What is the value of 𝐾 for this aqueous reaction at 298 K? A+B↽−−⇀C+DΔ𝐺°=28.87 kJ/mol A + B ↽ − − ⇀ C + D Δ G ° = 28.87 kJ / mol

Answer 1

A + B <==> C + D ... ∆Gº = 28.87 kJ/mol

∆Gº = -RT ln K

∆Gº = 28.87 kJ/mol

R = 0.008314 kJ/K mol

T = 298K

K = equilibrium constant = ?

28.87 kJ/mol = -(0.008314 kJ/K mol)(298K) ln K

ln K = 28.87 / - 2.48 = -11.6

K = 9.2x10^-6 (this agrees with the positive value of ∆Gº meaning the reaction is not spontaneous)