First-Order Reactions

A first-order rate law, $\text{rate} = k[\mathrm{A}]$, says that the reactant disappears at a rate proportional to how much is left. That is the same rule as continuous exponential decay, so the concentration follows an exponential curve.

Key idea

For a first-order reaction, $$\ln[\mathrm{A}] = \ln[\mathrm{A}]_0 - kt \qquad\text{or}\qquad [\mathrm{A}] = [\mathrm{A}]_0\,e^{-kt},$$ and the half-life $t_{1/2} = \dfrac{\ln 2}{k} = \dfrac{0.693}{k}$ is the same no matter how much reactant there is.

Why it works

The rate law says $-\dfrac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]$: the fraction that reacts in each small time interval is always the same, $k\,dt$. Separating and integrating, $\int \dfrac{d[\mathrm{A}]}{[\mathrm{A}]} = -k\int dt$ gives $\ln[\mathrm{A}] - \ln[\mathrm{A}]_0 = -kt$. For the half-life, set $[\mathrm{A}] = \tfrac{1}{2}[\mathrm{A}]_0$: $\ln\tfrac{1}{2} = -kt_{1/2}$, so $t_{1/2} = \ln 2 / k$. The starting amount cancels out, and that is why the half-life never changes.

A first-order curve with half-life markers. Each half-life is the same length: from 1.0 to 0.50 M takes as long as from 0.50 to 0.25 M.

The same data as $\ln[\mathrm{A}]$ against time: a straight line with slope $-k$.

Worked examples

Worked example: Half-life and amount left

A first-order reaction has $k = 0.0200\ \text{min}^{-1}$ and $[\mathrm{A}]_0 = 0.400\ \text{M}$. Find the half-life and $[\mathrm{A}]$ after $50.0\ \text{min}$.

$t_{1/2} = \dfrac{0.693}{0.0200\ \text{min}^{-1}} = 34.7\ \text{min}$.
$kt = (0.0200)(50.0) = 1.00$, so $$[\mathrm{A}] = 0.400\,e^{-1.00} = (0.400)(0.368) = 0.147\ \text{M}$$

Answer: $t_{1/2} = 34.7\ \text{min}$; $[\mathrm{A}] = 0.147\ \text{M}$.

Worked example: How long to reach 10%

For the same reaction, how long until 10.0% of A is left?

Rearrange: $t = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]}$, and $\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]} = \dfrac{100}{10.0} = 10.0$.
$$t = \dfrac{\ln 10.0}{0.0200\ \text{min}^{-1}} = \dfrac{2.303}{0.0200} = 115\ \text{min}$$ That is 3.32 half-lives, as it should be: $(\tfrac{1}{2})^{3.32} = 0.100$.

Answer: $115\ \text{min}$.

Worked example: Fractions of a half-life

A reactant has a half-life of $12.0\ \text{min}$. What fraction is left after $30.0\ \text{min}$?

$30.0 / 12.0 = 2.5$ half-lives.
Fraction left $= \left(\tfrac{1}{2}\right)^{2.5} = 0.177$. This form is the same law: $\left(\tfrac{1}{2}\right)^{t/t_{1/2}} = e^{-kt}$ because $k = \ln 2/t_{1/2}$.

Answer: $0.177$ (17.7%).

Common mistake

Thinking the half-life gets longer (or shorter) as the reactant runs down. For first order the fraction that reacts per minute is fixed, so every half-life is the same length. Check: if a problem shows half-lives that change, the reaction is not first order.

Two measurements are enough to find $k$ for a first-order reaction: $k = \dfrac{1}{t_2 - t_1}\ln\dfrac{[\mathrm{A}]_1}{[\mathrm{A}]_2}$. Because only the ratio of concentrations enters, you can use anything proportional to concentration, such as absorbance or the pressure of a gas, without converting it first. Many real processes behave this way: the decomposition of $\mathrm{N_2O_5}$, the loss of many drugs from the bloodstream, and every radioactive decay.

Radioactive decay is the most famous first-order process, which is why half-lives of isotopes are constants you can look up.

First-Order Reactions

Worked examples

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