Topic of the Question:
This question belongs to the topic of chemical equilibrium, specifically focusing on the calculation of the equilibrium constant ($K_c$) using stoichiometry and initial concentrations of reactants.
Step 1 : Understanding the Question:
We are given that $1.1\text{ moles of reactant A}$ and $2.2\text{ moles of reactant B}$ are mixed in a $1\text{ L}$ flask. They react according to the equation: $\text{A} + 2\text{B} \rightleftharpoons 2\text{C} + \text{D}$. At equilibrium, the flask contains $0.2\text{ moles of product C}$. We need to find the value of the equilibrium constant $K_c$.
Step 2 : Key Formulas and Approach:
The equilibrium constant ($K_c$) for the given reaction is expressed as: $K_c = \frac{[\text{C}]^2[\text{D}]}{[\text{A}][\text{B}]^2}$.
We use an ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentrations of all species in terms of a change variable $x$.
Since the volume of the container is $1\text{ L}$, the molar concentration of each species is numerically equal to its number of moles: $[\text{X}] = \frac{\text{moles of X}}{1\text{ L}}$.
Step 3 : Detailed Explanation:
We construct the ICE table to track the moles of each reactant and product throughout the reaction:
Initial state: Moles of $\text{A} = 1.1$, Moles of $\text{B} = 2.2$, Moles of $\text{C} = 0$, Moles of $\text{D} = 0$.
Change state: Let $x$ moles of $\text{A}$ react. According to stoichiometry, $2x$ moles of $\text{B}$ react, producing $2x$ moles of $\text{C}$ and $x$ moles of $\text{D}$.
Equilibrium state: Moles of $\text{A} = 1.1 - x$, Moles of $\text{B} = 2.2 - 2x$, Moles of $\text{C} = 2x$, Moles of $\text{D} = x$.
We are given that the moles of $\text{C}$ at equilibrium is $0.2$. Therefore, we can write: $2x = 0.2 \implies x = 0.1$.
We substitute $x = 0.1$ back into our expressions to find the equilibrium moles of each species:
Moles of $\text{A} = 1.1 - 0.1 = 1.0\text{ mole}$.
Moles of $\text{B} = 2.2 - 2(0.1) = 2.0\text{ moles}$.
Moles of $\text{C} = 0.2\text{ mole}$.
Moles of $\text{D} = 0.1\text{ mole}$.
Since the volume of the flask is $1\text{ L}$, these mole values are equal to the molar concentrations: $[\text{A}] = 1.0\text{ M}$, $[\text{B}] = 2.0\text{ M}$, $[\text{C}] = 0.2\text{ M}$, and $[\text{D}] = 0.1\text{ M}$.
We substitute these concentrations into the $K_c$ expression: $K_c = \frac{(0.2)^2 \times (0.1)}{(1.0) \times (2.0)^2}$.
Simplifying the numerator and denominator: $K_c = \frac{0.04 \times 0.1}{1.0 \times 4.0} = \frac{0.004}{4.0} = 0.001$.
Step 4 : Final Answer:
The calculated value of the equilibrium constant $K_c$ is $0.001$, which corresponds to Option (C).