Gaseous `N_(2)O_(4)` dissociates into gaseous `NO_(2)` according to the reaction :
`[N_(2)O_(4)(g)hArr2NO_(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 :

Options

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

Explanation

To find the density of the equilibrium mixture of gases resulting from the dissociation of \( N_2O_4 \) into \( NO_2 \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The dissociation reaction is:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

### Step 2: Define the degree of dissociation
Let the degree of dissociation be represented by \( \alpha \). Given that \( \alpha = 0.2 \), this means that 20% of \( N_2O_4 \) dissociates.

### Step 3: Calculate the moles at equilibrium
Initially, we have 1 mole of \( N_2O_4 \). At equilibrium:
- Moles of \( N_2O_4 \) remaining = \( 1 - \alpha = 1 - 0.2 = 0.8 \) moles
- Moles of \( NO_2 \) produced = \( 2\alpha = 2 \times 0.2 = 0.4 \) moles

Total moles at equilibrium:
\[ \text{Total moles} = \text{Moles of } N_2O_4 + \text{Moles of } NO_2 = 0.8 + 0.4 = 1.2 \text{ moles} \]

### Step 4: Calculate the average molar mass of the mixture
The molar mass of \( N_2O_4 \) is 92 g/mol, and the molar mass of \( NO_2 \) is 46 g/mol.

The average molar mass of the mixture can be calculated as:
\[ \text{Average molar mass} = \frac{(0.8 \text{ moles} \times 92 \text{ g/mol}) + (0.4 \text{ moles} \times 46 \text{ g/mol})}{1.2 \text{ moles}} \]

Calculating the numerator:
\[ (0.8 \times 92) + (0.4 \times 46) = 73.6 + 18.4 = 92 \text{ g} \]

Now, dividing by the total moles:
\[ \text{Average molar mass} = \frac{92 \text{ g}}{1.2} = 76.67 \text{ g/mol} \]

### Step 5: Calculate the density of the gas mixture
Using the ideal gas law, the density \( D \) can be calculated using the formula:
\[ D = \frac{P \times M}{R \times T} \]
where:
- \( P = 1 \text{ atm} \)
- \( M = 76.67 \text{ g/mol} \)
- \( R = 0.0821 \text{ L atm/(K mol)} \)
- \( T = 300 \text{ K} \)

Substituting the values:
\[ D = \frac{1 \text{ atm} \times 76.67 \text{ g/mol}}{0.0821 \text{ L atm/(K mol)} \times 300 \text{ K}} \]

Calculating the denominator:
\[ 0.0821 \times 300 = 24.63 \text{ L atm/mol} \]

Now calculating the density:
\[ D = \frac{76.67}{24.63} \approx 3.11 \text{ g/L} \]

### Final Answer
The density of the equilibrium mixture is approximately \( 3.11 \text{ g/L} \).

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