For the gas-phase reaction A(g) 2 B(g) , the equilibrium constant K_p is 1 at 300 K. If the standard internal energy change ( U^ ) for the reaction is 3.51 kJ mol ⁻¹ , what is the standard entropy change ( S^ ) in J K ⁻¹ mol ⁻¹ ? (Given: R = 8.3 J K ⁻¹ mol ⁻¹ )

Step-by-step solution

Since the equilibrium constant K_p = 1 , the standard Gibbs free energy change G^ = -RT K_p = 0 . From the relation G^ = H^ - T S^ , when G^ = 0 , we get H^ = T S^ . First, calculate the standard enthalpy change ( H^ ) using the relation H^ = U^ + n_g RT . For the reaction A(g) 2 B(g) , the change in the number of moles of gas is n_g = 2 - 1 = 1 . Calculate the work term n_g RT : n_g RT = 1 8.3 J K ⁻¹ mol ⁻¹ 300 K = 2490 J mol ⁻¹ = 2.49 kJ mol ⁻¹ Now, find H^ : H^ = 3.51 kJ mol ⁻¹ + 2.49 kJ mol ⁻¹ = 6.00 kJ mol ⁻¹