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 = -18.3 kJ/mol?

Answer 1

∆G = RT ln (Q/K)

∆G = RT ln Q - ln K

-18.3 kJ/mol = (0.008314 kJ/molK)(298K) lnQ - ln 14.7

-18.3 = 6.66 ln Q

ln Q = - 2.75

Q = 0.064