For the reaction `PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)`, the equation connecting the degree of dissociation `(alpha)` of `PCl_(5)(g)` with the equilibrium constant `K_(p)` is

Options

A. `alpha=(K_(P)//P)/(4+K_(P)//P)`
B. `alpha=[(K_(p)//P)/(K_(p)//P+1)]^(1//2)`
C. `alpha=[(K_(P)//P)/(K_(P)//P+1)]`
D. `alpha=[(K_(P)//P)/(4+K_(P)//P)]^(1//2)`

Answer

To derive the equation connecting the degree of dissociation (α) of PCl₅ with the equilibrium constant (Kₚ) for the reaction:

\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]

we will follow these steps:

### Step 1: Define Initial Conditions
Assume we start with 1 mole of PCl₅. At equilibrium, let α be the degree of dissociation of PCl₅.

### Step 2: Determine Moles at Equilibrium
- Moles of PCl₅ that dissociate = α
- Moles of PCl₅ remaining = \( 1 - \alpha \)
- Moles of PCl₃ formed = α
- Moles of Cl₂ formed = α

Thus, at equilibrium, the total number of moles is:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

### Step 3: Calculate Partial Pressures
Assuming the total pressure is P, we can calculate the partial pressures of each component:
- Partial pressure of PCl₅:
\[ P_{\text{PCl}_5} = \frac{(1 - \alpha)}{(1 + \alpha)} P \]
- Partial pressure of PCl₃:
\[ P_{\text{PCl}_3} = \frac{\alpha}{(1 + \alpha)} P \]
- Partial pressure of Cl₂:
\[ P_{\text{Cl}_2} = \frac{\alpha}{(1 + \alpha)} P \]

### Step 4: Write the Expression for Kₚ
The equilibrium constant Kₚ for the reaction is given by:
\[ K_p = \frac{P_{\text{PCl}_3} \cdot P_{\text{Cl}_2}}{P_{\text{PCl}_5}} \]

Substituting the expressions for the partial pressures:
\[ K_p = \frac{\left(\frac{\alpha}{(1 + \alpha)} P\right) \cdot \left(\frac{\alpha}{(1 + \alpha)} P\right)}{\frac{(1 - \alpha)}{(1 + \alpha)} P} \]

### Step 5: Simplify the Expression
This simplifies to:
\[ K_p = \frac{\alpha^2 P^2}{(1 + \alpha)^2} \cdot \frac{(1 + \alpha)}{(1 - \alpha)} \]
\[ K_p = \frac{\alpha^2 P}{(1 - \alpha)(1 + \alpha)} \]

### Step 6: Rearranging for α
Rearranging the equation gives:
\[ \alpha^2 = \frac{K_p (1 - \alpha)(1 + \alpha)}{P} \]

### Step 7: Final Expression
From the above expression, we can derive α:
\[ \alpha = \sqrt{\frac{K_p}{P + K_p}} \]

### Conclusion
Thus, the equation connecting the degree of dissociation (α) of PCl₅ with the equilibrium constant (Kₚ) is:
\[ \alpha = \frac{K_p}{P + K_p} \]