In the dissociation of `PCl_(5)` as
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`
If the degree of dissociation is `alpha` at equilibrium pressure P, then the equilibrium constant for the reaction is

Options

A. `K_(p)=alpha^(2)/(1+alpha^(2)P)`
B. `K_(p)=(alpha^(2)P^(2))/(1-alpha^(2))`
C. `K_(p)=P^(2)/(1-alpha^(2))`
D. `K_(p)=(alpha^(2)P)/(1-alpha^(2))`

Explanation

To solve the problem regarding the dissociation of \( PCl_5 \) into \( PCl_3 \) and \( Cl_2 \), we will follow these steps:

### Step 1: Write the Reaction
The dissociation reaction is given as:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

### Step 2: Define the Degree of Dissociation
Let the degree of dissociation of \( PCl_5 \) be represented by \( \alpha \). This means that if we start with 1 mole of \( PCl_5 \), then \( \alpha \) moles will dissociate.

### Step 3: Determine Moles at Equilibrium
At equilibrium, the moles of each species will be:
- Moles of \( PCl_5 \) = \( 1 - \alpha \)
- Moles of \( PCl_3 \) = \( \alpha \)
- Moles of \( Cl_2 \) = \( \alpha \)

### Step 4: Calculate Total Moles at Equilibrium
The total number of moles at equilibrium is:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

### Step 5: Calculate Mole Fractions
The mole fractions of each species are:
- Mole fraction of \( PCl_5 \):
\[ \chi_{PCl_5} = \frac{1 - \alpha}{1 + \alpha} \]
- Mole fraction of \( PCl_3 \):
\[ \chi_{PCl_3} = \frac{\alpha}{1 + \alpha} \]
- Mole fraction of \( Cl_2 \):
\[ \chi_{Cl_2} = \frac{\alpha}{1 + \alpha} \]

### Step 6: Calculate Partial Pressures
Using the total pressure \( P \), the partial pressures can be calculated as:
- Partial pressure of \( PCl_5 \):
\[ P_{PCl_5} = \chi_{PCl_5} \cdot P = \frac{1 - \alpha}{1 + \alpha} \cdot P \]
- Partial pressure of \( PCl_3 \):
\[ P_{PCl_3} = \chi_{PCl_3} \cdot P = \frac{\alpha}{1 + \alpha} \cdot P \]
- Partial pressure of \( Cl_2 \):
\[ P_{Cl_2} = \chi_{Cl_2} \cdot P = \frac{\alpha}{1 + \alpha} \cdot P \]

### Step 7: Write the Expression for Equilibrium Constant \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]

### Step 8: Substitute the Partial Pressures into the \( K_p \) Expression
Substituting the expressions for the partial pressures:
\[ K_p = \frac{\left(\frac{\alpha}{1 + \alpha} \cdot P\right) \cdot \left(\frac{\alpha}{1 + \alpha} \cdot P\right)}{\frac{1 - \alpha}{1 + \alpha} \cdot P} \]

### Step 9: Simplify the Expression
This simplifies to:
\[ K_p = \frac{\alpha^2 \cdot P^2}{(1 + \alpha)^2 \cdot \frac{1 - \alpha}{1 + \alpha} \cdot P} \]
Cancelling \( P \) and simplifying further:
\[ K_p = \frac{\alpha^2 \cdot P}{1 - \alpha^2} \]

### Final Result
Thus, the equilibrium constant \( K_p \) for the reaction is:
\[ K_p = \frac{\alpha^2 P}{1 - \alpha^2} \]