At a certain temperature 2 moles of carbonmonoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction
`CO+Cl_(2)hArrCOCl_(2)` in a 5 lit vessel. At equilibrium if one mole of CO is present then equilibrium constant for the reaction is :

Options

A. 2
B. 2.5
C. `3.0`
D. 4

Answer

To solve the problem, we need to find the equilibrium constant (Kc) for the reaction:

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

### Step 1: Write the initial amounts of reactants and products
Initially, we have:
- Moles of CO = 2
- Moles of Cl2 = 3
- Moles of COCl2 = 0

### Step 2: Set up the change in moles at equilibrium
Let \( x \) be the amount of CO that reacts at equilibrium. Then at equilibrium, the moles will be:
- Moles of CO = \( 2 - x \)
- Moles of Cl2 = \( 3 - x \)
- Moles of COCl2 = \( x \)

### Step 3: Use the information given in the problem
We are given that at equilibrium, 1 mole of CO is present. Therefore:
\[ 2 - x = 1 \]
Solving for \( x \):
\[ x = 2 - 1 = 1 \]

### Step 4: Calculate the moles of each substance at equilibrium
Using \( x = 1 \):
- Moles of CO = \( 2 - 1 = 1 \)
- Moles of Cl2 = \( 3 - 1 = 2 \)
- Moles of COCl2 = \( 1 \)

### Step 5: Calculate the concentrations
The volume of the vessel is 5 liters, so we can find the concentrations:
- Concentration of CO = \( \frac{1 \text{ mole}}{5 \text{ L}} = \frac{1}{5} \text{ M} \)
- Concentration of Cl2 = \( \frac{2 \text{ moles}}{5 \text{ L}} = \frac{2}{5} \text{ M} \)
- Concentration of COCl2 = \( \frac{1 \text{ mole}}{5 \text{ L}} = \frac{1}{5} \text{ M} \)

### Step 6: Write the expression for the equilibrium constant (Kc)
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[\text{COCl}_2]}{[\text{CO}][\text{Cl}_2]} \]

### Step 7: Substitute the equilibrium concentrations into the Kc expression
Substituting the values we found:
\[ K_c = \frac{\left(\frac{1}{5}\right)}{\left(\frac{1}{5}\right)\left(\frac{2}{5}\right)} \]

### Step 8: Simplify the expression
Calculating the denominator:
\[ K_c = \frac{\frac{1}{5}}{\frac{1 \times 2}{5 \times 5}} = \frac{\frac{1}{5}}{\frac{2}{25}} = \frac{1}{5} \times \frac{25}{2} = \frac{25}{10} = 2.5 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is:
\[ K_c = 2.5 \]

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