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I understand what the standard molar entropy is, and how to use it in calculations, but I'm interested in understanding exactly how it's defined and measured.

My recollection from a long time ago is that it goes roughly like this:

  • for each element, assume that its entropy goes to zero at 0 kelvin when it's in some standard state

  • then, for the compound of interest, calculate and/or measure the entropy difference between the compound and its constituent elements at 0 K.

However, I'm not even sure that's exactly correct, and there are a lot of details that need to be filled in if it is right. Most of the following issues will be covered by the exact definition, but to be more elaborate, these immediately come to my mind: what exactly is the standard state for each element? Do all elements form a perfect crystal at 0 K, and how can we prove this? Should I think of the third law (zero entropy at zero kelvin) as being a definition that sets an otherwise arbitrary baseline, or can it be proven that the quantum mechanical von Neumann entropy really is zero for pure elements at 0 K?

An internet search turned up plenty of information about the values of standard molar entropy for various compounds and how to use them, but I wasn't able to find a resource with a detailed description of its definition.

What is the precise definition (preferably with reference) of standard molar entropy?

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    $\begingroup$ That's kinda too broad. "Do all elements form a perfect crystal at 0K, and how can we prove this?" kinda doesn't make sense - they don't, as "perfect crystal" is a model. $\endgroup$
    – Mithoron
    Jan 20, 2020 at 22:51
  • $\begingroup$ Well yes, but if standard molar entropy is defined in terms of such a model then, in order to make it, we must know something about how reality diverges from the model. $\endgroup$
    – N. Virgo
    Jan 21, 2020 at 8:51
  • $\begingroup$ Sure en.wikipedia.org/wiki/Residual_entropy $\endgroup$
    – Mithoron
    Jan 21, 2020 at 21:59
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    $\begingroup$ This is a great question. I can conceive of a few ways in standard molar entropies might be determined, the most obvious of which is to integrate from zero to an experimentally-accessible temperature using the the Debye $T^3$ law (which models the constant-volume heat capacity of solids as they approach absolute zero) and then measure the additional heat flow calorimetrically. But that's just one possibility. I did look at the CODATA tables for standard molar entropies, but wasn't able to find details on how they are determined. I'd thus be interested to see an answer on this. $\endgroup$
    – theorist
    Jan 22, 2020 at 2:13
  • $\begingroup$ @Mithoron the question is about how that's taken into account when defining and measuring the standard molar entropy, as found in thermodynamic tables. $\endgroup$
    – N. Virgo
    Jan 22, 2020 at 11:27

1 Answer 1

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Not quite an answer, but too long for a comment:

Here's a link to the official CODATA site showing the values: codata.info (via the Internet Archive). It references the following source: Cox, J. D., Wagman, D. D., and Medvedev, V. A., CODATA Key Values for Thermodynamics, Hemisphere Publishing Corp., New York, 1989. Unfortunately, that source isn't readily viewable online (all I found was what appears to be a German version, at Berichte der Bunsengesellschaft für physikalische Chemie 1990, 94 (1), 93–93 (It's behind a paywall.) But if you could get your hands on it (interlibrary loan, say), it should contain a detailed description of how the values were actually calculated.

FWIW, here's a link showing how standard molar entropies might be calculated (but I don't know if the procedure described here is what is used to determine the official CODATA values): Standard Molar Entropy of Solid Aluminum Oxide, Comments to Tandy Grubbs, Stetson University (via the Internet Archive). Essentially, it mentions using the Debye extrapolation I mentioned in an earlier comment up to $\approx\pu{15 K}$, and then using a differential scanning calorimeter to determine the reversible heat flow between $\pu{15 K}$ and $\pu{298.15 K}$

And here's a purely theoretical approach (unfortunately, everything except the abstract is behind a paywall):
Pascal, T. A.; Lin, S.; Goddard Iii, W. A. Thermodynamics of liquids: standard molar entropies and heat capacities of common solvents from 2PT molecular dynamics. Phys. Chem. Chem. Phys. 2011, 13 (1), 169–181. DOI: 10.1039/C0CP01549K.

Also, some general comments on the issue of the entropy of a substance at absolute zero:

It is possible the question of whether substances need to be perfect crystals to have zero entropy at $\pu{0 K}$ does not enter into the calculation of standard molar entropies. Consider the following:

  1. Any calculation of standard molar entropy requires determining the initial reversible heat flow starting at $\pu{0 K}$, at standard pressure: $$ S^\circ(\pu{298.15 K}) = \int_{0}^{\pu{298.15 K}}\frac{\mathrm{đ}q_\mathrm{rev}}{T} $$ The initial part of this integral (starting at $\pu{0 K}$) can't be done experimentally, and thus has to be based upon a model.

  2. We may not know how to model whether a substance being a perfect crystal at $\pu{0 K}$ will affect the amount of reversible heat flow needed to increase its temperature. In addition, we may not know how to predict the actual state of substances at $\pu{0 K}$ (i.e., the extent to which they are not perfect crystals–and to the extent that they are not, what the distribution of their forms actually is).

  3. Further, even if there were an effect on heat flow from not being a perfect crystal, and we did know how to model it, the effect may be small compared to the experimental errors that are present in determining standard molar entropies.

  4. It's still an open question whether a substance needs to be a perfect crystal to have zero entropy at $\pu{0 K}$. Some in the field argue it does, some argue it doesn't. Those in the latter camp argue that, given that no other states are accessible at $\pu{0 K}$, the fact that the system is locked into only one microstate (even if it's not a perfect crystal) gives it zero entropy.

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    $\begingroup$ Please avoid editing your answer multiple times in short succession. It will cause an automatic flag eventually, which will unnecessarily have to be investigated. (Also, I am trying to improve this post now for about an hour now.) $\endgroup$
    – Martin - マーチン ♦
    Jan 26, 2020 at 20:48
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    $\begingroup$ @Martin-マーチン Understood, I'll leave it alone for now. Though I'd argue the automatic flag is a problem with the system, not with the practice. I.e., we're all volunteers, and if a volunteer is willing to spend extra time to repeatedly refine his/her post to improve it, that's to be encouraged, not discouraged. Yes, one could argue the volunteer should find a work-around that avoids the flag, but that may be inconvenient to the volunteer (when reviewing, I like to see the entire post w/o the editing screen), and we don't want to create inconveniences for those wishing to optimize their ans. $\endgroup$
    – theorist
    Jan 26, 2020 at 20:57
  • $\begingroup$ @Martin-マーチン Incidentally, I was thinking of flagging you in a comment to bring this question to your attention, since I know you have an interest in and knowledge of international standard-setting, and I thus suspected you might know the answer to this. Also, wrt the automatic flagging: The SE sites are undergoing continuous improvement, and this is just my opinion on one way the site might be improved. I know the auto flag is there to prevent gaming the system, but it has its downsides. $\endgroup$
    – theorist
    Jan 26, 2020 at 21:04
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    $\begingroup$ The automatic flag is necessary to safeguard everybody from roll-back wars, vandalism, reputation gaming, etc.. This is all quite important. Draft versions will be saved, there is a live preview of the full text, so posting it doesn't have any benefit. I do not wish to discourage you from improving your answer, I am simply pointing out the things that are going to happen if you continue with this approach. I am also a volunteer here, but unfortunately I don't have that much time with the site anymore, so I prefer having no false positive flags. $\endgroup$
    – Martin - マーチン ♦
    Jan 26, 2020 at 21:13
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    $\begingroup$ Thanks @Nathaniel. I haven't given up on this. I'll either see if I can get my hands on Cox/Wagman/Medvedev's CODATA Key Values for Thermodynamics, or email NIST. $\endgroup$
    – theorist
    Feb 1, 2020 at 19:13

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