Gaseous $$N_2O_4$$ dissociates into gaseous $$NO_2$$ according to the reaction $$N_2O_4(g) \rightleftharpoons 2NO_2(g)$$. At 300 K and 1 atm pressure, the degree of dissociation of $$N_2O_4$$ is 0.2. If one mole of $$N_2O_4$$ gas is contained in a vessel, then the density of the equilibrium mixture is:

We start with 1 mole of $$N_2O_4$$ gas in a vessel. The reaction is $$N_2O_4(g) \rightleftharpoons 2NO_2(g)$$. The degree of dissociation, $$\alpha$$, is given as 0.2 at 300 K and 1 atm pressure. This means 20% of the $$N_2O_4$$ dissociates.

Initial moles of $$N_2O_4$$ = 1 mole, and initial moles of $$NO_2$$ = 0 moles.

At equilibrium:

• Moles of $$N_2O_4$$ dissociated = $$\alpha \times 1 = 0.2$$ moles.
• Moles of $$N_2O_4$$ left = initial moles - dissociated moles = $$1 - 0.2 = 0.8$$ moles.
• Since each mole of $$N_2O_4$$ produces 2 moles of $$NO_2$$, moles of $$NO_2$$ formed = $$2 \times 0.2 = 0.4$$ moles.

Total moles at equilibrium = moles of $$N_2O_4$$ + moles of $$NO_2$$ = $$0.8 + 0.4 = 1.2$$ moles.

The total mass remains constant because the reaction involves only rearrangement of atoms. The molar mass of $$N_2O_4$$ is $$(2 \times 14) + (4 \times 16) = 28 + 64 = 92$$ g/mol. Since we started with 1 mole of $$N_2O_4$$, the total mass is $$1 \times 92 = 92$$ grams.

To find the density, we need the volume of the gas mixture at equilibrium. Using the ideal gas law: $$PV = nRT$$, where:

• $$P = 1$$ atm (pressure)
• $$V = ?$$ (volume in liters)
• $$n = 1.2$$ moles (total moles at equilibrium)
• $$R = 0.0821$$ L·atm·K⁻¹·mol⁻¹ (gas constant)
• $$T = 300$$ K (temperature)

Rearranging for volume: $$V = \frac{nRT}{P}$$.

Substitute the values: $$V = \frac{(1.2) \times (0.0821) \times (300)}{1}$$.

First, compute $$RT$$: $$0.0821 \times 300 = 24.63$$ L·atm·mol⁻¹.

Then, $$n \times RT = 1.2 \times 24.63 = 29.556$$ L·atm.

Since $$P = 1$$ atm, $$V = \frac{29.556}{1} = 29.556$$ liters.

Density is mass per unit volume: $$\text{Density} = \frac{\text{Total mass}}{\text{Volume}} = \frac{92}{29.556}$$.

Calculate the division: $$92 \div 29.556 \approx 3.1127$$ g/L, which rounds to 3.11 g/L.

Hence, the correct answer is Option A.