Question

At 444°C , the equilibrium constant K for the reaction 2 AB ⇌ A 2 + B 2 is 1 64 , The degree of dissociation of AB will be

Answer 1

0.9624

Answer 2

For the reaction $2 \text{AB} \rightleftharpoons \text{A}_2 + \text{B}_2$, $\Delta n_g = 0$, so $K_p = K_c$. Let initial moles of AB be $n_0$ and degree of dissociation be $\alpha$. At equilibrium: AB: $n_0(1-\alpha)$, A$_2$: $n_0\alpha/2$, B$_2$: $n_0\alpha/2$. $K = \frac{[\text{A}_2][\text{B}_2]}{[\text{AB}]^2} = \frac{(\alpha/2)^2}{(1-\alpha)^2} = \frac{\alpha^2}{4(1-\alpha)^2}$. $\sqrt{K} = \frac{\alpha}{2(1-\alpha)} \implies \alpha = \frac{2\sqrt{K}}{1+2\sqrt{K}}$. Given $K = 164$, $\alpha = \frac{2\sqrt{164}}{1+2\sqrt{164}} \approx 0.9624$.