One mole each of A?(g) and B?(g) are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K. A?(g) + B?(g) ? 2AB(g)

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Solution

Q.
Use the following data :

Substance

$\frac{Δ_{f} H^{⊝} (500 K)}{kJ mol⁻¹}$
$\frac{S^{⊝} (500 K)}{J K⁻¹ mol⁻¹}$

AB(g)

32

222

$A_{2}$ (g)

6

146

$B_{2}$ (g)

x

280

One mole each of $A_{2} (g) and B_{2} (g)$ are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K.

$A_{2} (g) + B_{2} (g) ⇌ 2 AB (g)$

The value of x $(in kJ mol⁻¹)$ is __________. (Nearest integer)

(Given : log K = 2.2 $R = 8.3 J K⁻¹ mol⁻¹)$ [2026]

Ans.
(70)

$A_{2} + B_{2} \overset{500 K}{\to} 2 AB \log K = 2.2$

$Δ H ° = (2 \times 32) - (6 + x) = (58 - x) kJ$

$Δ S ° = (2 \times 222) - (146 + 280) = 18 J$

$Δ G ° = - R T \ln K$

$Δ G ° = - \frac{8.314 \times 500 \times 2.2 \times 2.303}{1000}$

$Δ G ° = - 21.06$

$Δ H ° - T Δ S ° = - 21.06$

$58 - x - 500 (\frac{18}{1000}) = - 21.06$

$x = 70.06 kJ/mol$

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