Use the following data : array|c|c|c| Substance & f H (500 K) kJ mol-1 & S (500 K) J K-1 mol-1 \ AB(g) & 32 & 222 \ A2(g) & 6 & 146 \ B2(g) & x & 280 \ array One mole each of A2(g) and B2(g) are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K. A2(g) + B2(g) 2AB(g) The value of x (in kJ mol-1) is . (Nearest integer) (Given : K = 2.2, R = 8.3 J K-1 mol-1)
Answer
70
Solution
For A2(g) + B2(g) 2AB(g):
r H = 2(32) - 6 - x = (58 - x) kJ mol-1
r S = 2(222) - 146 - 280 = 18 J K-1 mol-1 = 0.018 kJ K-1 mol-1
r G = -2.303 RT K
= -2.303 8.3 500 2.2 J mol-1
= -21026 J mol-1 = -21.03 kJ mol-1
r G = r H - T r S
-21.03 = (58 - x) - 500 0.018
-21.03 = 49 - x
x = 70.03 70
Answer: 70