One lit of $S{O}_3$ was placed in a two litre vessels of a certain temperature. The following equilibrium was established in the vessel $2{\mathrm{SO}}_{3\left(\mathrm{g}\right)}\rightleftharpoons 2{\mathrm{SO}}_{2\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}$ the equilibrium mixture reacted with 0.2 mole ${\mathrm{KMnO}}_4$ in acidic medium. $K_c$ value is ______.

It is given to calculate the value of $K_C$.

 $K_{(}C)=\frac{\left[S{O}_2{]}^2\right[O_2]}{[S{O}_3{]}^2}$

 $2S{O}_{3\left(g\right)}\rightleftharpoons 2S{O}_{2\left(g\right)}+O_{2\left(g\right)}$

$KMn{O}_4+S{O}_2\to M{n}^{2+}+S{O}_3$

0.2 moles of $KMn{O}_4$ in acidic medium oxidize 0.5 moles of $S{O}_2$.

So, equivalent of $KMn{O}_4$ = no. of equivalent of $S{O}_2$

$0.2\times 5=2x\times 2$

x = 0.25

$K_{(}C)=\frac{[2\times 0.25{]}^2[0.25]}{[1-2(0.25){]}^2}$

$K_{(}C)$ = 0.125

Hence, the answer is 0.125.