At 460 K, ∆G = -13.5 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 Kp

First, we will find the value of Kp, and the substitute it into the above equation and solve for ∆Gº.

2A(g) ==> B(g)

Kp = (B) / (A)²

Kp = (0.420) / (7.20)² = 0.420 / 51.84

Kp = 8.10x10^-3

∆G = ∆Gº + RT ln Kp

-13.5 kJ/mol = ∆Gº + (0.008314 kJ/molK)(460K) ln (8.10x10^-3)

-13.5 kJ/mol = ∆Gº + (3.82) (-4.82)

-13.5 kJ/mol = ∆Gº - 18.40

∆Gº = 4.9 kJ / mol