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BALGANESH

PCl5 = PCl3 + Cl2

initially let 1 mole of pcl5 be present since degree of dissociation is 20

% .at equilibrium we get

0.8 mole =pcl5

0.2 mole = pcl2

0.2 mole =cl2

total no of moles at equi = 1.2

now partial pressure = mole fraction . total pressure

therefroe pp (partial pressure ) of pcl5 =0.8/1.2 = 2/3

pp of pcl2 = pp of cl2 = 0.2/1.2 = 1/6

now Kp ( eq constant ) = pp of pcl2 . pp of cl2 / pp of pcl5

solving we get Kp =1/24

rate if useful