For the reaction $\mathrm{A}+\mathrm{B}\to 1$ Products, it is observed that:

(i) On doubling the initial concentration of A only, the rate of reaction is also doubled.

(ii)On doubling the initial concentration of both A and B, there is a change by a factor of 8 in the rate of the reaction. The rate of the reaction is given by:

We know that, rate

But        $\begin{array}{r}\mathrm{R}=\mathrm{k}\left[\mathrm{A}{]}^{\mathrm{m}}\right[\mathrm{B}{]}^{\mathrm{n}}\left(\mathrm{i}\right)\\ 2\mathrm{R}=\mathrm{k}\left[2\mathrm{A}{]}^{\mathrm{m}}\right[\mathrm{B}{]}^{\mathrm{n}}\left(\mathrm{ii}\right)\\ 8\mathrm{R}=\mathrm{k}\left[2\mathrm{A}{]}^{\mathrm{m}}\right[2\mathrm{B}{]}^{\mathrm{n}}\left(\mathrm{iii}\right)\end{array}$

Dividing equation (i) by (ii),we get:

                      $\begin{array}{r}\frac{\mathrm{R}}{2\mathrm{R}}=\frac{\mathrm{k}\times [\mathrm{A}{]}^{\mathrm{m}}\times [\mathrm{B}{]}^{\mathrm{n}}}{\mathrm{k}\times 2^{\mathrm{m}}\times [\mathrm{A}{]}^{\mathrm{m}}\times [\mathrm{B}{]}^{\mathrm{n}}}\\ \frac{1}{2^1}=\frac{1}{2^{\mathrm{m}}}\cdot \text{ So, }\mathrm{m}=1\end{array}$

Dividing equation (ii) by (iii),we get

                              $\begin{array}{l}\frac{2\mathrm{R}}{8\mathrm{R}}=\frac{\mathrm{k}\times 2^{\mathrm{m}}\times [\mathrm{A}{]}^{\mathrm{m}}\times [\mathrm{B}{]}^{\mathrm{n}}}{\mathrm{k}\times 2^{\mathrm{m}}\times [\mathrm{A}{]}^{\mathrm{m}}\times 2^{\mathrm{n}}\times [\mathrm{B}{]}^{\mathrm{n}}};\\ \frac{1}{4}=\frac{1}{2^{\mathrm{n}}};\frac{1}{2^2}=\frac{1}{2^{\mathrm{n}}};\mathrm{n}=2\end{array}$

Thus, rate $=\mathrm{k}\left[\mathrm{A}\right][\mathrm{B}{]}^2$