At `250^(@)C` and one atmosphere `PCl_(5)` is `40%` dissociated The equilibrium constant `(K_(c))` for dissociation of `PCl_(5)` is

Options

A. 0.26
B. 2.6
C. 1.9
D. 0.19

Answer

To find the equilibrium constant \( K_c \) for the dissociation of \( PCl_5 \) at \( 250^\circ C \) and one atmosphere, we can follow these steps:

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

### Step 2: Define initial conditions
At the start (initial state), we assume we have 1 mole of \( PCl_5 \) and no \( PCl_3 \) or \( Cl_2 \):
- Initial moles of \( PCl_5 \) = 1 mole
- Initial moles of \( PCl_3 \) = 0 moles
- Initial moles of \( Cl_2 \) = 0 moles

### Step 3: Determine the change in moles at equilibrium
Given that \( PCl_5 \) is 40% dissociated, we can calculate the moles that dissociate:
\[ \text{Moles dissociated} = 40\% \text{ of } 1 \text{ mole} = 0.4 \text{ moles} \]

### Step 4: Calculate equilibrium moles
At equilibrium, the moles of each species will be:
- Moles of \( PCl_5 \) remaining = \( 1 - 0.4 = 0.6 \) moles
- Moles of \( PCl_3 \) formed = \( 0 + 0.4 = 0.4 \) moles
- Moles of \( Cl_2 \) formed = \( 0 + 0.4 = 0.4 \) moles

### Step 5: Calculate equilibrium concentrations
Assuming the volume of the system is 1 liter (since volume is not specified):
- Concentration of \( PCl_5 \) at equilibrium:
\[ [C_{PCl_5}] = \frac{0.6 \text{ moles}}{1 \text{ L}} = 0.6 \text{ M} \]
- Concentration of \( PCl_3 \) at equilibrium:
\[ [C_{PCl_3}] = \frac{0.4 \text{ moles}}{1 \text{ L}} = 0.4 \text{ M} \]
- Concentration of \( Cl_2 \) at equilibrium:
\[ [C_{Cl_2}] = \frac{0.4 \text{ moles}}{1 \text{ L}} = 0.4 \text{ M} \]

### Step 6: 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 7: Substitute the equilibrium concentrations into the expression
Substituting the concentrations we found:
\[ K_c = \frac{(0.4)(0.4)}{0.6} \]

### Step 8: Calculate \( K_c \)
Calculating the value:
\[ K_c = \frac{0.16}{0.6} = 0.2667 \]

### Final Result
Thus, the equilibrium constant \( K_c \) for the dissociation of \( PCl_5 \) at \( 250^\circ C \) is approximately:
\[ K_c \approx 0.267 \]