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:

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

we will follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_C \)

The equilibrium constant \( K_C \) is defined as:

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

For our reaction, this becomes:

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

### Step 2: Determine the concentrations of the species at equilibrium

We are given the number of moles of each species at equilibrium:

- Moles of \( \text{COCl}_2 = 0.2 \) moles
- Moles of \( \text{CO} = 0.1 \) moles
- Moles of \( \text{Cl}_2 = 0.1 \) moles

The volume of the vessel is 500 mL, which we convert to liters:

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

Now we can calculate the concentrations:

- Concentration of \( \text{COCl}_2 \):

\[ [\text{COCl}_2] = \frac{0.2 \, \text{moles}}{0.5 \, \text{L}} = 0.4 \, \text{M} \]

- Concentration of \( \text{CO} \):

\[ [\text{CO}] = \frac{0.1 \, \text{moles}}{0.5 \, \text{L}} = 0.2 \, \text{M} \]

- Concentration of \( \text{Cl}_2 \):

\[ [\text{Cl}_2] = \frac{0.1 \, \text{moles}}{0.5 \, \text{L}} = 0.2 \, \text{M} \]

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

Now we can substitute the concentrations into the \( K_C \) expression:

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

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

Calculating the denominator:

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

Now substituting back into the equation:

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

### Final Answer

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

\[ K_C = 10 \]

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