Two moles of `PCl_(5)` were heated to `327^(@)C` in a closed two-litre vessel, and when equilibrium was achieved, `PCl_(5)` was found to be `40%` dissociated into `PCl_(3)` and `Cl_(2)`. Calculate the equilibrium constant `K_(p)` and `K_(c)` for this reaction.

Explanation

`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 mol L^(-1) xx 0.0821" L atm " K^(-1) mol^(-1) xx 600 K = 13.15 " atm "`