How will the rate $\mathrm{r}=\mathrm{K}\left[{\mathrm{SO}}_2\right]\left[{\mathrm{O}}_2\right]$ of a reaction,  $2{\mathrm{SO}}_2+{\mathrm{O}}_2\to 2{\mathrm{SO}}_3$ change if the volume of the reaction vessel is tripled

For a given reaction, the rate is

$\mathrm{r}=\mathrm{K}\left[{\mathrm{SO}}_2\right]\left[{\mathrm{O}}_2\right]\to \mathrm{eqn}\left(1\right)$

If the volume of the reaction vessel to be tripled , then

${\mathrm{r}}_1=\mathrm{K}\left[\frac{\mathrm{n}}{\mathrm{v}}\right]\left[\frac{\mathrm{n}}{\mathrm{v}}\right] {\mathrm{r}}_2=\mathrm{K}\left[\frac{\mathrm{n}}{3\mathrm{v}}\right]\left[\frac{\mathrm{n}}{3\mathrm{v}}\right]$

$\mathrm{Then},\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{\mathrm{K}\left[{\displaystyle \frac{\mathrm{n}}{\mathrm{v}}}\right]\left[{\displaystyle \frac{\mathrm{n}}{\mathrm{v}}}\right]}{\mathrm{K}\left[{\displaystyle \frac{\mathrm{n}}{3\mathrm{v}}}\right]\left[{\displaystyle \frac{\mathrm{n}}{3\mathrm{v}}}\right]}=9$

$\Rightarrow {\mathrm{r}}_1=9{\mathrm{r}}_2$

$\therefore {\mathrm{r}}_2=\frac{1}{9}{\mathrm{r}}_1$

So, the final rate of reaction is $\frac{1}{9}th of its initial rate$