Question

1.1 mole of A mixed with 2.2 moles of B and the mixture is kept in a 1 L flask and the equilibrium, \(\text{A} + 2\text{B} \rightleftharpoons 2\text{C} + \text{D}\) is reached. If at equilibrium 0.2 mole of C is formed then the value of \(\text{K}_c\) will be

• A. \( 0.1 \)
• B. \( 0.01 \)
• C. \( 0.001 \)
• D. \( 1 \)

Hint:

Always verify container volume variations early. When the volume \(V = 1\text{ L}\), your calculation speed increases because moles directly equal molarities. If \(V \neq 1\text{ L}\), forgetting to divide individual equilibrium moles by the volume is a common pitfall that alters the final power values.

Answer 1

The Correct Option is C

Concept: The equilibrium constant (\(K_c\)) for a generalized reversible chemical reaction is expressed as the ratio of the product of the equilibrium molar concentrations of the products to that of the reactants, with each concentration term raised to a power equal to its stoichiometric coefficient in the balanced chemical equation. For the reaction: \[ \text{A} + 2\text{B} \rightleftharpoons 2\text{C} + \text{D} \] The equilibrium constant expression is: \[ K_c = \frac{[\text{C}]^2[\text{D}]}{[\text{A}][\text{B}]^2} \] Where \([\text{X}] = \frac{\text{moles of X at equilibrium}}{\text{Volume of flask in Liters}}\). Step 1: Setting up the equilibrium table in terms of degree of dissociation (\(x\)).
Let \(2x\) moles of C be formed at equilibrium based on its stoichiometric coefficient. The ICE (Initial, Change, Equilibrium) setup for the moles is as follows: \[ \begin{array}{lccccc} \text{Reaction:} & \text{A} & + & 2\text{B} & \rightleftharpoons & 2\text{C} & + & \text{D} \text{Initial (moles):} & 1.1 & & 2.2 & & 0 & & 0 \text{Change (moles):} & -x & & -2x & & +2x & & +x \text{Equilibrium (moles):} & 1.1 - x & & 2.2 - 2x & & 2x & & x \end{array} \] Given that the number of moles of C at equilibrium is \(0.2\): \[ 2x = 0.2 \quad \Rightarrow \quad x = 0.1 \]

Step 2: Calculating equilibrium concentrations and computing \(K_c\).
Substitute the value of \(x = 0.1\) into the equilibrium expressions to find the final moles:
• Moles of A \(= 1.1 - 0.1 = 1.0\text{ mole}\)
• Moles of B \(= 2.2 - 2(0.1) = 2.0\text{ moles}\)
• Moles of C \(= 0.2\text{ mole}\)
• Moles of D \(= 0.1\text{ mole}\) Since the volume of the container is \(V = 1\text{ L}\), the active masses (molarities) are identical to the number of moles: \[ [\text{A}] = 1.0\text{ M}, \quad [\text{B}] = 2.0\text{ M}, \quad [\text{C}] = 0.2\text{ M}, \quad [\text{D}] = 0.1\text{ M} \] Substitute these molarities into the \(K_c\) expression: \[ K_c = \frac{(0.2)^2 \times (0.1)}{(1.0) \times (2.0)^2} \] \[ K_c = \frac{0.04 \times 0.1}{1.0 \times 4.0} = \frac{0.004}{4} = 0.001 \]