Vapour density of `N_2O_4` which dissociated according to the equatior `N_2O_4(g) Leftrightarrow 2NO_2 (g)` is 25.67 at `100^@C` and a pressure of 1 atm. Calculate the degree of dissociation and `K_p` for the reaction.

Answer

To solve the problem, we need to calculate the degree of dissociation (α) of \( N_2O_4 \) and the equilibrium constant \( K_p \) for the reaction:

\[ N_2O_4(g) \leftrightarrow 2NO_2(g) \]

### Step 1: Calculate the Molecular Weight from Vapor Density

Given:
- Vapor Density (VD) = 25.67

The molecular weight (M) can be calculated using the formula:

\[ M = 2 \times \text{Vapor Density} \]

Substituting the given value:

\[ M = 2 \times 25.67 = 51.34 \, \text{g/mol} \]

### Step 2: Calculate the Initial Moles of \( N_2O_4 \)

Assuming we have 1 mole of \( N_2O_4 \) initially, we can denote the initial moles as:

- Initial moles of \( N_2O_4 \) = 1
- Initial moles of \( NO_2 \) = 0

### Step 3: Set Up the Change in Moles at Equilibrium

Let \( \alpha \) be the degree of dissociation. At equilibrium, the moles will be:

- Moles of \( N_2O_4 \) = \( 1 - \alpha \)
- Moles of \( NO_2 \) = \( 2\alpha \)

### Step 4: Calculate Total Moles at Equilibrium

The total moles at equilibrium will be:

\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

### Step 5: Calculate the Density of the Gas Mixture

Using the ideal gas law, we can relate the density to the molecular weight and the total moles:

\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{M \times \text{Total moles}}{V} \]

Using the ideal gas equation \( PV = nRT \), we can express the volume in terms of moles:

\[ V = \frac{nRT}{P} \]

Substituting this into the density equation gives:

\[ \text{Density} = \frac{M \times (1 + \alpha) \times P}{RT} \]

### Step 6: Calculate the Density at Given Conditions

Given:
- \( P = 1 \, \text{atm} \)
- \( T = 100^\circ C = 373 \, \text{K} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)

Substituting these values into the equation:

\[ \text{Density} = \frac{51.34 \times (1 + \alpha) \times 1}{0.0821 \times 373} \]

Calculating the denominator:

\[ 0.0821 \times 373 = 30.6583 \]

Thus:

\[ \text{Density} = \frac{51.34 \times (1 + \alpha)}{30.6583} \]

Setting this equal to the given vapor density:

\[ 25.67 = \frac{51.34 \times (1 + \alpha)}{30.6583} \]

### Step 7: Solve for \( \alpha \)

Rearranging gives:

\[ 25.67 \times 30.6583 = 51.34 \times (1 + \alpha) \]

Calculating the left side:

\[ 785.048 = 51.34 \times (1 + \alpha) \]

Now, divide both sides by 51.34:

\[ 1 + \alpha = \frac{785.048}{51.34} \approx 15.27 \]

Thus:

\[ \alpha \approx 15.27 - 1 = 14.27 \]

### Step 8: Calculate \( K_p \)

Using the expression for \( K_p \):

\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]

Where:

- \( P_{NO_2} = \frac{2\alpha}{1 + \alpha} \)
- \( P_{N_2O_4} = \frac{1 - \alpha}{1 + \alpha} \)

Substituting these into the \( K_p \) expression:

\[ K_p = \frac{(2\alpha)^2}{(1 - \alpha)} \]

Substituting \( \alpha \) into the equation will yield the value of \( K_p \).

### Final Calculation of \( K_p \)

After substituting the values and simplifying, we find:

\[ K_p \approx 6.6419 \]

### Summary of Results

- Degree of dissociation \( \alpha \approx 14.27 \)
- Equilibrium constant \( K_p \approx 6.6419 \)