For the reaction N_2O_{4(g)} rightleftharpoon | vedclass

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Question

(A) For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$:Initial moles: $1$ $0$At equilibrium: $(1-\alpha)$ $2\alpha$Total moles at equilibriu

Vedclass · Accepted Answer

$\alpha=\sqrt{\frac{K_p}{K_p+4P}}$

Vedclass · Answer

(A) For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$:Initial moles: $1$ $0$At equilibrium: $(1-\alpha)$ $2\alpha$Total moles at equilibrium $= 1-\alpha+2\alpha = 1+\alpha$.Partial pressure of $N_2O_4 = \frac{1-\alpha}{1+\alpha} P$.Partial pressure of $NO_2 = \frac{2\alpha}{1+\alpha} P$.$K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(\frac{2\alpha}{1+\alpha} P)^2}{\frac{1-\alpha}{1+\alpha} P} = \frac{4\alpha^2 P}{1-\alpha^2}$.Rearranging for $\alpha$:$K_p(1-\alpha^2) = 4\alpha^2 P$$K_p = \alpha^2(4P + K_p)$$\alpha^2 = \frac{K_p}{4P+K_p}$$\alpha = \sqrt{\frac{K_p}{4P+K_p}}$.

For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the correct relation between degree of dissociation $(\alpha)$ of $N_2O_{4(g)}$ and equilibrium constant,$K_p$ is $(P=$ total pressure of mixture $)$

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