If the equilibrium constant for the reaction, $2 \mathrm{SO}_2+\mathrm{O}_2 ightleftharpoons 2…
If the equilibrium constant for the reaction, $2 \mathrm{SO}_2+\mathrm{O}_2 ightleftharpoons 2 \mathrm{SO}_3$ is 64 at $500 \mathrm{~K}$, then the equilibrium constant for the reaction $\mathrm{SO}_3 ightleftharpoons \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2$ at the same temperature is
$8$
$\frac{1}{8}$
$32$
$\frac{1}{64}$
Solution
Given, $2 \mathrm{SO}_2+\mathrm{O}_2 ightleftharpoons 2 \mathrm{SO}_3, K=64$
$\mathrm{SO}_3 ightleftharpoons \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2, K^{\prime}=?$
$\because$ New $(K)=[K]^{1 / n}$ i.e. $K^{\prime}$
where, $n=$ factor to the new equilibrium constant, which is $n$th root of the previous value and
$K^{\prime}=\left[\frac{1}{K}ight]^n,\left(n=\frac{1}{2}ight)$
$\therefore$ Hence, new $K$ i.e. $\left(K^{\prime}ight)=\left[\frac{1}{64}ight]^{1 / 2}=\frac{1}{8} \quad\left(\because n=\frac{1}{2}ight)$
$\therefore$ Option (b) is the correct answer.