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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?

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  • 1
    $\begingroup$ Forget integrals, just use $p\Delta V$. $\endgroup$ Feb 23, 2019 at 9:05
  • $\begingroup$ I just came to know the equation $$ dG = VdP - SdT$$ is only applicable when no reaction is taking place , so my expression is incorrect. $\endgroup$
    – Starboy
    Feb 23, 2019 at 11:21
  • $\begingroup$ My comment still applies. $\endgroup$ Feb 23, 2019 at 11:47
  • $\begingroup$ 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). $\endgroup$ May 5 at 12:22

1 Answer 1

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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

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

This leads to

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

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 $p_{eq}$:

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

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  • $\begingroup$ 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? $\endgroup$
    – satan 29
    Aug 10, 2020 at 9:56
  • $\begingroup$ @satan29 In this problem T is held constant. $\endgroup$
    – Buck Thorn
    Aug 10, 2020 at 10:42
  • $\begingroup$ but what about the UdN term? $\endgroup$
    – satan 29
    Aug 10, 2020 at 11:09
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    $\begingroup$ 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. $\endgroup$
    – Buck Thorn
    Aug 10, 2020 at 17:05
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    $\begingroup$ @SarveshMaheshwari For each phase you can show that $\mathrm{d}\mu _i = V_i \mathrm{d}p + S_i \mathrm{d}T$. $\endgroup$
    – Buck Thorn
    Jan 22 at 19:22

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