The degree of dissociation of `PCl_(5) (alpha)` obeying the equilibrium, `PCl_(5)hArrPCl_(3)+Cl_(2)` is related to the pressure at equilibrium by :

Options

A. `a prop P`
B. `alpha prop (1)/(sqrtP)`
C. `alpha prop (1)/(p^(2))`
D. `alpha prop (1)/(p^(4))`

Answer

To solve the problem of relating the degree of dissociation (α) of \( PCl_5 \) at equilibrium to the pressure, we can follow these steps:

### Step 1: Write the equilibrium reaction
The equilibrium reaction is given by:
\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]

### Step 2: Define initial conditions
Initially, we have:
- \( n(PCl_5) = 1 \) mole
- \( n(PCl_3) = 0 \) moles
- \( n(Cl_2) = 0 \) moles

### Step 3: Define the degree of dissociation
Let the degree of dissociation of \( PCl_5 \) be \( \alpha \). At equilibrium:
- Moles of \( PCl_5 \) remaining = \( 1 - \alpha \)
- Moles of \( PCl_3 \) formed = \( \alpha \)
- Moles of \( Cl_2 \) formed = \( \alpha \)

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

### Step 5: Express partial pressures
Assuming the total pressure at equilibrium is \( P \), the partial pressures can be expressed as:
- \( P_{PCl_5} = \frac{(1 - \alpha)}{(1 + \alpha)} P \)
- \( P_{PCl_3} = \frac{\alpha}{(1 + \alpha)} P \)
- \( P_{Cl_2} = \frac{\alpha}{(1 + \alpha)} P \)

### Step 6: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} = \frac{\left(\frac{\alpha}{(1 + \alpha)} P\right) \cdot \left(\frac{\alpha}{(1 + \alpha)} P\right)}{\frac{(1 - \alpha)}{(1 + \alpha)} P} \]

### Step 7: Simplify the expression for \( K_p \)
Substituting the partial pressures into the equation:
\[ K_p = \frac{\frac{\alpha^2}{(1 + \alpha)^2} P^2}{\frac{(1 - \alpha)}{(1 + \alpha)} P} \]
This simplifies to:
\[ K_p = \frac{\alpha^2 P}{(1 - \alpha)(1 + \alpha)} \]

### Step 8: Assume \( \alpha \) is small
For small values of \( \alpha \), we can approximate \( 1 - \alpha \approx 1 \) and \( 1 + \alpha \approx 1 \). Thus:
\[ K_p \approx \frac{\alpha^2 P}{1} \]

### Step 9: Solve for \( \alpha \)
From the equation \( K_p = \alpha^2 P \), we can express \( \alpha \) as:
\[ \alpha^2 = \frac{K_p}{P} \]
Taking the square root gives:
\[ \alpha = \sqrt{\frac{K_p}{P}} \]

### Step 10: Relate \( \alpha \) and \( P \)
From the derived equation, we can conclude that:
\[ \alpha \propto \frac{1}{\sqrt{P}} \]

### Final Answer
Thus, the degree of dissociation \( \alpha \) is inversely proportional to the square root of the pressure \( P \).

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