1) For the reaction $H_{2}(g) + \frac{1}{2}O_{2}(g)\rightarrow H_{2}O(l)$.ΔH = - 285.8 kJ/mol, ΔS = - 0.163kJ/molK.What is the value of free energy change at 27°C for the reaction? A) -236.9kJ/mol B) -281.4kJ/mol C) -334.7kJ/mol D) +334.7kJ/mol Answer: Option AExplanation:ΔG = ΔH - TΔS T = 27+273 = 300K ΔG = (-285.8) - 300(-0.163) kJ/mol = -236.9kJ/mol
