Initially 0.8 mole of PCl5  and 0.2 mole of PCl3  are mixed in one litre vessel.  At equilibrium 0.4 mole of PCl3  is present.  The value of Kc  for the reactionPCl5(g)⇔PCl3(g)+Cl2(g)

PCl5⇌PCl3+Cl2                                       Given (0.2+x)=0.4 0.8                0.2         0                                                            ⇒x=0.2     x                  x            x_ (0.8−x)     0.2+x   x         ∴Kc=0.4×0.20.6                   =0.133 M =0.133 mole.lit−1