A Level H1 Chemistry Kinetics Equilibrium Quiz

A-Level Chemistry H1 Quiz - Kinetics Equilibrium

ANSWER KEY AND MARKING SCHEME

Total Marks: 40

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Section A: Short Answer and Structured Response (10 marks)

1. Define the term rate of reaction. [1]

Answer: The rate of reaction is the change in concentration of a reactant or product per unit time. [1]

Accept: Amount of reactant consumed or product formed per unit time.

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2. State two factors that affect the rate of a chemical reaction, other than concentration of reactants. [2]

Answer: Any two from: [1 mark each]

• Temperature
• Pressure (for gases)
• Surface area / particle size (for solids)
• Presence of a catalyst
• Light (for photochemical reactions)

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3. Explain why an increase in temperature increases the rate of a chemical reaction. [2]

Answer:

• More molecules/particles have energy greater than or equal to the activation energy [1]
• This results in a greater frequency of effective/successful collisions [1]

Accept: Reference to Maxwell-Boltzmann distribution showing larger area under curve beyond Ea at higher temperature.

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4. State Le Chatelier's principle. [1]

Answer: If a system at dynamic equilibrium is subjected to a change in conditions, the position of equilibrium shifts to oppose/minimise/counteract the change. [1]

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5. A reaction has the rate equation: rate = k[A][B]². State the overall order of this reaction. [1]

Answer: 3 (third order) [1]

Order with respect to A = 1, order with respect to B = 2; overall = 1 + 2 = 3.

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6. Define the term activation energy. [1]

Answer: Activation energy is the minimum energy that colliding particles must possess for a reaction to occur / for effective collisions to take place. [1]

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7. State the effect of a catalyst on the activation energy of a reaction. [1]

Answer: A catalyst lowers the activation energy / provides an alternative reaction pathway with a lower activation energy. [1]

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8. Write the expression for the equilibrium constant, Kc, for the following reaction:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) [1]

Answer: Kc = [SO₃]² / ([SO₂]² [O₂]) [1]

Accept any equivalent correct expression. Square brackets must be used.

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Section B: Calculation and Data Interpretation (18 marks)

9. The decomposition of hydrogen peroxide, H₂O₂, was studied at 298 K.

(a) Calculate the average rate of decomposition of H₂O₂ over the first 200 seconds. State the units. [2]

Answer:

• Change in concentration = 0.20 − 0.40 = −0.20 mol dm⁻³ (decrease of 0.20 mol dm⁻³) [1]
• Rate = 0.20 / 200 = 0.0010 mol dm⁻³ s⁻¹ [1]

Accept: 1.0 × 10⁻³ mol dm⁻³ s⁻¹. Award [1] for correct calculation, [1] for correct units. Negative sign not required if magnitude stated as rate of decomposition.

(b) Using the data, explain how the rate of reaction changes with time. [2]

Answer:

• The rate of reaction decreases with time [1]
• This is because the concentration of H₂O₂ decreases, so there are fewer reactant particles per unit volume, resulting in fewer effective collisions per unit time [1]

Accept: Reference to decreasing gradient of concentration-time graph / calculation showing rate decreases (e.g., first 100 s: 0.0012 mol dm⁻³ s⁻¹; last 100 s: 0.0004 mol dm⁻³ s⁻¹).

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10. The reaction between nitrogen monoxide and hydrogen.

(a) Determine the order of reaction with respect to NO. Explain your reasoning. [2]

Answer:

• Order with respect to NO = 2 [1]
• Reasoning: Comparing Experiments 1 and 3, [NO] doubles (0.10 → 0.20) while [H₂] is constant; the rate increases by a factor of 4 (2.0 × 10⁻⁴ → 8.0 × 10⁻⁴), so rate ∝ [NO]² [1]

(b) Determine the order of reaction with respect to H₂. Explain your reasoning. [2]

Answer:

• Order with respect to H₂ = 1 [1]
• Reasoning: Comparing Experiments 1 and 2, [H₂] doubles (0.10 → 0.20) while [NO] is constant; the rate doubles (2.0 × 10⁻⁴ → 4.0 × 10⁻⁴), so rate ∝ [H₂] [1]

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

Answer: rate = k[NO]²[H₂] [1]

(d) Calculate the value of the rate constant, k, stating its units. [2]

Answer:

• Using Experiment 1: k = rate / ([NO]²[H₂]) = (2.0 × 10⁻⁴) / ((0.10)² × 0.10) [1]
• k = (2.0 × 10⁻⁴) / (0.0010) = 0.20 [1]
• Units: dm⁶ mol⁻² s⁻¹ [1]

Accept: 2.0 × 10⁻¹ dm⁶ mol⁻² s⁻¹. Award [1] for correct value, [1] for correct units. Allow calculation from any experiment.

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11. Consider the equilibrium: N₂O₄(g) ⇌ 2NO₂(g) ΔH = +58 kJ mol⁻¹

(a) Predict and explain the effect on the position of equilibrium when the temperature is increased. [2]

Answer:

• Position of equilibrium shifts to the right / favours the forward reaction [1]
• The forward reaction is endothermic (ΔH is positive); increasing temperature favours the endothermic reaction to absorb the added heat, opposing the change (Le Chatelier's principle) [1]

(b) Predict and explain the effect on the position of equilibrium when the total pressure is increased. [2]

Answer:

• Position of equilibrium shifts to the left / favours the backward reaction [1]
• There are fewer moles of gas on the left (1 mole) than on the right (2 moles); increasing pressure favours the side with fewer gas molecules to reduce the pressure, opposing the change (Le Chatelier's principle) [1]

(c) State what would be observed if the pressure of the system is increased at constant temperature. [1]

Answer: The mixture becomes paler / the brown colour fades / becomes lighter. [1]

Reason: Equilibrium shifts left, producing more colourless N₂O₄ and less brown NO₂.

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Section C: Application and Analysis (12 marks)

12. The Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹

(a) State the conditions of temperature and pressure typically used in the Haber process. [2]

Answer:

• Temperature: approximately 400–500 °C (or 673–773 K) [1]
• Pressure: approximately 200–300 atm (or 20–30 MPa) [1]

Accept any values within these ranges.

(b) Explain why a temperature higher than that which gives the maximum equilibrium yield is used in practice. [2]

Answer:

• The forward reaction is exothermic, so a lower temperature would give a higher equilibrium yield of ammonia [1]
• However, a higher temperature is used to increase the rate of reaction, so ammonia is produced more quickly / a reasonable rate is achieved, making the process economically viable [1]

Accept: Reference to compromise between yield and rate.

(c) Explain why a very high pressure is not used in the Haber process, despite it favouring the forward reaction. [2]

Answer:

• Very high pressures are expensive to generate and maintain / require expensive, reinforced equipment [1]
• There are safety risks associated with operating at very high pressures / increased risk of explosion or leaks [1]

Accept: High energy costs for compression; thicker-walled reaction vessels needed; balancing economic factors.

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

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

Answer: 2 (second order) [1]

(b) Suggest why the reaction is unlikely to occur in a single step. [2]

Answer:

• The stoichiometric equation involves 3 reactant ions (1 S₂O₈²⁻ + 2 I⁻) [1]
• A single-step (termolecular) reaction would require three particles to collide simultaneously with correct orientation and sufficient energy, which is statistically very unlikely / has very low probability [1]

Accept: The rate equation shows orders that do not match the stoichiometric coefficients (rate ∝ [I⁻] not [I⁻]²), suggesting a multi-step mechanism.

(c) Explain whether the proposed mechanism is consistent with the rate equation. [2]

Answer:

• Yes, the mechanism is consistent [1]
• The slow step (Step 1) is the rate-determining step; the rate equation derived from the slow step is rate = k[S₂O₈²⁻][I⁻], which matches the experimentally determined rate equation [1]

Accept: The molecularity of the slow step matches the orders of the reactants in the rate equation.

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14. Methanol manufacture: CO(g) + 2H₂(g) ⇌ CH₃OH(g) ΔH = −91 kJ mol⁻¹

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

Answer: Kc = [CH₃OH] / ([CO] [H₂]²) [1]

Accept any equivalent correct expression. Square brackets must be used.

(b) State and explain the effect of adding a catalyst on the value of Kc. [2]

Answer:

• No effect / Kc remains unchanged [1]
• A catalyst increases the rate of both the forward and backward reactions equally; it does not affect the position of equilibrium or the equilibrium constant; Kc is only affected by temperature [1]

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Section D: Data Analysis and Extended Reasoning (10 marks)

15. Maxwell-Boltzmann distribution.

(a) Sketch the Maxwell-Boltzmann distribution curve for a gas at two different temperatures, T₁ and T₂, where T₂ > T₁. Label the activation energy, Ea, on your sketch. [2]

Answer:

• Two curves drawn with correct shape (starting at origin, rising to a peak, then decreasing asymptotically) [1]
• Curve for T₂ shifted to the right with a lower peak and broader distribution; Ea labelled as a vertical line to the right of both peaks; area under curve beyond Ea visibly larger for T₂ [1]

Accept: Clear labels of T₁ and T₂; Ea marked on the energy axis.

(b) Use your sketch to explain why a small increase in temperature can lead to a large increase in the rate of reaction. [2]

Answer:

• The area under the curve to the right of Ea represents the number/proportion of molecules with energy ≥ Ea [1]
• At T₂, this area is significantly larger than at T₁, meaning a much larger proportion of molecules can undergo effective collisions, leading to a large increase in rate [1]

Accept: Reference to exponential relationship / more molecules exceeding Ea.

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16. Reaction: BrO₃⁻(aq) + 5Br⁻(aq) + 6H⁺(aq) → 3Br₂(aq) + 3H₂O(l)
Rate equation: rate = k[BrO₃⁻][Br⁻][H⁺]²

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

Answer: 4 (fourth order) [1]

Overall order = 1 + 1 + 2 = 4.

(b) Deduce the effect on the initial rate if the concentration of H⁺ is halved while all other concentrations are kept constant. [1]

Answer: The rate decreases by a factor of 4 / becomes one-quarter of the original rate. [1]

Reason: rate ∝ [H⁺]²; halving [H⁺] gives (½)² = ¼ of the original rate.

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17. Reaction: A + 2B → C

Experiment	[A] / mol dm⁻³	[B] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	1.0 × 10⁻³
2	0.20	0.10	2.0 × 10⁻³
3	0.10	0.20	4.0 × 10⁻³

(a) Determine the order of reaction with respect to A and B. [2]

Answer:

• Order with respect to A = 1: Comparing Expt 1 and 2, [A] doubles, [B] constant, rate doubles [1]
• Order with respect to B = 2: Comparing Expt 1 and 3, [B] doubles, [A] constant, rate quadruples [1]

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

Answer: rate = k[A][B]² [1]

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18. Contact process: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −197 kJ mol⁻¹

(a) State and explain the effect of increasing temperature on the equilibrium yield of SO₃. [2]

Answer:

• Yield of SO₃ decreases / equilibrium shifts left [1]
• The forward reaction is exothermic; increasing temperature favours the endothermic backward reaction to absorb the added heat (Le Chatelier's principle) [1]

(b) State and explain the effect of increasing pressure on the equilibrium yield of SO₃. [2]

Answer:

• Yield of SO₃ increases / equilibrium shifts right [1]
• There are fewer moles of gas on the right (2 moles) than on the left (3 moles); increasing pressure favours the side with fewer gas molecules to reduce the pressure (Le Chatelier's principle) [1]

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19. Decomposition of hydrogen peroxide with a catalyst.

(a) Sketch a graph of volume of oxygen against time for the reaction with and without a catalyst on the same axes. Label both curves. [2]

Answer:

• Both curves start at the origin and level off at the same final volume of oxygen [1]
• Curve with catalyst has a steeper initial gradient and levels off sooner; both curves clearly labelled [1]

(b) Explain, in terms of activation energy, why the initial rate is faster with a catalyst. [1]

Answer: A catalyst provides an alternative reaction pathway with a lower activation energy, so a greater proportion of molecules have sufficient energy to react, leading to more frequent effective collisions and a faster initial rate. [1]

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20. Equilibrium: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) Kc = 4.0 × 10⁻² dm³ mol⁻¹

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

Answer: Kc = [PCl₃][Cl₂] / [PCl₅] [1]

Accept any equivalent correct expression. Square brackets must be used.

(b) At equilibrium, [PCl₅] = 0.20 mol dm⁻³ and [PCl₃] = 0.10 mol dm⁻³. Calculate the equilibrium concentration of Cl₂. [2]

Answer:

• Kc = [PCl₃][Cl₂] / [PCl₅] → 4.0 × 10⁻² = (0.10 × [Cl₂]) / 0.20 [1]
• [Cl₂] = (4.0 × 10⁻² × 0.20) / 0.10 = 0.080 mol dm⁻³ [1]

Accept: 8.0 × 10⁻² mol dm⁻³. Award [1] for correct substitution, [1] for correct answer with units.

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END OF ANSWER KEY