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A-Level · Chemistry

AC4 Kinetics and Equilibria

AQA 7405 · Calculator allowed · about 160 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question is about the collision theory model of reaction rates.

(a)State two conditions that must both be met for a collision between reactant particles to result in a successful (reactive) collision.(2)

(b)Use the Maxwell-Boltzmann distribution of molecular energies to explain why increasing the temperature of a reaction mixture increases the rate of reaction. You should refer to the effect on the shape of the distribution and on the number of particles with energy greater than or equal to the activation energy.(4)

(c)Define the term activation energy, Ea.(1)

(Total for Question 1 is 7 marks)

This question tests your understanding of reaction orders and how they can be identified from graphs.

(a)Define what is meant by the order of reaction with respect to a particular reactant.(1)

(b)State the units of the rate of a chemical reaction.(1)

(c)In an experiment monitoring the concentration of reactant X over time, a graph of [X] against time is a straight line with a constant negative gradient for as long as X remains present in the mixture. State the order of reaction with respect to X, and explain how the graph shows this.(2)

(d)In a separate experiment, a graph of initial rate against the initial concentration of reactant Y is a straight line passing through the origin. State the order of reaction with respect to Y, and explain how the graph shows this.(2)

(e)In a third experiment, a graph of initial rate against the initial concentration of reactant Z is a curve passing through the origin, such that doubling [Z] causes the initial rate to increase by a factor of four. State the order of reaction with respect to Z, and explain how this observation shows this.(2)

(Total for Question 2 is 8 marks)

A reaction between reactants A and B was investigated using the initial rates method. The table shows the results of three experiments carried out at the same constant temperature.

Experiment	[A] (mol dm-3)	[B] (mol dm-3)	Initial rate (mol dm-3 s-1)
1	0.10	0.10	2.0 x 10^-3
2	0.20	0.10	4.0 x 10^-3
3	0.20	0.20	1.6 x 10^-2

(a)Using experiments 1 and 2, determine the order of reaction with respect to A. Show your reasoning.(2)

(b)Using experiments 2 and 3, determine the order of reaction with respect to B. Show your reasoning.(2)

(c)Deduce the overall order of reaction and write the rate equation for this reaction.(2)

(d)Using the data from Experiment 1, calculate the rate constant, k, for this reaction, including its units.(3)

(Total for Question 3 is 9 marks)

A student investigates the rate of the reaction between excess calcium carbonate (marble chips) and dilute hydrochloric acid:
CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g)
The total volume of CO2 gas produced is measured at regular time intervals using a gas syringe (continuous monitoring), and plotted as a graph of gas volume against time. The total volume of the reaction mixture (acid plus chips) is 50.0 cm3. This question is based on the required practical activity on measuring reaction rates.

(a)Suggest two reasons why the calcium carbonate is used in excess in this experiment.(2)

(b)As the reaction proceeds, the gradient of the volume-time graph continually decreases, until it becomes zero. Explain, in terms of collision theory, why the rate of reaction decreases over time.(2)

(c)A tangent drawn to the volume-time graph at t = 0 has a gradient of 0.90 cm3 s-1. Given that the molar volume of a gas is 24000 cm3 mol-1 at room temperature and pressure, and that the total volume of the reaction mixture is 50.0 cm3, calculate the initial rate of this reaction in mol dm-3 s-1.(4)

(d)In an alternative experiment to find the initial rate of a different, colourless-to-blue-black reaction, a small, fixed volume of sodium thiosulfate solution and starch indicator are added at the start to the reaction mixture, and the time, t, taken for the blue-black colour to appear is measured (an iodine clock reaction). Explain why 1/t is taken to be directly proportional to the initial rate of the reaction being studied.(2)

(e)State one variable, other than the concentrations being investigated, that must be kept constant between repeat experiments in this type of rate investigation, and explain why it must be controlled.(2)

(Total for Question 4 is 12 marks)

The concentration of reactant A was monitored during its decomposition at a constant temperature. The results are shown in the table.

Time (min)	0	10	20	30	40
[A] (mol dm-3)	0.800	0.400	0.200	0.100	0.0500

You may use: for a first order reaction, k = ln2 / t(1/2); [A]t = [A]0 x e^-kt, where [A]t is the concentration remaining after time t.

(a)Show that these data are consistent with the reaction being first order with respect to A.(3)

(b)Calculate the rate constant, k, for this reaction in s-1, giving your answer to 3 significant figures.(3)

(c)Using your value of k (or k = 1.16 x 10^-3 s-1 if you did not obtain an answer to (b)), calculate the concentration of A remaining after 25.0 minutes have elapsed, giving your answer to 3 significant figures.(2)

(Total for Question 5 is 8 marks)

The reaction between nitrogen dioxide and carbon monoxide has the overall equation:
NO2(g) + CO(g) -> NO(g) + CO2(g)
The experimentally determined rate equation is rate = k[NO2]^2. This is consistent with the following two-step mechanism:
Step 1: NO2 + NO2 -> NO3 + NO
Step 2: NO3 + CO -> NO2 + CO2

(a)Identify the reaction intermediate in this mechanism, and explain how you can recognise it as an intermediate.(2)

(b)Deduce which step, 1 or 2, is the rate-determining step, using the given rate equation to support your answer.(3)

(c)Explain why the rate equation for a multi-step reaction cannot be predicted directly from the stoichiometric coefficients of the overall balanced equation, and must instead be found by experiment.(2)

(d)A student suggests an alternative single-step mechanism in which one molecule of NO2 and one molecule of CO collide directly and simultaneously with a second molecule of NO2, in a single three-body (termolecular) collision. Suggest why this proposed mechanism is very unlikely.(1)

(Total for Question 6 is 8 marks)

The rate constant for a first order reaction is 2.00 x 10^-5 s-1 at 300 K and 3.51 x 10^-4 s-1 at 350 K.
You may use: ln(k2/k1) = (Ea/R)(1/T1 - 1/T2); k = A x e^-Ea/(RT); R = 8.31 J K-1 mol-1.

(a)Calculate the activation energy, Ea, of this reaction in kJ/mol. Give your answer to 3 significant figures.(5)

(b)Using your value of Ea (or Ea = 50.0 kJ/mol if you did not obtain an answer to (a)), calculate the Arrhenius constant, A, for this reaction, stating its units.(3)

(Total for Question 7 is 8 marks)

The Arrhenius equation can be written in the linear form:
ln k = -(Ea/R)(1/T) + ln A
so a graph of ln k (y-axis) against 1/T (x-axis) gives a straight line. For a particular first order reaction, this graph has a gradient of -9640 K and a y-intercept of 28.5. (R = 8.31 J K-1 mol-1.)

(a)State the units of the quantity plotted on each axis of this graph.(2)

(b)Use the gradient of the graph to calculate the activation energy, Ea, of the reaction in kJ/mol. Give your answer to 3 significant figures.(3)

(c)Use the y-intercept of the graph to calculate the Arrhenius constant, A, for this reaction, stating its units.(3)

(Total for Question 8 is 8 marks)

This question is about the equilibrium constant, Kc, for homogeneous equilibria.
H2(g) + I2(g) ≤> 2HI(g)

(a)Write the expression for Kc for this equilibrium.(1)

(b)State the units of Kc for this equilibrium.(1)

(c)At a particular temperature, the equilibrium concentrations in a sealed 2.00 dm3 container are [H2] = 0.100 mol dm-3, [I2] = 0.0500 mol dm-3 and [HI] = 0.400 mol dm-3. Calculate Kc at this temperature.(3)

(d)Explain why changing the concentration of H2 at constant temperature does not change the value of Kc, even though it does change the position of equilibrium.(2)

(e)State what is meant by a dynamic equilibrium.(1)

(f)Which expression correctly gives Kc for the equilibrium N2(g) + 3H2(g) ≤> 2NH3(g)?(1)

A) Kc = [N2][H2]³ / [NH3]²
B) Kc = [NH3]² / ([N2][H2]³)
C) Kc = [NH3] / ([N2][H2])
D) Kc = 2[NH3] / ([N2] x 3[H2])

(Total for Question 9 is 9 marks)

A sealed container holds an equilibrium mixture of N2O4(g) and NO2(g):
N2O4(g) ≤> 2NO2(g)
At equilibrium, the total pressure is 2.00 atm, and the mole fractions are 0.400 for N2O4 and 0.600 for NO2.

(a)State what is meant by the mole fraction of a gas in a mixture.(1)

(b)Calculate the partial pressure of N2O4 and of NO2 in this equilibrium mixture.(2)

(c)Write the expression for Kp for this equilibrium, and use your answers to (b) to calculate its value, stating units.(4)

(d)State and explain the effect of increasing the total pressure of this equilibrium mixture, at constant temperature, on the position of equilibrium.(2)

(Total for Question 10 is 9 marks)

2.00 mol of PCl5 gas is placed in a sealed 5.00 dm3 container and allowed to reach equilibrium at a constant temperature:
PCl5(g) ≤> PCl3(g) + Cl2(g)
At equilibrium, 1.20 mol of PCl5 remains.

(a)Calculate the number of moles of PCl3 and of Cl2 present at equilibrium.(2)

(b)Calculate the equilibrium concentration, in mol dm-3, of each of the three species.(3)

(c)Write the expression for Kc for this equilibrium, and calculate its value, stating units.(4)

(Total for Question 11 is 9 marks)

In an experiment to determine the equilibrium constant for esterification, 0.100 mol of ethanoic acid, CH3COOH, and 0.100 mol of ethanol, C2H5OH, together with a small (catalytic) amount of concentrated sulfuric acid, are sealed in a flask and left at a constant temperature until equilibrium is reached:
CH3COOH(l) + C2H5OH(l) ≤> CH3COOC2H5(l) + H2O(l)
A sample of the equilibrium mixture is then titrated with 1.00 mol dm-3 sodium hydroxide solution, using phenolphthalein indicator, to find the amount of unreacted ethanoic acid. The titration requires 40.0 cm3 of the sodium hydroxide solution. This question is based on practical techniques for determining an equilibrium constant.

(a)Suggest why the flask is sealed (stoppered) throughout the time the mixture is left to reach equilibrium.(1)

(b)Suggest why the titration should be carried out quickly (without delay).(1)

(c)Calculate the number of moles of ethanoic acid present in the sample at equilibrium.(2)

(d)Hence deduce the number of moles of ethanol, ethyl ethanoate and water present at equilibrium.(3)

(e)Write the expression for Kc for this equilibrium, and use your answers to calculate its value. Explain why Kc has no units for this equilibrium.(4)

(Total for Question 12 is 11 marks)

Ammonia is manufactured industrially by the Haber process:
N2(g) + 3H2(g) ≤> 2NH3(g) deltaH = -92 kJ/mol
Industrially, a compromise temperature of about 450 degrees C and a pressure of about 200 atm are used, together with an iron catalyst.

(a)State and explain the effect of increasing the pressure on the position of this equilibrium, and on the equilibrium yield of ammonia.(2)

(b)Explain, in terms of the frequency of collisions, why increasing the pressure also increases the rate at which this equilibrium is reached.(2)

(c)The iron catalyst is essential for making this process economically viable. Explain why the catalyst has no effect on the position of equilibrium or on the equilibrium yield of ammonia.(2)

(d)Explain why a compromise temperature of about 450 degrees C, rather than a much lower temperature, is used industrially in the Haber process, given that the forward reaction is exothermic. Your answer should refer to the effect of temperature on both the position of equilibrium and the rate of reaction.(6)

(Total for Question 13 is 12 marks)

A catalytic converter in a car exhaust system uses a solid platinum/palladium catalyst to catalyse the following reaction between two exhaust gases:
2CO(g) + 2NO(g) -> 2CO2(g) + N2(g)

(a)Describe how the reaction profile (enthalpy against reaction progress) for this reaction would differ with the catalyst present compared with the profile without the catalyst, stating clearly what changes and what stays the same.(2)

(b)Using the Maxwell-Boltzmann distribution of molecular energies, explain why lowering the activation energy increases the rate of reaction at a fixed temperature.(2)

(c)Outline the steps involved when the solid platinum catalyst increases the rate of this reaction, referring to adsorption and desorption.(2)

(d)This reaction is, in principle, reversible, with the reverse reaction having a different (higher) activation energy from the forward reaction. State and explain the effect of the platinum catalyst on the equilibrium constant, Kc, for this reaction.(2)

(Total for Question 14 is 8 marks)

Mark scheme · AC4 Kinetics and Equilibria

Question 1

(a) B1 the colliding particles must have a combined energy greater than or equal to the activation energy, Ea, of the reaction oe
(a) B1 the particles must collide with the correct orientation (geometry) oe
(a) Answer: Particles must collide with energy ≥ activation energy, and with the correct orientation.
(b) B1 at higher temperature the Maxwell-Boltzmann distribution curve becomes flatter and broader, with its peak shifted to a higher energy oe
(b) B1 the total area under the curve stays the same (the total number of particles is unchanged) oe
(b) B1 a greater proportion (larger area) of particles now has energy greater than or equal to the activation energy, Ea oe
(b) B1 this means a greater proportion of collisions are successful (result in reaction), so the rate of reaction increases oe
(b) Answer: Higher temperature flattens and broadens the Maxwell-Boltzmann distribution and shifts it to higher energy; more particles have energy ≥ Ea, so more collisions are successful and the rate increases.
(c) B1 the minimum energy that colliding particles must possess in order to react oe
(c) Answer: The minimum energy required for a collision to result in a reaction.

Question 2

(a) B1 the power to which the concentration of that reactant is raised in the (experimentally determined) rate equation oe
(a) Answer: The power to which that reactant's concentration is raised in the rate equation.
(b) B1 mol dm^-3 s^-1 (or another valid concentration/time combination, e.g. mol dm^-3 min^-1) oe
(b) Answer: mol dm-3 s-1
(c) B1 zero order
(c) B1 the gradient (rate) is constant and does not change as [X] falls, so the rate is independent of [X] oe
(c) Answer: Zero order; a constant gradient shows the rate does not depend on [X].
(d) B1 first order
(d) B1 rate is directly proportional to [Y] (shown by the straight line through the origin) oe
(d) Answer: First order; rate is directly proportional to [Y].
(e) B1 second order
(e) B1 rate is proportional to [Z]², so doubling [Z] increases the rate by a factor of 2² = 4, consistent with the observation oe
(e) Answer: Second order; rate is proportional to [Z]², so doubling [Z] quadruples the rate.

Question 3

(a) M1 [A] doubles (0.10 to 0.20) with [B] constant, and the initial rate doubles (2.0x10^-3 to 4.0x10^-3) oe
(a) A1 order with respect to A = 1 cao
(a) Answer: Order with respect to A = 1
(b) M1 [B] doubles (0.10 to 0.20) with [A] constant, and the initial rate increases by a factor of 4 (4.0x10^-3 to 1.6x10^-2) oe
(b) A1 order with respect to B = 2 cao
(b) Answer: Order with respect to B = 2
(c) M1 overall order = 1 + 2 = 3
(c) A1 rate = k[A][B]²
(c) Answer: Overall order = 3; rate = k[A][B]²
(d) M1 k = rate / ([A][B]²) = 2.0x10^-3 / (0.10 x 0.10²), ft from (c)
(d) A1 k = 2.0 cao
(d) A1 units: mol^-2 dm⁶ s^-1
(d) Answer: k = 2.0 mol^-2 dm⁶ s^-1

Question 4

(a) B1 ensures hydrochloric acid is the limiting reagent, so the volume of gas produced depends only on the amount (concentration) of acid used oe
(a) B1 ensures the reaction does not stop early through the marble chips running out, so the full effect of the acid on rate can be measured oe
(a) Answer: Excess CaCO3 makes HCl the limiting reagent, so gas volume depends only on the acid, and the chips do not run out before the acid is used up.
(b) B1 the concentration of HCl decreases as the reaction proceeds (it is used up) oe
(b) B1 there are fewer acid particles per unit volume, so the frequency of successful collisions between acid and carbonate particles decreases, decreasing the rate oe
(b) Answer: As [HCl] falls, collisions between HCl and CaCO3 particles become less frequent, so the rate decreases.
(c) M1 initial rate of gas production (mol s-1) = 0.90 / 24000
(c) A1 = 3.75 x 10^-5 mol s-1
(c) M1 divides by the volume of the reaction mixture in dm3 (50.0 cm3 = 0.0500 dm3)
(c) A1 initial rate = 7.50 x 10^-4 mol dm-3 s-1 cao
(c) Answer: 7.50 x 10^-4 mol dm-3 s-1
(d) B1 the same small, fixed amount of product (iodine) must form before all of the thiosulfate is used up and the colour change is triggered, in every repeat oe
(d) B1 only a small/negligible extent of reaction has occurred by this point (reactant concentrations have barely changed), so t is inversely proportional to the (approximately constant) initial rate, giving rate proportional to 1/t oe
(d) Answer: A fixed small amount of product must form to trigger the colour change each time, so t is inversely proportional to the (near-initial) rate, giving rate proportional to 1/t.
(e) B1 temperature
(e) B1 temperature affects the rate constant, k (and hence the rate), so if it were not controlled it would not be possible to attribute any change in rate solely to the change in concentration oe
(e) Answer: Temperature must be kept constant, because it affects k and therefore the rate, independently of concentration.

Question 5

(a) M1 identifies successive half-lives: 0.800 to 0.400 takes 10 min; 0.400 to 0.200 takes a further 10 min; 0.200 to 0.100 takes a further 10 min; 0.100 to 0.0500 takes a further 10 min
(a) A1 the half-life is constant at 10 min throughout the reaction
(a) A1 a constant half-life (independent of concentration) is characteristic of first order kinetics, so the data are consistent with the reaction being first order with respect to A cso
(a) Answer: Half-life is constant at 10 min throughout, confirming first order kinetics with respect to A.
(b) M1 t(1/2) = 10 min = 600 s
(b) M1 k = ln2 / 600
(b) A1 k = 1.16 x 10^-3 s-1 awrt
(b) Answer: k = 1.16 x 10^-3 s-1
(c) M1 converts t = 25.0 min to 1500 s and substitutes into [A]t = [A]0 x e^-kt, ft from (b)
(c) A1 [A]t = 0.141 mol dm-3 awrt
(c) Answer: 0.141 mol dm-3

Question 6

(a) B1 NO3
(a) B1 it is formed in step 1 and used up (reacts) in step 2, so it does not appear in the overall equation oe
(a) Answer: NO3 is the intermediate; it is produced in step 1 and consumed in step 2, so it cancels out of the overall equation.
(b) M1 the rate equation, rate = k[NO2]², involves only NO2, to the power 2 oe
(b) M1 step 1 involves two molecules of NO2 colliding, matching the species and the order (2) shown in the rate equation oe
(b) A1 step 1 is the rate-determining (slow) step; step 2 (involving CO) must be fast, since CO does not appear in the rate equation oe
(b) Answer: Step 1 is the rate-determining step, because the rate equation (rate = k[NO2]²) matches the two NO2 molecules reacting in step 1; CO's absence from the rate equation shows step 2 (fast) occurs after the RDS.
(c) B1 the rate equation depends only on the species (and their powers) involved in the rate-determining (slowest) step of the mechanism, not on the overall equation oe
(c) B1 the overall balanced equation gives no information about the individual steps of the mechanism, so the rate equation cannot be deduced from it and must be found experimentally oe
(c) Answer: The rate equation reflects only the rate-determining step, which cannot be identified from the overall equation alone, so it must be found experimentally.
(d) B1 a simultaneous, correctly orientated collision between three separate particles at once, each also possessing sufficient energy, is statistically far less probable than a two-body collision oe
(d) Answer: Simultaneous three-particle collisions with correct orientation and sufficient energy are far less probable than two-particle collisions.

Question 7

(a) M1 calculates the ratio k2/k1 = 3.51x10^-4 / 2.00x10^-5 = 17.6 awrt
(a) M1 ln(17.6) = 2.87 awrt
(a) M1 calculates 1/T1 - 1/T2 = 1/300 - 1/350 = 4.76 x 10^-4 K-1 awrt
(a) M1 rearranges to Ea = R x ln(k2/k1) / (1/T1 - 1/T2) and substitutes values
(a) A1 Ea = 50.0 kJ/mol awrt
(a) Answer: Ea = 50.0 kJ/mol
(b) M1 rearranges k = A e^-Ea/RT to A = k x e^Ea/RT, using the data at T = 300 K, ft from (a)
(b) M1 Ea/(RT) = 50000/(8.31x300) = 20.1 awrt; A = 2.00x10^-5 x e^20.1
(b) A1 A = 1.03 x 10⁴ s-1 awrt (units s-1, same as k, since the reaction is first order)
(b) Answer: A = 1.03 x 10⁴ s-1

Question 8

(a) B1 y-axis (ln k): no units (dimensionless)
(a) B1 x-axis (1/T): K-1
(a) Answer: ln k has no units; 1/T has units K-1.
(b) M1 Ea = -gradient x R = 9640 x 8.31
(b) A1 Ea = 80100 J/mol awrt
(b) A1 Ea = 80.1 kJ/mol cao
(b) Answer: Ea = 80.1 kJ/mol
(c) M1 A = e^intercept = e^28.5
(c) A1 A = 2.38 x 10¹² awrt
(c) B1 units: s-1 (since the reaction is first order, A has the same units as k)
(c) Answer: A = 2.38 x 10¹² s-1

Question 9

(a) B1 Kc = [HI]² / ([H2][I2])
(a) Answer: Kc = [HI]² / ([H2][I2])
(b) B1 no units (Kc is dimensionless for this equilibrium)
(b) Answer: No units.
(c) M1 Kc = (0.400)² / (0.100 x 0.0500), ft from (a)
(c) M1 = 0.160 / 0.00500
(c) A1 Kc = 32.0 cao
(c) Answer: Kc = 32.0
(d) B1 Kc is a constant that depends only on temperature (for a given equilibrium) oe
(d) B1 although adding more H2 shifts the position of equilibrium (more HI forms, by Le Chatelier's principle), the equilibrium concentrations adjust so that the value given by the Kc expression returns to the same value at that temperature oe
(d) Answer: Kc depends only on temperature; concentrations re-adjust as the equilibrium shifts so that the value of the Kc expression is unchanged.
(e) B1 in a closed system, the forward and reverse reactions occur at the same (equal) rate, so the concentrations of reactants and products remain constant over time oe
(e) Answer: Forward and reverse reaction rates are equal, so concentrations remain constant, in a closed system.
(f) B1 B
(f) Answer: B

Question 10

(a) B1 the number of moles of that gas divided by the total number of moles of gas present in the mixture oe
(a) Answer: Moles of that gas / total moles of gas in the mixture.
(b) M1 partial pressure = mole fraction x total pressure, for each gas
(b) A1 P(N2O4) = 0.800 atm; P(NO2) = 1.20 atm
(b) Answer: P(N2O4) = 0.800 atm; P(NO2) = 1.20 atm
(c) B1 Kp = (P(NO2))² / P(N2O4)
(c) M1 Kp = (1.20)² / 0.800, ft from (b)
(c) A1 Kp = 1.80 awrt
(c) B1 units: atm
(c) Answer: Kp = 1.80 atm
(d) B1 position of equilibrium shifts towards N2O4 (to the left, towards fewer moles of gas)
(d) B1 the system opposes the increase in pressure (Le Chatelier's principle) by favouring the side with fewer gas molecules (1 mol N2O4 compared with 2 mol NO2), reducing the total moles of gas and so the pressure oe
(d) Answer: Equilibrium shifts left, towards N2O4, since this side has fewer gas moles, partially opposing the pressure increase.

Question 11

(a) M1 moles of PCl5 reacted = 2.00 - 1.20 = 0.80 mol
(a) A1 moles PCl3 = moles Cl2 = 0.80 mol (1:1:1 stoichiometry)
(a) Answer: PCl3 = Cl2 = 0.80 mol
(b) M1 divides each number of moles by the volume, 5.00 dm3, ft from (a)
(b) A1 [PCl5] = 1.20/5.00 = 0.240 mol dm-3
(b) A1 [PCl3] = [Cl2] = 0.80/5.00 = 0.160 mol dm-3 (each)
(b) Answer: [PCl5] = 0.240 mol dm-3; [PCl3] = [Cl2] = 0.160 mol dm-3
(c) B1 Kc = [PCl3][Cl2] / [PCl5]
(c) M1 Kc = (0.160 x 0.160) / 0.240, ft from (b)
(c) A1 Kc = 0.107 awrt
(c) B1 units: mol dm-3
(c) Answer: Kc = 0.107 mol dm-3

Question 12

(a) B1 to prevent the loss (by evaporation) of any of the volatile reactants or products (e.g. ethanol or ethyl ethanoate), which would introduce error into the amounts present oe
(a) Answer: Sealing prevents evaporation of volatile reactants/products, which would introduce error.
(b) B1 adding sodium hydroxide removes the free ethanoic acid from the system, which could disturb/shift this reversible equilibrium; titrating quickly minimises the extent to which the equilibrium shifts during titration, so the result better reflects the true equilibrium composition oe
(b) Answer: Quick titration minimises the extent to which removing the acid disturbs the equilibrium, keeping the result close to the true equilibrium composition.
(c) M1 moles NaOH = (40.0/1000) x 1.00
(c) A1 = 0.0400 mol, so moles CH3COOH at equilibrium = 0.0400 mol (1:1 neutralisation) cao
(c) Answer: 0.0400 mol CH3COOH
(d) M1 moles of CH3COOH reacted = 0.100 - 0.0400 = 0.0600 mol, ft from (c)
(d) A1 moles ethanol remaining = 0.100 - 0.0600 = 0.0400 mol
(d) A1 moles ethyl ethanoate = moles water = 0.0600 mol each (1:1:1:1 stoichiometry)
(d) Answer: Ethanol = 0.0400 mol; ethyl ethanoate = water = 0.0600 mol each
(e) B1 Kc = [CH3COOC2H5][H2O] / ([CH3COOH][C2H5OH])
(e) M1 since all species share the same fixed volume, the volume terms cancel, so Kc = (0.0600 x 0.0600) / (0.0400 x 0.0400), ft from (d)
(e) A1 Kc = 2.25 cao
(e) B1 no units, because there are two concentration terms on the top and two on the bottom of the expression, so the units cancel exactly oe
(e) Answer: Kc = 2.25 (no units)

Question 13

(a) B1 the position of equilibrium shifts towards NH3 (to the right), because this side has fewer moles of gas (4 mol of gaseous reactants forming 2 mol of gaseous product) oe
(a) B1 so increasing the pressure increases the equilibrium yield of ammonia oe
(a) Answer: Equilibrium shifts right (towards NH3, the side with fewer gas moles), increasing the yield of ammonia.
(b) B1 increasing the pressure increases the concentration of the gas particles (they are pushed closer together in the same volume) oe
(b) B1 this increases the frequency of collisions between particles, increasing the rate of both the forward and reverse reactions oe
(b) Answer: Higher pressure increases gas particle concentration, so collisions happen more often, increasing both forward and reverse reaction rates.
(c) B1 the catalyst provides an alternative reaction pathway of lower activation energy, increasing the rate of both the forward and reverse reactions by the same factor oe
(c) B1 equilibrium is reached more quickly, but the equilibrium position (and therefore the equilibrium yield) is unchanged oe
(c) Answer: The catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached faster but its position (and yield) is unaffected.
(d) L3 (5-6): A detailed, coherent and logically structured explanation that correctly links the effect of temperature on the position of equilibrium (via Le Chatelier's principle, since the forward reaction is exothermic) with its effect on the rate of reaction (via collision theory/the Maxwell-Boltzmann distribution), and clearly explains why a compromise is needed for economic reasons. Accurate scientific terminology is used throughout.
(d) L2 (3-4): Some relevant scientific points relating to either the equilibrium yield or the rate of reaction are made, with some correct reasoning, but the discussion is not fully developed, may focus mainly on one aspect (yield or rate) with limited linkage between them, or contains minor inaccuracies.
(d) L1 (1-2): Basic, isolated relevant statements are made (e.g. only that the reaction is exothermic, or only that a lower temperature gives a slower rate), with little structure, linkage or use of scientific reasoning.
(d) L0 (0): No relevant content, or the answer does not relate to the question.
(d) Answer: Levels-of-response answer; see indicative content.

Question 14

(a) B1 the activation energy, Ea, is lower with the catalyst present (the peak of the profile is lower) oe
(a) B1 the overall enthalpy change, deltaH, of the reaction (the relative energies of reactants and products) is unchanged, with or without the catalyst oe
(a) Answer: The catalyst lowers the activation energy (lower peak); the overall enthalpy change is unchanged.
(b) B1 at a fixed temperature the Maxwell-Boltzmann distribution is unchanged, but because the activation energy is now lower, a greater proportion (area under the curve) of particles has energy greater than or equal to this new, lower activation energy oe
(b) B1 a greater proportion of collisions are therefore successful (result in reaction) per unit time, increasing the rate of reaction oe
(b) Answer: At fixed temperature, more particles now exceed the (lower) activation energy, so more collisions are successful, increasing the rate.
(c) B1 gas molecules of CO and NO are adsorbed onto active sites on the surface of the solid catalyst, weakening their bonds and holding the reactant molecules close together in a favourable orientation, lowering the activation energy oe
(c) B1 the product molecules (CO2 and N2) then desorb from the catalyst surface, freeing up the active sites for further reactant molecules oe
(c) Answer: CO and NO are adsorbed onto active sites (weakening bonds, favourable orientation); products desorb, freeing sites for further reaction.
(d) B1 the catalyst has no effect on Kc oe
(d) B1 it lowers the activation energy of both the forward and reverse reactions by the same amount, increasing both rates equally; equilibrium is reached faster, but the equilibrium position (and hence Kc) is unchanged oe
(d) Answer: The catalyst has no effect on Kc; it speeds up the forward and reverse reactions equally, so equilibrium is reached faster but its position (and Kc) is unaffected.

Kinetics and Equilibria - Worksheets, Questions and Revision