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°C and one atmosphere, 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
Assuming we start with 1 mole of \( PCl_5 \):
- Initial moles of \( PCl_5 \) = 1
- Initial moles of \( PCl_3 \) = 0
- Initial moles of \( Cl_2 \) = 0

### Step 3: Calculate moles at equilibrium
Given that \( PCl_5 \) is 40% dissociated, we can calculate the moles at equilibrium:
- Moles of \( PCl_5 \) that dissociate = \( 0.4 \times 1 = 0.4 \)
- Moles of \( PCl_5 \) remaining = \( 1 - 0.4 = 0.6 \)
- Moles of \( PCl_3 \) formed = \( 0.4 \)
- Moles of \( Cl_2 \) formed = \( 0.4 \)

### Step 4: Set up equilibrium concentrations
Since we are assuming the volume of the system is 1 liter, the equilibrium concentrations will be:
- Concentration of \( PCl_5 \) = \( 0.6 \, \text{mol/L} \)
- Concentration of \( PCl_3 \) = \( 0.4 \, \text{mol/L} \)
- Concentration of \( Cl_2 \) = \( 0.4 \, \text{mol/L} \)

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

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

### Step 7: Calculate \( K_c \)
Now, we can calculate \( K_c \):
\[ K_c = \frac{0.16}{0.6} = 0.2667 \]

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