Lesson 12: HL: Rate Expressions and Mechanisms

Higher Level kinetics replaces description with law. The rate expression (rate law) for a reaction states rate = k[A]^m[B]^n, where m and n are the orders with respect to each reactant — determined only by experiment, never read from the balanced equation — and k is the rate constant, whose value depends on temperature and catalyst but not on concentrations. The overall order is m + n. This lesson (Reactivity 2.2, HL) builds the complete method: deducing orders from initial-rate data, calculating k and its chameleon units, reading order from graphs and half-lives, extracting activation energy with the Arrhenius equation, and proposing mechanisms whose rate-determining step is consistent with the observed law.

Initial rates: vary one thing at a time

The method of initial rates runs several experiments with different starting concentrations and measures the initial rate (tangent at t = 0, before concentrations drift). Compare two runs where only [A] changes: if doubling [A] doubles the rate, the reaction is first order in A; if it quadruples the rate, second order; if nothing changes, zero order. The rate constant then falls out of any single run, and its units must be derived, never guessed: rearranging, k = rate ÷ [A]^m[B]^n. For an overall second-order reaction k has units mol⁻¹ dm³ s⁻¹; for third order, mol⁻² dm⁶ s⁻¹; for first order, simply s⁻¹. Zero-order reactions have a constant rate (often surface-saturated catalysis), and on a concentration–time graph give a straight line whose gradient is −k.

Half-life and the Arrhenius equation

For a first-order reaction the half-life t₁/₂ = ln 2 ÷ k is constant — the fingerprint that identifies first order from raw data: successive half-lives are equal. Radioactive decay and many decompositions live here. Temperature enters through the Arrhenius equation, k = A·e^(−E_a/RT): the exponential factor is the fraction of collisions with energy ≥ E_a, and A (the frequency factor) packages collision frequency and orientation. In linear form, ln k = −(E_a ÷ R)(1 ÷ T) + ln A, so a plot of ln k against 1/T is a straight line with gradient −E_a ÷ R. Two-point calculations use ln(k₂ ÷ k₁) = −(E_a ÷ R)(1 ÷ T₂ − 1 ÷ T₁). Because k rises exponentially, a modest E_a makes rates fiercely temperature-sensitive — the quantitative skeleton under Lesson 11's "doubling per 10 K" rule of thumb.

Worked example 1: orders, k, and its units from data

For A + B → products, initial rates (mol dm⁻³ s⁻¹): run 1: [A] = 0.100, [B] = 0.100, rate = 4.0 × 10⁻⁵. Run 2: [A] = 0.200, [B] = 0.100, rate = 8.0 × 10⁻⁵. Run 3: [A] = 0.100, [B] = 0.200, rate = 1.6 × 10⁻⁴. Runs 1 → 2: [A] doubles, rate doubles → first order in A. Runs 1 → 3: [B] doubles, rate ×4 → second order in B. Rate = k[A][B]². From run 1: k = 4.0 × 10⁻⁵ ÷ (0.100 × 0.100²) = 4.0 × 10⁻⁵ ÷ 1.0 × 10⁻³ = 4.0 × 10⁻² mol⁻² dm⁶ s⁻¹. Check with run 3: 0.040 × 0.100 × 0.0400 = 1.6 × 10⁻⁴ ✓. Verifying k against a second run is a two-second habit that catches most order errors before they metastasise through a Paper 2 question.

Worked example 2: an iodine-clock Arrhenius investigation (practical)

In the iodine clock, thiosulfate mops up iodine as it forms; when it is exhausted, the starch turns blue-black — a sharp, low-judgement endpoint, which is why this experiment survives into serious coursework. Hold concentrations fixed and run at five temperatures (water baths, 20–60 °C); rate ∝ 1 ÷ t, and since concentrations are identical in every run, k ∝ 1 ÷ t, so ln(1 ÷ t) plotted against 1 ÷ T has gradient −E_a ÷ R. Suppose the gradient is −6500 K: E_a = 6500 × 8.31 = 54.0 kJ mol⁻¹. The evaluation is where this becomes IA material: plot maximum and minimum gradient lines through the error bars (temperature ±0.5 K from the thermometer, time ±0.2 s on the clock change) to give E_a with an uncertainty, not just a value; note that 1 ÷ t is only proportional to k because the *extent* of reaction at the endpoint is identical each run — if thiosulfate amounts varied, the proportionality collapses; and record that the water-bath temperature is the reactants' temperature only after equilibration, so pre-warm the solutions separately before mixing. A student who can raise those three points unprompted is writing at distinction level.

Mechanisms: the rate law as witness

A mechanism is a sequence of elementary steps; the slowest — the rate-determining step (RDS) — sets the rate law, whose reactants are exactly the species colliding in that step (and any species formed in a fast pre-equilibrium before it). The test of a proposed mechanism is consistency: its RDS must reproduce the experimental orders, and its steps must sum to the overall equation. NO₂ + CO → NO + CO₂ is first order in NO₂ squared and zero order in CO, explained by two steps: 2NO₂ → NO₃ + NO (slow), then NO₃ + CO → NO₂ + CO₂ (fast) — CO enters only after the RDS, hence its absence from the law. Intermediates (here NO₃) are produced and consumed within the mechanism and never appear in the overall equation; catalysts are consumed early and regenerated late. If your proposed intermediate also appears in the rate law via a pre-equilibrium, its concentration must be substituted out using the equilibrium — the origin of fractional and inverse orders that occasionally appear in HL questions.

Nature of Science — and the real world

Rate laws are pure empiricism: no amount of theory reading a balanced equation can predict an order, and the history of kinetics is full of beautiful mechanisms demolished by one clean initial-rates data set. The Arrhenius equation (1889) pre-dated the collision theory that later explained it — mathematics first, mechanism after. In the real world, this mathematics sets drug dosages (first-order elimination half-lives), tunes combustion engines, models atmospheric ozone destruction (whose steps were deduced exactly this way), and underlies every "use within 28 days" printed on a reagent bottle.

Common mistakes

• Reading orders from stoichiometric coefficients. Orders are experimental; the equation is silent.
• Quoting k without units, or with first-order units for every reaction. Derive them from the rate law each time.
• Using Celsius temperatures in the Arrhenius equation. T is always kelvin.
• Proposing mechanisms whose steps do not sum to the overall equation, or that feature termolecular collisions of three particles as an elementary step without comment (vanishingly unlikely).

IB Exam Experience

Question 1. For the reaction X + Y → Z, doubling [X] at constant [Y] doubles the rate, and doubling [Y] at constant [X] quadruples the rate. What is the rate expression?

• (A) Rate = k[X][Y]
• (B) Rate = k[X]²[Y]
• (C) Rate = k[X][Y]²
• (D) Rate = k[X]²[Y]²

Question 2. A first-order reaction has a half-life of 120 s. What fraction of the reactant remains after 360 s?

• (A) 1/2
• (B) 1/4
• (C) 1/6
• (D) 1/8

Question 3. What are the units of the rate constant for a reaction that is overall third order?

• (A) s⁻¹
• (B) mol dm⁻³ s⁻¹
• (C) mol⁻¹ dm³ s⁻¹
• (D) mol⁻² dm⁶ s⁻¹

Answer key. 1. (C) — first order in X, second order in Y. 2. (D) — 360 s is three half-lives: (1/2)³ = 1/8 remains. 3. (D) — k = rate ÷ concentration³ = (mol dm⁻³ s⁻¹) ÷ (mol³ dm⁻⁹) = mol⁻² dm⁶ s⁻¹.

Structured question (Paper 2 style). The gas-phase reaction 2NO(g) + O₂(g) → 2NO₂(g) is studied by the initial-rates method.

(a) Deduce the orders with respect to NO and O₂, and write the rate expression. [3] (b) Calculate the value of the rate constant, including its units. [2] (c) A proposed mechanism is: NO + O₂ ⇌ NO₃ (fast equilibrium); NO₃ + NO → 2NO₂ (slow). State whether this mechanism is consistent with the rate expression, giving your reasoning. [2] Marking notes. (a) Runs 1→2: [NO] doubles, rate ×4 → second order in NO [1]; runs 1→3: [O₂] doubles, rate ×2 → first order in O₂ [1]; rate = k[NO]²[O₂] [1]. (b) k = 1.0 × 10⁻⁴ ÷ (0.010² × 0.010) = 1.0 × 10² mol⁻² dm⁶ s⁻¹ [1 for value, 1 for units]. (c) Yes — the slow step uses NO₃ and NO; the fast pre-equilibrium makes [NO₃] ∝ [NO][O₂], so the RDS rate ∝ [NO][O₂] × [NO] = [NO]²[O₂] [2 for the substitution reasoning; a bare "yes" scores 0].

IA idea: Use an iodine-clock (or magnesium–acid) system to determine the order of reaction with respect to one reactant from initial-rate data, with a full log–log analysis and uncertainty treatment.

A two-point Arrhenius calculation, and the zero-order corner

Two-point Arrhenius, executed cleanly: a reaction's rate constant triples between 298 K and 308 K (k₂ ÷ k₁ = 3). ln(k₂ ÷ k₁) = −(E_a ÷ R)(1 ÷ T₂ − 1 ÷ T₁) ln 3 = 1.0986; (1 ÷ 298 − 1 ÷ 308) = 1.089 × 10⁻⁴ K⁻¹ (note the sign handling: 1 ÷ T₂ − 1 ÷ T₁ is negative, cancelling the leading minus). E_a = 1.0986 × 8.31 ÷ 1.089 × 10⁻⁴ = 83.8 kJ mol⁻¹. A tripling per 10 K near room temperature corresponds to this substantial barrier — compare the ~54 kJ mol⁻¹ behind the commoner doubling rule. The zero-order corner deserves its paragraph: when a catalyst's surface is saturated, or a reagent is in such excess that its concentration is effectively constant, rate stops depending on concentration — the concentration–time graph is a straight line of gradient −k, and the half-life shrinks as the reaction proceeds (t₁/₂ = [A]₀ ÷ 2k). Enzyme catalysis at high substrate concentration lives here (the bridge into biochemistry's Michaelis–Menten world), as does the decomposition of ammonia on hot tungsten. In exams, zero order is recognised by constant rate in the data or by the straight concentration–time plot — and by the giveaway that changing the concentration changes nothing, the result students least trust and most often need.