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 dissociation of \( PCl_5 \), we can follow these steps:

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

### Step 2: Determine initial moles and changes
Initially, we have 2 moles of \( PCl_5 \) and no \( PCl_3 \) or \( Cl_2 \):
- Initial moles of \( PCl_5 = 2 \)
- Initial moles of \( PCl_3 = 0 \)
- Initial moles of \( Cl_2 = 0 \)

Given that 40% of \( PCl_5 \) dissociates, we can calculate the moles that dissociate:
\[ \text{Moles dissociated} = 0.4 \times 2 = 0.8 \text{ moles} \]

### Step 3: Calculate moles at equilibrium
At equilibrium, the moles of each species will be:
- Moles of \( PCl_5 \) remaining:
\[ 2 - 0.8 = 1.2 \text{ moles} \]
- Moles of \( PCl_3 \) formed:
\[ 0 + 0.8 = 0.8 \text{ moles} \]
- Moles of \( Cl_2 \) formed:
\[ 0 + 0.8 = 0.8 \text{ moles} \]

### Step 4: Calculate concentrations at equilibrium
The volume of the vessel is given as 2 L. We can now calculate the concentrations:
- Concentration of \( PCl_5 \):
\[ \text{Concentration of } PCl_5 = \frac{1.2 \text{ moles}}{2 \text{ L}} = 0.6 \text{ mol/L} \]
- Concentration of \( PCl_3 \):
\[ \text{Concentration of } PCl_3 = \frac{0.8 \text{ moles}}{2 \text{ L}} = 0.4 \text{ mol/L} \]
- Concentration of \( Cl_2 \):
\[ \text{Concentration of } Cl_2 = \frac{0.8 \text{ moles}}{2 \text{ L}} = 0.4 \text{ mol/L} \]

### 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 concentrations into the \( K_c \) expression
Substituting the values we calculated:
\[ K_c = \frac{(0.4)(0.4)}{0.6} \]

### Step 7: Calculate \( K_c \)
Now, we can perform the calculation:
\[ K_c = \frac{0.16}{0.6} = 0.2667 \text{ mol/L} \]

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

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