Entropy of diamond and graphite at 0K

[wangasu]

Hi, all. Does anybody know the 0K entropy of diamond and graphite? According to the third law of thermodynamics, the entropy of one of carbon polymorphs which should be called perfect crystal goes to 0. How about diamond and graphite? Thanks.

 

[Curl]

I think they're both zero since you have complete information of the state of the structure at 0K

 

[RedX]

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]

I think they're both zero since you have complete information of the state of the structure at 0K

The temperature-dependent entropy will go to zero for both of them. But if one is defined as 'perfect' , others will have the configurational entropy at 0K, which will not be zero, so the total entropy will not be zero any more. Does this make sense?

 

[wangasu]

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.

4.56 J/mol-K?

 

[RedX]

4.56 J/mol-K?

Yeah.

I'm used to setting Boltzmann's constant equal to 1, so I forgot to add it.

But if you add it, you get 4.56 J/mol-K.

I'm not a chemist, so I have no clue if that's right or not.

You might get a better answer in the condensed matter physics section, rather than the classical physics section.

But I gave my reasoning.

Basically in graphite each carbon is connected to 3 other carbons, so one of the connections must be a double bond. So my guess is that the entropy comes from which of the bonds to choose to be a double bond.

Diamond seems to be perfect in that each carbon is connected to 4 other carbons, all single bonds, so there is no choice.

 

[alxm]

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.

 

[wangasu]

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.

How could the interplanar (van der waals) and intraplanar (covalent) bonds be equivalent?

 

[Cthugha]

Except for the small effect of crystal defects, the entropy of any crystalline substance is zero at 0 K.

Is ice considered as non-crystalline in this respect? I always thought ice was the prime example for residual entropy.

 

[alxm]

How could the interplanar (van der waals) and intraplanar (covalent) bonds be equivalent?

Van der Waals interactions aren't normally considered to be 'bonds'. Point was that there are no single or double (covalent) bonds, just as there are no dragons and unicorns.
The actual bond in benzene or graphene rings http://wps.prenhall.com/wps/media/objects/724/741576/Instructor_Resources/Chapter_07/Text_Images/FG07_01-03UN.JPG" , if you will.

Is ice considered as non-crystalline in this respect? I always thought ice was the prime example for residual entropy.

Well, I should've said "perfect crystal" (or you could consider ice to have lots of 'defects'). Amorphous and otherwise disordered crystals (such as water) do have entropy at 0 K.

 

[DrDu]

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.

The various valence bond structures are in resonance (that is they form a qm superposition) with each other so that a unique ground state results. In chemistry, the most prominent example for resonance is the benzene molecule. In a molecular orbital picture, the interaction of the p_z orbitals leads to the formation of a band which is partly filled.

 

[daumphys]

At 0K, I do not know the entropy difference between big single graphite and divided many pieces of graphite. Does anyone have precise information?

 

[RedX]

Okay, since everything is in resonance in graphite, then I guess it has no entropy at 0K.

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.

 

[wangasu]

Does that mean that any crystal has zero entropy at 0K, independent of simple or complex crystals? In case residual entropy is not zero, can we call it configurational entropy different from vibrational one?

 

[alxm]

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.

I don't really know how you arrived at that; it's not what I said.

Let's try this again. Boltzmann: S = k*log W, where W is the number of energetically equivalent 'microstates' that can make up the macro-state. If a system only has one possible state, it has zero entropy; log 1 = 0.

If everything was classical, then a system at 0 K would be entirely stationary, so you have always have zero entropy, because even if other microstates with the same energy exist, they're not "available", you can't get from the current state to the other ones, since that'd involve moving. It's frozen. Quantum-mechanically, things remain in motion even at absolute zero. And they can move from one point to another without changing their energy. 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]

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.

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.

If a crystal is finite in size, then there's no way it can be infinite in size, so shouldn't the entropy of a finite size crystal still be zero at 0 K since an infinite size is not accessible? Technically I guess the finite sized crystal could rearrange itself so that's it's longer in a given direction, but at 0 K that would involve motion.

 

[DrDu]

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.

I think this entropy value is also an approximation which goes back to Pauling.
However, I would guess that at sufficiently low temperatures even the entropy of ice goes to zero due to quantum mechanical tunneling effects between different hydrogen bonded structures.