1.1 moles of A are mixed with 2.2 moles of B and the mixture is kept in a one litre flask till the equilibrium
` A+ 2B hArr 2C + D `
is reached. At equilibrium, 0.2 moles of C are formed. The equilibrium constant of the reaction is

Options

A. `0.001`
B. ` 0.002`
C. ` 0.0033`
D. ` 0.004`

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 chemical equation
The balanced chemical equation is:
\[ A + 2B \rightleftharpoons 2C + D \]

### Step 2: Determine initial moles and changes at equilibrium
Initially, we have:
- Moles of \( A = 1.1 \)
- Moles of \( B = 2.2 \)
- Moles of \( C = 0 \)
- Moles of \( D = 0 \)

Let \( x \) be the amount of \( A \) that reacts at equilibrium. According to the stoichiometry of the reaction:
- \( A \) will decrease by \( x \)
- \( B \) will decrease by \( 2x \)
- \( C \) will increase by \( 2x \)
- \( D \) will increase by \( x \)

### Step 3: Use the information given to find \( x \)
We know from the problem that at equilibrium, \( 0.2 \) moles of \( C \) are formed. Therefore:
\[ 2x = 0.2 \]
\[ x = 0.1 \]

### Step 4: Calculate moles at equilibrium
Now we can calculate the moles of each substance at equilibrium:
- Moles of \( A \) at equilibrium:
\[ A = 1.1 - x = 1.1 - 0.1 = 1.0 \]
- Moles of \( B \) at equilibrium:
\[ B = 2.2 - 2x = 2.2 - 0.2 = 2.0 \]
- Moles of \( C \) at equilibrium:
\[ C = 2x = 0.2 \]
- Moles of \( D \) at equilibrium:
\[ D = x = 0.1 \]

### Step 5: Write the expression for the equilibrium constant \( K \)
The equilibrium constant \( K \) is given by:
\[ K = \frac{[C]^2[D]}{[A][B]^2} \]
Since the reaction is in a 1-liter flask, the concentrations are equal to the number of moles.

### Step 6: Substitute the equilibrium values into the expression
Substituting the equilibrium values:
\[ K = \frac{(0.2)^2(0.1)}{(1.0)(2.0)^2} \]
Calculating the numerator:
\[ (0.2)^2 = 0.04 \quad \text{and} \quad 0.04 \times 0.1 = 0.004 \]
Calculating the denominator:
\[ (2.0)^2 = 4 \quad \text{and} \quad 1.0 \times 4 = 4 \]
Thus,
\[ K = \frac{0.004}{4} = 0.001 \]

### Final Answer
The equilibrium constant \( K \) is \( 0.001 \).

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