For the reaction, X2O4(l)⟶2XO2(g)ΔU=2.1kcal,ΔS=20calK-1at 300 K Hence, ΔG is

ΔH=ΔU+ΔngRT Given, ΔU=2.1kcal,Δng=2R=2×10-3kcal,T=300 K∴  ΔH=2.1+2×2×10-3×300=3.3kcal Again, ΔG=ΔH-TΔS Given, ΔS=20×10-3kcalK-1 On putting the value of ΔH in the equation, we get ΔG=3.3-300×20×10-3=3.3-6×103×10-3=-2.7kcal