This question involves equilibria and the calculation of equilibrium concentrations after a change in conditions. Here's the step-by-step solution:
- Initially, the equilibrium condition is given for the reaction \( \text{A(g)} \leftrightarrow \text{B(g)} \).
- Initial concentrations at equilibrium are:
- Concentration of A: \([A] = 0.5 \, \text{M}\)
- Concentration of B: \([B] = 0.375 \, \text{M}\)
- Calculate the equilibrium constant \( K_c \). Since the reaction is at equilibrium:
- \(K_c = \frac{[B]}{[A]} = \frac{0.375}{0.5} = 0.75\)
- Now, 0.1 moles of A is added to the flask. The new concentration of A is:
- New [A] = Initial [A] + Additional [A] = \(0.5 + \frac{0.1}{1}\, \text{M} = 0.6 \, \text{M}\)
- Upon reheating to temperature T, the system will re-establish equilibrium with the same \( K_c \). Let the change in concentration of A upon reaching new equilibrium be \( x \).
- Equilibrium concentrations will be:
- \([A] = 0.6 - x\)
- \([B] = 0.375 + x\)
- Using \( K_c \):
- \(\frac{0.375 + x}{0.6 - x} = 0.75\)
- Solve for \( x \):
- \(0.375 + x = 0.75(0.6 - x)\)
- \(0.375 + x = 0.45 - 0.75x\)
- \(1.75x = 0.45 - 0.375 = 0.075\)
- \(x = \frac{0.075}{1.75} = 0.0429 \, \text{M} (approx.)\)
- Substitute \( x \) back to find new equilibrium concentrations:
- New [A] = 0.6 - 0.0429 ≈ 0.557 M
- New [B] = 0.375 + 0.0429 ≈ 0.418 M
Therefore, after re-establishing equilibrium, the correct concentrations should be 0.557 M for A and 0.418 M for B. However, the provided correct answer in the prompt is 0.742 M for A and 0.557 M for B, which suggests a mistake. Thus, this final calculation needs verification or possibly reconsideration of the initial assumptions, perhaps with constraint on rounding errors or interpretation of given conditions.