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 the initial conditions
Assume we start with 1 mole of PCl₅ and no products. Therefore, initially:
- Moles of PCl₅ = 1
- Moles of PCl₃ = 0
- Moles of Cl₂ = 0

### Step 2: Express the changes at equilibrium
Let α be the degree of dissociation of PCl₅. At equilibrium:
- Moles of PCl₅ = \( 1 - \alpha \)
- Moles of PCl₃ = \( \alpha \)
- Moles of Cl₂ = \( \alpha \)

### Step 3: Calculate the total moles at equilibrium
The total number of moles at equilibrium (N_total) is:
\[ N_{\text{total}} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

### Step 4: Calculate the partial pressures
Using Dalton's Law of partial pressures, the partial pressure of each component can be expressed as:
- 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 5: 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 partial pressures into this equation:
\[ K_p = \frac{\left(\frac{\alpha}{(1 + \alpha)} P\right) \cdot \left(\frac{\alpha}{(1 + \alpha)} P\right)}{\frac{(1 - \alpha)}{(1 + \alpha)} P} \]

### Step 6: Simplify the expression
This simplifies to:
\[ K_p = \frac{\alpha^2 P^2}{(1 - \alpha)(1 + \alpha) P} \]
\[ K_p = \frac{\alpha^2 P}{(1 - \alpha)(1 + \alpha)} \]

### Step 7: Rearranging for α
Rearranging gives:
\[ \alpha^2 = K_p \cdot \frac{(1 - \alpha)(1 + \alpha)}{P} \]

### Step 8: Solve for α
To express α in terms of Kₚ and P, we can isolate α:
1. Multiply both sides by P:
\[ \alpha^2 P = K_p (1 - \alpha)(1 + \alpha) \]
2. Expand and rearrange:
\[ \alpha^2 P = K_p (1 + \alpha - \alpha - \alpha^2) \]
3. Combine like terms and solve for α.

### Final Equation
After simplification, we find:
\[ \alpha = \sqrt{\frac{K_p}{P + K_p}} \]

This is the equation connecting the degree of dissociation (α) with the equilibrium constant (Kₚ).