A reaction has a standard free‑energy change of −17.00 kJ mol−1(−4.063 kcal mol−1). Calculate the equilibrium constant for the reaction at 25 °C.

Answer 1

∆Gº = -RT ln K

∆Gº = -17.00 kJ / mol

R = 0.008314 kJ / Kmol

T = 25 + 273 = 298K

K = ?

-17 kJ/mol = -(0.008314 kJ/Kmol)(298K) ln K

ln K = -17 / -2.478

ln K = 6.86

K = 954