9701 Chemistry · Topics 8 & 26 · AS + A Level
Reaction Kinetics and Rate Equations Cheat Sheet — A Level Chemistry 9701
Kinetics is where a clear diagram earns marks that words cannot. These sheets cover the Boltzmann distribution and exactly what changes when you raise the temperature or add a catalyst, then the A2 material: determining orders from initial-rate data and from concentration–time graphs, constant half-life as evidence of first order, and using the order to deduce the rate-determining step.
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What’s on this cheat sheet
Reaction Kinetics — AS
01 · Collision theory
A reaction occurs only when particles collide with energy ≥ Ea and in the correct orientation. Rate depends on the frequency of successful collisions.
Activation energy — the minimum energy colliding particles need for a reaction to occur.
02 · Factors affecting rate
| Factor | Why rate rises |
|---|---|
| concentration ↑ | more particles per volume → more frequent collisions |
| pressure ↑ (gas) | same effect as concentration |
| surface area ↑ | more particles exposed |
| temperature ↑ | more particles exceed Ea; collisions also more frequent |
| catalyst | alternative route of lower Ea |
Temperature matters most: the fraction of particles with E ≥ Ea rises sharply, which outweighs the small rise in collision frequency.
03 · Boltzmann distribution
Starts at the origin (no particle has zero energy) and never touches the axis. Raising T flattens and shifts the peak right; the area under the curve is constant — it is the total number of particles — but the shaded area beyond Ea grows.
04 · Catalysis
A catalyst increases rate by providing an alternative route of lower Ea. It is unchanged in mass and chemical composition at the end and does not change ΔH or the position of equilibrium — it speeds both directions equally.
Homogeneous — same phase as the reactants (Cl• radicals in ozone depletion, H⁺ in esterification). Heterogeneous — different phase, works by adsorption onto active sites (Fe in Haber, V₂O₅ in Contact, Pt/Rh in catalytic converters).
05 · Following a reaction
Choose a property that changes measurably: gas volume in a syringe, mass loss on a balance, colour by colorimetry, pH, or titrating quenched samples at intervals.
Rate at any instant = gradient of the tangent to a concentration–time graph. Initial rate is the gradient at t = 0.
06 · Reaction pathway diagrams
ΔH is the gap between reactant and product levels and is unchanged by a catalyst — only the barrier height falls. Label both axes and mark Ea from the reactants to the peak.
07 · Worked example — initial rate
rate = 48 ÷ 20 = 2.4 cm³ s⁻¹
In moles: 48 ÷ 24 000 = 2.0 × 10⁻³ mol
rate = 1.0 × 10⁻⁴ mol s⁻¹
The curve flattens as HCl is used up and its concentration falls; the reaction ends when the limiting reactant is gone.
08 · Reading a rate graph
The curve is steepest at the start, where the concentration is highest, and flattens as reactants are used up.
A more concentrated solution, a higher temperature or a powdered solid all give a steeper initial gradient. If the same amount of limiting reactant is used, the curves level off at the same final volume or mass.
A catalysed reaction reaches that same plateau sooner — the yield is unchanged.
09 · Catalysis in the atmosphere
Chlorine radicals from CFCs catalyse ozone breakdown:
Cl• + O₃ → ClO• + O₂ then ClO• + O → Cl• + O₂. The radical is regenerated, so one atom destroys many ozone molecules.
NO from engines catalyses the same reaction. In a catalytic converter Pt/Rh instead promotes 2CO + 2NO → 2CO₂ + N₂ on its surface.
10 · Industrial catalysts
| Process | Catalyst |
|---|---|
| Haber (NH₃) | Fe |
| Contact (SO₃) | V₂O₅ |
| Ostwald (HNO₃) | Pt/Rh |
| hydrogenation | Ni |
| cracking | zeolite |
A catalyst lets a plant run at a lower temperature for the same rate, saving energy and — for an exothermic reaction — improving the yield as well.
Marks lost here
— Saying a catalyst “lowers the activation energy” of the reaction rather than providing a route with a lower Ea.
— Drawing a Boltzmann curve that starts above the origin or touches the x‑axis.
— Claiming higher temperature works mainly by more frequent collisions.
— Saying a catalyst increases yield; it only shortens the time to reach the same equilibrium.
Rate Equations — A2
11 · Rate equation
m and n are the orders — found only by experiment, never from the stoichiometric equation. Overall order = m + n.
Units of k: 1st order s⁻¹, 2nd order mol⁻¹ dm³ s⁻¹, zero order mol dm⁻³ s⁻¹. Work them out by rearranging, don’t memorise.
12 · Recognising the order
| Order | conc–time | rate–conc |
|---|---|---|
| zero | straight line, constant gradient | horizontal |
| first | curve, constant half‑life | straight through origin |
| second | steeper curve, half‑life doubles | upward curve |
Constant half‑life is the signature of first order — check two successive half‑lives from the graph.
13 · Half‑life and k
t½ is the time for the concentration to fall to half its value. For first order it is independent of the starting concentration.
14 · Worked example — initial rates
| Exp | [A] | [B] | rate |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻⁴ |
| 3 | 0.20 | 0.20 | 8.0 × 10⁻⁴ |
2 → 3: [B] ×2, rate unchanged → zero order in B
rate = k[A]²
k = 2.0 × 10⁻⁴ ÷ (0.10)² = 0.020 mol⁻¹ dm³ s⁻¹
15 · Mechanism and rate‑determining step
The rate‑determining step is the slowest step. Only species involved in or before it appear in the rate equation, and their coefficients there are the orders.
A species that is zero order takes part only after the rate‑determining step. A catalyst can appear in the rate equation even though it is not in the overall equation.
rate = k[(CH₃)₃CBr] → OH⁻ absent → SN1, slow step forms a carbocation, tertiary halogenoalkane.
16 · Worked example — half‑life
Half‑life constant → first order
k = 0.693 ÷ 50 = 0.0139 s⁻¹
rate = k[A]
At [A] = 0.30: rate = 0.0139 × 0.30 = 4.2 × 10⁻³ mol dm⁻³ s⁻¹
17 · Catalysis by transition metals
Variable oxidation states let a transition metal accept and donate electrons, providing a lower‑energy route.
Fe³⁺ catalysing S₂O₈²⁻ + 2I⁻: Fe³⁺ oxidises I⁻ to I₂ and is reduced to Fe²⁺, then S₂O₈²⁻ re‑oxidises Fe²⁺. Both steps involve oppositely charged ions, unlike the uncatalysed reaction between two anions.
18 · Finding an order experimentally
Initial rates — repeat the run changing one concentration at a time and compare the starting gradients.
Continuous monitoring — follow one run to completion, plot concentration against time, then read successive half‑lives or take tangents and plot rate against concentration.
Clock reaction — time a fixed small change, such as the appearance of the blue starch–iodine colour. Then 1 ÷ t is proportional to the initial rate.
19 · Worked example — checking a mechanism
step 1 (slow) 2NO₂ → NO₃ + NO
step 2 (fast) NO₃ + CO → NO₂ + CO₂
Only step 1 is rate‑determining, and it involves two NO₂:
rate = k[NO₂]²
CO does not appear, so the mechanism predicts zero order in CO. If experiment shows first order in CO, the mechanism is wrong.
20 · Temperature and k
Only k changes with temperature; the orders do not. A rise of about 10 K roughly doubles k for many reactions near room temperature.
The reason is the Boltzmann distribution: a small rise in T sharply increases the fraction of molecules with energy ≥ Ea. A reaction with a large Ea is the more sensitive to temperature.
21 · Autocatalysis
A product catalyses the reaction that makes it, so the rate rises after the start instead of falling, then drops away as the reactants run out.
MnO₄⁻ oxidising ethanedioate is the standard case: the Mn²⁺ formed catalyses the next round, which is why the titration starts slowly and then speeds up. On a concentration–time graph the curve is S‑shaped.
Marks lost here
— Reading orders off the balanced equation instead of the data.
— Changing two concentrations at once when comparing experiments.
— Omitting the units of k, or quoting them from memory for the wrong order.
— Forgetting that a zero‑order species still takes part in the reaction, just not in the slow step.
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Reaction Kinetics and Rate Equations — Frequently Asked Questions
How do you find the order of a reaction from initial rate data?
Compare two experiments in which only one concentration changes. If doubling that concentration leaves the rate unchanged the order is zero; if the rate doubles it is first order; if the rate quadruples it is second order.
What does a constant half-life tell you?
A half-life that stays the same as the reaction proceeds is the signature of first-order kinetics with respect to that reactant. For first order, k = ln 2 / t½.
How does the rate equation reveal the rate-determining step?
The species that appear in the rate equation, with their orders as coefficients, are the species present in or before the rate-determining step. Anything that appears in the overall equation but not in the rate equation joins after that step.
Why does a small temperature rise increase rate so much?
Raising the temperature shifts the Boltzmann distribution so that a much larger fraction of molecules exceed the activation energy. The increase in the frequency of collisions is comparatively minor — it is the fraction of successful collisions that matters.
