Two moles of `PCl_(5)` is heated in a closed vessel of 2 L capacity. When the equilibrium is attained 40 % of it has been found to be dissociated. What is the Kc in `mol//dm^(3)` ?

Options

A. 0.532
B. 0.266
C. 0.133
D. 0.174

Explanation

To find the equilibrium constant \( K_c \) for the reaction of \( PCl_5 \) dissociating into \( PCl_3 \) and \( Cl_2 \), we can follow these steps:

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

### Step 2: Set up the initial conditions
We start with 2 moles of \( PCl_5 \) in a 2 L vessel. The initial concentration of \( PCl_5 \) can be calculated as:
\[ \text{Initial concentration of } PCl_5 = \frac{2 \text{ moles}}{2 \text{ L}} = 1 \text{ mol/L} \]

### Step 3: Determine the change at equilibrium
Given that 40% of \( PCl_5 \) dissociates, we can calculate the amount that dissociates:
\[ \text{Amount dissociated} = 0.4 \times 2 \text{ moles} = 0.8 \text{ moles} \]

### Step 4: Calculate the equilibrium concentrations
At equilibrium, the concentrations of the species will be:
- For \( PCl_5 \):
\[ \text{Concentration of } PCl_5 = \frac{2 - 0.8}{2} = \frac{1.2}{2} = 0.6 \text{ mol/L} \]
- For \( PCl_3 \) and \( Cl_2 \) (since they are produced in a 1:1 ratio):
\[ \text{Concentration of } PCl_3 = \text{Concentration of } Cl_2 = \frac{0.8}{2} = 0.4 \text{ mol/L} \]

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

### Step 6: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the values we found:
\[ K_c = \frac{(0.4)(0.4)}{0.6} \]

### Step 7: Calculate \( K_c \)
Calculating the above expression:
\[ K_c = \frac{0.16}{0.6} \approx 0.2667 \text{ mol/dm}^3 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) is approximately:
\[ K_c \approx 0.267 \text{ mol/dm}^3 \]

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