Determine the temperature of a reaction if K = 1.2 x 10⁻⁶ when ∆G° = +28.3 kJ/mol.

Answer 1

∆Gº = -RT ln K

28.3 kJ/mol = - 0.008314 kJ/mol-K (T) * ln 1.2x10^-6

T = 28.3 kJ/mol ÷ (-0.008314 kJ/mol-K)(ln 1.2x10^-6)

T = 28.3 kJ/mol / (-0.008314)(-13.6) = 28.3 / 0.113

T = 250K