a' moles of `PCl_(5)` are heated in a closed container to equilibriate `PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)` at pressure of p atm . If x moles of `PCl_(5)` dissociate at equilibrium , then

Options

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

Answer

To solve the problem step by step, we will analyze the dissociation of \( PCl_5 \) and derive the relationship between the moles that dissociate and the equilibrium constant.

### Step 1: Write the equilibrium reaction
The dissociation of \( PCl_5 \) can be represented as:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

### Step 2: Define initial moles and changes
Let:
- \( A \) = initial moles of \( PCl_5 \)
- \( x \) = moles of \( PCl_5 \) that dissociate at equilibrium

At equilibrium:
- Moles of \( PCl_5 \) remaining = \( A - x \)
- Moles of \( PCl_3 \) formed = \( x \)
- Moles of \( Cl_2 \) formed = \( x \)

### Step 3: Calculate total moles at equilibrium
The total moles at equilibrium can be calculated as:
\[ \text{Total moles} = (A - x) + x + x = A + x \]

### Step 4: Define the degree of dissociation
The degree of dissociation \( \alpha \) is defined as:
\[ \alpha = \frac{x}{A} \]

### Step 5: Express pressures in terms of total pressure \( P \)
The total pressure \( P \) is given. The partial pressures can be expressed as follows:
- Pressure due to \( PCl_5 \):
\[ P_{PCl_5} = \frac{(1 - \alpha)P}{(1 + \alpha)} \]
- Pressure due to \( PCl_3 \):
\[ P_{PCl_3} = \frac{\alpha P}{(1 + \alpha)} \]
- Pressure due to \( Cl_2 \):
\[ P_{Cl_2} = \frac{\alpha P}{(1 + \alpha)} \]

### Step 6: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]
Substituting the pressures:
\[ K_p = \frac{\left(\frac{\alpha P}{(1 + \alpha)}\right) \cdot \left(\frac{\alpha P}{(1 + \alpha)}\right)}{\frac{(1 - \alpha)P}{(1 + \alpha)}} \]
This simplifies to:
\[ K_p = \frac{\alpha^2 P}{(1 - \alpha)(1 + \alpha)} \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ K_p(1 - \alpha)(1 + \alpha) = \alpha^2 P \]
Expanding and rearranging:
\[ K_p + K_p \alpha - \alpha^2 K_p = \alpha^2 P \]
This leads to:
\[ K_p = \alpha^2 \left( \frac{P + K_p}{K_p} \right) \]

### Step 8: Solve for \( \alpha \)
From the above equation, we can express \( \alpha \) in terms of \( K_p \) and \( P \):
\[ \alpha = \sqrt{\frac{K_p}{K_p + P}} \]

### Step 9: Relate \( x \) to \( A \)
Since \( \alpha = \frac{x}{A} \), we can write:
\[ \frac{x}{A} = \sqrt{\frac{K_p}{K_p + P}} \]
Thus:
\[ x = A \sqrt{\frac{K_p}{K_p + P}} \]

### Final Answer
The relationship between the moles of \( PCl_5 \) that dissociate and the equilibrium constant is:
\[ x = A \sqrt{\frac{K_p}{K_p + P}} \]