The degree of dissociation of `SO_(3)` is `alpha` at equilibrium pressure `P_(0)`.
`K_(P)` for `2SO_(3)(g)hArr2SO_(2)(g)+O_(2)(g)`is

Options

A. `[(P_(0)alpha^(3))//2(1-alpha)^(3)]`
B. `[(P_(0)alpha^(3))//(2+alpha)(1-alpha)^(2)]`
C. `[(P_(0)alpha^(2))//2(1-alpha)^(2)]`
D. None of these

Explanation

To find the equilibrium constant \( K_p \) for the reaction

\[ 2 \text{SO}_3(g) \rightleftharpoons 2 \text{SO}_2(g) + \text{O}_2(g) \]

given that the degree of dissociation of \( \text{SO}_3 \) is \( \alpha \) at equilibrium pressure \( P_0 \), we can follow these steps:

### Step 1: Write the Initial and Equilibrium Conditions
Assume we start with 2 moles of \( \text{SO}_3 \). At equilibrium, the degree of dissociation \( \alpha \) indicates how much of \( \text{SO}_3 \) has dissociated.

- **Initial moles:**
- \( \text{SO}_3 = 2 \)
- \( \text{SO}_2 = 0 \)
- \( \text{O}_2 = 0 \)

- **Change in moles:**
- \( \text{SO}_3 \) dissociates: \( -2\alpha \)
- \( \text{SO}_2 \) forms: \( +2\alpha \)
- \( \text{O}_2 \) forms: \( +\alpha \)

- **Equilibrium moles:**
- \( \text{SO}_3 = 2 - 2\alpha \)
- \( \text{SO}_2 = 2\alpha \)
- \( \text{O}_2 = \alpha \)

### Step 2: Calculate Total Moles at Equilibrium
The total moles at equilibrium can be calculated as:

\[ \text{Total moles} = (2 - 2\alpha) + (2\alpha) + \alpha = 2 + \alpha \]

### Step 3: Determine Partial Pressures
The partial pressures of each species can be expressed in terms of the total pressure \( P_0 \):

- **Partial pressure of \( \text{SO}_3 \):**
\[ P_{\text{SO}_3} = \left(\frac{2 - 2\alpha}{2 + \alpha}\right) P_0 \]

- **Partial pressure of \( \text{SO}_2 \):**
\[ P_{\text{SO}_2} = \left(\frac{2\alpha}{2 + \alpha}\right) P_0 \]

- **Partial pressure of \( \text{O}_2 \):**
\[ P_{\text{O}_2} = \left(\frac{\alpha}{2 + \alpha}\right) P_0 \]

### Step 4: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) is given by the expression:

\[ K_p = \frac{(P_{\text{SO}_2})^2 \cdot P_{\text{O}_2}}{(P_{\text{SO}_3})^2} \]

Substituting the partial pressures we calculated:

\[ K_p = \frac{\left(\left(\frac{2\alpha}{2 + \alpha}\right) P_0\right)^2 \cdot \left(\frac{\alpha}{2 + \alpha}\right) P_0}{\left(\left(\frac{2 - 2\alpha}{2 + \alpha}\right) P_0\right)^2} \]

### Step 5: Simplify the Expression
Now, simplifying the expression:

\[ K_p = \frac{\left(\frac{4\alpha^2}{(2 + \alpha)^2} P_0^2\right) \cdot \left(\frac{\alpha}{2 + \alpha}\right) P_0}{\left(\frac{(2 - 2\alpha)^2}{(2 + \alpha)^2} P_0^2\right)} \]

This simplifies to:

\[ K_p = \frac{4\alpha^3 P_0}{(2 - 2\alpha)^2} \cdot \frac{(2 + \alpha)^2}{(2 + \alpha)^2} \]

The \( P_0^2 \) cancels out, leading to:

\[ K_p = \frac{4\alpha^3 P_0}{(2 - 2\alpha)^2} \]

### Step 6: Final Expression for \( K_p \)
Thus, the final expression for \( K_p \) is:

\[ K_p = \frac{P_0 \alpha^3}{(2 + \alpha)(1 - \alpha)^2} \]