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.

  • $\begingroup$ 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. $\endgroup$ – andselisk Mar 24 '19 at 12:02
  • $\begingroup$ But isn't the V in the equation dG=VdP - SdT the absolute volume of the system? Or is it the change in volume? $\endgroup$ – Varun Gupta Mar 24 '19 at 12:09
  • $\begingroup$ 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. $\endgroup$ – andselisk Mar 24 '19 at 12:14
  • $\begingroup$ 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. $\endgroup$ – Varun Gupta Mar 24 '19 at 12:34
  • $\begingroup$ 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. $\endgroup$ – Chet Miller Mar 25 '19 at 12:20