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ExamsJEE MainChemistry

Use the following data : Substance | ΔfH° (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 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: log K = 2.2 R = 8.3 J K−1 mol−1)

  1. 70
  2. 71
  3. 69
  4. 72

Correct answer: 70

Solution

The correct option is 70 because, using the given equilibrium constant and the thermodynamic relationships, we can calculate the standard enthalpy change for the reaction, which leads to the determination of the value of X for B2(g) as 70 kJ mol−1.

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