Gaseous N2O4 dissociates into gaseous NO2 according to the reaction N2O4(g)⇌2NO2(g)At 300 K and 1 atm pressure the degree of dissociation of N2O4 is 0.2. If one mole of N2O4 is contained in a vessel, then the density of equilibrium mixture is

A
$1.56 g / L$
B
$3.11 g / L$
C
$4.56 g / L$
D
$6.22 g / L$

Solution:

We have

$\begin{aligned} N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g) \\ n_{0} (1 - α) n_{0} (2 α) \end{aligned}$

Total amount of gases, $n = n_{0} (1 - α) + n_{0} (2 α) = n_{0} (1 + α) = (1 mol) (1 + 0.2) = 1.2 mol$

The density of equilibrium mixture is

$\begin{array}{l} ρ = \frac{mass of 1 mol of N_{2} O_{4}}{V} = \frac{(92 g {mol}^{- 1})}{(1.2 mol) R T / p} \\ = \frac{(92 g {mol}^{- 1}) p}{(1.2 mol) (R T)} = \frac{(92 g {mol}^{- 1}) (101.325 kPa)}{(1.2 mol) (8.314 kPa {dm}^{3} K^{- 1} {mol}^{- 1}) (300 K)} = 3.11 g {dm}^{- 3} \end{array}$

Gaseous $N_{2} O_{4}$ dissociates into gaseous ${NO}_{2}$ according to the reaction $N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g)$
At 300 K and 1 atm pressure the degree of dissociation of $N_{2} O_{4}$ is 0.2. If one mole of $N_{2} O_{4}$ is contained in a vessel, then the density of equilibrium mixture is

Solution:

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