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A Level H1 Chemistry Kinetics Equilibrium Quiz

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A Level H1 Chemistry AI Generated Generated by Gemma 4 31B Updated 2026-08-17

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Questions

A-Level Chemistry H1 Quiz - Kinetics Equilibrium

Name: ____________________
Class: ____________________
Date: ____________________
Score: ________ / 50

Duration: 60 Minutes
Total Marks: 50

Instructions:

Answer all questions in the spaces provided.
Show all working for calculations.
Use a scientific calculator where necessary.
State the units for all final numerical answers.

Section A: Reaction Kinetics (Questions 1–10)

Define the term rate of reaction. [1]
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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 B but proportional to the square of the concentration of A. Write the rate equation for this reaction. [1]
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Explain, in terms of collision theory, why an increase in temperature typically increases the rate of a chemical reaction. [2]
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A reaction is second-order with respect to reactant X. If the concentration of X is tripled, by what factor does the initial rate of reaction increase? [1]
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(a) What is the unit for the rate constant $k$ for a first-order reaction? [1] \

(b) How does the unit for $k$ change for a second-order reaction? [1] \
Draw a Maxwell-Boltzmann distribution curve for a set of molecules at temperature $T_1$ . On the same axes, draw the curve for the same molecules at a higher temperature $T_2$ . Label the activation energy $E_a$ . [3]

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The decomposition of $\text{N}_2\text{O}_5(\text{g})$ is a first-order reaction. If the initial concentration is $0.10\text{ mol dm}^{-3}$ and the rate constant is $5.0 \times 10^{-4}\text{ s}^{-1}$ , calculate the initial rate of reaction. [2]
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Explain why a catalyst increases the rate of reaction without being consumed in the process. [2]
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In a reaction $\text{P} + \text{Q} \rightarrow \text{R}$ , the following data was obtained:
- Exp 1: $[\text{P}] = 0.1, [\text{Q}] = 0.1, \text{Rate} = 2.0 \times 10^{-3}$
- Exp 2: $[\text{P}] = 0.2, [\text{Q}] = 0.1, \text{Rate} = 8.0 \times 10^{-3}$
- Exp 3: $[\text{P}] = 0.1, [\text{Q}] = 0.2, \text{Rate} = 4.0 \times 10^{-3}$ Determine the order of reaction with respect to P and Q. [2]
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Describe how the rate of reaction is affected when the surface area of a solid reactant is increased. Justify your answer using collision theory. [2]
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Section B: Chemical Equilibrium (Questions 11–20)

State the conditions necessary for a system to reach a state of dynamic equilibrium. [2]
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For the equilibrium: $\text{PCl}_5(\text{g}) \rightleftharpoons \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g})$ , write the expression for the equilibrium constant $K_c$ . [1]
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Consider the reaction: $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \quad \Delta H = -92\text{ kJ mol}^{-1}$ . Predict the effect of increasing the temperature on: (a) The position of equilibrium [1] (b) The value of $K_c$ [1]
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Explain why increasing the pressure of a gaseous system shifts the equilibrium toward the side with fewer moles of gas, according to Le Chatelier's Principle. [2]
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For the reaction $\text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g})$ , the $K_c$ is $50.0$ at $448^\circ\text{C}$ . If $[\text{H}_2] = 0.10\text{ mol dm}^{-3}$ and $[\text{I}_2] = 0.10\text{ mol dm}^{-3}$ at equilibrium, calculate the equilibrium concentration of $\text{HI}$ . [3]
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What is the effect of adding a catalyst to a system already at equilibrium? Explain your answer. [2]
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In the Haber process, a compromise temperature is used. Explain why a very low temperature, despite favoring the forward reaction, is not used industrially. [2]
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For the equilibrium $\text{CO}(\text{g}) + \text{H}_2\text{O}(\text{g}) \rightleftharpoons \text{CO}_2(\text{g}) + \text{H}_2(\text{g})$ , how does the addition of an inert gas (like Argon) at constant volume affect the position of equilibrium? Explain. [2]
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A reaction has $K_c = 1.5 \times 10^{-5}$ . Does this value indicate that the equilibrium mixture consists mainly of reactants or products? Justify your answer. [2]
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For the reaction $\text{A}(\text{s}) + \text{B}(\text{g}) \rightleftharpoons \text{C}(\text{g})$ , write the expression for $K_c$ and explain why the concentration of A is omitted. [2]
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Answers

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

Section A: Reaction Kinetics

Definition: The change in concentration of a reactant or product per unit time. [1]
Rate Equation: $\text{Rate} = k[\text{A}]^2$ [1]
Explanation:
- Higher temperature increases the average kinetic energy of molecules. [1]
- A greater fraction of molecules possess energy $\ge E_a$ , leading to more frequent successful collisions. [1]
Factor: $3^2 = 9$ times. [1]
Units:
- (a) $\text{s}^{-1}$ [1]
- (b) $\text{mol}^{-1}\text{dm}^3\text{s}^{-1}$ [1]
Diagram:
- X-axis: Energy; Y-axis: Number of molecules. [1]
- $T_2$ curve: Flatter peak, shifted right, higher tail. [1]
- $E_a$ line: Vertical line on the right side of the curve. [1]
Calculation:
- $\text{Rate} = k[\text{N}_2\text{O}_5]^1$
- $\text{Rate} = (5.0 \times 10^{-4})(0.10) = 5.0 \times 10^{-5}\text{ mol dm}^{-3}\text{s}^{-1}$ [2]
Catalyst:
- Provides an alternative reaction pathway with a lower activation energy. [1]
- It is regenerated at the end of the mechanism, so it is not consumed. [1]
Orders:
- P: Exp 1 $\to$ 2: $[\text{P}]$ doubles, rate quadruples ( $2^2=4$ ). Order = 2. [1]
- Q: Exp 1 $\to$ 3: $[\text{Q}]$ doubles, rate doubles ( $2^1=2$ ). Order = 1. [1]
Surface Area:
- Rate increases. [1]
- More particles are exposed at the surface, increasing the frequency of collisions per unit time. [1]

Section B: Chemical Equilibrium

Conditions:
- Closed system (no matter enters or leaves). [1]
- Constant temperature and pressure. [1]
Expression: $K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}$ [1]
Temperature Effect:
- (a) Shifts to the left (reactants) as the reaction is exothermic. [1]
- (b) $K_c$ decreases. [1]
Pressure:
- Increasing pressure increases the frequency of collisions. [1]
- The system shifts to the side with fewer gas moles to reduce the pressure/stress. [1]
Calculation:
- $50.0 = \frac{[\text{HI}]^2}{(0.10)(0.10)}$
- $[\text{HI}]^2 = 50.0 \times 0.01 = 0.5$
- $[\text{HI}] = \sqrt{0.5} = 0.707\text{ mol dm}^{-3}$ [3]
Catalyst in Equilibrium:
- No effect on the position of equilibrium. [1]
- It increases the rate of both forward and reverse reactions equally. [1]
Haber Process:
- Low temperature makes the reaction rate too slow for industrial viability. [2]
Inert Gas:
- No effect. [1]
- At constant volume, the partial pressures/concentrations of the reacting species remain unchanged. [1]
Analysis:
- Mainly reactants. [1]
- $K_c \ll 1$ indicates the equilibrium lies far to the left. [1]
Expression & Logic:
- $K_c = \frac{[\text{C}]}{[\text{B}]}$ [1]
- The concentration of a pure solid is constant and is incorporated into the $K_c$ value. [1]