One mole each of A₂(g) and B₂(g) are taken in a 1 L closed flask and allowed to establish equilibrium at 500 K:

A₂(g) + B₂(g) ⇌ 2AB(g)

Find the value of x (in kJ mol⁻¹).
(Given: log K = 2.2, R = 8.3 J K⁻¹ mol⁻¹)

Concept Used

ΔG° = ΔH° − TΔS°
ΔG° = −2.303 RT log K

Solution

Step 1: Calculate ΔG°

ΔG° = −2.303 × R × T × log K
= −2.303 × 8.3 × 500 × 2.2
≈ −21000 J ≈ −21 kJ

Step 2: Calculate ΔH°

ΔH° = [2 × ΔfH°(AB)] − [ΔfH°(A₂) + ΔfH°(B₂)]

= (2 × 32) − (6 + x)
= 64 − 6 − x
= 58 − x

Step 3: Calculate ΔS°

ΔS° = [2 × S°(AB)] − [S°(A₂) + S°(B₂)]

= (2 × 222) − (146 + 280)
= 444 − 426
= 18 J K⁻¹

Step 4: Apply ΔG° = ΔH° − TΔS°

−21 = (58 − x) − (500 × 18 / 1000)

−21 = (58 − x) − 9

−21 = 49 − x

x = 70 kJ mol⁻¹

Final Answer:

x = 70 kJ mol⁻¹