Rate equations and orders

A rate equation shows how the rate of a reaction depends on the concentration of each reactant: rate = k[A]ᵐ[B]ⁿ. The powers are the orders of reaction, and they can only be found by experiment. The rate constant, k, changes with temperature, as the Arrhenius equation describes.

Key facts

  • Order 0: the rate does not depend on that concentration. Order 1: doubling the concentration doubles the rate. Order 2: doubling it makes the rate four times as fast.
  • The overall order is the sum of the orders. The units of k depend on it: mol dm⁻³ s⁻¹ for zero order, s⁻¹ for first order, dm³ mol⁻¹ s⁻¹ for second order.
  • Orders come from initial-rates experiments (change one concentration at a time) or from concentration–time graphs: a straight line for zero order, a constant half-life for first order.
  • For a first-order reaction the half-life is constant and k = ln 2 ÷ t½.
  • The rate-determining step is the slowest step in the mechanism. The species in the rate equation take part in it, or in a step before it.
  • The Arrhenius equation: k = Ae^(−Ea/RT). A graph of ln k against 1/T is a straight line with gradient −Ea/R.

Worked examples

Example 1

Initial rates for A + B → products: [A] 0.10, [B] 0.10, rate 2.0 × 10⁻⁴; [A] 0.20, [B] 0.10, rate 8.0 × 10⁻⁴; [A] 0.10, [B] 0.20, rate 4.0 × 10⁻⁴ (concentrations in mol dm⁻³, rates in mol dm⁻³ s⁻¹). Find the rate equation and k.

  1. Experiments 1 and 2: [A] doubles, [B] stays the same, the rate goes up 4 times. 2² = 4, so second order in A
  2. Experiments 1 and 3: [B] doubles, [A] stays the same, the rate doubles. first order in B
  3. Write the rate equation. rate = k[A]²[B]
  4. Use experiment 1 to find k. k = 2.0 × 10⁻⁴ ÷ (0.10² × 0.10) = 0.20 dm⁶ mol⁻² s⁻¹

Answer: rate = k[A]²[B]; k = 0.20 dm⁶ mol⁻² s⁻¹

Example 2

A first-order reaction has a half-life of 30 s. Find k, and the fraction of the reactant left after 90 s.

  1. For first order the half-life is constant. k = ln 2 ÷ t½ = 0.693 ÷ 30 = 0.0231 s⁻¹
  2. 90 s is three half-lives. 90 ÷ 30 = 3
  3. Each half-life halves what is left. ½ × ½ × ½ = ⅛

Answer: k = 0.0231 s⁻¹; one eighth is left

Example 3

A reaction has k = 1.0 × 10⁻³ s⁻¹ at 300 K and 4.0 × 10⁻³ s⁻¹ at 320 K. Calculate the activation energy. (R = 8.31 J K⁻¹ mol⁻¹)

  1. Taking logs of the Arrhenius equation at two temperatures gives: ln(k₂ ÷ k₁) = (Ea ÷ R)(1/T₁ − 1/T₂)
  2. Work out each side. ln 4.0 = 1.386; 1/300 − 1/320 = 2.083 × 10⁻⁴
  3. Rearrange for Ea. Ea = 8.31 × 1.386 ÷ (2.083 × 10⁻⁴) = 55 300 J/mol = 55.3 kJ/mol

Answer: Ea = 55.3 kJ/mol