For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -15.9 kJ/mol?

Answer 1

∆G = -RT ln Q

-15.9 kJ/mol = -(0.008314 kJ/mol-K)(298K)(ln Q)

ln Q = -15.9 / - 2.48 = 6.41

Q = 608

note that since ∆G is negative, the reaction would be spontaneous and a large value of Q indicates the same thing, so it is a reasonable answer.