The degree of dissociation of `PCl_(5)` at `1` atm pressure is `0.2`. Calculate the pressure at which `PCl_(5)` is dissociated to `50%`?

Explanation

To solve the problem of calculating the pressure at which \( PCl_5 \) is dissociated to 50%, we will follow these steps:

### Step 1: Understand the Reaction
The dissociation of \( PCl_5 \) can be represented as:
\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]

### Step 2: Define Initial Conditions
Let’s assume we start with 1 mole of \( PCl_5 \) initially. Therefore:
- Initial moles of \( PCl_5 = 1 \)
- Initial moles of \( PCl_3 = 0 \)
- Initial moles of \( Cl_2 = 0 \)

### Step 3: Define Degree of Dissociation
Let \( \alpha \) be the degree of dissociation. Given that at 1 atm pressure, \( \alpha = 0.2 \).

### Step 4: Calculate Moles at Equilibrium
After dissociation, the moles at equilibrium will be:
- Moles of \( PCl_5 = 1 - \alpha \)
- Moles of \( PCl_3 = \alpha \)
- Moles of \( Cl_2 = \alpha \)

Substituting \( \alpha = 0.2 \):
- Moles of \( PCl_5 = 1 - 0.2 = 0.8 \)
- Moles of \( PCl_3 = 0.2 \)
- Moles of \( Cl_2 = 0.2 \)

### Step 5: Total Moles at Equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha = 1 + 0.2 = 1.2 \]

### Step 6: Calculate \( K_p \)
Using the expression for \( K_p \):
\[ K_p = \frac{(P_{PCl_3})(P_{Cl_2})}{(P_{PCl_5})} \]
At equilibrium, the pressures can be expressed as:
\[ P_{PCl_5} = \frac{(1 - \alpha)}{(1 + \alpha)} \cdot P \]
\[ P_{PCl_3} = \frac{\alpha}{(1 + \alpha)} \cdot P \]
\[ P_{Cl_2} = \frac{\alpha}{(1 + \alpha)} \cdot P \]

Substituting these into the \( K_p \) expression:
\[ K_p = \frac{\left(\frac{\alpha}{(1 + \alpha)} P\right) \left(\frac{\alpha}{(1 + \alpha)} P\right)}{\left(\frac{(1 - \alpha)}{(1 + \alpha)} P\right)} \]

This simplifies to:
\[ K_p = \frac{\alpha^2 P^2}{(1 - \alpha)(1 + \alpha) P} = \frac{\alpha^2 P}{(1 - \alpha)(1 + \alpha)} \]

### Step 7: Substitute Known Values
Substituting \( \alpha = 0.2 \) and \( P = 1 \):
\[ K_p = \frac{(0.2)^2 \cdot 1}{(1 - 0.2)(1 + 0.2)} = \frac{0.04}{0.8 \cdot 1.2} = \frac{0.04}{0.96} = 0.04167 \, \text{atm} \]

### Step 8: Calculate Pressure for 50% Dissociation
Now, we need to find the pressure \( P' \) when \( \alpha' = 0.5 \):
Using the same \( K_p \) expression:
\[ K_p = \frac{(0.5)^2 P'}{(1 - 0.5)(1 + 0.5)} \]
\[ 0.04167 = \frac{0.25 P'}{0.5 \cdot 1.5} \]
\[ 0.04167 = \frac{0.25 P'}{0.75} \]
\[ P' = \frac{0.04167 \cdot 0.75}{0.25} = \frac{0.03125}{0.25} = 0.125 \, \text{atm} \]

### Final Answer
The pressure at which \( PCl_5 \) is dissociated to 50% is approximately:
\[ P' \approx 0.125 \, \text{atm} \]

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