Two mole ofPCl_(5)" were heated to" 327^(@)Cin a closed two litre vessel and when equilibrium was achieved, PCl_(5) was found to be 40% dissociated intoPCl_(3) and Cl_(2).Calculate the equilibrium constants K_(p) and K_(c) for this reaction.

Two mole ofPCl_(5)" were heated to" 327^(@)Cin a closed two litre vess

1.	Two mole ofPCl_(5)" were heated to" 327^(@)Cin a closed two litre vessel and when equilibrium was achieved, PCl_(5) was found to be 40% dissociated intoPCl_(3) and Cl_(2).Calculate the equilibrium constants K_(p) and K_(c) for this reaction.
Answer» Solution :`PCl_(5) " dissociates as " :PCl_(5) hArr PCl_(3) + Cl_(2)` IntiaL amount of `PCl_(5) = 2" moles (Given)"` `% " age dissociation at equilibrium " = 40%` ` :. PCl_(5) " dissociated at equilibrium " = 40/100 xx 0.8 ` mole `:. " Amounts of " PCl_(5), Pcl_(3) and Cl_(2) " at equilibrium will be " ` `Pcl_(5) = 2- 0.8 = 1.2 ` mole `PCl _(3) = 0.8 ` `Cl_(2)-0.8 " " [:' 1 " mole of " PCl_(5) " on dissociation GIVES 1 mole of "PCl_(3) "and 1 mole of "Cl_(2) ]` Since the volume of the vessel is 2 litres is , therefore , the molar concentrations at equilibrium will be `[ PCl_(5)] = 1.2/2 = 0.6 " mol " L^(-1), [ PCl_(3)] = 0.8/2 = 0.4 " mol " L^(-1) and [ Cl_(2) ] = 0.8/2 = 0.4 "mol" L^(-1)` Applying the LAW of chemical equilibrium to the dissociation equilibrium, we get `K_(c)= ([PCl_(3)] [ Cl_(2)])/ ([PCl_(5)]) = (.4 " mol " L^(-1) xx 0.4" mol " L^(-1))/(0.6 " mol " L^(-1))= 0.267 "mol" L^(-1)` `K_(p) = K_(c) (RT)^(Delta N)` Here `Delta n = n_(p) - n_(r) = 2- 1 = 1 " mol " :. K_(p) = K_(c) (RT)` But `T= 327 + 273 = 600 K " (Given) "` ` R= 0.0821" L ATM"K^(-1) mol^(-1) :. K_(p) = 0.267 molL^(-1) xx 0.0821" L atm "K^(-1) mol^(-1) xx 600 K = 13.15 " atm"`