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A Level H1 Chemistry Kinetics Equilibrium Quiz
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Questions
A-Level Chemistry H1 Quiz - Kinetics Equilibrium
Name: ________________________
Class: ________________________
Date: ________________________
Score: ______ / 40
Duration: 50 minutes
Total Marks: 40
Instructions:
- Answer ALL questions in the spaces provided.
- Show all working for calculation questions.
- State units where appropriate.
- A Data Booklet is provided.
- Marks for each question are indicated in brackets.
Section A: Short Answer and Structured Response (10 marks)
Answer all questions in this section.
1. Define the term rate of reaction. [1]
2. State two factors that affect the rate of a chemical reaction, other than concentration of reactants. [2]
3. Explain why an increase in temperature increases the rate of a chemical reaction. [2]
4. State Le Chatelier's principle. [1]
5. A reaction has the rate equation: rate = k[A][B]². State the overall order of this reaction. [1]
6. Define the term activation energy. [1]
7. State the effect of a catalyst on the activation energy of a reaction. [1]
8. Write the expression for the equilibrium constant, Kc, for the following reaction:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) [1]
Section B: Calculation and Data Interpretation (18 marks)
Answer all questions in this section.
9. The decomposition of hydrogen peroxide, H₂O₂, was studied at 298 K. The concentration of H₂O₂ was measured at different times. The results are shown in the table below.
| Time / s | [H₂O₂] / mol dm⁻³ |
|---|---|
| 0 | 0.40 |
| 100 | 0.28 |
| 200 | 0.20 |
| 300 | 0.14 |
| 400 | 0.10 |
(a) Calculate the average rate of decomposition of H₂O₂ over the first 200 seconds. State the units. [2]
(b) Using the data, explain how the rate of reaction changes with time. [2]
10. The reaction between nitrogen monoxide and hydrogen is shown below.
2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g)
The following kinetic data were obtained at a constant temperature.
| Experiment | [NO] / mol dm⁻³ | [H₂] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.10 | 0.20 | 4.0 × 10⁻⁴ |
| 3 | 0.20 | 0.10 | 8.0 × 10⁻⁴ |
(a) Determine the order of reaction with respect to NO. Explain your reasoning. [2]
(b) Determine the order of reaction with respect to H₂. Explain your reasoning. [2]
(c) Write the rate equation for this reaction. [1]
(d) Calculate the value of the rate constant, k, stating its units. [2]
11. Consider the following equilibrium system:
N₂O₄(g) ⇌ 2NO₂(g) ΔH = +58 kJ mol⁻¹
The equilibrium mixture is a pale brown colour at room temperature.
(a) Predict and explain the effect on the position of equilibrium when the temperature is increased. [2]
(b) Predict and explain the effect on the position of equilibrium when the total pressure is increased. [2]
(c) State what would be observed if the pressure of the system is increased at constant temperature. [1]
Section C: Application and Analysis (12 marks)
Answer all questions in this section.
12. The Haber process for the manufacture of ammonia is represented by the following equation:
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]
(b) Explain why a temperature higher than that which gives the maximum equilibrium yield is used in practice. [2]
(c) Explain why a very high pressure is not used in the Haber process, despite it favouring the forward reaction. [2]
13. The reaction between persulfate ions and iodide ions is shown below.
S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq)
The rate equation for this reaction is: rate = k[S₂O₈²⁻][I⁻]
(a) State the overall order of this reaction. [1]
(b) Suggest why the reaction is unlikely to occur in a single step. [2]
(c) A student proposed the following two-step mechanism for this reaction:
Step 1: S₂O₈²⁻ + I⁻ → SO₄²⁻ + SO₄I⁻ (slow)
Step 2: SO₄I⁻ + I⁻ → SO₄²⁻ + I₂ (fast)
Explain whether this proposed mechanism is consistent with the rate equation. [2]
14. Methanol can be manufactured by the reaction of carbon monoxide with hydrogen.
CO(g) + 2H₂(g) ⇌ CH₃OH(g) ΔH = −91 kJ mol⁻¹
At 500 K, the equilibrium constant Kc for this reaction is 14.5 dm⁶ mol⁻².
(a) Write the expression for Kc for this reaction. [1]
(b) State and explain the effect of adding a catalyst on the value of Kc. [2]
Section D: Data Analysis and Extended Reasoning (10 marks)
Answer all questions in this section.
15. The Maxwell-Boltzmann distribution curve shows the distribution of molecular energies in a gas at a given temperature.
(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]
(b) Use your sketch to explain why a small increase in temperature can lead to a large increase in the rate of reaction. [2]
16. The reaction between bromate(V) ions and bromide ions in acidic solution is shown below.
BrO₃⁻(aq) + 5Br⁻(aq) + 6H⁺(aq) → 3Br₂(aq) + 3H₂O(l)
The rate equation is: rate = k[BrO₃⁻][Br⁻][H⁺]²
(a) State the overall order of this reaction. [1]
(b) Deduce the effect on the initial rate if the concentration of H⁺ is halved while all other concentrations are kept constant. [1]
17. For the reaction: A + 2B → C, the following data were obtained.
| 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]
(b) Write the rate equation for this reaction. [1]
18. 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 and explain the effect of increasing temperature on the equilibrium yield of SO₃. [2]
(b) State and explain the effect of increasing pressure on the equilibrium yield of SO₃. [2]
19. A student investigates the effect of a catalyst on the decomposition of hydrogen peroxide. The student measures the volume of oxygen gas produced over time.
(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]
(b) Explain, in terms of activation energy, why the initial rate is faster with a catalyst. [1]
20. The equilibrium constant, Kc, for the reaction below is 4.0 × 10⁻² dm³ mol⁻¹ 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 0.10 mol dm⁻³. Calculate the equilibrium concentration of Cl₂. [2]
END OF QUIZ
Check your answers carefully.
Answers
A-Level Chemistry H1 Quiz - Kinetics Equilibrium
ANSWER KEY AND MARKING SCHEME
Total Marks: 40
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.
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)
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.
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]
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.
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]
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]
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.
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).
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.
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.
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.
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.
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]
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.
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.
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]
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]
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]
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.
END OF ANSWER KEY