138 gm of `N_2O_4(g)` is placed in 8.2L container at 300 K.The equilibrium vapour density of mixture was found to be 30.67.Then (R=0.082 L atm `"mol"^(-1) K^(-1)`)

Options

A. `alpha`=degree of dissociation of `N_2O_4=0.25`
B. `K_p` of `N_2O_4 hArr 2NO_2` (g) will be 6 atm
C. Total pressure at equilibrium =4.5 atm
D. The total no. of moles at equilibrium is 1.5

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Given Data:
- Mass of \( N_2O_4 = 138 \, \text{g} \)
- Volume of container \( V = 8.2 \, \text{L} \)
- Temperature \( T = 300 \, \text{K} \)
- Observed vapor density \( d = 30.67 \)
- Gas constant \( R = 0.082 \, \text{L atm} \, \text{mol}^{-1} \, \text{K}^{-1} \)

### Step 1: Calculate the Molar Mass of \( N_2O_4 \)
The molar mass of \( N_2O_4 \) can be calculated as follows:
\[ \text{Molar mass of } N_2O_4 = (2 \times 14) + (4 \times 16) = 28 + 64 = 92 \, \text{g/mol} \]

### Step 2: Calculate the Initial Moles of \( N_2O_4 \)
Using the mass and molar mass, we can find the initial number of moles:
\[ n_{initial} = \frac{\text{mass}}{\text{molar mass}} = \frac{138 \, \text{g}}{92 \, \text{g/mol}} = 1.5 \, \text{mol} \]

### Step 3: Determine the Theoretical Vapor Density
The theoretical vapor density \( D \) can be calculated using the molar mass:
\[ D = \frac{\text{molar mass}}{2} = \frac{92}{2} = 46 \, \text{g/mol} \]

### Step 4: Calculate the Degree of Dissociation \( \alpha \)
Using the formula for degree of dissociation:
\[ \alpha = \frac{1}{n - 1} \left( D - d \right) \div D \]
Where:
- \( n = 2 \) (since 1 mole of \( N_2O_4 \) produces 2 moles of \( NO_2 \))
- \( D = 46 \, \text{g/mol} \)
- \( d = 30.67 \, \text{g/mol} \)

Substituting the values:
\[ \alpha = \frac{1}{2 - 1} \left( 46 - 30.67 \right) \div 46 = \frac{1}{1} \left( 15.33 \right) \div 46 \approx 0.333 \]

### Step 5: Calculate Total Moles at Equilibrium
The total number of moles at equilibrium can be calculated as:
\[ n_{total} = (1 - \alpha) + 2\alpha = 1 + \alpha = 1 + 0.333 \approx 1.333 \, \text{mol} \]

### Step 6: Calculate Total Pressure at Equilibrium
Using the ideal gas law:
\[ P = \frac{nRT}{V} \]
Substituting the values:
\[ P = \frac{1.333 \times 0.082 \times 300}{8.2} \approx 4.0 \, \text{atm} \]

### Step 7: Calculate \( K_p \)
Using the expression for \( K_p \):
\[ K_p = \frac{P_{NO_2}^2}{P_{N_2O_4}} = \frac{(2\alpha P)^2}{(1 - \alpha)P} \]
Substituting the values:
\[ K_p = \frac{(2 \times 0.333 \times 4.0)^2}{(1 - 0.333) \times 4.0} = \frac{(2.664)^2}{0.667 \times 4.0} \approx 6.0 \, \text{atm} \]

### Final Answers:
- Degree of dissociation \( \alpha \approx 0.333 \)
- Total pressure at equilibrium \( P \approx 4.0 \, \text{atm} \)
- Total number of moles at equilibrium \( n_{total} \approx 1.333 \, \text{mol} \)
- \( K_p \approx 6.0 \, \text{atm} \)