For the reaction H2(g) + 12O2(g) → H2O(l),ΔH = −285.8kJmol−1 ΔS = −0.163kJmol−1K−1. What is the value of free energy change 27∘C for the reaction.
ΔG = ΔT − TΔS,T = 27 + 273 = 300K
ΔG = (−285.8) − (300)(−0.163) = −236.9kJmol−1