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

∆G = ∆Gº + RT ln K

∆G = -12.8 kJ/mol

∆Gº = ?

R = gas constant = 0.008315 kJ/K mol

T = temp in K = 460 K

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

∆Gº = ∆G - RT ln K = -12.8 kJ/mol - (0.008314 kJ/K mol)(460 K) (-4.82)

∆Gº = -12.8 + 0.04

∆Gº = -12.76 kJ/mol

(please check my math)