In a 500 ml capacity vessel CO and `Cl_(2)` are mixed to form `COCl_(2)`. At equilibrium, it contains 0.2 moles of `COCl_(2)` and 0.1 mole of each of CO and `Cl_(2)`. The equilibrium constant `K_(c)` for the reaction `CO+Cl_(2) hArr COCl_(2)` is

Options

A. 5
B. 10
C. 15
D. 20

Answer

To find the equilibrium constant \( K_c \) for the reaction:

\[ \text{CO} + \text{Cl}_2 \rightleftharpoons \text{COCl}_2 \]

we will follow these steps:

### Step 1: Identify the moles at equilibrium
From the problem, we know:
- Moles of \( COCl_2 \) at equilibrium = 0.2 moles
- Moles of \( CO \) at equilibrium = 0.1 moles
- Moles of \( Cl_2 \) at equilibrium = 0.1 moles

### Step 2: Convert moles to concentration
The volume of the vessel is 500 mL, which is equivalent to 0.5 L. The concentration (C) can be calculated using the formula:

\[ C = \frac{\text{moles}}{\text{volume (L)}} \]

Calculating the concentrations:

- Concentration of \( COCl_2 \):
\[ [C] = \frac{0.2 \, \text{moles}}{0.5 \, \text{L}} = 0.4 \, \text{mol/L} \]

- Concentration of \( CO \):
\[ [CO] = \frac{0.1 \, \text{moles}}{0.5 \, \text{L}} = 0.2 \, \text{mol/L} \]

- Concentration of \( Cl_2 \):
\[ [Cl_2] = \frac{0.1 \, \text{moles}}{0.5 \, \text{L}} = 0.2 \, \text{mol/L} \]

### Step 3: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:

\[ K_c = \frac{[\text{products}]}{[\text{reactants}]} \]

For our reaction:

\[ K_c = \frac{[COCl_2]}{[CO][Cl_2]} \]

### Step 4: Substitute the concentrations into the \( K_c \) expression
Now substituting the concentrations we calculated:

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

### Step 5: Calculate \( K_c \)
Calculating the denominator:

\[ (0.2)(0.2) = 0.04 \]

Now substituting back into the equation:

\[ K_c = \frac{0.4}{0.04} = 10 \]

### Step 6: Conclusion
Thus, the equilibrium constant \( K_c \) for the reaction is:

\[ K_c = 10 \]

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