Question:
For a particular reversible reaction at temperature T, Δ H and Δ S were found to be both positive. If Tₑ is the temperature at equilibrium, then the reaction would be spontaneous when:
For a particular reversible reaction at temperature T, Δ H and Δ S were found to be both positive. If Tₑ is the temperature at equilibrium, then the reaction would be spontaneous when:
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For Δ H>0 and Δ S>0:
• Low T: non-spontaneous
• High T: spontaneous
Updated On: Mar 19, 2026
- \(T_e > T\)
- \(T > T_e\)
- \(T_e = 5T\)
- T = Tₑ
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The Correct Option is B
Solution and Explanation
Step 1: Gibbs free energy change: Δ G=Δ H-TΔ S
Step 2: For spontaneity: Δ G<0 ⟹ T>(Δ H)/(Δ S)
Step 3: At equilibrium: Tₑ=(Δ H)/(Δ S) ⟹ Reaction is spontaneous when T>Tₑ
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