The dissociation of PCl₅(g) into PCl₃(g) and Cl₂(g) is observed at 300 K and a total pressure of 2 atm . If the observed vapor density of the equilibrium mixture is 69.5 , the standard free energy change ( G^ ) for the reaction is ________ J mol ⁻¹ . [Given : Atomic masses: P = 31 , Cl = 35.5 ; R = 8.3 J K ⁻¹ mol ⁻¹ ; (1.5) = 0.40 ]

Step-by-step solution

Molar mass of PCl₅ = 31 + 5 35.5 = 208.5 g mol ⁻¹ Theoretical vapor density ( D ) = 208.5 2 = 104.25 Observed vapor density ( d ) = 69.5 Degree of dissociation ( ) for PCl₅(g) PCl₃(g) + Cl₂(g) is given by: = D - d d (since 1 mole of reactant gives 2 moles of products) = 104.25 - 69.5 69.5 = 34.75 69.5 = 0.5 The equilibrium constant K_p is: K_p = ^2 P 1- ^2 Substitute = 0.5 and P = 2 atm : K_p = (0.5)^2 2 1 - (0.5)^2 = 0.25 2 0.75 = 2 3 Standard free energy change ( G^ ) is: G^ = -RT K_p G^ = -8.3 300 ( 2 3 ) G^ = 2