`2` mole of `PCI_(5)` were heated in a `5` litre vessel. It dissociated. `80%` at equilibrium find out the value of equilibrium constatn. Report your answer as `K_(C)xx50`.

Answer

To solve the problem, we will follow these steps:

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

### Step 2: Set up the initial conditions
We start with 2 moles of PCl₅ in a 5-liter vessel. Therefore, the initial concentrations are:
- \[ [\text{PCl}_5] = \frac{2 \text{ moles}}{5 \text{ L}} = 0.4 \text{ M} \]
- \[ [\text{PCl}_3] = 0 \text{ M} \]
- \[ [\text{Cl}_2] = 0 \text{ M} \]

### Step 3: Determine the change in concentration at equilibrium
According to the problem, 80% of PCl₅ dissociates at equilibrium. This means:
- Moles of PCl₅ that dissociate = \( 2 \text{ moles} \times 0.8 = 1.6 \text{ moles} \)
- Moles of PCl₅ remaining = \( 2 - 1.6 = 0.4 \text{ moles} \)

At equilibrium, the concentrations will be:
- \[ [\text{PCl}_5] = \frac{0.4 \text{ moles}}{5 \text{ L}} = 0.08 \text{ M} \]
- \[ [\text{PCl}_3] = \frac{1.6 \text{ moles}}{5 \text{ L}} = 0.32 \text{ M} \]
- \[ [\text{Cl}_2] = \frac{1.6 \text{ moles}}{5 \text{ L}} = 0.32 \text{ M} \]

### Step 4: Write the expression for the equilibrium constant (Kc)
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \]

### Step 5: Substitute the equilibrium concentrations into the Kc expression
Substituting the values we found:
\[ K_c = \frac{(0.32)(0.32)}{0.08} \]

### Step 6: Calculate Kc
Calculating the above expression:
\[ K_c = \frac{0.1024}{0.08} = 1.28 \]

### Step 7: Report the answer as Kc multiplied by 50
Now, we need to report the answer as \( K_c \times 50 \):
\[ K_c \times 50 = 1.28 \times 50 = 64 \]

### Final Answer
Thus, the value of the equilibrium constant \( K_c \) is:
\[ K_c = 64 \]

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