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Characteristics of Chemical Equilibrium

Gaseous N2O...

Question

Gaseous $N_{2} O_{4}$ dissociates into gaseous $N O_{2}$ according to the reaction $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ at 300 K and 1 atm pressure, the degree of dissociation of $N_{2} O_{4}$ is 0.2. If one mole of $N_{2} O_{4}$ gas is contained in a vessel, then the density of the equilibrium mixture is :

3.11 g/L

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4.56 g/L

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1.56 g/L

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6.22 g/L

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Solution

The correct option is B 3.11 g/L
$N_{2} O_{4} ⟶ 2 N O_{2}$

at $t = 0$ , moles of $N_{2} O_{4} = 1$ , moles of $N O_{2} = 0$

at $t = e q u i l i b r i u m$ , mole of $N_{2} O_{4} = 1 - a$ , mole of $N O_{2} = 2 a$

$a$ =degree of dissociation.

Molecular weight of mixture $= \frac{(1 - a) \times molar mass of N_{2} O_{4} + 2 a \times molar mass of N O_{2}}{(1 - a + 2 a)}$
$= \frac{(1 - 0.2) (28 + 64) + 2 \times 0.2 \times (14 + 32)}{1 + 0.2}$
$M = 76.66$

$P = 1 a t m, T = 300 K,$

$d = P M / R T$

$= \frac{1 \times 76.66}{0.082 \times 300} = 3.11 g m / l i t$ .

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Similar questions

Q. According to the reaction gaseous N

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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 gas is contained in a vessel, then the density of the equilibrium mixture is :