The equilibrium of formation of phosgene is represented as :
`CO(g)+Cl_(2)(g)hArrCOCl_(2)(g)`
The reaction is carried out in a `500 mL` flask. At equilibrium, `0.3` mol of phosgene, `0.1 mol` of `CO`, and `0.1` mol of `Cl_(2)` are present.
The equilibrium constant of the reaction is

Options

A. `30`
B. `15`
C. `5`
D. `25`

Explanation

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

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

Given:
- Volume of the flask = 500 mL = 0.5 L
- At equilibrium:
- Moles of COCl2 = 0.3 mol
- Moles of CO = 0.1 mol
- Moles of Cl2 = 0.1 mol

### Step 1: Calculate the concentrations of the species at equilibrium

The concentration \( C \) is calculated using the formula:

\[ C = \frac{\text{moles}}{\text{volume in liters}} \]

1. For COCl2:
\[ [\text{COCl}_2] = \frac{0.3 \, \text{mol}}{0.5 \, \text{L}} = 0.6 \, \text{mol/L} \]

2. For CO:
\[ [\text{CO}] = \frac{0.1 \, \text{mol}}{0.5 \, \text{L}} = 0.2 \, \text{mol/L} \]

3. For Cl2:
\[ [\text{Cl}_2] = \frac{0.1 \, \text{mol}}{0.5 \, \text{L}} = 0.2 \, \text{mol/L} \]

### Step 2: Write the expression for the equilibrium constant \( K_c \)

The equilibrium constant \( K_c \) for the reaction is given by:

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

### Step 3: Substitute the concentrations into the \( K_c \) expression

Substituting the calculated concentrations into the expression:

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

### Step 4: Calculate \( K_c \)

Calculating the denominator:

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

Now substituting back into the equation:

\[ K_c = \frac{0.6}{0.04} = 15 \]

### Final Answer

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

\[ \boxed{15} \]

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