For a reversible endothermic reaction at equilibrium, how does the equilibrium conversion change as temperature decreases?
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AIt decreases (less products are favored).
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BIt increases (more products are favored).
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CIt remains unaffected by temperature.
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DIt increases linearly with temperature drop.
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EIt first increases then decreases regardless of enthalpy sign.
Answer
Correct Answer: It decreases (less products are favored).
Explanation
Introduction / Context:The temperature dependence of equilibrium conversion is governed by Le Chatelier's principle and the van’t Hoff relation. Recognizing how endothermicity affects conversion is essential for reactor design and optimization.
Given Data / Assumptions:
- Reaction is reversible and endothermic (ΔH > 0).
- System is at equilibrium.
Concept / Approach:For endothermic reactions, increasing temperature favors products to absorb heat; decreasing temperature favors reactants. The van’t Hoff relation shows that ln K increases with T when ΔH > 0.
Step-by-Step Solution:1) Endothermic: ΔH > 0.2) van’t Hoff: d(ln K)/dT = ΔH/(R*T^2) > 0 ⇒ K increases with T.3) Lowering T decreases K, shifting equilibrium toward reactants, reducing conversion to products.
Verification / Alternative check:Le Chatelier's principle: remove heat (lower T) and the equilibrium shifts to the exothermic direction (here, the reverse), decreasing product conversion.
Why Other Options Are Wrong:Option b inverts the temperature effect for endothermic reactions; c and d ignore the thermodynamic relation; e is unsupported without specific complex mechanisms.
Common Pitfalls:Confusing endothermic with exothermic behavior; assuming kinetics rather than equilibrium control.
Final Answer:It decreases.