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Is the entropy of vaporization greater than entropy of fusion, for elements? And in which case they can be same, if it so happens?

I have asked this here because I couldn't find any satisfactory answers anywhere on the Internet and most of them were very limited in their description.

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  • $\begingroup$ We define fusion here as transition from solid to liquid. $\endgroup$
    – Shishir Maharana
    Commented Sep 2, 2020 at 5:54
  • $\begingroup$ What is the issue with using latent heat, since thermodynamically dS is defined as dQ/T? $\endgroup$
    – Safdar Faisal
    Commented Sep 2, 2020 at 5:55
  • $\begingroup$ Yes, you are free to use latent heat, but my premise was that, can this dilemma only be solved by using latent heat? Let me edit this question. Now it looks better. $\endgroup$
    – Shishir Maharana
    Commented Sep 2, 2020 at 5:56
  • $\begingroup$ We expect the entropy of a solid to be less than that of its liquid to be less than that of its vapour. The change in values is also generally greater in that order, for example to melt ice $\Delta S = 22$ J/K/mol and to vaporise $109$ J/K/mol. Many liquids obey Troutons's rule of $\approx 85$ J/K/mol for vaporisation. $\endgroup$
    – porphyrin
    Commented Sep 2, 2020 at 9:35
  • $\begingroup$ Of course, there is some temperature near the critical where the heat of vaporization at that temperature is equal to the heat of fusion at the fusion temperature. $\endgroup$
    – Chet Miller
    Commented Sep 2, 2020 at 12:21

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Here I'm assuming you are interested in comparing the magnitude of $\Delta S^{\circ}_{fus}$ at $T_{fus}$ with that of $\Delta S^{\circ}_{vap}$ at $T_{vap}$, for the elements in their standard states (so, for instance, hydrogen would be $\ce{H_2}$ rather than $\ce{H}$).

Like you, I was unable to find a comparative tabulation of these values on the internet. Fortunately, one can readily generate such a table using Wolfram Mathematica's chemical database.

For each element:

$$\Delta S^{\circ}_{fus} \text{ at } T_{fus} = \frac{\Delta H^{\circ}_{fus}}{T_{fus}}$$ $$\Delta S^{\circ}_{vap} \text{ at } T_{vap} = \frac{\Delta H^{\circ}_{vap}}{T_{vas}}$$

Wolfram has the above data for all of elements 1–93 (hydrogen through neptunium), except for helium (which can't be solidified at standard pressure, which is 1 bar), astatine, and francium.*

Here I've plotted $|\Delta S^{\circ}_{vap}|$ vs. $|\Delta S^{\circ}_{fus}|$* for these 90 elements, and added a y = x line. From the placement of the points relative to this line, you can see that all except one of the elements have $|\Delta S^{\circ}_{vap}| > |\Delta S^{\circ}_{fus}|$.

That single exception is hydrogen, for which:

$$|\Delta S^{\circ}_{fus}| \text{ at } T_{fus} = 39.8 \frac{J}{mol K}$$ $$|\Delta S^{\circ}_{vap}| \text{ at } T_{vap} = 22.3 \frac{J}{mol K}$$

enter image description here

*Note, however, this complication: Most, but not all, of these measurements were done at standard pressure (1 bar). For instance:

"When heated at standard atmospheric pressure, arsenic changes directly from a solid to a gas, or sublimates, at a temperature of 887 K. In order to form liquid arsenic, the atmospheric pressure must be increased. At 28 times standard atmospheric pressure, arsenic melts at a temperature of 1090 K. If it were also measured at a pressure of 28 atmospheres, arsenic's boiling point would be higher than its melting point, as you would expect."

https://education.jlab.org/itselemental/ele033.html

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