Chemistry II Partial Pressure

Question

For the reaction A (g) → 3 B (g), Kp = 0.783 at 298 K. What is the value of ∆G for this reaction at 298 K when the partial pressures of A and B are 5.70 atm and 0.250 atm?

Answer 1

A (g) → 3 B (g)

Kp = 0.783 = (B)³ / (A) = (0.250)³ / (5.70)

Kp = 0.00274

∆G = -RT ln K

∆G = -(0.008314kJ/Kmol)(298K) ln 0.00274

∆G =-(0.008314)(298)(-5.899)

∆G = 14.6 kJ/mol