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A Level H2 Chemistry Kinetics Equilibrium Quiz

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A-Level Chemistry H2 Quiz - Kinetics Equilibrium

Name: ________________________
Class: ________________________
Date: ________________________
Score: ______ / 60

Duration: 60 minutes
Total Marks: 60

Instructions:

Answer all questions.
Write your answers in the spaces provided.
The use of a scientific calculator is allowed.
A Data Booklet is provided for reference.
Marks are indicated in brackets [ ] at the end of each question or part question.

Section A: Reaction Kinetics (Questions 1–10)

1. The reaction between nitrogen monoxide and hydrogen proceeds according to the following equation: $2NO(g) + 2H_2(g) \rightarrow N_2(g) + 2H_2O(g)$

The rate equation for this reaction is determined to be: $\text{Rate} = k[NO]^2[H_2]$

(a) State the overall order of the reaction. [1]

(b) Determine the units of the rate constant, $k$ , for this reaction. [2]

(c) If the concentration of $NO$ is tripled and the concentration of $H_2$ is halved, calculate the factor by which the initial rate changes. [2]

2. The decomposition of hydrogen peroxide is catalyzed by iodide ions: $2H_2O_2(aq) \xrightarrow{I^-} 2H_2O(l) + O_2(g)$

The following data was obtained at constant temperature:

Experiment	$[H_2O_2]$ / mol dm⁻³	$[I^-]$ / 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	$4.0 \times 10^{-4}$

(a) Deduce the order of reaction with respect to $H_2O_2$ and $I^-$ . [2]

(b) Write the rate equation for this reaction. [1]

3. The Arrhenius equation relates the rate constant $k$ to the temperature $T$ : $\ln k = -\frac{E_a}{R} \left(\frac{1}{T}\right) + \ln A$

A student plotted $\ln k$ against $1/T$ and obtained a straight line with a gradient of $-5500 \text{ K}$ .

(a) Calculate the activation energy, $E_a$ , for this reaction. ( $R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}$ ) [2]

(b) Explain, in terms of collision theory, why increasing the temperature increases the rate of reaction. [2]

4. Consider the following two-step mechanism for the reaction between $NO_2$ and $CO$ : Step 1: $NO_2 + NO_2 \rightarrow NO_3 + NO$ (slow) Step 2: $NO_3 + CO \rightarrow NO_2 + CO_2$ (fast)

(a) Write the overall balanced equation for the reaction. [1]

(b) Identify the reaction intermediate. [1]

5. The hydrolysis of an ester, $RCOOR'$ , in acidic conditions is a first-order reaction with respect to the ester. The half-life of the reaction is 200 seconds.

(a) Calculate the rate constant, $k$ , for this reaction. [2]

(b) Calculate the time required for 75% of the ester to hydrolyze. [2]

6. Which of the following factors affects the value of the rate constant, $k$ ? A. Concentration of reactants B. Pressure of gaseous reactants C. Temperature D. Presence of a catalyst (changes mechanism)

Select the correct option(s) and explain your choice. [2]

7. A catalyst increases the rate of a reaction by: A. Increasing the frequency of collisions. B. Increasing the average kinetic energy of the molecules. C. Providing an alternative reaction pathway with lower activation energy. D. Shifting the position of equilibrium to the right.

Select the correct option. [1]

8. The reaction $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$ is first order with respect to $N_2O_5$ . Sketch a graph showing how the concentration of $N_2O_5$ changes with time. Label the axes and indicate the half-life, $t_{1/2}$ . [3]

9. Explain why the order of reaction with respect to a reactant cannot always be deduced from the stoichiometric coefficient in the balanced chemical equation. [2]

10. In an experiment to determine the order of reaction with respect to reactant A, the initial rate was measured at different concentrations of A, while keeping other reactants in large excess. The graph of $\log(\text{initial rate})$ against $\log[A]$ gave a straight line with a gradient of 1.5. What is the order of reaction with respect to A? [1]

Section B: Chemical Equilibria (Questions 11–15)

11. Consider the equilibrium system: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \quad \Delta H = -92 \text{ kJ mol}^{-1}$

State and explain the effect of each of the following changes on the yield of ammonia and the value of the equilibrium constant, $K_c$ .

(a) Increasing the pressure. [2]

(b) Increasing the temperature. [2]

12. At a certain temperature, the equilibrium constant $K_c$ for the reaction: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ is 50.

In a 1 dm³ vessel, 0.5 mol of $H_2$ and 0.5 mol of $I_2$ are mixed. Calculate the number of moles of $HI$ present at equilibrium. [4]

13. The dissociation of dinitrogen tetroxide is represented by: $N_2O_4(g) \rightleftharpoons 2NO_2(g)$

At equilibrium, the partial pressure of $N_2O_4$ is 0.4 atm and the partial pressure of $NO_2$ is 0.8 atm.

(a) Calculate the value of $K_p$ for this reaction. Include units. [2]

(b) If the total pressure of the system is increased by compressing the vessel, state and explain the effect on the mole fraction of $NO_2$ . [2]

14. Write the expression for the equilibrium constant, $K_c$ , for the following heterogeneous equilibrium: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$

Explain why the concentrations of the solids are not included in the expression. [2]

15. For the reaction $A(g) + B(g) \rightleftharpoons C(g)$ , $K_c = 10$ at 300 K. If the initial concentrations are $[A] = 0.1 \text{ mol dm}^{-3}$ , $[B] = 0.1 \text{ mol dm}^{-3}$ , and $[C] = 0.5 \text{ mol dm}^{-3}$ , determine the direction in which the reaction will proceed to reach equilibrium. Show your working. [3]

Section C: Integrated Concepts (Questions 16–20)

16. The Contact Process involves the equilibrium: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \quad \Delta H < 0$

Industrial conditions are typically 450°C and 1–2 atm. (a) Explain why a temperature of 450°C is used instead of a lower temperature, despite the forward reaction being exothermic. [2]

(b) Explain why a pressure of 1–2 atm is used instead of a much higher pressure. [2]

17. Ethanoic acid ( $CH_3COOH$ ) is a weak acid with $K_a = 1.7 \times 10^{-5} \text{ mol dm}^{-3}$ . (a) Write the expression for $K_a$ . [1]

(b) Calculate the pH of a 0.10 mol dm⁻³ solution of ethanoic acid. [3]

18. A buffer solution is prepared by mixing 50 cm³ of 0.10 mol dm⁻³ ethanoic acid with 50 cm³ of 0.10 mol dm⁻³ sodium ethanoate. (a) Calculate the pH of this buffer solution. [2]

(b) Explain how this buffer solution resists changes in pH when a small amount of strong acid ( $H^+$ ) is added. [2]

19. The solubility product, $K_{sp}$ , of magnesium hydroxide, $Mg(OH)_2$ , is $1.8 \times 10^{-11} \text{ mol}^3 \text{ dm}^{-9}$ at 25°C. (a) Write the expression for $K_{sp}$ of $Mg(OH)_2$ . [1]

(b) Calculate the solubility of $Mg(OH)_2$ in mol dm⁻³. [3]

20. Consider the equilibrium: $Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons [Fe(SCN)]^{2+}(aq)$ (Colorless) (Colorless) (Blood Red)

(a) State what would be observed if solid $KSCN$ is added to the equilibrium mixture. [1]

(b) Explain this observation using Le Chatelier’s Principle. [2]

Answers

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

1. (a) Overall order = $2 + 1 = 3$ (Third order). [1] (b) Rate = $k[NO]^2[H_2]$ . Units: $\text{mol dm}^{-3} \text{s}^{-1} = k (\text{mol dm}^{-3})^2 (\text{mol dm}^{-3})$ . $k = \frac{\text{mol dm}^{-3} \text{s}^{-1}}{\text{mol}^3 \text{dm}^{-9}} = \text{mol}^{-2} \text{dm}^6 \text{s}^{-1}$ . [2] (c) New Rate $\propto (3[NO])^2 (0.5[H_2]) = 9 \times 0.5 \times [NO]^2[H_2] = 4.5 \times \text{Original Rate}$ . Factor = 4.5. [2]

2. (a) Comparing Exp 1 and 2: $[I^-]$ constant, $[H_2O_2]$ doubles, Rate doubles $\rightarrow$ Order w.r.t $H_2O_2$ is 1. Comparing Exp 1 and 3: $[H_2O_2]$ constant, $[I^-]$ doubles, Rate doubles $\rightarrow$ Order w.r.t $I^-$ is 1. [2] (b) Rate $= k[H_2O_2][I^-]$ . [1] (c) Using Exp 1: $2.0 \times 10^{-4} = k(0.10)(0.10)$ . $k = \frac{2.0 \times 10^{-4}}{0.01} = 0.02 \text{ or } 2.0 \times 10^{-2}$ . Units: $\frac{\text{mol dm}^{-3} \text{s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})} = \text{mol}^{-1} \text{dm}^3 \text{s}^{-1}$ . [2]

3. (a) Gradient $= -E_a/R$ . $-5500 = -E_a / 8.31$ . $E_a = 5500 \times 8.31 = 45,705 \text{ J mol}^{-1} = 45.7 \text{ kJ mol}^{-1}$ . [2] (b) Higher temperature increases the average kinetic energy of molecules. [1] A larger proportion of molecules possess energy greater than or equal to the activation energy ( $E \ge E_a$ ), leading to more frequent effective collisions. [1]

4. (a) Add steps: $2NO_2 + NO_3 + CO \rightarrow NO_3 + NO + NO_2 + CO_2$ . Cancel intermediates/common terms: $NO_2 + CO \rightarrow NO + CO_2$ . [1] (b) $NO_3$ . [1] (c) Rate depends on the slow step (Step 1): Rate $= k[NO_2]^2$ . [1]

5. (a) For 1st order: $k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{200} = 3.47 \times 10^{-3} \text{ s}^{-1}$ . [2] (b) 75% hydrolyzed means 25% remains. This is 2 half-lives ( $100\% \rightarrow 50\% \rightarrow 25\%$ ). Time $= 2 \times 200 = 400 \text{ s}$ . [2] (Alternative: $t = \frac{1}{k} \ln(\frac{[A]_0}{[A]_t}) = \frac{1}{0.00347} \ln(4) \approx 400 \text{ s}$ )

6. C and D. [1 for selection, 1 for explanation] Temperature changes the energy distribution (Arrhenius). Catalysts change the mechanism/ $E_a$ . Concentration and Pressure affect the rate but not the constant $k$ (at constant T).

7. C. [1]

8. Graph: Y-axis: $[N_2O_5]$ , X-axis: Time. [1] Curve: Exponential decay starting from initial concentration, approaching zero asymptotically. [1] Label: Indicate constant time intervals for $t_{1/2}$ (e.g., $t_{1/2}$ , $2t_{1/2}$ ) showing concentration halving each time. [1]

9. The order of reaction is determined experimentally and depends on the reaction mechanism (specifically the rate-determining step). [1] The stoichiometric coefficient represents the overall mole ratio, which may involve multiple steps where reactants are consumed in fast steps after the RDS or in parallel pathways. [1]

10. Order = 1.5. [1] (Gradient of log-log plot equals the order).

11. (a) Yield: Increases. [1] Explanation: Forward reaction reduces moles of gas (4 to 2). High pressure favors the side with fewer moles. [1] (b) Yield: Decreases. [1] Explanation: Forward reaction is exothermic. High temperature favors the endothermic (reverse) direction to absorb heat. $K_c$ decreases. [1] (c) Yield: No change. [1] Explanation: Catalyst speeds up both forward and reverse rates equally. Equilibrium position is unchanged. $K_c$ is unchanged. [1]

12. Let $x$ be moles of $H_2$ reacted. Equilibrium moles: $H_2 = 0.5-x$ , $I_2 = 0.5-x$ , $HI = 2x$ . Volume = 1 dm³, so concentrations equal moles. $K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(2x)^2}{(0.5-x)(0.5-x)} = 50$ . $\frac{2x}{0.5-x} = \sqrt{50} \approx 7.07$ . $2x = 7.07(0.5-x) \Rightarrow 2x = 3.535 - 7.07x$ . $9.07x = 3.535 \Rightarrow x \approx 0.39$ . Moles of $HI = 2x = 0.78 \text{ mol}$ . [4]

13. (a) $K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(0.8)^2}{0.4} = \frac{0.64}{0.4} = 1.6 \text{ atm}$ . [2] (b) Mole fraction of $NO_2$ decreases. [1] Explanation: Increasing pressure shifts equilibrium to the side with fewer gas moles (left, $N_2O_4$ ). Thus, amount of $NO_2$ decreases relative to $N_2O_4$ . [1]

14. $K_c = [CO_2]$ . [1] Explanation: The concentration (or active mass) of pure solids is constant and is incorporated into the equilibrium constant value. [1]

15. $Q_c = \frac{[C]}{[A][B]} = \frac{0.5}{(0.1)(0.1)} = \frac{0.5}{0.01} = 50$ . [1] $Q_c (50) > K_c (10)$ . [1] The system has too much product. Reaction proceeds to the left (reverse) to reach equilibrium. [1]

16. (a) Lower temperature would increase yield (exothermic) but significantly decrease the rate of reaction. 450°C is a compromise temperature to ensure a commercially viable rate while maintaining an acceptable yield. [2] (b) Higher pressure would increase yield (fewer moles on right) and rate. However, 1-2 atm is used because the yield is already high enough at this pressure, and higher pressures require expensive, reinforced equipment and high energy costs for compression. [2]

17. (a) $K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$ . [1] (b) Assume $[H^+] = [CH_3COO^-]$ and $[CH_3COOH]_{eq} \approx 0.10$ . $1.7 \times 10^{-5} = \frac{x^2}{0.10}$ . $x^2 = 1.7 \times 10^{-6}$ . $x = [H^+] = \sqrt{1.7} \times 10^{-3} \approx 1.30 \times 10^{-3} \text{ mol dm}^{-3}$ . $\text{pH} = -\log(1.30 \times 10^{-3}) = 2.89$ . [3]

18. (a) Moles acid = $0.05 \times 0.1 = 0.005$ . Moles salt = $0.05 \times 0.1 = 0.005$ . Ratio [Salt]/[Acid] = 1. $\text{pH} = pK_a + \log(1) = pK_a$ . $pK_a = -\log(1.7 \times 10^{-5}) = 4.77$ . $\text{pH} = 4.77$ . [2] (b) Added $H^+$ reacts with the conjugate base ( $CH_3COO^-$ ) to form weak acid ( $CH_3COOH$ ). [1] $CH_3COO^- + H^+ \rightarrow CH_3COOH$ . This removes most of the added $H^+$ , keeping pH relatively constant. [1]

19. (a) $K_{sp} = [Mg^{2+}][OH^-]^2$ . [1] (b) Let solubility be $s$ mol dm⁻³. $[Mg^{2+}] = s$ , $[OH^-] = 2s$ . $K_{sp} = (s)(2s)^2 = 4s^3$ . $1.8 \times 10^{-11} = 4s^3$ . $s^3 = 4.5 \times 10^{-12}$ . $s = \sqrt[3]{4.5 \times 10^{-12}} \approx 1.65 \times 10^{-4} \text{ mol dm}^{-3}$ . [3]

20. (a) The solution becomes darker red / more intense red. [1] (b) Adding $KSCN$ increases $[SCN^-]$ . According to Le Chatelier’s Principle, the system shifts to the right (forward) to remove the excess $SCN^-$ . [1] This produces more $[Fe(SCN)]^{2+}$ , which is blood red. [1]

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