# Graphite and diamond

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 5.24 – 5.27.

Two of the main forms of elemental carbon are graphite and diamond. We can get some insight into the stability of these two forms by considering the Gibbs free energy. From the thermodynamic potential for the Gibbs free energy

$\displaystyle dG=-S\;dT+V\;dP+\mu\;dN \ \ \ \ \ (1)$

we have

$\displaystyle \left(\frac{\partial G}{\partial P}\right)_{T,N}=V \ \ \ \ \ (2)$

Using data given in Schroeder’s book, the molar volume of graphite is ${V_{g}=5.31\times10^{-6}\mbox{ m}^{3}}$ and of diamond is ${V_{d}=3.42\times10^{-6}\mbox{ m}^{3}}$. If we ignore compression effects over a wide range of pressures, we can integrate 2 to get a linear relationshipe between ${G}$ and ${P}$. If we take the Gibbs energy of graphite to be ${G_{g}=0}$ at ${T=298\mbox{ K}}$ and ${P=10^{5}\mbox{ Pa}}$ then for diamond under the same conditions we have ${G_{d}=2900\mbox{ J}}$, so the relations are (for one mole of carbon atoms):

 $\displaystyle G_{g}$ $\displaystyle =$ $\displaystyle V_{g}P=5.31\times10^{-6}P\ \ \ \ \ (3)$ $\displaystyle G_{d}$ $\displaystyle =$ $\displaystyle V_{d}P=2900+3.42\times10^{-6}P \ \ \ \ \ (4)$

As the diamond line starts above the graphite line but has a smaller slope, it will intersect the graphite line at a pressure ${P_{gd}}$ determined by

 $\displaystyle 5.31\times10^{-6}P_{gd}$ $\displaystyle =$ $\displaystyle 2900+3.42\times10^{-6}P_{gd}\ \ \ \ \ (5)$ $\displaystyle P_{gd}$ $\displaystyle =$ $\displaystyle 1.534\times10^{9}\mbox{ Pa}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 15.34\mbox{ kbar} \ \ \ \ \ (7)$

using the conversion ${10^{5}\mbox{ Pa}=1\mbox{ bar}}$.

If we include the effect of compressibility, things get a bit more complicated. The isothermal compressibility of graphite (given by Schroeder) is about

$\displaystyle \kappa_{T}=-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T,N}=3\times10^{-6}\mbox{ bar}^{-1}=3\times10^{-11}\mbox{ Pa}^{-1} \ \ \ \ \ (8)$

Assuming ${\kappa_{T}}$ is constant over the range of pressures we’re considering, we can integrate this to get

 $\displaystyle \frac{dV}{V}$ $\displaystyle =$ $\displaystyle -\kappa_{T}dP\ \ \ \ \ (9)$ $\displaystyle \ln\frac{V}{V_{0}}$ $\displaystyle =$ $\displaystyle \kappa_{T}\left(P_{0}-P\right) \ \ \ \ \ (10)$

where ${V_{0}}$ and ${P_{0}}$ are some reference volume and pressure, which we can take as the molar volume and a pressure of 1 bar.

Thus the volume is a function of the pressure:

$\displaystyle V=V_{0}e^{\kappa_{T}P_{0}}e^{-\kappa_{T}P} \ \ \ \ \ (11)$

For graphite, this gives

 $\displaystyle V_{g}$ $\displaystyle =$ $\displaystyle \left(5.31\times10^{-6}\right)e^{3\times10^{-6}}e^{-\kappa_{T}P}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 5.31\times10^{-6}e^{-3\times10^{-11}P} \ \ \ \ \ (13)$

since ${e^{3\times10^{-6}}\approx1}$. The Gibbs energy for graphite can then be found from 2 as

 $\displaystyle G_{g}$ $\displaystyle =$ $\displaystyle -\frac{5.31\times10^{-6}}{3\times10^{-11}}e^{-3\times10^{-11}P}+G_{0}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -1.77\times10^{5}e^{-3\times10^{-11}P}+G_{0} \ \ \ \ \ (15)$

The integration constant ${G_{0}}$ can be found from the condition that ${G_{g}\left(10^{5}\mbox{ Pa}\right)=0}$, so ${G_{0}=1.77\times10^{5}\mbox{ J}}$ and we get

$\displaystyle G_{g}=1.77\times10^{5}\left(1-e^{-3\times10^{-11}P}\right) \ \ \ \ \ (16)$

As ${\kappa_{T}}$ for diamond is about ten times smaller than ${\kappa_{T}}$ for graphite, we can approximate ${G_{d}}$ by the same straight line that we used above. The pressure at which diamond becomes more stable than graphite is found by equating ${G_{g}}$ and ${G_{d}}$ as before, although this time we get a transcendental equation (with ${P}$ found both in an exponent and as a linear factor), so we can solve the equation numerically using Maple:

 $\displaystyle 1.77\times10^{5}\left(1-e^{-3\times10^{-11}P_{gd}}\right)$ $\displaystyle =$ $\displaystyle 2900+3.42\times10^{-6}P_{gd}\ \ \ \ \ (17)$ $\displaystyle P_{gd}$ $\displaystyle =$ $\displaystyle 1.647\times10^{9}\mbox{ Pa}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 16.47\mbox{ kbar} \ \ \ \ \ (19)$

The Gibbs energies look like this, with the blue curve being graphite and the red curve being diamond:

The blue curve is actually slightly concanve downwards, as can be seen by plotting it for higher pressures, but in the pressure range here, it’s approximately linear. The red curve is actually a straight line.

There are a couple of rather odd features about the graphite-diamond transition. First, from the Gibbs energy plot, we see that graphite is actually the more stable form at atmospheric pressure and temperature, so that should mean that diamonds will spontaneously degrade into graphite over time. Fortunately for owners of diamonds, this process takes a very long time (although it should also be noted that diamonds, being carbon, do burn, so don’t throw your wedding ring into a fire!).

Another odd fact is that because of its more ordered crystal structure, diamond has a lower entropy than graphite (for 1 mole, the values are ${S_{d}=2.38\mbox{ J K}^{-1}}$ and ${S_{g}=5.74\mbox{ J K}^{-1}}$). Schroeder asks us to explain how the conversion of graphite to diamond at high pressure can result in an overall increase of the environment’s entropy. It’s not clear to me why this should be, though perhaps because ${G_{d} at high pressure, the energy transferred away from the graphite into the environment causes an overall increase in entropy. Comments welcome.

Finally, geochemists use a bizarre unit of volume equal to ${\mbox{kJ kbar}^{-1}}$. The conversion to the more standard cubic metre is given by

$\displaystyle 1\mbox{ kJ kbar}^{-1}=1\mbox{ J bar}^{-1}=10^{-5}\mbox{ J Pa}^{-1}=10^{-5}\mbox{ m}^{3} \ \ \ \ \ (20)$