For the reaction, X2O₄(l) → 2XO₂(g) ∆U = 2.1 kcal, ∆S = 20 cal K–1 at 300 K Hence, ∆G is (2014)

∆H = ∆U + ∆ngRT Given, ∆U = 2.1 kcal, ∆ng = 2, R = 2 × 10–3 kcal, T = 300 K ∆H = 2.1 + 2 × 2 × 10–3 × 300 = 3.3 kcal Again, ∆G = ∆H – T∆S Given, ∆S = 20 × 10–3 kcal K–1 On putting the values of ∆H and ∆S in the equation, we get ∆G = 3.3 – 300 × 20 × 10–3 = 3.3 – 6 × 103 × 10–3 = – 2.7 kcal