9.2 g of N_2O_{4(g)} is taken in a closed 1 | vedclass

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(C) $1$. Initial moles of $N_2O_4 = \frac{9.2 \ g}{92 \ g \ mol^{-1}} = 0.1 \ mol$. $2$. Since the volume of the vessel is $1 \ L$,the initial concent

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$0.2$

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(C) $1$. Initial moles of $N_2O_4 = \frac{9.2 \ g}{92 \ g \ mol^{-1}} = 0.1 \ mol$. $2$. Since the volume of the vessel is $1 \ L$,the initial concentration $[N_2O_4]_0 = 0.1 \ mol \ L^{-1}$. $3$. At equilibrium,$50\%$ of $N_2O_4$ is dissociated,so the amount dissociated is $0.1 	imes 0.5 = 0.05 \ mol$. $4$. The equilibrium concentrations are: $[N_2O_4]_{eq} = 0.1 - 0.05 = 0.05 \ mol \ L^{-1}$ and $[NO_2]_{eq} = 2 	imes 0.05 = 0.1 \ mol \ L^{-1}$. $5$. The equilibrium constant $K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(0.1)^2}{0.05} = \frac{0.01}{0.05} = 0.2 \ mol \ L^{-1}$.

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