# Why is the standard enthalpy of formation of diamond not zero?

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?

• Gives new meaning to the advertising "diamonds are forever". (No, not thermodynamically..) – Geoff Hutchison Jun 15 '15 at 15:34

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.