OCR GCSE Chemistry A: Controlling reactions (J248)

Sub-topic C5.2 Controlling reactions sits in Topic C5 Monitoring and controlling chemical reactions of the OCR GCSE (9–1) Chemistry A (Gateway Science) J248 specification (Version 4.0, August 2026). Every part of it is taught here: learning outcomes C5.2a to C5.2i, the mathematical learning outcomes CM5.2i to CM5.2iv, and the practical suggestions linked to PAG C1, PAG C7 and PAG C8. No statement in C5.2 is in bold type, so every outcome here is assessed at both Foundation and Higher Tier. According to the assessment overview, Topic C5 is examined in Paper 2 if you sit Foundation Tier, or Paper 4 if you sit Higher Tier.

Once you finish, move on to the revision notes and the practice questions. The OCR GCSE Chemistry hub holds the whole course, the J248 checklist lists every statement, and a free 10-minute diagnostic shows where to start.

Statements in C5.2

Statement	Skill you need	Tier
C5.2a	Suggest practical methods for finding the rate of a given reaction	Both tiers
C5.2b	Interpret rate graphs, including 1/t as a measure of rate and gradients (not order of reaction)	Both tiers
C5.2c	Describe how temperature, concentration, pressure and surface area change the rate	Both tiers
C5.2d	Explain the effects of temperature, concentration and pressure using the frequency and energy of collisions	Both tiers
C5.2e	Explain the effect of the size of solid pieces using surface area to volume ratio	Both tiers
C5.2f	Describe what catalysts are like and how they change the rate	Both tiers
C5.2g	Identify catalysts in reactions	Both tiers
C5.2h	Explain catalytic action using activation energy and reaction profiles	Both tiers
C5.2i	Recall that enzymes are catalysts in biological systems	Both tiers
CM5.2i–iv	Arithmetic and ratio for rates; drawing and reading graphs; gradients; proportionality	Both tiers

The specification flags two misconceptions here: misreading rate graphs, and thinking catalysts run out or get used up.

What “rate” means

The rate of reaction is how quickly reactants are used up or products are formed. You find it by measuring an amount that changes and dividing by the time taken.

What you measure	Rate calculation	Unit
Volume of gas made	volume ÷ time	cm³/s
Mass lost from the flask	mass ÷ time	g/s
Time for a fixed change (cross disappears, ribbon dissolves)	1 ÷ time	s⁻¹ (also written /s)

Change ÷ time gives a mean rate over that interval.

Measuring rate in practice (C5.2a, PAG C8)

Pick a method that tracks something that changes in the reaction you are given.

1. Collect the gas. Use a gas syringe, or an upturned measuring cylinder or burette full of water. Read the volume at regular times.
2. Measure mass loss. Stand the flask on a balance with a loose cotton wool plug in the neck. The plug lets gas out but keeps acid spray in.
3. Time a colour change or a precipitate. In the disappearing cross experiment, sodium thiosulfate solution reacts with hydrochloric acid and the mixture clouds as sulfur, a pale yellow solid, precipitates. You time it until a cross drawn beneath the flask can no longer be seen.
4. Time a solid disappearing. Time how long a measured length of magnesium ribbon takes to react completely.

Worked example 1: choosing and using a method (C5.2a, CM5.2i)

A student wants to follow the reaction of magnesium ribbon with dilute hydrochloric acid.

Why is mass loss a poor choice? Hydrogen is very light: 0.10 g of magnesium makes only about 0.008 g of it, too little for a school balance to track well. A gas syringe is better.

The syringe collects 66 cm³ of hydrogen in the first 24 s. Mean rate = 66 ÷ 24 = 2.75 cm³/s.

Carbon dioxide from marble chips and acid is much heavier, so mass loss works there. A fall of 0.84 g in 120 s gives 0.84 ÷ 120 = 0.007 g/s.

Fair test. Change one variable only; keep acid volume and concentration, solid mass and size, and temperature fixed. Repeat, discard anomalies and take a mean.

Interpreting rate graphs (C5.2b, CM5.2ii, CM5.2iii)

Plot time along the horizontal axis and the amount of product (or flask mass) up the vertical axis.

• The steeper the line, the faster the reaction.
• The curve is steepest at the start and gets less steep as reactants are used up.
• The line goes flat when one reactant has run out: the reaction has stopped.
• The final height (total product) depends on the amount of limiting reactant, not on the rate.

The rate at one moment is the gradient of a tangent: a straight line that just touches the curve at that time. Pick two points far apart on it and calculate rise ÷ run.

Worked example 2: reading a gas volume curve

Marble chips react with excess dilute hydrochloric acid, and a gas syringe collects the CO₂ given off.

Time (s)	0	25	50	75	100	125	150	175
Volume of CO₂ (cm³)	0	30	50	63	72	77	80	80

(a) When did the reaction stop? The volume first stops changing at 150 s.

(b) Mean rate over the first 50 s: 50 ÷ 50 = 1.0 cm³/s.

(c) Mean rate from 50 s to 100 s: (72 − 50) ÷ 50 = 22 ÷ 50 = 0.44 cm³/s. The rate falls as reactants are used up.

(d) Rate at 50 s. A tangent drawn at 50 s passes through (0 s, 18 cm³) and (100 s, 82 cm³). Gradient = (82 − 18) ÷ (100 − 0) = 64 ÷ 100 = 0.64 cm³/s.

1/t as a measure of rate

When every run ends at the same fixed point (cross gone, ribbon gone), rate is proportional to 1/t. Halving the time doubles the rate.

Worked example 3: proportionality from the disappearing cross (CM5.2iv)

Different volumes of sodium thiosulfate solution are made up to 50 cm³ with water, so the volume of thiosulfate is a measure of its concentration. The same volume of acid is added each time.

Thiosulfate (cm³)	50	40	30	20
Time for cross to vanish (s)	36	45	60	90
1/t (s⁻¹)	0.0278	0.0222	0.0167	0.0111

Going from 20 cm³ to 40 cm³ doubles the concentration. The time halves from 90 s to 45 s, so 1/t doubles from 0.0111 to 0.0222 s⁻¹. 1/t ÷ volume is about 0.00056 every time, so the rate is directly proportional to concentration: a graph of 1/t against volume is a straight line through the origin.

Factors that change the rate (C5.2c, C5.2d)

Particles must collide with at least the activation energy (the minimum energy needed) to react. More frequent successful collisions mean a faster reaction.

Change	Effect on rate	Explanation
Higher temperature	Increases	Particles move faster, so collisions are more frequent and a bigger fraction have at least the activation energy.
Higher concentration (solutions)	Increases	Reacting particles are more crowded, so they collide more often
Higher pressure (gases)	Increases	Gas particles squeezed into a smaller volume, so more frequent collisions
Larger surface area (smaller pieces)	Increases	More solid particles exposed, so more frequent collisions

Of the four, only temperature alters collision energy as well as frequency, so a temperature answer must mention activation energy.

The suggested practical reacts magnesium with acid warmed in a water bath to different temperatures, timing each run.

Surface area to volume ratio (C5.2e, CM5.2iv)

Only surface particles of a solid can be hit. Cutting a lump up keeps the volume but exposes more surface, so the surface area to volume ratio, and the rate, go up.

Worked example 4: one cube or many

A marble cube has sides of 6 mm. Surface area = 6 faces × 6 mm × 6 mm = 216 mm²; volume = 6 × 6 × 6 = 216 mm³. Ratio = 216 : 216 = 1 : 1.

Cut it into 2 mm cubes. There are 3 × 3 × 3 = 27 small cubes. Each has surface area 6 × 2 × 2 = 24 mm², so the total is 27 × 24 = 648 mm². The volume is still 216 mm³. Ratio = 648 : 216 = 3 : 1.

The ratio has tripled, so more particles are exposed and the initial rate is greater. A powder is fastest. The suggested practical compares marble chip sizes with hydrochloric acid at the same total mass.

Catalysts (C5.2f, C5.2g)

A catalyst speeds up a reaction and is still there, unchanged, when the reaction ends. Its characteristics:

• It speeds up the reaction.
• It is chemically unchanged, and the same mass is left, at the end.
• It does not change the amount of product.
• Only a small amount is needed, because it is not used up.
• It is specific: a catalyst that works for one reaction may do nothing for another.

Identifying a catalyst needs two pieces of evidence: the reaction is faster with it, and the same mass is recovered at the end.

Worked example 5: which black powder is a catalyst? (C5.2g)

Hydrogen peroxide slowly decomposes to water and oxygen; the specification suggests testing black powders, including manganese(IV) oxide, with it. A student stirs 0.40 g of a powder into peroxide solution, times how long the mixture takes to give 20 cm³ of oxygen, then filters, dries and reweighs the powder. Each powder is tested on fresh solution from one bottle.

Powder	Time for 20 cm³ of oxygen (s)	Mass recovered (g)
None	over 600	–
A	18	0.40
B	over 600	0.40
C	75	0.33

• A is a catalyst: 20 cm³ in 18 s instead of over 600 s, and all 0.40 g recovered.
• B is not a catalyst for this reaction: it is no faster than no powder at all.
• C is not acting as a catalyst: it speeds things up, but 0.07 g of it has been used up, so it is reacting.

Likewise, copper powder makes zinc react faster with dilute sulfuric acid and is recovered unchanged.

How catalysts work (C5.2h)

A catalyst gives the reaction an alternative pathway whose activation energy is lower. Without any change in temperature, a larger share of collisions now clear the energy barrier.

On a reaction profile (energy on the y-axis, progress of reaction on the x-axis):

• the reactant and product levels stay exactly where they were, so the overall energy change is the same;
• the catalysed hump is lower than the uncatalysed hump;
• activation energy is the height of the hump measured upwards from the reactants’ energy.

Worked example 6. Uncatalysed activation energy 75 kJ/mol, catalysed 58 kJ/mol: the catalyst lowers it by 75 − 58 = 17 kJ/mol. The energy change is unaffected.

Enzymes (C5.2i)

Enzymes are catalysts in biological systems: they speed up reactions in living cells without being used up. Catalase, for example, speeds up the breakdown of hydrogen peroxide in cells.

Common errors

• Saying a flat graph means “the reaction is at a constant rate”. Flat means it has stopped.
• Reading tangent points off the curve instead of the tangent line.
• Thinking a longer time means a faster reaction; rate is proportional to 1/t.
• Writing “more collisions” instead of “more frequent collisions”.
• Explaining temperature without activation energy.
• Claiming a catalyst energises the particles. What it actually does is lower the barrier.
• Saying a catalyst is used up or changes the amount of product.
• Calling a substance a catalyst because the rate went up, without checking that its mass is unchanged.
• Saying smaller pieces have less surface area. Total surface area goes up.

Next steps

Summarise with the Controlling reactions revision notes, then try the Controlling reactions practice questions. Reaction profiles were first met in the Energetics study guide. Related units: Monitoring chemical reactions and Equilibria.

Official syllabus

OCR GCSE (9–1) Chemistry A (Gateway Science) J248 specification, Version 4.0 (August 2026), for first assessment in 2018, published by OCR. Sub-topic C5.2 Controlling reactions, learning outcomes C5.2a–C5.2i, mathematical learning outcomes CM5.2i–CM5.2iv and the linked practical suggestions (PAG C1, PAG C7, PAG C8).