OCR A Level Chemistry A: Reaction rates (H432)

This study guide teaches 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. Within Module 3, 3.2.2 has lettered outcomes (a) to (g), and this page teaches each one. H432 is the full A level. OCR’s assessment overview puts Module 3 content into Periodic table, elements and physical chemistry (01), and also into Unified chemistry (03), which samples every module.

The chemistry hub for OCR gathers the course, and the printable H432 checklist lets you mark progress. Pair this guide with the Reaction rates revision notes and practice questions, and try the free diagnostics.

Outcomes in 3.2.2

Statement	Skill	Specification guidance
3.2.2(a)	Explain how concentration, including gas pressure, affects rate via collision frequency	–
3.2.2(b)	Calculate rate from gradients of quantity-time graphs	M3.1, M3.2, M3.5; e.g. concentration, gas volume, mass
3.2.2(c)	Explain how a catalyst (i) raises rate without being used up and (ii) gives a lower-Ea route, on enthalpy profiles	Details of processes not required
3.2.2(d)	(i) Homogeneous and heterogeneous catalysts; (ii) economic importance and sustainability benefits	HSW9,10: benefits weighed against toxicity of some catalysts
3.2.2(e)	Techniques for investigating rates: mass, gas volumes, time	PAG9; HSW4
3.2.2(f)	Explain the Boltzmann distribution qualitatively and its link to Ea	M3.1; see 3.2.1(c)
3.2.2(g)	Use Boltzmann distributions to explain the effect of (i) temperature and (ii) a catalyst on the proportion of molecules exceeding Ea, and so the rate	M3.1; HSW1,2,5

Activation energy, Ea, comes from 3.2.1(c): the minimum energy that colliding particles need in order to react.

3.2.2(a): Concentration, pressure and collision frequency

Particles must collide to react, and only collisions with energy of at least Ea succeed.

Concentration. A more concentrated solution has more reactant particles in each dm³. They collide more often, so more collisions per second have energy of at least Ea. The rate increases.

Pressure of gases. At constant temperature, a higher gas pressure squeezes the same molecules into a smaller volume, which raises the concentration. Collisions become more frequent and the rate increases.

The energy of each collision does not change, because the temperature is fixed. Only the frequency changes.

Worked example 1: compressing a gas mixture

A syringe holds 0.060 mol of a reacting gas mixture in 2.0 dm³. The plunger is pushed in until the volume is 1.0 dm³ at the same temperature. Explain the effect on the rate.

1. Concentration rises from 0.060 ÷ 2.0 = 0.030 to 0.060 ÷ 1.0 = 0.060 mol dm⁻³. Pressure and concentration have both doubled.
2. Molecules collide more frequently, so more collisions per second have energy of at least Ea and the rate increases.

3.2.2(b): Rates from graph gradients

Rate is the change in amount of a reactant or product per unit time. Follow any quantity that changes as the reaction goes, such as concentration, gas volume or mass.

• Mean rate over an interval = change in the quantity ÷ time taken.
• Rate at one instant = gradient of a tangent drawn at that time (M3.5).
• Initial rate = tangent gradient where t = 0.

The units come from the axes. Gas volume against time gives cm³ s⁻¹. Mass against time gives g s⁻¹. Concentration against time gives mol dm⁻³ s⁻¹. A falling quantity gives a negative gradient; quote the rate as a positive value.

For a tangent, place the ruler so it touches the curve at one point, extend the line well across the grid, and read two points far apart on it.

Worked example 2: gas volume

Magnesium ribbon reacts with dilute ethanoic acid, and the hydrogen is pushed into a gas syringe. The acid is the limiting reactant.

t / s	0	20	40	60	80	100	120
V(H₂) / cm³	0	31	54	70	80	85	86

(i) Mean rate over the first 40 s = 54 ÷ 40 = 1.35 cm³ s⁻¹.

(ii) A tangent drawn at t = 60 s passes through (20 s, 46 cm³) and (100 s, 94 cm³).

Gradient = (94 − 46) ÷ (100 − 20) = 48 ÷ 80 = 0.60 cm³ s⁻¹.

Check: 46 + 0.60 × 40 = 70 cm³ at 60 s, so the line does touch the curve there.

(iii) In moles, using 24.0 dm³ mol⁻¹ (24 000 cm³ mol⁻¹) at RTP: 0.60 ÷ 24 000 = 2.5 × 10⁻⁵ mol s⁻¹.

The rate falls over time because the acid concentration drops, so collisions become less frequent.

Worked example 3: mass loss

Potassium hydrogencarbonate reacts with excess dilute hydrochloric acid in a flask on a balance. Carbon dioxide escapes, so the mass falls. A tangent at t = 0 drops from 120.00 g to 118.90 g over 50 s.

1. Initial rate = 1.10 g ÷ 50 s = 0.022 g s⁻¹.
2. The mass lost is carbon dioxide (Mr 44.0), so 0.022 ÷ 44.0 = 5.0 × 10⁻⁴ mol s⁻¹ of CO₂.

For a concentration graph the same method gives mol dm⁻³ s⁻¹. If the tangent at t = 0 for a reactant A falls from 0.0100 mol dm⁻³ to zero over 400 s, the initial rate is 0.0100 ÷ 400 = 2.5 × 10⁻⁵ mol dm⁻³ s⁻¹.

3.2.2(c): What a catalyst does

A catalyst increases the rate without being used up by the overall reaction. It may take part in a step, but it is regenerated. It provides a different route with a lower activation energy. On an enthalpy profile diagram:

• Reactants and products sit at the same enthalpy levels with or without the catalyst. ΔH does not change.
• The catalysed curve has a lower peak. The catalysed Ea is the height from reactants to that lower peak.
• The reverse reaction’s Ea is lowered by the same amount, so the catalyst speeds up both directions.

Details of processes are not required.

Worked example 4: reading an enthalpy profile

An exothermic reaction has ΔH = −60 kJ mol⁻¹. The uncatalysed Ea is 180 kJ mol⁻¹ and the catalysed Ea is 105 kJ mol⁻¹.

1. The catalyst lowers Ea by 180 − 105 = 75 kJ mol⁻¹.
2. For the reverse reaction, add back the 60 kJ mol⁻¹ released going forward: uncatalysed reverse Ea = 180 + 60 = 240 kJ mol⁻¹.
3. Catalysed reverse Ea = 105 + 60 = 165 kJ mol⁻¹. This is also 75 kJ mol⁻¹ lower.

To sketch it, label the axes enthalpy and progress of reaction, put products below reactants, and draw two humps with arrows for each Ea and for ΔH.

3.2.2(d): Types of catalyst and why they matter

Homogeneous catalyst: in the same physical state as the reactants, such as acid catalysing ester formation in solution, or chlorine radicals breaking down ozone in the gas phase.

Heterogeneous catalyst: in a different physical state from the reactants, usually a solid with gases or liquids passing over it. Examples are iron in ammonia production, nickel for alkene hydrogenation and platinum-group metals in catalytic converters.

Economic and sustainability benefits. A catalyst lets an industrial process run at a useful rate at a lower temperature. Less fossil fuel is burned to supply heat, which cuts fuel costs and carbon dioxide emissions.

Weighing it up (HSW9,10). Some catalysts are toxic, so they must be contained, recovered and disposed of with care. Weigh the energy and CO₂ savings against these hazards.

Worked example 5: CO₂ saved by a catalyst

Switching to a new catalyst lets a plant run cooler, saving 4.45 × 10⁹ kJ of heat per day. That heat came from burning methane. Take ΔcH of methane as −890 kJ mol⁻¹. CH₄ + 2O₂ → CO₂ + 2H₂O.

1. Methane saved: 4.45 × 10⁹ ÷ 890 = 5.0 × 10⁶ mol.
2. One mole of CO₂ forms per mole of CH₄, so 5.0 × 10⁶ mol of CO₂ is not released.
3. Mass: 5.0 × 10⁶ × 44.0 = 2.2 × 10⁸ g = 220 tonnes of CO₂ per day.

3.2.2(e): Techniques for following a rate (PAG9)

PAG9 is a continuous monitoring method, including measurement of time and use of software to process data. Its example activity, finding a half-life, links to 5.1.1 (see How fast?).

Method	How it works	Points to watch
Mass loss	Flask on a balance; escaping gas lowers the mass	Cotton wool plug stops spray loss; suits CO₂, but hydrogen is too light
Gas syringe	Read the volume at set times	Plunger must move freely; replace the bung quickly
Over water	Gas displaces water from an inverted burette or cylinder	Unsuitable for gases that dissolve readily
Timing	Time a fixed change, such as a cross being hidden	1/time is proportional to the mean rate

Change one variable and control the rest: temperature (water bath), volumes, concentrations, and the mass and surface area of any solid. Start the clock as the reactants meet. Data loggers and spreadsheets can record and plot readings.

3.2.2(f): The Boltzmann distribution

Molecules in a gas have a spread of energies. The Boltzmann distribution plots number of molecules against energy. Learn its features:

• It begins at the origin, since no molecule has zero energy.
• Its single peak marks the most probable energy.
• It has a long tail on the high-energy side and approaches the energy axis without touching it. There is no maximum energy.
• The area under the curve is the total number of molecules.

Ea sits far out on the tail. Molecules in the region beyond Ea have enough energy to react on collision; they are a small proportion of the total.

3.2.2(g): Using the distribution to explain rate changes

Temperature

At a higher temperature the curve flattens and moves to the right:

• The peak is lower and at a higher energy.
• Total area is unchanged: heating adds no molecules.
• A much larger area now lies beyond Ea.

So a greater proportion of molecules have energy of at least Ea, more collisions succeed each second and the rate increases. Collisions also become a little more frequent, but the rise in the proportion beyond Ea is the main reason.

Catalyst

A catalyst leaves the curve unchanged, because the temperature is the same. It moves Ea to a lower value. More of the area now lies beyond the catalysed Ea, so a greater proportion of molecules can react on collision and the rate increases.

Worked example 6: building a full explanation

Explain, using a Boltzmann distribution, why warming a gas mixture from 300 K to 320 K increases the rate.

1. Sketch both curves with the same area; the 320 K curve has a lower peak, further right. Mark Ea.
2. The area beyond Ea is larger at 320 K.
3. So a greater proportion of molecules have energy of at least Ea, more collisions per second succeed, and the rate is higher.

Common errors

• Saying a higher concentration gives collisions more energy. It gives more frequent collisions.
• Claiming a catalyst changes ΔH.
• A Boltzmann curve that misses the origin or meets the energy axis.
• Drawing a higher-temperature curve with a taller peak or a larger area.
• Moving the curve to show a catalyst. Only the Ea line moves.
• Leaving units off a rate, or giving cm³ s⁻¹ when moles per second were asked for.
• Calling a solid catalyst with gaseous reactants homogeneous.

Where next

For quick recall use the revision notes, then try the practice questions. Revisit gas volume and mass conversions if the mole steps felt shaky. For PAG9 and planning, see practical skills for the written papers and PAG and endorsement notes, then look again at the PAG9 table above.

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.