From Real Exams Quiz

A Level H2 Chemistry Kinetics Equilibrium Quiz

Free A Level H2 Chemistry Kinetics Equilibrium quiz, LongCat Exam version, with questions, answers, and A Level-style practice for Singapore students.

These static practice materials are generated from the site's syllabus and paper-generation workflow, with source and model context shown so students and parents can evaluate the material before use.

A Level H2 Chemistry From Real Exams Generated by LongCat 2.0 LLM Updated 2026-08-17
Back to Subject Browse Free Exam Papers

Questions

Free quiz and exam paper access

Enter your details to view this paper

Your access is remembered on this device.

Answers

A-Level Chemistry H2 Quiz - Kinetics Equilibrium

Answer Key

Section A: Multiple Choice

1. Answer: C

Explanation:
The rate equation is rate=k[X]2[Y]\text{rate} = k[\text{X}]^2[\text{Y}]. The overall order = 2 + 1 = 3 (third order overall), so A is incorrect. Doubling [X] would increase the rate by a factor of 22=42^2 = 4 (quadruples), so B is incorrect. The order with respect to Y is 1, so D is incorrect.

For units of kk:
rate=k[X]2[Y]\text{rate} = k[\text{X}]^2[\text{Y}]
mol dm−3 s−1=k×(mol dm−3)2×(mol dm−3)\text{mol dm}^{-3}\text{ s}^{-1} = k \times (\text{mol dm}^{-3})^2 \times (\text{mol dm}^{-3})
mol dm−3 s−1=k×mol3 dm−9\text{mol dm}^{-3}\text{ s}^{-1} = k \times \text{mol}^3\text{ dm}^{-9}
k=mol dm−3 s−1mol3 dm−9=mol−2 dm6 s−1k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{\text{mol}^3\text{ dm}^{-9}} = \text{mol}^{-2}\text{ dm}^{6}\text{ s}^{-1} ✓

Common mistake: Students often add orders incorrectly or confuse the effect of concentration changes on rate.

[2]

2. Answer: C

Explanation:
The reaction is exothermic (ΔH=−197\Delta H = -197 kJ mol⁻¹). KcK_c depends only on temperature. For an exothermic reaction, decreasing the temperature shifts the equilibrium to the right (favouring products), which increases the value of KcK_c. Adding a catalyst does not change KcK_c (it only speeds up the rate at which equilibrium is reached). Increasing pressure or concentration shifts the position but does not change KcK_c.

Common mistake: Students often confuse changes in position of equilibrium with changes in the value of KcK_c. Only temperature changes affect KcK_c.

[2]

3. Answer: A

Explanation:
The activation energy for the catalysed reaction is the energy difference between the reactants (80 kJ mol⁻¹) and the catalysed transition state (110 kJ mol⁻¹):

Ea=110−80=30E_a = 110 - 80 = 30 kJ mol⁻¹

The diagram shows the catalysed pathway with a lower peak. The activation energy is measured from the energy level of the reactants up to the transition state of the catalysed pathway.

Common mistake: Students may read the absolute energy value at the peak (110 kJ mol⁻¹) instead of calculating the difference from the reactant energy level.

[2]

4. Answer: C

Explanation:
For the reaction N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g):

Kp=(pNH3)2(pN2)(pH2)3K_p = \frac{(p_{\text{NH}_3})^2}{(p_{\text{N}_2})(p_{\text{H}_2})^3}

Units of Kp=atm2atm×atm3=atm2atm4=atm−2K_p = \frac{\text{atm}^2}{\text{atm} \times \text{atm}^3} = \frac{\text{atm}^2}{\text{atm}^4} = \text{atm}^{-2}

Common mistake: Students forget to raise partial pressures to their stoichiometric coefficients when determining units.

[2]

5. Answer: B

Explanation:
For a first-order reaction, a plot of ln⁡[A]\ln[\text{A}] against time gives a straight line with gradient =−k= -k.
So k=0.025k = 0.025 s⁻¹.

The half-life for a first-order reaction is:

t1/2=ln⁡2k=0.6930.025=27.7 st_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{0.025} = 27.7 \text{ s}

Common mistake: Students may use the zero-order or second-order half-life formula instead of the first-order formula.

[2]

Section B: Structured Questions

6.

(a) Kc=[N2O4][NO2]2K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}

[1] — 1 mark for correct expression with products over reactants and correct powers.

(b)
[NO2]=0.0402.0=0.020[\text{NO}_2] = \frac{0.040}{2.0} = 0.020 mol dm⁻³
[N2O4]=0.0602.0=0.030[\text{N}_2\text{O}_4] = \frac{0.060}{2.0} = 0.030 mol dm⁻³

Kc=0.030(0.020)2=0.0304.0×10−4=75 mol−1 dm3K_c = \frac{0.030}{(0.020)^2} = \frac{0.030}{4.0 \times 10^{-4}} = 75 \text{ mol}^{-1}\text{ dm}^{3}

[2] — 1 mark for correct concentrations, 1 mark for correct KcK_c value (75) with correct units.

(c)
The reaction is exothermic (ΔH=−57.2\Delta H = -57.2 kJ mol⁻¹). Increasing the temperature adds heat to the system. By Le Chatelier's principle, the equilibrium shifts to the left (towards reactants) to absorb the added heat. This decreases the value of KcK_c because [N2O4][\text{N}_2\text{O}_4] decreases and [NO2][\text{NO}_2] increases.

[2] — 1 mark for stating KcK_c decreases, 1 mark for correct explanation linking temperature increase to endothermic direction (reverse) and the effect on KcK_c.

(d)
Increasing the pressure shifts the position of equilibrium to the right (towards fewer moles of gas — 2 moles → 1 mole). This increases the yield of N2O4\text{N}_2\text{O}_4. However, KcK_c does not change because KcK_c depends only on temperature, not pressure.

[2] — 1 mark for stating equilibrium shifts right (towards products), 1 mark for stating KcK_c does not change because it is temperature-dependent only.

[Total: 7]

7.

(a)
Comparing Experiments 1 and 2: when [H2O2][\text{H}_2\text{O}_2] doubles from 0.10 to 0.20, the rate doubles from 1.2×10−41.2 \times 10^{-4} to 2.4×10−42.4 \times 10^{-4}.
2.4×10−41.2×10−4=2\frac{2.4 \times 10^{-4}}{1.2 \times 10^{-4}} = 2 and 0.200.10=2\frac{0.20}{0.10} = 2, so 2=2n2 = 2^n, giving n=1n = 1.

The reaction is first order with respect to H2O2\text{H}_2\text{O}_2.

[2] — 1 mark for correct comparison showing rate doubles when concentration doubles, 1 mark for stating first order.

(b)
rate=k[H2O2]\text{rate} = k[\text{H}_2\text{O}_2]

[1]

(c)
Using data from Experiment 1:
k=rate[H2O2]=1.2×10−40.10=1.2×10−3k = \frac{\text{rate}}{[\text{H}_2\text{O}_2]} = \frac{1.2 \times 10^{-4}}{0.10} = 1.2 \times 10^{-3}

Units: mol dm−3 s−1mol dm−3=s−1\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{\text{mol dm}^{-3}} = \text{s}^{-1}

k=1.2×10−3 s−1k = 1.2 \times 10^{-3} \text{ s}^{-1}

[2] — 1 mark for correct value, 1 mark for correct units.

(d)
rate=1.2×10−3×0.50=6.0×10−5\text{rate} = 1.2 \times 10^{-3} \times 0.50 = 6.0 \times 10^{-5} mol dm⁻³ s⁻¹

[1]

[Total: 6]

8.

(a)
Order with respect to NO2\text{NO}_2 = 2. Overall order = 2.

[1]

(b)
Δ[NO2]=0.020−0.0121.0=0.008\Delta[\text{NO}_2] = \frac{0.020 - 0.012}{1.0} = 0.008 mol dm⁻³

Average rate of disappearance of NO2=0.00850=1.6×10−4\text{NO}_2 = \frac{0.008}{50} = 1.6 \times 10^{-4} mol dm⁻³ s⁻¹

From the stoichiometry:
rate of reaction=12×rate of disappearance of NO2=1.6×10−42=8.0×10−5\text{rate of reaction} = \frac{1}{2} \times \text{rate of disappearance of NO}_2 = \frac{1.6 \times 10^{-4}}{2} = 8.0 \times 10^{-5} mol dm⁻³ s⁻¹

[2] — 1 mark for correct change in concentration over time, 1 mark for dividing by 2 for the stoichiometric coefficient.

(c)
As the reaction proceeds, the concentration of NO2\text{NO}_2 decreases. According to collision theory, the rate of reaction depends on the frequency of effective collisions between reactant molecules. With fewer NO2\text{NO}_2 molecules per unit volume, the collision frequency decreases, so the rate of reaction decreases.

[2] — 1 mark for linking decreasing concentration to fewer collisions, 1 mark for explaining that fewer collisions means lower rate.

[Total: 5]

9.

(a)
The rate-determining step is Step 1 because it is the slowest step. The slowest step in a reaction mechanism controls the overall rate of reaction.

[1]

(b)
The rate equation is based on the rate-determining step:
rate=k[NO]2\text{rate} = k[\text{NO}]^2

[1]

(c)
N2O2\text{N}_2\text{O}_2 is an intermediate. It is produced in Step 1 and consumed in Step 2, so it does not appear in the overall equation. An intermediate is a species that is formed in one step and used up in a subsequent step.

[1]

[Total: 3]

10.

(a)
Let xx = amount of N2O4\text{N}_2\text{O}_4 that dissociates.

N2O4⇌2NO2\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2

N2O4\text{N}_2\text{O}_4NO2\text{NO}_2
Initial / mol0.200
Change / mol−x-x+2x+2x
Equilibrium / mol0.20−x0.20 - x2x2x

Equilibrium concentrations (in 1.0 dm³): [N2O4]=0.20−x[\text{N}_2\text{O}_4] = 0.20 - x, [NO2]=2x[\text{NO}_2] = 2x

Kc=(2x)20.20−x=0.14K_c = \frac{(2x)^2}{0.20 - x} = 0.14

4x20.20−x=0.14\frac{4x^2}{0.20 - x} = 0.14

4x2=0.14(0.20−x)=0.028−0.14x4x^2 = 0.14(0.20 - x) = 0.028 - 0.14x

4x2+0.14x−0.028=04x^2 + 0.14x - 0.028 = 0

Using the quadratic formula: x=−0.14+(0.14)2+4(4)(0.028)2(4)=−0.14+0.0196+0.4488=−0.14+0.46768=−0.14+0.6848=0.5448=0.068x = \frac{-0.14 + \sqrt{(0.14)^2 + 4(4)(0.028)}}{2(4)} = \frac{-0.14 + \sqrt{0.0196 + 0.448}}{8} = \frac{-0.14 + \sqrt{0.4676}}{8} = \frac{-0.14 + 0.684}{8} = \frac{0.544}{8} = 0.068

[N2O4]=0.20−0.068=0.132[\text{N}_2\text{O}_4] = 0.20 - 0.068 = 0.132 mol dm⁻³ ≈ 0.13 mol dm⁻³
[NO2]=2(0.068)=0.136[\text{NO}_2] = 2(0.068) = 0.136 mol dm⁻³ ≈ 0.14 mol dm⁻³

[3] — 1 mark for correct ICE table setup, 1 mark for correct quadratic solution, 1 mark for correct equilibrium concentrations.

(b)
The forward reaction is endothermic (ΔH=+58.0\Delta H = +58.0 kJ mol⁻¹). Increasing the temperature favours the endothermic (forward) direction. The equilibrium shifts to the right, producing more NO2\text{NO}_2. The value of KcK_c increases because the ratio [NO2]2[N2O4]\frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} increases.

[2] — 1 mark for stating KcK_c increases, 1 mark for correct explanation.

(c)
NO2\text{NO}_2 is a brown gas while N2O4\text{N}_2\text{O}_4 is colourless. On heating, the equilibrium shifts to the right (towards more NO2\text{NO}_2), so the concentration of brown NO2\text{NO}_2 increases, causing the brown colour to deepen.

[1]

[Total: 6]

11.

(a)
The forward reaction is exothermic, so a lower temperature would favour a higher equilibrium yield of ammonia. However, at low temperatures, the rate of reaction is extremely slow. A temperature of 450 °C is a compromise: it provides a reasonable rate of reaction (acceptable kinetics) while still giving a moderate yield. The iron catalyst also helps to increase the rate at this temperature.

[2] — 1 mark for explaining the rate-yield compromise, 1 mark for mentioning that 450 °C balances rate and yield.

(b)
The forward reaction proceeds with a decrease in the number of moles of gas (4 moles → 2 moles). By Le Chatelier's principle, increasing the pressure shifts the equilibrium to the side with fewer moles of gas, i.e., to the right, increasing the yield of ammonia.

[2] — 1 mark for stating equilibrium shifts to the right, 1 mark for linking to fewer moles of gas.

(c)
The iron catalyst increases the rate of both the forward and reverse reactions equally by providing an alternative reaction pathway with a lower activation energy. It does not change the position of equilibrium or the yield of ammonia because it affects the kinetics, not the thermodynamics. The catalyst allows equilibrium to be reached faster.

[2] — 1 mark for stating catalyst increases rate, 1 mark for explaining it does not affect yield (lowers EaE_a for both directions equally).

[Total: 6]

12.

(a)
ln⁡k2k1=EaR(1T1−1T2)\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

ln⁡8.76×10−33.46×10−3=Ea8.31(1298−1318)\ln\frac{8.76 \times 10^{-3}}{3.46 \times 10^{-3}} = \frac{E_a}{8.31}\left(\frac{1}{298} - \frac{1}{318}\right)

ln⁡(2.532)=Ea8.31(3.356×10−3−3.145×10−3)\ln(2.532) = \frac{E_a}{8.31}\left(3.356 \times 10^{-3} - 3.145 \times 10^{-3}\right)

0.929=Ea8.31×2.11×10−40.929 = \frac{E_a}{8.31} \times 2.11 \times 10^{-4}

Ea=0.929×8.312.11×10−4=7.7202.11×10−4=3.66×104E_a = \frac{0.929 \times 8.31}{2.11 \times 10^{-4}} = \frac{7.720}{2.11 \times 10^{-4}} = 3.66 \times 10^{4} J mol⁻¹

Ea=36.6E_a = 36.6 kJ mol⁻¹

[3] — 1 mark for correct substitution, 1 mark for correct rearrangement, 1 mark for correct answer with units.

(b)
The Maxwell-Boltzmann distribution shows that at higher temperatures, a greater proportion of molecules possess energy equal to or greater than the activation energy. A small increase in temperature shifts the distribution curve to the right and flattens it, significantly increasing the area under the curve beyond EaE_a. This means many more molecules can overcome the energy barrier, leading to a significant increase in the rate constant.

[2] — 1 mark for describing the shift in distribution, 1 mark for linking increased proportion of molecules with E≥EaE \geq E_a to increased rate constant.

[Total: 5]

13.

(a)
Kc=[CH3OH][CO][H2]2=0.25(0.15)(0.30)2=0.250.15×0.09=0.250.0135=18.5K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2} = \frac{0.25}{(0.15)(0.30)^2} = \frac{0.25}{0.15 \times 0.09} = \frac{0.25}{0.0135} = 18.5

Units: mol dm−3(mol dm−3)(mol dm−3)2=mol dm−3mol3 dm−9=mol−2 dm6\frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \frac{\text{mol dm}^{-3}}{\text{mol}^3\text{ dm}^{-9}} = \text{mol}^{-2}\text{ dm}^{6}

Kc=18.5K_c = 18.5 mol⁻² dm⁶ (or 19 to 2 s.f.)

[2] — 1 mark for correct expression and substitution, 1 mark for correct value with units.

(b)
(i) Halving the volume doubles all concentrations. The reaction quotient QcQ_c becomes:

Qc=2×0.25(2×0.15)(2×0.30)2=0.500.30×0.36=0.500.108=4.63Q_c = \frac{2 \times 0.25}{(2 \times 0.15)(2 \times 0.30)^2} = \frac{0.50}{0.30 \times 0.36} = \frac{0.50}{0.108} = 4.63

Since Qc<KcQ_c < K_c (4.63 < 18.5), the equilibrium shifts to the right (towards products) to restore equilibrium.

Alternatively: there are 3 moles of gas on the left and 1 mole on the right. Increasing pressure (by reducing volume) shifts equilibrium to the side with fewer moles of gas (to the right).

(ii) KcK_c does not change because it depends only on temperature, which has not changed.

[2] — 1 mark for stating equilibrium shifts right, 1 mark for stating KcK_c unchanged.

(c)
Adding helium at constant volume does not change the concentrations (or partial pressures) of the reacting gases, because the total volume is unchanged and helium does not participate in the reaction. Since the concentrations of reactants and products remain the same, the position of equilibrium is unaffected, and KcK_c remains unchanged.

[2] — 1 mark for stating no effect on equilibrium position, 1 mark for correct explanation (concentrations unchanged at constant volume).

[Total: 6]

14.

(a)
Order with respect to CH3CH2Br\text{CH}_3\text{CH}_2\text{Br}:
Compare Experiments 1 and 2 ( [OH−][\text{OH}^-] constant):
1.0×10−45.0×10−5=2.0\frac{1.0 \times 10^{-4}}{5.0 \times 10^{-5}} = 2.0 and 0.200.10=2.0\frac{0.20}{0.10} = 2.0
2.0=2.0n⇒n=12.0 = 2.0^n \Rightarrow n = 1 → first order

Order with respect to OH−\text{OH}^-:
Compare Experiments 1 and 3 ( [CH3CH2Br][\text{CH}_3\text{CH}_2\text{Br}] constant):
1.0×10−45.0×10−5=2.0\frac{1.0 \times 10^{-4}}{5.0 \times 10^{-5}} = 2.0 and 0.200.10=2.0\frac{0.20}{0.10} = 2.0
2.0=2.0n⇒n=12.0 = 2.0^n \Rightarrow n = 1 → first order

[2] — 1 mark for each order with correct reasoning.

(b)
rate=k[CH3CH2Br][OH−]\text{rate} = k[\text{CH}_3\text{CH}_2\text{Br}][\text{OH}^-]
Overall order = 1 + 1 = 2 (second order)

[1]

(c)
Using Experiment 1:
k=5.0×10−5(0.10)(0.10)=5.0×10−50.010=5.0×10−3k = \frac{5.0 \times 10^{-5}}{(0.10)(0.10)} = \frac{5.0 \times 10^{-5}}{0.010} = 5.0 \times 10^{-3}

Units: mol dm−3 s−1(mol dm−3)2=mol−1 dm3 s−1\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2} = \text{mol}^{-1}\text{ dm}^{3}\text{ s}^{-1}

k=5.0×10−3k = 5.0 \times 10^{-3} mol⁻¹ dm³ s⁻¹

[2] — 1 mark for correct value, 1 mark for correct units.

(d)
The rate equation shows that both CH3CH2Br\text{CH}_3\text{CH}_2\text{Br} and OH−\text{OH}^- appear in the rate equation, meaning both species are involved in the rate-determining step. This is consistent with an SN2\text{S}_\text{N}2 mechanism, which is a bimolecular nucleophilic substitution where the nucleophile attacks the substrate in a single concerted step. An SN1\text{S}_\text{N}1 mechanism would show first order only in the substrate (the nucleophile would not appear in the rate equation).

[2] — 1 mark for suggesting SN2\text{S}_\text{N}2, 1 mark for correct reasoning linking rate equation to mechanism.

[Total: 7]

15.

(a)
Kc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]}

[1]

(b)

CH3COOH\text{CH}_3\text{COOH}C2H5OH\text{C}_2\text{H}_5\text{OH}CH3COOC2H5\text{CH}_3\text{COOC}_2\text{H}_5H2O\text{H}_2\text{O}
Initial / mol0.500.5000
Change / mol−0.33−0.33+0.33+0.33
Equilibrium / mol0.170.170.330.33

In 1.0 dm³, concentrations = moles:

Kc=(0.33)(0.33)(0.17)(0.17)=0.10890.0289=3.77≈3.8K_c = \frac{(0.33)(0.33)}{(0.17)(0.17)} = \frac{0.1089}{0.0289} = 3.77 \approx 3.8

[3] — 1 mark for correct ICE table, 1 mark for correct equilibrium concentrations, 1 mark for correct KcK_c value.

(c)
Adding a catalyst (concentrated H2SO4\text{H}_2\text{SO}_4) does not change the value of KcK_c. A catalyst increases the rates of the forward and reverse reactions equally, allowing equilibrium to be reached faster, but it does not change the position of equilibrium or the equilibrium constant. KcK_c depends only on temperature.

[2] — 1 mark for stating KcK_c does not change, 1 mark for correct explanation.

[Total: 6]

Section C: Data Interpretation

16.

(a)
As temperature increases from 500 K to 600 K, KcK_c increases from 4.17×10−24.17 \times 10^{-2} to 1.561.56. By Le Chatelier's principle, increasing temperature favours the endothermic direction. Since KcK_c increases (more products at equilibrium), the forward reaction must be endothermic.

[2] — 1 mark for stating endothermic, 1 mark for correct reasoning.

(b)
Let xx = amount of PCl5\text{PCl}_5 that dissociates.

PCl5⇌PCl3+Cl2\text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2

PCl5\text{PCl}_5PCl3\text{PCl}_3Cl2\text{Cl}_2
Initial / mol dm⁻³0.2500
Change / mol dm⁻³−x-x+x+x+x+x
Equilibrium / mol dm⁻³0.25−x0.25 - xxxxx

Kc=x20.25−x=4.17×10−2K_c = \frac{x^2}{0.25 - x} = 4.17 \times 10^{-2}

x2=4.17×10−2(0.25−x)=1.0425×10−2−4.17×10−2xx^2 = 4.17 \times 10^{-2}(0.25 - x) = 1.0425 \times 10^{-2} - 4.17 \times 10^{-2}x

x2+4.17×10−2x−1.0425×10−2=0x^2 + 4.17 \times 10^{-2}x - 1.0425 \times 10^{-2} = 0

x=−4.17×10−2+(4.17×10−2)2+4(1.0425×10−2)2x = \frac{-4.17 \times 10^{-2} + \sqrt{(4.17 \times 10^{-2})^2 + 4(1.0425 \times 10^{-2})}}{2}

x=−0.0417+1.739×10−3+4.170×10−22=−0.0417+4.344×10−22=−0.0417+0.20842=0.16672=0.0834x = \frac{-0.0417 + \sqrt{1.739 \times 10^{-3} + 4.170 \times 10^{-2}}}{2} = \frac{-0.0417 + \sqrt{4.344 \times 10^{-2}}}{2} = \frac{-0.0417 + 0.2084}{2} = \frac{0.1667}{2} = 0.0834

[PCl5]=0.25−0.0834=0.167[\text{PCl}_5] = 0.25 - 0.0834 = 0.167 mol dm⁻³ ≈ 0.17 mol dm⁻³
[PCl3]=[Cl2]=0.0834[\text{PCl}_3] = [\text{Cl}_2] = 0.0834 mol dm⁻³ ≈ 0.083 mol dm⁻³

[3] — 1 mark for correct ICE table, 1 mark for correct quadratic solution, 1 mark for correct equilibrium concentrations.

(c)
Percentage dissociation =0.08340.25×100=33.4%≈33%= \frac{0.0834}{0.25} \times 100 = 33.4\% \approx 33\%

[1]

[Total: 6]

17.

(a)
Zero order with respect to I2\text{I}_2 means that the concentration of I2\text{I}_2 does not affect the rate of reaction. Changing [I2][\text{I}_2] has no effect on the rate. The rate equation does not include [I2][\text{I}_2].

[1]

(b)
Step 1 is the slow (rate-determining) step and involves CH3COCH3\text{CH}_3\text{COCH}_3 and H+\text{H}^+. The rate equation based on Step 1 would be: rate=k[CH3COCH3][H+]\text{rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+], which matches the experimentally determined rate equation. I2\text{I}_2 is not involved in the rate-determining step, consistent with zero order in I2\text{I}_2. Therefore, the mechanism is consistent with the rate equation.

[2] — 1 mark for identifying Step 1 as rate-determining, 1 mark for showing consistency with rate equation.

(c)
At a higher temperature, the rate constant kk increases. This is because more molecules possess energy equal to or greater than the activation energy. The Maxwell-Boltzmann distribution shifts so that a greater proportion of molecules have sufficient energy to overcome the energy barrier, leading to more frequent effective collisions and a larger rate constant.

[2] — 1 mark for stating kk increases, 1 mark for explanation in terms of activation energy and molecular energy distribution.

[Total: 5]

18.

(a)
Kp=(pSO3)2(pSO2)2(pO2)K_p = \frac{(p_{\text{SO}_3})^2}{(p_{\text{SO}_2})^2(p_{\text{O}_2})}

[1]

(b)
Kp=(0.925)2(0.050)2(0.025)=0.85562.5×10−3×0.025=0.85566.25×10−5=1.37×104K_p = \frac{(0.925)^2}{(0.050)^2(0.025)} = \frac{0.8556}{2.5 \times 10^{-3} \times 0.025} = \frac{0.8556}{6.25 \times 10^{-5}} = 1.37 \times 10^{4}

Units: atm2atm2×atm=atm−1\frac{\text{atm}^2}{\text{atm}^2 \times \text{atm}} = \text{atm}^{-1}

Kp=1.37×104K_p = 1.37 \times 10^{4} atm⁻¹

[2] — 1 mark for correct substitution, 1 mark for correct value with units.

(c)
The forward reaction is exothermic. A lower temperature (e.g., 400 K) would give a higher equilibrium yield of SO3\text{SO}_3. However, at 400 K, the rate of reaction would be too slow for industrial production, even with a catalyst. A temperature of 700 K is a compromise between a reasonable yield and a fast enough rate of reaction. The V2O5\text{V}_2\text{O}_5 catalyst helps increase the rate at this temperature.

[2] — 1 mark for explaining the rate-yield compromise, 1 mark for mentioning the role of the catalyst.

(d)
The forward reaction proceeds with a decrease in moles of gas (3 moles → 2 moles), so a higher pressure would increase the yield. However, the yield at 1 atm is already very high (92.5% SO3\text{SO}_3 at equilibrium). The cost of building and operating equipment to withstand high pressures would not be justified by the small additional increase in yield. Additionally, a high-pressure process poses greater safety risks.

[2] — 1 mark for stating that yield is already high at 1 atm, 1 mark for mentioning cost/safety considerations.

[Total: 7]

19.

(a)
Equilibrium is first established at approximately t = 40 s (where the concentrations become constant/plateau).

[1]

(b)
From the graph:
Equilibrium [A]=0.40[\text{A}] = 0.40 mol dm⁻³
Equilibrium [B]=0.80[\text{B}] = 0.80 mol dm⁻³

[1]

(c)
Kc=[B]2[A]=(0.80)20.40=0.640.40=1.6 mol dm−3K_c = \frac{[\text{B}]^2}{[\text{A}]} = \frac{(0.80)^2}{0.40} = \frac{0.64}{0.40} = 1.6 \text{ mol dm}^{-3}

[2] — 1 mark for correct expression, 1 mark for correct value.

(d)
When the volume is halved at t=50t = 50 s, the concentrations of both A and B instantly double: [A][\text{A}] jumps from 0.40 to 0.80 mol dm⁻³, and [B][\text{B}] jumps from 0.80 to 1.60 mol dm⁻³. Since there are more moles of gas on the right (2 moles of B vs 1 mole of A), the equilibrium shifts to the left (towards reactants). Over time, [A][\text{A}] increases from 0.80 and [B][\text{B}] decreases from 1.60 until a new equilibrium is established. Both curves level off at new constant values, with the new [A][\text{A}] higher than 0.40 and new [B][\text{B}] higher than 0.80 (but lower than 1.60).

[2] — 1 mark for describing the instantaneous doubling of concentrations, 1 mark for describing the shift to the left and new equilibrium.

[Total: 6]

20.

(a)
The rate-determining step is Step 1 (slow step). The rate equation based on this step is:
rate=k[NO2]2\text{rate} = k[\text{NO}_2]^2

[1]

(b)
NO3\text{NO}_3 is the intermediate. It is produced in Step 1 and consumed in Step 2, so it does not appear in the overall equation. An intermediate is formed in one elementary step and used up in a subsequent step.

[1]

(c)
The proposed mechanism gives a rate equation of rate=k[NO2]2\text{rate} = k[\text{NO}_2]^2, which matches the experimentally determined rate equation exactly. Therefore, the proposed mechanism is consistent with the experimental rate equation.

[2] — 1 mark for stating the mechanism gives rate=k[NO2]2\text{rate} = k[\text{NO}_2]^2, 1 mark for confirming consistency with experiment.

(d)
The energy profile diagram must show:

  • Two peaks (two humps) representing the two steps, with the first peak higher than the second (Step 1 is rate-determining, so it has the higher activation energy).
  • A trough (valley) between the two peaks representing the intermediate NO3\text{NO}_3.
  • Reactants at a higher energy level than products (exothermic reaction, ΔH<0\Delta H < 0).
  • The activation energy EaE_a shown as the energy difference between reactants and the first (higher) transition state.
  • ΔH\Delta H shown as a downward arrow from reactants to products (negative value).

[2] — 1 mark for describing two peaks with first higher, 1 mark for describing intermediate trough and exothermic overall.

[Total: 6]

END OF ANSWER KEY

Total Marks: 60

Mark Distribution Summary:

SectionQuestionsMarks
A: Multiple Choice1–510
B: Structured Questions6–1532
C: Data Interpretation16–2018
Total20 questions60