0.12 mol of $PC{I}_3$ are mixed with 0.9 mol of $PC{I}_5$ in a one litre vessel. The vessel contains 0.72 mol of $PC{I}_5$ at equilibrium. The value of $K_c$ for the reaction is :

Initial no. of mol of ${\mathrm{PCl}}_5=0.90;{\mathrm{PCl}}_3=0.12$ mol; no. mol of ${\mathrm{PCl}}_5$ at equilibrium $=0.72$

$\therefore$ At equilibrium (i) no. of mol of ${\mathrm{PCl}}_3=0.12+(0.90-0.72)=0.12+0.18=0.30$ mol

(ii) no. of mol of ${\mathrm{Cl}}_2=0.90-0.72=0.18$ mol

$\therefore {\mathrm{K}}_{\mathrm{c}}$ for the reaction, ${\mathrm{PCl}}_5\rightleftharpoons {\mathrm{PCl}}_3+{\mathrm{Cl}}_2$ is:

$\begin{array}{l}{K}_c=\frac{\left[{\mathrm{PCl}}_3\right]\left[{\mathrm{Cl}}_2\right]}{\left[{\mathrm{PCl}}_5\right]}=\frac{0.3\times 0.18}{0.72}\\ ={\mathbf{0}\mathbf{.}\mathbf{075}\mathbf{molL}}^{-}\end{array}$