Four moles of `PCl_(5)` are heated in a closed 4 `dm^(3)` container to reach equilibrium at 400 K. At equilibrium 50% of `PCl_(5)` is dissociated. What is the value of `K_(c)` for the dissociation of `PCl_(5)` into `PCl_(3) "and" Cl_(2)` at 400 K ?

Options

A. 0.5
B. 1
C. 1.15
D. 0.05

Explanation

To find the equilibrium constant \( K_c \) for the dissociation of \( PCl_5 \) into \( PCl_3 \) and \( Cl_2 \) at 400 K, 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: Determine initial moles and changes at equilibrium
Initially, we have 4 moles of \( PCl_5 \) in a closed container.

- Initial moles of \( PCl_5 \) = 4 moles
- At equilibrium, it is given that 50% of \( PCl_5 \) is dissociated. Therefore, the moles of \( PCl_5 \) that dissociate is:
\[ \text{Dissociated moles} = 0.5 \times 4 = 2 \text{ moles} \]

### Step 3: Calculate the moles at equilibrium
At equilibrium, the moles of each substance will be:
- Moles of \( PCl_5 \) = \( 4 - 2 = 2 \) moles
- Moles of \( PCl_3 \) = \( 2 \) moles (produced)
- Moles of \( Cl_2 \) = \( 2 \) moles (produced)

### Step 4: Calculate the concentrations
The volume of the container is given as 4 dm³ (which is equivalent to 4 liters). The concentrations can be calculated using the formula:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume (L)}} \]

Calculating the concentrations:
- Concentration of \( PCl_5 \):
\[ [C_{PCl_5}] = \frac{2 \text{ moles}}{4 \text{ L}} = 0.5 \text{ M} \]
- Concentration of \( PCl_3 \):
\[ [C_{PCl_3}] = \frac{2 \text{ moles}}{4 \text{ L}} = 0.5 \text{ M} \]
- Concentration of \( Cl_2 \):
\[ [C_{Cl_2}] = \frac{2 \text{ moles}}{4 \text{ L}} = 0.5 \text{ M} \]

### 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 concentrations into the \( K_c \) expression
Substituting the values we calculated:
\[ K_c = \frac{(0.5)(0.5)}{0.5} = \frac{0.25}{0.5} = 0.5 \]

### Final Answer
Thus, the value of \( K_c \) for the dissociation of \( PCl_5 \) at 400 K is:
\[ \boxed{0.5} \]