Kinetics
Collision theory, Maxwell–Boltzmann, rate equations, Arrhenius
Maxwell–Boltzmann distribution
The Maxwell–Boltzmann distribution shows the spread of molecular energies. The area under the curve equals the total number of molecules, and it stays constant.
At higher temperature the peak shifts right and sits lower, so many more molecules exceed Ea — that is why rate rises so steeply. A catalyst provides an alternative route with a lower Ea: the curve itself doesn't change, but the Ea line moves left.
Rate equations (A-level year 2)
rate = k[A]m[B]nThe orders m and n come only from experiment — never from the coefficients of the chemical equation. The rate-determining step contains the species that appear in the rate equation.
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
- Comparing experiments 1 and 2: [A] doubles while [B] stays constant, and the rate doubles — so the reaction is first order in A.
- Comparing experiments 2 and 3: [B] doubles while [A] stays constant, and the rate goes up ×4 — second order in B.
- So rate = k[A][B]², giving an overall order of 3. The units of k are rate ÷ (mol dm⁻³)³ = dm⁶ mol⁻² s⁻¹.
- Find k by substituting any one experiment's numbers into the rate equation.
3.1.9.1Concentration–time vs rate–concentration graphs
On a [X]–time graph, zero order gives a straight line down and first order gives a curve with a constant half-life. The rate at any time t is the gradient of the tangent at that point.
On a rate–[X] graph, zero order is a horizontal line, first order is a straight line through the origin, and second order is an upward curve.
3.1.9.2Using the Arrhenius equation
Because ln k = ln A − Ea/(RT), a plot of ln k (y) against 1/T (x) gives a gradient of −Ea/R and an intercept of ln A. The gradient has units of K: multiply it by −R (8.31 J K⁻¹ mol⁻¹) to get Ea in J mol⁻¹, then divide by 1000 for kJ mol⁻¹.
Deep dive📚 The rest of the chapter, in full
What each change does to the Maxwell–Boltzmann curve
| Change | Curve | Ea line | Why rate rises |
|---|---|---|---|
| Raise temperature | Flattens, peak moves right; area constant | Fixed | Far more molecules exceed Ea; also more frequent collisions (minor) |
| Add catalyst | Unchanged | Moves left | Alternative route with lower Ea — more of the existing distribution qualifies |
| Raise concentration/pressure | Taller (more molecules), same shape | Fixed | More 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.
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
Follow gas volume with a syringe, or mass loss from an open flask on a balance, plotted against time: the gradient of the tangent gives the rate at that moment, and the tangent at t = 0 gives the initial rate. Colorimetry suits coloured species (e.g. the iodine or bromine concentration falling) — it is continuous and non-invasive. Clock methods (RP3/7) time a fixed visible event, and the initial rate ∝ 1/t.
Extended🎓 Beyond the standard course
Molecularity ≠ order. Molecularity counts the particles in one elementary step and is always a whole number; order is experimental and can be zero or fractional. The two coincide only for single-step reactions.
Mechanism consistency test. NO₂ + CO → NO + CO₂ has rate = k[NO₂]², so the slow step must use two NO₂ (NO₂ + NO₂ → NO₃ + NO), with CO mopping up NO₃ in a fast step afterwards. CO is absent from the rate law because it enters after the rate-determining step.
Intermediates vs transition states. An intermediate sits in an energy dip and is isolable in principle. A transition state is the summit — a bond-breaking/forming instant with no lifetime.
Heterogeneous catalysis in three verbs: adsorb (reactants bond to the surface and their bonds weaken) → react (via the lower-Ea pathway) → desorb. Catalyst poisoning (Pb on catalytic converters) blocks the adsorption sites permanently.
Autocatalysis (MnO₄⁻/C₂O₄²⁻) gives an S-shaped concentration–time curve: a slow start, acceleration as the catalytic product accumulates, then a tail-off as the substrate runs out.
Mastery vault🏛 Every remaining spec point, banked
Reading the two graph families
On a concentration–time graph, zero order gives a straight line down with constant gradient. First order gives a curve with a constant half-life — the definitive test: measure two successive half-lives, and if they are equal the order is 1. Second order gives a curve with lengthening half-lives.
On a rate–concentration graph, zero order is a horizontal line, first order is a straight line through the origin (gradient = k), and second order is an upward curve — plot rate against concentration² to straighten it. Know which graph you have been given before saying anything: mixing the families is the classic self-inflicted wound.
Initial-rates table, deduced properly
| Expt | [X] | [Y] | Rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻⁴ |
| 3 | 0.20 | 0.30 | 2.4 × 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
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
The observed orders count how many of each species appear in steps up to and including the RDS. Given rate = k[NO]²[O₂], a plausible mechanism is 2NO ⇌ N₂O₂ (fast) then N₂O₂ + O₂ → 2NO₂ (slow) — the slow step's effective composition matches the rate law.
Species appearing AFTER the RDS never feature in the rate equation. Catalysts CAN feature, because they act before or at the RDS, and intermediates must not appear in the overall equation. The reverse test works too: a proposed mechanism whose slow step disagrees with the observed orders is simply wrong — say so and why.
Half-life, k units, and deducing mechanisms
This section digs into consequences of the rate law you have already met: the constant half-life of first-order reactions, how the units of k are fixed by the overall order, and how a proposed mechanism is tested against a rate equation.
Half-life (t½): the time taken for the concentration of a reactant to fall to half its value. For a first-order reaction t½ is constant — independent of starting concentration. This constancy is the diagnostic signature of first order.
Because a first-order rate is proportional to concentration, halving the concentration halves the rate, so each successive halving takes the same time: read three or four equal successive half-lives off a concentration–time curve and the order is 1. For zero order the half-life shortens each time (rate is constant, so it takes less time to lose the next half); for second order it lengthens.
Worked: t½ → k. A first-order decomposition has a measured half-life of 120 s. Since k = ln2 / t½, k = 0.693 / 120 = 5.8 × 10⁻³ s⁻¹. Check the units: first-order k always carries s⁻¹. If the concentration is 0.80 mol dm⁻³ at that instant, the rate = k[A] = 5.8 × 10⁻³ × 0.80 = 4.6 × 10⁻³ mol dm⁻³ s⁻¹, which has the correct rate units.
Deducing the units of k. Rate always has units mol dm⁻³ s⁻¹. Rearranging rate = k[A]ᵐ[B]ⁿ and cancelling gives k units of (mol dm⁻³)(1−overall order) s⁻¹:
| Overall order | Rate law example | Units of k |
| 0 | rate = k | mol dm⁻³ s⁻¹ |
| 1 | rate = k[A] | s⁻¹ |
| 2 | rate = k[A]² | mol⁻¹ dm³ s⁻¹ |
| 3 | rate = k[A]²[B] | mol⁻² dm⁶ s⁻¹ |
Fast unit rule: start from mol dm⁻³ s⁻¹ and remove one factor of mol dm⁻³ for every order above zero. Each removed (mol dm⁻³) flips to mol⁻¹ dm³.
Mechanism must match the rate equation. Only species involved up to and including the rate-determining step (RDS) appear in the rate law; anything that reacts only after the RDS does not. For the reaction with experimental rate = k[NO₂]², a proposed two-step mechanism with a slow first step 2NO₂ → NO₃ + NO followed by a fast NO₃ + CO → NO₂ + CO₂ is consistent: the slow step involves two NO₂ and no CO, matching the observed zero order in CO. A one-step mechanism NO₂ + CO → NO + CO₂ would predict rate = k[NO₂][CO], which disagrees, so it is rejected.
Temperature and k, quantified. A rough rule is that rate roughly doubles per 10 °C rise. Via Arrhenius k = Ae^(−Eₐ/RT), going from 298 K to 308 K increases the fraction of molecules with energy ≥ Eₐ sharply; for a typical Eₐ ≈ 50 kJ mol⁻¹ this multiplies k by about 1.9 — hence the "doubling" guideline. It is k, not the order, that changes with temperature.
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