Gibbs Free Energy And Spontaneity
For a reaction at 500 K, (\Delta H = -40) kJ/mol and (\Delta S = -100) J/mol·K. Which statement correctly describes the spontaneity of this reaction?
Select the correct option:
Solution
Spontaneous only below 400 K
The Gibbs free energy change (\Delta G = \Delta H - T\Delta S) determines spontaneity: if (\Delta G < 0), the reaction is spontaneous. Here (\Delta H = -40{,}000) J/mol and (\Delta S = -100) J/mol·K. For spontaneity: (-40{,}000 - T(-100) < 0 \Rightarrow -40{,}000 + 100T < 0 \Rightarrow T < 400) K. At temperatures below 400 K, the exothermic enthalpy term dominates, making (\Delta G < 0). Above 400 K, the negative entropy penalty ((-T\Delta S)) becomes larger, making (\Delta G > 0). Option 'Spontaneous at all temperatures' is wrong because with (\Delta S < 0), increasing temperature always penalizes (\Delta G). Option 'Non-spontaneous at all temperatures' is incorrect — at low T the reaction is indeed spontaneous. Option 'Spontaneous only above 400 K' contradicts the calculation since high temperatures amplify the unfavorable entropy term. This classic JEE pattern tests understanding of the temperature crossover point. Plausibility check: the crossover temperature (T = \Delta H / \Delta S = 40{,}000/100 = 400) K, which is dimensionally consistent.
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About This Question
- Subject
- chemistry
- Chapter
- chemical thermodynamics
- Topic
- gibbs free energy and spontaneity
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
Spontaneous only below 400 K
The Gibbs free energy change (\Delta G = \Delta H - T\Delta S) determines spontaneity: if (\Delta G < 0), the reaction is spontaneous. Here (\Delta H = -40{,}000) J/mol and (\Delta S = -100) J/mol·K. For spontaneity: (-40{,}000 - T(-100) < 0 \Rightarrow -40{,}000 + 100T < 0 \Rightarrow T < 400) K. At temperatures below 400 K, the exothermic enthalpy term dominates, making (\Delta G < 0). Above 400 K, the negative entropy penalty ((-T\Delta S)) becomes larger, making (\Delta G > 0). Option 'Spontaneous at all temperatures' is wrong because with (\Delta S < 0), increasing temperature always penalizes (\Delta G). Option 'Non-spontaneous at all temperatures' is incorrect — at low T the reaction is indeed spontaneous. Option 'Spontaneous only above 400 K' contradicts the calculation since high temperatures amplify the unfavorable entropy term. This classic JEE pattern tests understanding of the temperature crossover point. Plausibility check: the crossover temperature (T = \Delta H / \Delta S = 40{,}000/100 = 400) K, which is dimensionally consistent.
This medium difficulty chemistry question is from the chapter chemical thermodynamics, covering the topic of gibbs free energy and spontaneity. It appeared in the 2025 exam.
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