At 444,^oC,the equilibrium constant K for the | vedclass

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(B) The reaction is $2AB_{(g)} \rightleftharpoons A_{2(g)} + B_{2(g)}$.Let the initial moles of $AB$ be $1$.At equilibrium,the moles are: $AB = (1 - \

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$20$

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(B) The reaction is $2AB_{(g)} ightleftharpoons A_{2(g)} + B_{2(g)}$.Let the initial moles of $AB$ be $1$.At equilibrium,the moles are: $AB = (1 - \alpha)$,$A_2 = \alpha/2$,$B_2 = \alpha/2$,where $\alpha$ is the degree of dissociation.The equilibrium constant expression is $K = \frac{[A_2][B_2]}{[AB]^2} = \frac{(\alpha/2)(\alpha/2)}{(1-\alpha)^2} = \frac{\alpha^2}{4(1-\alpha)^2}$.Given $K = \frac{1}{64}$,we have $\frac{\alpha^2}{4(1-\alpha)^2} = \frac{1}{64}$.Taking the square root on both sides: $\frac{\alpha}{2(1-\alpha)} = \frac{1}{8}$.Simplifying: $\frac{\alpha}{1-\alpha} = \frac{2}{8} = \frac{1}{4}$.$4\alpha = 1 - \alpha$ $\Rightarrow 5\alpha = 1$ $\Rightarrow \alpha = 0.2$.Therefore,the degree of dissociation is $0.2 	imes 100 = 20\%$.

At $444\,^oC$,the equilibrium constant $K$ for the reaction $2AB_{(g)} \rightleftharpoons A_{2(g)} + B_{2(g)}$ is $\frac{1}{64}$. The degree of dissociation of $AB$ will be .....$\%$

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