The entropy of diamond and graphite at absolute zero (0K) is a topic of discussion rooted in the third law of thermodynamics. It is established that both diamond and graphite, as perfect crystals, have zero entropy at 0K. However, graphite's entropy can be expressed as S = (N/2) ln(3), where N represents the number of carbon atoms, due to its hexagonal bonding structure. The discussion highlights that any crystalline substance, including diamond and graphite, theoretically has zero entropy at absolute zero, barring any defects or impurities.
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Curl said:I think they're both zero since you have complete information of the state of the structure at 0K
RedX said:I'm not a chemist, but my guess is that diamond has zero entropy, and graphite has:
S=\frac{N}{2} \ln(3)
where N is the number of graphite atoms.
I'm basing this on the fact that diamond is tetrahedral, so all 4 electrons are bonding, whereas with graphite only 3 are bonding in a hexagonal honeycomb structure, and the 4th electron has a choice of which 3 neighbors to make a double bond with (hence the ln(3)). But once you choose which neighbor, that neighbor no longer has a choice. So hence the N/2.
wangasu said:4.56 J/mol-K?
alxm said:Except for the small effect of crystal defects, the entropy of any crystalline substance is zero at 0 K. That includes graphite and diamond.
Graphite doesn't have a degenerate electronic ground state. All bonds are equivalent.
alxm said:Except for the small effect of crystal defects, the entropy of any crystalline substance is zero at 0 K.
wangasu said:How could the interplanar (van der waals) and intraplanar (covalent) bonds be equivalent?
Cthugha said:Is ice considered as non-crystalline in this respect? I always thought ice was the prime example for residual entropy.
RedX said:I'm not a chemist, but my guess is that diamond has zero entropy, and graphite has:
S=\frac{N}{2} \ln(3)
where N is the number of graphite atoms.
I'm basing this on the fact that diamond is tetrahedral, so all 4 electrons are bonding, whereas with graphite only 3 are bonding in a hexagonal honeycomb structure, and the 4th electron has a choice of which 3 neighbors to make a double bond with (hence the ln(3)). But once you choose which neighbor, that neighbor no longer has a choice. So hence the N/2.
RedX said:So any crystal would have zero entropy at 0K. If a crystal has defects, it can't even exist at 0K, because only the ground state is occupied at 0K, and the ground state is without imperfections.
alxm said:So if you imagine a simplified water 'crystal' as a ring of four or five hydrogen-bonded water molecules, they have (at least) two equivalent states since they can swap protons back and forth along the hydrogen bonds. So ice would have entropy at 0 K.
Everything has entropy at 0 K. The only thing that has exactly zero entropy would be a perfect crystal, which is a theoretical construct of an infinitely large crystal that's wholly symmetric in such a way that there's only one conformation. Anything that's finitely-sized or has any kind of irregularities or impurities that'd break the symmetry will have entropy, since that creates equivalent states. But a perfect crystal of graphite or diamond is a prefect crystal by most ways of looking at it, so neither would have entropy at absolute zero.
At any real, non-zero temperature, graphite has a higher entropy than diamond.
RedX said:According to Wikipedia, the residual entropy of ice is Nk*log[3/2], which is less than two equivalent states which would produce Nk*Log[2]. I don't think swapping the hydrogens in a single molecule of H20 counts as a different arrangement. I think you actually have to destroy a hydrogen bond and reform it with a different neighbor, and in doing so that neighbor has to get a bond destroyed and reform it with another, etc.