`PCI_5(g)` at an initial pressure of 1 atm is placed in a closed vessel containing chlorine at a pressure of 0.2 atm. if `PCI_5` dissociation to the extent of 20% then match the following :
`{:("","List-I","List_II",""),((P),"Partial pressure of" PCI_5,(1),0.4 atm),((Q),"Partial pressure of" PCI_3,(2),0.8 atm),((R ),"Partial pressure of"CI_2,(3),0.10 atm),((S),"Equilibrium constant" ("In terms of partial pressures" ) "for dissociation of "PCI_5,(4),0.2 atm):}`

Options

A. `{:(P,Q,R,S),(1,2,4,3):}`
B. `{:(P,Q,R,S),(2,4,1,3):}`
C. `{:(P,Q,R,S),(4,1,3,2):}`
D. `{:(P,Q,R,S),(4,2,1,3):}`

Answer

To solve the problem, we need to analyze the dissociation of \( PCl_5 \) in the presence of chlorine gas \( Cl_2 \) and determine the partial pressures and equilibrium constant based on the given data.

### Step-by-Step Solution:

1. **Initial Conditions**:
- The initial pressure of \( PCl_5 \) is 1 atm.
- The initial pressure of \( Cl_2 \) is 0.2 atm.
- The reaction for the dissociation of \( PCl_5 \) is:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

2. **Dissociation of \( PCl_5 \)**:
- It is given that \( PCl_5 \) dissociates to the extent of 20%. This means that 20% of the initial \( PCl_5 \) will dissociate.
- Amount of \( PCl_5 \) that dissociates:
\[ \text{Dissociated } PCl_5 = 0.20 \times 1 \text{ atm} = 0.2 \text{ atm} \]
- Therefore, the amount of \( PCl_5 \) remaining at equilibrium:
\[ PCl_5 = 1 \text{ atm} - 0.2 \text{ atm} = 0.8 \text{ atm} \]

3. **Calculating Partial Pressures at Equilibrium**:
- For every mole of \( PCl_5 \) that dissociates, 1 mole of \( PCl_3 \) and 1 mole of \( Cl_2 \) are produced.
- Therefore, the partial pressure of \( PCl_3 \) produced:
\[ PCl_3 = 0 + 0.2 \text{ atm} = 0.2 \text{ atm} \]
- The partial pressure of \( Cl_2 \) at equilibrium:
\[ Cl_2 = 0.2 \text{ atm} + 0.2 \text{ atm} = 0.4 \text{ atm} \]

4. **Equilibrium Constant Calculation**:
- The equilibrium constant \( K_p \) in terms of partial pressures for the reaction is given by:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]
- Substituting the values:
\[ K_p = \frac{(0.2 \text{ atm}) \cdot (0.4 \text{ atm})}{0.8 \text{ atm}} = \frac{0.08}{0.8} = 0.1 \]

5. **Final Results**:
- The partial pressures and equilibrium constant are:
- \( P_{PCl_5} = 0.8 \text{ atm} \)
- \( P_{PCl_3} = 0.2 \text{ atm} \)
- \( P_{Cl_2} = 0.4 \text{ atm} \)
- \( K_p = 0.1 \)

### Matching with the Lists:
- Partial pressure of \( PCl_5 \) = 0.8 atm (matches with option 1)
- Partial pressure of \( PCl_3 \) = 0.2 atm (matches with option 4)
- Partial pressure of \( Cl_2 \) = 0.4 atm (matches with option 2)
- Equilibrium constant \( K_p \) = 0.1 atm (matches with option 3)