For the reaction,N_2 + O_2 rightleftharpoons | vedclass

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(C) Let the initial moles of $N_2$ and $O_2$ be $1$ mole each.Let $\alpha$ be the degree of dissociation for both $N_2$ and $O_2$.The reaction is $N_2

Vedclass · Accepted Answer

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

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(C) Let the initial moles of $N_2$ and $O_2$ be $1$ mole each.Let $\alpha$ be the degree of dissociation for both $N_2$ and $O_2$.The reaction is $N_2 + O_2 \rightleftharpoons 2NO$.At equilibrium,moles are: $N_2 = (1 - \alpha)$,$O_2 = (1 - \alpha)$,$NO = 2\alpha$.The equilibrium constant $K_c$ is given by: $K_c = \frac{[NO]^2}{[N_2][O_2]} = \frac{(2\alpha)^2}{(1 - \alpha)(1 - \alpha)} = \frac{(2\alpha)^2}{(1 - \alpha)^2}$.Given $K_c = 2$,we have $2 = \left(\frac{2\alpha}{1 - \alpha}\right)^2$.Taking the square root on both sides: $\sqrt{2} = \frac{2\alpha}{1 - \alpha}$.$\sqrt{2}(1 - \alpha) = 2\alpha$.$\sqrt{2} - \alpha\sqrt{2} = 2\alpha$.$\sqrt{2} = \alpha(2 + \sqrt{2})$.$\alpha = \frac{\sqrt{2}}{2 + \sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}(\sqrt{2} + 1)} = \frac{1}{1 + \sqrt{2}}$.

For the reaction,$N_2 + O_2 \rightleftharpoons 2NO$,the equilibrium constant is $K_c = 2$. If both reactants have the same initial moles,what is the degree of dissociation of $N_2$ and $O_2$?

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