For the dissociation reaction `N_(2)O_($) (g)hArr 2NO_(2)(g)`, the degree of dissociation `(alpha)`interms of `K_(p)` and total equilibrium pressure P is:

Options

A. `alpha=sqrt((4P+K_(p))/(K_(P)))`
B. `alpha=sqrt((K_(P))/(4P+K_(p)))`
C. `alpha=sqrt((K_(P))/(4P))`
D. none of these

Answer

To derive the expression for the degree of dissociation (α) in terms of \( K_p \) and total equilibrium pressure \( P \) for the dissociation reaction:

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

we will follow these steps:

### Step 1: Define Initial Conditions
At the start of the reaction (before any dissociation occurs), we have:
- Initial moles of \( N_2O_4 = 1 \)
- Initial moles of \( NO_2 = 0 \)

### Step 2: Define Change in Moles
Let \( \alpha \) be the degree of dissociation of \( N_2O_4 \). At equilibrium:
- Moles of \( N_2O_4 \) remaining = \( 1 - \alpha \)
- Moles of \( NO_2 \) produced = \( 2\alpha \)

### 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 Dalton's Law, the partial pressures can be expressed as:
- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \left(\frac{2\alpha}{1 + \alpha}\right) P \]
- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \left(\frac{1 - \alpha}{1 + \alpha}\right) P \]

### 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:
\[ K_p = \frac{\left(\frac{2\alpha}{1 + \alpha} P\right)^2}{\frac{(1 - \alpha)}{(1 + \alpha)} P} \]
This simplifies to:
\[ K_p = \frac{4\alpha^2 P^2}{(1 + \alpha)^2} \cdot \frac{(1 + \alpha)}{(1 - \alpha)} = \frac{4\alpha^2 P^2}{(1 - \alpha)(1 + \alpha)} \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ K_p (1 - \alpha)(1 + \alpha) = 4\alpha^2 P^2 \]

### Step 8: Final Expression
This can be further simplified to find \( \alpha \):
\[ K_p = \frac{4\alpha^2 P^2}{(1 - \alpha)(1 + \alpha)} \]
From this equation, we can express \( \alpha \) in terms of \( K_p \) and \( P \).

### Conclusion
The expression for the degree of dissociation \( \alpha \) in terms of \( K_p \) and total equilibrium pressure \( P \) is:
\[ \alpha = \sqrt{\frac{K_p}{4P + K_p}} \]