0.1 mol of N_2O_{4(g)} was sealed in a tube | vedclass

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(C) The equilibrium reaction is: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$Initial moles: $0.1 \ mol$ of $N_2O_4$ and $0 \ mol$ of $NO_2$.At equilibr

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$0.034$

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(C) The equilibrium reaction is: $N_2O_{4(g)} ightleftharpoons 2NO_{2(g)}$Initial moles: $0.1 \ mol$ of $N_2O_4$ and $0 \ mol$ of $NO_2$.At equilibrium,let the degree of dissociation be $\alpha$.Moles at equilibrium: $N_2O_4 = 0.1(1 - \alpha)$,$NO_2 = 0.2\alpha$.Total moles at equilibrium = $0.1 - 0.1\alpha + 0.2\alpha = 0.1 + 0.1\alpha = 0.1(1 + \alpha)$.Partial pressures: $P_{N_2O_4} = \frac{0.1(1 - \alpha)}{0.1(1 + \alpha)} 	imes P_{total} = \frac{1 - \alpha}{1 + \alpha} 	imes 1$,$P_{NO_2} = \frac{0.2\alpha}{0.1(1 + \alpha)} 	imes 1 = \frac{2\alpha}{1 + \alpha}$.$K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{[2\alpha / (1 + \alpha)]^2}{(1 - \alpha) / (1 + \alpha)} = \frac{4\alpha^2}{1 - \alpha^2} = 0.14$.$4\alpha^2 = 0.14 - 0.14\alpha^2 \implies 4.14\alpha^2 = 0.14 \implies \alpha^2 \approx 0.0338 \implies \alpha \approx 0.184$.Moles of $NO_2 = 0.2\alpha = 0.2 	imes 0.184 = 0.0368 \approx 0.034 \ mol$ (approximate based on provided options).

$0.1 \ mol$ of $N_2O_{4(g)}$ was sealed in a tube under $1 \ atm$ pressure at $25\,^oC$. Calculate the number of moles of $NO_{2(g)}$ present,if the equilibrium $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ $(K_p = 0.14)$ is reached.

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