In 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(g)+Cl(g)hArrCOCl_(2)(g)` is

Options

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

Answer

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

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

given the equilibrium concentrations, we can follow these steps:

### Step 1: Identify the equilibrium concentrations
From the problem, we have:
- Moles of \( \text{COCl}_2 \) at equilibrium = 0.2 moles
- Moles of \( \text{CO} \) at equilibrium = 0.1 moles
- Moles of \( \text{Cl}_2 \) at equilibrium = 0.1 moles

### Step 2: Convert the volume from mL to L
The volume of the vessel is given as 500 mL. To convert this to liters, we divide by 1000:

\[ \text{Volume in L} = \frac{500 \, \text{mL}}{1000} = 0.5 \, \text{L} \]

### Step 3: Calculate the equilibrium concentrations (molarity)
To find the molarity (concentration in moles per liter), we use the formula:

\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in L}} \]

Calculating each concentration:

1. For \( \text{COCl}_2 \):
\[ [\text{COCl}_2] = \frac{0.2 \, \text{moles}}{0.5 \, \text{L}} = 0.4 \, \text{M} \]

2. For \( \text{CO} \):
\[ [\text{CO}] = \frac{0.1 \, \text{moles}}{0.5 \, \text{L}} = 0.2 \, \text{M} \]

3. For \( \text{Cl}_2 \):
\[ [\text{Cl}_2] = \frac{0.1 \, \text{moles}}{0.5 \, \text{L}} = 0.2 \, \text{M} \]

### Step 4: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by the formula:

\[ K_c = \frac{[\text{Products}]}{[\text{Reactants}]} \]

For our reaction, this becomes:

\[ K_c = \frac{[\text{COCl}_2]}{[\text{CO}][\text{Cl}_2]} \]

### Step 5: Substitute the equilibrium concentrations into the expression
Now we can substitute the calculated concentrations:

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

### Step 6: Calculate \( K_c \)
Calculating the denominator:

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

Now substituting back into the equation for \( K_c \):

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

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is:

\[ \boxed{10} \]