# Conversion of graphite into diamond [duplicate]

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The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at $$T = \pu{298 K}$$ are $$\pu{0 kJ mol-1}$$ and $$\pu{2.9 kJ mol-1}$$, respectively.

The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by $$\pu{2e-6 m3 mol-1}$$.

If C(graphite) is converted to C(diamond) isothermally at $$\pu{T = 298 K}$$, the pressure at which C(graphite) is in equilibrium with C(diamond), is

(A) $$\pu{14501 bar}$$
(B) $$\pu{58001 bar}$$
(C) $$\pu{1450 bar}$$
(D) $$\pu{29001 bar}$$

We know that

$$\mathrm{d}G = V\mathrm{d}P - S\mathrm{d}T$$

At constant temperature, $$\mathrm{d}T = 0$$ and we are left with

$$\mathrm{d}G = V\mathrm{d}P$$

But as the volume is also changing in conversion of graphite to diamond and the pressure has also to be increased, how do I integrate this equation? Please tell me if I'm going wrong.

## marked as duplicate by Community♦Mar 24 at 17:14

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• Why not just use $ΔV$ instead of $V$ since the contraction takes place? You then integrate from 1 bar to the unknown pressure, assuming that $ΔG = 0$ at equilibrium. – andselisk Mar 24 at 12:02
• But isn't the V in the equation dG=VdP - SdT the absolute volume of the system? Or is it the change in volume? – Varun Gupta Mar 24 at 12:09
• That equation is derived for the constant volume (partial derivative with constant volume). In your case volume changes, so this has to be taken into account too. – andselisk Mar 24 at 12:14
• Sorry but I'm not getting it, can you please ellaborate this equation a bit because I don't know that there are partial derivatives in this equation too. – Varun Gupta Mar 24 at 12:34
• You need to apply the equation dG=VdP to the graphite and the diamond separately. So, for example, for the graphite, $G_g-G^0_g=V_gP$. At equilibrium, the free energy of the graphite is equal to the free energy of the diamond. – Chet Miller Mar 25 at 12:20