Given that the equilibrium constant for the reaction:

$2{\mathrm{SO}}_2\left(g\right)+{\mathrm{O}}_2\left(g\right)\rightleftharpoons 2{\mathrm{SO}}_3\left(g\right)$ has a value of 278 at a particular temperature. What is the value of equilibrium constant for the following reaction at the same temperature?

${\mathrm{SO}}_3\left(g\right)\rightleftharpoons {\mathrm{SO}}_2\left(g\right)+\frac{1}{2}{\mathrm{O}}_2\left(g\right)$

For $2{\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{SO}}_3\left(\mathrm{g}\right)\dots \left(i\right),K=278$

On reversing the above equation, we get:

$2{\mathrm{SO}}_3\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) ... \left(ii\right)$

K' for this reaction= 1/K = 1/278

On dividing equation (ii) by 2, we get:

${\mathrm{SO}}_3\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{SO}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right) ... \left(iii\right)$

$\therefore$ The value of K for this reaction will be:

$K^{\prime \prime }=(K{)}^{1/2}={\left(\frac{1}{K}\right)}^{1/2}={\left(\frac{1}{278}\right)}^{1/2}=0.06=6\times {10}^{-2}.$