OCR A Level Chemistry A: Reaction rates (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

Section 3.2.2, Reaction rates, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, is the focus here; every outcome from (a) to (g) is tested. H432 is the full A level. Where needed, use 24.0 dm³ mol⁻¹ for one mole of gas at RTP.

Learn the content from the study guide and revision notes. Other H432 topics sit on the chemistry hub for this course; a checklist to print and the free diagnostics are there too.

Questions

1. Which change increases the proportion of molecules with energy of at least Ea?

A. Increasing the pressure of a gas at constant temperature B. Increasing the concentration of a solution C. Increasing the temperature D. Using a larger flask for the same reaction mixture [1]

2. Zinc powder reacts with aqueous copper(II) sulfate.

(a) State what is meant by the term activation energy. [1] (b) Explain why the reaction is faster with 0.50 mol dm⁻³ copper(II) sulfate than with 0.20 mol dm⁻³ copper(II) sulfate. [2]

3. Excess iron powder reacts with dilute sulfuric acid. The hydrogen is collected in a gas syringe.

t / s	0	30	60	90	120	150	180
V(H₂) / cm³	0	42	72	92	104	110	112

(a) Calculate the mean rate over the first 60 s. [1] (b) A tangent to the curve at 90 s passes through (30 s, 59 cm³) and (150 s, 125 cm³). Calculate the rate at 90 s. [2] (c) Convert your answer to (b) into mol s⁻¹. [2] (d) Explain why the gas is produced more slowly as time goes on. [2] (e) The experiment is repeated with a different concentration of acid. State one variable, other than temperature, that must be kept the same. [1]

4. Zinc carbonate powder, ZnCO₃, is added to excess dilute sulfuric acid in a conical flask on a balance.

ZnCO₃(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂O(l) + CO₂(g)

(a) The mass falls from 85.40 g to 85.04 g in the first 20 s. Calculate the mean rate of CO₂ production in mol s⁻¹. [2] (b) State why a plug of cotton wool is placed in the neck of the flask. [1] (c) Explain why this method would be unsuitable for following the reaction of zinc metal with the same acid. [1] (d) 1.881 g of zinc carbonate was used. Calculate the total mass loss when the reaction is complete. Mr of ZnCO₃ = 125.4. [2]

5. A reaction has ΔH = +48 kJ mol⁻¹ and an uncatalysed activation energy of 210 kJ mol⁻¹. A catalyst lowers the activation energy by 75 kJ mol⁻¹.

(a) Sketch a labelled enthalpy profile diagram showing the catalysed and uncatalysed routes. [3] (b) Calculate the activation energy of the catalysed reverse reaction. [2] (c) State the value of ΔH for the catalysed reaction. [1]

6.

(a) Explain the difference between a homogeneous catalyst and a heterogeneous catalyst. [2] (b) Classify each catalyst as homogeneous or heterogeneous. (i) Iron in the reaction of nitrogen gas with hydrogen gas (ii) Sulfuric acid in a solution of ethanol and ethanoic acid (iii) Nickel in the reaction of gaseous ethene with hydrogen (iv) Chlorine radicals in the breakdown of ozone in the stratosphere [2]

7. A sample of gas is heated from temperature T₁ to a higher temperature T₂.

(a) Sketch Boltzmann distributions for the gas at T₁ and T₂ on the same axes. Label the axes. [3] (b) Use your sketch to explain why the rate of reaction increases. [3]

8. Explain, with reference to a Boltzmann distribution, why adding a catalyst makes a reaction faster. [3]

9. A new catalyst lets a process run cooler, saving 1.11 × 10⁹ kJ of heat per day that was supplied by burning propane. Use these values: ΔcH of propane = −2220 kJ mol⁻¹; Mr of CO₂ = 44.0.

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

(a) Calculate the mass of CO₂, in tonnes, that is no longer released each day. [3] (b) The new catalyst contains a toxic metal compound. Evaluate whether switching to it improves the sustainability of the process. [2]

10. Plan an experiment to find out how the concentration of nitric acid affects the rate of its reaction with copper(II) carbonate powder, by following the loss in mass. [6]

11. In a gas-phase reaction, the concentration of product P was measured over time.

t / s	0	50	100	150	200	300
[P] / mol dm⁻³	0	0.0240	0.0410	0.0530	0.0610	0.0700

(a) Calculate the mean rate of formation of P over the first 100 s. [1] (b) A tangent at t = 0 passes through (0 s, 0) and (60 s, 0.0330 mol dm⁻³). Calculate the initial rate. [2] (c) The experiment is repeated at the same temperature but at twice the pressure. Predict and explain the effect on the initial rate. [2] (d) The experiment is repeated at the original pressure but at a higher temperature. Explain, with reference to the Boltzmann distribution, the effect on the initial rate. [3]

Answers

1. C [1] Examiner insight: At a fixed temperature the energy distribution is unchanged, so A, B and D alter only collision frequency.

2. (a) The minimum energy colliding particles need in order to react [1] (b) The more concentrated solution has more Cu²⁺ ions per unit volume [1], so the ions hit the zinc surface more often and more successful collisions happen per second [1]. Examiner insight: Saying the particles “have more energy” in (b) contradicts the chemistry; concentration does not change collision energy.

3. (a) 72 ÷ 60 = 1.2 cm³ s⁻¹ [1] (b) Gradient = (125 − 59) ÷ (150 − 30) [1] = 66 ÷ 120 = 0.55 cm³ s⁻¹ [1] (c) 0.55 ÷ 24 000, using 24 000 cm³ mol⁻¹ [1] = 2.3 × 10⁻⁵ mol s⁻¹ (2.29 × 10⁻⁵) [1] (d) The sulfuric acid is consumed, lowering its concentration [1]; collisions with the iron happen less often [1]. (e) Any one: mass of iron powder, its particle size, or the volume of acid [1]. Examiner insight: In (c), dividing by 24.0 instead of 24 000 gives an answer a thousand times too large.

4. (a) 0.36 g ÷ 20 s = 0.018 g s⁻¹ [1]; 0.018 ÷ 44.0 = 4.1 × 10⁻⁴ mol s⁻¹ [1] (b) It lets CO₂ out but stops acid spray escaping and adding to the mass loss [1]. (c) The gas would be hydrogen, whose small Mr gives too small a mass change to measure precisely [1]. (d) n(ZnCO₃) = 1.881 ÷ 125.4 = 0.0150 mol = n(CO₂) [1]; mass = 0.0150 × 44.0 = 0.660 g [1] Examiner insight: Stopping at 0.018 g s⁻¹ in (a) answers only half the question.

5. (a) Axes labelled enthalpy and progress of reaction, with products drawn above reactants [1]; two humps, the catalysed one lower [1]; Ea, Ea(cat) and ΔH shown with correctly directed arrows [1]. (b) Catalysed forward Ea = 210 − 75 = 135 kJ mol⁻¹ [1]; reverse = 135 − 48 = 87 kJ mol⁻¹ [1] (c) +48 kJ mol⁻¹ [1] Examiner insight: Each Ea arrow runs from the reactant level to a peak; one drawn from the products shows the reverse Ea.

6. (a) Homogeneous: the catalyst is in the same physical state as the reactants [1]. Heterogeneous: the catalyst is in a different physical state from the reactants [1]. (b) (i) heterogeneous; (ii) homogeneous; (iii) heterogeneous; (iv) homogeneous. All four correct [2]; two or three correct earn one mark. Examiner insight: “Physical state” or “phase” is needed; “different substance” does not explain the distinction.

7. (a) Axes: number of molecules (y) and energy (x) [1]; both curves start at the origin and do not meet the energy axis at high energy [1]; the T₂ curve has a lower peak displaced to higher energy, with the same area [1]. (b) Ea marked: the area to the right of Ea is larger at T₂ [1], so a greater proportion of molecules have energy of at least Ea [1], giving more successful collisions per second [1]. Examiner insight: A T₂ curve lying wholly above the T₁ curve shows more molecules, not a higher temperature.

8. With the catalyst the reaction follows a different pathway of lower activation energy, so Ea sits at a lower energy on the same curve [1]. A greater proportion of molecules have energy of at least the catalysed Ea [1]. More collisions per second are successful, so the rate increases [1]. Examiner insight: Redrawing the curve for a catalyst is wrong; the temperature has not changed, so only the Ea marker moves.

9. (a) n(C₃H₈) = 1.11 × 10⁹ ÷ 2220 = 5.0 × 10⁵ mol [1]; n(CO₂) = 3 × 5.0 × 10⁵ = 1.5 × 10⁶ mol [1]; mass = 1.5 × 10⁶ × 44.0 = 6.6 × 10⁷ g = 66 tonnes [1] (b) Benefit: less fuel burned, so less CO₂ released and fewer fossil resources used [1]. Against: the toxic catalyst must be contained, recovered and disposed of safely, so sustainability improves only if its escape is prevented [1]. Examiner insight: An “evaluate” answer that lists only benefits is one-sided; give a point on each side and a conclusion.

10. Six points such as:

• Weigh the same mass of copper(II) carbonate powder, of the same particle size, and measure the same volume of acid for every run [1].
• Make several nitric acid concentrations by diluting a stock solution with distilled water [1].
• Keep the temperature constant, for example by standing the acid in a water bath first and checking with a thermometer [1].
• Stand a conical flask of acid on a balance; tip in the carbonate, add a cotton wool plug and start the timer at once [1].
• Record the mass at regular intervals, such as every 15 s, until it stops changing [1].
• Plot mass lost against time, draw a tangent at t = 0 for each initial rate, and compare with concentration, repeating runs to check reliability [1].

Examiner insight: State how each control variable is controlled; “keep it a fair test” alone earns nothing.

11. (a) 0.0410 ÷ 100 = 4.1 × 10⁻⁴ mol dm⁻³ s⁻¹ [1] (b) 0.0330 ÷ 60 [1] = 5.5 × 10⁻⁴ mol dm⁻³ s⁻¹ [1] (c) The initial rate increases [1]: doubling the pressure doubles the gas concentrations, so collisions are more frequent [1]. (d) At the higher temperature, the Boltzmann curve is flatter and shifted to higher energy [1]. A greater proportion of molecules have energy of at least Ea [1]. More successful collisions per second, so the initial rate increases [1]. Examiner insight: In (c) the temperature is fixed, so answers about Ea or the Boltzmann curve gain nothing.

Where marks are usually lost

• Giving concentration or pressure an effect on collision energy, as in questions 2 and 11.
• Taking a tangent gradient from points on the curve instead of on the drawn line.
• Mixing cm³ and dm³ when converting a gas volume rate to moles, as in question 3.
• Stopping at g s⁻¹ when mol s⁻¹ was asked for.
• Moving the Boltzmann curve for a catalyst, or giving the hotter curve a higher peak.
• Using the wrong mole ratio for CO₂ from a fuel, as in question 9.
• Listing control variables in a plan without saying how each is controlled.

Next steps

• Rules in brief: Reaction rates revision notes
• Teaching and examples: Reaction rates study guide
• Book a free trial class
• A wider check: all free 10-minute diagnostics
• The H432 course hub
• Tick-off list: printable OCR checklist

Official syllabus

Source: section 3.2.2 Reaction rates, learning outcomes (a) to (g), in the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, published by OCR.