Phosphorus pentachloride dissociates as follows, ina closed reaction vessel, `PC1_(5(g))hArrPC1_(3(g)) + C1_(2(g))` If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of `PC1_(5)` is x, the partial pressure of `PC1_(3)` will be:

Options

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

Explanation

To solve the problem, we need to find the partial pressure of PCl3 at equilibrium when phosphorus pentachloride (PCl5) dissociates into PCl3 and Cl2. We will follow these steps:

### Step-by-Step Solution:

1. **Write the balanced equation for the dissociation:**
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]

2. **Define the initial conditions:**
- Let the initial number of moles of PCl5 be 1 mole.
- Therefore, the initial moles of PCl3 and Cl2 are both 0.

3. **Define the degree of dissociation:**
- Let the degree of dissociation of PCl5 be \( x \).
- At equilibrium, the moles of each substance will be:
- Moles of PCl5 = \( 1 - x \)
- Moles of PCl3 = \( x \)
- Moles of Cl2 = \( x \)

4. **Calculate the total number of moles at equilibrium:**
\[ \text{Total moles} = (1 - x) + x + x = 1 + x \]

5. **Express the total pressure at equilibrium:**
- Let the total pressure at equilibrium be \( P \).
- The total pressure can also be expressed in terms of the number of moles and the ideal gas law.

6. **Calculate the mole fraction of PCl3:**
\[ \text{Mole fraction of PCl3} = \frac{\text{Moles of PCl3}}{\text{Total moles}} = \frac{x}{1 + x} \]

7. **Calculate the partial pressure of PCl3:**
- The partial pressure of a gas can be calculated using its mole fraction and the total pressure:
\[ \text{Partial pressure of PCl3} = \text{Mole fraction of PCl3} \times P = \frac{x}{1 + x} \times P \]

8. **Final expression for the partial pressure of PCl3:**
\[ \text{Partial pressure of PCl3} = \frac{xP}{1 + x} \]

### Final Answer:
The partial pressure of PCl3 at equilibrium is given by:
\[ \text{Partial pressure of PCl3} = \frac{xP}{1 + x} \]