between the gas molecules. The gas moves through this very small opening into a vacuum.

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Kinetics and Chemical Equilibrium Video: Introduction to Kinetics Video: Collision Theory Video: Elementary rate laws Video: Rate Law and Reaction order Video: Mechanisms and the rate-determining step Video: Catalysts Video: Reactions in equilibrium Video: Heterogeneous equilibrium Video: Le Chatelier’s principle

The Collision Theory

The collision theory provides a useful way to visualise and understand chemical reactions. The reacting molecules must collide in order for the chemical reaction to occur.

Fig. 29 - The collision theory.

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In order for a collision to create new molecules in a reaction, the following requirements must occur: 1. The kinetic energies of the colliding molecules must pass the threshold energy called the activation energy.

Fig. 30 - Reaction threshold energy.

2. The colliding molecules must have the correct spatial orientation. Note that the rate of a reaction increases with temperature mostly because the frequency of collisions, with sufficient relative kinetic energy, increases.

Equations for Reaction Rates

The rate of reaction represents the speed at which the concentration of a reactant or product is changing. The rate of a reaction is given in mol L-1 s-1. The rate of a reaction can be affected pressure, temperature, concentration of certain substances, surface area of reactants, and the presence of a catalyst. First we will discuss the collision theory. This theory states that, for a reaction to occur, particles must collide with the correct orientation and with sufficient energy for a reaction to occur. Many factors affect the rate of reaction by affecting the frequency of particle collisions, and/or the propagation of collisions that have enough energy to react.

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We will use this collision theory to explain the effect of a variety of factors on the rate of a chemical reaction.

Increase concentration of reactants

Increasing the concentration of reactants in solution increases the rate of reaction, as there are a greater number of particles available to react. As a result the frequency of collisions between particles in increased.

Increased temperature of reaction

Increasing the temperature of a reaction increases the kinetic energy of the reacting particles. This increases the frequency of particle collisions, and a greater proportion of collisions will have the energy required to react, resulting in a greater number of successful collisions.

Increase surface area of reactants

Increasing the surface area of solid reactants increases the number of particles that are exposed and available to react, and as a consequence this increases the frequency of particle collisions. An increase in the frequency of particle collisions results in an increase in reaction rate.

Increase pressure of reaction

Increasing the pressure of a reaction involving gases forces the gas particles closer together. This increases the frequency of particle collisions, and therefore increases the rate of reaction.

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Use a catalyst in the reaction

A catalyst provides an alternative route for the reaction to proceed, with a lower activation energy. This means that particle collisions need less energy in order for a reaction to occur, increasing the reaction rate. Chemical reaction:

aA + bB → cC + dD

In the above reaction, the lower case letters are the stoichiometric coefficients of the balanced equation. A rate law can be used to represent a forward reaction:

Ratef = Kf [A]a[B]b Kf is the rate constant for the forward reaction. The superscripts a and b are the order of each respective reactant. The sum of the superscripts a and b are the overall order of the reaction.

Table 6. - Rate law

Students must be able to determine the rate law from a table of trials. It is common that students are required to determine the order of the reaction with respect to each reactant using a table of trials. www.acegamsat.com

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Determining the Rate Law by Experiment

The value of the rate constant and the order of reactants must be determined by experiment. Rate data was obtained for the following reaction:

A + 2B → C + 2D

To find the order of each reactant, we can compare the rates between two experiments in which only the concentration of one of the reactants changes. For instance, when comparing experiment 1 to experiment 3, the initial concentration of B is tripled and the concentration of A stays the same. The reaction rate stays the same. Thus, the rate of reaction is not dependent on the concentration of B, and B is not in the rate law expression. Comparing experiment 2 and experiment 3, the initial concentration of A is tripled and the concentration of B stays the same. The reaction rate is also tripled. Thus, the rate of reaction is directly proportional to the concentration of A. Since the concentration of A and the rate of the reaction both triple, the reaction is first order with respect to A.

Rate law: Rate = k[A]

Remember: Rate law is determined by experiment.

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Reaction orders

Students are commonly tested on their knowledge of reaction orders in the GAMSAT. In previous years, students were asked to identify the correct graph (out of 4 given graphs) that represents a specific reaction order.

Fig. 31 - Characteristics of the different reaction orders.

Each reaction order carries different characteristics. The following reaction orders apply to reactions where no reverse reaction takes place:

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Zeroth order: Plotting [A] with respect to time (t) results in a straight line with a negative slope (-k). First order: Plotting ln[A] with time (t) gives a straight line with a negative slope (-k). A first order reaction takes on the form:

A → products, and rate = k[A]. Second order: Plotting 1/[A] with time (t) gives a straight line with a positive slope (k).

2A → products, and rate = k[A]2 Another form of a second order reaction is as follows, but note this does not produce the same graph as the single reactant second order reaction.

A + B → products, and rate = k[A][B] Rates of Reversible Reactions

A complex reaction can be broken down into elementary steps.

Table 7 - Elementary steps and their rate laws.

The overall rate of the reaction is determined by the rate of the slowest elementary step. This step is known as the rate-determining step.

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If the slow step is the first step, the rate law will apply to this step only. Let’s look at the following reaction:

NO2(g) + CO(g)→NO(g) + CO2(g)

This reaction occurs in two elementary steps:

NO2 + NO2→NO + NO3 (slow) NO3 + CO→NO2 + CO2 (fast)

The first elementary step in this example is the slower reaction and is thus the ratedetermining step. The rate equation for this reaction is:

rate=k1[NO2][NO2]=k1[NO2]2 Remember that the rate law assumes negligible contribution from the reverse reaction.

Catalysis A catalyst is a substance that increases the rate of a reaction without being consumed. Most catalysts work by lowering the activation energy.

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Fig. 32 - Decreased activation energy with catalyst.

When the activation energy is lowered, more collisions will have sufficient relative kinetic energy to create a reaction. As a result, more reactions will occur and there will be an increase in the overall reaction rate. A catalyst does not alter the equilibrium constant as it increases both the forward and reverse reactions. Students should be aware of the two different types of catalysts:

Heterogeneous catalyst: In a different phase than the reactants or products. These catalysts are usually solids, and the reactants and products are usually gases or liquids.

Homogeneous catalyst: In the same phase as the reactants and products. These catalysts are usually in the liquid or gas phase. Aqueous acids and bases usually act as homogeneous catalysts.

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Equilibrium

Many chemical reactions are reversible, and the forward and backward reactions can occur at the same time. The conditions when the rate of the forward reaction is equal to the rate of the backward reaction are called chemical equilibrium. At chemical equilibrium, there is no change in the concentration of products or reactants. A constant K characterises the state of equilibrium:

aA + bB ←→ cC + dD where a, b, c, and d are the corresponding stoichiometric coefficients.

K = [C]c [D]d [A]a [B]b

Example: Calculating K from Known Equilibrium Amounts

1. Write the equilibrium expression for the reaction.

2. Determine the molar concentrations of each the species involved. 3. Then substitute into the equilibrium expression and solve for K. Calculate the value of the equilibrium constant, Kc , for the given system, if 0.20 moles of CO2, 0.08 moles of H2, 0.01 moles of CO, and 0.01 moles of H2O vapor were present in a 2.00 L reaction vessel were present at equilibrium.

CO2(g) + H2(g) ←→ CO(g) + H2O(g) •

Write the equilibrium expression for the reaction system.

Kc = [CO][ H2O] / [CO2][ H2]

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•

Since Kc is being determined, check to see if the given equilibrium amounts are expressed in moles per liter (molarity).  In this example, they are not; conversion of each is required. [CO2] = 0.20 mol CO2/2.00 L = 0.10 M  [H2] = 0.08 mol H2/2.00 L = 0.04 M

[CO] = 0.01 mol CO/2.00 L = 0.005 M 

[H2O] = 0.01 mol H2O/2.00 L = 0.005 M

• Substitute each concentration into the equilibrium expression and calculate the value of the equilibrium constant. Kc = [0.005][0.005] / [0.10][0.04] Kc = 0.00625 or 6.25 x 10-3

Fig. 33 - Equilibrium expression.

The equilibrium constant has a given value at a given temperature. The value of K changes if the temperature changes. Also, if we change the concentration of A, B, C, or D, the system will evolve in such a way as to reestablish the value of K. Remember that when finding K, we use products divided by reactants—products over reactants.

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The Reaction Quotient

For reactions that are not in equilibrium, the following equation is used:

Q is the reaction quotient, which is used to predict the direction in which a reaction will proceed. Since reactions always move toward equilibrium, Q will change toward K at all times. Q and K can thus be compared for a reaction to determine which direction the reaction will occur.

The three following conditions should be memorised for the GAMSAT: 1. If Q is equal to K, the reaction is at equilibrium 2. If Q is greater than K, then the ratio of products to reactants is greater than when at equilibrium. This shows that the reverse reaction rate is favoured and will be greater than the forward rate. 3. If Q is less than K, then the ratio of products to reactants is less than when at equilibrium. This shows that the forward reaction rate is favoured and will be greater than the reverse reaction rate.

Fig. 34 - Reaction quotient conditions.

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Le Chatelier’s Principle

Le Chatelier’s principle can be applied to systems at equilibrium. The principle states that when a system at equilibrium is stressed, the system will shift in the direction that will reduce stress.

The following types of stresses obey Le Chatelier’s principle: - Addition or removal of a product or reactant - Heating or cooling the system - Altering the pressure of the system

Table 8 - Le Chatelier’s principle.

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The following reaction is known as the Haber process. This process is use for the manufacture of ammonia from nitrogen and hydrogen. An iron catalyst is used with temperatures between 400-450 degrees Celsius, and at a pressure of 200 atm.

N2(g) + 3H2(g)  2NH3(g) + Heat

Fig. 35 - The Haber Process.

Questions involving this process and Le Chatelier’s principle have been previously examined in the GAMSAT. We will now examine what can happen as a result of different stresses on the Haber process. The process is an exothermic reaction, so it creates heat (heat on right hand side of reaction). If the temperature is raised, the reaction will shift to the left (reverse reaction) and become endothermic. This will cause the extra heat on the right side of the reaction to be decreased. As a result of this shift, there will be a decrease in NH3 and an increase in both N2 and H2.

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Fig. 36 - Effect of temperature on product formation.

A container holds N2, H2, and NH3 gas at equilibrium. If N2 gas is added, the system will attempt to compensate for this increased concentration of N2 by reducing the partial pressure of N2 with the forward reaction. This forward reaction will decrease both the partial pressure of N2 and also H2. NH3 and heat will be increased as a result (forward reaction).

Fig. 37 - Changes in equilibrium concentration with addition of extra NH2.

If the temperature of a container is constant and size of the container is reduced, total pressure increases. There are 4 gas molecules on the left side of the reaction and 2 gas molecules on the right. The equilibrium will shift to the side with the least number of gas molecules to try and bring to pressure closer to equilibrium. In this case, there will be an increase in NH3 and heat will be generated, as the equilibrium will shift to the right side. www.acegamsat.com

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For this to be an economic viable procedure scientists must choose a temperature that is low enough for reasonable yield, but high enough for a fast reaction. The pressure must be low enough to not need expensive reinforced equipment, but high enough to give a reasonable yield.

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Thermodynamics Video: First law of Thermodynamics Video: Enthalpy Video: Heat of formation Video: Hess’s law and reaction enthalpy change Video: Hess’s law example Video: Gibbs free energy and spontaneity Video: Gibbs free energy example Thermodynamics is the study of energy and its relationship to chemical systems. To solve problems in thermodynamics, we need to define a system and its surroundings. The system is the object that experiences a thermodynamic transformation. Any part of the universe that is in direct contact with the system is its surroundings. There are three types of systems: open, closed, and isolated.

Fig. 38 - Thermodynamic systems

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1. Open systems exchange both mass and heat/energy with their surroundings. 2. Closed systems exchange heat/energy but not mass. 3. Isolated systems do not exchange heat/energy or mass.

The First Law of Thermodynamics

Students need to be familiar with the following concepts to understand the first law of thermodynamics. These concepts include heat, work, and internal energy. In chemistry, we look at work as any energy transfer that is not heat. Internal energy is the average total mechanical energy (kinetic + potential) of the particles that make up the system. Heat is the transfer of energy from a warmer body to a cooler body. There are three forms of heat transfer that are usually tested in the GAMSAT—conduction, convection, and radiation.

Fig. 39 - Convection, conduction, and radiation.

Conduction is thermal energy transfer via molecular collisions, which requires direct physical contact. The process involves higher energy molecules of one system transferring

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some of their energy to the lower energy molecules of the other system via molecular collisions. Convection is thermal energy transfer via fluid movements. Warm fluid moves in the direction of cooler fluid due to differences in pressure or density. Radiation is thermal energy transfer via electromagnetic waves. For example, when metal is heated and glowing red, it radiates visible electromagnetic waves. This first law states that energy of the system and its surroundings is always conserved. This means that any change to a system must equal the heat flow in the system, plus the work done on the system.

ΔU = Q + W ΔU = Change in internal energy

Q = Quantity of heat supplied to the system by its surroundings W = net work done by the system

This law considers all net energy transfers to the system as positive and all net energy transfers from the system as negative. These important points regarding internal energy should be memorised: - Work done on system = + W - Work done by system = - W - Heat added to system = + Q - Heat given off by system = - Q

The Second Law of Thermodynamics

The second law of thermodynamics demonstrates that heat cannot be completely changed to work in a cycle-like process. This law allows for the determination of the preferred direction of a given transformation – for example, we know that heat flows from a hot source to a cold source.

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Fig. 40 - Second law of thermodynamics.

Thermodynamic functions

In terms of the GAMSAT, students should be familiar with the following seven state functions and their SI units:

1. Internal energy

U - joule

2. Temperature

T - kelvin

3. Pressure

P - pascal

4. Volume

V – m3

5. Enthalpy

H – kJ mol-1 (kilojoules per mole)

6. Entropy

S –J K-1 mol-1 (joules per kelvin per mole)

7. Gibbs energy

G – kJ mol-1 (kilojoules per mole)

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Internal Energy

The questions in the GAMSAT will not directly ask about internal energy; however, it is important to have a general idea about this concept. Internal energy is the collective energy of molecules. For a reaction in a system with constant volume, there is no work of any kind and the change in internal energy is equal to the heat.

ΔU = q

Temperature

Two bodies in thermal equilibrium share a thermodynamic property called temperature. An increase in thermal energy results in an increase in temperature. For a fluid, temperature is directly proportional to the translational kinetic energy of its molecules. For a gas, the greater the random translational energy per mole of gas, the greater the temperature. In the GAMSAT, two measurement systems for temperature can be used: degrees Celsius (C°) and Kelvin.

Students should know the following two points: 1. To convert degree Celsius to Kelvin, simply add 273.15. 2. An increase of 1 K is equivalent to an increase of 1°C.

Example:

If a pot of water has a temperature of 30 °C its temperature in kelvin will be 273.15 + 30 = 303.15. Now if we raise the temperature of the pot by 1°C the final temperature will be 31°C or 304.15 kelvin. Notice the equivalent increase of 1 kelvin when the temperature is increase by 1°C.

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Pressure

For the GAMSAT, students should understand that the greater the random translational kinetic energy of gas molecules per volume, the greater the pressure.

Enthalpy

Enthalpy is a man-made property and is described as an equation:

H = U + PV The standard enthalpy of formation, ΔH°f, is the change in enthalpy for a reaction that creates one mole of that compound from its raw elements. In many reactions, enthalpy is used to approximate heat. The change in enthalpy from reactants to products can be referred to as the heat of reaction:

ΔH°reaction = ΔHf° products - ΔHf° reactants Hess’ law states that when adding reactions, their enthalpies can be added because enthalpy is a state function. Example: Given a simple chemical equation with the variables A, B, and C representing different compounds:

A+B ⇋ C and the standard enthalpy of formation values: • • •

ΔHfo[A] = 433 KJ/mol ΔHfo[B] = -256 KJ/mol ΔHfo[C] = 523 KJ/mol

the equation for the standard enthalpy change of formation is:

ΔHreactiono = ΔHfo[C] - (ΔHfo[A] + ΔHfo[B])

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ΔHreactiono = (1 mol)(523 kJ/mol) - ((1 mol)(433 kJ/mol) + (1 mol)(-256 kJ/mol)\) Because there is one mole each of A, B, and C, the standard enthalpy of formation of each reactant and product is multiplied by 1 mole, which eliminates the mol denominator:

ΔHreactiono = 346 kJ

The result is 346 kJ, which is the standard enthalpy change of formation for the creation of variable C. Note: A system that releases heat to the surroundings, an exothermic reaction, has a negative ΔH by convention, because the enthalpy of the products is lower than the enthalpy of the reactants of the system.

A system of reactants that absorbs heat from the surroundings in an endothermic reaction has a positive ΔH, because the enthalpy of the products is higher than the enthalpy of the reactants of the system.

Entropy

Entropy (S) is also a state function that measures the degree of disorder in a system. For example, the entropy of water is higher than the entropy of ice. This is because ice is an organised crystalline structure, and water is a fluid with randomly moving molecules. Gas has higher entropy than water, since the gas molecules exhibit an increased random motion of movement.

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For the GAMSAT, students should remember that entropy increases as we go from a solid to a liquid to a gas.

Gibbs Free Energy

The Gibbs free energy equation can be used to determine if a reaction is spontaneous or not, or if the reaction is at equilibrium. It is crucial that students learn the following as it appears in almost every GAMSAT examination.

ΔG = ΔH - TΔS Where:

H is the enthalpy of the system (kJ mol-1 – kilojoules per mole) T is the temperature (kelvin). S is the entropy of the system (J K-1 mol-1 - joules per kelvin per mole)

Example: Calculate ∆G at 310 K for the following reaction:

2NO(g)+O2(g)→2NO2(g Given:

ΔH = - 150 kJ and ΔS – 180 JK-1

ΔS = - 180 j K-1 = - 0.180 kJ K-1 ∆G = ΔH - TΔS

∆G = - 150 kJ - (310 K)(-0.180 kJ K-1) ∆G = - 150 kJ - (310 x -0.180 kJ) ∆G = - 150 kJ + 55.8 kJ

∆G = - 94.2 kJ

Since the ∆G is negative, this reaction will be spontaneous.

A reaction carried out at constant pressure is spontaneous if ΔG < 0.

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A reaction carried out at constant pressure is NOT spontaneous if ΔG > 0.

ΔG = 0 if the reaction is spontaneous in both directions and thus in a state of equilibrium.

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Solutions

Video: Solubility and intermolecular forces

A solution is defined as a homogeneous mixture of two or more compounds in a single phase. The solvent is the compound that is in greater abundance and the solute is the compound in lesser abundance.

Fig. 42 - Solute, solvent, and solutions.

For the GAMSAT, students should remember the general rule of “like dissolves like.” This means that polar solvents dissolve polar solutes and nonpolar solvents dissolve nonpolar solutes. When ionic compounds dissolve, they break apart into their respective cations and anions. This process is called solvation. Water is a good solvent for ionic substances. The water surrounds and attaches to one side of an ionic compound, 3overcomes the strong ionic bond, and breaks the compound apart. This process is termed hydration. If something becomes hydrated, it is said to be in an aqueous phase. When an ionic compound breaks down and forms ions in aqueous solution, the solution is able to conduct electricity. A compound that forms ions in aqueous solution is termed an electrolyte.

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Fig. 43 - Different types of electrolytes.

Units of Concentration

Molarity (M) is the number of moles of the solute divided by the volume of the solution. Units are mol/L. Molality (m) is the number of moles of solute divided by kilograms of solvent. Units are mol/kg. The mole fraction is the number of moles of a compound divided by the total moles of all species in the solution. It is a ratio, so it has no units.

Example:

If a mixture of gases contains 7.50 g H2, 3.25 g O2, and 5.55 g N2, what is the mole fraction of oxygen?

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n1 nt

= X1 (Remember: mole fraction is a ratio of moles, not grams)

First we will find the moles of each gas. 7.5 g H2 x

1 mol H2 2.02 g H2

3.25 g O2 x

5.55 g N2 x

= 3.71 mol H2

1 mol O2 32.0 g O2 1 mol N2 28.0 g N2

= 0.102 mol O2

= 0.198 mol N2

nt = mol H2 + mol O2 + mol N2 nt = 3.71 mol H2 + 0.102 mol O2 + 0.198 mol N2 nt = 4.01 mole fraction of oxygen

0.102 mol O2 4.01 mol total

= 0.0254

Vapor Pressure

Vapor pressure is the pressure of a vapor in thermodynamic equilibrium with its condensed phases in a closed container. All liquids and solids have a tendency to evaporate into a gaseous form, and all gases have a tendency to condense back to their liquid or solid form. The vapor pressure of the components in a solution behave as follows:

Pv = Xv(Pv)pure Pv is the vapor pressure of component v in equilibrium with the solution Xv is the mole fraction of component v in the liquid (Pv)pure is the vapor pressure of pure component v at the same temperature www.acegamsat.com

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Looking at this equation, the vapor pressure of any component of a mixture is lowered by the presence of other components in the mixture. Dissolving a nonvolatile solute in a solvent will produce a vapor pressure of solution that is lower than that of the pure solvent. The extent of this lowering can be determined by the mole fraction of the solvent in solution (X).

P = PoX P is the vapor pressure of solution Po is the vapor pressure of pure solvent at same temperature as P X is the mole fraction of the solvent in solution

Example:

4 moles of cordial are added to a pitcher containing 1 liter of water on at 25o C. The vapor pressure of water alone is 23.8 mm Hg at 25 o C. What is the new vapor pressure of cordial? The equation, P = PoX can be rewritten as:

Pcordial = XH2O(PH2O)pure PH2O = 23.8 mm Hg

To solve for the mole fraction, you must first convert the 1 L of water into moles: 1 L = 1000 mL = 1000 g

Knowing this, you can convert the mass of water (1000 g) into moles: 1000 g / 18.02 g (molar mass of water) = 55.49 moles H2O. Solve for the mole fraction, XH2O. XH2O. = moles H2O / total moles

= 55.49 moles / (55.49 moles + 4 moles)

= 55.49 / 59.49 = 0.93

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Finally, apply the equation: Pcordial = XH2O. x 23.8mm Hg

Pcordial = 0.93 x 23.88 mm Hg = 22.21 mm Hg

Solubility

Solubility is defined as the limit of solute that can be dissolved in a given amount of solvent at equilibrium. On the GAMSAT, the solute will usually be a salt and the solvent will most likely be water. A salt will dissolve into its respective ions. These ions can reattach and form the salt again in a process termed precipitation. When the rate of dissolution and precipitation are equal, the solution is saturated.

Fig. 44 - Solubility.

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The solubility product (Ksp) is used to create an equilibrium expression. Ksp is set to equal the products over reactants raised to the power of their coefficients in the balanced equation. Liquids and solids are left out of the expression.

Example: Calculate the solubility product constant for lead(II) chloride, if 50.0 mL of a saturated solution of lead(II) chloride was found to contain 0.2207 g of lead(II) chloride dissolved in it. 1. Write the equation for the dissolving of lead(II) chloride and the equilibrium expression for the dissolving process. Note: PBCl2 is not included in the expression as it is a solid.

PbCl2(s) → Pb2+(aq) + 2 Cl-(aq) Ksp = [Pb2+][Cl-]2 2. Convert the amount of dissolved lead(II) chloride into moles per liter. (0.2207 g PbCl2)(1/50.0 mL solution)(1000 mL/1 L)(1 mol PbCl2/278.1 g PbCl2) = 0.0159 M PbCl2 3.

Create an "ICE" table—(initial, change, and equilibrium).

4. Substitute the equilibrium concentrations into the equilibrium expression and solve for Ksp.

Ksp = [0.0159][0.0318]2 = 1.61 x 10-5

Students should know the difference between solubility and solubility product as these can easily be confused. Solubility product (Ksp) is a constant found in a textbook, and solubility is the maximum number of moles of solute that can dissolve in solution.

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Remember that as the temperature increases, the solubility of salts generally increases. The opposite is seen with gases, as the solubility decreases as temperature increases.

Phase Diagrams

Temperature and pressure are important properties that determine the phase of a particular substance. Understanding phase diagrams is important as they arise in almost every GAMSAT examination. A phase diagram is used to show the phases of a substance at different temperatures and pressures. Each section of a phase diagram is representative of a different phase. The lines marking the boundaries represent the equilibrium between the different phases.

Fig. 45 - Phase diagram.

The points along the line connecting points C and A in the phase diagram represent all combinations of temperature and pressure at which the solid is in equilibrium with the gas. At these temperatures and pressures, the rate at which the solid sublimes to form a gas is equal to the rate at which the gas condenses to form a solid. www.acegamsat.com

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CA line: rate at which the solid sublimes to gas is equal to the rate at which gas condenses to form a solid The solid line between points A and B is identical to the plot of temperature dependence of the vapor pressure of the liquid. It contains all the combinations of temperature and pressure at which the liquid boils. At every point along this line, the liquid boils to form a gas and the gas condenses to form a liquid at the same rate. AB line: rate at which the liquid boils to form a gas is equal to the rate at which gas condenses to form a liquid The solid line between points A and D contains the combinations of temperature and pressure at which the solid and liquid are in equilibrium. At every point along this line, the solid melts at the same rate at which the liquid freezes. AD line: rate at which solid melts to form a liquid is equal to the rate at which liquid freezes to form a solid

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Acids and Bases It is important to study the following 12 short videos to refine an understanding of acids and bases. These videos are found here:

12 videos for acids and bases

Definitions of Acids

For the GAMSAT, students should be familiar with the following definitions of acids— Bronsted and Lowry, Arrhenius, and Lewis. 1. The Bronsted and Lowry definition for acids and bases is as follows: An acid is anything that donates a proton. A base is anything that accepts a proton.

Fig. 46 - Demonstration of Bronsted and Lowry’s definition of acids and bases.

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2. The Arrhenius definition states that: An Arrhenius acid is a substance that produces hydrogen ions in aqueous solution.

HCl + H2O → H3O+ + ClAn Arrhenius base is a substance that produces hydroxide ions in aqueous solution.

NH3 + H2O ⇌ NH4+ OH3. The Lewis definition states that: An acid is anything that accepts a pair of electrons, and a base is anything that donates a pair of electrons.

Fig. 47 - Demonstration of the Lewis definition of acids and bases.

For ease in the GAMSAT, just think of an acid as H+ and a base as OH-.

pH

pH is used to measure the hydrogen ion concentration. If measuring the hydrogen ion concentration in moles per liter or molarity (M), pH is given by:

pH = -log[H+] Where the brackets [ ] indicate concentration. H+

The pH scale generally runs from 0 -14, but any pH value is possible since any concentration is possible.

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Fig. 48 - The pH scale.

At 25°C: A pH of 7 is neutral A pH less than 7 is acidic A pH greater than 7 is basic

It is essential that students understand the pH scale for the GAMSAT.

Each point of the pH scale corresponds to a tenfold difference in H+ concentration. For example, an acid with a pH of 3 produces 10 times as many hydrogen ions as an acid with a pH of 4, and 100 times as many hydrogen ions as an acid with a pH of 5.

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Fig. 49 - Concentration of hydrogen ions at different pH levels.

For the GAMSAT, students must be able to identify conjugate acids and bases. The following hypothetical acid-base reaction in aqueous solution will be used to highlight the different conjugates.

HA + H2O → A- + H3O+ HA is the acid (donates proton) and water is the base (accepts proton). The reactants are referred to as the acid and base, and the products as the conjugate acid and conjugate base.

Fig. 50 - Conjugate acid-base pairs. www.acegamsat.com

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The acid will have its conjugate base, and the base will have its conjugate acid. A- is the conjugate base of the acid HA. H3O+ is the conjugate acid of the base H2O. Remember that the stronger the acid, the weaker its conjugate base, and the stronger the base, the weaker its conjugate acid. For the purpose of the GAMSAT, we assume that a strong base or acid completely dissociates in water. In acid and base reactions, assume H3O+ and H+ are the same thing. Some acids are called polyprotic acids. These are acids that can donate more than one proton. For example: H2SO4 → HSO42- + 2H+

Fig. 51 - Examples of polyprotic acids.

Equilibrium Constants for Acid-Base Reactions

Pure water can ionise with itself to form hydronium and hydroxide ions:

H2O + H2O ⇄ H3O+ + OH− The equilibrium constant for this reaction is Kw.

Kw = [H+][OH-] = 1.0 x 10-14 An acid dissociation constant (Ka) is a quantitative measure of strength of an acid in solution. The dissociation constant is commonly written as a quotient of the equilibrium concentrations (in mol/L). The Ka is used in the Henderson-Hasselbalch equation when determining the pH of an acid-base reaction at equilibrium. www.acegamsat.com

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Fig. 52 - The Henderson-Hasselbalch Equation There is a Kb value that corresponds to every Ka value. Kb is the equilibrium constant for the reaction of the conjugate base with water.

Table 9 - Ka and Kb value

The following equations should be memorised for the GAMSAT: www.acegamsat.com

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Ka Kb = Kw pKa + pKb = pKw pKa + pKb = 14 pKa = -log Ka pKb = -log Kb

Logarithms

Students must have a basic understanding of logarithms to solve pH and rate-law problems in the GAMSAT. Students must memorise the following rules:

Table 10 - Logarithms

Logarithm product rule

The logarithm of the multiplication of x and y is the sum of logarithm of x and logarithm of y.

logb(x ∙ y) = logb(x) + logb(y) For example:

log10(3 ∙ 7) = log10(3) + log10(7)

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Logarithm quotient rule

The logarithm of the division of x and y is the difference of logarithm of x and logarithm of y.

logb(x / y) = logb(x) - logb(y) For example:

log10(3 / 7) = log10(3) - log10(7)

Logarithm power rule

The logarithm of x raised to the power of y is y times the logarithm of x.

logb(x y) = y ∙ logb(x) For example:

log10(28) = 8∙ log10(2)

Logarithm of 1

The base b logarithm of one is zero:

logb(1) = 0

For example, the base two logarithm of one is zero:

log2(1) = 0

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10 power log rule

10log10n = n 10log105 = 5

Titrations

The following videos (Buffers and titrations) are important for the GAMSAT section on buffers and titrations. A titration is the drop-by-drop mixing of a base and an acid. The purpose of a titration is usually to determine the concentration of a given sample of base or acid, which is reacted with an equivalent amount of strong base or acid of known concentration (titrant).

Fig. 53 - Titration setup.

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The changing pH of the unknown as the basic or acidic titrant is added is represented graphically as a sigmoidal curve.

Fig. 54 - The equivalence point.

The end point or equivalence point is reached when a certain amount of titrant has been added. The end point is detected using an indicator that changes colour when this point is reached.

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The following should be memorised for the GAMSAT: Strong acids and strong bases have an equivalence point at a pH of 7. The acid can be added to the base or the base can be added to the acid. This can be seen using NaOH and HCL in the following figures. 1. Strong acid added to strong base

2. Strong base added to strong acid

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A strong acid and weak base have an equivalence point of less than 7. HCl (strong acid) and ammonia (weak base) are used in the following figures.

3. Adding strong acid to weak base

4. Adding weak base to strong acid

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A weak acid and strong base have an equivalence point of more than 7. Ethanoic acid (weak acid) and NaOH (strong base) are used in the following figures.

5. Adding weak acid to strong base.

6. Adding strong base to weak acid

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The following is a summary of all the acid/base titrations:

Fig. 55 - Acid-base titrations summary.

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Buffers A buffer is defined as a solution that resists change in pH when a small amount of base or acid is added. A buffer solution is composed of a mixture of a weak base and its salt or a weak acid and its salt.

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Electrochemistry Video: Introduction to redox reactions Video: Redox reaction example Video: Introduction to Galvanic cells Video: Electrodes and voltage of Galvanic cell Video: Standard cell potentials Video: Introduction to electrolysis

Oxidation-Reduction

In an oxidation-reduction reaction (redox reaction), electrons are transferred from one ionic species to another.

Fig. 56 - Electron transfer in redox reactions.

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The reductant is the reducing agent, which is oxidised (loses electrons). The oxidant is the oxidising agent, which is reduced (gains electrons). Redox reactions are covered in almost every GAMSAT exam. Students need to know that the species that loses electrons is oxidised, and the species that gains electrons is reduced.

Potentials

Electrons are transferred in a redox reaction. Due to their charge, an electric potential (E) is created. The potentials for the oxidation and reduction components are shown table.

Table 11 - Standard reduction potential table.

Each component is called a half reaction. Two half reactions must occur—reduction and oxidation. The reverse of a reduction half reaction is an oxidation half reaction.

The standard reduction potential table shows the half reactions all as reduction potentials. In order to find the oxidation potential, simply reverse the half equation and the sign of the reduction potential. This table does not need to be memorised, as it will be

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provided in the exam. Students should just know to reverse the half equation and the sign of the reduction potential for the oxidation reaction.

Galvanic Cells

A galvanic cell turns chemical energy into electrical energy. Specifically, it uses a spontaneous redox reaction to produce electricity. The following spontaneous reaction (redox equation) can be used to create a galvanic cell (battery):

Zn(s) + Cu+2(aq)  Zn2+(aq) + Cu(s) According to the galvanic cell, oxidation and reduction parts are separated and carried in two separate containers. Zinc is placed in a ZnSO4 solution and copper is placed in a CuSO4 solution. The metallic rods in a galvanic cell are known as electrodes. Zn metal dissolves in zinc sulfate solution by freeing electrons, which climb up the electrode and travel through the external wire, go through the voltmeter to register the voltage, enter the copper sulfate solution through the copper electrode, and then reduce the Cu+2 ions to copper metal. Therefore, the oxidation occurs at the Zn electrode and reduction at the Cu electrode. The Zn electrode is known as an anode. The Cu electrode is the cathode. The anode is the electrode where oxidation (loss of electrons) takes place, and the cathode is where reduction (gain of electrons) takes place. Each reaction is one half of the entire cell reaction and hence is called a half-cell reaction. The half-cell reactions are:

Anode reaction (Zn electrode): Zn(s)  Zn+2 (aq) + 2eCathode reaction (Cu electrode): Cu+2(aq) + 2e-  Cu(s)

Finding the potential of the cell:

As mentioned previously, the cell potential, Ecell, is calculated from the half-cell reduction potentials. The following formula is used to find the cell potential:

Ecell = Ecathode - Eanode

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Based on the values for the standard reduction potentials for the two half-cells in the above equations, (–0.76 V for zinc anode; +0.34 V for copper cathode) the standard cell potential, E°cell, for the galvanic cell would be:

E°cell = +0.34 V – (–0.76 V) = +1.10 V

Fig. 57 - Galvanic cell.

Note: A salt bridge is needed to prevent the buildup of positive charges in the anode. The salt bridge contains an inert electrolyte solution such as KCl, whose ions do not react with other ions.

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Electrolytic cell

In a galvanic cell, a spontaneous redox reaction is used to produce a current, whereas in an electrolytic cell, current is used to drive a non-spontaneous reaction.

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Appendix Atoms and Molecules Video: Elements and Atoms Video: Atomic number, Mass number, and Isotopes Video: The mole and Avogadro's number Video: Empirical, molecular, and structural formulas Video: Ionic, Covalent, and Metallic bonds Video: Naming Ionic and Covalent Molecular Compounds Video: Chemical Reactions Introduction

Molecules and Interactions Video: Intermolecular and Intramolecular forces

Empirical Formula Video: Empirical formula from mass composition Gases Video: Ideal Gas Equation Video: Ideal gas equation example 1 Video: Ideal gas equation example 2 Video: Ideal gas equation example 3 Video: Ideal gas equation example 4 Video: Partial pressure example Video: Vapor pressure example

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Kinetics and Chemical Equilibrium Video: Introduction to Kinetics Video: Collision Theory Video: Elementary rate laws Video: Rate Law and Reaction order Video: Mechanisms and the rate-determining step Video: Catalysts Video: Reactions in equilibrium Video: Heterogeneous equilibrium Video: Le Chatelier’s principle

Thermodynamics Video: First law of Thermodynamics Video: Enthalpy Video: Heat of formation Video: Hess’s law and reaction enthalpy change Video: Hess’s law example Video: Gibbs free energy and spontaneity Video: Gibbs free energy example Solutions Video: Solubility and intermolecular forces Acids and Bases 12 videos for acids and bases

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Titrations Buffers and titrations

Electrochemistry Video: Introduction to redox reactions Video: Redox reaction example Video: Introduction to Galvanic cells Video: Electrodes and voltage of Galvanic cell Video: Standard cell potentials Video: Introduction to electrolysis

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