This mini-lesson covers AQA 3.1.5 Kinetics and 3.1.9 Rate equations: collision theory and the Maxwell–Boltzmann distribution, rate = k[A]m[B]n, finding orders from data and graphs, the rate-determining step, and the Arrhenius equation.
Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Everything here is A-level standard — the maths is done properly, not skipped. Press Start when you're ready.
Kinetics · collision theory
Collision theory and the Maxwell–Boltzmann distribution
For a reaction to happen, particles must collide with at least the activation energy Eaand with the correct orientation. At any moment only a tiny fraction of molecules have enough energy.
The Maxwell–Boltzmann distribution plots the number of molecules against their energy. It starts at the origin (no molecule has zero energy), rises to a peak (the most probable energy), and has a long tail that never touches the axis. Only the molecules in the shaded area to the right of Ea can react.
Increasing temperature shifts the peak to the right and makes it lower and broader. The total area is unchanged (the number of molecules is the same), but the area beyond Ea increases dramatically — this is the main reason rate rises so steeply with temperature, far more than the small increase in collision frequency.
A catalyst does not change the distribution at all. It provides an alternative reaction route with a lower Ea, so a much larger fraction of the existing molecules can react.
Increasing concentration or pressure gives more frequent collisions per second — the distribution shape is unchanged.
Quick check
Quick check
?A reaction is heated from 300 K to 310 K and the rate roughly doubles. Which explanation is correct?
Kinetics · rate equations
The rate equation and orders of reaction
rate = k [A]m [B]nm = order with respect to A · n = order with respect to B · overall order = m + n
The single most important fact: the orders cannot be deduced from the balanced equation. They must be found experimentally, by changing one concentration at a time (the initial rates method) and seeing what happens to the rate.
Zero order — double [A], rate is unchanged. A concentration–time graph is a straight line.
First order — double [A], rate doubles. The half-life is constant, and a graph of ln[A] against t is a straight line.
Second order — double [A], rate goes up × 4; triple it and rate goes up × 9.
The units of k depend on the overall order. First order overall: s⁻¹. Second order overall: mol⁻¹ dm³ s⁻¹. Third order overall: mol⁻² dm⁶ s⁻¹. Work them out by rearranging k = rate ÷ [A]ᵐ[B]ⁿ and cancelling.
Calculate
Your turn
1In an experiment, [A] is kept constant while [B] is doubled, and the rate increases by a factor of 4. State the order of the reaction with respect to B.
Hint: If doubling gives ×4, then 2ᵐ = 4.
Calculate
Your turn
2The rate equation is rate = k[A][B]². When [A] = 0.20 mol dm⁻³ and [B] = 0.10 mol dm⁻³, the rate is 2.4 × 10⁻³ mol dm⁻³ s⁻¹. Calculate k.
3For that same reaction, rate = k[A][B]². State the overall order.
Hint: Overall order = m + n = 1 + 2.
Sort it
Sort each clue by the order it points to
Tap a piece of evidence, then tap the order in X that it indicates.
🟩 Zero order in X
🟪 First order in X
🟦 Second order in X
Kinetics · mechanism
The rate-determining step
Most reactions happen in a series of steps. The slowest step is the rate-determining step (RDS) — it acts as a bottleneck, so the overall rate depends only on what happens up to and including it.
The rule that links the rate equation to the mechanism: the species that appear in the rate equation — and the number of each, given by its order — are the species involved in the RDS (or in a fast step before it).
Worked example
For NO₂ + CO → NO + CO₂ the experimental rate equation is rate = k[NO₂]².
CO does not appear, so CO is not in the rate-determining step. Two NO₂ molecules are, so a consistent mechanism is:
Step 1 (slow): 2NO₂ → NO + NO₃
Step 2 (fast): NO₃ + CO → NO₂ + CO₂
Adding the steps gives the overall equation, and the slow step contains two NO₂ — exactly as the rate equation requires.
Zero order = not in the RDS. If a reagent is zero order, changing its concentration cannot change the rate, because it only takes part after the bottleneck.
Quick check
Quick check
?For the reaction A + 2B + C → products, the rate equation is rate = k[A][C]. What can you deduce?
Kinetics · Arrhenius
The Arrhenius equation
k = A e−Ea/RT → ln k = ln A − Ea / (RT)A = pre-exponential factor · R = 8.31 J K⁻¹ mol⁻¹ · T in K · E_a in J mol⁻¹
The logarithmic form is the useful one: it is the equation of a straight line y = mx + c if you plot ln k (y) against 1/T (x).
Gradient = −Ea / R, so Ea = −gradient × R (the gradient is negative, so Ea comes out positive).
Intercept = ln A.
Answers come out in J mol⁻¹ — divide by 1000 for kJ mol⁻¹.
Reading the equation: a larger Ea or a lower T makes the exponent more negative, so k is smaller and the reaction is slower. A catalyst lowers Ea, which increases k — the rate constant really does change, whereas concentration changes never alter k.
Calculate
Your turn
4A plot of ln k against 1/T gives a straight line of gradient −6.50 × 10³ K. Calculate the activation energy in kJ mol⁻¹. (R = 8.31 J K⁻¹ mol⁻¹)
kJ mol⁻¹
Hint: E_a = −gradient × R = 6.50 × 10³ × 8.31 = 54 015 J mol⁻¹; then ÷ 1000.
Match it
Match each symbol to its meaning
Tap a symbol on the left, then its meaning on the right.
Symbol
Meaning
Quick check
Quick check
?Adding a catalyst to a reaction at constant temperature changes which of these?
Quick check
Quick check
?The rate equation for a reaction is rate = k[A]²[B]. What are the units of k?
Kinetics · exam traps
Where students throw away rate marks
You cannot read the orders off the equation. Orders come from experiment only. A reagent with a coefficient of 2 can still be zero order.
Zero order does not mean "does not react". It means the rate does not depend on that concentration — the reagent reacts after the rate-determining step.
Units of k must be derived, not memorised blindly: rearrange k = rate ÷ [A]ᵐ[B]ⁿ and cancel.
A catalyst changes k, not the orders. Concentration changes never change k; only temperature (and a catalyst, via Ea) do.
First-order half-life is constant. That is the fingerprint: if [A] takes the same time to fall from 0.80 to 0.40 as from 0.40 to 0.20, the reaction is first order in A. Use it to identify the order straight from a concentration–time graph.
Quick check
Quick check
?A reaction is first order overall. What are the units of the rate constant k?
Calculate
Your turn
5A first-order reaction has [A] falling from 0.80 to 0.20 mol dm⁻³ in 40 s. Calculate its half-life.
s
Hint: 0.80 → 0.40 → 0.20 is TWO half-lives, and they take 40 s in total.
Recap
The big ideas to know
Collision theory: a reaction needs collisions with E ≥ E_a AND the correct orientation
Maxwell–Boltzmann: raising temperature shifts the curve right and flattens it — many more particles exceed E_a
Catalyst: provides an alternative route with a lower E_a; it is not used up
Rate equation: rate = k[A]m[B]n — orders come from EXPERIMENT, not from the stoichiometry
Orders: zero (no effect) · first (×2 → ×2, constant half-life) · second (×2 → ×4)
Mechanism: the rate equation shows the species in (or before) the rate-determining step, and how many of each
Arrhenius: ln k = ln A − E_a/RT — plot ln k against 1/T; gradient = −E_a/R
That is the whole of AQA 3.1.5 and 3.1.9. Press Finish.
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