1. Δ H 0 and Δ S 0 4. Δ H < 0 and Δ S < 0

Answer: 3. Δ H 0 Key idea: Δ G=Δ H-TΔ S . A reaction is spontaneous at all temperatures when Δ G 0 (so -TΔ S is also negative). That is option 3. In option 4 the -TΔ S term turns positive at high T; option 2 is never spontaneous.

Topic: Second law, entropy and spontaneity· Chemical Thermodynamics

The correct thermodynamic conditions for the spontaneous reaction at all temperatures is :

1. \(\Delta H < 0\) and \(\Delta S = 0\)

2. \(\Delta H > 0\) and \(\Delta S < 0\)

3. \(\Delta H < 0\) and \(\Delta S > 0\)

4. \(\Delta H < 0\) and \(\Delta S < 0\)

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Answer: 3. \(\Delta H < 0\) and \(\Delta S > 0\)

Key idea: \(\Delta G=\Delta H-T\Delta S\). A reaction is spontaneous at all temperatures when \(\Delta G<0\) for every T.

This needs \(\Delta H<0\) (negative) and \(\Delta S>0\) (so \(-T\Delta S\) is also negative). That is option 3. In option 4 the \(-T\Delta S\) term turns positive at high T; option 2 is never spontaneous.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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The correct thermodynamic conditions for the spontaneous reaction at all temperatures is :

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