For the reaction, $\mathrm{SO}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{SO}_{3}$ eq

Question

For the reaction, $\mathrm{SO}_{2}+\frac{1}{2} \mathrm{O}_{2} ightleftharpoons \mathrm{SO}_{3}$ equilibrium constant is K .Hence for the reaction $2 \mathrm{SO}_{3} ightarrow 2 \mathrm{SO}_{2}+\mathrm{O}_{2}$  ---- equilibrium constant $\mathrm{K}^{\prime}$, expressed in terms of $K$ is\

Professor PC Thomas Classes · Answer

$\mathrm{SO}_{2}+\frac{1}{2} \mathrm{O}_{2} ightarrow \mathrm{SO}_{3} \ldots-----\mathrm{K}$\ $\because \mathrm{SO}_{3} ightarrow \mathrm{SO}_{2}+\frac{1}{2} \mathrm{O}_{2}----\frac{1}{\mathrm{~K}}$\ $2 \mathrm{SO}_{3} ightarrow 2 \mathrm{SO}_{2}+\mathrm{O}_{2}$\ equilibrium constant $=\frac{1}{\mathrm{~K}^{2}}$