Initially, 0.8 mole of `PCl_(5)` and 0.2 mol of `PCl_(3` 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. `"0.13 mol L"^(-1)`
B. `"0.66 mol L"^(-1)`
C. `"0.013 mol L"^(-1)`
D. `"0.05 mol L"^(-1)`

Answer

To solve the problem, we need to determine the equilibrium constant \( K_c \) for the reaction:

\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

### Step 1: Write 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 2: Set up the change in concentrations
Let \( x \) be the amount of \( PCl_5 \) that dissociates at equilibrium. The changes in concentrations will be:
- \( [PCl_5] = 0.8 - x \)
- \( [PCl_3] = 0.2 + x \)
- \( [Cl_2] = x \)

### Step 3: Use the equilibrium information
At equilibrium, we are given that \( [PCl_3] = 0.4 \, \text{mol/L} \). Therefore, we can set up the equation:

\[ 0.2 + x = 0.4 \]

### Step 4: Solve for \( x \)
From the equation above, we can solve for \( x \):

\[ x = 0.4 - 0.2 = 0.2 \]

### Step 5: Calculate the equilibrium concentrations
Now we can find the equilibrium concentrations:
- \( [PCl_5] = 0.8 - x = 0.8 - 0.2 = 0.6 \, \text{mol/L} \)
- \( [PCl_3] = 0.4 \, \text{mol/L} \) (given)
- \( [Cl_2] = x = 0.2 \, \text{mol/L} \)

### Step 6: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:

\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

### Step 7: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the values we found:

\[ K_c = \frac{(0.4)(0.2)}{0.6} \]

### Step 8: Calculate \( K_c \)
Now, calculate \( K_c \):

\[ K_c = \frac{0.08}{0.6} = \frac{8}{60} = \frac{2}{15} \approx 0.1333 \]

### Final Answer
Thus, the value of \( K_c \) for the reaction is approximately \( 0.1333 \).

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