Pressure above which graphite will spontaneously change into diamond

[anisotropic]

Homework Statement

Calculate the pressure at which graphite spontaneously changes to diamond.

T = 298.15 K (isothermal)
V_diamond - V_graphite = -2 x 10^-6m³/mol

At 273.15K and 1 bar (10⁵ Pa):
ΔG_{f graphite} = 0
ΔG_{f diamond} = 2.9 kJ/mol
i.e. graphite is the more stable and preferred form of carbon in the above conditions.

Homework Equations

Not really sure here... Could be anything.

Perhaps useful is: G = H - TS

But I don't see where P fits in.

The Attempt at a Solution

Well, I know G will be 0 at the pressure I am trying to identify (i.e. graphite and diamond will be in equilibrium, with equal tendency for the carbon to form either).

 

[Mapes]

G = H - TS is definitely relevant if you expand H by using its definition.

 

[Blueshift5]

So, using the equation G = H - TS, I can rearrange to solve for P:

P = (H - G)/T

At 298.15K, the enthalpy change from graphite to diamond is 2.9 kJ/mol, and the entropy change is 0 (since both forms have the same number of atoms and therefore the same entropy). Plugging these values in, I get:

P = (2.9 kJ/mol - 0)/298.15K

P = 9.73 kJ/mol

Therefore, the pressure at which graphite will spontaneously change into diamond is approximately 9.73 kJ/mol. This calculation assumes ideal conditions and does not take into account any kinetic barriers that may affect the transformation. It is also important to note that this pressure will vary at different temperatures and may not be accurate for all conditions.