The third law of thermodynamics states that the entropy of a perfect crystal at 0 $\mathrm{K}$ is zero. In Appendix $4, \mathrm{F}^{-}(a q), \mathrm{OH}^{-}(a q)$ and $\mathrm{S}^{2-}(a q)$ all have negative standard entropy values. How can $S^{\circ}$ values be less than zero?

Name: The third law of thermodynamics states that the entropy of a perfect crystal at 0 K is zero. In Appendix 4, F^-(a q), OH^-(a q) and S^2-(a q) all have negative standard entropy values. How can S^∘ values be less than zero?
Uploaded: 2022-09-09T08:00:44-08:00
Duration: 2 min 36 s
Channel: Adi S
Description: The third law of thermodynamics states that the entropy of a perfect crystal at 0 K is zero. In Appendix 4, F^-(a q), OH^-(a q) and S^2-(a q) all have negative standard entropy values. How can S^∘ values be less than zero?

          The third law of thermodynamics states that the entropy of a perfect crystal at 0 $\mathrm{K}$ is zero. In Appendix $4, \mathrm{F}^{-}(a q), \mathrm{OH}^{-}(a q)$ and $\mathrm{S}^{2-}(a q)$ all have negative standard entropy values. How can $S^{\circ}$ values be less than zero?

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Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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09/09/2022

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The third law of thermodynamics states that the entropy of a perfect crystal at 0 $\mathrm{K}$ is zero. In Appendix $4, \mathrm{F}^{-}(a q), \mathrm{OH}^{-}(a q)$ and $\mathrm{S}^{2-}(a q)$ all have negative standard entropy values. How can $S^{\circ}$ values be less than zero?