What factors affect the rate of reaction in Chemistry?

In Chemistry, particularly under the Edexcel IGCSE curriculum, the rate of reaction is influenced by several factors: temperature, concentration of reactants, surface area of solid reactants, presence of a catalyst, and pressure (for reactions involving gases). Each of these factors can increase or decrease the rate at which reactants are converted into products.

How does temperature influence the rate of reaction?

Temperature affects the rate of reaction by increasing the kinetic energy of the particles involved. As temperature rises, particles move faster, leading to more frequent and energetic collisions. This increase in collision frequency and energy results in a higher rate of reaction, as more particles have the necessary energy to overcome the activation energy barrier.

Why is a catalyst important in chemical reactions?

A catalyst is crucial in chemical reactions because it provides an alternative pathway with a lower activation energy. This means that more reactant particles have enough energy to react, increasing the rate of reaction without being consumed in the process. Catalysts are especially important in industrial processes where they help increase efficiency and reduce costs.

How can the rate of reaction be measured in a laboratory setting?

In a laboratory, the rate of reaction can be measured by monitoring changes in concentration of reactants or products over time. Common methods include measuring the volume of gas produced, changes in mass, or changes in color or turbidity. For example, in the reaction between hydrochloric acid and magnesium, the rate can be measured by the volume of hydrogen gas produced over time.

What is the role of concentration in the rate of reaction?

Concentration plays a significant role in the rate of reaction. A higher concentration of reactants leads to more particles in a given volume, increasing the likelihood of collisions between reactant particles. This increase in collision frequency typically results in a faster reaction rate. Conversely, a lower concentration results in fewer collisions and a slower reaction rate.

Why is "Explaining temperature effect on rate using ONLY the Ea reason" a common mistake in Rates of Reaction ?

Why it happens: Students remember the Ea reason but forget the frequency reason. How to avoid it: BOTH effects are needed for full marks: (1) faster particles → more frequent collisions; (2) more particles with E ≥ Ea → more successful collisions. A 6-mark question expects both. (Note: the Ea effect is the bigger one in reality.) Source: 4CH1 Examiner Reports 2022-2024

Why is "Saying a catalyst 'gives the particles more energy' or 'lowers the energy of the reactants'" a common mistake in Rates of Reaction ?

Why it happens: Misunderstanding what a catalyst does. How to avoid it: A catalyst does NOT change the energy of the reactants or products (ΔH is unchanged). It provides an ALTERNATIVE PATHWAY with a LOWER ACTIVATION ENERGY. Use exactly that phrase. Source: 4CH1 Examiner Reports 2023-2024

Why is "Quoting rate without units" a common mistake in Rates of Reaction ?

Why it happens: Treating rate as a 'plain number'. How to avoid it: Rate ALWAYS has units. Depending on what's measured: mol/dm³/s, cm³/s, g/s, or just 1/s for fixed-change experiments. Quote the unit.

Why is "Confusing average rate (over a time interval) with instantaneous rate (gradient at a point)" a common mistake in Rates of Reaction ?

Why it happens: Both involve dividing change by time, but with different time windows. How to avoid it: AVERAGE rate = total change ÷ total time (a chord of the curve). INSTANTANEOUS rate at time t = gradient of the TANGENT at that point. INITIAL rate = instantaneous rate at t = 0 (the steepest tangent). For a 'rate' question, check carefully whether they want average or initial. Source: 4CH1 Examiner Reports 2022-2024

Why is "Saying the reaction has 'stopped because the rate is zero'" a common mistake in Rates of Reaction ?

Why it happens: Confusing 'rate' with the underlying reaction. How to avoid it: When a curve plateaus, one of the reactants has been COMPLETELY USED UP — there are no more reactant particles to collide → no more reaction → rate = 0. This is the OBSERVED rate that's zero, not a 'paused' reaction. The plateau value tells you the maximum amount of product the reaction can produce.

Why is "Saying 'pressure increases concentration' as the only explanation" a common mistake in Rates of Reaction ?

Why it happens: Simplifying the mechanism. How to avoid it: Pressure effects only apply to GAS reactions. Increasing pressure compresses the gas into a smaller volume → higher number density of particles → more frequent collisions. For aqueous solutions, pressure has essentially no effect — use 'concentration' as the variable instead.

Physical ChemistryPearson Edexcel IGCSE Chemistry 4CH1 Higher

Rates of Reaction

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Rates of Reaction — Pearson Edexcel International GCSE Chemistry 4CH1 Study Notes (2026 onwards, Spec Issue 4)

How fast a reaction goes. Collision theory: for reaction, particles must collide with E ≥ Ea and correct orientation. Five factors affect rate: concentration, pressure, surface area, temperature, catalyst. Experimental methods: gas syringe, mass loss, disappearing-cross, colorimeter. Rate = gradient of concentration-time or volume-time graph.

At a glance

Collision theory: particles must (i) collide, (ii) energy ≥ Ea, (iii) correct orientation → 'successful collision'.
Rate of reaction ∝ number of successful collisions per second. Units: mol/dm³/s, cm³/s, g/s.
Concentration ↑ → more particles per volume → more frequent collisions → faster rate.
Pressure (gas) ↑ → particles closer together → more collisions → faster rate.
Surface area ↑ (powder vs lump) → more particles exposed → more collisions at surface → faster rate.
Temperature ↑ → TWO effects: (1) faster particles, more frequent collisions; (2) MORE particles with E ≥ Ea, more SUCCESSFUL collisions. Rule of thumb: rate doubles per 10 °C.
Catalyst → provides ALTERNATIVE PATHWAY with LOWER Ea → faster rate WITHOUT being used up.
Measuring rate: gas-syringe (V vs t), balance (m vs t), disappearing-cross (1/t), colorimeter (colour).

What you’ll learn

Mapped to the Cambridge IGCSE 4CH1 syllabus (2026 onwards).

3.9 — Describe experiments to investigate the effect of changes in surface area of a solid, concentration of solutions, temperature, and the use of a catalyst on the rate of a reaction.
3.10 — Understand that the rate of reaction is the change in concentration (or amount) of a reactant or product per unit time.
3.11 — Explain the effects of changes in surface area, concentration, temperature and the use of a catalyst on the rate of reaction in terms of particle collision theory.
3.12 — Recall that a catalyst is a substance that speeds up the rate of a reaction without being chemically changed at the end of the reaction; a catalyst lowers the activation energy of a reaction.
3.13 — Recall that enzymes are biological catalysts and that enzymes are used in the production of alcoholic drinks by fermentation.
3.14 — Calculate rates of reaction from graphical data, including the gradient of the tangent at a specified point on a curve.
3.15 — Carry out an investigation to study the effect of variables on the rate of reaction, including measuring gas volume / mass loss / time for a fixed change.

Collision theory — the foundation (spec 3.10, 3.11)

Particles must collide with E ≥ Ea and correct orientation.

Collision theory is the central idea explaining how chemical reactions occur and what controls their rate.

The three requirements for a reaction.

For a chemical reaction to occur between two reactant particles (atoms, ions, molecules), they must:

COLLIDE with each other — they must physically meet. Two particles can't react if they never come close.
Collide with sufficient ENERGY — at least the activation energy (Ea). Each pair must bring enough kinetic energy to the collision to break the necessary bonds and form the transition state.
Collide with the CORRECT ORIENTATION — the molecules must approach each other in a geometry that allows the reactive parts to interact. For example, if H atoms have to swap positions, the H atoms need to be facing each other on impact.

A collision that satisfies ALL THREE conditions is called a SUCCESSFUL COLLISION (or 'effective collision'). Most collisions are unsuccessful — particles just bounce off each other without reacting.

A reaction only happens when particles collide with energy ≥ Ea AND the correct orientation.

Rate of reaction ∝ number of successful collisions per second.

This is the fundamental link between collision theory and the observed rate. If you want a reaction to go faster, you must INCREASE the rate of successful collisions. There are exactly two ways to do this:

(a) Increase the FREQUENCY of collisions (collisions per second). (b) Increase the PROPORTION of collisions that are successful (those with enough energy / correct orientation).

The five factors that affect rate (concentration, pressure, surface area, temperature, catalyst) all work through one or both of these mechanisms.

A snapshot of particles in a gas.

At room temperature, gas molecules move at hundreds of m/s and collide ~10⁹ times per second. But for the reaction H₂ + I₂ → 2HI, only about 1 in 10¹⁰ collisions is successful at room T — the rate is incredibly slow. Heat the gas to 300 °C and many more particles exceed Ea → reaction is fast.

Why activation energy is needed.

Even when the OVERALL reaction is exothermic (releases energy), the molecules have to first BREAK some bonds in the reactants. This bond-breaking is the 'uphill' part — it requires energy IN. Once the bonds are broken (transition state), new bonds form and release more energy than was put in, giving a net exothermic process. But the initial 'energy hump' (Ea) must be cleared.

Cubes of sugar in tea analogy. A sugar cube in cold tea dissolves slowly: the sugar molecules at the surface collide with water molecules, but only a tiny fraction have enough energy to escape into the water. Stir → more collisions per second → faster dissolving. Warm the tea → molecules have more kinetic energy → more break free per second → faster dissolving. Powder the sugar → more surface area → more sugar molecules at the surface where collisions can happen → faster dissolving. Same principles as chemical reactions.

Three requirements: collide + E ≥ Ea + correct orientation = successful collision.
Rate ∝ number of successful collisions per second.
Two ways to increase rate: more frequent collisions OR higher proportion successful.
All five rate factors work through one (or both) of these mechanisms.

See the full worked example for rates of reaction →

Concentration and pressure (spec 3.10, 3.11)

Both increase particle density → more frequent collisions → faster rate.

Concentration of a reactant in solution.

INCREASING concentration → INCREASES rate.

Mechanism: more concentrated solution = more reactant particles per unit volume (cm³ or dm³). With more particles in the same space, particles are CLOSER TOGETHER on average → they encounter each other more frequently → more COLLISIONS PER SECOND → more successful collisions per second → faster rate.

The temperature and Ea are UNCHANGED, so the PROPORTION of collisions that are successful (have energy ≥ Ea) is the same as before. Only the TOTAL NUMBER of collisions per second has increased.

Quantitative observation. For many reactions, doubling the concentration of one reactant DOUBLES the initial rate. For Mg + 2HCl → MgCl₂ + H₂:

1.0 mol/dm³ HCl gives a rate of ~ 1 cm³/s H₂ release.
2.0 mol/dm³ HCl gives a rate of ~ 2 cm³/s H₂ release.

(This is 'first-order' dependence on HCl; some reactions are 'second-order' where doubling concentration QUADRUPLES rate, but that's beyond IGCSE.)

Pressure of a gas.

INCREASING pressure of a gas → INCREASES rate (for gas-phase reactions only).

Mechanism: higher pressure means the gas particles are SQUASHED into a SMALLER VOLUME. The number of particles is the same, but they're packed into less space → higher number density → particles closer together → more frequent collisions per second. Identical mechanism to concentration in solution.

Pressure changes have NO effect on rates of:

Reactions of solids (no gas to compress).
Reactions in aqueous solution (liquids don't compress meaningfully).
Reactions between a solid and a solution (pressure doesn't change concentration in solution).

Industrial example. The Haber process (N₂ + 3H₂ ⇌ 2NH₃) is operated at ~200 atm — partly to shift equilibrium right (Le Chatelier — fewer moles of gas) and partly to INCREASE THE RATE (faster collisions between N₂ and H₂).

Graphical comparison.

Imagine three reactions of Mg + HCl with HCl concentrations 0.5, 1.0, 2.0 mol/dm³:

All three plot 'volume of H₂' vs time as curves.
2.0 mol/dm³ has the STEEPEST initial gradient (fastest initial rate).
1.0 mol/dm³ has intermediate gradient.
0.5 mol/dm³ has the SHALLOWEST gradient.

All three plateau at the SAME TOTAL VOLUME OF H₂ (the maximum determined by the moles of Mg, which is the same in all three runs). The rate at any point is the gradient of the curve; all curves get less steep over time as reactants are consumed.

Predict-and-test exam question pattern. A typical Paper 1 question: 'Predict what would happen if the HCl concentration is halved' → expected answer: rate would HALF; final volume of H₂ would still be the same (Mg in excess) OR final volume would HALF (if HCl is the limiting reactant).

Concentration ↑ → more particles per volume → more collisions → faster rate.
Pressure (gas) ↑ → particles closer together → same effect.
Pressure has NO effect on reactions of solids or solutions.
Steeper initial gradient on a graph = faster rate.
Final volume / mass at plateau depends on TOTAL moles of limiting reactant, not concentration.

See the full worked example for rates of reaction →

Surface area of a solid (spec 3.10, 3.11)

Powder vs lump: more atoms exposed → more frequent collisions → faster rate.

When a SOLID reacts with a GAS or SOLUTION, the reaction happens at the SURFACE of the solid. The interior atoms can't react (they're surrounded by other solid atoms).

Effect of surface area. INCREASING surface area → INCREASES rate.

Mechanism: more solid particles EXPOSED to the other reactant → more sites where collisions can occur per unit time → more frequent collisions at the surface → more successful collisions → faster rate.

Surface area vs particle size. Smaller pieces of solid have a LARGER total surface area for the same total mass.

Consider a 1 cm cube vs eight 0.5 cm cubes (same total volume):

1 cm cube: surface area = 6 × 1² = 6 cm². Volume = 1 cm³.
Eight 0.5 cm cubes: total surface area = 8 × 6 × 0.5² = 8 × 1.5 = 12 cm². Total volume = 8 × 0.125 = 1 cm³.

Cutting the cube into 8 smaller cubes DOUBLED the surface area without changing the mass or volume. Continuing to subdivide gives ever more surface area — a fine powder of the same mass has a HUGE surface area (sometimes thousands of times larger than a single lump).

Breaking a solid into smaller pieces increases the total surface area exposed, so more collisions can happen and the rate rises.

Classic 4CH1 examples.

Marble chips + HCl: large chips reach slowly; small chips fast; powder almost instantaneous.
Magnesium ribbon vs Mg powder + HCl: ribbon fizzes for a minute; powder reacts in seconds.
Iron filings vs iron block + HCl: filings react fast; block slow.

Why dust explosions happen. Coal dust, flour dust, sugar dust, and powdered metals (Mg, Al) can EXPLODE if a flame is introduced. The huge surface area means the entire mass of fuel can burn essentially simultaneously → enormous heat release in a fraction of a second → explosion. Sealed-flour mills, sugar refineries, coal mines, and metal-grinding workshops have all suffered fatal dust explosions historically. Modern safety procedures include explosion vents, dust collectors, and prohibiting open flames.

Required practical investigation.

The 4CH1 syllabus expects you to be able to describe an experiment investigating the effect of surface area on rate, typically using marble chips + dilute HCl:

Same mass (e.g. 5.0 g) of CaCO₃ in 3 different forms: large chips, small chips, powder.
Same volume (50 cm³) and concentration (1.0 mol/dm³) of HCl in a conical flask.
Plug the flask neck with COTTON WOOL (lets CO₂ escape but prevents acid spray).
Place on an electronic balance.
Record mass at fixed intervals (every 10 s) — mass DECREASES as CO₂ escapes.
Plot mass loss vs time on the same axes for the three runs.
Initial gradient (rate) is STEEPEST for powder; intermediate for small chips; shallowest for large chips.
All three curves eventually reach the SAME total mass loss (same total moles of CaCO₃ react with excess HCl → same total CO₂ produced).

Control variables (fair test):

Same mass of CaCO₃.
Same volume of HCl.
Same concentration of HCl.
Same temperature of HCl.
Same atmospheric pressure.
Same shape/size of flask.

ONLY the surface area is varied.

Reaction occurs at the SURFACE of the solid.
Smaller pieces → larger total surface area → more atoms exposed → faster rate.
Same mass: 1 cm cube has SA 6 cm²; 8 × 0.5 cm cubes have SA 12 cm².
Examples: marble chips/powder + HCl; Mg ribbon/powder + HCl.
Dust explosions: huge SA of powder + spark → all fuel burns at once.
Required practical: marble chips + HCl, measure mass loss vs time.

See the full worked example for rates of reaction →

Temperature — the BIG factor (spec 3.10, 3.11)

TWO effects: more frequent collisions AND more particles exceed Ea.

Temperature has the BIGGEST effect on rate of reaction. A 10 °C rise typically DOUBLES the rate. A 30 °C rise gives ~ 8× the rate. This is a much larger effect than the other factors give.

Why temperature matters so much — TWO distinct mechanisms.

Temperature is a measure of the AVERAGE KINETIC ENERGY of particles. At higher T, particles have more KE on average — they move faster.

Mechanism 1: More frequent collisions.

Faster-moving particles cover more distance per second → encounter each other more often → MORE COLLISIONS PER SECOND. The collision frequency increases roughly as $\sqrt{T}$ (where T is in kelvin), so doubling absolute temperature increases frequency by ~ 1.4×. This is a MODEST effect.

Mechanism 2: Higher proportion of successful collisions (the BIGGER effect).

The energy of any one particle isn't fixed — particles have a DISTRIBUTION of energies (some slow, some fast). At higher T, the distribution shifts to higher energies → many more particles have energy ≥ Ea.

So out of all the collisions occurring per second, a MUCH HIGHER FRACTION are now successful. This is the bigger effect — it can easily explain a 2× increase from a 10 °C rise even though the collision frequency itself only goes up by ~ 2%.

At higher temperature the energy distribution shifts right, so a far larger fraction of particles have energy ≥ Ea and collisions succeed.

Why this matters for full marks.

In a 6-mark question on 'why does temperature affect rate?', you need BOTH:

Particles move faster → more frequent collisions.
More particles have E ≥ Ea → higher proportion of successful collisions.

The full answer: 'At higher T, particles have more KE → move faster → collide more frequently. ALSO, a greater proportion of particles have energy ≥ Ea, so a higher fraction of collisions are successful. Combined effect: many more successful collisions per second → faster rate.'

Mentioning only one of the two effects caps at 4 of 6 marks.

Rule of thumb: rate doubles per 10 °C.

This is a useful approximation, though the exact ratio depends on the reaction:

20 °C → 30 °C: rate × 2.
20 °C → 40 °C: rate × 4.
20 °C → 50 °C: rate × 8.

For some reactions the doubling-per-10-°C is over-estimated; for others under-estimated. But it's a good order-of-magnitude guide.

Required practical — disappearing-cross experiment.

The standard 4CH1 way to investigate the effect of temperature on rate:

Reaction: Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl + SO₂ + S(s) + H₂O. The yellow sulfur precipitate makes the solution cloudy.
Apparatus: conical flask + paper with a BLACK CROSS drawn on it.
Method: warm the Na₂S₂O₃ solution to a known temperature (e.g. 20, 30, 40, 50, 60 °C). Place flask on cross. Add fixed volume of HCl; start stopwatch. Look DOWN through the solution at the cross. STOP stopwatch when cross is no longer visible.
Plot 1/time (s⁻¹) on y-axis vs temperature (°C) on x-axis.
Expected result: 1/time increases approximately exponentially with T → rate doubles approximately every 10 °C.

Control variables: same concentration of Na₂S₂O₃, same volume + concentration of HCl, same flask shape, same observer (subjective endpoint).

Why we cool reactions in everyday life.

Refrigerators keep food cold to SLOW DOWN the rate of bacterial decay and enzymatic browning.
Freezers at −18 °C slow rates ~ 2¹⁵ = ~ 30,000× compared to 20 °C (estimated by 10-°C rule extrapolated below 20 °C).
Domestic heating systems use cooler water in summer to slow corrosion.

Why we heat reactions in industry.

Haber process: 450 °C — high enough to make the rate practical (without the temperature, reaction would take days).
Cracking: 500-900 °C — needed to break C–C bonds.
Industrial cooking (Maillard reactions): high T accelerates browning reactions.

Temperature is the biggest factor — rate roughly doubles per 10 °C rise.
TWO mechanisms: (1) faster particles → more collisions; (2) more particles exceed Ea → more successful collisions.
Mechanism 2 is the bigger effect (typically 80%+ of the rate increase).
Full-mark answers MUST mention both effects.
Disappearing-cross required practical: vary T, measure 1/time.
Refrigeration slows reactions (preserves food); industrial heating speeds them up.

See the full worked example for rates of reaction →

Catalysts — speed up without being used up (spec 3.12, 3.13)

Lower Ea via alternative pathway. Not consumed. Examples: enzymes, Fe (Haber), Pt/Rh (cars), V₂O₅ (contact), MnO₂ (H₂O₂).

A catalyst is a substance that SPEEDS UP a chemical reaction WITHOUT itself being chemically changed or used up.

The catalyst is present in the same amount and same chemical form AT THE END of the reaction as at the start. Catalysts are usually needed only in small amounts because they are continuously regenerated.

How catalysts work — the mechanism.

Catalysts provide an ALTERNATIVE REACTION PATHWAY with a LOWER ACTIVATION ENERGY than the uncatalysed pathway.

This is the standard mark-scheme phrase — use it exactly: 'alternative pathway with lower activation energy'.

Why a lower Ea gives a faster rate.

With a lower Ea, a GREATER PROPORTION of colliding particles have enough energy to react (E ≥ Ea). The mean kinetic energy of the particles hasn't changed (T is the same) — but the energy threshold for success has been lowered. So a higher fraction of the same number of collisions is now successful → more successful collisions per second → faster rate.

A catalyst does NOT:

Give the particles more energy. (Temperature does that.)
Make the products more energetic. (ΔH is unchanged.)
Change the position of equilibrium for a reversible reaction. (Both forward and backward rates increase EQUALLY → equilibrium composition unchanged.)
Get used up. (It's regenerated.)

On a reaction profile diagram.

The catalysed pathway has a SMALLER PEAK (lower Ea) but the reactants and products are at the SAME levels (ΔH unchanged). Drawn as two overlaid humps:

Tall hump = uncatalysed Ea.
Short hump = catalysed Ea.

A catalyst provides an alternative pathway with lower Ea; the reactant and product energies (and so ΔH) are unchanged.

Industrial catalysts — important examples.

Reaction	Catalyst	Application
Haber process: N₂ + 3H₂ ⇌ 2NH₃	Iron (Fe)	Ammonia for fertilisers
Contact process: 2SO₂ + O₂ ⇌ 2SO₃	Vanadium(V) oxide (V₂O₅)	Sulfuric acid manufacture
Catalytic cracking	Aluminium oxide (Al₂O₃) / zeolite	Convert long alkanes → shorter
Hydrogenation: alkene + H₂ → alkane	Nickel (Ni)	Make margarine from oils
Hydration: alkene + H₂O → alcohol	Phosphoric acid (H₃PO₄)	Industrial ethanol
Decomposition: 2H₂O₂ → 2H₂O + O₂	Manganese(IV) oxide (MnO₂)	Lab demo
Catalytic converter (cars)	Platinum + rhodium	Reduce CO + NOₓ emissions

Enzymes — biological catalysts.

Enzymes are PROTEINS that catalyse metabolic reactions in living things. They are extremely efficient and very specific — each enzyme catalyses only one (or a few) reactions. Examples:

Amylase (in saliva) breaks down starch → maltose during digestion.
Pepsin (in stomach) breaks down proteins → peptides.
Catalase (in cells) decomposes H₂O₂ → H₂O + O₂ — protects cells from oxidative damage.
Yeast enzymes (zymase) ferment sugars to ethanol — used in brewing and bioethanol production: $\text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{yeast}} 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2$ . Optimum T ≈ 30-35 °C; killed by high T or strong acid.

Why enzymes are 'specific'. Each enzyme has a 3D 'active site' shape that matches only one (or a few) substrate molecules — the 'lock and key' model. Lock + correct key = reaction; lock + wrong key = no reaction.

Investigating catalysis — required practical.

A common experiment: investigate the effect of MnO₂ on the rate of H₂O₂ decomposition.

Add MnO₂ powder to 20 cm³ of dilute H₂O₂(aq) in a flask connected to a gas syringe.
Measure volume of O₂ collected over time.
Compare with a control (no MnO₂).
Expected: with MnO₂, large amount of O₂ collected in 30 seconds; without MnO₂, almost none in the same time.
The MnO₂ powder can be FILTERED OFF at the end — its mass is UNCHANGED → confirms it's not consumed.

Reusability. Industrial catalysts are reused continuously for months or years. They can be 'poisoned' (deactivated by contaminants like sulfur), in which case they must be replaced or regenerated.

Catalyst = speeds up reaction WITHOUT being used up.
Mechanism: ALTERNATIVE PATHWAY with LOWER activation energy.
Lower Ea → more particles have E ≥ Ea → more successful collisions → faster rate.
Does NOT change ΔH or equilibrium position.
Examples: Fe (Haber), V₂O₅ (Contact), Pt/Rh (cars), MnO₂ (H₂O₂), Ni (hydrogenation), enzymes.
Enzymes = biological catalysts; specific via lock-and-key 3D shape; used in fermentation.

See the full worked example for rates of reaction →

Measuring rate experimentally (spec 3.14, 3.15)

Gas-syringe, balance, disappearing cross. Rate = gradient OR 1/time.

There are several practical methods for measuring the rate of a reaction. The choice depends on what's changing during the reaction.

Method 1: Gas-syringe — volume of gas evolved.

Suitable for: any reaction producing a gas (Mg + HCl → H₂; CaCO₃ + HCl → CO₂; H₂O₂ + MnO₂ → O₂).
Apparatus: conical flask sealed with a bung; delivery tube to a gas-syringe (or upturned measuring cylinder over water).
Procedure: add reactants to flask; start stopwatch; record V(gas) every 10-30 s for 3-5 min.
Plot: volume on y-axis vs time on x-axis. The curve rises and plateaus when one reactant is exhausted.
Rate: gradient of the curve at a chosen time (instantaneous rate); or (V_final − V_initial)/(t_final − t_initial) for average rate.
Pros: direct, accurate. Cons: gas-syringe needed (~£20 each); if too much gas, exceeds capacity.

Method 2: Mass-loss on electronic balance.

Suitable for: reactions producing a gas that ESCAPES (e.g. CaCO₃ + HCl → CO₂). NOT for H₂ (too light to register mass change).
Apparatus: conical flask + cotton-wool plug (lets gas escape, prevents acid spray) + 2-3 decimal-place balance + stopwatch.
Procedure: place flask on balance; tare; add reactants; record mass at fixed intervals.
Plot: mass LOSS (g) vs time (s). Mass decreases over time as CO₂ escapes.
Rate: gradient (g/s).
Pros: simple, cheap, continuous. Cons: only works for CO₂-producing reactions; cotton wool needed for safety.

Method 3: Disappearing-cross (turbidity).

Suitable for: reactions producing a precipitate (cloudy product). Classic: Na₂S₂O₃ + HCl → S (yellow ppt).
Apparatus: conical flask placed on top of a piece of paper with a BLACK CROSS drawn on it.
Procedure: mix reactants in flask; start stopwatch; look DOWN through the solution at the cross. STOP stopwatch when cross no longer visible (yellow S has obscured it).
Single endpoint: not a continuous curve — just one time per run.
Rate: rate ∝ 1/time. Plot 1/t vs the variable being studied (concentration, temperature).
Pros: very simple, no special apparatus. Cons: subjective endpoint (depends on observer); only one point per run.

Method 4: Colorimeter — colour intensity.

Suitable for: reactions where a coloured species changes intensity (e.g. iodine clock, KMnO₄ reduction).
Apparatus: colorimeter (with a specific filter for the coloured wavelength) + cuvette.
Procedure: measure light absorbance through the reaction mixture at intervals.
Pros: continuous, accurate, less subjective than the cross. Cons: requires the right equipment.

Reading a rate from a graph.

For a volume-vs-time graph of gas produced:

INITIAL rate (often what's asked) = gradient at t = 0 = the steepest part of the curve.
AVERAGE rate over time interval = (V_final − V_initial) / (t_final − t_initial).
INSTANTANEOUS rate at time t = gradient of the TANGENT to the curve at time t.

To find a gradient: pick two points on the curve (or its tangent), calculate (Δy / Δx). Always include units (cm³/s for volume; g/s for mass; etc.).

Shape of typical rate-vs-time curves.

Initial gradient is steepest: most reactant, fastest rate.
Gradient decreases over time: reactant being used up → less to collide with → fewer successful collisions per second.
Plateau at the end: ONE reactant has been completely used up → no more reaction → rate = 0.
Plateau value depends on TOTAL moles of LIMITING reagent (not concentration).

Comparing different conditions on one graph.

If you plot Mg + HCl at different acid concentrations on the same axes:

All curves start at (0, 0).
HIGHER concentration → STEEPER initial gradient (faster initial rate).
All curves PLATEAU at the same total volume of H₂ (if Mg is the limiting reagent).
Higher-concentration runs reach the plateau SOONER.

Gas-syringe: V(gas) vs t; rate = gradient.
Balance: mass loss vs t; rate = gradient (g/s).
Disappearing cross: time for cross to disappear; rate ∝ 1/t.
Initial rate = steepest gradient at t = 0.
Plateau at end = limiting reagent exhausted.
Higher concentration / smaller particles / higher T → steeper initial gradient.

See the full worked example for rates of reaction →

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Rate of reaction = change in concentration (or amount) per unit time. Units: mol/dm³/s, cm³/s, g/s.
Collision theory: 'successful collision = particles collide with E ≥ Ea AND correct orientation'.
Catalyst phrase: 'provides an ALTERNATIVE PATHWAY with a LOWER ACTIVATION ENERGY'.
Temperature effect (BOTH parts): (1) faster particles → more frequent collisions; (2) more particles with E ≥ Ea → more successful collisions.
Catalysts speed up forward AND backward reactions EQUALLY → equilibrium position unchanged.
Examples of catalysts: Fe (Haber); V₂O₅ (Contact); Pt/Rh (cars); MnO₂ (H₂O₂); Ni (hydrogenation); enzymes.
Disappearing-cross experiment: Na₂S₂O₃ + HCl → S (yellow ppt obscures cross).

How it’s examined

Rates of reaction appears on EVERY 4CH1 paper, typically worth 6-10 marks per session combined. Paper 1 (4CH1/1C) items: (a) describe an experiment investigating one of the factors (5-6 marks); (b) explain the effect of one factor using collision theory (4-6 marks — temperature is the most common, asking for BOTH effects); (c) calculate rate from graph data (3-5 marks). Paper 2 (4CH1/2C) items: deeper collision-theory explanations; catalysts in industrial processes (Haber, Contact); enzymes in fermentation. The 'two reasons temperature affects rate' question is high-frequency — make sure you explain BOTH frequency AND fraction-exceeding-Ea effects for full marks.

Sources: Pearson Edexcel International GCSE Chemistry 4CH1 specification — Issue 4, September 2024 (valid for 2026 onwards); Pearson Edexcel 4CH1 Examiner Reports 2022-2024; 4CH1/1C May/Jun 2024 question paper and mark scheme; 4CH1/1C January 2024 question paper and mark scheme; 4CH1/2C May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-12.

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Step-by-step worked examples — Rates of Reaction

Step-by-step solutions to past-paper-style questions on rates of reaction , written exactly the way a tutor would explain them at the board.

1Explain in terms of collision theory why concentration affects rate (5 marks, Paper 1)
Core• Adapted from 4CH1/1C May/Jun 2024 Q5• collision theory, concentration, spec-3.10, spec-3.11
Question
Magnesium ribbon is added to hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. Explain in terms of collision theory why DOUBLING the concentration of the HCl approximately DOUBLES the initial rate of reaction. Include in your answer the effect on (a) the number of acid particles in a given volume, (b) the frequency of collisions, and (c) the proportion of successful collisions. (5 marks)
Step-by-step solution
1. Step 1
  Collision theory — the foundation (1 M). For a reaction to occur, reactant particles must COLLIDE with each other with (i) sufficient energy ≥ activation energy (Ea) and (ii) correct orientation. Such a collision is called a 'successful collision' or 'effective collision'. The rate of reaction is proportional to the NUMBER OF SUCCESSFUL COLLISIONS PER SECOND.
2. Step 2
  (a) Effect on particle density (1 A). When the concentration of HCl is DOUBLED, the number of H⁺ (and Cl⁻) particles per unit volume of solution DOUBLES. The solid Mg ribbon experiences a doubled density of acid particles around it.
3. Step 3
  (b) Effect on collision frequency (1 A). Twice as many particles in the same volume means there are TWICE AS MANY COLLISIONS PER SECOND between Mg surface atoms and H⁺ ions. The FREQUENCY of collisions doubles.
4. Step 4
  (c) Effect on the proportion that are successful (1 B). The energy of the particles does NOT change (temperature is unchanged) → the PROPORTION of collisions with energy ≥ Ea is UNCHANGED. So out of double the collisions, double the number are successful.
5. Step 5
  Conclusion — rate doubles (1 B). Twice as many successful collisions per second → rate doubles. Observable: a Mg ribbon dropped into 2.0 mol/dm³ HCl fizzes about twice as vigorously as one in 1.0 mol/dm³ HCl. (Strictly, the relationship rate ∝ [HCl] is only first-order; for more complicated reactions, doubling concentration may quadruple the rate — but at IGCSE, this proportional behaviour is expected.)
Answer
Doubling HCl concentration doubles the number of H⁺ particles per unit volume → DOUBLES the frequency of collisions between H⁺ and Mg surface atoms. Temperature unchanged → proportion of successful collisions (energy ≥ Ea) unchanged. So twice as many successful collisions per second → rate approximately doubles.
Examiner tip
Edexcel 5-mark mark scheme: 1 for definition of successful collision (energy ≥ Ea, correct orientation); 1 for doubled concentration = doubled particle density per volume; 1 for doubled frequency of collisions; 1 for unchanged proportion of successful collisions (temperature dependent, not concentration dependent); 1 for conclusion: rate doubles. Common error: saying 'more particles have more energy' — concentration changes density not particle energies.
2Temperature effect on rate: two reasons (6 marks, Paper 1)
Core• Adapted from 4CH1/1C Jan 2024 Q7• collision theory, temperature, spec-3.10, spec-3.11
Question
Explain in terms of particle behaviour why INCREASING the temperature increases the rate of a reaction. Your answer should give TWO distinct reasons. (6 marks)
Step-by-step solution
1. Step 1
  Set up — kinetic energy and collisions (1 M). Temperature is a measure of the average kinetic energy of particles. At a higher temperature, particles have MORE kinetic energy on average. We need two distinct effects flowing from this.
2. Step 2
  Reason 1 — particles move FASTER (1 A). With more KE, particles move faster. Faster-moving particles cover more distance per second → they ENCOUNTER each other more often → the FREQUENCY OF COLLISIONS increases.
3. Step 3
  Reason 1 continued — leads to more rate (1 A). More collisions per second → more chances per second for a successful (energetic + correctly orientated) collision → rate increases.
4. Step 4
  Reason 2 — more particles exceed Ea (1 B). At a higher temperature, the kinetic energy distribution shifts to higher energies — a GREATER PROPORTION of particles now have energy ≥ activation energy (Ea). So more of the collisions that DO occur are SUCCESSFUL (have enough energy to break the necessary bonds).
5. Step 5
  Reason 2 continued — leads to more rate (1 B). Higher proportion of successful collisions → for the same number of collisions, more produce products → rate increases.
6. Step 6
  Why this is more important than reason 1 (1 B). Both reasons contribute, but the EFFECT ON Ea (reason 2) is MUCH larger than the effect on frequency (reason 1). A 10 °C rise in temperature typically DOUBLES the rate — but the KE increase only raises particle speed by ~ 1-2 %. The big effect comes from many more particles now having E ≥ Ea — this is why the 'rule of thumb' is reason 2's contribution.
Answer
REASON 1: Higher T → particles have more KE → move faster → collide MORE FREQUENTLY → more chances for a successful collision → rate increases. REASON 2 (the bigger effect): Higher T → a GREATER PROPORTION of particles have energy ≥ Ea → a higher fraction of collisions are SUCCESSFUL → rate increases. Rule of thumb: rate roughly doubles per 10 °C rise.
Examiner tip
Edexcel 6-mark mark scheme: 1 for temperature → KE link; 1 for more frequent collisions; 1 for linking to rate (reason 1 conclusion); 1 for more particles with E ≥ Ea; 1 for more successful collisions / linking to rate; 1 for any extra detail (10 °C rule, or noting reason 2 is the bigger effect). Both reasons MUST be given for full marks — answering with only frequency or only Ea caps at 4/6.
3Explain how a catalyst works (energy profile diagram) (5 marks, Paper 1)
Core• Adapted from 4CH1/1C May/Jun 2024 Q6• catalyst, Ea, spec-3.13, spec-3.14
Question
Manganese(IV) oxide is a catalyst for the decomposition of hydrogen peroxide: 2H₂O₂ → 2H₂O + O₂. (a) Describe what is meant by a catalyst. (b) Explain how a catalyst speeds up a reaction using the idea of activation energy. (c) Sketch a reaction profile diagram showing the activation energy for the catalysed and uncatalysed reaction. (5 marks)
Step-by-step solution
1. Step 1
  (a) Definition of a catalyst (1 M). A catalyst is a substance that speeds up a chemical reaction without itself being chemically used up in the reaction. It is present in the same amount and chemical form at the end as at the start. Catalysts are usually needed only in SMALL amounts.
2. Step 2
  (b) How catalysts speed up reactions (1 A). A catalyst provides an ALTERNATIVE PATHWAY (route) for the reaction with a LOWER ACTIVATION ENERGY than the uncatalysed pathway. The reactants are not made more energetic — but the energy 'barrier' they need to cross is lowered.
3. Step 3
  (b) Effect on number of successful collisions (1 A). With a lower Ea, a GREATER PROPORTION of colliding particles have enough energy to react. So at the same temperature, MORE successful collisions per second occur → rate increases. The catalyst itself does not appear in the overall balanced equation.
4. Step 4
  (c) Reaction profile (1 B). A reaction profile for an EXOTHERMIC reaction (H₂O₂ decomposition is exothermic, ΔH ≈ −98 kJ/mol): • Flat reactants line (left, higher energy). • Two HUMPS overlaid: a TALL hump for the uncatalysed pathway (labelled 'Ea uncatalysed'), and a SHORTER hump for the catalysed pathway (labelled 'Ea catalysed'). • Flat products line (right, lower energy — exothermic). • ΔH is the SAME for both pathways (the catalyst doesn't change the energy difference between reactants and products).
5. Step 5
  (c) Key takeaway from the diagram (1 B). The catalyst lowers Ea (smaller hump) → easier for particles to reach the top → more successful collisions → faster rate. But ΔH (vertical distance from reactants to products) is UNCHANGED — the catalyst does NOT alter the overall energy change of the reaction, only the speed at which equilibrium is reached.
Answer
(a) A catalyst speeds up a reaction without being chemically used up (recovered unchanged at the end). (b) It provides an ALTERNATIVE PATHWAY with a LOWER activation energy → a GREATER proportion of particles have E ≥ Ea → more successful collisions per second → faster rate. (c) Reaction profile shows two humps over the same reactants→products energy levels: the catalysed pathway has a SMALLER hump (lower Ea). ΔH is the same for both pathways.
Examiner tip
Edexcel 5-mark mark scheme: 1 for catalyst definition (speeds up + not used up); 1 for 'alternative pathway' OR 'lower Ea'; 1 for linking lower Ea → more particles with E ≥ Ea → more successful collisions; 1 for reaction profile correctly drawn (two humps, same ΔH); 1 for noting ΔH unchanged. The phrase 'alternative pathway with lower activation energy' is the mark-scheme phrasing — use exactly.
4Investigating the effect of surface area on rate (6 marks, Paper 1)
Core• Adapted from 4CH1/1C Jan 2024 Q9• required practical, surface area, spec-3.12, spec-3.15
Question
Describe an experiment to investigate the effect of surface area on the rate of reaction between marble chips (CaCO₃) and dilute hydrochloric acid: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Include the apparatus, method, measurements taken, control variables, and how you would calculate and compare the rates. Sketch a typical graph. (6 marks)
Step-by-step solution
1. Step 1
  Apparatus (1 M). Conical flask, electronic balance (± 0.01 g), cotton wool plug (lets gas escape but prevents acid spray), 100 cm³ measuring cylinder, stopwatch, samples of marble chips of different surface area (e.g. large chips, small chips, powder — all SAME total mass).
2. Step 2
  Method (1 A). 1. Weigh out the SAME mass of marble chips for each test (e.g. 5.0 g of large chips, 5.0 g of small chips, 5.0 g of powder). 2. Add 50 cm³ of HCl(aq) of the SAME concentration (e.g. 1.0 mol/dm³) and SAME temperature to the conical flask. 3. Place the flask on the balance with a cotton wool plug in the neck. Record initial mass. 4. Start the stopwatch as soon as the marble chips are added. 5. Record the mass at regular intervals (e.g. every 10 s for 3 min). 6. Repeat with each surface-area sample.
3. Step 3
  Why mass DECREASES (1 A). CO₂ gas is released and ESCAPES through the cotton wool plug. The mass of the flask + contents DROPS over time. Mass loss = mass of CO₂ released.
4. Step 4
  Control variables (1 B). To make a FAIR test, keep the following SAME: (i) mass of CaCO₃ (5.0 g in each); (ii) volume of acid (50 cm³); (iii) concentration of acid (1.0 mol/dm³); (iv) temperature of acid (e.g. room T 20 °C); (v) shape of flask. ONLY the surface area is changed.
5. Step 5
  Calculating rate (1 B). Plot mass loss (g) on the y-axis against time (s) on the x-axis. The GRADIENT of the curve at any point = rate of reaction (g/s). Compare the initial rates (gradient at t = 0) for the three runs. Alternatively, compare TIME TAKEN for a fixed mass loss (e.g. 1.0 g) — rate ∝ 1/time.
6. Step 6
  Expected result (1 B). Powder reacts FASTEST (steepest gradient at t = 0); small chips intermediate; large chips slowest. Explanation: smaller pieces have a GREATER total surface area (per same mass) → more CaCO₃ atoms exposed to acid → more frequent collisions between HCl and the surface → higher rate. All three lines eventually PLATEAU at the same total mass loss (because the same mass of CaCO₃ produces the same mass of CO₂ when fully reacted).
Answer
Apparatus: conical flask + cotton wool + electronic balance + stopwatch. Method: same mass CaCO₃ (5 g) in different surface-area forms (large chips, small chips, powder); add 50 cm³ of 1.0 mol/dm³ HCl(aq); record mass every 10 s for 3 min. Control variables: mass of CaCO₃, volume + concentration + temperature of HCl. Plot mass loss vs time. Rate = gradient. POWDER fastest (steepest curve); LARGE CHIPS slowest. All curves plateau at the same final mass loss (same total reaction).
Examiner tip
Edexcel 6-mark mark scheme: 1 for valid apparatus (balance + cotton wool, OR gas syringe + flask); 1 for keeping mass + acid volume/conc the same; 1 for measuring at regular intervals; 1 for control variables explicit list; 1 for graph + rate as gradient; 1 for expected result + explanation in terms of surface area. The cotton-wool plug is essential — it prevents acid spray escaping (which would lose mass NOT due to CO₂).
5Calculate rate from a volume-time graph (5 marks, Paper 1)
Core• Adapted from 4CH1/1C May/Jun 2024 Q4• rate calculation, graph gradient, spec-3.9, spec-3.15
Question
A student measured the volume of hydrogen gas (cm³) produced over time (s) when 0.10 g of magnesium powder reacted with excess dilute HCl. The volume readings were: t = 0 s → 0 cm³; t = 10 s → 40 cm³; t = 20 s → 65 cm³; t = 30 s → 80 cm³; t = 60 s → 100 cm³ (plateau). (a) Calculate the AVERAGE rate of reaction over the first 30 seconds. (b) Calculate the INITIAL rate (over the first 10 s). (c) Explain why the rate slows down over time. (5 marks)
Step-by-step solution
1. Step 1
  (a) Average rate over 0-30 s (1 M). Change in volume = 80 − 0 = 80 cm³. Time = 30 − 0 = 30 s.
  $\text{rate} = \dfrac{\Delta V}{\Delta t} = \dfrac{80}{30} = 2.67\ \text{cm}^3/\text{s}$
2. Step 2
  (a) Round and quote (1 A). Rate ≈ 2.67 cm³/s (3 sf). Unit MUST be quoted.
3. Step 3
  (b) Initial rate (over 0-10 s) (1 A). Change in volume = 40 − 0 = 40 cm³. Time = 10 s.
  $\text{initial rate} = \dfrac{40}{10} = 4.0\ \text{cm}^3/\text{s}$
4. Step 4
  (c) Why rate slows (1 B). As the reaction proceeds, the concentration of HCl DECREASES (acid is being consumed) AND the amount of Mg DECREASES (it dissolves). With fewer reactant particles per unit volume, there are FEWER COLLISIONS per second between Mg surface atoms and H⁺ ions → fewer successful collisions per second → rate slows down.
5. Step 5
  (c) Why the curve plateaus at 100 cm³ (1 B). Eventually one of the reactants is COMPLETELY USED UP (here the Mg, since the acid is in excess). When no Mg is left, no more H₂ can be produced → rate = 0 → curve becomes horizontal. The MAXIMUM volume (100 cm³) corresponds to the total H₂ from 0.10 g of Mg.
Answer
(a) Average rate over 0-30 s = 80 cm³ / 30 s = 2.67 cm³/s. (b) Initial rate over 0-10 s = 40 cm³ / 10 s = 4.0 cm³/s. (c) The rate slows because [HCl] and Mg surface area decrease as reactants are used up → fewer collisions per second → slower rate. Plateau at 100 cm³ because all Mg has reacted (acid in excess).
Examiner tip
Edexcel 5-mark mark scheme: 1 each for parts (a) and (b) numerical answers with unit; 1 for noting [reactants] decreases; 1 for fewer collisions → slower rate; 1 for plateau = reactant used up. The INITIAL rate is the gradient at t = 0 (steepest part of the curve). Always quote units — cm³/s or g/s.

Key Formulae — Rates of Reaction

The formulae you need to memorise for rates of reaction on the Pearson Edexcel IGCSE 4CH1 paper, with every variable defined in plain English and a note on when to use it.

Rate of reaction (general definition)
$\text{rate} = \dfrac{\Delta\text{quantity}}{\Delta\text{time}}$
$\Delta\text{quantity}$
Change in amount of reactant or product (mol, g, cm³, or another measurable property)
$\Delta\text{time}$
Time taken for the change, in seconds (s)
When to use
Use to calculate the rate from data of amount vs time. Units depend on what's measured: cm³/s for gas volume, g/s for mass loss, mol/dm³/s for concentration. The 'average rate' uses the total change over the whole time; the 'initial rate' uses only the steepest part (around t = 0).
Example
80 cm³ of H₂ released in 30 s: $\text{rate} = 80/30 = 2.67\ \text{cm}^3/\text{s}$ .
Rate from a concentration-time / volume-time graph
$\text{rate} = \text{gradient of the curve}$
$\text{gradient}$
Slope of the tangent to the curve at any point (rise/run = Δy/Δx)
When to use
Read the graph: pick two points on the tangent line and compute (y₂ − y₁) / (x₂ − x₁). The INITIAL RATE is the gradient AT t = 0 (the steepest point, when there is most reactant). The rate DECREASES as reactants are used up — the curve flattens. The rate reaches 0 at the plateau (one reactant exhausted).
Example
Tangent at t = 0 passes from (0, 0) to (5, 30): initial rate = (30 − 0)/(5 − 0) = 6 cm³/s.
Rate from a fixed-change experiment (e.g. disappearing cross)
$\text{rate} \propto \dfrac{1}{t}$
$t$
Time taken for a fixed amount of change (e.g. cross to disappear, fixed mass loss, fixed colour change). Units: s.
When to use
When you measure 'time for X to happen' (a single endpoint) at different conditions (concentration, temperature, etc.), plot 1/t against the variable. If the relationship is linear and passes through the origin, rate is directly proportional to that variable.
Example
Disappearing cross: time 20 s for [HCl] = 0.5 mol/dm³ → 1/t = 0.05 s⁻¹. Time 10 s for [HCl] = 1.0 mol/dm³ → 1/t = 0.10 s⁻¹. Doubling [HCl] doubles 1/t → rate ∝ [HCl].

Model Answers — Rates of Reaction

High-scoring sample answers for rates of reaction on the Pearson Edexcel IGCSE 4CH1 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1

4CH1/1C-style — collision theory overview6 marks

State collision theory. Use it to list and briefly explain the FIVE factors that affect the rate of a chemical reaction. (6 marks)

Model answer

Collision theory. For a chemical reaction to occur, reactant particles must (a) COLLIDE with each other; (b) collide with energy GREATER THAN OR EQUAL TO the activation energy (Ea); and (c) collide with the CORRECT ORIENTATION. A collision that meets all three conditions is called a successful collision or 'effective collision'. The rate of reaction is proportional to the number of successful collisions per second.

Factor 1: Concentration of a reactant (in solution).

Effect: increasing concentration INCREASES the rate.
Explanation: more reactant particles per unit volume → particles are CLOSER together → more frequent collisions → more successful collisions per second → faster rate. (The proportion of collisions that are successful is unchanged — temperature is the same.)
Example: 2.0 mol/dm³ HCl reacts faster with Mg than 1.0 mol/dm³ HCl.

Factor 2: Pressure (for gas reactions).

Effect: increasing pressure INCREASES the rate.
Explanation: pressure ↑ means the gas particles are squashed into a smaller volume → higher concentration of gas particles per unit volume → same effect as concentration → more frequent collisions → faster rate.
Example: in the Haber process, NH₃ forms faster at 200 atm than at 1 atm.

Factor 3: Surface area of a solid reactant.

Effect: increasing surface area (using powder or smaller pieces rather than a lump) INCREASES the rate.
Explanation: a powder has a much GREATER total surface area than the same mass as a single lump → MORE solid particles are exposed to the other reactant → more frequent collisions at the surface → faster rate.
Example: marble chips (large) react slowly with HCl; marble powder reacts vigorously with the same acid.

Factor 4: Temperature.

Effect: increasing temperature INCREASES the rate. Rule of thumb: rate roughly DOUBLES per 10 °C rise.
Explanation: TWO effects working together:
1. Higher T → particles have more kinetic energy → move FASTER → more frequent collisions.
2. Higher T → a GREATER PROPORTION of particles have energy ≥ Ea → a higher fraction of collisions are SUCCESSFUL. (This is the larger effect.)
Combined: many more successful collisions per second → big rate increase.
Example: HCl + sodium thiosulfate reacts twice as fast at 30 °C as at 20 °C (disappearing-cross experiment).

Factor 5: Catalyst.

Effect: adding a catalyst INCREASES the rate without being chemically used up.
Explanation: provides an ALTERNATIVE PATHWAY for the reaction with a LOWER activation energy → a greater proportion of colliding particles have E ≥ Ea → more successful collisions → faster rate. ΔH is unchanged (the catalyst doesn't alter the energy of reactants or products — only the size of the energy barrier between them).
Examples: MnO₂ for H₂O₂ → H₂O + O₂; Fe for the Haber process (N₂ + H₂ → NH₃); Pt/Rh in car catalytic converters; enzymes (biological catalysts) for digestion.

Summary table.

Factor	Direction	Why (in 4 words)
[reactant] ↑	rate ↑	more frequent collisions
Pressure (gas) ↑	rate ↑	particles closer together
Surface area ↑	rate ↑	more atoms exposed
Temperature ↑	rate ↑	more frequent + more successful
Catalyst added	rate ↑	lower activation energy

Why this scores

Edexcel 6-mark mark scheme: 1 for collision theory definition (collide + Ea + orientation); 5 for each of the five factors (1 mark each for naming AND giving a correct mechanism — typically 'more collisions' or 'lower Ea'). Light bonus marks for examples. Most-common error: stating temperature only via the Ea reason (full marks need BOTH frequency and Ea). 'Pressure' is a separate factor from concentration even though the mechanism is identical.

Question 2
4CH1/2C-style — catalysts in detail6 marks
Explain in detail how a catalyst speeds up a chemical reaction. Use an energy profile diagram (described in words) and a labelled discussion of activation energy. Give THREE industrial examples of catalysts and the reaction each catalyses. (6 marks)
Model answer
Definition of a catalyst. A substance that increases the rate of a chemical reaction WITHOUT itself being chemically changed or used up over the course of the reaction. It is present in the same amount and same chemical form at the start AND the end of the reaction. Catalysts are typically needed only in small amounts because they are continuously regenerated.

How a catalyst speeds up a reaction.

The catalyst does NOT make the reactants more energetic. Instead, it provides an ALTERNATIVE PATHWAY (route) for the reaction with a LOWER ACTIVATION ENERGY (Ea) than the uncatalysed pathway.

Step-by-step mechanism:

Reactant molecules bind to the catalyst surface (or reactive sites in an enzyme) — this is called adsorption for solid catalysts.

On the catalyst, bonds are weakened / reorientated, making it easier to break them.

The reaction proceeds through a transition state of LOWER energy than the uncatalysed transition state.

Products are released from the catalyst — the catalyst is freed up to bind another set of reactants.

The catalyst does not appear in the overall stoichiometric equation; it is sometimes written above or below the arrow:

$\text{N}_2 + 3\text{H}_2 \xrightarrow{\text{Fe}} 2\text{NH}_3$

Reaction profile (described in words).

For an EXOTHERMIC reaction with a catalyst:

Flat reactants line (left, high).

TWO humps overlaid: a TALL hump for the uncatalysed pathway (labelled 'Ea uncatalysed'); a SHORTER hump for the catalysed pathway (labelled 'Ea catalysed').

Flat products line (right, low — exothermic).

The two humps meet the SAME starting and ending lines (the catalyst does not change ΔH, only the height of the barrier).

Key insight: with a lower Ea, a much GREATER PROPORTION of the colliding reactant particles have energy ≥ Ea at any given temperature → more successful collisions per second → faster rate.

Why catalysts don't change ΔH or the position of equilibrium.

ΔH is the energy difference between reactants and products. The catalyst doesn't change either — only the BARRIER between them.

For reversible reactions, catalysts speed up BOTH the forward and backward reactions equally → the position of equilibrium is reached FASTER, but at the SAME final composition. (See Topic 3.3 — Le Chatelier.)

Three industrial examples.

Iron (Fe) in the Haber process. Catalyses the reaction $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ at 450 °C, 200 atm. Without iron, the reaction would be too slow to be economic. Used to manufacture ammonia for fertilisers.

Vanadium(V) oxide (V₂O₅) in the Contact process. Catalyses $2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3$ at 450 °C, 1-2 atm. SO₃ is then converted to H₂SO₄ — the most-produced industrial chemical.

Platinum + rhodium in catalytic converters. Convert CO and NOₓ in car exhaust to harmless CO₂ and N₂: $2\text{CO} + 2\text{NO} \rightarrow 2\text{CO}_2 + \text{N}_2$ . Reduces air pollution.

Bonus example — biological. Enzymes are biological catalysts (proteins) that catalyse metabolic reactions. Examples: amylase (breaks down starch), pepsin (breaks down protein in stomach), catalase (breaks down H₂O₂ → H₂O + O₂ in cells).
Why this scores
Edexcel 6-mark mark scheme: 1 for catalyst definition (speeds up + not used up); 1 for alternative pathway / lower Ea; 1 for linking Ea reduction to more particles with E ≥ Ea; 1 for reaction profile description (two humps, same ΔH); 1-2 for industrial examples (any three from Fe Haber, V₂O₅ Contact, Pt/Rh catalytic converter, MnO₂ + H₂O₂, enzymes). Use the exact phrase 'alternative pathway with lower activation energy' for full marks.
Question 3
4CH1/1C-style — experimental rate-measuring methods6 marks
Describe THREE different experimental methods you could use to follow the rate of a chemical reaction. For each method, state what is being measured, give a suitable reaction, and describe the apparatus. (6 marks)
Model answer
Method 1: Gas-syringe (or upturned measuring cylinder) — measuring volume of gas evolved over time.

What is measured: VOLUME of gas (cm³) at regular intervals.

Suitable reaction: any reaction producing a gas, e.g. magnesium + dilute HCl → MgCl₂ + H₂(g); or calcium carbonate + dilute HCl → CaCl₂ + H₂O + CO₂(g); or hydrogen peroxide decomposition catalysed by MnO₂ → H₂O + O₂(g).

Apparatus: conical flask sealed with a bung connected by a delivery tube to a gas syringe (or to an upturned measuring cylinder over water). Add the reactants to the flask and start the stopwatch. Record the gas volume at fixed time intervals (every 10-30 s for 3-5 min).

Plotting: volume (cm³) on y-axis vs time (s) on x-axis. Rate = gradient.

Pros: direct measurement of gas; gas-syringe is accurate (± 1 cm³). Cons: gas-syringe expensive; if too much gas, may exceed syringe capacity.

Method 2: Mass-loss on electronic balance — measuring mass over time.

What is measured: TOTAL MASS of the reaction flask + contents (g), which DECREASES as gas escapes.

Suitable reaction: any reaction producing a gas that can escape, e.g. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ (CO₂ escapes via cotton-wool plug). NOT suitable for H₂ (escapes too easily and is too light to measure mass loss accurately).

Apparatus: conical flask + cotton-wool plug + 2-3 decimal place electronic balance + stopwatch. Place flask on balance, add reactants, record mass at fixed time intervals.

Plotting: mass loss (g) on y-axis vs time (s) on x-axis. Rate = gradient (g/s).

Pros: simple, cheap, continuous monitoring. Cons: only works for CO₂ (heavy enough) or similar; cotton wool prevents spray escaping.

Method 3: Disappearing-cross (turbidity) — measuring TIME for a fixed change.

What is measured: time taken for a precipitate to obscure a marked cross/X drawn on paper under the flask.

Suitable reaction: sodium thiosulfate + hydrochloric acid → sodium chloride + water + sulfur (yellow precipitate) + sulfur dioxide: $\text{Na}_2\text{S}_2\text{O}_3(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{SO}_2 + \text{S}(\text{s})$ .

Apparatus: conical flask placed on a piece of paper with a BLACK CROSS drawn on it. Mix Na₂S₂O₃ + HCl in the flask; start stopwatch. Look DOWN through the solution at the cross. STOP the stopwatch when the cross is no longer visible (yellow sulfur has obscured it).

Plotting / calculating rate: rate ∝ 1/time (1/s). Often plotted as 1/time vs concentration (or temperature) — a straight line confirms a first-order dependence.

Pros: very simple, no expensive apparatus. Cons: subjective endpoint (depends on observer's eyesight); only one data point per run (not a continuous curve).

Other valid methods (any three from above + colorimeter, conductivity meter, pH meter): a colorimeter measures the intensity of a coloured species over time (e.g. iodine clock — bluish-black starch-iodide); conductivity changes can indicate ion concentration changes; pH meter follows neutralisation rate.
Why this scores
Edexcel 6-mark mark scheme: 2 marks per method (1 for apparatus + suitable reaction, 1 for what's measured + how rate is found). Three valid methods needed for full marks. The disappearing-cross is a 4CH1 'required practical' — know it in detail. Common error: confusing 'mass loss' (works for CaCO₃ + acid) with 'gas volume' (works for any gas) — list which reactions are suitable for which method.
Question 4
4CH1/2C-style — concentration vs rate experiment5 marks
Describe how you would investigate the effect of HCl concentration on the rate of reaction between sodium thiosulfate and dilute hydrochloric acid using the disappearing-cross method. State the control variables, expected results, and how the rate is calculated. (5 marks)
Model answer
The reaction. Sodium thiosulfate + hydrochloric acid produces sulfur as a yellow precipitate that progressively turns the colourless solution cloudy:

$\text{Na}_2\text{S}_2\text{O}_3(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{SO}_2(\text{g}) + \text{S}(\text{s})$

Apparatus.

5-6 conical flasks (each containing 50 cm³ of Na₂S₂O₃(aq) at a fixed concentration, e.g. 0.10 mol/dm³).

Measuring cylinders for HCl and water.

Stopwatch.

A piece of white paper with a BLACK CROSS drawn on it (or 'X').

A range of HCl concentrations (e.g. 2.0, 1.6, 1.2, 0.8, 0.4 mol/dm³ — prepared by mixing stock HCl with water).

Method.

Place a flask of 50 cm³ Na₂S₂O₃(aq) on top of the paper with the cross.

Add 10 cm³ of HCl (e.g. 2.0 mol/dm³) and START the stopwatch immediately.

Look DOWN through the solution at the cross from above. STOP the stopwatch when the cross is no longer visible (the yellow S precipitate has obscured it).

Record the time (s).

Repeat steps 1-4 with HCl concentrations of 1.6, 1.2, 0.8, 0.4 mol/dm³.

Repeat each run 2-3 times and take a MEAN time.

Control variables (fair test).

VOLUME of Na₂S₂O₃ (50 cm³) — same each time.

CONCENTRATION of Na₂S₂O₃ (0.10 mol/dm³) — same each time.

VOLUME of HCl + water mixture (10 cm³ total) — same each time.

TEMPERATURE (room T ≈ 20 °C) — keep same.

VOLUME of solution in the flask (60 cm³ total) — same depth of liquid above the cross.

The SAME observer (because endpoint is subjective).

Calculating rate.

The 'fixed change' is the SAME amount of sulfur needed to obscure the cross. Time inversely measures rate:

$\text{rate} \propto \dfrac{1}{\text{time}}$

So plot 1/time (1/s) on the y-axis vs [HCl] (mol/dm³) on the x-axis.

Expected result.

A STRAIGHT LINE through the origin → 1/time is proportional to [HCl] → rate is proportional to [HCl]. As [HCl] doubles, the rate doubles (and the time halves).

Explanation. With double the H⁺ concentration, there are TWICE as many H⁺ particles per unit volume → twice the collision frequency with Na₂S₂O₃ → twice the rate. (Collision theory.)
Why this scores
Edexcel 5-mark mark scheme: 1 for the method (paper with cross + mix + time disappearance); 1 for valid concentration range (3+ concentrations); 1 for control variables list; 1 for plot of 1/time vs [HCl]; 1 for expected straight line + explanation in terms of collisions. Mark scheme penalises forgetting 'same volume of HCl + water' (you must vary CONCENTRATION not total volume). Diluting stock HCl with water keeps total volume constant.

Key Definitions and Keywords — Rates of Reaction

Definitions to memorise and the exact keywords mark schemes credit for rates of reaction answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4CH1 sitting.

Rate of reaction
Examiner keyword
The change in concentration (or amount) of a reactant or product per unit time. Units: mol/dm³/s, g/s, cm³/s. A measure of HOW FAST the reaction goes.
Collision theory
Examiner keyword
The theory that for a reaction to occur, reactant particles must (i) collide, (ii) collide with energy ≥ activation energy Ea, and (iii) collide with the correct orientation. The rate of reaction is proportional to the number of SUCCESSFUL collisions per second.
Activation energy (Ea)
Examiner keyword
The minimum energy that colliding particles must have for a reaction to occur — the energy needed to break the bonds in the reactants and form the transition state.
Successful (effective) collision
Examiner keyword
A collision between reactant particles that has BOTH sufficient energy (≥ Ea) AND correct orientation, so that a reaction actually occurs. Only a small proportion of all collisions are successful at any given temperature.
Catalyst
Examiner keyword
A substance that increases the rate of a reaction WITHOUT being chemically used up. It provides an alternative pathway with a LOWER activation energy. Recovered unchanged at the end. Examples: MnO₂ (H₂O₂ decomposition), Fe (Haber), V₂O₅ (Contact), Pt/Rh (catalytic converters), enzymes.
Enzyme
A biological catalyst — a protein that catalyses a specific metabolic reaction in living things. Highly efficient and very specific. Examples: amylase (digests starch), pepsin (digests protein), catalase (decomposes H₂O₂).
Effect of concentration on rate
Examiner keyword
Increasing the concentration of a reactant in solution increases the rate. Reason: more reactant particles per unit volume → more frequent collisions → more successful collisions per second.
Effect of temperature on rate
Examiner keyword
Increasing temperature increases the rate via TWO effects: (1) particles have more KE → move faster → more frequent collisions; (2) a greater proportion of particles have E ≥ Ea → more successful collisions. Rule of thumb: rate roughly doubles per 10 °C rise.
Effect of surface area on rate
Examiner keyword
Increasing surface area of a solid reactant (e.g. using powder rather than a lump) increases the rate. Reason: more solid particles exposed → more frequent collisions at the surface → faster rate. The same mass of CaCO₃ as powder reacts much faster with acid than as a single lump.
Effect of pressure on rate (gas reactions)
Examiner keyword
Increasing pressure of gases increases the rate. Reason: same effect as concentration — gas particles are forced into a smaller volume → higher particle density → more frequent collisions.
Disappearing-cross experiment
Examiner keyword
A classic 4CH1 required practical for measuring rate: sodium thiosulfate + dilute HCl → yellow S precipitate that obscures a black cross drawn on paper beneath the flask. The time for the cross to disappear is measured; rate ∝ 1/t.

Common Mistakes and Misconceptions — Rates of Reaction

The traps other students keep falling into on rates of reaction questions — taken from recent Pearson Edexcel IGCSE 4CH1 examiner reports and mark schemes — and how to avoid them.

✕Explaining temperature effect on rate using ONLY the Ea reason
4CH1 Examiner Reports 2022-2024
Why it happens
Students remember the Ea reason but forget the frequency reason.
How to avoid it
BOTH effects are needed for full marks: (1) faster particles → more frequent collisions; (2) more particles with E ≥ Ea → more successful collisions. A 6-mark question expects both. (Note: the Ea effect is the bigger one in reality.)
✕Saying a catalyst 'gives the particles more energy' or 'lowers the energy of the reactants'
4CH1 Examiner Reports 2023-2024
Why it happens
Misunderstanding what a catalyst does.
How to avoid it
A catalyst does NOT change the energy of the reactants or products (ΔH is unchanged). It provides an ALTERNATIVE PATHWAY with a LOWER ACTIVATION ENERGY. Use exactly that phrase.
✕Quoting rate without units
Why it happens
Treating rate as a 'plain number'.
How to avoid it
Rate ALWAYS has units. Depending on what's measured: mol/dm³/s, cm³/s, g/s, or just 1/s for fixed-change experiments. Quote the unit.
✕Confusing average rate (over a time interval) with instantaneous rate (gradient at a point)
4CH1 Examiner Reports 2022-2024
Why it happens
Both involve dividing change by time, but with different time windows.
How to avoid it
AVERAGE rate = total change ÷ total time (a chord of the curve). INSTANTANEOUS rate at time t = gradient of the TANGENT at that point. INITIAL rate = instantaneous rate at t = 0 (the steepest tangent). For a 'rate' question, check carefully whether they want average or initial.
✕Saying the reaction has 'stopped because the rate is zero'
Why it happens
Confusing 'rate' with the underlying reaction.
How to avoid it
When a curve plateaus, one of the reactants has been COMPLETELY USED UP — there are no more reactant particles to collide → no more reaction → rate = 0. This is the OBSERVED rate that's zero, not a 'paused' reaction. The plateau value tells you the maximum amount of product the reaction can produce.
✕Saying 'pressure increases concentration' as the only explanation
Why it happens
Simplifying the mechanism.
How to avoid it
Pressure effects only apply to GAS reactions. Increasing pressure compresses the gas into a smaller volume → higher number density of particles → more frequent collisions. For aqueous solutions, pressure has essentially no effect — use 'concentration' as the variable instead.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Rates of Reaction — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Rates of Reaction

Short Study Notes in the form of Slides

Short Study Notes — Rates of Reaction

Detailed Notes

What you’ll learn

How it’s examined

1Explain in terms of collision theory why concentration affects rate (5 marks, Paper 1)

2Temperature effect on rate: two reasons (6 marks, Paper 1)

3Explain how a catalyst works (energy profile diagram) (5 marks, Paper 1)

4Investigating the effect of surface area on rate (6 marks, Paper 1)

5Calculate rate from a volume-time graph (5 marks, Paper 1)

Rate of reaction (general definition)

Rate from a concentration-time / volume-time graph

Rate from a fixed-change experiment (e.g. disappearing cross)

Rate of reaction

Collision theory

Activation energy (Ea)

Successful (effective) collision

Catalyst

Enzyme

Effect of concentration on rate

Effect of temperature on rate

Effect of surface area on rate

Effect of pressure on rate (gas reactions)

Disappearing-cross experiment

✕Explaining temperature effect on rate using ONLY the Ea reason

✕Saying a catalyst 'gives the particles more energy' or 'lowers the energy of the reactants'

✕Quoting rate without units

✕Confusing average rate (over a time interval) with instantaneous rate (gradient at a point)

✕Saying the reaction has 'stopped because the rate is zero'

✕Saying 'pressure increases concentration' as the only explanation

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Rates of Reaction — frequently asked questions