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 find 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
The initial concentration of \( PCl_5 \) is given as 1 mol/L. At equilibrium, let \( x \) moles/L of \( PCl_5 \) be consumed.

- Initial concentration of \( PCl_5 \) = 1 mol/L
- Change in concentration of \( PCl_5 \) = -x
- At equilibrium, concentration of \( PCl_5 \) = \( 1 - x \) mol/L

### Step 2: Determine Equilibrium Concentrations
From the stoichiometry of the reaction, for every mole of \( PCl_5 \) that dissociates, one mole of \( PCl_3 \) and one mole of \( Cl_2 \) are produced.

- Concentration of \( PCl_3 \) at equilibrium = \( x \) mol/L
- Concentration of \( Cl_2 \) at equilibrium = \( x \) mol/L

### Step 3: Calculate Total Moles at Equilibrium
The total number of moles at equilibrium can be calculated as follows:

\[ \text{Total moles} = \text{moles of } PCl_5 + \text{moles of } PCl_3 + \text{moles of } Cl_2 = (1 - x) + x + x = 1 + x \]

### Step 4: Calculate Partial Pressures
The partial pressure of each component can be expressed in terms of the total pressure \( P \):

- Partial pressure of \( PCl_5 \):
\[ P_{PCl_5} = \left( \frac{1 - x}{1 + x} \right) P \]

- Partial pressure of \( PCl_3 \):
\[ P_{PCl_3} = \left( \frac{x}{1 + x} \right) P \]

- Partial pressure of \( Cl_2 \):
\[ P_{Cl_2} = \left( \frac{x}{1 + x} \right) P \]

### Step 5: Write the Expression for \( K_p \)
The expression for \( K_p \) is given by the ratio of the product of the partial pressures of the products to the partial pressure of the reactants:

\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]

Substituting the partial pressures into the equation:

\[ K_p = \frac{\left( \frac{x}{1 + x} P \right) \cdot \left( \frac{x}{1 + x} P \right)}{\left( \frac{1 - x}{1 + x} P \right)} \]

### Step 6: Simplify the Expression
Now, simplify the expression:

\[ K_p = \frac{\frac{x^2}{(1 + x)^2} P^2}{\frac{1 - x}{1 + x} P} \]

This simplifies to:

\[ K_p = \frac{x^2 P}{(1 + x)(1 - x)} \]

### Final Expression for \( K_p \)
Thus, the expression for \( K_p \) is:

\[ K_p = \frac{x^2 P}{(1 + x)(1 - x)} \]