1.1 mole of A and 2.2 moles of B reach an cquilibrium in I lit container according to the reaction. `A + 2B hArr `2C + D. If at equilibrium 0.1 mole of D is present, the equilibrium constant is:

Options

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

Answer

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

### Step 1: Set up the initial concentrations
We start with the initial moles of reactants and products:
- Initial moles of \( A = 1.1 \) moles
- Initial moles of \( B = 2.2 \) moles
- Initial moles of \( C = 0 \) moles
- Initial moles of \( D = 0 \) moles

Since the reaction occurs in a 1-liter container, the initial concentrations are:
- \([A] = 1.1\) M
- \([B] = 2.2\) M
- \([C] = 0\) M
- \([D] = 0\) M

### Step 2: Define the change in concentration at equilibrium
Let \( x \) be the change in moles 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 \( D \).

At equilibrium, the changes will be:
- \( A \) will decrease by \( x \)
- \( B \) will decrease by \( 2x \)
- \( C \) will increase by \( 2x \)
- \( D \) will increase by \( x \)

### Step 3: Write the equilibrium concentrations
At equilibrium, the concentrations will be:
- \([A] = 1.1 - x\)
- \([B] = 2.2 - 2x\)
- \([C] = 2x\)
- \([D] = x\)

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

### Step 5: Substitute \( x \) into the equilibrium concentrations
Now, substituting \( x = 0.1 \):
- \([A] = 1.1 - 0.1 = 1.0\) M
- \([B] = 2.2 - 2(0.1) = 2.0\) M
- \([C] = 2(0.1) = 0.2\) M
- \([D] = 0.1\) M

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

### Step 7: Substitute the equilibrium concentrations into the expression
Now substituting the equilibrium concentrations:
\[ K_c = \frac{(0.2)^2(0.1)}{(1.0)(2.0)^2} \]

### Step 8: Calculate \( K_c \)
Calculating the numerator:
\[ (0.2)^2(0.1) = 0.04 \times 0.1 = 0.004 \]
Calculating the denominator:
\[ (1.0)(2.0)^2 = 1.0 \times 4.0 = 4.0 \]
Now, substituting these values into the equation for \( K_c \):
\[ K_c = \frac{0.004}{4.0} = 0.001 \]

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

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