For the reaction: `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`, if the initial concentration of `PCl_(5)=1` mol/`l` and x moles/litre of `PCl_(5)` are consumed at equilibrium, the correct expression of `K_(p)` if P is total pressure is

Options

A. `(x^(2)P)/(1-x^(2))`
B. `((1+x)^(2)P)/((1-x)^(2))`
C. `(x^(2)P)/((2+x)^(2))`
D. `(x^(2)P)/(2(1-x)^(2))`

Answer

To solve the problem, we need to derive the expression for the equilibrium constant \( K_p \) for the reaction:

\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

### Step 1: Define Initial Concentrations
We start with the initial concentration of \( PCl_5 \):
- Initial concentration of \( PCl_5 = 1 \, \text{mol/L} \)
- Initial concentrations of \( PCl_3 \) and \( Cl_2 \) are both \( 0 \, \text{mol/L} \).

### Step 2: Change in Concentrations at Equilibrium
Let \( x \) be the amount of \( PCl_5 \) that dissociates at equilibrium. Therefore, at equilibrium:
- Concentration of \( PCl_5 = 1 - x \, \text{mol/L} \)
- Concentration of \( PCl_3 = x \, \text{mol/L} \)
- Concentration of \( Cl_2 = x \, \text{mol/L} \)

### Step 3: Calculate Total Pressure at Equilibrium
The total pressure \( P \) at equilibrium can be expressed as:
\[ P_{total} = P_{PCl_5} + P_{PCl_3} + P_{Cl_2} \]
Using the ideal gas law, we can relate concentrations to pressure:
- The pressure contribution from \( PCl_5 \) is proportional to its concentration: \( P_{PCl_5} = (1 - x) \)
- The pressure contribution from \( PCl_3 \) is: \( P_{PCl_3} = x \)
- The pressure contribution from \( Cl_2 \) is: \( P_{Cl_2} = x \)

Thus, the total pressure becomes:
\[ P_{total} = (1 - x) + x + x = 1 + x \]

### Step 4: 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 we found:
\[ K_p = \frac{x \cdot x}{1 - x} = \frac{x^2}{1 - x} \]

### Step 5: Substitute Total Pressure into the Expression
We can express \( K_p \) in terms of total pressure \( P_{total} \):
Since \( P_{total} = 1 + x \), we can rearrange this to find \( x \):
\[ x = P_{total} - 1 \]
Now substituting \( x \) back into the expression for \( K_p \):
\[ K_p = \frac{(P_{total} - 1)^2}{1 - (P_{total} - 1)} = \frac{(P_{total} - 1)^2}{2 - P_{total}} \]

### Final Expression for \( K_p \)
Thus, the final expression for \( K_p \) in terms of total pressure \( P_{total} \) is:
\[ K_p = \frac{(P_{total} - 1)^2}{2 - P_{total}} \]