Question

For the reaction \( \mathrm{A \rightleftharpoons B} \), the number of moles of A and B at equilibrium in a \(1\,\text{L}\) vessel are \(0.50\) and \(0.375\), respectively. If \(0.10\) mol of A is added further, determine the number of moles of A and B at the new equilibrium.

• A. \(0.557,\ 0.557\)
• B. \(0.418,\ 0.557\)
• C. \(0.33,\ 0.56\)
• D. \(0.6,\ 0.2\)

Hint:

After adding a reactant at equilibrium, always recalculate using the {same equilibrium constant}. ICE tables make such problems quick and systematic.

Answer 1

The Correct Option is A

Concept:
For a reversible reaction at constant temperature:

The equilibrium constant \(K\) remains unchanged.
After a disturbance (like adding reactant), the system readjusts to re-establish equilibrium.

Step 1: Calculate the Equilibrium Constant
For reaction: \[ \mathrm{A \rightleftharpoons B} \] \[ K = \frac{[B]}{[A]} \] Given equilibrium concentrations (1 L vessel): \[ [A] = 0.50,\quad [B] = 0.375 \] \[ K = \frac{0.375}{0.50} = 0.75 \]
Step 2: Concentrations After Adding A
Added A \(= 0.10\) mol \[ [A] = 0.50 + 0.10 = 0.60,\quad [B] = 0.375 \] Let at new equilibrium: \[ [A] = 0.60 - x,\quad [B] = 0.375 + x \]
Step 3: Apply Equilibrium Constant
\[ K = \frac{0.375 + x}{0.60 - x} = 0.75 \] \[ 0.375 + x = 0.45 - 0.75x \] \[ 1.75x = 0.075 \Rightarrow x = 0.0429 \]
Step 4: Final Equilibrium Amounts
\[ [A] = 0.60 - 0.0429 \approx 0.557 \] \[ [B] = 0.375 + 0.0429 \approx 0.557 \] \[ \boxed{A = 0.557,\quad B = 0.557} \]