For a reversible reaction at 298 K, Delta H° = -50 kJ/mol and Delta S° = -100 J/(mol*K). At what temperature does the reaction transition from spontaneous to non-spontaneous?

Answer and explanation

Correct answer: D. 500 K

  1. Step 1: Recall Gibbs free energy equation: Delta G° = Delta H° - T * Delta S°.
  2. Step 2: At equilibrium transition between spontaneity and non-spontaneity, Delta G° = 0.
  3. Step 3: Set Delta H° - T * Delta S° = 0 => T = Delta H° / Delta S°.
  4. Step 4: Convert units: Delta H° = -50,000 J/mol; Delta S° = -100 J/(mol*K).
  5. Step 5: Compute T: T = -50,000 / -100 = 500 K.
Answer
500 K.
Why not the others: '250 K' divides by -200; '300 K' evaluates at ambient temperature; '298 K' assumes standard temperature.

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