An isomerization reaction A( g ) B( g ) is conducted in a closed vessel at a constant temperature of 300 K and 1 atm pressure. The standard free energy change ( G^ ) of the reaction is -3436.2 J mol ⁻¹ . If initially 1 mole of A was taken, the percentage of A converted to B at equilibrium is ____. (Given: R = 8.3 J K ⁻¹ mol ⁻¹ , 10 = 2.3 , 2 = 0.3 )

Step-by-step solution

For the reaction A( g ) B( g ) , the standard Gibbs free energy change is given by: G^ = -2.303 RT K_p Using the given approximations: -3436.2 = -2.3 8.3 300 K_p -3436.2 = -5727 K_p K_p = 3436.2 5727 = 0.6 Since 4 = 2 2 = 2(0.3) = 0.6 , we have K_p = 4 . Let be the degree of dissociation. Initial moles: 1 mole of A , 0 moles of B . Moles at equilibrium: (1 - ) moles of A , moles of B . Since the number of moles does not change, the partial pressures are proportional to the number of moles. K_p = P_B P_A = 1 - Subst