HookWhy the 4 pm flat white is still with you at 2 am
Caffeine leaves a healthy adult's bloodstream with a half-life of roughly five hours. Drink a 200 mg flat white at 4 pm and about 100 mg is still circulating at 9 pm, and about 50 mg at 2 am — which is why the late coffee wrecks sleep so reliably. The revealing detail is what the half-life doesn't depend on: the dose. Halving takes the same five hours whether you start from 200 mg or from 50. A constant half-life is the fingerprint of a first-order process — rate directly proportional to concentration — and it is the same mathematics that governs radioactive decay and half the reactions in this section.
3.1.9 is kinetics grown up. AS collision theory (3.1.5) explained qualitatively why reactions speed up; this section turns rate into a precise algebraic claim — the rate equation — whose exponents are discovered by experiment, never copied from the balanced equation. The prize is bigger than prediction: because only the steps up to and including the slowest one control the rate, the orders you measure are evidence about a mechanism you can never watch directly. Kinetics is chemistry's espionage, and Required Practical 7 is the fieldcraft.
ModelThe rate equation — written by experiment, never copied
For a reaction involving species A and B, the rate equation has the form \(\text{rate} = k[\text{A}]^m[\text{B}]^n\). The powers \(m\) and \(n\) are the orders with respect to each reactant, their sum is the overall order, and \(k\) is the rate constant — fixed for a given reaction at a given temperature. Rate itself is the change in concentration per unit time, measured in \(\text{mol dm}^{-3}\,\text{s}^{-1}\).
The creed of the whole section: orders bear no necessary relation to the balanced equation's coefficients. They are determined only by experiment. (The contrast is deliberate — in the \(K_c\) expression of 3.1.6 the powers are the coefficients, and examiners set questions specifically to catch students transferring the habit.) The three orders you meet: zero order means concentration is irrelevant to rate; first order means doubling the concentration doubles the rate; second order means doubling it quadruples the rate.
Because the orders vary, \(k\) has whatever units make the rate equation dimensionally honest — so its units change with overall order and must be re-derived every time, not remembered.
Derive the units of \(k\) for an overall second-order reaction, say \(\text{rate} = k[\text{A}][\text{B}]\). Rearrange and substitute units: \[ k = \frac{\text{rate}}{[\text{A}][\text{B}]} = \frac{\text{mol dm}^{-3}\,\text{s}^{-1}}{\text{mol}^2\,\text{dm}^{-6}} = \text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1} \] The same algebra gives zero order \(\text{mol dm}^{-3}\,\text{s}^{-1}\), first order \(\text{s}^{-1}\), third order \(\text{mol}^{-2}\,\text{dm}^{6}\,\text{s}^{-1}\). Pause on first order: \(k\) is a pure "per second", with no concentration in it at all — which is the deep reason a first-order half-life is constant, and why caffeine's five hours never changes with the dose.
MechanismInitial-rate tables — the doubling game
The initial rate is the gradient of a concentration–time curve at \(t = 0\), before products accumulate and complicate things. Given a table of initial-rate experiments, the deduction method is mechanical: find two experiments in which only one concentration changes, then compare the factor by which the rate changed with the factor by which that concentration changed. Rate unchanged: zero order. Same factor: first order. Factor squared (doubling gives \(\times 4\)): second order.
Once every order is pinned, calculate \(k\) by substituting any complete row into the rate equation. The marks here are AO3 — analysis — and they are awarded for the reasoning sentence, not the bare answer: name the two experiments, state what was held constant, state both factors, conclude the order.
Peroxodisulfate oxidises iodide: \(\text{S}_2\text{O}_8^{2-} + 2\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_2\). Initial-rate data at 25 °C: experiment 1 has \([\text{S}_2\text{O}_8^{2-}] = 0.020\), \([\text{I}^-] = 0.10\) (both \(\text{mol dm}^{-3}\)) and rate \(1.4 \times 10^{-5}\ \text{mol dm}^{-3}\,\text{s}^{-1}\); experiment 2: \(0.040\), \(0.10\), rate \(2.8 \times 10^{-5}\); experiment 3: \(0.040\), \(0.20\), rate \(5.6 \times 10^{-5}\). Comparing 1 and 2: peroxodisulfate doubles, iodide constant, rate doubles — first order in \(\text{S}_2\text{O}_8^{2-}\). Comparing 2 and 3: iodide doubles, rate doubles — first order in \(\text{I}^-\). So \(\text{rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\), overall order 2. From experiment 1: \[ k = \frac{1.4 \times 10^{-5}}{0.020 \times 0.10} = 7.0 \times 10^{-3}\ \text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1} \] Notice the balanced equation's \(2\text{I}^-\) did not make the reaction second order in iodide. The data outranks the stoichiometry, every time.
ModelGraph shapes — orders you can see
Continuous monitoring produces a concentration–time curve, and its shape declares the order. Zero order: a straight line sloping down — constant gradient, because the rate ignores the falling concentration. First order: exponential decay with a constant half-life — successive halvings take equal times, caffeine's five hours again. Second order: also a decay curve, but deeper-and-flatter, with the half-life lengthening as concentration falls. Rate at any moment is the gradient of a tangent; for a reactant the rate is minus that gradient, and the initial rate is the tangent ruled at \(t = 0\).
Measure several tangents and you can re-plot as a rate–concentration graph, which is even more diagnostic: a horizontal line is zero order; a straight line through the origin is first order (gradient \(= k\)); an upward curve suggests second order — confirmed if plotting rate against \([\text{X}]^2\) straightens it. Exam questions hand you one representation and ask for another, so practise moving between the two.
MechanismThe rate-determining step — reading a mechanism from the orders
Most reactions are sequences of elementary steps, and the overall rate is throttled by the slowest — the rate-determining step (RDS). Hence the rule that turns kinetics into espionage: the rate equation contains exactly the species involved up to and including the RDS. The order with respect to a species equals the number of its particles participating by that point; a species that is zero order enters the mechanism after the slow step.
Worked case: \(\text{NO}_2 + \text{CO} \rightarrow \text{NO} + \text{CO}_2\) is observed (at lower temperatures) to obey \(\text{rate} = k[\text{NO}_2]^2\) — carbon monoxide is missing and two \(\text{NO}_2\) appear. A consistent mechanism: slow step \(2\text{NO}_2 \rightarrow \text{NO}_3 + \text{NO}\), then fast step \(\text{NO}_3 + \text{CO} \rightarrow \text{NO}_2 + \text{CO}_2\). Check it two ways: the slow step uses two \(\text{NO}_2\) and no CO (matching the orders), and adding the steps cancels the \(\text{NO}_3\) intermediate to recover the overall equation. Any proposed mechanism must pass both tests.
The same logic runs through organic chemistry. Acid-catalysed iodination of propanone is zero order in iodine — iodine strikes only after the slow step. And hydrolysis of a tertiary halogenoalkane shows \(\text{rate} = k[\text{RBr}]\): the C–Br bond breaks by itself in the slow step, with the hydroxide arriving later — kinetic evidence telling apart two substitution mechanisms that give identical products.
ModelArrhenius — how temperature gets inside k
\(k\) is only constant at constant temperature, and the Arrhenius equation says exactly how it moves: \(k = Ae^{-E_a/RT}\). The exponential factor is the fraction of collisions with energy at least the activation energy \(E_a\); the pre-exponential factor \(A\) reflects how often suitably oriented collisions occur. Two consequences drop out. Raising \(T\) grows \(k\) exponentially — the origin of the rough rule that rate doubles for a 10 K rise near room temperature. And a catalyst grows \(k\) by shrinking \(E_a\), attacking the exponent directly.
Take natural logs and the equation straightens: \[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \] Plot \(\ln k\) against \(1/T\) and you get a straight line with gradient \(-E_a/R\) and intercept \(\ln A\), with \(R = 8.31\ \text{J K}^{-1}\,\text{mol}^{-1}\). This plot is how real activation energies are measured, and AQA examines it both graphically and by direct substitution.
A student measures \(k\) at five temperatures and plots \(\ln k\) against \(1/T\); the gradient comes out at \(-9.65 \times 10^{3}\ \text{K}\). Then \[ E_a = -\text{gradient} \times R = 9.65 \times 10^{3} \times 8.31 = 8.02 \times 10^{4}\ \text{J mol}^{-1} = 80.2\ \text{kJ mol}^{-1} \] Two habits protect the marks: the gradient of this plot is always negative (the line slopes down), and the two minus signs cancel to give a positive \(E_a\); and because \(R\) is in joules, the answer arrives in \(\text{J mol}^{-1}\) — convert to \(\text{kJ mol}^{-1}\) at the end, not before. The exam also runs the equation forwards: given \(E_a\), \(A\) and \(T\), evaluate the exponent \(-E_a/RT\) first and write it down — that line is the method mark — then raise \(e\) to it for \(k\).
CaseRequired practical 7 — the clock and the curve
RP7 asks you to measure rate two complementary ways. Initial rates — the iodine clock. Hydrogen peroxide oxidises iodide in acid: \(\text{H}_2\text{O}_2 + 2\text{I}^- + 2\text{H}^+ \rightarrow \text{I}_2 + 2\text{H}_2\text{O}\). The flask also contains starch and a small, fixed amount of sodium thiosulfate, which silently recycles iodine back to iodide (\(2\text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\)) until it runs out — then free iodine meets starch and the mixture snaps blue-black. Because every run makes the same amount of iodine before the flash, the initial rate is proportional to \(1/t\). Method discipline: change one concentration between runs and top up with water to the same total volume, so every other species' concentration stays constant — that named trick is the control-variable mark. Run in a thermostatted water bath: temperature is the control that matters most, because \(k\) responds to it exponentially. Plot \(1/t\) against concentration and read the order from the shape.
Continuous monitoring — propanone and iodine. With an acid catalyst, propanone slowly strips the brown colour from iodine. Follow it in a colorimeter (blue filter — the complement of the solution's colour — and a calibration curve converting absorbance to \([\text{I}_2]\)), or withdraw timed samples and quench each in sodium hydrogencarbonate — killing the \(\text{H}^+\) catalyst freezes the composition at the sampling moment — then titrate the remaining iodine with thiosulfate. Either way you get a concentration–time curve, tangents, and rates.
Errors, honestly stated. The clock's \(1/t\) treats rate as constant across the timed interval — only fair if just a few per cent of reactant is consumed before the flash. The eye-judged colour change adds reaction-time error (mitigated by the sharpness of the clock endpoint); temperature drift between runs skews everything; and a colorimeter with a data logger beats the eye on both counts, sampling continuously without disturbing the flask.
In an iodine-clock run, doubling \([\text{I}^-]\) (with water making the volume back up) cuts the time to blue-black from 48 s to 24 s. Then \(1/t\) rises from \(1/48 = 0.021\ \text{s}^{-1}\) to \(1/24 = 0.042\ \text{s}^{-1}\) — the rate has doubled, so the reaction is first order in iodide. Prediction test: with \([\text{I}^-]\) tripled, rate should triple, so \(t \approx 48/3 = 16\ \text{s}\). If the stopwatch says 16 s, the order stands; if it drifts to 20 s or more, suspect the water bath before you suspect the chemistry — a few degrees of cooling suppresses \(k\) more than most pipetting errors ever could.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Re-derive the units of \(k\) every single time — they depend on overall order, and AQA awards them as a separate mark. When deducing orders from a table (AO3), write the full comparison sentence: "between experiments 1 and 2, \([\text{I}^-]\) doubles while \([\text{S}_2\text{O}_8^{2-}]\) is constant, and the rate doubles, so the reaction is first order in \(\text{I}^-\)" — the reasoning carries the mark, and an asserted order without the comparison drops it.
Calculations follow the M1–M2–M3 rhythm: rearranged expression, substitution with units visible, evaluated answer. On Arrhenius questions convert at the end (\(R\) is in joules, so \(E_a\) lands in \(\text{J mol}^{-1}\)), state that the gradient equals \(-E_a/R\), and expect the reverse task of reading \(\ln A\) from the intercept. On graphs, rule tangents — a freehand tangent at \(t = 0\) is the most commonly lost practical-skills mark — and label axes with quantity and unit.
RP7 questions are method questions in disguise, often a 6-mark extended response: give the sequence (fixed volumes, thermostatted bath, start clock on mixing), name the one variable changed and the water top-up that holds every other concentration constant, justify the quench (removes the catalyst, freezing composition) and justify \(1/t\) as a rate proxy (same iodine made each run, small fraction reacted). Finally, connect forward: the rate-determining-step logic reappears in halogenoalkane mechanisms (3.3.3) and in catalysis (3.2.5.6), and examiners reward candidates who use it there unprompted.