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A Level H2 Chemistry Kinetics Equilibrium Quiz

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Questions

A-Level Chemistry H2 Quiz - Kinetics Equilibrium

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

Duration: 60 minutes
Total Marks: 40
Instructions:

Answer all questions.
Write your answers in the spaces provided.
You may use a scientific calculator.
The Data Booklet is relevant to some questions.

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

1. Which of the following statements correctly describes the effect of a catalyst on the equilibrium constant, $K_c$ , and the rate of reaction?
[1]
A. $K_c$ increases; Rate increases
B. $K_c$ remains unchanged; Rate increases
C. $K_c$ decreases; Rate decreases
D. $K_c$ remains unchanged; Rate remains unchanged

2. For the reaction $2NO_2(g) \rightleftharpoons N_2O_4(g)$ , $\Delta H < 0$ . Which change will increase the value of the equilibrium constant $K_c$ ?
[1]
A. Increasing the pressure
B. Decreasing the pressure
C. Increasing the temperature
D. Decreasing the temperature

3. The rate equation for a reaction is given by: Rate $= k[A][B]^2$ .
If the concentration of $A$ is doubled and the concentration of $B$ is halved, what is the effect on the initial rate?
[1]
A. The rate is halved.
B. The rate remains unchanged.
C. The rate doubles.
D. The rate quadruples.

4. Consider the equilibrium: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ .
At equilibrium, the partial pressures are $p(H_2) = 0.20$ atm, $p(I_2) = 0.20$ atm, and $p(HI) = 0.80$ atm.
Calculate the value of $K_p$ .
[1]
A. 4
B. 16
C. 20
D. 80

5. A student plots $\ln k$ against $1/T$ for a reaction and obtains a straight line with a gradient of $-5000$ K.
Given $R = 8.31$ J mol $^{-1}$ K $^{-1}$ , calculate the activation energy, $E_a$ , in kJ mol $^{-1}$ .
[1]
A. 4.16
B. 41.6
C. 602
D. 5000

Section B: Kinetics Concepts (Questions 6-10)

6. Explain why increasing the temperature increases the rate of a reaction, referring to the Maxwell-Boltzmann distribution.
[2]

7. Define the term half-life of a reaction.
[1]

8. State the units of the rate constant, $k$ , for a second-order reaction where Rate $= k[A]^2$ .
[1]

9. For the equilibrium $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ , explain the effect of adding an inert gas at constant volume on the position of equilibrium.
[1]

10. Distinguish between the order of reaction with respect to a reactant and the stoichiometric coefficient of that reactant in the balanced equation.
[1]

Section C: Structured Questions - Kinetics (Questions 11-15)

11. The reaction between propanone and iodine in acidic solution is represented by the equation:
$CH_3COCH_3(aq) + I_2(aq) \xrightarrow{H^+} CH_3COCH_2I(aq) + H^+(aq) + I^-(aq)$

The following initial rate data were obtained at constant temperature:

Experiment	$[CH_3COCH_3]$ / mol dm $^{-3}$	$[I_2]$ / mol dm $^{-3}$	$[H^+]$ / mol dm $^{-3}$	Initial Rate / mol dm $^{-3}$ s $^{-1}$
1	0.10	0.10	0.10	$2.0 \times 10^{-5}$
2	0.20	0.10	0.10	$4.0 \times 10^{-5}$
3	0.10	0.20	0.10	$2.0 \times 10^{-5}$
4	0.10	0.10	0.20	$4.0 \times 10^{-5}$

Deduce the order of reaction with respect to Propanone, $CH_3COCH_3$ .
[1]

12. Using the data in Question 11, deduce the order of reaction with respect to Iodine, $I_2$ .
[1]

13. Using the data in Question 11, deduce the order of reaction with respect to Hydrogen ions, $H^+$ .
[1]

14. Write the rate equation for this reaction and calculate the value of the rate constant, $k$ , including its units.
[4]

Answer space

15. A proposed mechanism for this reaction is:
Step 1: $CH_3COCH_3 + H^+ \rightleftharpoons CH_3C(OH^+)CH_3$ (fast equilibrium)
Step 2: $CH_3C(OH^+)CH_3 \rightarrow CH_3C(OH)=CH_2 + H^+$ (slow)
Step 3: $CH_3C(OH)=CH_2 + I_2 \rightarrow CH_3COCH_2I + H^+ + I^-$ (fast)

Explain whether this mechanism is consistent with the rate equation you derived in Question 14.
[2]

Section D: Structured Questions - Equilibrium (Questions 16-20)

16. Methanol is manufactured industrially by the reaction of carbon monoxide and hydrogen:
$CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g) \quad \Delta H = -91 \text{ kJ mol}^{-1}$

Write the expression for the equilibrium constant, $K_p$ , for this reaction.
[1]

17. In a specific experiment, 1.00 mol of $CO$ and 2.00 mol of $H_2$ were placed in a closed vessel. At equilibrium, at a total pressure of 10.0 atm, 0.40 mol of $CH_3OH$ was present.
Calculate the amount, in moles, of $CO$ and $H_2$ present at equilibrium.
[2]

Answer space

18. Using the equilibrium amounts from Question 17, calculate the mole fraction and partial pressure of each gas at equilibrium.
[4]

Answer space

19. Calculate the value of $K_p$ for this equilibrium and state its units.
[3]

Answer space

20. The manufacturer operates the process at 500 K and 100 atm.
(a) Explain why a lower temperature is not used, despite the forward reaction being exothermic.
(b) If the volume of the vessel is halved at constant temperature, state and explain the effect on the yield of methanol.
[3]

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Answers

A-Level Chemistry H2 Quiz Answers - Kinetics Equilibrium

1. B
Catalysts lower activation energy, increasing rate, but do not affect thermodynamics ( $K_c$ ).

2. D
Reaction is exothermic ( $\Delta H < 0$ ). Decreasing T shifts equilibrium to the right (products), increasing $K_c$ .

3. A
Rate $\propto [A][B]^2$ . New Rate $\propto (2[A])(0.5[B])^2 = 2 \times 0.25 \times [A][B]^2 = 0.5 \times$ Original Rate.

4. B
$K_p = \frac{p(HI)^2}{p(H_2)p(I_2)} = \frac{0.80^2}{0.20 \times 0.20} = \frac{0.64}{0.04} = 16$ .

5. B
Gradient $= -E_a/R$ . $-5000 = -E_a/8.31 \Rightarrow E_a = 5000 \times 8.31 = 41550$ J mol $^{-1} = 41.6$ kJ mol $^{-1}$ .

Increasing temperature increases the average kinetic energy of particles. [1]
A larger proportion of particles have energy greater than or equal to the activation energy ( $E \ge E_a$ ), leading to more frequent successful collisions. [1]

7.
The time taken for the concentration of a reactant to decrease to half of its initial value. [1]

8.
mol $^{-1}$ dm $^3$ s $^{-1}$ [1]
(Rate is mol dm $^{-3}$ s $^{-1}$ , $[A]^2$ is mol $^2$ dm $^{-6}$ . $k = \text{Rate}/[A]^2$ )

9.
No effect. [1]
Adding inert gas at constant volume does not change the partial pressures of the reacting gases, so the position of equilibrium remains unchanged.

10.
Order of reaction is determined experimentally from rate data, whereas stoichiometric coefficient is from the balanced chemical equation. They are only equal if the reaction is elementary. [1]

11.
1 (Doubling [Propanone] doubles rate) [1]

12.
0 (Doubling [Iodine] has no effect on rate) [1]

13.
1 (Doubling [H+] doubles rate) [1]

14.
Rate $= k[CH_3COCH_3][H^+]$ [1]
$k = \frac{\text{Rate}}{[CH_3COCH_3][H^+]}$
$k = \frac{2.0 \times 10^{-5}}{0.10 \times 0.10} = \frac{2.0 \times 10^{-5}}{0.01} = 2.0 \times 10^{-3}$ [1 for value]
Units: $\frac{\text{mol dm}^{-3} \text{s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})} = \text{mol}^{-1} \text{dm}^3 \text{s}^{-1}$ [1 for units]
Answer: $2.0 \times 10^{-3} \text{ mol}^{-1} \text{dm}^3 \text{s}^{-1}$ [1 for final answer with units]

15.
The rate equation depends on $[CH_3COCH_3]$ and $[H^+]$ . Step 2 is the rate-determining step (slow step). The reactants in Step 2 are derived from Propanone and H+ (via Step 1 equilibrium). Iodine is involved in Step 3 (fast), which is after the RDS, so it does not appear in the rate equation. This is consistent. [2]

16.
$K_p = \frac{p(CH_3OH)}{p(CO) \cdot p(H_2)^2}$ [1]

17.
Moles $CH_3OH$ formed = 0.40
Moles $CO$ reacted = 0.40 $\Rightarrow$ Equil $CO = 1.00 - 0.40 = 0.60$ mol [1]
Moles $H_2$ reacted = $2 \times 0.40 = 0.80$ $\Rightarrow$ Equil $H_2 = 2.00 - 0.80 = 1.20$ mol [1]

18.
Total moles at equilibrium = $0.60 + 1.20 + 0.40 = 2.20$ mol
Mole fraction $CO = 0.60 / 2.20 = 0.273$ [0.5]
Mole fraction $H_2 = 1.20 / 2.20 = 0.545$ [0.5]
Mole fraction $CH_3OH = 0.40 / 2.20 = 0.182$ [0.5]
Partial Pressure = Mole Fraction $\times$ Total Pressure (10.0 atm)
$p(CO) = 0.2727 \times 10.0 = 2.73$ atm [0.5]
$p(H_2) = 0.5454 \times 10.0 = 5.45$ atm [0.5]
$p(CH_3OH) = 0.1818 \times 10.0 = 1.82$ atm [0.5]
[1 for correct set of mole fractions, 1 for correct set of partial pressures]

19.
$K_p = \frac{1.818}{(2.727)(5.454)^2}$
$K_p = \frac{1.818}{2.727 \times 29.75} = \frac{1.818}{81.13} = 0.0224$ [2 for calculation]
Units: $\frac{\text{atm}}{\text{atm} \cdot \text{atm}^2} = \text{atm}^{-2}$ [1 for units]

20.
(a) At lower temperatures, the rate of reaction is too slow to be economically viable / equilibrium is reached too slowly. [1]
(b) Yield increases. Halving volume increases pressure. Equilibrium shifts to side with fewer moles of gas (RHS: 1 mol vs LHS: 3 mol) to oppose change. [2]