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LearnYCH11WCH14 · Unit 4: Rates, Equilibria and Further Organic ChemistryKineticsRate equations and orders
Kinetics · Subtopic 1 of 1

Rate Equations and Orders

The rate equation rate=k[A]m[B]n\text{rate}=k[A]^m[B]^n has orders m,nm,n (0, 1 or 2) found by experiment, not from the equation. Initial-rates method: change one concentration at a time; doubling [A] doubles the rate (1st order), quadruples it (2nd order), or leaves it unchanged (zero order). Graphs: on a concentration–time graph, zero order is a straight line, first order has a constant half-life (t1/2=ln⁡2kt_{1/2}=\frac{\ln2}{k}). k increases with temperature (Arrhenius: k=Ae−Ea/RTk=Ae^{-E_a/RT}; ln⁡k=ln⁡A−EaRT\ln k=\ln A-\frac{E_a}{RT}). The rate-determining step (slowest step) contains the species in the rate equation.

Cambridge 9701 Paper 4Edexcel YCH11 WCH14AQA 7405 Paper 1≈ 55 minRevision 1 · updated 2026-10-30
Section 1 of 14

Learning objectives

  • Deduce orders from initial-rate data and graphs.
  • Calculate k and its units.
  • Use half-life for first-order reactions.
  • Link rate equations to mechanisms; use the Arrhenius equation.
Pause and think

2-bromo-2-methylpropane hydrolyses with rate = k[RBr]. Why doesn't [OH⁻] appear?

Section 2 of 14

Core idea

Start hereThe slowest step decides

A queue moves only as fast as its slowest checkout. A multi-step reaction is the same: the rate equation reveals the slow step.

The big idea

rate = k[A]^m[B]^n, with orders from experiment. Orders tell you which species are in the rate-determining step.

1. Orders from initial rates

Worked exampleInitial-rate data

Exp 1: [A] 0.10, [B] 0.10, rate 2.0×10⁻⁴. Exp 2: [A] 0.20, [B] 0.10, rate 4.0×10⁻⁴. Exp 3: [A] 0.10, [B] 0.30, rate 1.8×10⁻³. Find the rate equation.

Goal
Orders.
Strategy
Change one concentration at a time.
  1. 1→2: [A]×2, rate×2⇒1st order in A\text{1→2: [A]}\times2,\ \text{rate}\times2\Rightarrow\text{1st order in A}
    2¹ = 2.
  2. 1→3: [B]×3, rate×9⇒2nd order in B\text{1→3: [B]}\times3,\ \text{rate}\times9\Rightarrow\text{2nd order in B}
    3² = 9.
  3. rate=k[A][B]2\text{rate}=k[A][B]^2
    Combine.
Answerrate = k[A][B]²; overall third order.
✓ Checkk from Exp 1: 2.0×10⁻⁴/(0.10×0.010) = 0.20 dm⁶ mol⁻² s⁻¹ ✓.
Recognising order from graphs

Rate–concentration and concentration–time shapes.

Arrhenius as a straight line
  1. k=Ae−Ea/RTk=Ae^{-E_a/RT}
    Arrhenius equation.
  2. ln⁡k=ln⁡A−EaR⋅1T\ln k=\ln A-\frac{E_a}{R}\cdot\frac{1}{T}
    Take natural logs.
  3. gradient=−EaR\text{gradient}=-\frac{E_a}{R}
    Compare with y = c + mx.

Summary

  • Orders from experiment.
  • t½ = 0.693/k (first order).
  • Slow step contains rate-equation species.
  • ln k vs 1/T gradient = −Ea/R.
Key words
Order
Power of a concentration term in the rate equation.
Rate constant
k; constant at fixed temperature.
Rate-determining step
Slowest step in a mechanism.
HistoryArrhenius. Svante Arrhenius proposed his rate–temperature equation in 1889; his PhD thesis on ions was nearly failed, but later won him the Nobel Prize.
Fun factRadiocarbon dating. Carbon-14 decay is first order with a half-life of 5730 years, which is why archaeologists can date ancient bones.
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