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Next exam on November 14 – 15, 2026:
All concepts

Rate of Reaction

化学反应速率

Expressing rate as a concentration change per unit time and calculating it, and how a catalyst and other conditions change that rate.

6 real-exam questions · 4 sittings5 in mock papers13 practice

Formulas

  • Rate of reaction

    Definition
    v=ΔcΔtv=\frac{\Delta c}{\Delta t}

    Applies when：An average rate over the interval, always positive; rates quoted for different species are in the ratio of their stoichiometric coefficients

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

  1. CSCA Chemistry December 2025 Past Paper
    In a 2 L closed container, the reaction 2SO₂(g) + O₂(g) → 2SO₃(g) occurs. In 10 s, the amount of SO₂ decreases from 1 mol to 0.6 mol. The reaction rate v(O₂) is ( )
    • A. 0.02 mol·L-1·s-1
    • B. 0.01 mol·L-1·s-1
    • C. 0.2 mol·L-1·s-1
    • D. 0.04 mol·L-1·s-1
    Show answer & solution

    Answer: B

    Answer: B In 10 s, SO₂ falls from 1.0 mol to 0.6 mol, a decrease of 0.4 mol; in a 2 L vessel Δc(SO₂) = 0.4 / 2 = 0.2 mol·L⁻¹. v(SO₂) = 0.2 / 10 = 0.02 mol·L⁻¹·s⁻¹. From the coefficients 2SO₂ : 1O₂, v(O₂) = v(SO₂) / 2 = 0.01 mol·L⁻¹·s⁻¹, option B.
    See it in the full CSCA Chemistry December 2025 Past Paper
  2. CSCA Chemistry March 2026 Past Paper
    Which statement about catalysts is correct?
    • A．A catalyst does not participate in the reaction
    • B．A catalyst lowers the activation energy
    • C．A catalyst can only speed up reactions
    • D．A catalyst's properties are unchanged before and after
    Show answer & solution

    Answer: B

    Catalysts lower activation energy ✓ . A: they participate. C: accelerate both directions. D: too broad. Answer: B.
    Take the full CSCA Chemistry March 2026 Past Paper — timed & graded
  3. CSCA Chemistry April 2026 Past Paper
    10 mg of adsorbent is added to 100 mL of a solution of pollutant T (molar mass 100 g/mol). If the adsorption capacity q per unit mass of adsorbent rises from 0 to 31 mg/g within 10 min, the rate of this adsorption reaction is ( ).
    • A. 1×10−31 \times 10^{-3} mol·L−1⋅min−1\mathrm{L^{-1}} \cdot \mathrm{min^{-1}}
    • B. 1×10−51 \times 10^{-5} mol·L−1⋅min−1\mathrm{L^{-1}} \cdot \mathrm{min^{-1}}
    • C. 1×10−61 \times 10^{-6} mol·L−1⋅min−1\mathrm{L^{-1}} \cdot \mathrm{min^{-1}}
    • D. 3.1×10−63.1 \times 10^{-6} mol·L−1⋅min−1\mathrm{L^{-1}} \cdot \mathrm{min^{-1}}
    Show answer & solution

    Answer: D

    Answer: D Mass of T adsorbed = 31 mg/g × 0.01 g = 0.31 mg = 3.1×10−43.1 \times 10^{-4} g; n(T) = 3.1×10−43.1 \times 10^{-4} g ÷ 100 g/mol = 3.1×10−63.1 \times 10^{-6} mol; concentration change = 3.1×10−63.1 \times 10^{-6} mol ÷ 0.1 L = 3.1×10−53.1 \times 10^{-5} mol/L; rate = 3.1×10−53.1 \times 10^{-5} ÷ 10 = 3.1×10−63.1 \times 10^{-6} mol·L−1⋅min−1\mathrm{L^{-1}} \cdot \mathrm{min^{-1}}.
    Take the full CSCA Chemistry April 2026 Past Paper — timed & graded

Real-exam evidence

Recorded past-paper appearances are listed below; online verification is labelled separately.

6 questions across 4 sittings1 of 4 sittings available to verify free online

5 questions across 4 mock papers — our own papers, so a coverage signal rather than real-exam evidence

13 related practice questions availablePractise free: Chemical Reaction Rate and Equilibrium

Sitting evidence comes from the CSCA past papers on record; questions whose original paper is missing are excluded, and sprint papers are deliberately not counted.

Related concepts

Chemical Equilibrium