A-Level H2 Chemistry: Reaction Kinetics — How to Nail Orders, Rate Constants and Mechanisms

Why Reaction Kinetics matters for your A-Level

Reaction Kinetics appears in almost every A-Level H2 Chemistry paper, and it is one of the few topics where you can score full marks reliably with practice. It rewards careful method rather than memorisation: examiners love asking you to work out the order of reaction, calculate the rate constant k with correct units, interpret a graph, or test whether a proposed mechanism is consistent with the rate equation. These are skills, and skills can be drilled.

The catch is that kinetics is also where careless students leak marks. The two biggest culprits are assuming the order equals the stoichiometric coefficient, and giving k without units or with the wrong units. This article explains the core ideas clearly, works through full exam-style examples, and flags the traps so you do not fall into them.

Core concept 1: Rate of reaction and how it is measured

The rate of reaction is the change in concentration of a reactant or product per unit time, with units of mol dm^-3 s^-1. We measure it by following any property that changes as the reaction proceeds, for example:

Gas volume or pressure when a gas is produced (e.g. a gas syringe).
Mass loss if a gas escapes (mass on a balance falls).
Colour intensity (colorimetry) if a coloured species is formed or used up.
Titration of samples withdrawn and quenched at known times.
Conductivity if the number or type of ions changes.

The initial rate is the gradient of the tangent to a concentration–time graph at t = 0. Initial rates are useful because at the very start the reverse reaction and product interference are negligible, so the data is cleanest.

Core concept 2: Collision theory, activation energy and the Maxwell–Boltzmann distribution

For a reaction to occur, particles must collide with sufficient energy and the correct orientation. The minimum energy a collision must have to lead to reaction is the activation energy, E_a. Only a small fraction of collisions are successful, which is why most reactions are not instantaneous.

The Maxwell–Boltzmann distribution shows how molecular energies are spread out in a sample. The curve starts at the origin, rises to a peak (the most probable energy), then tails off; the area under the curve equals the total number of particles. Only molecules with energy greater than or equal to E_a (the area to the right of the E_a line) can react.

Factors affecting rate

Concentration (and pressure for gases): higher concentration means more particles per unit volume, so the frequency of effective collisions increases and rate rises. This does not change the shape of the Maxwell–Boltzmann curve.
Temperature: raising temperature shifts the Maxwell–Boltzmann curve to the right and flattens it, so a much larger fraction of molecules now exceed E_a. Particles also move faster, increasing collision frequency. The dominant effect is the larger fraction of successful collisions, which is why a modest rise in temperature can roughly double the rate.
Surface area: for a heterogeneous reaction, breaking a solid into smaller pieces exposes more surface, giving more sites for collisions and a faster rate.
Catalysts: a catalyst provides an alternative reaction pathway with a lower E_a. On the Maxwell–Boltzmann diagram, the E_a line moves left, so a greater fraction of molecules now have enough energy to react. A catalyst does not change the enthalpy change of the reaction or shift the position of equilibrium; it speeds up forward and reverse reactions equally.

Core concept 3: The rate equation, order and the rate constant

The rate equation relates rate to the concentrations of reacting species:

rate = k[A]^m[B]ⁿ

Here k is the rate constant, m is the order with respect to A, and n is the order with respect to B. The overall order is m + n. Crucially, m and n are determined experimentally; you cannot read them from the balanced equation.

What each order means

Zero order in A: changing [A] has no effect on rate. rate = k[A]⁰ = k for that term.
First order in A: doubling [A] doubles the rate.
Second order in A: doubling [A] quadruples the rate (2² = 4); tripling it gives 9 times the rate.

Working out the units of k

The units of k depend on the overall order. Rearrange the rate equation and substitute the units (rate is mol dm^-3 s^-1, each concentration is mol dm^-3):

Overall zero order: rate = k, so k has units mol dm^-3 s^-1.
Overall first order: k = rate / [A], so k has units s^-1.
Overall second order: k = rate / [A]², so k has units mol^-1 dm³ s^-1.
Overall third order: k has units mol^-2 dm⁶ s^-1.

A reliable shortcut: for overall order N, the units of k are (mol dm^-3)^1-N s^-1. Always derive the units rather than memorising blindly, and always state them.

Core concept 4: Determining order from data and graphs

From initial-rates data

Compare experiments in which only one concentration changes at a time, and see how the rate responds. This isolates the order for each reactant.

From concentration–time graphs

Zero order: a graph of [A] against time is a straight line with constant negative gradient; rate is constant until the reactant runs out.
First order: [A] against time is a curve with a constant half-life — the time for [A] to halve is the same no matter where you start. This constant half-life is the classic diagnostic for first order.
Second order: [A] against time is also a curve, but the half-life increases as the reaction proceeds.

From rate–concentration graphs

Zero order: rate against [A] is a horizontal line.
First order: rate against [A] is a straight line through the origin (rate is proportional to [A]); the gradient is k.
Second order: rate against [A] is an upward curve; a straight line through the origin only appears when you plot rate against [A]².

Half-life of first-order reactions

For a first-order reaction the half-life is constant and is given by t_1/2 = ln 2 / k = 0.693 / k. This means k and t_1/2 are interchangeable for first-order reactions, and a constant half-life on a graph immediately tells you the reaction is first order in that reactant.

Worked example 1: Find the order and rate constant from initial-rates data

For the reaction A + B → products, the following initial rates were measured at constant temperature:

Expt 1: [A] = 0.10, [B] = 0.10, initial rate = 2.0 × 10^-4 mol dm^-3 s^-1
Expt 2: [A] = 0.20, [B] = 0.10, initial rate = 8.0 × 10^-4 mol dm^-3 s^-1
Expt 3: [A] = 0.20, [B] = 0.20, initial rate = 8.0 × 10^-4 mol dm^-3 s^-1

(All concentrations in mol dm^-3.)

Step 1 — order with respect to A. Compare Expts 1 and 2, where [B] is constant. [A] doubles (0.10 → 0.20) and the rate goes from 2.0 × 10^-4 to 8.0 × 10^-4, a factor of 4. Since 4 = 2², the reaction is second order in A.

Step 2 — order with respect to B. Compare Expts 2 and 3, where [A] is constant. [B] doubles (0.10 → 0.20) but the rate is unchanged (factor of 1 = 2⁰). The reaction is zero order in B.

Step 3 — rate equation and overall order. rate = k[A]². The overall order is 2.

Step 4 — calculate k with units. Substitute Expt 1: k = rate / [A]² = (2.0 × 10^-4) / (0.10)² = (2.0 × 10^-4) / (1.0 × 10^-2) = 2.0 × 10^-2 mol^-1 dm³ s^-1.

Always check k against a second experiment. Using Expt 2: k = (8.0 × 10^-4) / (0.20)² = (8.0 × 10^-4) / (4.0 × 10^-2) = 2.0 × 10^-2. The values agree, confirming the rate equation.

Worked example 2: Half-life and graph interpretation

The decomposition of a compound X was followed, and [X] was recorded over time:

t = 0 s: [X] = 0.80 mol dm^-3
t = 100 s: [X] = 0.40 mol dm^-3
t = 200 s: [X] = 0.20 mol dm^-3
t = 300 s: [X] = 0.10 mol dm^-3

Step 1 — find successive half-lives. [X] halves from 0.80 to 0.40 in 100 s; from 0.40 to 0.20 in another 100 s; from 0.20 to 0.10 in another 100 s. The half-life is constant at 100 s.

Step 2 — deduce the order. A constant half-life is the diagnostic for a first-order reaction. So rate = k[X].

Step 3 — calculate k from t_1/2. k = 0.693 / t_1/2 = 0.693 / 100 = 6.93 × 10^-3 s^-1. Note the units (s^-1) are correct for an overall first-order reaction.

Step 4 — predict. After a further 100 s (t = 400 s), [X] would halve again to 0.05 mol dm^-3. This kind of prediction is a common follow-up question.

Worked example 3: Mechanism and the rate-determining step

Most reactions occur in a sequence of steps called the mechanism. The slowest step is the rate-determining step (RDS), and the rate equation is determined by the species (and their numbers) involved up to and including the RDS. A proposed mechanism is only acceptable if it is consistent with the experimentally determined rate equation.

Suppose the overall reaction is NO₂ + CO → NO + CO₂, and experiment gives rate = k[NO₂]². A student proposes:

Step 1 (slow): NO₂ + NO₂ → NO₃ + NO
Step 2 (fast): NO₃ + CO → NO₂ + CO₂

Is it consistent? The slow step (RDS) involves two NO₂ molecules and no CO, so the predicted rate equation is rate = k[NO₂]². This matches the experimental rate equation, so the mechanism is consistent. Note that CO does not appear in the rate equation even though it is in the overall balanced equation, because it only reacts after the RDS. This is exactly why order cannot be read off the stoichiometry.

Catalysts: homogeneous and heterogeneous

A homogeneous catalyst is in the same phase as the reactants (e.g. an aqueous catalyst in an aqueous reaction). It works by forming an intermediate, providing a new pathway of lower E_a, and is then regenerated. A classic example is the catalysis of the peroxodisulfate–iodide reaction by Fe²⁺/Fe³⁺ ions, where the catalyst shuttles between oxidation states.

A heterogeneous catalyst is in a different phase, typically a solid catalysing a gas-phase reaction. Reactants adsorb onto active sites on the surface, which weakens bonds and brings molecules together in favourable orientations, then products desorb. The Haber process (iron catalyst) and catalytic converters (platinum/rhodium) are standard examples. In both cases the key idea on a Maxwell–Boltzmann diagram is the same: the catalyst lowers E_a, so a larger fraction of molecules can react.

Common exam traps and mistakes

Assuming order = stoichiometric coefficient. Orders are experimental. In NO₂ + CO above, CO has a coefficient of 1 but is zero order. Never read orders from the balanced equation.
Wrong or missing units for k. Derive the units from the overall order every time. A first-order k is s^-1; a second-order k is mol^-1 dm³ s^-1. Stating k with no units loses marks.
Misreading half-life. Constant half-life means first order; increasing half-life means second order. Check several successive half-lives, not just one.
Changing more than one variable at a time when deducing orders from a table. Always pick experiments where only one concentration changes.
Confusing rate–concentration with concentration–time graphs. A straight line through the origin on a rate-against-[A] plot means first order; a straight line on a [A]-against-time plot means zero order. They are not the same thing.
Saying a catalyst is "used up" or changes ΔH. A catalyst is regenerated and does not change the enthalpy change or the position of equilibrium; it only lowers E_a.
Forgetting that temperature changes the shape of the Maxwell–Boltzmann curve, while concentration does not. Be precise about which factor does what on the diagram.
Quoting k without checking it against a second data point. A quick second calculation catches arithmetic slips.

How to practise and revise

Drill initial-rates tables until the "double the concentration, see the factor" logic is automatic: factor 2 means first order, factor 4 means second order, factor 1 means zero order.
Practise deriving k units from scratch for zero, first, second and third overall order. Make it muscle memory.
Sketch all three graph types (concentration–time and rate–concentration) from memory and label what each shape tells you about order.
Test mechanisms against rate equations: write the rate equation predicted by the slow step and compare it with the given experimental one.
Memorise the diagnostics: constant half-life = first order; horizontal rate–concentration line = zero order; straight line through origin on rate vs [A] = first order.
Always state units and check with a second data point. These two habits alone recover a surprising number of marks.

Master these patterns and Reaction Kinetics becomes one of the most dependable mark-scorers in your H2 Chemistry paper — and if you would like structured, exam-focused guidance, our tutors at Intuitional are happy to help you build that fluency.