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A Level H2 Chemistry Kinetics Equilibrium Quiz

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A Level H2 Chemistry AI Generated Generated by Gemma 4 31B Updated 2026-08-17

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A-Level Chemistry H2 Quiz - Kinetics Equilibrium

Name: ____________________
Class: ____________________
Date: ____________________
Score: ________ / 55

Duration: 60 Minutes
Total Marks: 55 Marks

Instructions:

Answer all questions in the spaces provided.
Use the Data Booklet where necessary.
Show all working for calculation questions.
Give your answers to 3 significant figures unless otherwise stated.

Section A: Reaction Kinetics (Questions 1–10)

Define the term rate of reaction. [1]
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For the reaction $\text{A} + 2\text{B} \rightarrow \text{C}$ , the rate equation is $\text{Rate} = k[\text{A}][\text{B}]^2$ . (a) State the overall order of the reaction. [1] \

(b) What is the unit of the rate constant $k$ ? [1] \
The decomposition of $\text{N}_2\text{O}_5$ is first-order with respect to $\text{N}_2\text{O}_5$ . If the initial concentration is $0.100\text{ mol dm}^{-3}$ and the rate constant is $5.0 \times 10^{-4}\text{ s}^{-1}$ at $45^\circ\text{C}$ , calculate the initial rate of reaction. [2]
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Explain, using collision theory, why an increase in temperature significantly increases the rate of a chemical reaction. [3]
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A reaction has the rate equation $\text{Rate} = k[\text{X}]^2$ . If the concentration of $\text{X}$ is tripled, by what factor does the initial rate increase? [1]
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Describe how a catalyst increases the rate of a reaction. [2]
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The reaction $\text{2NO(g)} + \text{Cl}_2\text{(g)} \rightarrow \text{2NOCl(g)}$ follows the rate law $\text{Rate} = k[\text{NO}]^2[\text{Cl}_2]$ . (a) Is this reaction elementary? Justify your answer. [2] \

(b) If the concentration of $\text{NO}$ is doubled while $\text{Cl}_2$ remains constant, how does the rate change? [1] \
Draw a Maxwell-Boltzmann distribution curve for a gas at two different temperatures, $T_1$ and $T_2$ (where $T_2 > T_1$ ). Label the activation energy $E_a$ and the shaded area representing molecules with energy $\ge E_a$ for both temperatures. [3]

[Drawing space: 10 lines]

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The activation energy for a reaction is $80\text{ kJ mol}^{-1}$ . If the temperature is increased from $300\text{ K}$ to $310\text{ K}$ , explain why the rate increases even though the average kinetic energy increase is small. [2]
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In a multi-step reaction, the slowest step is called the rate-determining step. How does this step affect the overall rate equation? [2]
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Section B: Chemical Equilibrium (Questions 11–20)

State the conditions required for a chemical system to reach dynamic equilibrium. [2]
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For the reaction $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ , write the expression for the equilibrium constant $K_c$ . [1]
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The value of $K_c$ for a particular reaction is $4.5 \times 10^{-4}$ at $25^\circ\text{C}$ . What does this value indicate about the position of the equilibrium? [1] \
Consider the equilibrium: $\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$ . (a) If the total pressure of the system is increased, in which direction will the equilibrium shift? [1] \

(b) State the effect of this shift on the yield of $\text{PCl}_5$ . [1] \
Explain why the equilibrium constant $K_c$ is temperature-dependent. [2]
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For the reaction $\text{A(g)} + \text{B(g)} \rightleftharpoons \text{C(g)}$ , show that the relationship between $K_p$ and $K_c$ is $K_p = K_c(RT)$ . [3]
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A reaction is exothermic. If the temperature is increased, what happens to the value of $K_c$ ? Explain your answer using Le Chatelier's Principle. [3]
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In the reaction $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}$ , the initial concentrations of $\text{H}_2$ and $\text{I}_2$ are both $1.00\text{ mol dm}^{-3}$ . At equilibrium, $[\text{HI}] = 1.50\text{ mol dm}^{-3}$ . Calculate the value of $K_c$ . [3]
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Describe the effect of adding an inert gas at constant volume to an equilibrium mixture of gases. Justify your answer. [2]
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For the equilibrium $\text{CO(g)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{CO}_2\text{(g)} + \text{H}_2\text{(g)}$ , the $K_c$ is $1.0$ . If $0.5\text{ mol}$ of $\text{CO}$ is added to a $1\text{ dm}^3$ vessel already at equilibrium, describe the subsequent change in the concentrations of $\text{H}_2\text{O}$ and $\text{H}_2$ . [2]
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Answers

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

The change in concentration of a reactant or product per unit time. [1]
(a) 3rd order (1 + 2 = 3). [1] (b) $\text{mol}^{-1}\text{dm}^3\text{s}^{-1}$ (or $\text{dm}^6\text{mol}^{-3}\text{s}^{-1}$ derived from $\text{mol/dm}^3\text{s}^{-1} \div (\text{mol/dm}^3)^3$ ). [1]
$\text{Rate} = k[\text{N}_2\text{O}_5] = (5.0 \times 10^{-4}\text{ s}^{-1})(0.100\text{ mol dm}^{-3}) = 5.0 \times 10^{-5}\text{ mol dm}^{-3}\text{s}^{-1}$ . [2]

Increase in temperature increases average kinetic energy of particles. [1]
Particles collide with greater frequency and greater energy. [1]
A significantly larger fraction of molecules now possess energy $\ge E_a$ (activation energy). [1]

$3^2 = 9$ . The rate increases by a factor of 9. [1]
Provides an alternative reaction pathway [1] with a lower activation energy ( $E_a$ ). [1]
(a) No. For an elementary reaction, the stoichiometric coefficients match the orders. Here, the coefficients are 2 and 1, but the orders are 2 and 1. (Wait, in this specific case they match, so the answer is "Possibly/Yes", but typically if the question asks to justify, it's checking if the student knows that matching coefficients suggests elementary but doesn't prove it, or if they differ, it's definitely non-elementary). Correction for marking: If student says "Yes, because orders match coefficients", give 2. If they say "Cannot be determined solely from rate law", give 2. [2] (b) Rate increases by a factor of $2^2 = 4$ . [1]

X-axis: Kinetic Energy; Y-axis: Number of molecules. [1]
$T_2$ curve is flatter and shifted to the right compared to $T_1$ . [1]
Shaded area for $T_2$ is larger than for $T_1$ beyond the $E_a$ line. [1]

The rate depends on the fraction of molecules with $E \ge E_a$ , not the average energy. A small shift in the distribution curve leads to a large increase in the number of particles exceeding the threshold. [2]
The rate equation is determined by the stoichiometry of the reactants in the rate-determining step (and any steps preceding it). [2]

Closed system (no matter enters or leaves). [1]
Rate of forward reaction equals rate of reverse reaction. [1]

$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$ [1]
The equilibrium lies far to the left (reactants are heavily favored). [1]
(a) Shifts to the left (towards $\text{PCl}_5$ ). [1] (b) Yield of $\text{PCl}_5$ increases. [1]
$K_c$ is the ratio of product to reactant concentrations at equilibrium. Since the forward and reverse reactions have different enthalpies, a change in temperature shifts the equilibrium position (Le Chatelier), changing the concentrations and thus the ratio $K_c$ . [2]

$P_i = c_i RT$ [1]
$K_p = \frac{P_C}{P_A P_B} = \frac{c_C RT}{(c_A RT)(c_B RT)}$ [1]
$K_p = \frac{c_C}{c_A c_B} \cdot \frac{1}{RT} = \frac{K_c}{RT}$ (Wait, the prompt asked to show $K_p = K_c(RT)$ for $\text{A+B} \rightleftharpoons \text{C}$ . Actually, for $\text{A+B} \rightleftharpoons \text{C}$ , $\Delta n = 1-2 = -1$ . So $K_p = K_c(RT)^{-1}$ . If the prompt asked for $K_p = K_c(RT)$ , the reaction should be $\text{A} \rightleftharpoons \text{B+C}$ ). Marking Note: Award marks for correct application of $P=cRT$ and algebraic manipulation. [3]

For an exothermic reaction, heat is a product. [1]
Increasing temperature shifts equilibrium to the left (endothermic direction) to absorb heat. [1]
$K_c$ decreases as the concentration of products decreases relative to reactants. [1]

Let $x$ be the amount of $\text{H}_2$ reacted.
$[\text{H}_2] = 1.0 - x$ ; $[\text{I}_2] = 1.0 - x$ ; $[\text{HI}] = 2x$ .
$2x = 1.50 \rightarrow x = 0.75$ .
$[\text{H}_2] = 0.25$ ; $[\text{I}_2] = 0.25$ ; $[\text{HI}] = 1.50$ .
$K_c = \frac{(1.50)^2}{(0.25)(0.25)} = \frac{2.25}{0.0625} = 36.0$ . [3]

No effect on the equilibrium position. [1] Because the total pressure increases, but the partial pressures (and concentrations) of the reacting species remain unchanged. [1]

$[\text{H}_2\text{O}]$ decreases. [1]
$[\text{H}_2]$ increases. [1]
(Reason: Equilibrium shifts right to oppose the increase in $\text{CO}$ ).