# Since the formation of diamond from Graphite is endothermic and the entropy decreases for this process, how is it possible to make diamond?

Surely G would always be positive and thus would make it impossible to make diamond. Clearly this is not the case. But how? Would an increase or decrease in temperature aid it's formation? I think a decrease in temperature would favor diamond but wouldn't this contradict le chatilier's principle?

Diamond is significantly denser than graphite.

atomic densities:

• $1.14 \times 10^{23}$ cm$^{-3}$ for graphite
• $1.77 \times 10^{23}$ cm$^{-3}$ for diamond

This suggests that higher pressure would favor the formation of diamond. Indeed, as the following phase diagram shows, diamond is the most stable allotrope of solid carbon at high pressures. Temperature has little effect, as long as the pressure is high, diamond is favored over graphite at all temperatures up to the point of liquefication. If you have a look at this phase diagram, you can see the main condition at which diamond is the favourable structure: high pressure. The Gibbs free energy is dependant on pressure, in addition to temperature: $$dG = -SdT + Vdp$$

With a little math and assuming the temperature is constant, the gibbs energy moving from one pressure to another: $$G(p_2)=G(p_1)+\int_{p_1}^{p_2}Vdp$$

In practice, one method of making synthetic diamonds relies on high pressure and high temperature: carbon dissolved in a molten metal precipitates out onto small seed diamonds.

It's not really necessary to calculate anything to rationalize why diamond would be the favoured allotrope at high pressure—diamond has a much higher density than graphite. At high pressure, the side of the equation with less volume will be favoured by Le Chatelier's principle.

• if you take the definition G=H-TS how is it possible for the reaction to happen because surely G is always positive? – RobChem Nov 23 '14 at 19:36
• Enthalpy is also dependant on pressure. ∆H of formation at standard state (which is what you'll find in a table) is not the same as at very high pressure. – Michael DM Dryden Nov 23 '14 at 19:49

Thermodynamics gives two possible solutions: you make $\Delta G$ positive through appropiate choice of reaction conditions (i.e. high pressure), or you couple the diamond-formation reaction with another reaction, so that the total $\Delta G$ becomes positive. There are processes for gas-phase deposition of diamond films on surfaces, and those don't depend on high pressure.

You're forgetting the key - enthalpy is defined at constant pressure. You need to be thinking about Helmholtz, not Gibbs. If we say that the reaction from graphite to diamond is caused by pressure, chances are the pressure will not be constant throughout.

• Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. At the moment this reads more like a comment than an actual answer - could you elaborate a little more. With a bit more rep, you will be able to post comments on any question/answer. – Martin - マーチン Jan 22 '16 at 3:49