A Level H2 Chemistry Kinetics Equilibrium Quiz

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A-Level Chemistry H2 Quiz - Kinetics Equilibrium

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: _______ / 40

Duration: 60 minutes
Total Marks: 40
Topic: Kinetics & Equilibrium (Reaction Kinetics and Chemical Equilibria)

Instructions:

• Answer all 20 questions.
• Show all working clearly where calculations are required.
• Use the Data Booklet if necessary.
• Write your answers in the spaces provided.

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Section A: Reaction Kinetics (Questions 1–7)

1. [2 marks] The rate of a reaction is found to increase by a factor of 8 when the concentration of reactant X is doubled. Deduce the order of reaction with respect to X.

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2. [3 marks] For the reaction 2NO(g) + O₂(g) → 2NO₂(g), the following initial rate data were obtained:

Experiment	[NO] / mol dm⁻³	[O₂] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	2.0 × 10⁻³
2	0.20	0.10	8.0 × 10⁻³
3	0.10	0.20	4.0 × 10⁻³

Determine the rate equation for the reaction.

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3. [2 marks] State two factors that increase the rate of a reaction by increasing the frequency of effective collisions.

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4. [3 marks] The hydrolysis of aspirin is catalysed by H⁺(aq). A proposed mechanism is:

Step 1: Aspirin + H⁺ ⇌ protonated aspirin (fast)
Step 2: Protonated aspirin + H₂O → products (slow)

Identify the rate-determining step and deduce the rate equation.

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5. [2 marks] Explain, using the Maxwell–Boltzmann distribution, why increasing temperature increases reaction rate.

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6. [3 marks] The first-order decomposition of N₂O₅ has a rate constant k = 5.0 × 10⁻⁴ s⁻¹ at 45 °C. Calculate the time required for the concentration to fall from 0.020 mol dm⁻³ to 0.0050 mol dm⁻³.

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7. [2 marks] A catalyst speeds up both forward and reverse reactions equally. Explain why a catalyst does not change the position of equilibrium.

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Section B: Chemical Equilibria (Questions 8–14)

8. [2 marks] Write the expression for K_c for the equilibrium: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

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9. [3 marks] At 450 °C, K_c = 0.50 dm⁶ mol⁻² for the above equilibrium. In a 1.0 dm³ vessel, the equilibrium mixture contains 0.20 mol N₂ and 0.60 mol H₂. Calculate the equilibrium amount, in moles, of NH₃.

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10. [2 marks] State Le Chatelier’s principle in one sentence.

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11. [3 marks] The equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is exothermic. Predict and explain the effect on the equilibrium yield of SO₃ when (a) temperature is increased, and (b) pressure is increased.

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12. [2 marks] For the heterogeneous equilibrium CaCO₃(s) ⇌ CaO(s) + CO₂(g), write the K_p expression.

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13. [3 marks] At 1000 K, K_p = 1.2 atm for the above equilibrium. If the total pressure is 2.0 atm and only CO₂ is gaseous, calculate the partial pressure of CO₂ at equilibrium.

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14. [2 marks] Explain why changing the amount of solid CaCO₃ has no effect on the equilibrium position.

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Section C: Integrated and Data-Based (Questions 15–20)

15. [2 marks] The reaction H₂(g) + I₂(g) ⇌ 2HI(g) has K_c = 50 at 700 K. If 1.0 mol H₂ and 1.0 mol I₂ are placed in a 1.0 dm³ vessel, calculate the equilibrium concentration of HI.

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16. [3 marks] For the reaction A + B → C, the rate doubles when [A] is doubled and quadruples when [B] is doubled. Write the rate equation and state the overall order.

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17. [3 marks] The graph below shows concentration vs time for a reactant in a first-order reaction.

Image pending generation: graph for Q17.

Using the graph, estimate the half-life of the reaction and the rate constant k.

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18. [2 marks] State and explain the effect of adding a catalyst on the activation energy and the value of K_c.

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19. [3 marks] The equilibrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) has K_c = 0.040 mol dm⁻³ at 250 °C. In a 2.0 dm³ vessel, 0.50 mol PCl₅ is heated. Calculate the equilibrium concentrations of all species.

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20. [2 marks] Explain why the equilibrium constant K_c changes with temperature but is unaffected by changes in concentration or pressure.

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A-Level Chemistry H2 Quiz - Kinetics Equilibrium (Answer Key)

Topic: Kinetics & Equilibrium
Total Marks: 40

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Section A: Reaction Kinetics

1. [2 marks]
Order with respect to X = 3 (third order).
Working: Rate ∝ [X]ⁿ. If [X] doubles, rate × 8 ⇒ 2ⁿ = 8 ⇒ n = 3.
Teaching note: The order is the power to which concentration must be raised in the rate equation. A factor of 8 increase from doubling means 2³.
Common mistake: Stating order = 8 or confusing factor with power.

2. [3 marks]
Rate equation: rate = k[NO]²[O₂].
Working:

• From Exp 1→2: [NO] doubles, [O₂] constant, rate ×4 ⇒ order in NO = 2.
• From Exp 1→3: [O₂] doubles, [NO] constant, rate ×2 ⇒ order in O₂ = 1.
• Overall: rate = k[NO]²[O₂].
  Mark breakdown: 1 mark for NO order, 1 mark for O₂ order, 1 mark for equation.
  Teaching note: Compare experiments where only one concentration changes.

3. [2 marks]
Any two of: increased concentration, increased pressure (for gases), increased temperature, increased surface area, use of catalyst. (Credit only those increasing collision frequency effectively: concentration, pressure, surface area.)
Teaching note: Effective collisions require correct orientation and sufficient energy; frequency is raised by more particles per unit volume.

4. [3 marks]
Rate-determining step = Step 2 (slow).
Rate = k[protonated aspirin][H₂O]. Since protonated aspirin formed fast from aspirin + H⁺, rate ≈ k'[aspirin][H⁺][H₂O].
Mark breakdown: 1 mark RDS, 2 marks rate equation.
Teaching note: The slow step governs rate; intermediates from fast pre-equilibria can be substituted.

5. [2 marks]
Increasing T shifts Maxwell–Boltzmann distribution to higher energies; more molecules exceed Eₐ, so more effective collisions per second ⇒ faster rate.
Teaching note: Only molecules with energy > activation energy react.

6. [3 marks]
First-order: t = (1/k) ln([A]₀/[A]).
t = (1 / 5.0×10⁻⁴) × ln(0.020 / 0.0050) = 2000 × ln(4) = 2000 × 1.386 = 2772 s ≈ 2.77 × 10³ s.
Mark breakdown: 1 mark formula, 1 mark substitution, 1 mark answer.
Common mistake: Using wrong order formula.

7. [2 marks]
Catalyst lowers Eₐ for forward and reverse equally, so forward and reverse rates increase by same factor; equilibrium constant unchanged ⇒ no shift in position.
Teaching note: Equilibrium position depends on ΔG°, not on rate.

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Section B: Chemical Equilibria

8. [2 marks]
K_c = [NH₃]² / ([N₂][H₂]³)
Teaching note: Products over reactants, powers = coefficients.

9. [3 marks]
K_c = [NH₃]² / ([N₂][H₂]³) = 0.50
[NH₃]² = 0.50 × 0.20 × (0.60)³ = 0.50 × 0.20 × 0.216 = 0.0216
[NH₃] = √0.0216 = 0.147 mol dm⁻³ ⇒ in 1.0 dm³, moles = 0.147 mol.
Mark breakdown: 1 mark substitution, 1 mark square root, 1 mark unit/moles.

10. [2 marks]
Le Chatelier’s principle: If a system at equilibrium is subjected to a change, the system shifts to oppose that change.

11. [3 marks]
(a) Increase T: equilibrium shifts left (endothermic direction) to absorb heat ⇒ yield SO₃ decreases.
(b) Increase P: shifts to side with fewer gas moles (right, 3→2) ⇒ yield SO₃ increases.
Mark breakdown: 1 mark per prediction, 1 mark explanation.

12. [2 marks]
K_p = P_CO₂ (solids omitted).

13. [3 marks]
Only CO₂ is gaseous, so P_CO₂ = K_p = 1.2 atm. Total pressure given is consistent (2.0 atm includes inert? Actually only CO₂ present ⇒ P_CO₂ = 1.2 atm at equilibrium).
Teaching note: For heterogeneous equilibria, solid activities = 1.

14. [2 marks]
Solids have constant activity (concentration); changing amount does not change their activity, so no effect on K_p or position.

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Section C: Integrated and Data-Based

15. [2 marks]
Let x = [HI] at eq. Then [H₂]=[I₂]=(1−x/2). K_c = x²/((1−x/2)²)=50. x/(1−x/2)=√50≈7.07 ⇒ x = 7.07−3.54x ⇒ 4.54x=7.07 ⇒ x≈1.56 (impossible >1). Recheck: initial 1.0 each, x max 2.0? Actually 2HI from 1 H₂+1 I₂, so x ≤2. Solve: x² = 50(1−x/2)² ⇒ x/(1−0.5x)=7.07 ⇒ x = 7.07 − 3.535x ⇒ 4.535x = 7.07 ⇒ x = 1.56 mol dm⁻³.
Teaching note: Use ICE table.

16. [3 marks]
Rate = k[A]¹[B]²; overall order = 3.
Working: Doubling A ⇒ ×2 ⇒ order A=1; doubling B ⇒ ×4 ⇒ order B=2.

17. [3 marks]
From graph: half-life ≈ time for 0.80→0.40 ≈ 230 s. k = ln2 / t½ = 0.693/230 = 3.0×10⁻³ s⁻¹.
Image requirement: Decay curve must show exponential drop; half-life read from plot.

18. [2 marks]
Catalyst lowers Eₐ; K_c unchanged (thermodynamic constant depends only on T).

19. [3 marks]
Let x = [PCl₃]=[Cl₂] at eq. [PCl₅]=0.25−x (0.50 mol/2 dm³).
K_c = x²/(0.25−x) = 0.040 ⇒ x² = 0.010 − 0.040x ⇒ x²+0.040x−0.010=0.
x = [−0.040 + √(0.0016+0.040)]/2 = [−0.040+0.203]/2 = 0.0815 mol dm⁻³.
[PCl₅]=0.1685, [PCl₃]=[Cl₂]=0.0815 mol dm⁻³.
Mark breakdown: 1 setup, 1 solve, 1 final conc.

20. [2 marks]
K_c = e^(−ΔG°/RT); ΔG° changes with T ⇒ K_c changes. Concentration/pressure changes shift position but not ΔG° ⇒ K_c constant.