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A Level H2 Chemistry Kinetics Equilibrium Quiz
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Questions
A-Level Chemistry H2 Quiz - Kinetics Equilibrium
Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50
Duration: 1 hour 15 minutes Total Marks: 50
Instructions:
- Answer ALL questions in the spaces provided.
- Show all working for calculation questions.
- Use of the Data Booklet is relevant to some questions.
- State symbols are required where appropriate.
- Marks are indicated in brackets [ ].
Section A: Multiple Choice & Short Answer (10 marks)
Answer all questions in this section.
1. The rate equation for a reaction is: rate = k[A]²[B].
What is the overall order of the reaction?
A. 1 B. 2 C. 3 D. 4
[1 mark]
Answer: ______
2. For the equilibrium: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹
Which change would increase the equilibrium yield of ammonia?
A. Increasing temperature B. Decreasing pressure C. Adding a catalyst D. Increasing pressure
[1 mark]
Answer: ______
3. A reaction has the rate equation: rate = k[P][Q]².
If the concentration of P is doubled and the concentration of Q is halved, by what factor does the rate change?
A. Rate is halved B. Rate remains unchanged C. Rate is doubled D. Rate is quartered
[1 mark]
Answer: ______
4. Define the term activation energy.
[2 marks]
5. State Le Chatelier's principle.
[2 marks]
Section B: Structured Questions (20 marks)
Answer all questions in this section.
6. The equilibrium constant, Kc, for a reaction at 298 K is 4.0 × 10⁻³. At 350 K, Kc is 2.5 × 10⁻².
Deduce whether the forward reaction is exothermic or endothermic. Explain your reasoning.
[3 marks]
7. Iodine monochloride, ICl, reacts with hydrogen according to the equation:
2ICl(g) + H₂(g) → 2HCl(g) + I₂(g)
The following kinetic data were obtained at a constant temperature.
| Experiment | [ICl] / mol dm⁻³ | [H₂] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 1.5 × 10⁻³ |
| 2 | 0.20 | 0.10 | 6.0 × 10⁻³ |
| 3 | 0.10 | 0.20 | 3.0 × 10⁻³ |
(a) Determine the order of reaction with respect to ICl. Show your working.
[2 marks]
(b) Determine the order of reaction with respect to H₂. Show your working.
[2 marks]
(c) Write the rate equation for the reaction.
[1 mark]
(d) Calculate the value of the rate constant, k, stating its units. Use data from Experiment 1.
[3 marks]
8. The decomposition of hydrogen peroxide is catalysed by manganese(IV) oxide:
2H₂O₂(aq) → 2H₂O(l) + O₂(g)
(a) On the axes below, sketch and label:
- A curve showing the Maxwell-Boltzmann distribution of molecular energies at temperature T₁.
- A second curve showing the distribution at a higher temperature T₂.
[3 marks]
Number of
molecules
^
|
|
|
|
|
+---------------------------------->
Energy
(b) Use your sketch to explain why the rate of reaction increases with temperature.
[3 marks]
9. Explain, using a labelled energy profile diagram, how a catalyst increases the rate of a reaction.
[4 marks]
10. Manganese(IV) oxide is a heterogeneous catalyst for the decomposition of hydrogen peroxide. Explain what is meant by the term heterogeneous catalyst.
[2 marks]
Section C: Data Interpretation & Extended Response (20 marks)
Answer all questions in this section.
11. The Haber process for the manufacture of ammonia is represented by the equilibrium:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹
The graph below shows how the percentage yield of ammonia at equilibrium varies with temperature and pressure.
% yield of NH₃
^
100 | 1000 atm
| /
80 | /
| / 500 atm
60 | /
| / 200 atm
40 | /
| / 100 atm
20 | /
| /
0 |_________________________________>
300 500 700 900 1100
Temperature / K
(a) State and explain the effect of increasing temperature on the equilibrium yield of ammonia.
[3 marks]
(b) State and explain the effect of increasing pressure on the equilibrium yield of ammonia.
[3 marks]
12. In practice, the Haber process is carried out at approximately 700 K and 250 atm in the presence of an iron catalyst. Explain why these conditions are chosen, even though they do not give the maximum equilibrium yield.
[4 marks]
13. The reaction between persulfate ions and iodide ions is:
S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq)
The rate equation for this reaction is: rate = k[S₂O₈²⁻][I⁻]
(a) What is the overall order of this reaction?
[1 mark]
(b) The following mechanism has been proposed for this reaction:
Step 1: S₂O₈²⁻ + I⁻ → SO₄²⁻ + SO₄⁻ + I (slow) Step 2: SO₄⁻ + I⁻ → SO₄²⁻ + I (fast) Step 3: I + I → I₂ (fast)
Explain why this mechanism is consistent with the rate equation.
[3 marks]
14. The activation energy for the reaction between persulfate ions and iodide ions is 52 kJ mol⁻¹. In the presence of Fe²⁺ ions as a catalyst, the activation energy is reduced to 38 kJ mol⁻¹.
Calculate the factor by which the rate constant increases at 298 K when the catalyst is used, assuming the Arrhenius factor, A, remains constant.
[The gas constant, R = 8.31 J K⁻¹ mol⁻¹]
[4 marks]
Section D: Data Analysis & Application (10 marks)
Answer all questions in this section.
15. The decomposition of dinitrogen pentoxide (N₂O₅) in carbon tetrachloride solution is a first-order reaction:
2N₂O₅ → 4NO₂ + O₂
The rate constant at 298 K is 3.4 × 10⁻⁵ s⁻¹.
Calculate the half-life of the reaction at this temperature.
[2 marks]
16. For a certain first-order reaction, the concentration of reactant drops from 0.80 mol dm⁻³ to 0.20 mol dm⁻³ in 40 minutes.
Calculate the rate constant for this reaction.
[3 marks]
17. The equilibrium below is established when ethanoic acid reacts with ethanol:
CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)
At 298 K, the equilibrium constant Kc = 4.0.
If 1.0 mol of ethanoic acid and 1.0 mol of ethanol are mixed, calculate the equilibrium amount of ethyl ethanoate formed.
[3 marks]
18. The contact process for sulfuric acid manufacture involves the equilibrium:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −197 kJ mol⁻¹
At a certain temperature, the equilibrium mixture contains 0.60 mol SO₂, 0.40 mol O₂, and 0.80 mol SO₃ in a vessel of volume 2.0 dm³.
Calculate the value of Kc at this temperature, stating its units.
[2 marks]
19. The rate of a chemical reaction doubles when the temperature is increased from 300 K to 310 K.
Calculate the activation energy of the reaction, assuming the Arrhenius factor remains constant.
[The gas constant, R = 8.31 J K⁻¹ mol⁻¹]
[3 marks]
20. The equilibrium constant Kp for the reaction:
CO(g) + 2H₂(g) ⇌ CH₃OH(g)
is 2.25 × 10⁻² atm⁻² at 500 K.
If the partial pressures of CO and H₂ at equilibrium are 2.0 atm and 4.0 atm respectively, calculate the partial pressure of CH₃OH at equilibrium.
[2 marks]
Answers
A-Level Chemistry H2 Quiz - Kinetics Equilibrium - ANSWER KEY
Total Marks: 50
Section A: Multiple Choice & Short Answer (10 marks)
1. C. 3 [1 mark] Overall order = 2 + 1 = 3
2. D. Increasing pressure [1 mark] 4 moles of gas on left, 2 moles on right. Increasing pressure favours side with fewer gas molecules (forward reaction).
3. A. Rate is halved [1 mark] New rate = k(2[P])(½[Q])² = k(2[P])(¼[Q]²) = ½ × k[P][Q]² = half original rate
4. Activation energy is the minimum energy that colliding particles must possess for a reaction to occur / for effective collisions to result in a chemical reaction. [2 marks] Award 1 mark for "minimum energy" and 1 mark for "for reaction to occur/effective collisions".
5. Le Chatelier's principle states that if a system at dynamic equilibrium is subjected to a change in conditions (temperature, pressure, or concentration), the position of equilibrium shifts to oppose/counteract the change. [2 marks] Award 1 mark for "system at equilibrium subjected to change" and 1 mark for "equilibrium shifts to oppose the change".
Section B: Structured Questions (20 marks)
6. The forward reaction is endothermic. [1 mark]
As temperature increases, Kc increases (from 4.0 × 10⁻³ to 2.5 × 10⁻²). [1 mark]
By Le Chatelier's principle, increasing temperature favours the endothermic direction. Since Kc increases with temperature, the forward reaction must be endothermic. [1 mark]
Accept: equilibrium shifts to absorb added heat, so forward reaction is endothermic.
7. (a) Order with respect to ICl: Comparing Experiments 1 and 2: [H₂] constant, [ICl] doubles from 0.10 to 0.20. Rate increases from 1.5 × 10⁻³ to 6.0 × 10⁻³, i.e., by a factor of 4. Since 2² = 4, the reaction is second order with respect to ICl. [2 marks] Award 1 mark for identifying experiments 1 and 2, 1 mark for correct order with reasoning.
(b) Order with respect to H₂: Comparing Experiments 1 and 3: [ICl] constant, [H₂] doubles from 0.10 to 0.20. Rate increases from 1.5 × 10⁻³ to 3.0 × 10⁻³, i.e., by a factor of 2. Since 2¹ = 2, the reaction is first order with respect to H₂. [2 marks] Award 1 mark for identifying experiments 1 and 3, 1 mark for correct order with reasoning.
(c) rate = k[ICl]²[H₂] [1 mark]
(d) Using Experiment 1: rate = k[ICl]²[H₂] 1.5 × 10⁻³ = k × (0.10)² × (0.10) 1.5 × 10⁻³ = k × 1.0 × 10⁻³ k = 1.5 dm⁶ mol⁻² s⁻¹ [3 marks] Award 1 mark for correct substitution, 1 mark for correct value, 1 mark for correct units. Units: rate/(concentration)³ = mol dm⁻³ s⁻¹ / (mol dm⁻³)³ = dm⁶ mol⁻² s⁻¹.
8. (a) Sketch showing two Maxwell-Boltzmann curves:
- Both curves start at origin, rise to a peak, and tail off asymptotically.
- T₂ curve has a lower peak, shifted to the right (higher energy).
- T₂ curve has a broader distribution with more molecules at higher energies.
- Both curves have the same area under them (same total number of molecules).
- Activation energy (Ea) line drawn vertically; area under T₂ curve beyond Ea is larger than area under T₁ curve beyond Ea. [3 marks] Award 1 mark for correct shape of both curves, 1 mark for T₂ shifted right with lower peak, 1 mark for Ea labelled and larger shaded area for T₂.
(b) At higher temperature, the Maxwell-Boltzmann distribution shifts so that a greater proportion of molecules have energy greater than or equal to the activation energy. This means a larger fraction of collisions are effective/energetic enough to result in reaction. Additionally, molecules move faster at higher temperature, increasing collision frequency. Both factors increase the rate of reaction. [3 marks] Award 1 mark for "greater proportion exceed Ea", 1 mark for "more effective collisions", 1 mark for "increased collision frequency".
9. Energy profile diagram:
- Y-axis: Energy/Potential energy; X-axis: Reaction coordinate/Progress of reaction.
- Reactants on left, products on right (products lower than reactants for exothermic decomposition).
- Uncatalysed pathway: single hump with high activation energy (Ea, uncat).
- Catalysed pathway: alternative route with lower hump (Ea, cat), possibly showing intermediate.
- Ea, cat < Ea, uncat clearly labelled.
Explanation: A catalyst provides an alternative reaction pathway with lower activation energy. This means a greater proportion of colliding molecules possess energy ≥ Ea, resulting in more effective collisions per unit time, hence a faster rate of reaction. The catalyst is chemically unchanged at the end of the reaction. [4 marks] Award 1 mark for correctly labelled axes, 1 mark for two pathways with different Ea values, 1 mark for Ea(cat) < Ea(uncat), 1 mark for explanation linking lower Ea to more effective collisions and increased rate.
10. A heterogeneous catalyst is one that is in a different phase (physical state) from the reactants. In this reaction, MnO₂ is a solid while the reactants (H₂O₂) are in aqueous solution/liquid phase. [2 marks] Award 1 mark for "different phase", 1 mark for identifying MnO₂(s) vs H₂O₂(aq/l).
Section C: Data Interpretation & Extended Response (20 marks)
11. (a) Increasing temperature decreases the equilibrium yield of ammonia. [1 mark]
The forward reaction is exothermic (ΔH = −92 kJ mol⁻¹). By Le Chatelier's principle, increasing temperature causes the equilibrium to shift in the endothermic direction to absorb the added heat. The endothermic direction is the reverse reaction (decomposition of NH₃), so the equilibrium shifts left, reducing the yield of ammonia. [2 marks] Award 1 mark for stating yield decreases, 1 mark for exothermic forward reaction, 1 mark for shift to endothermic/reverse direction.
(b) Increasing pressure increases the equilibrium yield of ammonia. [1 mark]
On the left side of the equation, there are 4 moles of gas (1 N₂ + 3 H₂). On the right side, there are 2 moles of gas (2 NH₃). By Le Chatelier's principle, increasing pressure causes the equilibrium to shift to the side with fewer gas molecules to reduce the pressure. This is the forward reaction, so the yield of ammonia increases. [2 marks] Award 1 mark for stating yield increases, 1 mark for 4 moles → 2 moles, 1 mark for shift to side with fewer gas molecules.
12. The chosen conditions (700 K, 250 atm, iron catalyst) represent a compromise between equilibrium yield, rate of reaction, and economic considerations. [1 mark]
-
Temperature (700 K): Although lower temperatures give a higher equilibrium yield (as the reaction is exothermic), the rate of reaction would be too slow at low temperatures. 700 K provides a reasonable rate while still giving an acceptable yield. [1 mark]
-
Pressure (250 atm): Higher pressures increase yield and rate, but very high pressures are expensive (stronger equipment needed, higher energy costs for compression). 250 atm is an economically viable compromise. [1 mark]
-
Catalyst (iron): The catalyst increases the rate of reaction by providing an alternative pathway with lower activation energy. It does not affect the equilibrium position or yield but allows equilibrium to be reached more quickly at the operating temperature. [1 mark]
Accept any reasonable economic/engineering justification. Award marks for addressing each of the three conditions.
13. (a) Overall order = 1 + 1 = 2 [1 mark]
(b) The rate-determining step (slow step) is Step 1: S₂O₈²⁻ + I⁻ → products. [1 mark]
The rate equation derived from the slow step would be: rate = k[S₂O₈²⁻][I⁻], which matches the experimentally determined rate equation. [1 mark]
Steps 2 and 3 are fast and do not affect the overall rate. The mechanism is consistent because the molecularity of the rate-determining step matches the orders in the rate equation. [1 mark]
Award 1 mark for identifying Step 1 as rate-determining, 1 mark for deriving rate equation from Step 1, 1 mark for matching with experimental rate equation.
14. Using the Arrhenius equation in logarithmic form: ln(k₂/k₁) = (Ea₁ − Ea₂) / (RT) Ea₁ = 52 kJ mol⁻¹ = 52000 J mol⁻¹ Ea₂ = 38 kJ mol⁻¹ = 38000 J mol⁻¹ T = 298 K, R = 8.31 J K⁻¹ mol⁻¹
ln(k₂/k₁) = (52000 − 38000) / (8.31 × 298) ln(k₂/k₁) = 14000 / 2476.38 ln(k₂/k₁) = 5.653
k₂/k₁ = e⁵·⁶⁵³ = 285 (approximately 290) [4 marks] Award 1 mark for correct formula, 1 mark for converting kJ to J, 1 mark for correct substitution, 1 mark for correct answer (accept 280–300).
Section D: Data Analysis & Application (10 marks)
15. For a first-order reaction: t₁/₂ = ln 2 / k t₁/₂ = 0.693 / (3.4 × 10⁻⁵) t₁/₂ = 2.04 × 10⁴ s (or 5.66 hours) [2 marks] Award 1 mark for correct formula, 1 mark for correct answer with units.
16. For a first-order reaction: ln([A]₀/[A]t) = kt [A]₀ = 0.80 mol dm⁻³, [A]t = 0.20 mol dm⁻³, t = 40 min
ln(0.80/0.20) = k × 40 ln 4 = 40k 1.386 = 40k k = 0.0347 min⁻¹ (or 5.78 × 10⁻⁴ s⁻¹) [3 marks] Award 1 mark for correct formula, 1 mark for correct substitution, 1 mark for correct answer with units.
17. CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O Initial: 1.0 mol, 1.0 mol, 0, 0 Change: −x, −x, +x, +x Equilibrium: (1.0 − x), (1.0 − x), x, x
Kc = [CH₃COOC₂H₅][H₂O] / ([CH₃COOH][C₂H₅OH]) = 4.0 x² / (1.0 − x)² = 4.0 x / (1.0 − x) = 2.0 x = 2.0 − 2x 3x = 2.0 x = 0.667 mol
Equilibrium amount of ethyl ethanoate = 0.667 mol (or 0.67 mol) [3 marks] Award 1 mark for ICE table/setup, 1 mark for solving quadratic, 1 mark for correct answer.
18. Concentrations at equilibrium: [SO₂] = 0.60/2.0 = 0.30 mol dm⁻³ [O₂] = 0.40/2.0 = 0.20 mol dm⁻³ [SO₃] = 0.80/2.0 = 0.40 mol dm⁻³
Kc = [SO₃]² / ([SO₂]²[O₂]) Kc = (0.40)² / ((0.30)² × 0.20) Kc = 0.16 / (0.09 × 0.20) Kc = 0.16 / 0.018 Kc = 8.89 dm³ mol⁻¹ (or 8.9 dm³ mol⁻¹) [2 marks] Award 1 mark for correct concentrations and substitution, 1 mark for correct value and units.
19. Using the Arrhenius equation: ln(k₂/k₁) = (Ea/R) × (1/T₁ − 1/T₂) k₂/k₁ = 2, T₁ = 300 K, T₂ = 310 K, R = 8.31 J K⁻¹ mol⁻¹
ln 2 = (Ea/8.31) × (1/300 − 1/310) 0.693 = (Ea/8.31) × (10/(300 × 310)) 0.693 = (Ea/8.31) × (10/93000) 0.693 = (Ea/8.31) × 1.075 × 10⁻⁴ Ea = 0.693 × 8.31 / (1.075 × 10⁻⁴) Ea = 53600 J mol⁻¹ = 53.6 kJ mol⁻¹ (accept 53–54 kJ mol⁻¹) [3 marks] Award 1 mark for correct formula, 1 mark for correct substitution, 1 mark for correct answer with units.
20. Kp = pCH₃OH / (pCO × (pH₂)²) 2.25 × 10⁻² = pCH₃OH / (2.0 × (4.0)²) 2.25 × 10⁻² = pCH₃OH / (2.0 × 16) 2.25 × 10⁻² = pCH₃OH / 32 pCH₃OH = 2.25 × 10⁻² × 32 = 0.72 atm [2 marks] Award 1 mark for correct expression and substitution, 1 mark for correct answer with units.