i. The initial pressure of `PCl_(5)` present in one litre vessel at `200 K` is `2` atm. At equilibrium the pressure increases to `3` atm with temperature increasing to `250`. The percentage dissociation of `PCl_(5)` at equilibrium is

Options

A. `30%`
B. `60%`
C. `0.2%`
D. `20%`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the Reaction
The dissociation reaction of phosphorus pentachloride (PCl₅) can be represented as:
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]

### Step 2: Initial Conditions
- Initial pressure of PCl₅ at 200 K = 2 atm
- Volume of the vessel = 1 L (constant)
- Initial moles of PCl₅ = 1 mole (since pressure is given in atm and volume is 1 L)

### Step 3: Final Conditions
- At equilibrium, the total pressure increases to 3 atm.
- Temperature increases to 250 K.

### Step 4: Calculate the Pressure of PCl₅ at 250 K
Using the ideal gas law relation (since volume is constant):
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Where:
- \( P_1 = 2 \, \text{atm} \)
- \( T_1 = 200 \, \text{K} \)
- \( P_2 = ? \)
- \( T_2 = 250 \, \text{K} \)

Substituting the known values:
\[ \frac{2}{200} = \frac{P_2}{250} \]
Cross-multiplying gives:
\[ P_2 = \frac{2 \times 250}{200} = 2.5 \, \text{atm} \]

### Step 5: Calculate the Change in Pressure
At equilibrium, the pressure of the system is 3 atm. Therefore, the change in pressure due to dissociation is:
\[ \Delta P = P_{\text{final}} - P_{\text{PCl}_5 \, \text{at} \, 250 \, \text{K}} = 3 \, \text{atm} - 2.5 \, \text{atm} = 0.5 \, \text{atm} \]

### Step 6: Relate Change in Pressure to Dissociation
Let \( \alpha \) be the degree of dissociation of PCl₅. The equilibrium pressures can be expressed as:
- Pressure of PCl₅ at equilibrium = \( 2.5 - 0.5 = 2 \, \text{atm} \)
- Pressure of PCl₃ at equilibrium = \( \alpha \, \text{atm} \)
- Pressure of Cl₂ at equilibrium = \( \alpha \, \text{atm} \)

### Step 7: Set Up the Equation
The total pressure at equilibrium is given by:
\[ P_{\text{total}} = P_{\text{PCl}_5} + P_{\text{PCl}_3} + P_{\text{Cl}_2} \]
Substituting the values:
\[ 3 = (2.5 - 0.5) + \alpha + \alpha \]
This simplifies to:
\[ 3 = 2 + 2\alpha \]
Solving for \( \alpha \):
\[ 2\alpha = 3 - 2 \\ 2\alpha = 1 \\ \alpha = 0.5 \, \text{atm} \]

### Step 8: Calculate the Percentage Dissociation
The percentage dissociation is given by:
\[ \text{Percentage Dissociation} = \left( \frac{\alpha}{\text{Initial Pressure}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage Dissociation} = \left( \frac{0.5}{2.5} \right) \times 100 = 20\% \]

### Final Answer
The percentage dissociation of PCl₅ at equilibrium is **20%**.

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