For the reaction system: $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2…
For the reaction system: $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_2(\mathrm{~g})$ volume is suddenly reduce to half its value by increasing the pressure on it. If the reaction is of first order with respect to $\mathrm{O}_2$ and second order with respect to $\mathrm{NO}$, the rate of reaction will
diminish to one-eighth of its initial value
increase to eight times of its initial value
increase to four times of its initial value
diminish to one-fourth of its initial value
Solution
$\mathrm{r}=\mathrm{k}\left[\mathrm{O}_2\right][\mathrm{NO}]^2$. When the volume is reduced to $1 / 2$, The conc. will double.
$\therefore$ New rate $=\mathrm{k}\left[2 \mathrm{O}_2\right][2 \mathrm{NO}]^2=8 \mathrm{k}\left[\mathrm{O}^2\right][\mathrm{NO}]^2$. The new rate increases by eight times.