For the following equilibrium N_2O_4 rightlef | vedclass

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(D) Let the initial moles of $N_2O_4$ be $1$ and $NO_2$ be $0$. At equilibrium,the moles are $N_2O_4 = (1 - \alpha)$ and $NO_2 = 2\alpha$,where $\alph

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$33.33$

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(D) Let the initial moles of $N_2O_4$ be $1$ and $NO_2$ be $0$. At equilibrium,the moles are $N_2O_4 = (1 - \alpha)$ and $NO_2 = 2\alpha$,where $\alpha$ is the degree of dissociation.Total moles at equilibrium = $(1 - \alpha) + 2\alpha = 1 + \alpha$.Since volume is proportional to the number of moles at constant pressure and temperature,the mole fraction is equal to the volume fraction.Given that $NO_2$ is $50\%$ of the total volume,we have $\frac{2\alpha}{1 + \alpha} = 0.5$.$2\alpha = 0.5(1 + \alpha) \implies 2\alpha = 0.5 + 0.5\alpha$.$1.5\alpha = 0.5 \implies \alpha = \frac{0.5}{1.5} = \frac{1}{3}$.Percentage of dissociation = $\alpha 	imes 100 = \frac{1}{3} 	imes 100 = 33.33\%$.

For the following equilibrium $N_2O_4 \rightleftharpoons 2NO_2$ in gaseous phase,$NO_2$ is $50\%$ of the total volume when equilibrium is set up. Hence,the percentage of dissociation of $N_2O_4$ is.......$\%$

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