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 step by step, we will follow the reaction equilibrium process and calculate the equilibrium constant \( K_c \).

### Step 1: Write the balanced chemical equation
The reaction given is:
\[ A + 2B \rightleftharpoons 2C + D \]

### Step 2: Set up the initial concentrations
We are given:
- Initial moles of \( A = 1.1 \) mol
- Initial moles of \( B = 1.2 \) mol
- Volume of the flask = 1 L

Since the volume is 1 L, the initial concentrations are:
- \([A]_{initial} = 1.1 \, \text{mol/L}\)
- \([B]_{initial} = 1.2 \, \text{mol/L}\)
- \([C]_{initial} = 0 \, \text{mol/L}\)
- \([D]_{initial} = 0 \, \text{mol/L}\)

### Step 3: Define the change in concentrations at equilibrium
Let \( x \) be the amount of \( A \) that reacts at equilibrium. According to the stoichiometry of the reaction:
- For every 1 mole of \( A \) that reacts, 2 moles of \( B \) react, producing 2 moles of \( C \) and 1 mole of \( D \).

Thus, the changes in concentration will be:
- \([A] = 1.1 - x\)
- \([B] = 1.2 - 2x\)
- \([C] = 2x\)
- \([D] = x\)

### Step 4: Use the information given in the problem
We know that at equilibrium, 0.1 mol of \( D \) is formed:
\[ x = 0.1 \]

### Step 5: Calculate the equilibrium concentrations
Substituting \( x = 0.1 \) into the expressions for equilibrium concentrations:
- \([A] = 1.1 - 0.1 = 1.0 \, \text{mol/L}\)
- \([B] = 1.2 - 2(0.1) = 1.2 - 0.2 = 1.0 \, \text{mol/L}\)
- \([C] = 2(0.1) = 0.2 \, \text{mol/L}\)
- \([D] = 0.1 \, \text{mol/L}\)

### Step 6: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C]^2[D]}{[A][B]^2} \]

### Step 7: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the values:
\[ K_c = \frac{(0.2)^2(0.1)}{(1.0)(1.0)^2} \]

### Step 8: Calculate \( K_c \)
Calculating the numerator:
\[ (0.2)^2 = 0.04 \]
Thus,
\[ K_c = \frac{0.04 \times 0.1}{1.0} = \frac{0.004}{1} = 0.004 \]

### Final Answer
The value of \( K_c \) is:
\[ \boxed{0.004} \]