Use the following data : Substance f –1 H (500K) kJmol –1 –1 S (500K) JK mol AB(g) 222 A2(g) 146 B2(g) X One mole each of A2(g) and B2(g) are taken in a 1L closed flask and allowed to establish the eq

Question

Use the following data : Substance f –1 H (500K) kJmol –1 –1 S (500K) JK mol AB(g) 222 A2(g) 146 B2(g) X One mole each of A2(g) and B2(g) are taken in a 1L closed flask and allowed to establish the equilibrium at 500K. A2(g) + B2(g) 2AB(g) The value of x (in kJ mol –1) is ……….. (Nearest integer) (Given: log K=2.2 R=8.3 JK –1 mol –1)

Prashant Kotian · Accepted Answer

Correct answer: .. 500K 2 A B 2AB   logK = 2.2 H° = (2 × 32) – (6 + x) = (58 – x) kJ S° = (2 × 222) – (146 + 280) = 18 Joule G° = –RTlnK 8.314 500 2.2 2.303 G     G° = –21.06 H° – TS° = –21.06 58 x 21.06          x = 70.06 KJ/mol

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