At 460 K, ∆G = -13.80 kJ/mol for the reaction 2 A (g) → B (g). If the partial pressures of A and B are 7.20 atm and 0.420 atm respectively, what is ∆G° for this reaction?

Answer 1

2A ==> B

Q = (B) / (A)²

Q = (0.420) / (7.20)² = 0.420 / 51.84

Q = 0.0081

∆G = ∆Gº+ RT ln Q

-13.80 kJ/mol = ∆Gº (0.008314 kJ/molK)(460) ln 0.0081

-13.80 = -18.42∆Gº

∆Gº = 0.749 kJ/mol