Physical Chemistry A2

Question 1

Enthalpy change.

Answer 1

1. ionic substances are made of charged particles which attract each other electrostatically.
2. lattice enthalpy - the enthalpy change when one mole of solid ionic lattice forms from its gaseous ions under standard conditions.
3. standard enthalpy of formation - the enthalpy chane when one mole of a compound is formed from its constituent elements in their standard states.
4. standard enthalpy of atomisation - the enthalpy change when one mole of gaseous atoms is formed from its elements in its standard state.
5. ionisations energies (1st and 2nd) - the enthalpy change when one mole of gaseous ions is formed.
6. electron affinities (1st and 2nd) - the enthalpy change when one mole of a gaseous ion is formed.

Question 2

Born - Haber cycles for dissolving.

Answer 2

1. there is an energy change when we dissolve a solid called the standard enthalpy change of solution.
2. this process can be broken down into two steps; lattice breakdown and hydration.
3. the size and charge of an ion affects the magnitude of the enthalpy change of the lattice enthalpy and the standard enthalpy change of hydration - the smaller and more highly charged the ions are the greater the magnitudes.
4. we can use Born-Haber cycles relating to solution and hydration.

Question 3

Predicting enthalpy changes.

Answer 3

1. enthalpy changes alone are not enough for predicting whether reactions will happen.
2. entropy must also be considered, it is a measure of the disorder and total entropy always increases.
3. ΔS(system) is given by: ΔS(system) = ΣS⦵(products) - ΣS⦵(reactants)
4. calculating ΔS(surroundings): ΔS(surroundings) = -ΔH / T
5. calculating ΔS(total): ΔS(total) = ΔS(system) + ΔS(surroundings)
6. we can predict entropy changes to the system based on increases or decreases in the number of ways we can organise the system and the energy in it (disorder).

Question 4

Free energy.

Answer 4

1. the Gibbs’ equation is given by: ΔG = ΔH - TΔS(system).
2. We can rearrange the Gibbs’ equation and show: - (ΔG / T) = ΔS(total)
3. So if ΔG is negative the reaction will be feasible.
4. When ΔG = 0 the system is in an equilibrium.
5. the Gibbs’ equation doesn’t tell us everything, kinetic factors like the activation energy and rate of reaction mean some reactions don’t happen spontaneously.

Question 5

Rate equations.

Answer 5

1. A rate of reaction tells us hoe fast a reaction is happening.
2. It is usually calculated by: rate of reaction = change in concentration of reactant or product / change in time.
3. the order of a reagent tells us how its concentration affects the rate of reaction.
4. we can express the relationship between the concentration of reagents and the rate of reaction with the rate equation: rate = k[A][B][C]*
5. the overall order of a reaction given by the sum of all te orders of the reagents.
6. we can use experimental data where we vary concentrations and measure the rate to find the orders of reagents.

Question 6

Rate constant.

Answer 6

1. we can qualitatively explain how temperature affects the rate of reaction and rate constant.
2. we have met the arrhenius equation which shows the exponential relationship between temperature and rate constant: k = Ae -(Ea / RT).
3. we can determine A and Ea graphically using: In(k) = In(A) -(Ea / RT).

Question 7

Concentration time graphs and half-life.

Answer 7

1. we can find the rate of reaction from a concentration-time graph by measuring the gradient.
2. if the concentration-time graph is a straight line, the reactant that is changing concentration is zero order.
3. if the concentration-time graph is a curved line, the reactant that is changing concentration is first order.
4. the half-life is the time taken for the concentration to half and is constant for the first order reactions.
5. we can determine the rate constant, k, from the half-life of a first order reaction using: k = (ln2 / t1/2).

Question 8

Clock reactions.

Answer 8

1. we know a clock reaction is a reaction with an obvious end point so we can measure the reaction time.
2. we’ve met the iodine clock reaction and understand how to carry this out.
3. we can use 1/time as a measure of the rate.

Question 9

Rate-concentration graphs.

Answer 9

1. we can recognise that rate-concentration graphs for reactions of order: 0, 1 and 2.
2. we can find the rate contant from a first order graph by taking the gradient.
3. we can use data from second order graphs and the rate equation to find the rate constant.

Question 10

Rate determining steps.

Answer 10

1. we know a reaction mechanism is the steps that make up our overall reaction.
2. we know an intermediate is a species created in the mechanism which is then used up again.
3. we know that rate-determining step is the slowest step in the reaction mechanism and controls the rate.
4. we can predict the rate-determining step.
5. we can suggest possible mechanisma consistent with the rate equation and overall reaction equation.

Question 11

Equilibrium and Kp.

Answer 11

1. partial pressures are the pressure a particular species of gas would exert on its own.
2. the sum of the partial pressures of all gases in a mixture gives the total pressure.
3. the amount of gas in a mixture is quantified by its mole fraction given by: XA = number of moles of substance A / total number of moles of all substances.
4. you can find the partial pressure of a gas in a mixture using: p(A) = XA x total pressure.
5. we can use a new equilibrium constant for equilibriums involving gases, Kp: Kp = p(C)^cp(D)^d / p(A)ap(B)b
6. we don’t include solids or pure liquids in the equation for Kp.

Question 12

Significance of k.

Answer 12

1. equilibrium constants tell us how far an equilibrium reaction has progressed, with high K meaning products are favoured and low K meaning reactants are favoured.
2. temperature changes push the reaction in the direction that will counteract the change.
3. K values are only affected by temperature, if we change the concentration or pressure, the equilibrium position will move to keep K constant.
4. catalysts do not affect the amount of reactants or products present, they just increase the rate.

Question 13

Half-cells.

Answer 13

1. if we can harness the flow of electrons in a redox equation we have a source of electrical energy.
2. Half-cells can be made with: metal in a solution of aqueous ions, gas and aqueous ions or two different aqueous ions.
3. standard electrode potentials tell us about a half cells tendency to donate or accept electrons.
4. the hydrogen half-cell is used as a reference with a standard cell potential of 0V so we can measure other standard cell potentials.

Question 14

Full cells.

Answer 14

1. constructing whole cells allows for the flow of electrons which generates electricity.
2. the electrochemical series is a list of all the half-cells standard electrode potentials.
3. standard cell potentials are the difference between the standard electrode potentials and tell us the extent to which electrons flow; E⦵cell = E⦵positive terminal - E⦵negative terminal
4. cell diagrams are a neater way to draw out cells.
5. as a cell generates electricity when a chemical reaction happens and we can work this reaction out using the electrochemical series.
6. the higher the cell potential the more thermodynamically feasible the reaction is.
7. the conclusions we draw from cell potentials can be limited as they only apply if we have standard conditions.

Question 15

Fuel cells.

Answer 15

1. we’ve seen that chemical cells are used for disposable batteries and rechargeable batteries.
2. alkaline hydrogen fuel cells combine hydrogen and oxygen which they receive from external sources and the reaction generates electricity.
3. acidic hydrogen fuel cells combine hydrogen and oxygen which they receive from external sources and the reaction generates electricity.

Question 16

Brownsted-lowry acids and bases.

Answer 16

1. the bronsted-lowry definition of acids and bases describes acids as proton donators and bases as proton acceptors.
2. conjugate pairs are species in an acid-base reaction which become each other by gaining or losing a proton.
3. Ionic equations provide a more simple way to look at acid-base reactions.
4. mono, bi and tribasic acids release one, two and three protons per molecule respectively, the polybasic acids release each proton in separate steps.

Question 17

Acid-base reactions

Answer 17

1. neutralisation isthe reaction of an acid and a base to form a salt and water.
2. acids react with carbonates to make a salt, carbon dioxide and water.
3. acids react with bases and alkalis to make a salt and water.
4. acids react with metals to make a salt, and hydrogen gas but this is a redox reaction rather than an acid base reaction.

Question 18

pH of strong and weak acids.

Answer 18

1. concentration of H+ ions gives a measure of the acidity of a solution.
2. because of the vast scale of possible [H+] we use pH to measure it; pH = log10([H+]).
3. strong acids dissociate fully so we may use [H+] = [acid] to calculate pH.
4. weak acids do not fully dissociate so we found the formula for [H+] was [H+(aq)] = /Ka x [HA(aq)].
5. we can use pH to kind Ka.
6. pKa is usually more useful than Ka because of the scale of possible values.

Question 19

Ionisation of water.

Answer 19

1. we’ve seen that the water is amphoteric which means that it can act as both an acid and a base.
2. we’ve looked at the dissociation of water and the ionic product of water. Kw = [OH-(aq)] x [H+(aq)].
3. we know Kw is constant at constant temperature and so can be used to find [H+] if we know [OH-].
4. we know that the strength of a base refers to how much it dissociates in water.
5. we can use the expression for Kw to calculate the pH for a strong base.

Question 20

Strong and weak acids and Ka.

Answer 20

1. Dissociation is when an acid releases its protons.
2. Strong acids dissociate 100% in water.
3. Weak acids dissociate less than 100% in water.
4. Ka is an acid equilibrium constant that tells us how much the acid dissociates and is given by; Ka = [A-(aq)] [H+(aq)] / [HA(aq)].

Question 21

Buffers.

Answer 21

1. A buffer solution minimises changes to the pH when we add small amounts of acid or base.
2. Acidic buffers are made from a weak acid and its conjugate base.
3. Buffers use an equilibrium to react with any extra H+ or OH- ions; weak acids(aq) ⇌ conjugate base(aq) + H+(aq)
4. Carbonic acid and hydrocarbonate ions form a buffer to maintain the pH of human blood.
5. We can use the expression for Ka of the acid in a buffer to find the pH; Ka = [A-(aq)] [H+(aq)] / [HA(aq)].