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 = -17.7 kJ/mol?

Answer 1

Use the relation ΔG = -RT lnK + RT lnQ and solve for Q and assuming the ΔG is per mole of A.

ln Q = ΔG + RT lnK / RT = -17.7 x 10³ J/mol + (8.314 J/ºK-mol x 298 K x ln 14.7) /( 8.314 J/K-mol x 298 K)

ln Q = -4.46 Take anitlog (base e)

Q = 0.0116 This indicates there is a preponderance of reactant over product.