One mole of `PCl_5` is heated in a 2L vessel. When equilibrium is attained, the vessel is found to contain 0.2 moles of `PCl_5` Calculate the equilibrium constant.

Answer

To solve the problem of calculating the equilibrium constant for the decomposition of \( PCl_5 \), we will follow these steps:

### Step 1: Write the balanced equation for the decomposition of \( PCl_5 \)

The balanced equation for the decomposition of phosphorus pentachloride is:
\[ PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g) \]

### Step 2: Determine the initial moles and concentrations

Initially, we have 1 mole of \( PCl_5 \) in a 2L vessel. Therefore, the initial concentration of \( PCl_5 \) is:
\[ \text{Initial moles of } PCl_5 = 1 \text{ mole} \]
\[ \text{Volume} = 2 \text{ L} \]
\[ \text{Initial concentration of } PCl_5 = \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M} \]

### Step 3: Set up the equilibrium expression

At equilibrium, we are given that there are 0.2 moles of \( PCl_5 \). Therefore, the concentration of \( PCl_5 \) at equilibrium is:
\[ \text{Equilibrium moles of } PCl_5 = 0.2 \text{ moles} \]
\[ \text{Equilibrium concentration of } PCl_5 = \frac{0.2 \text{ moles}}{2 \text{ L}} = 0.1 \text{ M} \]

### Step 4: Determine the change in moles

Let \( x \) be the amount of \( PCl_5 \) that decomposes. Initially, we have:
- \( PCl_5 \): \( 0.5 \) moles
- \( PCl_3 \): \( 0 \) moles
- \( Cl_2 \): \( 0 \) moles

At equilibrium, we have:
- \( PCl_5 \): \( 0.5 - x \)
- \( PCl_3 \): \( x \)
- \( Cl_2 \): \( x \)

Since we know that at equilibrium \( PCl_5 = 0.2 \) moles, we can set up the equation:
\[ 0.5 - x = 0.1 \]

### Step 5: Solve for \( x \)

Rearranging the equation gives:
\[ x = 0.5 - 0.1 = 0.4 \]

### Step 6: Calculate the equilibrium concentrations of \( PCl_3 \) and \( Cl_2 \)

At equilibrium:
- \( PCl_3 = x = 0.4 \) moles
- \( Cl_2 = x = 0.4 \) moles

Now, we can find the equilibrium concentrations:
\[ \text{Equilibrium concentration of } PCl_3 = \frac{0.4 \text{ moles}}{2 \text{ L}} = 0.2 \text{ M} \]
\[ \text{Equilibrium concentration of } Cl_2 = \frac{0.4 \text{ moles}}{2 \text{ L}} = 0.2 \text{ M} \]

### Step 7: Write the expression for the equilibrium constant \( K_c \)

The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

### Step 8: Substitute the equilibrium concentrations into the expression

Substituting the equilibrium concentrations we found:
\[ K_c = \frac{(0.2)(0.2)}{0.1} \]
\[ K_c = \frac{0.04}{0.1} = 0.4 \]

### Final Answer

The equilibrium constant \( K_c \) for the decomposition of \( PCl_5 \) is:
\[ K_c = 0.4 \]