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Determining reaction orders by the method of initial rates
The method of initial rates determines the concentration exponents in a rate law by comparing experiments that begin with different reactant concentrations.
Suppose the unknown rate law is
$$r=k[A]^m[B]^n.$$
If two experiments differ only in $[A]$, then dividing their initial-rate equations cancels $k$ and the unchanged concentration of B:
$$\frac{r_2}{r_1} =\left(\frac{[A]_2}{[A]_1}\right)^m.$$
The exponent $m$ can then be inferred from how the rate changes. A second comparison that changes only B determines $n$.
Example
Consider these initial-rate data at one temperature:
| Experiment | $[A]_0$ (M) | $[B]_0$ (M) | Initial rate (M s$^{-1}$) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | $2.0\times10^{-3}$ |
| 2 | 0.20 | 0.10 | $4.0\times10^{-3}$ |
| 3 | 0.20 | 0.20 | $1.6\times10^{-2}$ |
Compare experiments 1 and 2. Only A changes, doubling from $0.10$ M to $0.20$ M, and the rate doubles:
$$\frac{4.0\times10^{-3}}{2.0\times10^{-3}} =2 =2^m,$$
so
$$m=1.$$
Now compare experiments 2 and 3. Only B doubles, while the rate increases by a factor of four:
$$\frac{1.6\times10^{-2}}{4.0\times10^{-3}} =4 =2^n,$$
so
$$n=2.$$
The rate law is therefore
$$\boxed{r=k[A][B]^2}.$$
Use any experiment to determine $k$. From experiment 1,
$$k=\frac{2.0\times10^{-3}}{(0.10)(0.10)^2} =2.0,\mathrm{M^{-2},s^{-1}}.$$
Initial-rate measurements are useful because they compare experiments before large composition changes or product accumulation complicate the interpretation.
The method determines an empirical rate law. Agreement between that law and a proposed molecular mechanism can support the mechanism, but the rate law by itself does not prove one unique sequence of elementary steps.