Question

Gaseous $N_2O_4$ dissociates into gaseous $NO_2$ according to the reaction $N _{2} O _{4}( g ) \rightleftharpoons 2 NO _{2}( g )$ 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 :

• A. 1.56 g/L
• B. 3.11 g/L
• C. 4.56 g/L
• D. 6.22 g/L

Answer 1

The Correct Option is B

To solve this problem, we need to determine the density of the equilibrium mixture of gases. The dissociation reaction is given as:

N _{2} O _{4}( g ) \rightleftharpoons 2 NO _{2}( g )

We are provided the following information:

• Initial moles of N_2O_4: 1 mole
• Degree of dissociation, \alpha = 0.2
• Temperature, T = 300 \, K
• Pressure, P = 1 \, atm

Step-by-Step Calculation:

1. When N_2O_4 dissociates, \alpha (degree of dissociation) of it converts to NO_2. Thus, if the initial moles of N_2O_4 are 1, then moles dissociated = 0.2 \times 1 \, \text{mole} = 0.2.
2. The reaction produces 2 moles of NO_2 for each mole of N_2O_4 dissociated. Therefore, moles of NO_2 formed = 2 \times 0.2 = 0.4.
3. Moles of N_2O_4 remaining = Initial moles - Moles dissociated = 1 - 0.2 = 0.8.
4. Total moles at equilibrium = Moles of N_2O_4 + Moles of NO_2 = 0.8 + 0.4 = 1.2.
5. The ideal gas equation, in terms of density, is given by: \rho = \frac{PM}{RT}, where \rho is the density, P is the pressure, M is the molar mass, R is the universal gas constant, and T is the temperature.
6. The average molar mass, M, of the gas mixture can be calculated as:
   M = \frac{\text{moles of } N_2O_4 \cdot \text{molar mass of } N_2O_4 + \text{moles of } NO_2 \cdot \text{molar mass of } NO_2}{\text{total moles}}
   Substituting the known values: M = \frac{0.8 \times 92 + 0.4 \times 46}{1.2} = \frac{73.6 + 18.4}{1.2} = \frac{92}{1.2} = 76.67 \, \text{g/mol}
7. Using \rho = \frac{1 \, \text{atm} \times 76.67 \, \text{g/mol}}{0.0821 \, \text{L atm K}^{-1} \text{ mol}^{-1} \times 300 \, \text{K}}, we get:
   \rho = \frac{76.67}{24.63} = 3.11 \, \text{g/L}

Hence, the density of the equilibrium mixture is 3.11 g/L. Therefore, the correct answer is 3.11 g/L.