Initially, 0.8 mole of `PCl_(5)` and 0.2 mol of `PCl_(30` are mixed in one litre vessel. At equilibrium, 0.4 mol of `PCl_(3)` is present. The value of `K_(c )` for the reaction
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`
would be

Options

A. `0.1 mol L^(-1)`
B. `0.05 mol L^(-1)`
C. `0.013 molL^(-1)`
D. ` 0.66 mol L^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation.
The reaction is given as:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

### Step 2: Set up the initial concentrations.
Initially, we have:
- \( [PCl_5] = 0.8 \, \text{mol/L} \)
- \( [PCl_3] = 0.2 \, \text{mol/L} \)
- \( [Cl_2] = 0 \, \text{mol/L} \)

### Step 3: Determine the change in concentrations at equilibrium.
At equilibrium, we are told that \( [PCl_3] = 0.4 \, \text{mol/L} \). This means that the change in concentration of \( PCl_3 \) is:
\[ \Delta[PCl_3] = 0.4 - 0.2 = 0.2 \, \text{mol/L} \]

Since 1 mole of \( PCl_5 \) produces 1 mole of \( PCl_3 \) and 1 mole of \( Cl_2 \), the change in concentration of \( PCl_5 \) will be the same as the increase in \( PCl_3 \):
\[ \Delta[PCl_5] = -0.2 \, \text{mol/L} \]
\[ \Delta[Cl_2] = +0.2 \, \text{mol/L} \]

### Step 4: Calculate the equilibrium concentrations.
Now we can calculate the equilibrium concentrations:
- \( [PCl_5] = 0.8 - 0.2 = 0.6 \, \text{mol/L} \)
- \( [PCl_3] = 0.4 \, \text{mol/L} \) (given)
- \( [Cl_2] = 0 + 0.2 = 0.2 \, \text{mol/L} \)

### 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 into the \( K_c \) expression.
Now substituting the equilibrium values:
\[ K_c = \frac{(0.4)(0.2)}{0.6} \]

### Step 7: Calculate \( K_c \).
Calculating the above expression:
\[ K_c = \frac{0.08}{0.6} = \frac{8}{60} = \frac{2}{15} \]

Thus, the value of \( K_c \) is:
\[ K_c = \frac{2}{15} \]

### Final Answer:
The value of \( K_c \) for the reaction is \( \frac{2}{15} \).

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