MCQ
At $444\,^oC$ , the equilibrium constant $K$ for the reaction $2A{B_{(g)}} \rightleftharpoons {A_{2(g)}} + {B_{2(g)}}$ is $\frac {1}{64}$ . The degree of dissociation of $AB$ will be .....$\%$
- A$10$
- ✓$20$
- C$30$
- D$50$
✓
Answer
Correct option: B.
$20$
b
$\mathop {\mathop {2AB}\limits_1 }\limits_{1 - \alpha } \rightleftharpoons \mathop {\mathop {{A_2}(g)}\limits_0 }\limits_{\frac{\alpha }{2}} + \mathop {\mathop {{B_2}(g)\,\,}\limits_0 }\limits_{\frac{\alpha }{2}} \,\,\,\,\,\,\,{K_p} = \frac{1}{{64}}$
$\mathop {\mathop {2AB}\limits_1 }\limits_{1 - \alpha } \rightleftharpoons \mathop {\mathop {{A_2}(g)}\limits_0 }\limits_{\frac{\alpha }{2}} + \mathop {\mathop {{B_2}(g)\,\,}\limits_0 }\limits_{\frac{\alpha }{2}} \,\,\,\,\,\,\,{K_p} = \frac{1}{{64}}$
$\mathrm{K}_{\mathrm{p}}=\frac{\left(\frac{\alpha}{2}\right)^{2}}{(\mathrm{p}-\alpha)^{2}}$
$\sqrt{\frac{1}{64}}=\frac{\alpha}{2(1-\alpha)}$
$\alpha=\frac{1}{5}=0.2$
$\alpha=20 \%$
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