What is the value of 𝐾 for this aqueous reaction at 298 K?

Question

A+B↽−−⇀C+D Δ𝐺°=15.85 kJ/mol

K=_______

Answer 1

A + B <==> C + D

K = [C][D] / [A][B]

∆Gº = -RT ln K

∆Gº = 15.85 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol

T = temp in K = 298K

Solving for K, we have...

ln K = ∆Gº / -RT = 15.85 / -(0.008314)(298) = 15.85 / -2.478

ln K = -6.396

K = 1.67x10^-3