Why doesn’t diamond have $\Delta H_\mathrm{f}^\circ=0$, when graphite does? Is it something to do with the definition – diamonds can’t really form at STP, even though it is naturally occurring?

  • $\begingroup$ Gives new meaning to the advertising "diamonds are forever". (No, not thermodynamically..) $\endgroup$ Jun 15, 2015 at 15:34

2 Answers 2


You are on the right track - diamond is not the thermodynamically stable carbon phase at STP. Taking two figures from A.T. Dinsdale, 'SGTE Data for Pure Elements', CALPHAD 15(4) 317-425 (1991) one sees:

Gibbs energy of phases of C relative to graphite


P-T phase diagram for C

Since graphite is the thermodynamically stable phase of carbon at STP, it is usually selected as the reference phase so it has $\Delta H^0_f = 0$. In the reference above, both absolute Gibbs free energies, as well as free energies with respect to the stable phase at STP, are given for most elements.


Carbon naturally exists as two allotropes, graphite and diamond. By definition, the most stable allotrope at STP (the one with the lowest heat of formation at STP) is assigned a heat of formation of zero. Graphite has the lower heat of formation and is assigned a heat of formation of zero, while diamond being slightly less stable has a heat of formation of 2.4kJ/mol.


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