Question

One mole of compound AB reacts with one mole of compound CD according to the equation: \[ {AB} + {CD} \rightleftharpoons {AD} + {CB} \] At equilibrium, it was found that \(\frac{3}{4}\) mole of AB and CD had been converted to AD and CB. There is no change in volume. The equilibrium constant for the reaction is:

• A. \( \frac{9}{16} \)
• B. \( \frac{1}{9} \)
• C. \( \frac{16}{9} \)
• D. 9

Hint:

When calculating equilibrium constants, the concentration of each species at equilibrium is crucial. Ensure that changes in concentration are correctly accounted for based on the stoichiometry of the reaction.

Answer 1

The Correct Option is D

Step 1: Set up the initial and equilibrium concentrations. Given that one mole each of AB and CD reacts, at equilibrium: - \({AB}_{{eq}} = {CD}_{{eq}} = 1 - \frac{3}{4} = \frac{1}{4}\) moles (remaining) - \({AD}_{{eq}} = {CB}_{{eq}} = \frac{3}{4}\) moles (formed) 
Step 2: Write the expression for the equilibrium constant (K). \[ K = \frac{[{AD}][{CB}]}{[{AB}][{CD}]} \] 
Step 3: Plug in the equilibrium concentrations. \[ K = \frac{\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)}{\left(\frac{1}{4}\right)\left(\frac{1}{4}\right)} = \frac{\frac{9}{16}}{\frac{1}{16}} = 9 \]