All question types · Kinetics

First order: concentration after time t

C3L6 · UKChO Round 1 · UKChO Round 2 · IChO · unreviewed

k = ln 2 / t½, then [A] = [A]₀ e^(−kt).

Practise this with fresh numbers

Worked example

A first-order reaction has a half-life of 42 min. Starting from [A]₀ = 0.41 mol dm⁻³, what is [A] after 110 min?

1. Half-life (first order): solve for k
$k = \dfrac{\ln 2}{t_{1/2}}$
= 2.751 × 10⁻⁴ s⁻¹
2. First-order integrated rate law: solve for [A]
$[\mathrm{A}] = [\mathrm{A}]_0\,e^{-kt}$
= 0.06674 mol dm⁻³

Answer: [A] = 0.0667 mol dm⁻³

Method

k = ln 2 / t½, then [A] = [A]₀ e^(−kt). Keep k and t in the same time units.

Common mistakes

That is the zero-order result. A constant half-life is the signature of first-order kinetics.
That is the second-order result. The question says first order: use [A] = [A]₀e^(−kt).

Formulas used

Half-life (first order)
$t_{1/2} = \dfrac{\ln 2}{k}$
First-order integrated rate law
$\ln\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]} = kt$

Practise this with fresh numbers

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