A quantity of `PCI _5` was heated in a 2 litre vessel at 525 K . It dissociates as `PCI_5 (g) hArr PCI _3 (g) + CI_2 (g)` At equilibrium 0.2 mol each of `PCI_5 , PCl_3and Cl_2` is found in the reaction mixture. The equilibrium constant `K_c` for the reaction is -

Options

A. 0.2
B. 0.5
C. 0.1
D. 0.05

Answer

To solve the problem, we need to calculate the equilibrium constant \( K_c \) for the dissociation reaction of \( PCl_5 \) into \( PCl_3 \) and \( Cl_2 \).

The reaction can be represented as:

\[ PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g) \]

### Step 1: Write down the expression for the equilibrium constant \( K_c \)

The equilibrium constant \( K_c \) for the reaction is given by the formula:

\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

### Step 2: Determine the concentrations at equilibrium

We are given that at equilibrium, there are 0.2 moles of each substance in a 2-liter vessel. We can calculate the concentrations as follows:

\[ [PCl_5] = \frac{0.2 \text{ moles}}{2 \text{ L}} = 0.1 \text{ M} \]

\[ [PCl_3] = \frac{0.2 \text{ moles}}{2 \text{ L}} = 0.1 \text{ M} \]

\[ [Cl_2] = \frac{0.2 \text{ moles}}{2 \text{ L}} = 0.1 \text{ M} \]

### Step 3: Substitute the concentrations into the \( K_c \) expression

Now we can substitute these values into the equilibrium constant expression:

\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{(0.1)(0.1)}{0.1} \]

### Step 4: Calculate \( K_c \)

Calculating the above expression gives:

\[ K_c = \frac{0.01}{0.1} = 0.1 \]

### Final Answer

Thus, the equilibrium constant \( K_c \) for the reaction is:

\[ K_c = 0.1 \]

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