At $298.15 \mathrm{~K}, \Delta G_f^{\circ}(\mathrm{C}$, graphite $)=0$, and $\Delta G_f^{\circ}(\mathrm{C}$, diamond $)=2.90 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Therefore, graphite is the more stable solid phase at this temperature at $P=P^{\circ}=1$ bar. Given that the densities of graphite and diamond are 2.25 and $3.52 \mathrm{~kg} / \mathrm{L}$, respectively, at what pressure will graphite and diamond be in equilibrium at $298.15 \mathrm{~K}$ ?
At equilibrium $\Delta G=G(\mathrm{C}$, graphite $)-G(\mathrm{C}$, diamond $)=0$. Using the pressure dependence of $G,\left(\partial G_m / \partial P\right)_T=V_m$, we establish the condition for equilibrium:
$$
\begin{gathered}
\Delta G=\Delta G_f^{\circ}(\mathrm{C}, \text { graphite })-\Delta G_f^{\circ}(\mathrm{C}, \text { diamond }) \\
+\left(V_m^{\text {graphite }}-V_m^{\text {diamond }}\right)(\Delta P)=0 \\
0=0-2.90 \times 10^3+\left(V_m^{\text {graphite }}-V_m^{\text {diamond }}\right)(P-1 \mathrm{bar}) \\
P=1 \mathrm{bar}+\frac{2.90 \times 10^3}{M_C\left(\frac{1}{\rho_{\text {graphite }}}-\frac{1}{\rho_{\text {diamond }}}\right)} \\
=1 \mathrm{bar}+\frac{2.90 \times 10^3}{12.00 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}^{-1} \times\left(\frac{1}{2.25 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}}-\frac{1}{3.52 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}}\right)}\\
=10^5 \mathrm{~Pa}+1.51 \times 10^9 \mathrm{~Pa}=1.51 \times 10^4 \mathrm{bar}
\end{gathered}
$$
Fortunately for all those with diamond rings, although the conversion of diamond to graphite at $1 \mathrm{bar}$ and $298 \mathrm{~K}$ is spontaneous, the rate of conversion is vanishingly small.
