Phosphorus pentachloride dissociates as follows in a closed reaction vessel
` PCl_5 (g) hArr PCl_3 (g) + Cl_2( g)`
If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of `PCl_5 ` is x, the partial pressure of `PCl_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 will follow these steps:

### Step 1: Write the reaction and define the variables
The dissociation of phosphorus pentachloride (PCl₅) can be represented as:
\[ PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g) \]

Let:
- The initial amount of PCl₅ = 1 mole
- Degree of dissociation of PCl₅ = \( x \)

### Step 2: Determine the moles at equilibrium
At equilibrium, the moles of each component will be:
- Moles of PCl₅ = \( 1 - x \) (since \( x \) moles have dissociated)
- Moles of PCl₃ = \( x \) (produced from the dissociation)
- Moles of Cl₂ = \( x \) (produced from the dissociation)

### Step 3: Calculate the total moles at equilibrium
The total moles at equilibrium can be calculated as:
\[ \text{Total moles} = \text{Moles of PCl₅} + \text{Moles of PCl₃} + \text{Moles of Cl₂} \]
\[ \text{Total moles} = (1 - x) + x + x = 1 + x \]

### Step 4: Use Dalton's Law of Partial Pressures
According to Dalton's Law, the partial pressure of a gas in a mixture is given by:
\[ P_i = \chi_i \cdot P \]
where \( P_i \) is the partial pressure of the gas, \( \chi_i \) is the mole fraction of the gas, and \( P \) is the total pressure.

### Step 5: Calculate the mole fraction of PCl₃
The mole fraction of PCl₃ (\( \chi_{PCl_3} \)) can be calculated as:
\[ \chi_{PCl_3} = \frac{\text{Moles of PCl₃}}{\text{Total moles}} = \frac{x}{1 + x} \]

### Step 6: Calculate the partial pressure of PCl₃
Now, we can find the partial pressure of PCl₃:
\[ P_{PCl_3} = \chi_{PCl_3} \cdot P = \left(\frac{x}{1 + x}\right) \cdot P \]

### Final Answer
Thus, the partial pressure of PCl₃ at equilibrium is:
\[ P_{PCl_3} = \frac{xP}{1 + x} \]

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