Chemical Equilibrium Part 2

Concept:For a reversible reaction at constant temperature, the equilibrium constant $K_c$ remains unchanged.Explanation:Step 1: Calculate $K_c$ from the initial equilibrium.$K_c = \frac{[B]}{[A]} = \frac{0.375}{0.5} = 0.75$Step 2: After adding 0.1 mol of $A$ to the 1 L flask, the new initial concentrations (before re-equilibrium) are $[A] = 0.6$ M and $[B] = 0.375$ M. The reaction will proceed in the forward direction to reduce $A$ and increase $B$.Step 3: Let $x$ be the decrease in concentration of $A$ and the increase in concentration of $B$.At the new equilibrium: $[A] = 0.6 - x$ M, $[B] = 0.375 + x$ M.Step 4: Apply the equilibrium constant expression:$\frac{0.375 + x}{0.6 - x} = 0.75$Cross‑multiply: $0.375 + x = 0.75(0.6 - x) = 0.45 - 0.75x$Combine like terms: $x + 0.75x = 0.45 - 0.375$$\Rightarrow 1.75x = 0.075$$\Rightarrow x = \frac{0.075}{1.75} = \frac{3}{70} \approx 0.0429$ MStep 5: Calculate the new equilibrium concentrations:$[A] = 0.6 - 0.0429 = 0.557$ M$[B] = 0.375 + 0.0429 = 0.418$ MAnswer:Option D: $0.557$ M and $0.418$ M.