Le Chatelier's principle for exothermic equilibrium

Steps:

• The forward reaction H₂(g) + I₂(g) ⇌ 2HI(g) is exothermic (ΔH < 0).
• At T₁ (500 K), the graph shows mole fraction of HI decreasing from 0.5 to 0.23 and plateauing, indicating equilibrium favors reactants due to smaller K.
• Lowering temperature to T₂ shifts equilibrium right (Le Chatelier's principle), increasing K and thus the equilibrium mole fraction of HI.
• The new graph approaches a higher HI mole fraction than 0.23 before plateauing.

Why B is correct:

• B shows the mole fraction of HI decreasing to a value above 0.23 and flattening, matching the increased K for exothermic reaction at lower T.

Why the others are wrong:

• A decreases to below 0.23 (ignores temperature effect on K).
• C shows increasing HI (inconsistent with initial conditions favoring decomposition).
• D remains flat at 0.5 (no shift toward higher product).

Final answer: B