Pressure at which graphite and diamond are in equilibrium

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

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

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

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

I applied

$$\Delta G_{(p,T)} =\Delta_\mathrm{f}G^\circ + \int_{p_1}^{p_2}V\,\mathrm dp,$$

and since the system is at equilibrium,

$$\Delta_\mathrm{f}G^\circ = -\int_{p_1}^{p_2}V\,\mathrm dp.$$

Now I am stuck. I have not been given any relation between pressure and volume. Is there any assumption I have to make to solve this integral?

• Forget integrals, just use $p\Delta V$. Feb 23, 2019 at 9:05
• I just came to know the equation $$dG = VdP - SdT$$ is only applicable when no reaction is taking place , so my expression is incorrect. Feb 23, 2019 at 11:21
• My comment still applies. Feb 23, 2019 at 11:47
• Allowing for the volune change to be $2.0×10^{-6}$ in the units given, we really have only two significant digits and thus the one bar ambient pressure is not significant. It should not be included in (a), (b) or (d). May 5 at 12:22

For each phase $$i$$ (graphite or diamond) you can show that

$$\mathrm{d}\mu _i = V_i \mathrm{d}p - S_i \mathrm{d}T$$

or after integration

$$\mu_i = \mu_i^\circ+\int_{p^\circ}^{p} V_{i} dp$$ at constant T (note that each phase consists of a pure substance and $$V_{i}$$ refers to the molar volume of phase $$i$$; below the subscript $$m$$ is used instead to refer to the molar volume of a pure phase).

We are asked to find the pressure $$p=p_{eq}$$ at which carbon coexists in the two phases (the Gibbs free energy is equal in both) so that

$$\mu(diamond) = \mu(graphite)$$

$$\Delta \mu^{\circ}=-\int_{p^\circ}^{p_{eq}}\Delta V_mdP$$

and, since $$\Delta \mu^{\circ}= \Delta_f G_m^{\circ}$$, ultimately to the expression you provided:

$$\Delta_f G_m^{\circ}=-\int_{p^\circ}^{p_{eq}}\Delta V_mdP$$

The approximation you are allowed to make at this point is that the solids are incompressible, such that their $$V_m$$ are constant with change of pressure. It follows that

$$\Delta_f G_m^{\circ}=-\Delta V_m\int_{p^\circ}^{p_{eq}}dp=-\Delta V_m(p_{eq}-p^\circ)$$

which can be solved for $$p_{eq}$$:

$$p_{eq}=-\frac{\Delta_f G_m^{\circ}}{\Delta V_m}+p^\circ$$

• The answers to(including yours) : chemistry.stackexchange.com/questions/138563/…. suggest That the general equation for dG is: dG=vdP-sdT +sum(UdN). However your first equation seems to only consider dG=vdP. Why then, did you end up with a correct answer? Aug 10, 2020 at 9:56
• @satan29 In this problem T is held constant. Aug 10, 2020 at 10:42
• but what about the UdN term? Aug 10, 2020 at 11:09
• There is no $\mu dN$ term. Think of it as "phase equilibrium" between graphite and diamond. You keep both phases at constant composition and look for the point where their chemical potential is equal. Aug 10, 2020 at 17:05
• @SarveshMaheshwari For each phase you can show that $\mathrm{d}\mu _i = V_i \mathrm{d}p + S_i \mathrm{d}T$. Jan 22 at 19:22