Each integrated rate law is a straight line in the right coordinates. So with one run of concentration against time, you can test all three orders at once. It is the same idea as fitting a linear model: the slope means something.
Key idea
Why it works
| $t$ (s) | 0 | 10 | 20 | 30 | 40 |
|---|---|---|---|---|---|
| $[\mathrm{A}]$ (M) | 0.500 | 0.333 | 0.250 | 0.200 | 0.167 |
| $1/[\mathrm{A}]$ (M⁻¹) | 2.00 | 3.00 | 4.00 | 5.00 | 6.00 |
Worked examples
Worked example: Which plot is straight?
- $[\mathrm{A}]$ falls by 0.167, then 0.083, then 0.050: not straight. $\ln[\mathrm{A}]$: $-0.693, -1.10, -1.39, -1.61$; the steps shrink, so not straight either.
- $1/[\mathrm{A}]$ rises by exactly 1.00 every 10 s: a straight line. Second order.
- $$k = \text{slope} = \dfrac{6.00 - 2.00\ \text{M}^{-1}}{40\ \text{s}} = 0.100\ \text{M}^{-1}\,\text{s}^{-1}$$
Answer: Second order, $k = 0.100\ \text{M}^{-1}\,\text{s}^{-1}$.
Worked example: From a slope to k and a half-life
- A straight $\ln[\mathrm{A}]$ plot means first order.
- $k = -\text{slope} = 0.0347\ \text{min}^{-1}$.
- $t_{1/2} = \dfrac{0.693}{0.0347} = 20.0\ \text{min}$.
Answer: First order, $k = 0.0347\ \text{min}^{-1}$, $t_{1/2} = 20.0\ \text{min}$.
Worked example: A zero-order run
- $[\mathrm{A}]$ itself drops by $0.10\ \text{M}$ every 20 s: the plain plot is already straight, so the reaction is zero order.
- Slope $= \dfrac{0.50 - 0.80}{60\ \text{s}} = -5.0 \times 10^{-3}\ \text{M/s}$, so $k = 5.0 \times 10^{-3}\ \text{M/s}$.
Answer: Zero order, $k = 5.0 \times 10^{-3}\ \text{M/s}$.
Real data scatter a little, so no plot is perfectly straight. Fit a line to each of the three plots and keep the one whose points hug the line best (the correlation closest to $\pm 1$). Collect data over two or three half-lives: early on, all three plots look nearly straight, because any smooth curve does over a short stretch.
Common mistake