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A Level H2 Chemistry Kinetics Equilibrium Quiz

Free A Level H2 Chemistry Kinetics Equilibrium quiz, Gemma31B Exam version, with questions, answers, and A Level-style practice for Singapore students.

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A Level H2 Chemistry From Real Exams Generated by Gemma 4 31B Updated 2026-08-17

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Questions

A-Level Chemistry H2 Quiz - Kinetics Equilibrium

Name: ____________________
Class: ____________________
Date: ____________________
Score: ________ / 65

Duration: 90 Minutes
Total Marks: 65

Instructions:

Answer all questions in the spaces provided.
Use the Data Booklet where necessary.
Show all working for calculations.
Maintain appropriate significant figures (usually 3 s.f.).

Section 1: Chemical Equilibrium (Questions 1–10)

Define the term dynamic equilibrium as applied to a chemical system. [2]

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For the reaction $2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g})$ , write the expression for the equilibrium constant $K_c$ in terms of concentrations. [1]

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The value of $K_c$ for a reaction is $4.5 \times 10^{-4}$ at $300\text{ K}$ and $1.2 \times 10^{-2}$ at $400\text{ K}$ . State and explain the effect of temperature on the equilibrium constant for this reaction. [2]

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Predict and explain the shift in equilibrium position for the reaction $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})$ when the total pressure of the system is increased at constant temperature. [3]

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A reaction has a $K_p$ value of $0.25\text{ atm}$ . If the volume of the container is halved, explain why the value of $K_p$ remains unchanged. [2]

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For the equilibrium $\text{PCl}_5(\text{g}) \rightleftharpoons \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g})$ , the initial concentration of $\text{PCl}_5$ is $0.50\text{ mol dm}^{-3}$ . At equilibrium, the concentration of $\text{PCl}_3$ is $0.20\text{ mol dm}^{-3}$ . Calculate the value of $K_c$ . [3]

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Explain why the addition of a catalyst does not change the position of equilibrium. [2]

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Consider the reaction $\text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g})$ . If the concentration of $\text{H}_2$ is doubled while other concentrations remain constant, describe the change in the reaction quotient $Q_c$ relative to $K_c$ and the subsequent direction of the shift. [3]

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Write the expression for $K_p$ for the reaction $\text{CaCO}_3(\text{s}) \rightleftharpoons \text{CaO}(\text{s}) + \text{CO}_2(\text{g})$ . Explain why only one species appears in the expression. [2]

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A system is at equilibrium. If the reaction is exothermic, explain using Le Chatelier's principle how an increase in temperature affects the yield of products. [3]

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Section 2: Reaction Kinetics (Questions 11–20)

Define the order of reaction with respect to a specific reactant. [2]

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For the reaction $\text{A} + \text{B} \rightarrow \text{C}$ , the rate is found to be independent of the concentration of $\text{B}$ but proportional to the square of the concentration of $\text{A}$ . Write the rate equation. [2]

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The rate constant $k$ for a reaction is $2.5 \times 10^{-3}\text{ mol}^{-1}\text{dm}^3\text{s}^{-1}$ . State the overall order of this reaction based on the units of $k$ . [2]

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Explain the difference between the rate-determining step and the overall reaction rate in a multi-step mechanism. [2]

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A reaction is first-order with respect to reactant $\text{X}$ . If the initial concentration of $\text{X}$ is doubled, how does the initial rate of reaction change? [1]

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Using the Arrhenius equation, explain why a small increase in temperature leads to a significant increase in the rate of reaction. [3]

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For the reaction $2\text{NO}(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow 2\text{NOCl}(\text{g})$ , the following data were obtained:
- Exp 1: $[\text{NO}] = 0.10, [\text{Cl}_2] = 0.10, \text{Rate} = 1.2 \times 10^{-4}\text{ mol dm}^{-3}\text{s}^{-1}$
- Exp 2: $[\text{NO}] = 0.20, [\text{Cl}_2] = 0.10, \text{Rate} = 4.8 \times 10^{-4}\text{ mol dm}^{-3}\text{s}^{-1}$
- Exp 3: $[\text{NO}] = 0.10, [\text{Cl}_2] = 0.20, \text{Rate} = 2.4 \times 10^{-4}\text{ mol dm}^{-3}\text{s}^{-1}$ Determine the rate equation for this reaction. [4]
  
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Draw a Maxwell-Boltzmann distribution curve for particles at temperature $T_1$ and $T_2$ (where $T_2 > T_1$ ). Label the activation energy $E_a$ and the area representing particles with energy $\ge E_a$ . [4]

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A reaction has an activation energy of $50\text{ kJ mol}^{-1}$ . If a catalyst is added that lowers the activation energy to $30\text{ kJ mol}^{-1}$ , explain in terms of collision theory why the rate increases. [3]

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The decomposition of $\text{N}_2\text{O}_4(\text{g}) \rightarrow 2\text{NO}_2(\text{g})$ is first-order. If the half-life is $100\text{ s}$ , calculate the rate constant $k$ in $\text{s}^{-1}$ . [3]

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Answers

Answer Key - Kinetics Equilibrium Quiz

A state where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant over time. (2)
$K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}$ (1)
$K_c$ increases as temperature increases. The reaction is endothermic; therefore, increasing temperature shifts the equilibrium to the right to oppose the change, increasing the concentration of products. (2)
Shift to the right (towards $\text{NH}_3$ ). Increasing pressure shifts the equilibrium to the side with fewer moles of gaseous molecules (4 moles $\rightarrow$ 2 moles) to reduce the pressure. (3)
$K_p$ is only dependent on temperature. While partial pressures change when volume changes, the ratio defined by $K_p$ remains constant at a constant temperature. (2)
- $\text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2$
- Initial: $0.50, 0, 0$
- Change: $-0.20, +0.20, +0.20$
- Equil: $0.30, 0.20, 0.20$
- $K_c = \frac{(0.20)(0.20)}{0.30} = \frac{0.04}{0.30} = 0.133\text{ mol dm}^{-3}$ (3)
A catalyst increases the rate of both the forward and reverse reactions equally by providing an alternative pathway with lower activation energy. (2)
$Q_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$ . If $[\text{H}_2]$ doubles, $Q_c < K_c$ . The system will shift to the right (forward direction) to restore equilibrium. (3)
$K_p = P(\text{CO}_2)$ . Pure solids ( $\text{CaCO}_3$ and $\text{CaO}$ ) have constant activity/concentration and are incorporated into the equilibrium constant. (2)
For an exothermic reaction, heat is a product. Increasing temperature shifts the equilibrium to the left (towards reactants) to absorb the added heat, decreasing the yield of products. (3)
The power to which the concentration of a reactant is raised in the rate equation. (2)
$\text{Rate} = k[\text{A}]^2$ (2)
Second order. Units $\text{mol}^{-1}\text{dm}^3\text{s}^{-1}$ correspond to $k = \text{Rate} / [\text{Reactant}]^2$ . (2)
The rate-determining step is the slowest step in a mechanism; its rate governs the overall rate of the reaction. (2)
The initial rate doubles. (1)
$k = Ae^{-E_a/RT}$ . An increase in $T$ increases the exponential term $e^{-E_a/RT}$ , meaning a larger fraction of molecules possess energy $\ge E_a$ , leading to more successful collisions per unit time. (3)
- Compare Exp 1 & 2: $[\text{NO}]$ doubles, rate increases $4\times$ ( $4.8/1.2$ ). Order w.r.t $\text{NO} = 2$ .
- Compare Exp 1 & 3: $[\text{Cl}_2]$ doubles, rate increases $2\times$ ( $2.4/1.2$ ). Order w.r.t $\text{Cl}_2 = 1$ .
- $\text{Rate} = k[\text{NO}]^2[\text{Cl}_2]$ (4)
- X-axis: Kinetic Energy, Y-axis: Number of particles.
- $T_2$ curve is flatter and shifted right.
- $E_a$ line marked on X-axis.
- Shaded area under $T_2$ curve to the right of $E_a$ is larger than for $T_1$ . (4)
Lowering $E_a$ increases the fraction of molecules that have energy $\ge E_a$ during collisions. This increases the frequency of successful collisions, thus increasing the rate. (3)
For first-order: $k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{100} = 6.93 \times 10^{-3}\text{ s}^{-1}$ (3)