What you'll learn
Reaction kinetics is the study of how fast chemical reactions happen and why. Rates matter everywhere: in how quickly food spoils in the Caribbean heat, how fast a catalytic converter cleans exhaust gases, and how efficiently industrial plants produce ammonia and methanol. This guide covers the meaning and measurement of reaction rate, collision theory and activation energy, the Maxwell–Boltzmann distribution, the effects of concentration, pressure, temperature, surface area and catalysts, rate equations, orders of reaction, the rate constant, half-life, and how kinetic data reveal reaction mechanisms through the rate-determining step.
CAPE kinetics questions involve interpreting graphs, deducing orders from experimental data, writing rate equations and using them to suggest mechanisms. Practise reading graphs carefully and explaining each factor using collision theory.
Key terms and definitions
Rate of reaction — The change in concentration of a reactant or product per unit time, in mol dm⁻³ s⁻¹.
Collision theory — Particles must collide with sufficient energy (at least the activation energy) and the correct orientation to react.
Activation energy, Eₐ — The minimum energy that colliding particles must have for a reaction to occur.
Catalyst — A substance that increases the rate of reaction by providing an alternative pathway with lower activation energy, without being used up.
Homogeneous catalyst — A catalyst in the same phase as the reactants.
Heterogeneous catalyst — A catalyst in a different phase from the reactants.
Rate equation — rate = k[A]ᵐ[B]ⁿ, showing how rate depends on concentrations.
Order of reaction — The power to which a concentration is raised in the rate equation.
Overall order — The sum of the individual orders.
Rate constant, k — The constant of proportionality in the rate equation; it changes with temperature.
Half-life, t½ — The time taken for the concentration of a reactant to fall to half its initial value.
Rate-determining step — The slowest step in a multi-step mechanism, which controls the overall rate.
Core concepts
Measuring rate
Rate can be followed by measuring a property that changes as the reaction proceeds:
- volume of gas produced (gas syringe or displacement of water);
- mass loss as gas escapes;
- colour change using a colorimeter;
- time for a precipitate to obscure a mark (as in sodium thiosulfate and acid);
- pH or conductivity changes;
- titration of samples taken at intervals and quenched.
On a concentration–time graph, the rate at a given time is the gradient of the tangent to the curve. The initial rate is the gradient at t = 0.
Collision theory and the Maxwell–Boltzmann distribution
For reaction, particles must collide with energy ≥ Eₐ and the correct orientation. The Maxwell–Boltzmann distribution shows how energies are spread among molecules in a gas: most have moderate energy, few have very high energy, and the curve starts at the origin and never touches the energy axis at high energies. The area under the curve to the right of Eₐ represents the proportion of molecules with enough energy to react.
Factors affecting rate
Concentration (or pressure for gases): more particles per unit volume → more frequent collisions → more successful collisions per second.
Temperature: particles move faster and collide more often, but more importantly, the Maxwell–Boltzmann curve shifts to the right and flattens, so a much greater proportion of molecules have energy ≥ Eₐ. As a rough guide, a 10 °C rise often roughly doubles the rate of many reactions.
Surface area: smaller pieces expose more particles to collisions, increasing rate — powdered calcium carbonate reacts faster than marble chips.
Catalysts: provide an alternative pathway with lower activation energy, so more molecules have enough energy to react. On the Maxwell–Boltzmann diagram, the Eₐ line moves to the left. Catalysts do not change ΔH or the position of equilibrium.
Catalysts in industry and the environment
- Heterogeneous: iron in the Haber process (ammonia at Point Lisas, Trinidad); platinum, palladium and rhodium in catalytic converters; nickel in hydrogenating vegetable oils; vanadium(V) oxide in the Contact process. Reactants are adsorbed onto the catalyst surface, bonds weaken, reaction occurs and products desorb.
- Homogeneous: acid catalysts in esterification; enzymes in biological systems; Fe²⁺/Fe³⁺ in the reaction between iodide and peroxodisulfate ions.
Catalysts save energy and cost by allowing reactions at lower temperatures, which also reduces environmental impact.
Rate equations and orders
For A + B → products, rate = k[A]ᵐ[B]ⁿ. Orders must be found experimentally; they cannot be deduced from the balanced equation.
- Zero order: rate does not depend on [A]. Concentration–time graph is a straight line; rate–concentration graph is horizontal.
- First order: rate ∝ [A]. Doubling [A] doubles the rate. Concentration–time graph is a curve with a constant half-life.
- Second order: rate ∝ [A]². Doubling [A] quadruples the rate. Half-life increases as concentration falls.
Units of k
- Zero order: mol dm⁻³ s⁻¹
- First order: s⁻¹
- Second order overall: dm³ mol⁻¹ s⁻¹
Determining orders from initial rates
Compare experiments where only one concentration changes. If doubling [A] doubles the rate, the order with respect to A is 1; if it quadruples, the order is 2; if there is no change, the order is 0.
Half-life
For a first-order reaction, the half-life is constant and independent of initial concentration — a key test for first-order kinetics. Radioactive decay is first order. The relationship k = 0.693 ÷ t½ applies to first-order reactions.
Mechanisms and the rate-determining step
Many reactions occur in several steps. The slowest step determines the overall rate, and the rate equation includes only species involved in (or before) that step.
For example, if the rate equation for the hydrolysis of a halogenoalkane is rate = k[RX], only RX is involved in the slow step, suggesting a mechanism in which RX first ionises (Sₙ1). If rate = k[RX][OH⁻], both are involved, suggesting a one-step mechanism (Sₙ2).
Worked examples
Example 1: Deducing orders
| Experiment | [A] / mol dm⁻³ | [B] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 4.0 × 10⁻⁴ |
| 3 | 0.10 | 0.20 | 8.0 × 10⁻⁴ |
From 1 → 2, [A] doubles and the rate doubles: first order in A. From 1 → 3, [B] doubles and the rate quadruples: second order in B. Rate = k[A][B]², overall third order.
k = rate ÷ ([A][B]²) = 2.0 × 10⁻⁴ ÷ (0.10 × 0.10²) = 2.0 × 10⁻⁴ ÷ 1.0 × 10⁻³ = 0.20 dm⁶ mol⁻² s⁻¹.
Example 2: Half-life
A first-order reaction has a half-life of 30 s. What fraction of reactant remains after 90 s?
90 s = 3 half-lives. Fraction remaining = (½)³ = 1/8.
Example 3: Explaining a temperature effect
Explain why a small increase in temperature causes a large increase in rate. (3 marks)
Particles gain kinetic energy and move faster (1). The Maxwell–Boltzmann distribution shifts so that a much larger proportion of molecules have energy equal to or greater than the activation energy (1). The frequency of successful collisions increases greatly (1).
Example 4: Rate from a graph
On a graph of [product] against time, how do you find the rate at 20 s?
Draw a tangent to the curve at t = 20 s and calculate its gradient (change in concentration ÷ change in time).
Example 5: Using half-lives to find order
In the decomposition of a compound X, its concentration falls from 0.80 to 0.40 mol dm⁻³ in 50 s, from 0.40 to 0.20 mol dm⁻³ in a further 50 s, and from 0.20 to 0.10 mol dm⁻³ in a further 50 s. What is the order with respect to X? Calculate k.
The half-life is constant (50 s), so the reaction is first order in X. k = 0.693 ÷ t½ = 0.693 ÷ 50 = 0.0139 s⁻¹.
Example 6: Recognising zero order
A graph of [Y] against time for a catalysed reaction is a straight line sloping downwards. What is the order with respect to Y, and why might this happen?
A straight-line decrease means the rate is constant, independent of [Y], so the reaction is zero order in Y. This often happens in heterogeneous catalysis when the catalyst surface is completely covered with reactant: adding more reactant cannot increase the rate, because the number of active sites, not the concentration, is limiting.
Example 7: Linking a rate equation to a mechanism
The reaction 2NO₂ + F₂ → 2NO₂F has the rate equation rate = k[NO₂][F₂]. Suggest a two-step mechanism consistent with this.
Step 1 (slow): NO₂ + F₂ → NO₂F + F Step 2 (fast): NO₂ + F → NO₂F
The rate equation contains one NO₂ and one F₂, matching the species in the slow, rate-determining step. Adding the two steps gives the overall equation.
Example 8: Interpreting a Maxwell–Boltzmann diagram
On one diagram, two curves are drawn for the same gas at T₁ and T₂, where T₂ > T₁. Describe how the curve for T₂ differs and explain the effect on rate.
The T₂ curve has a lower, broader peak shifted to higher energy, while the total area under both curves is the same (same number of molecules). A larger area lies to the right of Eₐ at T₂, so a greater proportion of molecules have sufficient energy to react, and the rate increases.
Identifying order from data: a summary
| Evidence | Zero order | First order | Second order |
|---|---|---|---|
| Doubling concentration | rate unchanged | rate doubles | rate quadruples |
| Concentration–time graph | straight line | curve, constant half-life | curve, half-life increases |
| Rate–concentration graph | horizontal line | straight line through origin | upward curve |
Using two independent pieces of evidence — such as initial rates and half-life — gives the most convincing conclusion about order.
Common mistakes and how to avoid them
- Deducing orders from the balanced equation — orders come only from experiment.
- Saying higher temperature works mainly by more collisions — the key reason is more molecules exceeding Eₐ.
- Saying catalysts increase the energy of molecules — they lower Eₐ.
- Drawing Maxwell–Boltzmann curves that touch the x-axis or do not start at the origin.
- Wrong units for k.
- Forgetting that catalysts do not change ΔH or equilibrium position.
Self-test questions
- A reaction has the rate equation rate = k[A]². What happens to the rate if [A] is tripled? Answer: The rate increases by a factor of 3² = 9.
- Give the units of k for a reaction that is first order in A and first order in B. Answer: Overall second order: dm³ mol⁻¹ s⁻¹.
- Explain why powdered zinc reacts faster with dilute acid than a single lump of the same mass. Answer: The powder has a much larger surface area, so more zinc atoms are exposed and the frequency of collisions with H⁺ ions is greater.
- Why does a catalyst not change the yield at equilibrium? Answer: It lowers the activation energy of the forward and reverse reactions equally, so both rates increase by the same factor and the position of equilibrium is unchanged.
How this links to your Internal Assessment
Kinetics provides excellent SBA investigations: the effect of concentration on the rate of reaction of sodium thiosulfate with acid, the effect of temperature on the reaction of magnesium with acid, or the effect of catalysts on the decomposition of hydrogen peroxide. High-scoring reports state a clear hypothesis, identify independent, dependent and controlled variables, repeat measurements, plot appropriate graphs (such as rate against concentration or 1/time against temperature), and evaluate errors like reaction-time judgement and temperature fluctuations.
Exam technique for Rates of reaction and kinetics
- Explain each factor with collision theory, mentioning frequency of collisions and the proportion with energy ≥ Eₐ.
- Label Maxwell–Boltzmann diagrams with axes ("number of molecules" and "energy") and Eₐ.
- In order questions, state which experiments you compared and what changed.
- Give units for k.
- Link rate equations to mechanisms using the rate-determining step.
Quick revision summary
- Rate = change in concentration per unit time; gradient of the tangent.
- Collisions need energy ≥ Eₐ and correct orientation.
- Concentration, pressure, surface area increase collision frequency; temperature increases the proportion with ≥ Eₐ; catalysts lower Eₐ.
- Heterogeneous (Fe, Pt, V₂O₅, Ni) and homogeneous (acid, enzymes) catalysts.
- rate = k[A]ᵐ[B]ⁿ; orders from experiment; zero, first and second order graphs.
- First-order half-life is constant; k = 0.693/t½.
- The rate-determining step links the rate equation to the mechanism.