Time for a second-order reaction to reach a concentration

Worked example

For the second-order decomposition of NO₂ (rate = k[NO₂]², k = 0.65 dm³ mol⁻¹ s⁻¹), find the time for the concentration to drop from 0.077 mol dm⁻³ to 0.025 mol dm⁻³.

1. 1. Second-order integrated rate law: solve for t

   $t = \dfrac{1}{k}\left(\dfrac{1}{[\mathrm{A}]} - \dfrac{1}{[\mathrm{A}]_0}\right)$

   = 41.56 s

Answer: t = 41.6 s

Method

Integrating rate = k[A]² gives 1/[A] = 1/[A]₀ + kt, so t = (1/[A] − 1/[A]₀) ÷ k. Unlike first order, the time to halve gets longer as the concentration falls.

Formulas used

• Second-order integrated rate law

  $\dfrac{1}{[\mathrm{A}]} = \dfrac{1}{[\mathrm{A}]_0} + kt$

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