Van't Hoff dependence: How does the equilibrium constant K change with temperature for an exothermic reaction?

Difficulty: Easy

Correct Answer: K decreases as temperature increases (exothermic reaction)

Explanation:

Introduction / Context:Temperature affects chemical equilibria according to the van't Hoff equation. Predicting the direction of K with temperature informs reactor operation, conversion targets, and separation strategies.

Given Data / Assumptions:

  • Exothermic reaction: ΔH < 0.
  • Ideal solution/gas assumptions suffice for qualitative trends.
  • No phase change complications.

Concept / Approach:The van't Hoff relation is d(ln K)/dT = ΔH / (R * T^2). For exothermic reactions (ΔH negative), the derivative is negative, so raising temperature lowers K. Le Chatelier's principle concurs: adding heat shifts equilibrium toward reactants in an exothermic system.

Step-by-Step Solution:Note ΔH < 0 → d(ln K)/dT < 0.Temperature increases → ln K decreases → K decreases.Hence, choose the statement that K decreases with increasing T.

Verification / Alternative check:Ammonia synthesis and methanol synthesis (both exothermic) show lower equilibrium conversions at higher temperatures, matching the predicted K trend.

Why Other Options Are Wrong:(b) Reverses the dependence; (c) describes endothermic reactions; (d) is unnecessary.

Common Pitfalls:

  • Confusing kinetic rate increases with equilibrium shifts.
  • Ignoring pressure effects that can partially compensate in practice.

Final Answer:K decreases as temperature increases (exothermic reaction)

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