AQA A-Level Chemistry: Kinetics (7405)

This guide teaches section 3.1.5 Kinetics of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams from June 2016 onwards. It covers every point from 3.1.5.1 to 3.1.5.5: collision theory, the Maxwell–Boltzmann distribution, the effect of temperature, concentration and pressure, and catalysts, plus Required practical 3. All of it is AS content, so nothing here is A-level only. The specification places 3.1.5 on Paper 2 at AS and at A-level (sections 3.1.2 to 3.1.6), and A-level Paper 3 can test any content and any practical skill.

For quick recall, use the revision notes. To test yourself, use the practice questions. The course hub is AQA A-Level Chemistry, and the printable checklist lists every specification point. Rate equations, orders and the rate constant are a separate topic (3.1.9, A-level only) and are not taught here.

What this unit covers

Spec point	What you must be able to do	Status
3.1.5.1 Collision theory	Define activation energy; explain why most collisions do not lead to a reaction	AS and A-level
3.1.5.2 Maxwell–Boltzmann distribution	Draw and interpret distribution curves for different temperatures	AS and A-level
3.1.5.3 Effect of temperature	State the meaning of rate of reaction; use the distribution to explain why a small temperature rise gives a large rise in rate	AS and A-level
Required practical 3	Investigate how the rate of a reaction changes with temperature	AS and A-level
3.1.5.4 Concentration and pressure	Explain how a change in concentration or gas pressure changes collision frequency and rate	AS and A-level
3.1.5.5 Catalysts	Define a catalyst; use a Maxwell–Boltzmann distribution to explain how a catalyst speeds up a gas reaction	AS and A-level

3.1.5.1 Collision theory

Reactions can only occur when particles collide with sufficient energy. Definition: the activation energy is the minimum energy that colliding particles must have for a reaction to occur.

Why most collisions do not lead to a reaction. Most collisions fail because most particles have less energy than the activation energy. When two low-energy particles collide, no bonds break and they simply move apart unchanged.

So the rate of a reaction depends on how often collisions happen and what fraction of those collisions have energy at or above Ea. Every factor in this topic works through one or both of those.

3.1.5.2 The Maxwell–Boltzmann distribution

The molecules in a gas do not all have the same energy. The Maxwell–Boltzmann distribution shows how many molecules have each energy.

How to draw it

• x-axis: energy of the molecules. y-axis: number (or fraction) of molecules with that energy.
• The curve starts at the origin: no molecule has zero energy.
• It rises to a single peak. The peak is the most probable energy.
• It falls away to the right and never touches the energy axis: there is no maximum energy.
• The curve is not symmetrical. It has a long tail at high energy, so the mean energy lies to the right of the peak.
• The area under the curve equals the total number of molecules.
• Mark Ea as a vertical line well to the right of the peak. The area under the curve to the right of Ea is the number of molecules that have at least the activation energy.

Interpreting curves at different temperatures

When you raise the temperature of the same sample:

• the peak moves to the right (higher most probable energy);
• the peak gets lower;
• the curve gets wider and the high-energy tail rises;
• the total area stays the same, because the number of molecules has not changed.

The two curves cross once. To the right of Ea, the higher-temperature curve lies above the lower-temperature one, so more molecules have energy at or above Ea.

Worked example 1 (interpreting curves). Two curves, A and B, are drawn for the same sample of gas. Curve B has its peak further right and lower than A. Which curve is at the higher temperature, and what can you say about the areas?

Step 1: a higher temperature shifts the peak to higher energy and lowers it, so B is at the higher temperature. Step 2: the sample is the same, so the number of molecules is the same. The total areas under A and B are equal. Step 3: B has the larger area to the right of Ea, so a greater proportion of collisions in B succeed.

A common drawing error is to make the hotter curve taller as well as shifted. That would mean more molecules, which is wrong.

3.1.5.3 Effect of temperature on reaction rate

Rate of reaction is the change in the amount or concentration of a reactant or product per unit time. With concentrations, the unit is mol dm⁻³ s⁻¹.

mean rate = change in concentration ÷ time taken

Worked example 2. The concentration of hydrochloric acid in a reaction falls from 0.50 mol dm⁻³ to 0.30 mol dm⁻³ in 40 s. Calculate the mean rate.

change in concentration = 0.50 − 0.30 = 0.20 mol dm⁻³
mean rate = 0.20 ÷ 40 = 0.0050 mol dm⁻³ s⁻¹  (5.0 × 10⁻³ mol dm⁻³ s⁻¹)

Why a small temperature rise gives a large rise in rate

When the temperature rises, the molecules gain kinetic energy. Two things follow:

1. They move faster, so they collide a little more often.
2. Much more important, the Maxwell–Boltzmann curve shifts right, so a much larger proportion of molecules have energy at or above Ea.

Ea sits in the tail of the curve. A small shift of the whole curve to the right raises the tail a lot, so the number of molecules with E ≥ Ea goes up by a large factor. That is why a rise of only a few degrees can increase the rate considerably. Make the second point the main one: the rise in collision frequency alone is far too small to explain the change.

Required practical 3: how rate changes with temperature

The specification lists Required practical 3 with apparatus and techniques AT a (measuring mass, time, volume and temperature), AT b (water bath or electric heater) and AT k (safe handling of chemicals). It suggests sodium thiosulfate with hydrochloric acid as one possible reaction, using an initial rate method:

Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + SO₂(g) + S(s) + H₂O(l)

The sulfur forms a cloudy precipitate. A typical method:

1. Measure fixed volumes of thiosulfate solution and acid.
2. Warm both separately in a water bath to the chosen temperature, and record it.
3. Mix them in a conical flask over a cross drawn on paper and start a stopwatch.
4. Stop the watch when the cross can no longer be seen from above. Record the time, t.
5. Repeat at several temperatures, keeping the volumes, concentrations and the cross the same.

The same amount of sulfur forms each time before the cross disappears, so rate is proportional to 1/t. Sulfur dioxide is toxic and irritates the lungs, so work in a well-ventilated room and keep the volumes small.

Worked example 3. A student obtains these times.

Temperature / °C	t / s	1/t / s⁻¹
20	120	0.00833
30	62	0.0161
40	33	0.0303
50	17	0.0588

Step 1: work out 1/t for each run, as shown. Plotting 1/t against temperature gives a curve that gets steeper. Step 2: compare 20 °C and 30 °C: 120 ÷ 62 = 1.94. A 10 °C rise nearly doubles the rate here. Step 3: compare 20 °C and 50 °C: 120 ÷ 17 = 7.06. The rate rises about sevenfold over 30 °C.

The explanation is the Maxwell–Boltzmann one above: many more molecules have E ≥ Ea at the higher temperature.

3.1.5.4 Effect of concentration and pressure

Concentration. Raising the concentration of a solution puts more reactant particles in the same volume. Particles collide more often, so there are more collisions with E ≥ Ea per second, and the rate rises.

Pressure of a gas. Raising the pressure of a gas at constant temperature puts more molecules in each unit of volume, so collisions are more frequent and the rate rises.

Neither change alters the proportion of molecules with energy at or above Ea. The Maxwell–Boltzmann curve keeps its shape and position; only the number of collisions per second changes. Do not say that higher concentration gives particles “more energy”.

The specification suggests calcium carbonate with hydrochloric acid, followed by continuous monitoring, as a way to see the effect of acid concentration:

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

You can collect the CO₂ in a gas syringe, or record the mass lost on a balance as the gas escapes.

Worked example 4 (gas syringe). The volume of CO₂ collected is 48 cm³ after 30 s and 72 cm³ after 60 s. A tangent to the curve at t = 0 passes through (0, 0) and (20 s, 64 cm³).

mean rate, 0–30 s  = 48 ÷ 30 = 1.6 cm³ s⁻¹
mean rate, 30–60 s = (72 − 48) ÷ 30 = 0.80 cm³ s⁻¹
initial rate = gradient of tangent = 64 ÷ 20 = 3.2 cm³ s⁻¹

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

3.1.5.5 Catalysts

Definition from the specification: a catalyst is a substance that increases the rate of a chemical reaction without being changed in chemical composition or amount.

Catalysts work by providing an alternative reaction route of lower activation energy. They do not give molecules more energy.

Using a Maxwell–Boltzmann distribution

For a gas reaction at fixed temperature, a catalyst does not change the curve at all. What changes is where Ea sits:

1. Draw one Maxwell–Boltzmann curve.
2. Mark Ea for the uncatalysed route.
3. Mark Ea(cat) for the catalysed route to the left of it.
4. The area to the right of Ea(cat) is larger than the area to the right of Ea.

So a greater proportion of molecules have enough energy to react by the new route, more collisions are successful each second, and the rate increases.

Worked example 5. A student says: “A catalyst raises the rate because it moves the Maxwell–Boltzmann curve to higher energy.” Correct the statement.

Step 1: the curve depends only on temperature, and the temperature has not changed, so the curve stays the same. Step 2: the catalyst gives an alternative route with a lower activation energy, so the Ea line moves left. Step 3: more molecules now have energy at or above the new Ea, so more collisions succeed per second.

Common errors

• Defining activation energy as “the energy needed to start a reaction”. Say minimum energy for a reaction to occur when particles collide.
• Drawing the higher-temperature curve with a taller peak. The peak must be lower, so the area stays equal.
• Explaining the temperature effect only by “more frequent collisions”. The main reason is the larger proportion of molecules with E ≥ Ea.
• Saying that higher concentration or pressure increases the proportion of successful collisions. It increases the frequency of collisions; the proportion with E ≥ Ea is unchanged.
• Saying a catalyst “lowers the activation energy” with no mention of an alternative route.
• Shifting the Maxwell–Boltzmann curve for a catalyst. Only the Ea marker moves.

Next steps

Condense this into the revision notes, then try the practice questions. The AQA A-Level Chemistry exam preparation guide covers paper structure, and the free 10-minute diagnostics show where to focus.

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.1.5 Kinetics (3.1.5.1 to 3.1.5.5), with Required practical 3.