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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 Chemistry H2 Quiz - Kinetics Equilibrium: Answer Key
Total Marks: 40
Topic: Kinetics Equilibrium
Section A: Short Answer
1. [1 mark]
rate =
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]
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 .
Teaching note: Catalyst speeds up forward and reverse equally; only temperature changes .
6. [1 mark]
First order (overall order = 1).
Teaching note: Units s 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 > (Maxwell–Boltzmann curve shifts right, area beyond increases). More successful collisions per unit time → higher rate.
Marking: 1 for distribution shift, 1 for greater fraction >, 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 = .
Teaching note: Compare experiments holding one variable constant.
12. [2 marks]
Catalyst provides alternate pathway with lower → 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]
(b) [1] Units = atm (or pressure units); since 2 – 1 = 1.
Teaching note: units = (pressure), Δn = 1.
Section C: Data Interpretation & Calculation
15. [3 marks]
First-order:
s (3 s.f.)
Marking: 1 formula, 1 substitution, 1 answer.
16. [4 marks total]
(a) [1] ,
(b) [3]
Assume small: (too large; solve quadratic).
Exact: → using solver mol dm⁻³.
Marking: 1 expression, 2 solving, 1 value.
17. [2 marks]
Fraction of molecules with 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] mol dm⁻³
(b) [2]
From image: curves plateau at given values; use equilibrium concentrations.
19. [4 marks total]
(a) [1] 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 = ; from fast eq → rate = .
Marking: 1 + 2.
