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A Level H1 Chemistry Kinetics Equilibrium Quiz

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A-Level Chemistry H1 Quiz - Kinetics Equilibrium

Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50

Duration: 1 hour Total Marks: 50

Instructions:

This quiz contains 20 questions on Kinetics and Equilibrium.
Answer ALL questions in the spaces provided.
Show all working for calculation questions.
A Data Booklet is provided.
Marks for each question are indicated in brackets.

Section A: Multiple Choice (5 × 1 mark = 5 marks)

Circle the correct answer for each question.

1. Which statement best describes a system at dynamic equilibrium?

A. The concentrations of reactants and products are equal. B. The forward and reverse reactions have stopped. C. The rate of the forward reaction equals the rate of the reverse reaction. D. The reaction has gone to completion.

[1]

2. For the reaction A + 2B → C, the rate equation is found to be: rate = k[A][B]². What is the overall order of the reaction?

A. 1 B. 2 C. 3 D. 4

[1]

3. A catalyst increases the rate of a reaction by:

A. Increasing the energy of the reactants. B. Lowering the activation energy. C. Increasing the enthalpy change of the reaction. D. Shifting the equilibrium position towards the products.

[1]

4. Consider the equilibrium: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ mol⁻¹. Which change will increase the equilibrium yield of ammonia?

A. Increasing temperature B. Decreasing pressure C. Adding a catalyst D. Increasing pressure

[1]

5. The Boltzmann distribution curve below shows the distribution of molecular energies at two temperatures, T₁ and T₂.

   Number of
   molecules
      |
      |     T₁
      |    /  \
      |   /    \____
      |  /           \___
      | /                \_____ T₂ (higher temp, flatter curve shifted right)
      |/_____________________________
      |          Ea
      +------------------------------
               Energy

Which statement is correct?

A. At T₂, more molecules have energy greater than Ea. B. The area under both curves is different. C. The activation energy, Ea, is lower at T₂. D. A catalyst would shift the curve to the right.

[1]

Section B: Structured Questions (45 marks)

Answer all questions in the spaces provided.

6. Methanol can be produced industrially by the reaction of carbon monoxide with hydrogen:

CO(g) + 2H₂(g) ⇌ CH₃OH(g) ΔH = –91 kJ mol⁻¹

(a) State what is meant by the term dynamic equilibrium. [2]

(b) The reaction is carried out at a pressure of 50 atm and a temperature of 250 °C in the presence of a copper-based catalyst.

(i) Explain, using Le Chatelier's principle, why a high pressure is used. [2]

(ii) Explain why a temperature of 250 °C is used rather than a lower temperature, despite the fact that the equilibrium yield of methanol would be higher at a lower temperature. [2]

(iii) State the effect of the catalyst on the equilibrium yield of methanol. Explain your answer. [2]

7. The decomposition of hydrogen peroxide, H₂O₂, in aqueous solution is catalysed by manganese(IV) oxide, MnO₂.

2H₂O₂(aq) → 2H₂O(l) + O₂(g)

The rate of this reaction can be followed by measuring the volume of oxygen gas produced at regular time intervals.

(a) Sketch a graph to show how the volume of oxygen gas produced changes with time. Label the axes clearly. [2]

Volume of O₂ / cm³

|
|
|
|
|
|
+-----------------------------------
              Time / s

(b) On the same axes, sketch a second curve to show the results obtained if the experiment is repeated using the same volume and concentration of H₂O₂ but with a larger mass of MnO₂ catalyst. Label this curve clearly. [1]

(c) Explain, using collision theory, why the rate of reaction decreases as the reaction proceeds. [2]

(d) State and explain the effect on the initial rate of reaction if the experiment is repeated using a more concentrated solution of H₂O₂, keeping all other conditions the same. [2]

8. The reaction between iodine and propanone in the presence of an acid catalyst can be represented by the equation:

CH₃COCH₃(aq) + I₂(aq) → CH₃COCH₂I(aq) + HI(aq)

The rate equation for this reaction is: rate = k[CH₃COCH₃][H⁺]

(a) State the order of reaction with respect to: (i) propanone, CH₃COCH₃ [1] (ii) iodine, I₂ [1] (iii) hydrogen ions, H⁺ [1]

(b) Deduce the overall order of the reaction. [1]

(c) The rate constant, k, for this reaction has a value of 2.5 × 10⁻⁵ dm³ mol⁻¹ s⁻¹ at a particular temperature.

Calculate the initial rate of reaction when the initial concentrations are: [CH₃COCH₃] = 0.40 mol dm⁻³ [H⁺] = 0.10 mol dm⁻³ [I₂] = 0.050 mol dm⁻³

State the units of the rate. [3]

9. The equilibrium between dinitrogen tetroxide and nitrogen dioxide is shown below:

N₂O₄(g) ⇌ 2NO₂(g) ΔH = +58 kJ mol⁻¹ colourless brown

(a) Write the expression for the equilibrium constant, Kc, for this reaction. [1]

(b) A sealed syringe containing an equilibrium mixture of N₂O₄ and NO₂ at room temperature appears pale brown.

(i) Predict what would be observed when the plunger of the syringe is pushed in rapidly, decreasing the volume. Explain your answer. [3]

(ii) Predict what would be observed when the syringe is placed in a beaker of hot water. Explain your answer. [3]

(c) At 60 °C, the value of Kc for this reaction is 0.36 mol dm⁻³. At equilibrium, the concentration of NO₂ is 0.30 mol dm⁻³. Calculate the equilibrium concentration of N₂O₄. [2]

10. The reaction between bromate(V) ions and bromide ions in acidic solution can be represented by the equation:

BrO₃⁻(aq) + 5Br⁻(aq) + 6H⁺(aq) → 3Br₂(aq) + 3H₂O(l)

A series of experiments was carried out to determine the rate equation for this reaction. The results are shown in the table below.

Experiment	[BrO₃⁻] / mol dm⁻³	[Br⁻] / mol dm⁻³	[H⁺] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	0.10	1.2 × 10⁻³
2	0.20	0.10	0.10	2.4 × 10⁻³
3	0.10	0.30	0.10	3.6 × 10⁻³
4	0.10	0.10	0.20	4.8 × 10⁻³

(a) Determine the order of reaction with respect to: (i) BrO₃⁻ [1] (ii) Br⁻ [1] (iii) H⁺ [1]

(b) Write the rate equation for this reaction. [1]

(c) Calculate the value of the rate constant, k, using the data from Experiment 1. State the units of k. [3]

11. The Contact Process for the manufacture of sulfuric acid involves the equilibrium:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = –197 kJ mol⁻¹

(a) State the effect of increasing temperature on: (i) the equilibrium position [1] (ii) the value of Kc [1]

(b) In the industrial process, a temperature of about 450 °C is used, along with a vanadium(V) oxide catalyst and a pressure slightly above atmospheric pressure.

Explain why: (i) a temperature of 450 °C is used rather than a lower temperature [2]

(ii) a pressure only slightly above atmospheric pressure is used, even though a higher pressure would increase the equilibrium yield of SO₃ [2]

12. The hydrolysis of an ester in alkaline solution can be represented by:

CH₃COOCH₂CH₃(aq) + OH⁻(aq) → CH₃COO⁻(aq) + CH₃CH₂OH(aq)

The rate equation for this reaction is: rate = k[CH₃COOCH₂CH₃][OH⁻]

(a) State the overall order of this reaction. [1]

(b) In an experiment, the initial concentration of the ester was 0.050 mol dm⁻³ and the initial concentration of OH⁻ was 0.050 mol dm⁻³. The initial rate of reaction was 2.0 × 10⁻⁴ mol dm⁻³ s⁻¹.

Calculate the value of the rate constant, k, and state its units. [2]

(c) The experiment was repeated at a higher temperature. State and explain the effect on: (i) the initial rate of reaction [2]

(ii) the value of the rate constant, k [1]

13. The equilibrium below exists in an aqueous solution of ethanoic acid:

CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq)

(a) Write the expression for the acid dissociation constant, Ka, for ethanoic acid. [1]

(b) A solution of ethanoic acid has a concentration of 0.10 mol dm⁻³ and a pH of 2.87.

Calculate the value of Ka for ethanoic acid. [3]

(c) A small amount of solid sodium ethanoate is added to the ethanoic acid solution. State and explain the effect on: (i) the pH of the solution [2]

(ii) the value of Ka [1]

14. The graph below shows the Maxwell-Boltzmann distribution of molecular energies for a gaseous reaction at temperature T.

   Number of
   molecules
      |
      |       /\
      |      /  \
      |     /    \
      |    /      \____
      |   /           \___
      |  /                \_____
      | /________________________
      |          Ea
      +------------------------------
               Energy

(a) On the graph, shade the area that represents the number of molecules with energy greater than or equal to the activation energy, Ea. [1]

(b) On the same axes, sketch the distribution curve for the same reaction at a higher temperature. Label this curve T₂. [1]

(c) Explain, with reference to the Maxwell-Boltzmann distribution, why the rate of reaction increases with increasing temperature. [2]

(d) State and explain the effect of adding a catalyst on the Maxwell-Boltzmann distribution curve. [2]

15. The decomposition of gaseous hydrogen iodide is a reversible reaction:

2HI(g) ⇌ H₂(g) + I₂(g) ΔH = +53 kJ mol⁻¹

(a) Write the expression for the equilibrium constant, Kc, for this reaction. [1]

(b) At a certain temperature, an equilibrium mixture was found to contain 0.40 mol of HI, 0.15 mol of H₂, and 0.15 mol of I₂ in a vessel of volume 2.0 dm³.

Calculate the value of Kc at this temperature. State the units, if any. [3]

(c) State and explain the effect on the value of Kc if the temperature is increased. [2]

16. The reaction between peroxodisulfate(VI) ions and iodide ions can be represented by:

S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq)

The rate equation for this reaction is: rate = k[S₂O₈²⁻][I⁻]

(a) State the order of reaction with respect to S₂O₈²⁻ and I⁻. [1]

(b) The reaction is first order with respect to each reactant. Suggest why the stoichiometric coefficient of I⁻ in the equation is 2, but the order with respect to I⁻ is only 1. [2]

(c) In an experiment, the initial rate of reaction was 4.0 × 10⁻⁶ mol dm⁻³ s⁻¹ when [S₂O₈²⁻] = 0.020 mol dm⁻³ and [I⁻] = 0.040 mol dm⁻³.

Calculate the value of the rate constant, k, and state its units. [2]

17. The Haber Process for the manufacture of ammonia involves the equilibrium:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ mol⁻¹

(a) State and explain the effect of increasing the pressure on the equilibrium position. [2]

(b) State and explain the effect of increasing the temperature on the value of Kc. [2]

(c) In the industrial process, the reaction is carried out at a pressure of about 200 atm and a temperature of about 450 °C in the presence of an iron catalyst.

Explain why these conditions are a compromise. [3]

18. The rate of the reaction between magnesium ribbon and dilute hydrochloric acid was investigated by measuring the volume of hydrogen gas produced over time.

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

(a) Sketch a graph of volume of hydrogen gas produced against time. [1]

Volume of H₂ / cm³

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|
|
|
|
|
+-----------------------------------
              Time / s

(b) On the same axes, sketch a second curve to show the results if the experiment is repeated using the same mass of magnesium but with a more concentrated solution of hydrochloric acid. Label this curve clearly. [1]

(c) Explain, using collision theory, why the initial rate of reaction is greater when a more concentrated acid is used. [2]

(d) Suggest why the final volume of hydrogen gas collected is the same in both experiments. [1]

19. The equilibrium constant, Kc, for the reaction below is 4.0 × 10⁻² mol dm⁻³ at a certain temperature.

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

(a) Write the expression for Kc for this reaction. [1]

(b) At equilibrium, the concentration of PCl₅ is 0.20 mol dm⁻³ and the concentration of PCl₃ is 0.040 mol dm⁻³. Calculate the equilibrium concentration of Cl₂. [2]

(c) State and explain the effect on the equilibrium position if the volume of the container is increased at constant temperature. [2]

20. A student investigated the kinetics of the reaction between sodium thiosulfate and hydrochloric acid:

Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + SO₂(g) + S(s) + H₂O(l)

The student measured the time taken for a cross drawn on a piece of paper to become obscured by the precipitate of sulfur.

(a) Suggest how the student could use this method to investigate the effect of concentration on the rate of this reaction. [2]

(b) The student found that when the concentration of sodium thiosulfate was doubled, the time taken for the cross to disappear was halved. Deduce the order of reaction with respect to sodium thiosulfate. Explain your answer. [2]

(c) State one source of error in this experiment and suggest how it could be minimised. [2]

END OF QUIZ

Check your answers carefully.

Answers

A-Level Chemistry H1 Quiz - Kinetics Equilibrium: Answer Key

Total Marks: 50

Section A: Multiple Choice (5 × 1 mark = 5 marks)

1. C – The rate of the forward reaction equals the rate of the reverse reaction. [1] Explanation: At dynamic equilibrium, the forward and reverse reactions continue at equal rates. Concentrations remain constant but are not necessarily equal (A is incorrect). Reactions do not stop (B is incorrect). The reaction does not go to completion (D is incorrect).

2. C – 3 [1] Explanation: Overall order = sum of individual orders = 1 + 2 = 3.

3. B – Lowering the activation energy. [1] Explanation: A catalyst provides an alternative reaction pathway with a lower activation energy. It does not change the energy of reactants, the enthalpy change, or the equilibrium position.

4. D – Increasing pressure [1] Explanation: There are 4 moles of gas on the left (1 + 3) and 2 moles on the right. Increasing pressure favours the side with fewer gas molecules (the product side), increasing ammonia yield.

5. A – At T₂, more molecules have energy greater than Ea. [1] Explanation: At higher temperature, the distribution curve flattens and shifts to the right, so a larger proportion of molecules have energy ≥ Ea. The area under both curves is the same (same number of molecules). Ea is unchanged by temperature. A catalyst lowers Ea but does not shift the curve.

Section B: Structured Questions (45 marks)

6. CO(g) + 2H₂(g) ⇌ CH₃OH(g) ΔH = –91 kJ mol⁻¹

(a) Dynamic equilibrium is the state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction [1] and the concentrations of reactants and products remain constant [1]. [2]

(b)(i) There are 3 moles of gas on the left (1 + 2) and 1 mole of gas on the right [1]. According to Le Chatelier's principle, increasing pressure shifts the equilibrium to the side with fewer gas molecules, i.e., to the right, increasing the yield of methanol [1]. [2]

(b)(ii) A lower temperature would give a higher equilibrium yield because the forward reaction is exothermic and is favoured by lower temperatures [1]. However, at lower temperatures, the rate of reaction is too slow to be economically viable. A temperature of 250 °C is a compromise that gives a reasonable rate while still producing an acceptable yield [1]. [2]

(b)(iii) The catalyst has no effect on the equilibrium yield [1]. A catalyst increases the rate of both the forward and reverse reactions equally, so the equilibrium position is unchanged. It only allows equilibrium to be reached more quickly [1]. [2]

7. 2H₂O₂(aq) → 2H₂O(l) + O₂(g)

(a) Graph showing volume of O₂ increasing rapidly at first, then levelling off to a horizontal line (asymptotic curve) [1]. Axes correctly labelled: y-axis "Volume of O₂ / cm³", x-axis "Time / s" [1]. [2]

(b) Second curve drawn on same axes: starts at origin, rises more steeply initially, but levels off at the same final volume as the first curve [1]. Curve clearly labelled (e.g., "with more MnO₂"). [1]

(c) As the reaction proceeds, the concentration of H₂O₂ decreases [1]. According to collision theory, with fewer H₂O₂ molecules per unit volume, the frequency of effective collisions decreases, so the rate of reaction decreases [1]. [2]

(d) The initial rate of reaction increases [1]. A more concentrated solution of H₂O₂ means there are more H₂O₂ molecules per unit volume. This increases the frequency of effective collisions between H₂O₂ molecules and the catalyst surface, leading to a higher rate of reaction [1]. [2]

8. CH₃COCH₃(aq) + I₂(aq) → CH₃COCH₂I(aq) + HI(aq) Rate equation: rate = k[CH₃COCH₃][H⁺]

(a)(i) First order / order 1 [1] (a)(ii) Zero order / order 0 [1] (a)(iii) First order / order 1 [1]

(b) Overall order = 1 + 0 + 1 = 2 [1]

(c) rate = k[CH₃COCH₃][H⁺] [1] rate = (2.5 × 10⁻⁵) × (0.40) × (0.10) [1] rate = 1.0 × 10⁻⁶ mol dm⁻³ s⁻¹ [1] Note: [I₂] is not included because the reaction is zero order with respect to I₂. [3]

9. N₂O₄(g) ⇌ 2NO₂(g) ΔH = +58 kJ mol⁻¹

(a) Kc = [NO₂]² / [N₂O₄] [1]

(b)(i) The mixture initially becomes darker brown, then becomes paler [1]. When the volume is decreased, the pressure increases. According to Le Chatelier's principle, the equilibrium shifts to the side with fewer gas molecules (the left, 1 mole vs 2 moles) to oppose the change [1]. This produces more colourless N₂O₄, so the mixture becomes paler. The initial darkening is due to the sudden increase in concentration of NO₂ before the equilibrium shifts [1]. [3]

(b)(ii) The mixture becomes darker brown [1]. The forward reaction is endothermic. Increasing temperature favours the endothermic forward reaction, shifting the equilibrium to the right [1]. More brown NO₂ is produced, so the colour intensifies [1]. [3]

(c) Kc = [NO₂]² / [N₂O₄] = 0.36 [1] [N₂O₄] = [NO₂]² / Kc = (0.30)² / 0.36 = 0.090 / 0.36 = 0.25 mol dm⁻³ [1] [2]

10. BrO₃⁻(aq) + 5Br⁻(aq) + 6H⁺(aq) → 3Br₂(aq) + 3H₂O(l)

(a)(i) Order 1 with respect to BrO₃⁻ [1] Reasoning: Comparing experiments 1 and 2, [BrO₃⁻] doubles while [Br⁻] and [H⁺] are constant; rate doubles (1.2 → 2.4), so first order.

(a)(ii) Order 1 with respect to Br⁻ [1] Reasoning: Comparing experiments 1 and 3, [Br⁻] triples while others are constant; rate triples (1.2 → 3.6), so first order.

(a)(iii) Order 2 with respect to H⁺ [1] Reasoning: Comparing experiments 1 and 4, [H⁺] doubles while others are constant; rate quadruples (1.2 → 4.8), so second order.

(b) rate = k[BrO₃⁻][Br⁻][H⁺]² [1]

(c) Using Experiment 1: k = rate / ([BrO₃⁻][Br⁻][H⁺]²) [1] k = (1.2 × 10⁻³) / (0.10 × 0.10 × (0.10)²) k = (1.2 × 10⁻³) / (1.0 × 10⁻⁴) [1] k = 12 dm⁹ mol⁻³ s⁻¹ [1] Units: (mol dm⁻³ s⁻¹) / (mol dm⁻³)⁴ = dm⁹ mol⁻³ s⁻¹ [3]

11. 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = –197 kJ mol⁻¹

(a)(i) Equilibrium shifts to the left (towards reactants) [1]. (a)(ii) Kc decreases [1].

(b)(i) A lower temperature would give a higher equilibrium yield because the forward reaction is exothermic [1]. However, at lower temperatures, the rate of reaction is too slow to be economically viable. 450 °C is a compromise temperature that gives a reasonable rate while still producing an acceptable yield [1]. [2]

(b)(ii) There are 3 moles of gas on the left and 2 moles on the right, so high pressure favours the forward reaction [1]. However, using very high pressures is expensive (stronger equipment needed, higher energy costs) and increases safety risks. A pressure slightly above atmospheric gives a satisfactory yield at a reasonable cost [1]. [2]

12. CH₃COOCH₂CH₃(aq) + OH⁻(aq) → CH₃COO⁻(aq) + CH₃CH₂OH(aq) Rate equation: rate = k[CH₃COOCH₂CH₃][OH⁻]

(a) Overall order = 2 [1]

(b) k = rate / ([ester][OH⁻]) [1] k = (2.0 × 10⁻⁴) / (0.050 × 0.050) k = (2.0 × 10⁻⁴) / (2.5 × 10⁻³) k = 0.080 dm³ mol⁻¹ s⁻¹ [1] [2]

(c)(i) The initial rate of reaction increases [1]. At a higher temperature, molecules have greater kinetic energy. A larger proportion of molecules have energy greater than or equal to the activation energy, and collisions are more frequent. This leads to a higher frequency of effective collisions [1]. [2]

(c)(ii) The value of k increases [1]. The rate constant increases with temperature because a greater proportion of molecules can overcome the activation energy barrier. [1]

13. CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq)

(a) Ka = [CH₃COO⁻][H₃O⁺] / [CH₃COOH] [1] Note: [H₂O] is omitted as it is approximately constant.

(b) [H₃O⁺] = 10⁻ᵖᴴ = 10⁻²·⁸⁷ = 1.35 × 10⁻³ mol dm⁻³ [1] For a weak acid: [CH₃COO⁻] ≈ [H₃O⁺] = 1.35 × 10⁻³ mol dm⁻³ [1] [CH₃COOH] ≈ 0.10 – 1.35 × 10⁻³ ≈ 0.09865 mol dm⁻³ Ka = (1.35 × 10⁻³)² / 0.09865 = 1.85 × 10⁻⁵ mol dm⁻³ [1] [3]

(c)(i) The pH increases (solution becomes less acidic) [1]. Adding sodium ethanoate increases [CH₃COO⁻]. By Le Chatelier's principle, the equilibrium shifts to the left, reducing [H₃O⁺], so pH increases [1]. [2]

(c)(ii) Ka is unchanged [1]. Ka is a constant at a given temperature and is not affected by changes in concentration. [1]

14. Maxwell-Boltzmann distribution

(a) Area under the curve to the right of Ea is shaded [1]. [1]

(b) Curve drawn with peak lower and shifted to the right, crossing the original curve. The area under the new curve should be the same as the original [1]. Labelled T₂. [1]

(c) At a higher temperature, the distribution curve flattens and shifts to the right [1]. A much larger proportion of molecules now have energy greater than or equal to the activation energy, Ea. This means a higher frequency of effective collisions, so the rate of reaction increases [1]. [2]

(d) Adding a catalyst does not change the Maxwell-Boltzmann distribution curve [1]. A catalyst provides an alternative reaction pathway with a lower activation energy. The distribution curve remains the same, but a larger area under the curve now lies to the right of the new, lower Ea [1]. [2]

15. 2HI(g) ⇌ H₂(g) + I₂(g) ΔH = +53 kJ mol⁻¹

(a) Kc = [H₂][I₂] / [HI]² [1]

(b) [HI] = 0.40 / 2.0 = 0.20 mol dm⁻³ [1] [H₂] = 0.15 / 2.0 = 0.075 mol dm⁻³ [I₂] = 0.15 / 2.0 = 0.075 mol dm⁻³ Kc = (0.075 × 0.075) / (0.20)² [1] Kc = 0.005625 / 0.04 = 0.14 (no units, as number of moles on both sides is equal) [1] [3]

(c) Kc increases [1]. The forward reaction is endothermic. Increasing temperature favours the endothermic reaction, shifting the equilibrium to the right. This increases [H₂] and [I₂] and decreases [HI], so Kc increases [1]. [2]

16. S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq) Rate equation: rate = k[S₂O₈²⁻][I⁻]

(a) First order with respect to S₂O₈²⁻ and first order with respect to I⁻ [1].

(b) The rate equation is determined experimentally, not from the stoichiometric equation [1]. The reaction occurs in a series of steps (a mechanism). The rate-determining step involves one S₂O₈²⁻ ion and one I⁻ ion. The second I⁻ ion reacts in a subsequent fast step [1]. [2]

(c) k = rate / ([S₂O₈²⁻][I⁻]) [1] k = (4.0 × 10⁻⁶) / (0.020 × 0.040) k = (4.0 × 10⁻⁶) / (8.0 × 10⁻⁴) k = 5.0 × 10⁻³ dm³ mol⁻¹ s⁻¹ [1] [2]

17. N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ mol⁻¹

(a) Increasing pressure shifts the equilibrium to the right (towards products) [1]. There are 4 moles of gas on the left and 2 moles on the right. According to Le Chatelier's principle, the equilibrium shifts to the side with fewer gas molecules to oppose the increase in pressure [1]. [2]

(b) Kc decreases [1]. The forward reaction is exothermic. Increasing temperature favours the endothermic reverse reaction, shifting the equilibrium to the left. This decreases [NH₃] and increases [N₂] and [H₂], so Kc decreases [1]. [2]

(c) High pressure favours the forward reaction (fewer gas molecules on the right) and increases yield [1]. However, very high pressures are expensive and pose safety risks. 200 atm is a compromise. Low temperature favours the exothermic forward reaction and increases yield [1]. However, at low temperatures, the rate is too slow. 450 °C is a compromise that gives a reasonable rate. The iron catalyst increases the rate without affecting the equilibrium position [1]. [3]

18. Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

(a) Graph showing volume of H₂ increasing rapidly at first, then levelling off to a horizontal line [1]. [1]

(b) Second curve drawn on same axes: starts at origin, rises more steeply initially, but levels off at the same final volume as the first curve [1]. Labelled "more concentrated HCl". [1]

(c) A more concentrated acid has more H⁺ ions per unit volume [1]. This increases the frequency of effective collisions between H⁺ ions and the magnesium surface, leading to a higher initial rate of reaction [1]. [2]

(d) The same mass of magnesium is used in both experiments [1]. Magnesium is the limiting reagent in both cases, so the same amount of H₂ is produced when all the magnesium has reacted. [1]

19. PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) Kc = 4.0 × 10⁻² mol dm⁻³

(a) Kc = [PCl₃][Cl₂] / [PCl₅] [1]

(b) Kc = [PCl₃][Cl₂] / [PCl₅] = 4.0 × 10⁻² [1] [Cl₂] = Kc × [PCl₅] / [PCl₃] [Cl₂] = (4.0 × 10⁻² × 0.20) / 0.040 [Cl₂] = 0.0080 / 0.040 = 0.20 mol dm⁻³ [1] [2]

(c) Equilibrium shifts to the right (towards products) [1]. There is 1 mole of gas on the left and 2 moles on the right. Increasing the volume decreases the pressure. According to Le Chatelier's principle, the equilibrium shifts to the side with more gas molecules to oppose the decrease in pressure [1]. [2]

20. Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + SO₂(g) + S(s) + H₂O(l)

(a) Keep the concentration and volume of HCl constant [1]. Vary the concentration of Na₂S₂O₃ by diluting it with water while keeping the total volume constant. Measure the time taken for the cross to disappear for each concentration. The rate can be compared using 1/time [1]. [2]

(b) First order with respect to Na₂S₂O₃ [1]. When the concentration is doubled, the rate doubles (time halves, so rate = 1/time doubles). This is characteristic of a first-order reaction [1]. [2]

(c) Source of error: The disappearance of the cross is judged subjectively by eye; different people may judge the endpoint differently [1]. Minimisation: Use the same person to judge the endpoint for all trials, or use a light sensor to detect when the solution reaches a specific opacity [1]. [2]

END OF ANSWER KEY