What will be the expression of `K_(p)` for the given reaction if the total pressure inside the vessel is P and degree of dissociation of the reactant is a? The reaction
`N_(2)O_(4)hArr 2NO_(2)`

Options

A. `(4a^(2)P)/(1+a^(2))`
B. `(4a^(2)P)/(1-a^(2))`
C. `(a^(2)P)/(1-a^(2))`
D. `(a^(2))/(1-a^(2))`

Answer

To derive the expression for \( K_p \) for the reaction \( N_2O_4 \rightleftharpoons 2NO_2 \) given the total pressure \( P \) and degree of dissociation \( \alpha \), we can follow these steps:

### Step 1: Write the Reaction and Initial Conditions
The reaction is:
\[ N_2O_4 \rightleftharpoons 2NO_2 \]

Let’s assume we start with \( x \) moles of \( N_2O_4 \) at equilibrium.

### Step 2: Define the Change in Moles
At equilibrium, if the degree of dissociation is \( \alpha \), the moles of \( N_2O_4 \) and \( NO_2 \) can be expressed as:
- Moles of \( N_2O_4 \) at equilibrium: \( x(1 - \alpha) \)
- Moles of \( NO_2 \) at equilibrium: \( 2x\alpha \)

### Step 3: Calculate Total Moles at Equilibrium
The total number of moles \( n_T \) at equilibrium is:
\[ n_T = \text{moles of } N_2O_4 + \text{moles of } NO_2 = x(1 - \alpha) + 2x\alpha = x(1 + \alpha) \]

### Step 4: Express Partial Pressures
Using the ideal gas law, the partial pressures can be expressed as:
- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \frac{(x(1 - \alpha))}{n_T} \cdot P = \frac{x(1 - \alpha)}{x(1 + \alpha)} \cdot P = \frac{(1 - \alpha)P}{(1 + \alpha)} \]
- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \frac{(2x\alpha)}{n_T} \cdot P = \frac{2x\alpha}{x(1 + \alpha)} \cdot P = \frac{2\alpha P}{(1 + \alpha)} \]

### Step 5: Write the Expression for \( K_p \)
The expression for \( K_p \) is given by the ratio of the product of the partial pressures of the products to the reactants:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{\left(\frac{2\alpha P}{(1 + \alpha)}\right)^2}{\frac{(1 - \alpha)P}{(1 + \alpha)}} \]

### Step 6: Simplify the Expression
Substituting the partial pressures into the equation:
\[ K_p = \frac{(2\alpha P)^2}{(1 - \alpha)(1 + \alpha)} = \frac{4\alpha^2 P^2}{(1 - \alpha)(1 + \alpha)} \]

### Step 7: Final Expression for \( K_p \)
Thus, the final expression for \( K_p \) is:
\[ K_p = \frac{4\alpha^2 P}{1 - \alpha^2} \]