At 560 K, ∆G° = -30.7 kJ/mol for the reaction A (g) + 2 B (g) → 3 C (g).

Question

If the partial pressures of A, B, and C are 0.420 atm, 0.110 atm, and 7.20 atm respectively, what is the free energy for this reaction?

Answer 1

∆G = ∆Gº + RT ln Q

∆G = ?

∆Gº = -30.7 kJ/mol

R = 0.008314 kJ/mol-K

T = 560K

Q = reaction quotient (see below for calculation of the value of Q)

Q = (C)³ / (A)(B)² = (7.20)³ / (0.420)(0.110)² = 73,445

∆G = ∆Gº + RT ln Q

∆G = -30.7 + (0.008314)(560) (ln 73,445)

∆G = -30.7 + 52.2

∆G = 21.5 kJ/mol

(please be sure to check the math)