One mole of ${\mathrm{PCl}}_5$  is heated in a closed vessel at volume V litre.  If x moles of ${\mathrm{PCl}}_5$ dissociates into ${\mathrm{PCl}}_3$ and ${\mathrm{Cl}}_2$ at equilibrium, then the expressions for ${\mathrm{K}}_{\mathrm{c}}$  and ${\mathrm{K}}_{\mathrm{P}}$ respectively are (P is the total pressure at equilibrium).

Solution:

$PC{I}_5\rightleftharpoons PC{I}_3+C{I}_2$

  1                0        0                 initially

$\frac{1-x}{v}$           $\frac{x}{v}$       $\frac{x}{v}$              at equilibrium, $K_c=\frac{\frac{x}{v},\frac{x}{v}}{\left(\frac{1-x}{v}\right)}=\frac{x^2}{(1-x)v}$

$PC{I}_5\rightleftharpoons PC{I}_3+C{I}_2$

  1               0          0                 initially

1 – x              x           x                  at equilibrium

$\frac{1-x}{1+x}p$     $\frac{x}{1+x}p$   $\frac{x}{1+x}p$     Partial pressure    :   total number of moles

= 1 – x + x + x = 1 + x     Partial pressure = mole fraction x total pressure (p)

$K_p=\frac{\frac{x}{1+x}p\frac{x}{1+x}p}{\left(\frac{1-x}{1+x}\right)p}$=$K_p=\frac{\frac{x}{1+x}p\frac{x}{1+x}p}{\left(\frac{1-x}{1+x}\right)p}$