For the dissociation reaction `N_(2)O_(4) (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 find the degree of dissociation (α) for the reaction \( N_2O_4 (g) \rightleftharpoons 2 NO_2 (g) \) in terms of \( K_p \) and total equilibrium pressure \( P \), we can follow these steps:

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

Assume we start with 1 mole of \( N_2O_4 \) at the beginning (initially):
- Moles of \( N_2O_4 \) = 1
- Moles of \( NO_2 \) = 0

### Step 2: Define the Degree of Dissociation (α)
Let \( \alpha \) be the degree of dissociation of \( N_2O_4 \). At equilibrium:
- Moles of \( N_2O_4 \) = \( 1 - \alpha \)
- Moles of \( NO_2 \) = \( 2\alpha \)

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

### Step 4: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]

### Step 5: Calculate Partial Pressures
The partial pressures can be expressed in terms of mole fractions and total pressure \( P \):
- Mole fraction of \( NO_2 \):
\[ \text{Mole fraction of } NO_2 = \frac{2\alpha}{1 + \alpha} \]
- Mole fraction of \( N_2O_4 \):
\[ \text{Mole fraction of } N_2O_4 = \frac{1 - \alpha}{1 + \alpha} \]

Thus, the partial pressures are:
- \( P_{NO_2} = \frac{2\alpha}{1 + \alpha} \cdot P \)
- \( P_{N_2O_4} = \frac{1 - \alpha}{1 + \alpha} \cdot P \)

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

### Step 7: Simplify the Expression
This simplifies to:
\[ K_p = \frac{4\alpha^2 P^2}{(1 + \alpha)^2} \cdot \frac{1 + \alpha}{1 - \alpha} \]
\[ K_p = \frac{4\alpha^2 P (1 + \alpha)}{1 - \alpha} \]

### Step 8: Rearrange the Equation
Rearranging gives:
\[ K_p (1 - \alpha) = 4\alpha^2 P (1 + \alpha) \]
Expanding and rearranging terms:
\[ K_p - K_p \alpha = 4\alpha^2 P + 4\alpha^3 P \]

### Step 9: Collect Terms Involving \( \alpha \)
Grouping terms involving \( \alpha \):
\[ 4\alpha^2 P + K_p \alpha + K_p = 4\alpha^3 P \]

### Step 10: Solve for \( \alpha \)
Factoring out \( \alpha \):
\[ \alpha^2 (4P + K_p) = K_p \]
Thus,
\[ \alpha^2 = \frac{K_p}{4P + K_p} \]

### Step 11: Take the Square Root
Finally, taking the square root gives us the degree of dissociation:
\[ \alpha = \sqrt{\frac{K_p}{4P + K_p}} \]

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

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