OxfordAQA A-Level Chemistry: Kinetics (9620)

This guide teaches section 3.1.6 Kinetics of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. It covers every point from 3.1.6.1 to 3.1.6.5: collision theory, the Maxwell–Boltzmann distribution, the effect of temperature, the effect of concentration and pressure, and catalysts. The whole section is International AS content, so nothing here is International A-level only. The specification assesses 3.1.6 in Unit 2 (Organic 1 and Physical 1), and the International A2 Unit 5 (practical and synoptic) covers all content.

For quick recall, use the kinetics revision notes. To test yourself, try the kinetics practice questions. The course hub is OxfordAQA A-Level Chemistry, and the printable checklist lists every specification point. If you are not sure where your gaps are, start with a free diagnostic.

What this unit covers

Spec point	What you must be able to do	Stage
3.1.6.1 Collision theory	Define activation energy; explain why most collisions do not lead to a reaction	International AS
3.1.6.2 Maxwell–Boltzmann distribution	Draw and interpret distribution curves for different temperatures	International AS
3.1.6.3 Effect of temperature	Know what rate of reaction means; use the distribution to explain why a small temperature rise can give a large rise in rate	International AS
3.1.6.4 Concentration and pressure	Explain how a change in concentration, or in the pressure of a gas, changes collision frequency and so the rate	International AS
3.1.6.5 Catalysts	Define a catalyst; use a Maxwell–Boltzmann distribution to explain how a catalyst speeds up a reaction involving a gas	International AS

Section 3.1.6 lists no required practical. Required practical 8 (measuring rates) sits with rate equations in 3.1.11 (International A-level only), along with orders, the rate constant and tangents on concentration–time graphs. None of that is taught here.

3.1.6.1 Collision theory

The specification states that reactions can only occur when collisions take place between particles having sufficient energy, and calls this energy the activation energy.

Definition. The activation energy, Ea, is the minimum energy that particles must have when they collide for a reaction to occur.

On an enthalpy profile (see energetics), Ea is the height of the barrier from the reactants to the top of the curve.

Why most collisions do not lead to a reaction. In any sample, most particles have less energy than Ea. When two of them collide, the collision does not supply enough energy to break the bonds that need breaking, so the particles bounce apart unchanged. Only the small fraction of collisions with energy at or above Ea are successful.

Two quantities therefore control the rate:

1. the collision frequency (collisions per second);
2. the fraction of those collisions that carry energy of at least Ea.

Temperature, concentration, pressure and catalysts each act on one or both of these.

Worked example 1. A mixture of methane and air can sit in a closed room at 25 °C for days without reacting, yet the same mixture burns fast once a flame is applied. Explain this.

Step 1: at 25 °C the molecules are colliding all the time, so collisions are happening. Step 2: almost none of these collisions have energy at or above the activation energy for combustion, so they are unsuccessful. Step 3: the flame gives some molecules energy above Ea. The exothermic reaction then heats nearby molecules above Ea too, so the reaction spreads.

3.1.6.2 The Maxwell–Boltzmann distribution

At a fixed temperature, gas molecules share out energy unevenly: a few are slow, most are moderate, a few are very fast. The Maxwell–Boltzmann distribution plots this spread.

Drawing it

• Label the axes energy (horizontal) and number of molecules (vertical).
• Begin the line at (0, 0), since zero-energy molecules do not exist.
• Climb to a single maximum, the most probable energy, then descend more slowly.
• The shape is lopsided: the stretched-out high-energy side drags the mean energy above the most probable value.
• Let the tail flatten towards the energy axis without ever reaching it, as no upper energy limit exists.
• Everything under the line stands for the whole sample of molecules.
• Ea is a vertical line far out in the tail. The area to the right of it represents the molecules with E ≥ Ea.

Curves for different temperatures. For the same sample at a higher temperature:

• the peak moves to the right (higher most probable energy);
• the peak is lower;
• the curve is broader;
• the total area is unchanged, because the number of molecules is unchanged.

The cooler curve is higher at low energies and the hotter curve is higher at high energies, with a single crossing between them. Ea lies in the region where the hotter curve is on top.

Worked example 2 (interpreting curves from a grid). A teacher draws two distribution curves for the same gas sample on squared paper. The total area under each curve is 1200 small squares. To the right of Ea, the curve at T₁ encloses 18 squares and the curve at T₂ encloses 45 squares.

Step 1: proportion with E ≥ Ea at T₁ = 18 ÷ 1200 × 100 = 1.5 %. Step 2: proportion at T₂ = 45 ÷ 1200 × 100 = 3.75 %. Step 3: T₂ has the larger area beyond Ea, so T₂ is the higher temperature. Step 4: the proportion with E ≥ Ea has risen by 45 ÷ 18 = 2.5 times, so roughly 2.5 times as many collisions can succeed, even before any change in collision frequency.

The totals are equal because the sample is the same.

3.1.6.3 Effect of temperature on reaction rate

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

mean rate = change in concentration ÷ time taken

Worked example 3. In the hydrolysis of 1-bromobutane by hydroxide ions,

CH₃CH₂CH₂CH₂Br + OH⁻ → CH₃CH₂CH₂CH₂OH + Br⁻

the hydroxide ion concentration falls from 0.250 mol dm⁻³ to 0.190 mol dm⁻³ in 300 s. Find the mean rate.

change in [OH⁻] = 0.250 − 0.190 = 0.060 mol dm⁻³
mean rate       = 0.060 ÷ 300   = 2.0 × 10⁻⁴ mol dm⁻³ s⁻¹

The equation is 1 : 1, so bromide ions form at the same rate. A rate is always quoted as positive.

Why a small temperature rise gives a large rise in rate

Heating raises the average kinetic energy of the molecules, and this has two consequences.

1. Faster molecules meet slightly more often.
2. The whole distribution slides towards higher energy, so a much larger proportion of molecules have E ≥ Ea.

The second effect dominates. Ea sits far out in the tail of the curve. When the curve shifts right, the small area beyond Ea grows by a large factor, so the number of successful collisions per second rises sharply.

Worked example 4 (why the energy effect dominates). For a reaction with Ea = 50 kJ mol⁻¹, calculations of the kind you meet later in 3.1.11 give these figures for a rise from 295 K to 305 K:

Quantity	Change from 295 K to 305 K
Mean molecular speed (so collision frequency)	rises by about 1.7 %
Proportion of collisions with E ≥ Ea	about 1.95 times larger

Step 1: the frequency effect alone would raise the rate by a factor of only about 1.017. Step 2: the energy effect raises it by about 1.95. Step 3: together the rate rises by about 1.017 × 1.95 ≈ 1.98, so it nearly doubles for a 10 K rise, and almost all of that comes from the larger proportion with E ≥ Ea.

You do not need this calculation at International AS, only the explanation it supports.

3.1.6.4 Effect of concentration and pressure

Concentration (solutions). Making a solution more concentrated means each cm³ holds more reactant particles. With less space between them, they meet more often, so the number of collisions reaching Ea each second goes up and so does the rate.

Pressure (gases). Compressing a gas at fixed temperature leaves the same molecules in a smaller volume, so their number per cm³ rises. They meet more often, and the rate goes up.

What does not change. Neither change alters the energy of the molecules. The Maxwell–Boltzmann curve keeps its shape and position, and the proportion of collisions with E ≥ Ea is the same. Only the number of collisions per second goes up.

Worked example 5 (linking pressure to concentration). A gas at 298 K is compressed from 100 kPa to 200 kPa. Use pV = nRT (from amount of substance) to show how the number of molecules per unit volume changes. R = 8.31 J K⁻¹ mol⁻¹.

n/V = p/(RT)
at 100 kPa: 100 000 ÷ (8.31 × 298) = 40.4 mol m⁻³ = 0.0404 mol dm⁻³
at 200 kPa: 200 000 ÷ (8.31 × 298) = 80.8 mol m⁻³ = 0.0808 mol dm⁻³

Doubling the pressure doubles the amount of gas per dm³, so molecules collide more often. Their energies are unchanged, because the temperature is the same.

3.1.6.5 Catalysts

Definition (as the specification gives it). 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 extra energy. On an enthalpy profile, the catalysed route is a lower hump between the same reactant and product levels, so the enthalpy change is unchanged.

A catalyst does not change the position of an equilibrium; that point belongs to 3.1.7 and is covered in equilibria and Le Chatelier’s principle.

Using a Maxwell–Boltzmann distribution for a gas reaction

At a fixed temperature the catalyst does not move the curve. Only the activation energy that matters changes.

Sketch a single curve, because the temperature is fixed. Put two vertical lines in its tail: Ea for the uncatalysed reaction, and a lower Ea(cat) nearer the peak. Shade the strip between the two lines. Those molecules could not react before but can react by the catalysed route, so the successful-collision rate rises.

Worked example 6. On a distribution curve with a total area of 900 squares, 2 squares lie beyond Ea for the uncatalysed reaction and 27 squares lie beyond Ea(cat). By what factor does the catalyst increase the proportion of molecules that can react?

Step 1: without the catalyst, 2 ÷ 900 × 100 = 0.22 % of molecules have E ≥ Ea. Step 2: with it, 27 ÷ 900 × 100 = 3.0 % have E ≥ Ea(cat). Step 3: factor = 27 ÷ 2 = 13.5. The curve and the temperature are unchanged; only the energy barrier has moved.

Common errors

• An activation energy definition that drops the word minimum.
• A hotter Maxwell–Boltzmann curve whose peak is higher, or which encloses more area than the cooler one.
• A temperature answer built only on faster, more frequent collisions; the energy fraction is the key idea.
• Crediting concentration or pressure with giving molecules extra energy. They change how often molecules meet, nothing else.
• A catalyst answer with no alternative route.
• A second, shifted curve drawn for a catalyst. The temperature is the same, so the curve stays put.

Next steps

Condense this guide with the revision notes, then try the practice questions.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.1.6 Kinetics (International AS), points 3.1.6.1 to 3.1.6.5.