The decomposition of PCl ₅( g ) into PCl ₃( g ) and Cl ₂( g ) is carried out at 300 K . The observed vapour density of the equilibrium mixture at a total pressure of 1.2 atm is 69.5 . The standard Gibbs free energy change ( G^ ) for the reaction at 300 K is ____ J mol ⁻¹ (nearest integer). (Given: Molar mass of PCl ₅ = 208.5 g mol ⁻¹ , R = 8.3 J K ⁻¹ mol ⁻¹ , 10 = 2.3 , 2 = 0.3 )

Step-by-step solution

Theoretical vapour density of PCl ₅ ( D ) = Molar mass 2 = 208.5 2 = 104.25 Degree of dissociation ( ) is given by: = D - d (n - 1)d Here, n = 2 (since 1 mole of PCl ₅ gives 2 moles of products). = 104.25 - 69.5 (2 - 1) 69.5 = 34.75 69.5 = 0.5 For the reaction PCl ₅( g ) PCl ₃( g ) + Cl ₂( g ) : Total moles at equilibrium = 1 + K_p = ^2 1 - ^2 P K_p = (0.5)^2 1 - (0.5)^2 1.2 = 0.25 0.75 1.2 = 1 3 1.2 = 0.4 Standard Gibbs free energy change: G^ = -2.303 RT K_p Using the given approximations: G^ = -2.3 8.3 300 (0.4)