For${\mathrm{N}}_2+{\mathrm{O}}_2\to 2\mathrm{NO}$, initially ${\mathrm{N}}_2\&{\mathrm{O}}_2$are at pressures $500 \mathrm{mm} \& 700 \mathrm{mm}$ at $\mathrm{t}=0$. If the pressure of ${\mathrm{N}}_2$ is $480 \mathrm{mm}$ at $\mathrm{t}=20 \min$ average rate of reaction is (in $\mathrm{mm}-{\min }^{-1}$ )

$$\begin{gathered} {\mathrm{N}}_2+{\mathrm{O}}_2⟶2\mathrm{NO} \\ \mathrm{at} \\ \mathrm{t}=0 500\mathrm{mm} 700\mathrm{mm} \\ \mathrm{t}=20 480\mathrm{mm} \end{gathered}$$

1 mole of ${\mathrm{N}}_2$ consume 1 mole of ${\mathrm{O}}_2$ to give 2 mole of $\mathrm{NO}$. If ${\mathrm{P}}_{{\mathrm{N}}_2}$ is decreased by $20 \mathrm{mm}$, then ${\mathrm{P}}_{{\mathrm{O}}_2}$ is decreased by $20 \mathrm{mm}$ and $(2\times 20)\mathrm{nm}=40 \mathrm{mm}$ of ${\mathrm{P}}_{\mathrm{N}0}$ is increased i.e. change in preseuse $=20 \mathrm{mm}$

$$\begin{gathered} \text{ i.e.}\mathrm{charge}\text{ pressure }=20\mathrm{mm} \\ \text{charge time }=20\min \end{gathered}$$

$$\begin{gathered} \mathrm{Avg}. \mathrm{Rate} \mathrm{of} \mathrm{r}\times \mathrm{n}=\frac{\mathrm{charge} \mathrm{in} \mathrm{pr}. \mathrm{of} \mathrm{reaction}}{\mathrm{charge} \mathrm{in} \mathrm{time} } \\ =\frac{20}{20}=1 \mathrm{mm} {\min }^{-1} \end{gathered}$$