By dissociation of 4g mol of of` PCl_(5)` produce 0.8 mol of `PCl_(3)` if
vol of container is 1 ltr the equilibrium constant :-

Options

A. 0.2
B. 0.1
C. 0.4
D. 1

Explanation

To find the equilibrium constant (Kc) for the dissociation of PCl5, we will follow these steps:

### Step 1: Write the balanced chemical equation
The dissociation of phosphorus pentachloride (PCl5) can be represented as:
\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]

### Step 2: Determine the initial moles and changes
Initially, we have 4 moles of PCl5 and no products (PCl3 and Cl2). The change in moles when the system reaches equilibrium is given as:
- PCl5: decreases by 0.8 moles
- PCl3: increases by 0.8 moles
- Cl2: increases by 0.8 moles

### Step 3: Set up the equilibrium concentrations
At equilibrium, we can summarize the moles as follows:
- Moles of PCl5 = \( 4 - 0.8 = 3.2 \) moles
- Moles of PCl3 = \( 0 + 0.8 = 0.8 \) moles
- Moles of Cl2 = \( 0 + 0.8 = 0.8 \) moles

### Step 4: Calculate the equilibrium concentrations
Since the volume of the container is 1 liter, the concentrations (in mol/L) are equal to the number of moles:
- Concentration of PCl5 = \( \frac{3.2 \text{ moles}}{1 \text{ L}} = 3.2 \, \text{M} \)
- Concentration of PCl3 = \( \frac{0.8 \text{ moles}}{1 \text{ L}} = 0.8 \, \text{M} \)
- Concentration of Cl2 = \( \frac{0.8 \text{ moles}}{1 \text{ L}} = 0.8 \, \text{M} \)

### Step 5: Write the expression for the equilibrium constant (Kc)
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

### Step 6: Substitute the equilibrium concentrations into the Kc expression
Substituting the values we found:
\[ K_c = \frac{(0.8)(0.8)}{3.2} \]

### Step 7: Calculate Kc
Calculating the above expression:
\[ K_c = \frac{0.64}{3.2} = 0.2 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) is:
\[ K_c = 0.2 \]