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It is known that graphite (5.7 J/K mol) has a higher standard molar entropy than diamond (2.4 J/K mol) (data from Physical Chemistry by Atkins). I understand that this must be due to the structural differences between the two allotropes, however I have not been able to find any textbooks or scholarly sources addressing them. A few qualitative possibilities are as follows:

  1. Diamond is a well-ordered 3D lattice, whereas graphite is a series of graphene sheets stacked on top of each other held together by van der Waals forces. Thus, each individual graphene sheet can slide, creating an additional degree of freedom (or, more possibilities) that diamond does not possess.

  2. Graphite's structure includes delocalized electrons, while diamond's bonds just have localized electrons. Thus, the increased possibilities of locations for the delocalized electrons cause graphite to have a higher entropy than diamond.

  3. The bonds in diamond are weaker than the bonds in graphite (in terms of bond order), so diamond's vibrational force constants are less than graphite's vibrational ones. Therefore, diamond's vibrational energy levels that are occupied are few and spaced apart, so there are fewer ways to distribute the atoms compared to graphite.

I would like clarity on which of these contributes to thermodynamic stability the most, ideally quantitatively.

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  • $\begingroup$ Your 1 is the main factor. $\endgroup$ Dec 6, 2023 at 22:54
  • $\begingroup$ Hmmm, #3 sounds like it's the wrong way around. Weaker bond implies smaller force constant implies smaller frequency and spacing and greater entropy. $\endgroup$
    – Buck Thorn
    Dec 7, 2023 at 4:06
  • $\begingroup$ Don’t forget the electron degrees of freedom (or bond degrees if you wish). The entropy is defined for high quality crystals, so #1 is out. $\endgroup$
    – Jon Custer
    Dec 7, 2023 at 14:12

1 Answer 1

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The measured heat capacity $C_V$ of graphite at a low temperature, say 100 K is greater than that of diamond. A characteristic measure of the heat capacity is the Debye temperature defined as $\Theta=h\nu/k_B$ where $\nu$ is the maximum phonon frequency and $k_B$ the Boltzmann constant. The value for diamond is $\sim 1900$ K and for Graphite $\sim 410$ K. This means that on average diamond has higher frequency vibrational quanta than graphite and this means that the entropy is greater for graphite than diamond because more energy levels are populated in graphite. As the entropy is $\int_0^T C_V/T dT$ the area under the curve of heat capacity divided by temperature is greater for graphite than diamond and thus so is the entropy.

The graphs below show the calculated heat capacity and $C_V/T$ for diamond and graphite. You can see that the area under the curve, say, to $300$ K is greater for graphite than diamond and so the entropy is greater.

debye CV diamond

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  • $\begingroup$ It's not just greater, it's a lot greater. I'm a little surprised. You'd think the stronger in-plane bonds of graphite would constrain degrees of freedom in that direction exceptionally well. $\endgroup$
    – Buck Thorn
    Dec 8, 2023 at 7:59
  • $\begingroup$ @Buck Thorn The picture is a bit misleading, when integrated up to $5000$ K the graphite is $\approx 1.7$ times that of diamond, not so different to the ratio in the question. I got $3.4$ J/mol/K for diamond and $5.7$ for the graphite. I guess the vibration frequency in graphite is smaller so more levels populated. $\endgroup$
    – porphyrin
    Dec 8, 2023 at 9:44

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