For the gas-phase dissociation reaction PCl ₅( g ) PCl ₃( g ) + Cl ₂( g ) the standard free energy change G^ at 500 K and 1 atm total pressure is 2324 J mol ⁻¹ . The percentage dissociation of PCl ₅ at equilibrium is ________. [Given: ( 4 3 ) = 0.28 and R = 8.3 J K ⁻¹ mol ⁻¹ ]

Step-by-step solution

The standard free energy change is related to the equilibrium constant by the equation: G^ = - RT K_ p Substituting the given values: 2324 = -8.3 500 K_ p K_ p = - 2324 4150 = -0.56 Since ( 4 3 ) = 0.28 , we can write: -0.56 = -2 0.28 = -2 ( 4 3 ) = ( 9 16 ) Thus, K_ p = 9 16 . For the dissociation of PCl ₅ : PCl ₅( g ) PCl ₃( g ) + Cl ₂( g ) Initial moles: 1 0 0 Equilibrium moles: 1- Total moles at equilibrium = 1 + The partial pressures are: P_ PCl ₅ = ( 1- 1+ ) P P_ PCl ₃ = ( 1+ ) P P_ Cl ₂ = ( 1+ ) P The equili