The entropy is defined as randomness of a substance or a system. The randomness increases as we go from solids to liquids to gases. Similarly from hard diamond to soft graphite. Since graphite is soft and slippery, the orderness would be less than diamond.

But acc. to the formula for entropy change during phase transition,
$\Delta S=Q\ \text{(at const. pressure)}/T$
$\Delta S=\Delta H/T\ \text{(at const. pressure)}$

But $\Delta H$ of reaction/process C graphite --> C diamond is positive Meaning $\Delta S$ is positive. But it should be negative due to less entropy of diamond.

Where is the problem?

  • $\begingroup$ I'm not sure the numbers are right in this quiz I found, but if you run the numbers on the entropy at different temperatures, the entropy does change sign, i.e. at phase transition temperatures, the entropy of diamond is higher than that of graphite (for some reason). $\endgroup$
    – ralk912
    Mar 22 '18 at 19:20
  • $\begingroup$ Why do you think hardness should be associated with entropy? Graphite actually has an interested crystal structure, its not just amorphous: en.wikipedia.org/wiki/Graphite#Structure $\endgroup$
    – ericksonla
    Mar 22 '18 at 19:30
  • 1
    $\begingroup$ Even if graphite has interesting structure, doesn't it will still be less ordered than diamond. Considering the tight packing of 4 bonded C atoms. I am confused. $\endgroup$
    – user539746
    Mar 22 '18 at 19:48
  • $\begingroup$ Maybe the standard states are not equilibrium states with one another. I would think that the corresponding equilibrium states would be at a much higher pressure than the standard state of 1 atm. $\endgroup$ Mar 22 '18 at 21:49
  • 1
    $\begingroup$ If $\Delta$S had been negative and as $\Delta$H is positive, the reaction wouldn’t occur, as $\Delta$G would be positive. $\endgroup$
    – MollyCooL
    Mar 23 '18 at 11:54

The standard free energy of formation of graphite is $0\ \mathrm{kJ/mol}$, and the standard free energy of formation of diamond is $2.9\ \mathrm{kJ/mol}$. So, at $25\ \mathrm{^\circ C}$ and $1\ \mathrm{bar}$, graphite is favored thermodynamically over diamond. Let's see what pressure would be required for $\Delta G$ to be zero at $25\ \mathrm{^\circ C}$. Since the temperature is constant, the change in $G$ is described by $\mathrm dG=v\,\mathrm dP$, where $v$ is the molar volume. The density of graphite is $2.26\ \mathrm{g/cm^3}$ and the density of diamond is $3.51\ \mathrm{g/cm^3}$. So the molar volumes are $5.31\times 10^{-6}\ \mathrm{m^3/mol}$ for graphite and $3.42\times 10^{-6}\ \mathrm{m^3/mol}$ for diamond. So, at pressure $P$ (in Pa), the Gibbs free energy for graphite and diamond are: $$G_\text{Graphite}=0+5.31\times 10^{-6}P$$ $$G_\text{Diamond}=2900+3.42\times 10^{-6}P$$ So, $\Delta G$ at pressure $P$ is $$\Delta G_\mathrm{G\rightarrow D}=2900-1.89\times 10^{-6}P$$So the change in free energy is zero at $P=1.53\times 10^{9}\ \mathrm{Pa}=15\,300\ \mathrm{bar}$

Diamonds are formed naturally at even higher pressures and higher temperatures. But this is the approximate order of magnitude.

So at $25\ \mathrm{^\circ C}$, graphite is highly thermodynamically favored over diamond. The actual presence of diamonds at $25\ \mathrm{^\circ C}$ must therefore be a matter of kinetics, rather than thermodynamics.

  • 1
    $\begingroup$ 2.9 kJ isn't particularly high value. $\endgroup$
    – Mithoron
    Mar 24 '18 at 0:14
  • 1
    $\begingroup$ @Mithoron true but takes even less to form graphite $\endgroup$
    – hyportnex
    Mar 29 '18 at 16:38

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