1.1 mole of A is mixed with 1.2 mol of B and the mixture is kept in a 1 L flask till the equilibrium `A+2B hArr 2C+D` is reached. At equilibrium 0.1 mol of D is formed . The `K_c` of the reaction

Options

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

Answer

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

\[ A + 2B \rightleftharpoons 2C + D \]

Given:
- Initial moles of \( A = 1.1 \, \text{mol} \)
- Initial moles of \( B = 1.2 \, \text{mol} \)
- Volume of the flask = 1 L
- At equilibrium, moles of \( D = 0.1 \, \text{mol} \)

### Step 1: Set up the initial concentrations
Since the volume of the flask is 1 L, the initial concentrations of \( A \) and \( B \) can be calculated as follows:

- Initial concentration of \( A \) = \( 1.1 \, \text{mol/L} \)
- Initial concentration of \( B \) = \( 1.2 \, \text{mol/L} \)
- Initial concentration of \( C \) = \( 0 \, \text{mol/L} \)
- Initial concentration of \( D \) = \( 0 \, \text{mol/L} \)

### Step 2: Define the change in concentration at equilibrium
Let \( x \) be the amount of \( D \) formed at equilibrium. According to the problem, \( x = 0.1 \, \text{mol} \).

From the stoichiometry of the reaction:
- For every 1 mole of \( D \) formed, 2 moles of \( C \) are formed.
- Therefore, at equilibrium:
- Moles of \( C = 2x = 2(0.1) = 0.2 \, \text{mol} \)
- Moles of \( A \) remaining = \( 1.1 - x = 1.1 - 0.1 = 1.0 \, \text{mol} \)
- Moles of \( B \) remaining = \( 1.2 - 2x = 1.2 - 2(0.1) = 1.0 \, \text{mol} \)

### Step 3: Write the equilibrium concentrations
Since the volume of the flask is 1 L, the equilibrium concentrations are:
- Concentration of \( A = 1.0 \, \text{mol/L} \)
- Concentration of \( B = 1.0 \, \text{mol/L} \)
- Concentration of \( C = 0.2 \, \text{mol/L} \)
- Concentration of \( D = 0.1 \, \text{mol/L} \)

### Step 4: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by the formula:

\[ K_c = \frac{[C]^2[D]}{[A][B]^2} \]

Substituting the equilibrium concentrations into the expression:

\[ K_c = \frac{(0.2)^2(0.1)}{(1.0)(1.0)^2} \]

### Step 5: Calculate \( K_c \)
Now, we calculate \( K_c \):

\[ K_c = \frac{0.04 \times 0.1}{1} = 0.004 \]

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

\[ K_c = 0.004 \]