Given, ΔH=−57.8kJmol−1 ΔS=−176JK−1mol−1 T=298K Using Gibb’s free energy relation ΔG=ΔH−TΔS where, ∆G = change in Gibb’s free energy ∆H = change in enthalpy T = temperature ∆S = change in entropy ΔG=57.8kJ∕mol−[298K×(−176Jk−1mol−1)] =57.8kJ∕mol−(298×