Full chapter
Reaction Kinetics
Extract a rate law from data, test a mechanism, and explain how conditions and catalysts change reaction speed.
A-Level 9476 (2026-2027)
A rate measures change per unit time
A rate law is experimental evidence about speed, not a rearranged balanced equation.
A reaction rate can be measured as the change in a reactant or product concentration per unit time. Reactant disappearance rates are reported positively: -Δ[A]/Δt; product appearance rates use +Δ[P]/Δt. A tangent to a concentration-time curve gives the instantaneous rate, while a chord gives an average rate over the interval. State which species is being measured because stoichiometric coefficients connect their different disappearance and appearance rates.
For rate = k[A]m[B]n, m and n are the orders with respect to A and B; m + n is the overall order. The rate constant k is constant for a given reaction under specified temperature and catalytic conditions. Changing concentrations changes the rate through the concentration terms, not k, within the conditions where that rate law applies.
| Order in A | Dependence on [A] | Effect of doubling [A] |
|---|---|---|
| Zero | The [A] factor is 1. | No change in rate. |
| First | Rate is proportional to [A]. | Rate doubles. |
| Second | Rate is proportional to [A] squared. | Rate becomes four times as large. |
Compare experiments that isolate one concentration
Divide the rate equations so the constant and unchanged factors cancel.
| Experiment | [A] / mol dm-3 | [B] / mol dm-3 | Initial rate / mol dm-3 s-1 |
|---|---|---|---|
| 1 | 0.100 | 0.100 | 2.00 × 10-4 |
| 2 | 0.200 | 0.100 | 8.00 × 10-4 |
| 3 | 0.200 | 0.200 | 1.60 × 10-3 |
Worked example
Determine the complete rate equation
Find the orders, k and the predicted initial rate when [A] = 0.150 and [B] = 0.300 mol dm-3.
- Compare experiments 1 and 2: [A] doubles at fixed [B], while rate quadruples. Therefore m = 2.
- Compare experiments 2 and 3: [B] doubles at fixed [A], while rate doubles. Therefore n = 1.
- k = (2.00 × 10-4)/(0.1002 × 0.100) = 0.200.
- Overall order is 3. Units of k = (mol dm-3 s-1)/(mol dm-3)3 = dm6 mol-2 s-1.
- New rate = 0.200(0.150)2(0.300) = 0.00135 mol dm-3 s-1.
Rate = k[A]2[B], with k = 0.200 dm6 mol-2 s-1; predicted rate = 1.35 × 10-3 mol dm-3 s-1.
If both concentrations change, first use another pair to find one order, then divide out its known contribution. For example, if doubling both A and B raises the rate eightfold and B is first order, the remaining factor of four comes from A, which is second order. Check the final equation against every row, not just the pair used to derive it.
| Overall order | Units of k |
|---|---|
| 0 | mol dm-3 s-1 |
| 1 | s-1 |
| 2 | dm3 mol-1 s-1 |
| 3 | dm6 mol-2 s-1 |
Zero order loses equal amounts; first order loses equal fractions
Only first-order half-life stays constant as concentration falls.
Zero-order concentration-time graph
Concentration falls in a straight line from 0.20 mol per cubic decimetre to zero over 60 seconds. The constant negative gradient means a constant disappearance rate.
First-order concentration-time graph
Concentration falls from 0.20 to 0.10, 0.050 and 0.025 mol per cubic decimetre at 0, 20, 40 and 60 seconds. The curve becomes less steep, but each halving takes 20 seconds.
The half-life is the time for a reactant concentration to fall to half its value at the start of the interval. For first order, each equal time interval removes the same fraction, so half-life is independent of the starting concentration. Look for several successive halvings on a graph; a single halving interval alone cannot establish first-order behaviour.
Worked example
Use repeated half-lives without an integrated rate equation
A first-order reactant has a half-life of 12.0 min. Starting at 0.160 mol dm-3, find its concentration after 36.0 min and the percentage reacted.
- 36.0/12.0 = 3 half-lives.
- Halve three times: 0.160 → 0.0800 → 0.0400 → 0.0200 mol dm-3.
- Fraction remaining = 1/8; fraction reacted = 7/8.
0.0200 mol dm-3 remains; 87.5% has reacted. The first-order model approaches zero but does not reach it after a finite number of half-lives.
When another reactant is maintained effectively constant, an observed simple decay can describe the dependence on the changing reactant under those conditions. State that constraint before attributing the entire reaction an overall order. Integrated rate equations are not required for this syllabus.
A mechanism must match both the overall equation and the rate law
Intermediates cancel from the overall equation; a proposed rate law must use measurable reactant concentrations.
A mechanism is a sequence of elementary steps. The rate-determining step is the slow step that controls the overall rate under the specified conditions. For an elementary slow step involving A and B, the simplest collision model gives rate = k[A][B]. A later fast step can consume another reactant without that reactant appearing in the observed rate law.
Worked example
Test a two-step mechanism
Proposed steps are A + B → I (slow), followed by I + B → P (fast). The observed rate is k[A][B]. Is the mechanism consistent?
- Add the steps and cancel intermediate I: A + 2B → P.
- The elementary slow step predicts first order in A and first order in B.
- This matches the observed rate law even though B has coefficient 2 in the overall equation.
It is consistent with both pieces of evidence. Agreement supports the mechanism but does not prove it is the only possible mechanism.
Worked example
Use a fast pre-equilibrium to remove an intermediate
A proposed mechanism is 2A ⇌ I (fast equilibrium), then I + B → P (slow). Predict the rate law.
- The slow elementary step gives rate = k2[I][B].
- For the preceding equilibrium, K = [I]/[A]2, so [I] = K[A]2.
- Substitute: rate = k2K[A]2[B]; at fixed temperature, combine k2K into the observed constant k.
Rate = k[A]2[B], second order in A and first order in B. This prediction relies on the stated fast-equilibrium approximation.
A proposed elementary step cannot have an arbitrary concentration exponent unrelated to the species participating in that step. Conversely, an observed zero order does not mean the substance is absent from the reaction: its concentration may not control the slow process over the measured range. Always use the mechanism supplied rather than inventing a slow step from the overall coefficients.
Choose a signal that follows a changing species
The method must be fast enough, selective enough and controlled enough for the question.
| Observable change | Possible method | Key control or limitation |
|---|---|---|
| Gas produced | Record gas-syringe volume against time, or loss of mass if gas escapes. | Check for leaks, gas solubility and temperature/pressure changes; account for mixing delay. |
| Coloured species consumed or formed | Record absorbance with a colorimeter at a suitable wavelength. | Calibrate absorbance against concentration; use a blank and avoid interference from other coloured species. |
| Acid, base or redox species changes | Withdraw timed aliquots, quench if needed, then titrate. | Quenching must stop the reaction without changing the amount being analysed. |
| A small fixed amount of product triggers a colour change | Use a clock method and measure time to the same endpoint. | 1/time compares initial rates only when the fixed threshold is small and consistent. |
To compare concentration effects, change one initial concentration while holding total volume, temperature and other reactant concentrations constant, often by replacing part of a solution volume with water. Use an appropriate initial gradient, because reactant concentrations change during a run. Keep solid particle size or surface area constant if a solid participates.
Worked example
Plan a rate measurement for calcium carbonate and acid
How could the effect of acid concentration be investigated using the carbon dioxide produced?
- Use a gas-tight flask connected to a gas syringe; use equal masses and a controlled size range of carbonate pieces.
- Use acid in excess and hold its total volume and temperature constant between runs.
- Start timing reproducibly at mixing and record gas volume at short intervals; use the initial tangent rather than the final gas volume.
- Repeat to assess scatter and check the apparatus for leaks.
Initial gas-volume gradients compare initial rates. The final volume mainly reflects the limiting amount, so it cannot by itself show which reaction was faster.
This is a valid theoretical planning method. The 9476 practical syllabus excludes gas collection by gas syringe or water displacement as a hands-on examination operation; interpreting supplied gas-volume data or discussing an appropriate method remains useful chemical reasoning.
Reaction needs both encounters and sufficient energy
Heating changes the energy distribution; concentration mainly changes collision frequency.
The activation energy Ea is the minimum energy barrier that must be overcome along the reaction pathway. Colliding particles need sufficient energy and an appropriate orientation to react. At higher reactant concentration, more particles occupy a given volume, usually producing more frequent relevant collisions; the measured rate law tells how strongly rate responds.
Heating changes the Boltzmann energy distribution
At higher temperature the energy distribution has a lower, broader peak and a larger fraction of particles beyond the same activation-energy threshold. Both distributions are normalised; temperature does not move the threshold.
The distribution begins at zero and has a long tail: particles do not all have the same energy, and no sharp maximum energy exists. Raising temperature shifts the population towards higher energies and greatly increases the fraction able to cross the barrier. Together with a moderate increase in collision frequency, this raises k and hence the rate at fixed concentrations. Do not say that heating lowers activation energy.
Adding more particles at the same temperature does not change their normalised energy distribution or the fraction above Ea. It increases the number of potentially successful collisions. A temperature change and a concentration change therefore operate differently even when both make a reaction faster.
A catalyst supplies another mechanism
More particles can react at the same temperature because the barrier is lower.
A catalyst changes the threshold, not the distribution
One energy distribution at a fixed temperature is crossed by an uncatalysed activation threshold and a lower catalysed threshold. The catalysed threshold admits a larger fraction of particles without changing their energies.
A catalyst participates in intermediate steps and is regenerated by the end of its cycle. It accelerates the approach to equilibrium in both directions and does not change ΔH, the equilibrium constant or the equilibrium composition at the same temperature. It can be deactivated or poisoned in practice even though it is not consumed in the ideal overall equation.
A homogeneous catalyst reacts through a regenerated intermediate
Catalyst and reactants occupy the same phase in the two specified cycles.
| Step | Equation and role |
|---|---|
| Oxidise sulfur dioxide | SO2(g) + NO2(g) → SO3(g) + NO(g) |
| Regenerate nitrogen dioxide | 2NO(g) + O2(g) → 2NO2(g) |
| Overall after doubling the first step | 2SO2(g) + O2(g) → 2SO3(g) |
NO and NO2 cycle between forms and cancel from the overall equation. This gas-phase scheme illustrates homogeneous catalysis of atmospheric SO2 oxidation; real atmospheric sulfur chemistry also includes other oxidants and aqueous pathways. Formation and hydration of sulfur oxides contributes to acidic deposition.
| Step | Balanced equation |
|---|---|
| Peroxodisulfate oxidises iron(II) | S2O82-(aq) + 2Fe2+(aq) → 2SO42-(aq) + 2Fe3+(aq) |
| Iron(III) oxidises iodide and regenerates iron(II) | 2Fe3+(aq) + 2I-(aq) → 2Fe2+(aq) + I2(aq) |
| Overall | S2O82-(aq) + 2I-(aq) → 2SO42-(aq) + I2(aq) |
The uncatalysed reaction requires encounters between two negative reactants, I- and S2O82-. The iron redox cycle offers a different pathway involving positive iron ions, with a lower effective activation barrier. Fe2+ is oxidised and then regenerated; its concentration may be small compared with the reactants.
A solid surface brings adsorbed reactants into a more reactive arrangement
Adsorption, surface reaction and desorption must all remain possible.
- Adsorb the gases
N2 and H2 interact with active sites on solid iron; adsorption weakens bonds and can dissociate the molecules.
- React on the surface
Adsorbed nitrogen and hydrogen species combine through a sequence of lower-barrier steps.
- Desorb ammonia
NH3 leaves the surface, freeing sites for further reactants. Overall: N2 + 3H2 ⇌ 2NH3.
The solid catalyst is in a different phase from the reacting gases. A larger exposed surface gives more accessible active sites, while a poison that binds strongly can block them. Binding must be strong enough to activate reactants but not so strong that products cannot leave. Catalysis permits a useful rate at a lower temperature than the uncatalysed route; it does not independently increase the equilibrium yield.
In a vehicle catalytic converter, gases contact solid precious-metal surfaces such as platinum and rhodium. Adsorption and surface reaction help remove nitrogen oxides, with carbon monoxide acting as a reducing agent. A representative reaction is 2NO(g) + 2CO(g) → N2(g) + 2CO2(g). Other catalytic reactions oxidise CO and unburnt hydrocarbons. The converter requires a suitable operating temperature; it is less effective immediately after a cold start.
Enzymes are highly specific protein catalysts
Their active sites select substrates and depend on temperature and pH.
- Substrate fits the active site
The substrate has a complementary shape and suitable chemical interactions with the enzyme active site.
- An enzyme-substrate complex forms
The active site provides a lower-barrier pathway for a particular chemical transformation.
- Products leave
The enzyme is available again: E + S ⇌ ES → E + P.
Specificity has two aspects: which substrate binds effectively and which reaction is catalysed. A substrate that does not fit or interact appropriately is not efficiently converted. The lock-and-key model captures this selectivity; it is a simplified model rather than a claim that proteins are perfectly rigid.
| Condition | Effect on activity |
|---|---|
| Temperature rises within a suitable range | More frequent energetic encounters increase the catalytic rate. |
| Temperature becomes too high | The active-site structure can be disrupted, reducing effective binding and catalysis; activity falls despite faster molecular motion. |
| pH moves away from the suitable range | Changes in ionisation of active-site groups and substrate can alter interactions and structure, reducing activity. |
| Different enzyme | Its useful temperature and pH ranges can differ; do not assume all enzymes have the same optimum. |
Check your understandingWhy can cooling an enzyme slow its reaction without having the same consequence as severe heating?Think it through, then reveal the answer
Quick revision
Revisit the essentials, then return to an explanation when you need it.
Orders come from rates or an explicit elementary mechanism, not overall coefficients. A zero-order concentration-time plot is straight; first order has constant successive half-lives. Derive k units from the measured rate units and overall order.
Concentration changes collision frequency. Temperature changes collision frequency and the energetic fraction, increasing k. A catalyst changes the pathway and lowers the barrier while leaving the same-temperature energy distribution and equilibrium constant unchanged. Enzymes add active-site specificity and temperature/pH sensitivity.
Scope and references
Learning outcomes and sources
8. Reaction Kinetics. Use the outcome map to find the explanation for a particular syllabus requirement.
See the learning outcome map
8(a) Explain the central kinetic terms.
- Rate, rate equation, order, rate constant
- Half-life
- Rate-determining step
- Activation energy
- Catalysis
A rate measures change per unit timeZero order loses equal amounts; first order loses equal fractionsA mechanism must match both the overall equation and the rate lawReaction needs both encounters and sufficient energyA catalyst supplies another mechanism
8(b) Construct and use simple rate equations.
- Orders 0, 1 or 2 for a species
- (i) Initial-rate deduction
- (ii) Zero/first-order concentration-time graphs
- (iii) Check a proposed mechanism
- (iv) Predict order from a mechanism
- (v) Calculate an initial rate from concentrations
- No integrated rate equations required
A rate measures change per unit timeCompare experiments that isolate one concentrationZero order loses equal amounts; first order loses equal fractionsA mechanism must match both the overall equation and the rate law
8(c) Understand and use first-order half-life.
- (i) Independence from concentration
- (ii) Half-life calculations
Zero order loses equal amounts; first order loses equal fractions
8(d) Calculate a rate constant from initial-rate data.
- Substitute experimental rate and concentrations
- Derive units from overall order
8(e) Devise a suitable rate experiment.
- Select a measurable signal from supplied reaction information
- Control conditions and identify measurement limitations
8(f) Explain concentration effects through collision frequency.
- More particles per volume
- Distinguish concentration from temperature effects
8(g) Explain activation energy using a Boltzmann distribution.
- Energy threshold
- Area representing the energetic fraction
8(h) Explain temperature effects on k and rate.
- Boltzmann distribution
- Collision frequency
- Activation energy unchanged for the same pathway
8(i) Explain catalytic lowering of a reaction barrier.
- (i) Different mechanism, lower activation energy, larger k
- (ii) Larger energetic fraction in the unchanged distribution
8(j) Outline homogeneous and heterogeneous catalytic action.
- (i) Haber process
- (ii) Vehicle-exhaust nitrogen-oxide removal
- (iii) Atmospheric NOx catalysis of SO2 oxidation
- (iv) Fe2+ catalysis of I-/S2O8^2-
- Catalyst regeneration and phase distinction
A homogeneous catalyst reacts through a regenerated intermediateA solid surface brings adsorbed reactants into a more reactive arrangement
8(k) Describe enzymes and their sensitivities.
- Protein biological catalysts
- Reaction and substrate specificity
- Lock-and-key model
- Temperature and pH sensitivity
- Protein structural levels and denaturation details not required
- SEAB H2 Chemistry 9476, examination 2026
Topic 8, printed pages 20-21. All 11 outcomes and every nested rate-law, catalyst and enzyme requirement inspected.