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A Level H1 Chemistry Kinetics Equilibrium Quiz

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Questions

A-Level Chemistry H1 Quiz - Kinetics Equilibrium

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 40

Duration: 45 minutes
Total Marks: 40

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all working for calculation questions.
The use of a scientific calculator is allowed.
Data Booklet is available for reference.

Section A: Multiple Choice & Short Concepts (Questions 1-5)

1. Which statement correctly describes the effect of a catalyst on the equilibrium constant ( $K_c$ ) and the rate of reaction?

	Effect on $K_c$	Effect on Rate
A	Increases	Increases
B	No change	Increases
C	No change	No change
D	Decreases	Increases

Answer: ______ [1]

2. For the reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ , $\Delta H = -196 \text{ kJ mol}^{-1}$ . Which change will increase the yield of $SO_3$ at equilibrium?

A. Increasing the temperature
B. Decreasing the pressure
C. Adding a catalyst
D. Removing $SO_3$ as it forms

Answer: ______ [1]

3. The rate equation for a reaction is given as: Rate $= k[A][B]^2$ . What are the units of the rate constant, $k$ , if concentration is in $\text{mol dm}^{-3}$ and time is in seconds?

A. $\text{s}^{-1}$
B. $\text{dm}^3 \text{mol}^{-1} \text{s}^{-1}$
C. $\text{dm}^6 \text{mol}^{-2} \text{s}^{-1}$
D. $\text{mol dm}^{-3} \text{s}^{-1}$

Answer: ______ [1]

4. Define the term activation energy.

_________________________________________________________________________ [1]

5. State Le Chatelier’s Principle.

_________________________________________________________________________ [1]

Section B: Kinetics Concepts & Graphs (Questions 6-10)

6. Explain why increasing the temperature increases the rate of a reaction, referring to particle collisions and energy.

_________________________________________________________________________ [2]

7. A reaction is zero order with respect to reactant X. Sketch the graph of concentration of X against time.

Graph space

Answer space

[2]

8. The decomposition of hydrogen peroxide is catalyzed by iodide ions: $H_2O_2(aq) + 2I^-(aq) + 2H^+(aq) \rightarrow I_2(aq) + 2H_2O(l)$

The following initial rate data were obtained at constant temperature:

Experiment	$[H_2O_2]$ / $\text{mol dm}^{-3}$	$[I^-]$ / $\text{mol dm}^{-3}$	$[H^+]$ / $\text{mol dm}^{-3}$	Initial Rate / $\text{mol dm}^{-3} \text{s}^{-1}$
1	0.10	0.10	0.10	$2.0 \times 10^{-4}$
2	0.20	0.10	0.10	$4.0 \times 10^{-4}$
3	0.10	0.20	0.10	$4.0 \times 10^{-4}$
4	0.10	0.10	0.20	$2.0 \times 10^{-4}$

Determine the order of reaction with respect to $H_2O_2$ .

_________________________________________________________________________ [1]

9. Using the data in Question 8, determine the order of reaction with respect to $I^-$ .

_________________________________________________________________________ [1]

10. Using the data in Question 8, determine the order of reaction with respect to $H^+$ .

_________________________________________________________________________ [1]

Section C: Kinetics Calculations & Theory (Questions 11-15)

11. Write the rate equation for the reaction in Question 8 based on the orders determined.

_________________________________________________________________________ [1]

12. Calculate the value of the rate constant, $k$ , using data from Experiment 1 in Question 8.

Answer space

Value: _______________ [2]

13. State the units of the rate constant, $k$ , for the reaction in Question 8.

_________________________________________________________________________ [1]

14. Explain, in terms of collision theory, why the rate of reaction increases when the concentration of $H_2O_2$ is increased.

_________________________________________________________________________ [2]

15. On the axes below, sketch the Maxwell-Boltzmann distribution curve for gas particles at temperature $T_1$ . On the same axes, sketch the curve for a higher temperature $T_2$ . Label the activation energy, $E_a$ .

Graph space

Answer space

[2]

Section D: Equilibrium Principles & Calculations (Questions 16-20)

16. Nitrogen and hydrogen react to form ammonia in the Haber Process: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \quad \Delta H = -92 \text{ kJ mol}^{-1}$

Write the expression for the equilibrium constant, $K_c$ , for this reaction.

$K_c =$ [1]

17. In a sealed vessel of volume $2.0 \text{ dm}^3$ , $1.0 \text{ mol}$ of $N_2$ and $3.0 \text{ mol}$ of $H_2$ are mixed. At equilibrium, $0.4 \text{ mol}$ of $NH_3$ is present. Calculate the equilibrium amount (in mol) of $N_2$ .

Amount of $N_2$ : _______________ mol [1]

18. Using the data in Question 17, calculate the equilibrium amount (in mol) of $H_2$ .

Amount of $H_2$ : _______________ mol [1]

19. Calculate the value of $K_c$ at this temperature using the equilibrium amounts from Questions 17 and 18.

Answer space

Value: _______________
Units: _______________ [3]

20. State and explain the effect of increasing the temperature on the value of $K_c$ for the Haber Process.

_________________________________________________________________________ [2]

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Answers

A-Level Chemistry H1 Quiz - Kinetics Equilibrium (Answer Key)

Total Marks: 40

Section A: Multiple Choice & Short Concepts

1. B [1]

Catalysts lower activation energy, increasing rate, but do not affect the position of equilibrium or $K_c$ .

2. D [1]

Reaction is exothermic ( $\Delta H < 0$ ). Lowering T favors forward reaction.
4 moles gas $\rightarrow$ 2 moles gas. Increasing P favors forward reaction.
Removing product ( $SO_3$ ) shifts equilibrium to the right to replace it.

3. C [1]

Rate = $\text{mol dm}^{-3} \text{s}^{-1}$ .
$[A][B]^2 = (\text{mol dm}^{-3})^3 = \text{mol}^3 \text{dm}^{-9}$ .
$k = \text{Rate} / [A][B]^2 = (\text{mol dm}^{-3} \text{s}^{-1}) / (\text{mol}^3 \text{dm}^{-9}) = \text{dm}^6 \text{mol}^{-2} \text{s}^{-1}$ .

4. Activation Energy: [1]

The minimum energy [1] required for a collision to result in a reaction / for bonds to break.

5. Le Chatelier’s Principle: [1]

If a system at equilibrium is subjected to a change in conditions [1], the position of equilibrium shifts to counteract the change.

Section B: Kinetics Concepts & Graphs

6. Effect of Temperature on Rate: [2]

Particles have higher kinetic energy [1].
More particles have energy $\ge E_a$ / Frequency of effective collisions increases [1].

7. Zero Order Graph: [2]

Straight line with negative gradient [1].
Starts at initial concentration and decreases linearly [1].

8. Order w.r.t $H_2O_2$ : [1]

1 (First order).
Exp 1 to 2: $[H_2O_2]$ doubles, Rate doubles.

9. Order w.r.t $I^-$ : [1]

1 (First order).
Exp 1 to 3: $[I^-]$ doubles, Rate doubles.

10. Order w.r.t $H^+$ : [1]

0 (Zero order).
Exp 1 to 4: $[H^+]$ doubles, Rate unchanged.

Section C: Kinetics Calculations & Theory

11. Rate Equation: [1]

Rate $= k[H_2O_2][I^-]$

12. Calculation of $k$ Value: [2]

$k = \text{Rate} / ([H_2O_2][I^-])$
$k = (2.0 \times 10^{-4}) / (0.10 \times 0.10) = 0.02$ [1]
Value: 0.02 (or $2.0 \times 10^{-2}$ ) [1]

13. Units of $k$ : [1]

$\text{dm}^3 \text{mol}^{-1} \text{s}^{-1}$

14. Collision Theory Explanation: [2]

Higher concentration means more particles per unit volume [1].
Higher frequency of collisions [1].

15. Maxwell-Boltzmann Sketch: [2]

Curve $T_2$ peak is lower and to the right of $T_1$ [1].
$E_a$ marked correctly on x-axis [1].

Section D: Equilibrium Principles & Calculations

16. $K_c$ Expression: [1]

$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$

17. Moles of $N_2$ at Equilibrium: [1]

Reaction: $N_2 + 3H_2 \rightleftharpoons 2NH_3$
Change in $NH_3 = +0.4$ . So change in $N_2 = -0.2$ .
$N_2 = 1.0 - 0.2 = \mathbf{0.8 \text{ mol}}$

18. Moles of $H_2$ at Equilibrium: [1]

Change in $H_2 = 3 \times (-0.2) = -0.6$ .
$H_2 = 3.0 - 0.6 = \mathbf{2.4 \text{ mol}}$

19. Calculate $K_c$ : [3]

Concentrations ( $V=2.0 \text{ dm}^3$ $V = 2.0 dm^{3}$ ):
- $[NH_3] = 0.2$ , $[N_2] = 0.4$ , $[H_2] = 1.2$
Substitute: $K_c = \frac{(0.2)^2}{(0.4)(1.2)^3}$ [1]
Calculation: $K_c = \frac{0.04}{0.6912} \approx 0.058$ [1]
Units: $\text{dm}^6 \text{mol}^{-2}$ [1]

20. Effect of Temperature on $K_c$ : [2]

$K_c$ decreases [1].
Forward reaction is exothermic; increasing T shifts equilibrium to the left (endothermic direction) [1].