Determine ∆G° for a reaction when ∆G = -152.7 kJ/mol and Q = 0.043 at 298 K. (R = 8.314 J/mol ･ K)

Answer 1

∆G = ∆G° + RTlnQ

Rearrange using algebra to isolate ∆G°:

∆G° = ∆G - RTlnQ = -152.7 kJ/mol - (8.314 J/mol･K)(298 K)ln(0.043)

Problem: ∆G has units of kJ/mol, but R has units of J/mol･K. Convert R to kJ/mol･K by dividing it by 1000:

∆G° = -152.7 kJ/mol - (8.314*10^-3 kJ/mol･K)(298 K)ln(0.043) = -152.7 kJ/mol + 7.7958 kJ/mol = -144.9 kJ/mol