Determine the temperature of a reaction if K = 1.20 x 10⁻⁶ when ∆G° = +20.90 kJ/mol.

Answer 1

∆Gº = -RT ln K

20.90 kJ/mol = -(0.008314 kJ/molK)(T) ln 1.20x10^-6

20.90 kJ/mol = -0.008314 kJmol T x -13.63

T = 20.90 / 0.1133

T = 1844K