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(C) For the reaction: $A_2 + B_2 \rightleftharpoons 2AB$Initial moles: $1$ mol of $A_2$ and $1$ mol of $B_2$ are taken.At equilibrium: $A_2 = (1 - \al

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$\frac{1}{1 + \sqrt{2}}$

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(C) For the reaction: $A_2 + B_2 \rightleftharpoons 2AB$Initial moles: $1$ mol of $A_2$ and $1$ mol of $B_2$ are taken.At equilibrium: $A_2 = (1 - \alpha)$,$B_2 = (1 - \alpha)$,$AB = 2\alpha$.Equilibrium constant $K_c = \frac{[AB]^2}{[A_2][B_2]} = \frac{(2\alpha)^2}{(1-\alpha)(1-\alpha)} = \frac{4\alpha^2}{(1-\alpha)^2}$.Given $K_c = 2$,so $\frac{4\alpha^2}{(1-\alpha)^2} = 2$.Taking square root on both sides: $\frac{2\alpha}{1-\alpha} = \sqrt{2}$.$2\alpha = \sqrt{2} - \sqrt{2}\alpha$.$\alpha(2 + \sqrt{2}) = \sqrt{2}$.$\alpha = \frac{\sqrt{2}}{2 + \sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}(\sqrt{2} + 1)} = \frac{1}{1 + \sqrt{2}}$.

The equilibrium constant at a certain temperature for the reaction $A_2 + B_2 \rightleftharpoons 2AB$ is $2$. Calculate the degree of dissociation of either $A_2$ or $B_2$.

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