Gaseous N_2O_4 dissociates into gaseous NO_2 | vedclass

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(C) Reaction: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$Initial moles $(t=0)$: $1 \ mol$ of $N_2O_4$ and $0 \ mol$ of $NO_2$.At equilibrium $(t=eq)$:

Vedclass · Accepted Answer

$3.11$

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(C) Reaction: $N_2O_{4(g)} ightleftharpoons 2NO_{2(g)}$Initial moles $(t=0)$: $1 \ mol$ of $N_2O_4$ and $0 \ mol$ of $NO_2$.At equilibrium $(t=eq)$: $(1-\alpha) \ mol$ of $N_2O_4$ and $2\alpha \ mol$ of $NO_2$,where $\alpha = 0.2$.Total moles at equilibrium: $n_{total} = (1-\alpha) + 2\alpha = 1 + \alpha = 1 + 0.2 = 1.2 \ mol$.Molar mass of $N_2O_4 = 92 \ g/mol$ and $NO_2 = 46 \ g/mol$.Average molar mass of the mixture $(M_{mix})$:$M_{mix} = \frac{(1 - \alpha) 	imes 92 + 2\alpha 	imes 46}{1 + \alpha} = \frac{0.8 	imes 92 + 0.4 	imes 46}{1.2} = \frac{73.6 + 18.4}{1.2} = \frac{92}{1.2} = 76.66 \ g/mol$.Using the ideal gas law $PV = nRT$ or $PM = dRT$:$d = \frac{P 	imes M_{mix}}{R 	imes T} = \frac{1 	imes 76.66}{0.0821 	imes 300} = \frac{76.66}{24.63} \approx 3.11 \ g/L$.

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