For the reaction: X 2 O 4 l → 2 X O 2 g ∆ U = 2 .1   k   cal ,   ∆ S = 20   cal   K - 1 at 300   K Hence, ∆ G is:

Step-by-step solution

∆ H = ∆ U + ∆ n g   RT = 2 .1 + 2 × 2 × 300 1000 = 3 . 3   k   cal ∆ G = ∆ H - T ∆ S = 3.3 - 300 × 20 1000 = 3.3 - 6 = - 2.7   k   c a l