What is the free energy for this reaction?

Question

At 565 K, ∆G° = -18.1 kJ/mol for the reaction A (g) + 2 B (g) → 3 C (g). If the partial pressures of A, B, and C are 1.21 atm, 3.25 atm, and 0.235 atm respectively, what is the free energy for this reaction?

Answer 1

A(g) + 2B(g) ==> 3C(g)

∆Gº = -RT ln K

-18.1 kJ/mol = -(0.008314 kJ/Kmol)(565) ln K

ln K = -18.1 / -4.697

ln K = 3.85

K = 47.2

Q = (P_C)³ / (P_A)(P_B)²

Q = (0.235)³ / (1.21)(3.25)²

Q = 1.02x10^-3

∆G = ∆Gº + RT ln Q

∆G = -18.1 + (0.008314 kJ/Kmol)(565) ln 1.02x10^-3

∆G = -18.1 + (-32.4)

∆G = -50.5 kJ/mol