For the reaction A (g) → 3 B (g), Kp = 0.623 at 298 K. What is the value of ∆G for this reaction at 298 K when the partial pressures of A and B are 4.60 atm and 0.120 atm?

Question

For the reaction A (g) → 3 B (g), Kp = 0.623 at 298 K. What is the value of ∆G for this reaction at 298 K when the partial pressures of A and B are 4.60 atm and 0.120 atm?

Answer 1

Is the reaction @ equilibrium or not?

Compare Q to Kp:

Q = (P_B)³ / (P_A) = (0.120)³ / 4/60 = 0.00037 which is much below the Kp of 0.623, so not at equilibrium

∆G = ∆Gº + RT ln Q

First, we must find ∆Gº:

∆Gº = -RT ln K = -(0.008314 kJ/Kmol)(298) ln 0.623 = 1.17 kJ/mol

∆G = ∆Gº + RT ln Q

∆G = 1.17 kJ/mol + (0.008314 kJ/Kmol)(298) ln 0.00037

∆G = 1.17 - 19.58

∆G = -18.4 kJ/mol