Gaseous `N_2O_4` dissociates into gaseous `NO_2` according to the reaction `N_2O_4(g) hArr 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 :

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the dissociation reaction
The dissociation of dinitrogen tetroxide (N₂O₄) into nitrogen dioxide (NO₂) is represented as:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

### Step 2: Define the degree of dissociation
The degree of dissociation (α) is given as 0.2. This means that 20% of the N₂O₄ will dissociate into NO₂.

### Step 3: Calculate the moles at equilibrium
Initially, we have 1 mole of N₂O₄. The dissociation can be calculated as follows:
- Moles of N₂O₄ that dissociate = α × initial moles = 0.2 × 1 = 0.2 moles
- Moles of N₂O₄ remaining = 1 - 0.2 = 0.8 moles
- Moles of NO₂ produced = 2 × moles of N₂O₄ that dissociate = 2 × 0.2 = 0.4 moles

Thus, at equilibrium:
- Moles of N₂O₄ = 0.8 moles
- Moles of NO₂ = 0.4 moles

### Step 4: Calculate the total moles at equilibrium
Total moles at equilibrium = moles of N₂O₄ + moles of NO₂
\[ \text{Total moles} = 0.8 + 0.4 = 1.2 \text{ moles} \]

### Step 5: Calculate the molar mass of the mixture
The molar mass of N₂O₄ = 92 g/mol (approximately).
The molar mass of NO₂ = 46 g/mol.

Now, we can calculate the average molar mass (M) of the mixture:
\[ M = \frac{(moles \, of \, N_2O_4 \times M_{N_2O_4}) + (moles \, of \, NO_2 \times M_{NO_2})}{\text{Total moles}} \]
\[ M = \frac{(0.8 \times 92) + (0.4 \times 46)}{1.2} \]
\[ M = \frac{73.6 + 18.4}{1.2} = \frac{92}{1.2} = 76.67 \, g/mol \]

### Step 6: Use the ideal gas equation to find density
The ideal gas equation is given by:
\[ P = \frac{dRT}{M} \]
Where:
- P = pressure (1 atm)
- d = density (g/L)
- R = ideal gas constant (0.0821 L·atm/(K·mol))
- T = temperature (300 K)
- M = molar mass (76.67 g/mol)

Rearranging for density (d):
\[ d = \frac{PM}{RT} \]
Substituting the values:
\[ d = \frac{(1 \, atm)(76.67 \, g/mol)}{(0.0821 \, L·atm/(K·mol))(300 \, K)} \]
\[ d = \frac{76.67}{24.63} \approx 3.11 \, g/L \]

### Final Answer
The density of the equilibrium mixture is approximately **3.11 g/L**.

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