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 \( Cl_2 \), we will 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: Determine the initial moles and the change at equilibrium
Initially, we have 4 moles of \( PCl_5 \) and no products:
- Initial: \( PCl_5 = 4 \) moles, \( PCl_3 = 0 \) moles, \( Cl_2 = 0 \) moles.

Since 50% of \( PCl_5 \) dissociates, we can calculate the moles that dissociate:
- Moles dissociated = \( 50\% \) of \( 4 \) moles = \( 2 \) moles.

At equilibrium:
- \( PCl_5 = 4 - 2 = 2 \) moles,
- \( PCl_3 = 2 \) moles (produced),
- \( Cl_2 = 2 \) moles (produced).

### Step 3: Calculate the concentrations at equilibrium
The volume of the container is \( 4 \, dm^3 \) (or \( 4 \, L \)). The concentration is calculated as:
\[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} \]

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

### Step 4: Write the expression for \( K_c \)
The expression for the equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.5)(0.5)}{0.5} = \frac{0.25}{0.5} = 0.5 \]

### Final Answer
The value of \( K_c \) at 400 K is \( 0.5 \).

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