A reaction has a standard free‑energy change of −10.20 kJ mol−1(−2.438 kcal mol−1). Calculate the equilibrium constant for the reaction at 25 °C.

Question

𝐾eq=______

Answer 1

∆Gº = -RT ln K

∆Gº = -10.20 kJ/mol

R = 8.314 J/Kmol = 0.008314 J/Kmol

T = 25C + 273 = 298K

-10.20 = -(0.008314)(298) ln K

ln K = -10.20 / -2.478 = 4.116

K = 61.3