A Level H1 Chemistry Kinetics Equilibrium Quiz

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A-Level Chemistry H1 Quiz - Kinetics Equilibrium

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

Duration: 60 minutes
Total Marks: 40

Instructions:

• Answer all 20 questions.
• Section A: Short Answer (1–7)
• Section B: Structured & Calculation (8–14)
• Section C: Data Interpretation & Extended Reasoning (15–20)
• Show all working for calculation questions. Use appropriate units and chemical notation.

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Section A: Short Answer (Questions 1–7)

1. State what is meant by the term rate of reaction. [1]

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2. Write the rate equation for a reaction where the rate is directly proportional to the concentration of A and independent of the concentration of B, given rate constant $k$. [1]

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3. Write an equilibrium expression, $K_c$, for the reaction:
$\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ [1]

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4. State Le Chatelier’s Principle. [1]

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5. What is meant by the term activation energy, $E_a$? [1]

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6. State the order of reaction with respect to a reactant if doubling its concentration doubles the rate. [1]

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7. Explain why a catalyst increases the rate of a reaction. [1]

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Section B: Structured & Calculation (Questions 8–14)

8. For the reaction $\text{A} + \text{B} \rightarrow \text{products}$, the following initial rates were obtained:

Experiment	[A] / mol dm⁻³	[B] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	$2.0 \times 10^{-4}$
2	0.20	0.10	$4.0 \times 10^{-4}$
3	0.10	0.20	$2.0 \times 10^{-4}$

(a) Deduce the order with respect to A. [1]
(b) Deduce the order with respect to B. [1]
(c) Write the rate equation. [1]

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9. The decomposition of $\text{N}_2\text{O}_5$ is first order with rate constant $k = 5.0 \times 10^{-4}\ \text{s}^{-1}$. Calculate the half-life of $\text{N}_2\text{O}_5$. [2]

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10. For the equilibrium:
$\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}$
At equilibrium, $[\text{H}_2] = 0.20\ \text{mol dm}^{-3}$, $[\text{I}_2] = 0.20\ \text{mol dm}^{-3}$, $[\text{HI}] = 0.60\ \text{mol dm}^{-3}$. Calculate $K_c$. [2]

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11. State and explain the effect on the equilibrium position of the reaction
$\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)} \quad \Delta H = -92\ \text{kJ mol}^{-1}$
when (a) temperature is increased, and (b) pressure is increased. [2]

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12. The rate constant for a reaction at 300 K is $1.2 \times 10^{-3}\ \text{s}^{-1}$ and at 310 K is $2.4 \times 10^{-3}\ \text{s}^{-1}$. Suggest why the rate constant doubles with a 10 K increase. [2]

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13. For the reaction $\text{2NO}_2\text{(g)} \rightleftharpoons 2\text{NO(g)} + \text{O}_2\text{(g)}$, write the $K_c$ expression. [1]
Hence state the units of $K_c$. [1]

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14. A first-order reaction has a half-life of 120 s. Calculate the fraction of reactant remaining after 360 s. [2]

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Section C: Data Interpretation & Extended Reasoning (Questions 15–20)

15. The graph below shows concentration–time curves for a reactant in two experiments at different temperatures.

Image pending generation: graph for Q15.

(a) Which curve represents the higher temperature? [1]
(b) Explain your answer using collision theory. [2]

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16. The Boltzmann distribution of molecular energies is shown below.

Image pending generation: graph for Q16.

(a) State the effect on the shaded area when a catalyst is added. [1]
(b) Explain how the catalyst changes the distribution. [2]

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17. The reaction $\text{CO(g)} + \text{NO}_2\text{(g)} \rightleftharpoons \text{CO}_2\text{(g)} + \text{NO(g)}$ is exothermic.
(a) State the effect of increasing temperature on $K_c$. [1]
(b) Explain your answer in terms of equilibrium position. [2]

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18. In the industrial contact process, $\text{2SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)} \quad \Delta H = -197\ \text{kJ mol}^{-1}$.
(a) State two conditions used to maximise $\text{SO}_3$ yield. [2]
(b) Explain why a compromise is needed between rate and equilibrium yield. [2]

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19. The table shows initial rates for $\text{P} + \text{Q} \rightarrow \text{R}$:

[P]	[Q]	Rate
0.05	0.05	$1.0 \times 10^{-5}$
0.10	0.05	$4.0 \times 10^{-5}$
0.05	0.10	$2.0 \times 10^{-5}$

(a) Determine order w.r.t. P and Q. [2]
(b) Calculate $k$ with units. [2]

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20. Explain, using kinetic and equilibrium principles, why increasing pressure increases the yield of $\text{NH}_3$ in the Haber process but does not change the rate constant. [3]

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

Total Marks: 40
Topic: Kinetics & Equilibrium

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Section A: Short Answer

1. [1 mark]
Rate of reaction is the change in concentration of a reactant or product per unit time.
Teaching note: Usually expressed as $\text{mol dm}^{-3}\ \text{s}^{-1}$. It measures how fast a reaction proceeds.

2. [1 mark]
$\text{rate} = k[\text{A}]^1[\text{B}]^0 = k[\text{A}]$
Teaching note: Order 0 in B means concentration of B does not affect rate; $[\text{B}]^0 = 1$.

3. [1 mark]
$K_c = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$
Teaching note: Products over reactants, each raised to stoichiometric coefficient. Solids/liquids omitted (all gases here).

4. [1 mark]
Le Chatelier’s Principle: When a system at equilibrium is subjected to a change in condition (concentration, pressure, temperature), the equilibrium shifts to oppose the change.
Common mistake: Saying “cancels out” instead of “opposes”.

5. [1 mark]
Activation energy is the minimum energy required for a collision between reactant molecules to result in a successful reaction.
Teaching note: Represented as $E_a$ on energy profile diagrams.

6. [1 mark]
First order.
Reasoning: Rate ∝ [reactant]$^1$; doubling concentration doubles rate.

7. [1 mark]
A catalyst provides an alternative reaction pathway with lower activation energy, increasing the fraction of molecules with sufficient energy to react.
Teaching note: Catalyst is not consumed; does not affect equilibrium.

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Section B: Structured & Calculation

8. [3 marks total]
(a) Order w.r.t. A = 1 (Exp1→2: [A] doubles, rate doubles). [1]
(b) Order w.r.t. B = 0 (Exp1→3: [B] doubles, rate unchanged). [1]
(c) $\text{rate} = k[\text{A}]^1[\text{B}]^0 = k[\text{A}]$. [1]

9. [2 marks]
For first order: $t_{1/2} = \dfrac{\ln 2}{k}$
$= \dfrac{0.693}{5.0 \times 10^{-4}\ \text{s}^{-1}} = 1.386 \times 10^3\ \text{s} \approx 1.4 \times 10^3\ \text{s}$ (or 1386 s). [2]
Marking: 1 for formula, 1 for correct value.

10. [2 marks]
$K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \dfrac{(0.60)^2}{(0.20)(0.20)} = \dfrac{0.36}{0.04} = 9.0$ [2]
Units: none (cancels). Mark: 1 for expression, 1 for value.

11. [2 marks]
(a) Increasing temperature shifts equilibrium left (towards reactants) because forward reaction is exothermic; system opposes by absorbing heat. [1]
(b) Increasing pressure shifts equilibrium right (towards fewer moles: 4 → 2) to reduce pressure. [1]

12. [2 marks]
Rate constant increases with temperature because molecules have more kinetic energy; more collisions exceed $E_a$ (collision theory). [2]
Note: Exact doubling not required to explain; link T to $k$ via Arrhenius concept.

13. [2 marks]
(a) $K_c = \dfrac{[\text{NO}]^2[\text{O}_2]}{[\text{NO}_2]^2}$ [1]
(b) Units: $\dfrac{(\text{mol dm}^{-3})^2(\text{mol dm}^{-3})}{(\text{mol dm}^{-3})^2} = \text{mol dm}^{-3}$ [1]

14. [2 marks]
360 s = 3 half-lives. Fraction remaining = $(1/2)^3 = 1/8 = 0.125$. [2]
Method: $n = t / t_{1/2} = 360/120 = 3$.

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Section C: Data Interpretation & Extended Reasoning

15. [3 marks]
(a) T2 (320 K) – steeper fall. [1]
(b) Higher temperature gives molecules greater average kinetic energy; larger fraction exceeds $E_a$; more frequent successful collisions; faster rate. [2]

16. [3 marks]
(a) Shaded area increases (more molecules beyond $E_a$). [1]
(b) Catalyst lowers $E_a$ (line moves left); same distribution shape but more molecules now have energy > new $E_a$. [2]

17. [3 marks]
(a) $K_c$ decreases. [1]
(b) Exothermic forward reaction; increasing T shifts equilibrium left (to absorb heat), reducing product concentration; $K_c$ (products/reactants) falls. [2]

18. [4 marks]
(a) Lower temperature (favours exothermic), higher pressure (fewer moles: 3→2). [2]
(b) Low T slows rate (fewer collisions); high P costly/equipment-limited; compromise (e.g. 450°C, 2 atm) gives acceptable rate and yield. [2]

19. [4 marks]
(a) P: Exp1→2 [P]×2, rate×4 → order 2. Q: Exp1→3 [Q]×2, rate×2 → order 1. [2]
(b) $\text{rate} = k[\text{P}]^2[\text{Q}]$; using Exp1: $1.0\times10^{-5} = k(0.05)^2(0.05) = k(1.25\times10^{-4})$; $k = 8.0\times10^{-2}\ \text{dm}^6\ \text{mol}^{-2}\ \text{s}^{-1}$. [2]

20. [3 marks]
Increasing pressure increases concentration of gases; more collisions per unit time → higher rate (kinetic). Equilibrium shifts to side with fewer moles (right, 4→2) → more NH₃ (Le Chatelier). Rate constant $k$ depends only on temperature, not pressure, so unchanged. [3]