${\mathrm{A}}_2+{\mathrm{B}}_2\to 2\mathrm{AB}; \mathrm{R}.\mathrm{O}.\mathrm{R}=\mathrm{k}{\left[{\mathrm{A}}_2\right]}^{\mathrm{a}}{\left[{\mathrm{B}}_2\right]}^{\mathrm{b}}$

Initial $\left[A_2\right]$	Initial $\left[B_2\right]$	R.O.R.$\left(r\right)$$M{s}^{-1}$
0.2 0.1 0.2	0.2 0.4 0.4	0.04 0.04 0.08

Order of reaction with respect to $A_2$and $B_2$are respectively:

$\mathrm{R}.\mathrm{O}.\mathrm{R}=\mathrm{k}{\left[{\mathrm{A}}_2\right]}^{\mathrm{a}}{\left[{\mathrm{B}}_2\right]}^{\mathrm{b}}$

When both $\left[A_2\right]$ and $\left[B_2\right]$ is doubled:

$\mathrm{R}.\mathrm{O}.\mathrm{R}=\mathrm{k}{\left[0.2\right]}^{\mathrm{a}}{\left[0.2\right]}^{\mathrm{b}}$.....(1)

When $\left[A_2\right]=0.1$ and $\left[B_2\right]=0.4$ is doubled:

$\mathrm{R}.\mathrm{O}.\mathrm{R}=\mathrm{k}{\left[0.1\right]}^{\mathrm{a}}{\left[0.4\right]}^{\mathrm{b}}$...(2)

When $\left[A_2\right]=0.2$ and $\left[B_2\right]=0.4$ is doubled:

$\mathrm{R}.\mathrm{O}.\mathrm{R}=\mathrm{k}{\left[0.2\right]}^{\mathrm{a}}{\left[0.4\right]}^{\mathrm{b}}$....(3)

Dividing (3) and (2)

$2={\left[2\right]}^{\mathrm{a}}, \mathrm{so} \mathrm{a}=1$

Dividing (3) and (1)

$2={\left[2\right]}^b, \mathrm{so} b=1$

Hence, a=1 and b=1.