Initially 0.8 mole of ${\mathrm{PCl}}_5$ and 0.2 mole of ${\mathrm{PCl}}_3$ are mixed in one litre vessel. At equilibrium 0.4 mole of ${\mathrm{PCl}}_3$ is present. The value of ${\mathrm{K}}_{\mathrm{C}}$for the reaction ${\mathrm{PCI}}_5\left(\mathrm{g}\right)\iff {\mathrm{PCI}}_3\left(\mathrm{g}\right)+{\mathrm{CI}}_2\left(\mathrm{g}\right) \mathrm{is}$

The chemical equilibrium is

${\mathrm{PCI}}_5\left(\mathrm{g}\right)\iff {\mathrm{PCI}}_3\left(\mathrm{g}\right)+{\mathrm{CI}}_2\left(\mathrm{g}\right)$ 

Initial :          0.8      0.2      0

Equilibrium: 0.8-x   0.2+x   x

At equilibrium, 0.4moles of ${\mathrm{PCl}}_3$ is present.

0.2 + x=0.4

x = 0.2moles

Number of moles of ${\mathrm{PCl}}_5 \mathrm{is}$

$\therefore {\mathrm{n}}_{{\mathrm{PCl}}_5}=0.8-\mathrm{x}=0.8-0.2$

$\therefore {\mathrm{n}}_{{\mathrm{PCl}}_5}=0.6 \mathrm{mole}$

Number of moles of ${\mathrm{Cl}}_2 \mathrm{is}$

${\mathrm{n}}_{{\mathrm{Cl}}_2}=\mathrm{x}=0.2 \mathrm{mole}$

${\mathrm{K}}_{\mathrm{C}}=\frac{\left[{\mathrm{PCl}}_3\right]\left[{\mathrm{Cl}}_2\right]}{\left[{\mathrm{PCl}}_5\right]}=\frac{0.4\times 0.2}{0.6}$

$\therefore {\mathrm{K}}_{\mathrm{C}}=0.13 \mathrm{mol}.{\mathrm{L}}^{-1}$