For the reaction $2\mathrm{A}+3\mathrm{B}⟶$ product, A is in excess and on changing the concentration of B from 0.1 M to 0.4 M, rate becomes doubled. Thus, rate law is :

Since A is present in great excess, it has no effect on the reaction's rate; instead, the reaction's rate will only depend on the concentration of B.

$\therefore \mathrm{r}=\mathrm{k}[\mathrm{B}{]}^{\mathrm{n}}$

$\mathrm{r}=\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}[0.1{]}^{\mathrm{n}}\dots .(1)$

${\mathrm{r}}^{\mathrm{\prime }}=2\mathrm{r}=\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}[0.4{]}^{\mathrm{n}}\dots \dots ($

Divide equation (ii) with (i)

$\frac{2\mathrm{r}}{\mathrm{r}}=\frac{\mathrm{k}[0.4{]}^{\mathrm{n}}}{\mathrm{k}[0.1{]}^{\mathrm{n}}}$

$\mathrm{n}=\frac{1}{2}$

The rate law formula is $\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}[\mathrm{B}{]}^{\frac{1}{2}}$