and with respect to $\mathrm{B}$ is $y$.

Thus,

rate $=\mathrm{k}[A]^{x}[B]^{y}$

($x$ and $y$ are stoichiometric coefficient)

For the given cases,

$I$. rate $=\mathrm{k}(0.1)^{x}(0.1)^{y}=6.0 \times 10^{-3}$

$II$. rate $=\mathrm{k}(0.3)^{x}(0.2)^{y}=7.2 \times 10^{-2}$

$III$. rate $=\mathrm{k}(0.3)^{x}(0.40)^{y}=2.88 \times 10^{-1}$

$IV$. rate $=\mathrm{k}(0.34)^{x}(0.1)^{y}=2.40 \times 10^{-2}$

Dividing Eq. $(I)$ by Eq. $(IV)$, we get

$\left(\frac{0.1}{0.4}\right)^{x}\left(\frac{0.1}{0.1}\right)^{y}$

$\frac{6.0 \times 10^{-3}}{2.4 \times 10^{-2}}$

$\left(\frac{1}{4}\right)^{x} =\left(\frac{1}{4}\right)^{1}$

$\therefore x=1$

On dividing Eq. $(II)$ by Eq. $(III)$, we get

$\left(\frac{0.3}{0.3}\right)^{x}\left(\frac{0.2}{0.4}\right)^{y}=\frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}}$

$\left(\frac{1}{2}\right)^{y}=\frac{1}{4}$

$\left(\frac{1}{2}\right)^{y}=\left(\frac{1}{2}\right)^{2}$

$\therefore y=2$

Thus, rate law is,

$\text {rate} =k[A]^{1}[B]^{2}$

$=k[A][B]^{2}$