For which of the following reaction the degree of dissociation `(alpha)` and equilibrium constant `(K_p)` are
related as `K_p =(4alpha^2P)/((1-alpha^2))` ?

Options

A. `N_2O_4(g) hArr2NO_2(g)`
B. `H_2(g) +I_2(g) hArr 2HI(g)`
C. `N_2(g)+3H_2(g) hArr 2NH_3(g)`
D. `PCI_3(g)+CI_2(g) hArr PCI_5(g)`

Answer

To determine the reaction for which the degree of dissociation (α) and equilibrium constant (Kp) are related as \( K_p = \frac{4 \alpha^2 P}{1 - \alpha^2} \), we will analyze the reaction \( N_2O_4 \rightleftharpoons 2 NO_2 \).

### Step-by-Step Solution:

1. **Write the Reaction**:
The reaction we are considering is:
\[ N_2O_4(g) \rightleftharpoons 2 NO_2(g) \]

2. **Initial Moles**:
Assume we start with 1 mole of \( N_2O_4 \) and 0 moles of \( NO_2 \):
- Initial moles of \( N_2O_4 \) = 1
- Initial moles of \( NO_2 \) = 0

3. **Change in Moles**:
Let \( \alpha \) be the degree of dissociation of \( N_2O_4 \). Therefore, at equilibrium:
- Moles of \( N_2O_4 \) = \( 1 - \alpha \)
- Moles of \( NO_2 \) = \( 2\alpha \)

4. **Total Moles at Equilibrium**:
The total moles at equilibrium will be:
\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

5. **Mole Fractions**:
- Mole fraction of \( N_2O_4 \):
\[ \text{Mole fraction of } N_2O_4 = \frac{1 - \alpha}{1 + \alpha} \]
- Mole fraction of \( NO_2 \):
\[ \text{Mole fraction of } NO_2 = \frac{2\alpha}{1 + \alpha} \]

6. **Partial Pressures**:
Assuming total pressure is \( P \):
- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \left(\frac{1 - \alpha}{1 + \alpha}\right) P \]
- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \left(\frac{2\alpha}{1 + \alpha}\right) P \]

7. **Equilibrium Constant \( K_p \)**:
The expression for \( K_p \) is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]
Substituting the expressions for partial pressures:
\[ K_p = \frac{\left(\frac{2\alpha}{1 + \alpha} P\right)^2}{\left(\frac{1 - \alpha}{1 + \alpha} P\right)} \]

8. **Simplifying \( K_p \)**:
\[ K_p = \frac{4\alpha^2 P^2}{(1 + \alpha)^2} \cdot \frac{1 + \alpha}{1 - \alpha} = \frac{4\alpha^2 P}{(1 - \alpha)(1 + \alpha)} \]
Now, using the identity \( (1 + \alpha)(1 - \alpha) = 1 - \alpha^2 \):
\[ K_p = \frac{4\alpha^2 P}{1 - \alpha^2} \]

9. **Conclusion**:
We have derived that:
\[ K_p = \frac{4\alpha^2 P}{1 - \alpha^2} \]
Therefore, the reaction \( N_2O_4 \rightleftharpoons 2 NO_2 \) satisfies the given relationship.