The degree of dissociation `alpha` of the reaction"
`N_(2)O_(4(g))hArr 2NO_(2(g))`
can be related to `K_(p)` as:

Options

A. `alpha=((K_(p))/(P))/(4+(K_(p))/(P))`
B. `alpha=(K_(p))/(4+K_(p))`
C. `alpha=[(K_(p)//P)/(4+K_p//P)]^(1//2)`
D. `alpha=[(K_(p))/(4+K_(p))]^(1//2)`

Answer

To relate the degree of dissociation (α) of the reaction \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \) to \( K_p \), we can follow these steps:

### Step-by-Step Solution:

1. **Define Initial Conditions:**
Let the initial amount of \( N_2O_4 \) be \( A \) moles and initially, there are 0 moles of \( NO_2 \).

2. **Establish Change in Moles at Equilibrium:**
At equilibrium, if \( \alpha \) is the degree of dissociation, then:
- Moles of \( N_2O_4 \) at equilibrium = \( A(1 - \alpha) \)
- Moles of \( NO_2 \) at equilibrium = \( 2A\alpha \)

3. **Calculate Total Moles at Equilibrium:**
Total moles at equilibrium = Moles of \( N_2O_4 \) + Moles of \( NO_2 \)
\[ n_{total} = A(1 - \alpha) + 2A\alpha = A(1 + \alpha) \]

4. **Express Partial Pressures:**
The partial pressure of each gas can be expressed in terms of the total pressure \( P \):
- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \frac{2A\alpha}{A(1 + \alpha)} \cdot P = \frac{2\alpha P}{1 + \alpha} \]
- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \frac{A(1 - \alpha)}{A(1 + \alpha)} \cdot P = \frac{(1 - \alpha) P}{1 + \alpha} \]

5. **Write the Expression for \( K_p \):**
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{\left(\frac{2\alpha P}{1 + \alpha}\right)^2}{\frac{(1 - \alpha) P}{1 + \alpha}} \]

6. **Simplify the Expression:**
Substituting the expressions for partial pressures into the \( K_p \) equation:
\[ K_p = \frac{4\alpha^2 P^2}{(1 + \alpha)^2} \cdot \frac{1 + \alpha}{(1 - \alpha) P} \]
Simplifying gives:
\[ K_p = \frac{4\alpha^2 P}{(1 + \alpha)(1 - \alpha)} \]

7. **Rearranging for \( \alpha \):**
Rearranging the equation to isolate \( \alpha \):
\[ K_p(1 + \alpha)(1 - \alpha) = 4\alpha^2 P \]
Expanding and rearranging yields:
\[ K_p - K_p\alpha^2 = 4\alpha^2 P \]
\[ K_p = \alpha^2(4P + K_p) \]
\[ \alpha^2 = \frac{K_p}{4P + K_p} \]

8. **Final Expression for \( \alpha \):**
Taking the square root gives:
\[ \alpha = \sqrt{\frac{K_p}{4P + K_p}} \]