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A Level H2 Chemistry Kinetics Equilibrium Quiz

Free A Level H2 Chemistry Kinetics Equilibrium quiz, HY3 Exam version, with questions, answers, and A Level-style practice for Singapore students.

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A Level H2 Chemistry From Real Exams Generated by Tencent HY3 Free Updated 2026-08-17

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Answers

A-Level Chemistry H2 Quiz - Kinetics Equilibrium: Answer Key

Total Marks: 40
Topic: Kinetics Equilibrium


Section A: Short Answer

1. [1 mark]
rate = k[X]k[\text{X}]
Teaching note: Rate proportional to [X] and independent of [Y] means order w.r.t. X = 1, w.r.t. Y = 0. Rate equation shows only X.

2. [1 mark]
The minimum energy required for colliding particles to react / for a successful collision.
Teaching note: Activation energy is the energy barrier above the average kinetic energy that reactants must overcome.

3. [1 mark]
Kc=[NH3]2[N2][H2]3K_c = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}
Teaching note: Products over reactants, each concentration raised to its stoichiometric coefficient.

4. [1 mark]
Equilibrium shifts to the side with fewer moles of gas (product side).
Teaching note: Higher pressure favours the direction reducing gas moles.

5. [1 mark]
No, it does not change KcK_c.
Teaching note: Catalyst speeds up forward and reverse equally; only temperature changes KcK_c.

6. [1 mark]
First order (overall order = 1).
Teaching note: Units s1^{-1} are characteristic of first-order rate constant.

7. [1 mark]
Yield of C decreases.
Teaching note: Exothermic forward reaction is opposed by temperature increase; equilibrium shifts left.


Section B: Structured Response

8. [3 marks total]
(a) [2] Mix known volumes of I⁻, S₂O₈²⁻, thiosulfate and starch. Reaction with thiosulfate keeps I₂ low; when thiosulfate used up, blue-black colour appears. Time to colour change gives rate via [S₂O₈²⁻] consumed.
(b) [1] Temperature / volume / concentration of thiosulfate.
Marking: 2 for correct method, 1 for valid constant variable.

9. [3 marks total]
(a) [1] Monitor conductivity (Br⁻ replaces OH⁻) or pH (OH⁻ decreases).
(b) [2] Plot [OH⁻] vs time; if linear, zero order; if half-life constant, first order; determine gradient/rate at different [OH⁻].
Marking: 1 + 2.

10. [3 marks]
At higher T, more molecules have energy > EaE_a (Maxwell–Boltzmann curve shifts right, area beyond EaE_a increases). More successful collisions per unit time → higher rate.
Marking: 1 for distribution shift, 1 for greater fraction >EaE_a, 1 for rate increase.

11. [3 marks total]
(a) [1] Order w.r.t. P = 1 (doubling [P] doubles rate: exp1→2).
(b) [1] Order w.r.t. Q = 2 (doubling [Q] quadruples rate: exp1→3).
(c) [1] rate = k[P][Q]2k[\text{P}][\text{Q}]^2.
Teaching note: Compare experiments holding one variable constant.

12. [2 marks]
Catalyst provides alternate pathway with lower EaE_a → both forward and reverse rates increase equally → equilibrium reached faster, position unchanged.
Marking: 1 for faster attainment, 1 for no position change.

13. [3 marks total]
(a) [1] Solution becomes deeper red.
(b) [2] Adding Fe³⁺ increases reactant concentration; equilibrium shifts right to consume Fe³⁺, forming more FeSCN²⁺ (red).
Marking: 1 + 2.

14. [2 marks total]
(a) [1] Kp=PNO22PN2O4K_p = \dfrac{P_{\text{NO}_2}^2}{P_{\text{N}_2\text{O}_4}}
(b) [1] Units = atm (or pressure units); since 2 – 1 = 1.
Teaching note: KpK_p units = (pressure)Δn^{Δn}, Δn = 1.


Section C: Data Interpretation & Calculation

15. [3 marks]
First-order: t=1kln[A]0[A]tt = \dfrac{1}{k} \ln\dfrac{[A]_0}{[A]_t}
=11.5×103ln0.0400.010=11.5×103ln4= \dfrac{1}{1.5\times10^{-3}} \ln\dfrac{0.040}{0.010} = \dfrac{1}{1.5\times10^{-3}} \ln 4
=666.7×1.386=924= 666.7 \times 1.386 = 924 s (3 s.f.)
Marking: 1 formula, 1 substitution, 1 answer.

16. [4 marks total]
(a) [1] [CO]=0.20x[\text{CO}] = 0.20 - x, [H2]=0.402x[\text{H}_2] = 0.40 - 2x
(b) [3] Kc=x(0.20x)(0.402x)2=4.8K_c = \dfrac{x}{(0.20-x)(0.40-2x)^2} = 4.8
Assume xx small: x4.8×0.20×0.16=0.154x \approx 4.8 \times 0.20 \times 0.16 = 0.154 (too large; solve quadratic).
Exact: (0.20x)(0.402x)2=x/4.8(0.20-x)(0.40-2x)^2 = x/4.8 → using solver x0.095x \approx 0.095 mol dm⁻³.
Marking: 1 expression, 2 solving, 1 value.

17. [2 marks]
Fraction of molecules with E>EaE > E_a increases exponentially with T (Arrhenius). Doubling rate implies significantly larger fraction above barrier at 310 K.
Marking: 1 for increase, 1 for link to rate.

18. [3 marks total]
(a) [1] [C]eq=0.30[\text{C}]_{eq} = 0.30 mol dm⁻³
(b) [2] Kc=0.300.20×0.20=7.5K_c = \dfrac{0.30}{0.20 \times 0.20} = 7.5
From image: curves plateau at given values; use equilibrium concentrations.

19. [4 marks total]
(a) [1] KpK_p decreases.
(b) [1] Yield of SO₃ increases.
(c) [2] Exothermic: T↑ shifts left, Kp↓. Fewer moles product (3→2): P↑ shifts right, yield↑.
Marking: 1+1+2.

20. [3 marks total]
(a) [1] Step 2 (slow).
(b) [2] rate = k[N2O2][O2]k[\text{N}_2\text{O}_2][\text{O}_2]; from fast eq [N2O2][NO]2[\text{N}_2\text{O}_2] \propto [\text{NO}]^2 → rate = k[NO]2[O2]k'[\text{NO}]^2[\text{O}_2].
Marking: 1 + 2.