The stability of diamond and graphite

[pentazoid]

Homework Statement

Go through the arithmetic to verify that diamond becomes more stable than graphite at approximately 15 kbar

How can diamond ever be more stable than graphite, when it has less entropy

Homework Equations

The Attempt at a Solution

(dG/dP)=V, T and N are fixed.

dP=15 kbar.

V((diamond)=3.42e-6 m^3
V((graphite))=5.31e-6 m^3

dG(diamond)=15 kbar(3.42e-6)
dG(graphite)=15 kbar(5.31e-6)

Do my calculations above represent the the procedure I would go through to prove that Diamond is more stable than graphite?

the answer to the second question is, . A gas has a much higher entropy than any solid, but at low temperatures solids and liquids are more favorable

gibbs free energy

G= U + PV - TS

Diamond has smaller radius than graphite, therefore it has a smaller radius . U(graphite)>U(diamond) . Atoms inside graphite have the autonomy to move around its lattice, therefore it has more entropy and therefore has higher temperatures.

 

[Jhero]

However, at high pressures, the entropy contribution from the lattice vibrations becomes less significant compared to the enthalpy contribution from the smaller radius of diamond. This results in diamond becoming more stable than graphite at high pressures, despite having lower entropy.

To verify this, we can look at the Gibbs free energy equation:

dG = dH - TdS

At high pressures, dH (enthalpy) is the dominant factor in determining the stability of a substance, as the entropy contribution (TdS) becomes less significant. Since the enthalpy of diamond is lower than that of graphite, diamond will have a lower Gibbs free energy and therefore be more stable.

In conclusion, the calculations you have done are a valid way to show that diamond becomes more stable than graphite at high pressures. However, the underlying reason for this is due to the enthalpy and entropy contributions to the Gibbs free energy equation.