1.1 mol of A is mixed with 2.2 mol of B and the mixture is kept in a one - itre flask till the equilibrium, `A+2BhArr 2C+D` is reached. At equilibrium 0.2 mol of C is formed. The equilibrium constant of the above reaction is

Options

A. 0.002
B. 0.004
C. 0.001
D. 0.003

Answer

To find the equilibrium constant \( K \) for the reaction \( A + 2B \rightleftharpoons 2C + D \), we will follow these steps:

### Step 1: Write the balanced equation and initial concentrations
The balanced equation is:
\[ A + 2B \rightleftharpoons 2C + D \]

Initially, we have:
- Moles of \( A = 1.1 \, \text{mol} \)
- Moles of \( B = 2.2 \, \text{mol} \)
- Moles of \( C = 0 \, \text{mol} \)
- Moles of \( D = 0 \, \text{mol} \)

### Step 2: Set up the change in concentrations at equilibrium
Let \( x \) be the amount of \( A \) that reacts at equilibrium. Based on the stoichiometry of the reaction:
- \( A \) decreases by \( x \)
- \( B \) decreases by \( 2x \)
- \( C \) increases by \( 2x \)
- \( D \) increases by \( x \)

At equilibrium, we have:
- Moles of \( A = 1.1 - x \)
- Moles of \( B = 2.2 - 2x \)
- Moles of \( C = 2x \)
- Moles of \( D = x \)

### Step 3: Use the information provided to find \( x \)
We are given that at equilibrium, \( 0.2 \, \text{mol} \) of \( C \) is formed. Since \( C = 2x \):
\[ 2x = 0.2 \]
\[ x = 0.1 \, \text{mol} \]

### Step 4: Calculate the equilibrium concentrations
Now we can calculate the equilibrium concentrations:
- Moles of \( A = 1.1 - 0.1 = 1.0 \, \text{mol} \)
- Moles of \( B = 2.2 - 2(0.1) = 2.0 \, \text{mol} \)
- Moles of \( C = 0.2 \, \text{mol} \)
- Moles of \( D = 0.1 \, \text{mol} \)

Since the reaction is in a 1-liter flask, the molarity is equal to the number of moles.

### Step 5: Write the expression for the equilibrium constant \( K \)
The equilibrium constant \( K \) is given by:
\[ K = \frac{[C]^2[D]}{[A][B]^2} \]

Substituting the equilibrium concentrations:
\[ K = \frac{(0.2)^2(0.1)}{(1.0)(2.0)^2} \]

### Step 6: Calculate \( K \)
Calculating the values:
\[ K = \frac{(0.04)(0.1)}{(1.0)(4.0)} = \frac{0.004}{4.0} = 0.001 \]

### Final Answer
The equilibrium constant \( K \) for the reaction is:
\[ K = 0.001 \]

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