`1.1` mole of A mixed with `2.2` mole of B and then the mixture is then kept in one litre flask till the equilibrium is attained A+2B⇌2C+D . At the equilibrium `0.2` mole of C is formed, then the value of `K_(c)` will be:

Options

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

Answer

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

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

### Step 1: Determine Initial Moles
We start with the initial moles of reactants:
- Moles of A = 1.1
- Moles of B = 2.2
- Moles of C = 0 (since it hasn't formed yet)
- Moles of D = 0 (since it hasn't formed yet)

### Step 2: Change in Moles at Equilibrium
At equilibrium, we are given that 0.2 moles of C are formed. Since the stoichiometry of the reaction shows that 2 moles of C are produced for every mole of A consumed and 2 moles of B consumed, we can denote the change in moles as follows:
- Let \( x \) be the change in moles of A consumed, then:
- Moles of C formed = \( 2x \)
- Moles of D formed = \( x \)

Given that \( 2x = 0.2 \), we can solve for \( x \):
\[ x = \frac{0.2}{2} = 0.1 \]

### Step 3: Calculate Moles at Equilibrium
Now we can find the moles of each substance at equilibrium:
- Moles of A at equilibrium = \( 1.1 - x = 1.1 - 0.1 = 1.0 \)
- Moles of B at equilibrium = \( 2.2 - 2x = 2.2 - 2(0.1) = 2.0 \)
- Moles of C at equilibrium = \( 0 + 0.2 = 0.2 \)
- Moles of D at equilibrium = \( 0 + x = 0 + 0.1 = 0.1 \)

### Step 4: Calculate Concentrations
Since the volume of the flask is 1 L, the concentrations (in mol/L) are the same as the number of moles:
- Concentration of A = \( 1.0 \, \text{mol/L} \)
- Concentration of B = \( 2.0 \, \text{mol/L} \)
- Concentration of C = \( 0.2 \, \text{mol/L} \)
- Concentration of D = \( 0.1 \, \text{mol/L} \)

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

### Step 6: Substitute the Concentrations into the Expression
Substituting the equilibrium concentrations into the expression:
\[ K_c = \frac{(0.2)^2(0.1)}{(1.0)(2.0)^2} \]

### Step 7: Calculate \( K_c \)
Calculating the numerator:
\[ (0.2)^2 = 0.04 \quad \text{and} \quad 0.04 \times 0.1 = 0.004 \]

Calculating the denominator:
\[ (2.0)^2 = 4 \quad \text{and} \quad 1.0 \times 4 = 4 \]

Putting it all together:
\[ K_c = \frac{0.004}{4} = 0.001 \]

### Final Answer
The value of \( K_c \) is \( 0.001 \).

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