26.1 Rate equations, orders and rate constants

A rate equation relates rate to concentrations, e.g. rate=k[A]^m[B]^n. m and n are reaction orders, their sum is overall order, and k is the rate constant at a stated temperature.

Half-life is the time for a concentration to halve; the rate-determining step is the slowest elementary step in a mechanism, while an intermediate is formed in one step and consumed in another.

For rate=k[A]²[B], the overall order is three. An intermediate appears in the mechanism but not in the net equation.

Reaction orders are experimental, not automatically stoichiometric coefficients, and the RDS is not always the first step.

In the initial-rates method, compare experiments where one concentration changes while others stay constant. A factor change in rate reveals the order; concentration–time and rate–concentration graphs provide complementary evidence.

For rate=k[A]^m[B]^n, solve m and n separately, then calculate k with units. A first-order concentration–time curve has a constant half-life.

Doubling [A] leaves rate unchanged → zero order; doubling it doubles rate → first order; quadrupling rate → second order.

Do not infer order from a single experiment or confuse a curved concentration–time plot with a rate–concentration graph.

For a first-order reaction, t½ = 0.693/k, so every successive half-life is the same at fixed temperature, regardless of the initial concentration.

Use repeated halving to estimate remaining concentration and use the equation to calculate k. The independence is a diagnostic of first-order behaviour, not a universal property of all reactions.

If 0.80 mol dm⁻³ falls to 0.40 in 10 s and then 0.20 in another 10 s, t½=10 s and k=0.0693 s⁻¹.

A constant half-life does not mean a constant rate; the rate decreases as concentration falls.

From rate=k[A]^m[B]^n, rearrange k=rate/([A]^m[B]^n). For a first-order reaction, k=0.693/t½.

Use the order determined by evidence, keep concentration units consistent, and report k units that make the rate equation dimensionally correct.

If rate=2.0×10⁻³ mol dm⁻³ s⁻¹ and [A]=0.10 mol dm⁻³ for a first-order reaction, k=2.0×10⁻² s⁻¹; t½=34.7 s.

Do not use 0.693/t½ for a zero- or second-order reaction, and do not calculate k before determining order.

For a multi-step mechanism, the slow rate-determining step controls the observed rate. A proposed mechanism must reproduce the overall equation and the experimentally observed rate equation.

Intermediates are formed then consumed; catalysts are used then regenerated. If a fast pre-equilibrium precedes the slow step, substitute its relationship to eliminate intermediates from the rate law.

If the slow step is A + I → products and a fast step makes I proportional to [B], the observed rate may be k[A][B], even when I is absent from the overall equation.

Do not copy overall stoichiometric coefficients into the rate equation automatically, and do not leave an intermediate in the final observable law.

Increasing temperature raises the rate constant k because the Maxwell–Boltzmann distribution has a greater fraction of particles with energy at least the activation energy.

The rate also depends on concentration and collision frequency, but the temperature effect on k is intrinsic to the energy barrier. A catalyst changes the barrier rather than temperature.

A small temperature increase can cause a large rate increase because the high-energy tail beyond Ea grows disproportionately.

Temperature does not merely make particles “move faster”; it changes the fraction successful enough to react.