A-level chemistry · Chapter 5

Kinetics

Collision theory, Maxwell–Boltzmann, rate equations, Arrhenius

Activation energy Ea — the minimum energy colliding particles need for a reaction to occur.

Maxwell–Boltzmann distribution

Rate equations (A-level year 2)

rate = k[A]m[B]n

Arrhenius equation

k = A e−Ea/RT  ·  ln k = ln A − Ea/(RT)

A plot of ln k against 1/T is a straight line with gradient −Ea/R — how Ea is measured (required practical 3 uses this idea).

Try the Maxwell–Boltzmann simulation with the temperature slider.

3.1.9.1Finding orders from initial-rate data

Worked example. For A + B → products:
  1. Expt 1→2: [A] doubles ([B] constant), rate doubles → first order in A.
  2. Expt 2→3: [B] doubles ([A] constant), rate ×4 → second order in B.
  3. rate = k[A][B]² — overall order 3. Units of k: rate ÷ (mol dm⁻³)³ = dm⁶ mol⁻² s⁻¹.
  4. Get k by substituting any one experiment's numbers.

3.1.9.1Concentration–time vs rate–concentration graphs

3.1.9.2Using the Arrhenius equation

Exam tip. "Explain how the graph shows first order" wants the constant half-life stated with two successive half-life readings from the graph — not just the word "curve".

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What each change does to the Maxwell–Boltzmann curve

ChangeCurveEa lineWhy rate rises
Raise temperatureFlattens, peak moves right; area constantFixedFar more molecules exceed Ea; also more frequent collisions (minor)
Add catalystUnchangedMoves leftAlternative route with lower Ea — more of the existing distribution qualifies
Raise concentration/pressureTaller (more molecules), same shapeFixedMore collisions per second

Never say a catalyst "lowers the activation energy of the reaction" — it provides an alternative route with a lower activation energy, and is unchanged at the end. Small temperature rises matter enormously: ~10 K roughly doubles rate because the high-energy tail grows exponentially.

Rate equations (Year 2 core)

rate = k[A]ᵐ[B]ⁿ — the orders m and n come only from experiment, never from the equation's coefficients. Total order = m + n. Units of k change with total order: mol dm⁻³ s⁻¹ ÷ (mol dm⁻³)ᵒʳᵈᵉʳ — first order gives s⁻¹, second order gives mol⁻¹ dm³ s⁻¹.

Initial-rates deduction. Doubling [A] quadruples the rate → second order in A. Doubling [B] leaves rate unchanged → zero order in B. rate = k[A]². Insert any experiment's numbers to evaluate k with correct units.

Arrhenius: getting Ea from experiment

k = Ae^(−Ea/RT), so ln k = ln A − Ea/RT. Plot ln k (or ln(1/t) from clock experiments) against 1/T: a straight line of gradient −Ea/R. With R = 8.31, a gradient of −5.4 × 10³ K gives Ea = 5.4 × 10³ × 8.31 ≈ 45 kJ mol⁻¹. Watch units — convert J to kJ at the end.

Mechanisms and the rate-determining step

The rate equation contains only species involved up to and including the slowest (rate-determining) step. If rate = k[(CH₃)₃CBr] with no [OH⁻], the slow step is the C–Br bond breaking alone (an SN1-style mechanism); OH⁻ attacks the carbocation in a later fast step. Reverse logic works too: propose mechanisms whose slow-step molecularity matches the orders.

Measuring rate in the lab

Catalysts in industry. Heterogeneous: Fe in the Haber process, V₂O₅ in the Contact process, Ni in hydrogenation, Pt/Pd/Rh in catalytic converters (adsorption → reaction on the surface → desorption; poisoned by lead). Homogeneous catalysts work via an intermediate species in the same phase — cheaper separation is the heterogeneous advantage.

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Reading the two graph families

First-order half-life. [A] falls 0.48 → 0.24 mol dm⁻³ in 200 s and 0.24 → 0.12 in the next 200 s → constant t½ → first order. k = ln2 ÷ t½ = 0.693 ÷ 200 = 3.5 × 10⁻³ s⁻¹.

Initial-rates table, deduced properly

Expt[X][Y]Rate / mol dm⁻³ s⁻¹
10.100.102.0 × 10⁻⁴
20.200.108.0 × 10⁻⁴
30.200.302.4 × 10⁻³

1→2: [X] ×2, rate ×4 → order 2 in X. 2→3: [Y] ×3, rate ×3 → order 1 in Y. rate = k[X]²[Y]; from expt 1, k = 2.0 × 10⁻⁴ ÷ (0.10² × 0.10) = 0.20 mol⁻² dm⁶ s⁻¹. Always finish with k's value AND units — half the marks sit there.

Arrhenius, both directions

From two temperatures. k doubles from 300 K to 310 K: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) → ln 2 = (Ea/8.31)(1/300 − 1/310) → Ea = 0.693 × 8.31 ÷ 1.075 × 10⁻⁴ ≈ 53.6 kJ mol⁻¹ — the origin of the "doubling per 10 K" folk rule (it is only exact near this Ea and temperature).

The pre-exponential factor A represents collision frequency with correct orientation; e^(−Ea/RT) is the fraction of collisions with enough energy. Raising T leaves A almost alone but grows the exponential dramatically.

Mechanism–rate detective work

Experimental design for k and orders. Continuous method: one run, sample or monitor over time (colorimeter/quenched titration), plot concentration–time, take half-lives or tangents. Initial-rate method: many runs varying one concentration, clock reactions for speed. State which method suits the chemistry: coloured species → colorimetry; gas evolved → syringe; sudden visual endpoint → clock.
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