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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?

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    $\begingroup$ Gives new meaning to the advertising "diamonds are forever". (No, not thermodynamically..) $\endgroup$ Commented Jun 15, 2015 at 15:34
  • $\begingroup$ Because it was not defined as zero and graphite is of slightly lower enthalpy and is defined as zero. The difference seems to be calculated from heats of combustion. atct.anl.gov/Thermochemical%20Data/version%201.118/species/… I wonder if this has been remeasured using pure synthetic diamond and graphene if such exist. $\endgroup$
    – jimchmst
    Commented Jan 15 at 21:24
  • $\begingroup$ If you set both to zero, you create a contradiction. The only element that is set to zero in two contexts is hydrogen, with dihydrogen and the aqueous hydrogen ion. $\endgroup$
    – Karsten
    Commented Jan 16 at 0:38

2 Answers 2

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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

and

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.

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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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