Exothermic equilibrium shifts right at lower temperature

Steps:

• The reaction H₂ + I₂ ⇌ 2HI is exothermic, so K increases as temperature decreases.
• At T₁, equal initial H₂ and I₂ yield equilibrium mole fraction of HI at 0.5, indicating 50% conversion.
• Lowering temperature to T₂ shifts equilibrium right per Le Chatelier's principle, increasing HI production beyond 0.5.
• The graph at T₂ shows slower approach to equilibrium (lower rate) but higher final HI mole fraction.

Why B is correct:

• Option B depicts a curve reaching >0.5 HI mole fraction more slowly, matching increased K and reduced rate for exothermic reaction.

Why the others are wrong:

• A: Reaches same 0.5 as T₁, ignoring temperature effect on K.
• C: Reaches <0.5, contradicting shift for exothermic forward reaction.
• D: Faster approach to higher HI, but lower temperature slows kinetics.

Final answer: B