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9701 · 8.1

Rate of reaction — practice questions

Practice and worked examples for 9701 Rate of reaction. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

The reaction between magnesium ribbon and excess hydrochloric acid produces hydrogen gas. The volume of hydrogen collected was recorded over time. A tangent drawn to the curve at t = 0 s had a gradient of +3.6 cm³ s⁻¹. Calculate the initial rate of reaction in mol s⁻¹. (Assume the experiment was conducted at room temperature and pressure, where the molar volume of a gas is 24,000 cm³ mol⁻¹).

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  1. Identify the given rate: The gradient of the volume-time graph gives the rate in cm³ s⁻¹. Initial rate = 3.6 cm³ s⁻¹.
  2. Identify the conversion factor: The molar volume of a gas at RTP is 24,000 cm³ mol⁻¹. This means 1 mole of H₂ occupies 24,000 cm³.
  3. Set up the calculation: To convert from a volume (cm³) to moles (mol), we divide by the molar volume. Rate in mol s⁻¹ = Rate in cm³ s⁻¹ / Molar Volume Rate = 3.6 cm³ s⁻¹ / 24,000 cm³ mol⁻¹
  4. Calculate the final answer: Rate = 0.00015 mol s⁻¹ In standard form, this is 1.5 x 10⁻⁴ mol s⁻¹.

Worked example 2

In the decomposition of hydrogen peroxide, 2H2O2(aq)→2H2O(l)+O2(g)2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g), the concentration of H2O2H_2O_2 was measured. At t = 60 s, the concentration was 0.500 mol dm⁻³. At t = 180 s, the concentration had fallen to 0.200 mol dm⁻³. Calculate the average rate of reaction over this interval.

Show solution outline
  1. State the formula for rate: Rate = - (Change in concentration of reactant) / Time taken Rate = - Δ[H2O2H_2O_2] / Δt (The negative sign is used for a reactant because its concentration decreases. The final rate is always given as a positive value).
  2. Calculate the change in concentration and time: Δ[H2O2H_2O_2] = Final concentration - Initial concentration = 0.200 - 0.500 = -0.300 mol dm⁻³ Δt = Final time - Initial time = 180 - 60 = 120 s
  3. Substitute values into the formula: Rate = - (-0.300 mol dm⁻³) / 120 s
  4. Calculate the final answer: Rate = 0.300 / 120 mol dm⁻³ s⁻¹ Rate = 0.0025 mol dm⁻³ s⁻¹ or 2.5 x 10⁻³ mol dm⁻³ s⁻¹.
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