For a very small extent of dissociation of `PCl_(5)` into `PCl_(3)` and `Cl_(2)` is a gaseous phase reaction then degree of dissociation , x `:`

Options

A. `x prop P`
B. `x prop (1)/(p)`
C. `x prop sqrtp`
D. `x prop sqrt((1)/(p))`

Answer

To solve the problem of finding the degree of dissociation (denoted as \( x \)) of \( PCl_5 \) into \( PCl_3 \) and \( Cl_2 \), we can follow these steps:

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

### Step 2: Set Up Initial Conditions
Assume we start with 1 mole of \( PCl_5 \) and no products at time \( t = 0 \):
- Initial moles of \( PCl_5 = 1 \)
- Initial moles of \( PCl_3 = 0 \)
- Initial moles of \( Cl_2 = 0 \)

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

### Step 4: Calculate Total Moles at Equilibrium
The total number of moles at equilibrium is:
\[ \text{Total moles} = (1 - x) + x + x = 1 + x \]

### Step 5: Calculate Mole Fractions
Now, we can calculate the mole fractions of each component:
- Mole fraction of \( PCl_5 \):
\[ \chi_{PCl_5} = \frac{1 - x}{1 + x} \]
- Mole fraction of \( PCl_3 \):
\[ \chi_{PCl_3} = \frac{x}{1 + x} \]
- Mole fraction of \( Cl_2 \):
\[ \chi_{Cl_2} = \frac{x}{1 + x} \]

### Step 6: Calculate Partial Pressures
Assuming the total pressure is \( P \):
- Partial pressure of \( PCl_5 \):
\[ P_{PCl_5} = \chi_{PCl_5} \cdot P = \frac{(1 - x)}{(1 + x)} \cdot P \]
- Partial pressure of \( PCl_3 \):
\[ P_{PCl_3} = \chi_{PCl_3} \cdot P = \frac{x}{(1 + x)} \cdot P \]
- Partial pressure of \( Cl_2 \):
\[ P_{Cl_2} = \chi_{Cl_2} \cdot P = \frac{x}{(1 + x)} \cdot P \]

### Step 7: 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 partial pressures:
\[ K_p = \frac{\left(\frac{x}{1 + x} \cdot P\right) \cdot \left(\frac{x}{1 + x} \cdot P\right)}{\frac{(1 - x)}{(1 + x)} \cdot P} \]
This simplifies to:
\[ K_p = \frac{x^2 P^2}{(1 - x)(1 + x)} \]

### Step 8: Approximation for Small \( x \)
Since the extent of dissociation is very small, we can assume \( 1 - x \approx 1 \) and \( 1 + x \approx 1 \):
\[ K_p \approx \frac{x^2 P^2}{1} \]
Thus:
\[ K_p = x^2 P^2 \]

### Step 9: Solve for \( x \)
Rearranging gives:
\[ x = \sqrt{\frac{K_p}{P}} \]

### Conclusion
The degree of dissociation \( x \) is given by:
\[ x = \sqrt{\frac{K_p}{P}} \]