Question

1.1 mol of A is mixed with 2.2 mol of B and the mixture is kept in one litre flask till the equilibrium is reached. At equilibrium, 0.2 mol of C is formed. If the equilibrium reaction is $A + 2B \rightarrow 2C + D,$ the value of equilibrium constant is

• A. 0.0002
• B. 0.0004
• C. 0.001
• D. 0.003

Answer 1

The Correct Option is C

To determine the equilibrium constant for the given reaction, you need to follow the steps below:

1. Identify the balanced chemical equation:
   A + 2B \rightarrow 2C + D
2. Use the initial and equilibrium conditions to find changes in concentration:
   Initially, we have:
   • 1.1 \, \text{mol of A}
   • 2.2 \, \text{mol of B}
   • 0 \, \text{mol of C}
   • 0 \, \text{mol of D}
   At equilibrium:
   • 0.2 mol of C is formed.
3. Calculate changes in mols for each reactant and product:
   From the stoichiometry of the reaction:
   • 2 mol of C is formed for every 1 mol of A reacted.
   • 1 mol of C is formed for every 0.5 mol of A reacted.
   Therefore, x = \frac{0.2}{2} = 0.1 \, \text{mol of A reacted}.
4. Determine equilibrium concentrations:
   
   • Equilibrium moles of A: 1.1 - 0.1 = 1.0 \, \text{mol}
   • Equilibrium moles of B: 2.2 - (2 \times 0.1) = 2.0 \, \text{mol}
   • Equilibrium moles of C: 0 + 0.2 = 0.2 \, \text{mol}
   • Equilibrium moles of D: 0 + 0.1 = 0.1 \, \text{mol}
   Since the reaction occurs in a 1-litre flask, the molar concentrations remain the same as the moles.
5. Write the expression for the equilibrium constant \( K_c \):
   K_c = \frac{[\text{C}]^2 \cdot [\text{D}]}{[\text{A}] \cdot [\text{B}]^2}
6. Substitute the equilibrium concentrations 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 \times 4}
   K_c = \frac{0.004}{4} = 0.001

Thus, the value of the equilibrium constant for the reaction is 0.001.