Lesson 23: Kinetics: Rate Equations (HL) HL ONLY

Lesson 12 built the HL kinetics toolkit: orders from initial rates, k and its units, the Arrhenius equation, mechanisms. This lesson (Reactivity 2.2, HL extension) turns that toolkit onto harder game: integrated rate laws that use the whole concentration–time curve rather than its first seconds, half-life fingerprints read directly from data, activation energies extracted from real plotted lines with honest uncertainties, and mechanisms involving pre-equilibria, where algebra — not intuition — produces the rate law. The unifying principle never changes: a kinetic claim is only as good as the data treatment behind it, and at HL the treatment is graphical and logarithmic.

Integrated rate laws: straight lines as verdicts

For a reaction A → products, the three simple orders integrate into three testable straight lines:

The workflow: plot all three candidate graphs; the one that straightens identifies the order, and its gradient gives k (with the units to match). Half-life behaviour is the quick diagnostic: constant half-life ⇒ first order, no exceptions for simple kinetics. Concentration–time data from absorbance (Beer–Lambert: absorbance ∝ concentration — the colorimeter is the kinetics workhorse), gas volume, or pH all feed the same plots. Note the strategic difference from initial rates: integrated laws use every data point (efficient, and sensitive to mechanism changes mid-reaction — if the "straight" line bends late, something changed: product inhibition, a reverse reaction approaching equilibrium, temperature drift), while initial rates sidestep those complications by never letting the reaction get old. HL questions expect you to choose between them with a reason.

Activation energy from plotted data — and its uncertainty

With k at several temperatures, ln k against 1 ÷ T gives gradient −E_a ÷ R and intercept ln A. Two-point spot calculations are the fast version: ln(k₂ ÷ k₁) = −(E_a ÷ R)(1 ÷ T₂ − 1 ÷ T₁). The HL difference is the error analysis: draw the steepest and shallowest lines consistent with the error bars (temperature ±0.5 K matters at the 1 ÷ T axis; k inherits the uncertainty of the rate measurement), and quote E_a as a value ± a range. An E_a of 54 ± 6 kJ mol⁻¹ from five honest points outranks a bare 54.0 from two — examiners are buying judgement, not arithmetic.

Worked example 1: half-life data — order and k without plotting

A reaction's concentration falls: 0.800 → 0.400 mol dm⁻³ in 120 s; → 0.200 in a further 120 s; → 0.100 in a further 120 s. Successive half-lives are equal (120 s each), so the reaction is first order in A. k = ln 2 ÷ t₁/₂ = 0.693 ÷ 120 = 5.78 × 10⁻³ s⁻¹. Now extend: time for 90% consumption means [A] ÷ [A]₀ = 0.100: t = ln(1 ÷ 0.100) ÷ k = 2.303 ÷ 5.78 × 10⁻³ = 398 s. And the concentration after 600 s: ln[A] = ln(0.800) − 5.78 × 10⁻³ × 600 = −0.223 − 3.47 → [A] = 0.800 × e^(−3.47) = 0.0249 mol dm⁻³. Three marks' worth of follow-through from one fingerprint observation — which is why half-life questions are HL staples.

Worked example 2: an Arrhenius investigation with real apparatus (practical)

The hydrolysis of 2-chloro-2-methylpropane (tert-butyl chloride) in aqueous ethanol is first order and conveniently measurable by conductivity or by titrating the HCl produced; run it in water baths at 15, 25, 35, 45 and 55 °C, extract k at each temperature from ln[RX]-against-t plots, then plot ln k against 1 ÷ T. Suppose the fitted gradient is −7840 K: E_a = 7840 × 8.31 = 65.1 kJ mol⁻¹. The practical intelligence lives in the controls: the substrate hydrolyses measurably fast, so solutions must be pre-equilibrated in the bath *separately* and mixed at temperature (mixing cold solutions and waiting for warm-up fabricates a slow start and bends every ln plot); the solvent composition must be identical in every run (the rate of an S_N1 hydrolysis is exquisitely solvent-sensitive — water content changes the mechanism's energetics, Lesson 25); and at the highest temperature, evaporation changes concentrations, so seal the vessels. Error bars: temperature ±0.5 K maps to ±0.5 ÷ T² on the 1 ÷ T axis — visible at this precision; k's uncertainty from the ln[A] plot's gradient spread compounds it; the final E_a should carry the steepest/shallowest-line range (perhaps ±5 kJ mol⁻¹). Stating that range *and* the dominant contributor is the analysis criterion, demonstrated.

Mechanisms with pre-equilibrium: the algebra of the rate law

When a fast equilibrium precedes the slow step, the intermediate's concentration is not measurable directly — eliminate it. Canonical example: 2NO + O₂ → 2NO₂ is third order overall (second order in NO). Mechanism: NO + O₂ ⇌ NO₃ (fast, equilibrium constant K), then NO₃ + NO → 2NO₂ (slow, k₂). From the equilibrium, [NO₃] = K[NO][O₂]; the rate = k₂[NO₃][NO] = k₂K[NO]²[O₂] — the observed third-order law, with k_obs = k₂K. Predictions follow: because K for an exothermic pre-equilibrium falls as T rises, some such reactions show a *negative* apparent activation energy — rate decreasing with temperature, as this one famously does — a result no single-step picture can explain and a gift of an exam discussion. The general discipline: write each step, identify the RDS, express every intermediate through equilibria before it, and check the final law against experiment; disagreement falsifies the mechanism, agreement merely acquits it — mechanisms are proven innocent, never guilty-free beyond doubt, and that asymmetry is proper Popperian science.

Nature of Science — and the real world

Chemical kinetics is where chemistry meets differential equations, and its history (van 't Hoff, Arrhenius, Lindemann's resolution of the unimolecular paradox) shows mathematics repeatedly rescuing mechanism. The Lindemann story deserves the telling: first-order gas reactions seemed to require collisions for activation yet showed first-order kinetics — resolved by separating activation (collisional) from reaction (unimolecular, delayed), an early triumph of mechanism over paradox. Modern stakes: atmospheric chemists model ozone and methane lifetimes with exactly these integrated laws; pharmacokinetics doses patients by half-lives; and every accelerated-ageing test in industry is an Arrhenius plot with money riding on the extrapolation.

Common mistakes

• Plotting ln k against T (not 1 ÷ T), or using Celsius anywhere in Arrhenius work.
• Claiming constant half-life "proves zero order" — it is the signature of first order.
• Forgetting that a second-order half-life depends on [A]₀, so quoting "the half-life" of a second-order reaction without its concentration is meaningless.
• Writing a rate law containing an intermediate — intermediates must be eliminated via the pre-equilibrium before comparison with experiment.

IB Exam Experience

Question 1. A reaction has a constant half-life of 40 s at a given temperature, independent of the starting concentration. What is the order, and the rate constant?

• (A) Zero order; k = 0.025 s⁻¹
• (B) First order; k = 0.0173 s⁻¹
• (C) First order; k = 40 s⁻¹
• (D) Second order; k = 0.0173 mol⁻¹ dm³ s⁻¹

Question 2. For a first-order reaction, which plot is linear?

• (A) [A] against t
• (B) 1 ÷ [A] against t
• (C) ln[A] against t
• (D) [A]² against t

Question 3. The rate constant of a reaction doubles when the temperature rises from 300 K to 310 K. Which estimate of E_a is consistent with this? (R = 8.31 J K⁻¹ mol⁻¹)

• (A) ≈ 5 kJ mol⁻¹
• (B) ≈ 54 kJ mol⁻¹
• (C) ≈ 540 kJ mol⁻¹
• (D) ≈ 0.54 kJ mol⁻¹

Answer key. 1. (B) — constant half-life ⇒ first order; k = ln 2 ÷ 40 = 0.0173 s⁻¹. 2. (C) — the integrated first-order law is ln[A] = ln[A]₀ − kt. 3. (B) — ln 2 = −(E_a ÷ 8.31)(1 ÷ 310 − 1 ÷ 300) = (E_a ÷ 8.31)(1.075 × 10⁻⁴) → E_a = 0.693 × 8.31 ÷ 1.075 × 10⁻⁴ ≈ 53 600 J mol⁻¹ ≈ 54 kJ mol⁻¹.

Structured question (Paper 2 style). The decomposition of N₂O₅ in solution is first order: 2N₂O₅ → 4NO₂ + O₂. At 45 °C the concentration of N₂O₅ falls from 0.0400 to 0.0300 mol dm⁻³ in 200 s. (a) Calculate the rate constant at 45 °C. [2] (b) Calculate the half-life of the reaction at this temperature. [1] (c) Calculate the concentration remaining after 600 s from the same start. [2] (d) A student claims the reaction must be elementary (a single step) because the rate law is first order in N₂O₅. Evaluate the claim. [1] Marking notes. (a) k = −ln([A] ÷ [A]₀) ÷ t = −ln(0.0300 ÷ 0.0400) ÷ 200 = 0.2877 ÷ 200 = 1.44 × 10⁻³ s⁻¹ [2]. (b) t₁/₂ = ln 2 ÷ k = 0.693 ÷ 1.44 × 10⁻³ = 481 s [1]. (c) [A] = 0.0400 × e^(−1.44 × 10⁻³ × 600) = 0.0400 × e^(−0.863) = 0.0400 × 0.422 = 0.0169 mol dm⁻³ [2]. (d) Not necessarily — a first-order law also arises from mechanisms whose rate-determining step is unimolecular after pre-equilibria; the overall equation (2N₂O₅) itself involves two molecules, so a single elementary step matching the stoichiometry is impossible: the law constrains but does not prove a mechanism [1].

IA idea: Determine the activation energy of the hydrolysis of 2-chloro-2-methylpropane by measuring k at five temperatures, presenting E_a with an uncertainty range from steepest/shallowest Arrhenius fits.

Order from rate–concentration curves: the second graphical route

Integrated laws are one route to order; the differential route plots measured rates directly. Take tangents to a concentration–time curve at several starting concentrations (or run initial-rate experiments), then plot rate against [A]: a straight line through the origin means rate ∝ [A] — first order; a parabola (rate quadrupling when [A] doubles) means second order, confirmed by plotting rate against [A]² to straighten it. A log–log plot does both jobs at once: ln(rate) against ln[A] has gradient equal to the order, fractional orders included — the method that catches the 0.5 and 1.5 orders that tidy integer methods miss, and the analysis expected when an IA's kinetics refuses to be first or second order. Worked sketch: rates 2.0 × 10⁻⁵, 8.0 × 10⁻⁵, 1.8 × 10⁻⁴ mol dm⁻³ s⁻¹ at [A] = 0.10, 0.20, 0.30: ratios 1 : 4 : 9 against concentration ratios 1 : 2 : 3 — rate ∝ [A]², second order, with k = rate ÷ [A]² = 2.0 × 10⁻³ mol⁻¹ dm³ s⁻¹ consistent across all three points. Consistency of k across the data set is the verdict; a drifting k is the data declining to fit your order.