The first-order equations came from integrating $\text{rate} = k[\mathrm{A}]$. Integrating the zero- and second-order rate laws in the same way gives two more straight-line laws, each with its own half-life.
Key idea
Zero order: $[\mathrm{A}] = [\mathrm{A}]_0 - kt$ and $t_{1/2} = \dfrac{[\mathrm{A}]_0}{2k}$. Second order: $\dfrac{1}{[\mathrm{A}]} = \dfrac{1}{[\mathrm{A}]_0} + kt$ and $t_{1/2} = \dfrac{1}{k[\mathrm{A}]_0}$. Only the first-order half-life is independent of concentration.
Why it works
Zero order: $-d[\mathrm{A}]/dt = k$ is a constant rate, so the concentration falls in a straight line. Second order: $-d[\mathrm{A}]/dt = k[\mathrm{A}]^2$; separating gives $\int [\mathrm{A}]^{-2}\,d[\mathrm{A}] = -k\int dt$, so $\dfrac{1}{[\mathrm{A}]} - \dfrac{1}{[\mathrm{A}]_0} = kt$. Put $[\mathrm{A}] = \tfrac{1}{2}[\mathrm{A}]_0$ into each to get the half-lives. A second-order reaction slows sharply as A runs down (the rate goes as the square), so each half-life is twice as long as the one before. A zero-order reaction keeps its pace, so each half-life is half as long as the one before.
| Zero order | First order | Second order | |
|---|---|---|---|
| Rate law | $k$ | $k[\mathrm{A}]$ | $k[\mathrm{A}]^2$ |
| Straight-line plot | $[\mathrm{A}]$ vs $t$ | $\ln[\mathrm{A}]$ vs $t$ | $1/[\mathrm{A}]$ vs $t$ |
| Slope | $-k$ | $-k$ | $+k$ |
| Half-life | $[\mathrm{A}]_0/2k$ | $0.693/k$ | $1/(k[\mathrm{A}]_0)$ |
| Units of k | $\text{M/s}$ | $\text{s}^{-1}$ | $\text{M}^{-1}\,\text{s}^{-1}$ |
Worked examples
Worked example: A second-order reaction
$\text{rate} = k[\mathrm{A}]^2$ with $k = 0.50\ \text{M}^{-1}\,\text{s}^{-1}$ and $[\mathrm{A}]_0 = 0.20\ \text{M}$. Find the first half-life, the second half-life and $[\mathrm{A}]$ at $30\ \text{s}$.
- First half-life: $t_{1/2} = \dfrac{1}{(0.50)(0.20)} = 10\ \text{s}$.
- The second half-life starts at $0.10\ \text{M}$: $\dfrac{1}{(0.50)(0.10)} = 20\ \text{s}$, twice as long.
- At $30\ \text{s}$: $$\dfrac{1}{[\mathrm{A}]} = \dfrac{1}{0.20} + (0.50)(30) = 5.0 + 15 = 20\ \text{M}^{-1}$$ so $[\mathrm{A}] = 0.050\ \text{M}$. That fits: 10 s plus 20 s is two half-lives, and $0.20 \to 0.10 \to 0.050$.
Answer: $10\ \text{s}$, $20\ \text{s}$ and $0.050\ \text{M}$.
Worked example: A zero-order reaction
A reaction on a catalyst surface is zero order with $k = 5.0 \times 10^{-3}\ \text{M/s}$ and $[\mathrm{A}]_0 = 0.20\ \text{M}$. Find the half-life and when A runs out.
- $t_{1/2} = \dfrac{0.20}{2(5.0 \times 10^{-3})} = 20\ \text{s}$.
- $[\mathrm{A}] = 0$ when $0.20 = (5.0 \times 10^{-3})t$: $t = 40\ \text{s}$. The second "half-life" (from 0.10 M to 0.050 M) takes only 10 s.
Answer: $20\ \text{s}$; A is gone at $40\ \text{s}$.
Common mistake
Using $t_{1/2} = 0.693/k$ for every reaction. That formula holds only for first order. Check the units of $k$: if $k$ is in $\text{M}^{-1}\,\text{s}^{-1}$, the reaction is second order and its half-life needs $[\mathrm{A}]_0$.
Zero-order behavior is common when a reaction happens on a surface or on an enzyme that is fully occupied: adding more reactant cannot speed it up, because every site is already busy.