The degree of dissociation `'alpha'` of the reaction
`N_(2)O_(4)(g)hArr 2NO_(2)(g)`
Can be related of `K_(p)` as [Given : Total pressure at equilibrium = P]

Options

A. `alpha=((Kp)/(p))/(4+(Kp)/(p))`
B. `alpha=(Kp)/(4+Kp)`
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 the equilibrium constant \( K_p \) given a total pressure \( P \) at equilibrium, we can follow these steps:

### Step 1: Define Initial Conditions
Assume we start with 1 mole of \( N_2O_4 \) initially. Therefore, the initial concentration of \( N_2O_4 \) is:
- Initial moles of \( N_2O_4 \) = 1 mole
- Initial moles of \( NO_2 \) = 0 moles

### Step 2: Define Change in Moles
Let \( \alpha \) be the degree of dissociation of \( N_2O_4 \). At equilibrium:
- Moles of \( N_2O_4 \) that dissociate = \( \alpha \)
- Moles of \( NO_2 \) produced = \( 2\alpha \)

Thus, at equilibrium:
- Moles of \( N_2O_4 \) = \( 1 - \alpha \)
- Moles of \( NO_2 \) = \( 2\alpha \)

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

### Step 4: Calculate Partial Pressures
Using the total pressure \( P \), we can express the partial pressures of the gases:
- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \frac{(1 - \alpha)}{(1 + \alpha)} P \]
- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \frac{2\alpha}{(1 + \alpha)} P \]

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

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

### Step 7: Rearranging for α
Rearranging gives:
\[ K_p (1 - \alpha)(1 + \alpha) = 4\alpha^2 P \]
Expanding and rearranging:
\[ K_p + K_p \alpha - 4\alpha^2 P = 0 \]

### Step 8: Solve for α
This is a quadratic equation in terms of \( \alpha \). Using the quadratic formula:
\[ \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = -4P, b = K_p, c = K_p \).

### Step 9: Final Expression for α
The solution will yield:
\[ \alpha = \frac{K_p}{4P + K_p} \]

### Conclusion
Thus, the degree of dissociation \( \alpha \) is related to \( K_p \) as:
\[ \alpha = \frac{K_p}{4P + K_p} \]