What is K for a reaction if ∆G° =150.8 kJ/mol at 25.0 °C?

Answer 1

∆Gº = -RT ln K

∆Gº = 150.8 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol

T = 25.0ºC + 273 = 298K

K = ?

150.8 = -(0.008314)(298) ln K

150.8 = -2.478 ln K

ln K = -60.89

K = 3.60x10^-27

Answer 2

K = e^-ΔGº/(RT)

Where R = 8.314 J/mole K ΔG° must be in J/mole and T in Kelvin.