Standard Enthelpy of formation of Graphene

[James Essig]

I am curious as to what the Standard Enthalpy of formation of Graphene is. I've noticed several highly technical papers on this subject but I got lost in the abstruse math and terminology. I am interested in the Standard Enthapies o both single layer and multi-layer graphene. I will be happy to provide more information about my work per your request.

 

[TeethWhitener]

I don’t think this is really a straightforward question to answer. The energy required to separate a single graphene sheet from bulk graphite is dependent on the surface area, which is not an intensive property.

 

[stefan r]

I am curious as to what the Standard Enthalpy of formation of Graphene is. I've noticed several highly technical papers on this subject but I got lost in the abstruse math and terminology. I am interested in the Standard Enthapies o both single layer and multi-layer graphene. I will be happy to provide more information about my work per your request.

I believe the enthalpy is zero. For example the enthalpy of formation of CO is determined by O₂and Graphene. Multi-layer differs from single layer by the surface tension. The edge of a graphene sheet has slightly higher energy.

 

[DrDu]

I don’t think this is really a straightforward question to answer. The energy required to separate a single graphene sheet from bulk graphite is dependent on the surface area, which is not an intensive property.

... and the number of sheets scales with the length, so that the enthalpy to dissociate a block of graphite into single graphene sheets should be extensive.

 

[TeethWhitener]

... and the number of sheets scales with the length, so that the enthalpy to dissociate a block of graphite into single graphene sheets should be extensive.

This is not true. The number of sheets scales with thickness perpendicular to the easy axis. This is dependent on the geometry of the graphite sample.

You could consider comparing the energy of a unit cell of graphene alone versus a unit cell of graphene in an infinite matrix of graphite.

 

[DrDu]

This is not true. The number of sheets scales with thickness perpendicular to the easy axis. This is dependent on the geometry of the graphite sample.

So if x is the direction of the easy axis, $\Delta x$ the spacing of the sheets in graphite, V the total volume of graphite and $A(x)$ the area of the sheets as a function of the position along the easy axis (which is perpendicular to the sheets), the total area is $A=\sum_i A(x_i) \approx \int dx A(x)/\Delta x=V/\Delta X$. Hence A scales like V and is therefore an extensive quantity.

 

[TeethWhitener]

Ok. This is fine if you’re assuming the limit of many sheets, where the fraction of graphene that sees graphite on both top and bottom is large. For separating a few-layer graphite sheet into individual graphene sheets, this won’t work (because, e.g, the top graphene sheet is more loosely bound than a middle graphene sheet). But maybe that’s not important to the OP.

 

[DrDu]

Ok. This is fine if you’re assuming the limit of many sheets, where the fraction of graphene that sees graphite on both top and bottom is large. For separating a few-layer graphite sheet into individual graphene sheets, this won’t work (because, e.g, the top graphene sheet is more loosely bound than a middle graphene sheet). But maybe that’s not important to the OP.

Yes, but this is not different from the situation of determining the standard enthalpy of formation of any substance. You always take the thermodynamical limit, where surface effects become negligible.

 

[TeethWhitener]

Yes, but this is not different from the situation of determining the standard enthalpy of formation of any substance. You always take the thermodynamical limit, where surface effects become negligible.

Right. This was the point of my post #5. But I thought about this a little more and intensivity actually fails in an even more interesting way that’s relevant at the nanoscale.

Consider a sample of graphite (arbitrary parallelpiped), and peel off one layer of graphene at a time:
$$(Graphene)_n \rightarrow (Graphene)_{n-1} + Graphene$$
$$(Graphene)_{n-1} + Graphene \rightarrow (Graphene)_{n-2} + 2{}Graphene$$
Etc. Now we assume that each reaction has the same energy which is only dependent on area (call it $\Delta E_{peel}(A)$). The total energy for complete exfoliation into $n$ graphene sheets is therefore $(n-1)\Delta E_{peel}(A)$ (the factor of $n-1$ comes from the fact that you can’t peel a single graphene sheet from itself). But the volume of the parallelpiped is $V=Anz$, where $n$ is number of layers and $z$ is lattice spacing between layers. This means that the total energy per unit volume to exfoliate this (or any) parallelpiped of graphene has a factor of $(n-1)/n$.

In the infinite limit, this factor becomes 1 and goes away, but as the graphite gets thinner and thinner, the enthalpy of formation deviates appreciably from intensivity. Not a particularly useful revelation, but it’s a pretty interesting demonstration of how familiar bulk properties are often limit cases which break down at the nanoscale. Could be a nice p chem exam question.

 

[stefan r]

...
Etc. Now we assume that each reaction has the same energy which is only dependent on area (call it $\Delta E_{peel}(A)$).

That would be the free energy. Temperature will effect the energy needed to peel a layer.

 

[TeethWhitener]

That would be the free energy. Temperature will effect the energy needed to peel a layer.

It’s internal energy. The easiest way to calculate free energy is probably to compare the partition function for each of the species in the setup I outlined in post 5.

 

[James Essig]

Thanks for the enthusiastic replies to my question. I have better grasp of the chemistry now that folks have had a chance to respond.