Calculate the value of ΔG in kJ at 298 K if the partial pressures of N2, H2 and NH3 are 8.553, 8.881, and 63 atm respectively.

Question

The value of ΔG^o for the reaction, N₂(g) + 3 H₂(g) <=> 2 NH₃(g) is -32.90 kJ at 298 K.

Calculate the value of ΔG in kJ at 298 K if the partial pressures of N₂, H₂ and NH₃ are 8.553, 8.881, and 63 atm respectively.

The correct answer is -33.9

Answer 1

∆G = ∆Gº + RT ln K

K = (NH₃)² / (N₂)(H₂)³ = (63)² / (8.553)(8.881)³ = 0.6625

∆G = -32.90 + (0.00314)(298) ln 0.6625 = -32.90 + -0.4118

∆G = -33.3 (you might want to check the math, but this is close to the answer given)