At 460 K, ∆G = -12.8 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

∆Gº = ∆G - RT ln K

K = (B)/(A)² = 0.420 / (7.20)² = 0.00810

∆Gº = -12.8 kJ/mol - (0.008314 kJ/Kmol)(460K) + ln 0.00810

∆Gº = -12.8 kJ/mol - 3.82 + (-4.82)

∆Gº = -21.44 kJ/mol

(be sure to check the math)