Question 6.13: At 298.15 K ΔG°f (C,graphite) =0, and ΔG°f (C,diamond) =2.9...
At 298.15 K \Delta G°_{f}\left( C,graphite \right)=0, and \Delta G°_{f}\left( C,diamond \right)=2.90 KJ mol^{-1} Therefore, graphite is the more stable solid phase at this temperature at P=P°=1 bar . Given that the densities of graphite and diamond are 2.25 and 3.52 kg L, respectively, at what pressure will graphite and diamond be in equilibrium at 298.15 K?
At equilibrium \Delta G =G\left( C, graphite \right)- G\left( C,diamond \right)=0. Using the pressure dependence of G, \left( \partial G _{m}/\partial P \right)_{T} =V_{m}, we establish the condition for equilibrium:
\Delta G=\Delta G°_{f}\left( C,graphite \right)-\Delta G°_{f}\left( C,diamond \right)+\left( \mathrm{V}_{m}^{graphite} -\mathrm{V}_{m}^{diamond}\right)\left( \Delta P \right)=0
0=0 -2.90 × 10^{3} + \left( \mathrm{V}_{m}^{graphite}-{V}_{m}^{diamond} \right)\left( P-1 bar \right)
P =1 bar + \frac{2.90 × 10^{3}}{M_{C}\left( \frac{1}{\rho_{graphite}}-\frac{1}{\rho_{diamond}} \right)}
=1 bar +\frac{2.90 × 10^{3}}{12.00 × 10^{-3} kg mol^{-1}×\left( \frac{1}{2.25 × 10^{3} kg m^{-3} }-\frac{1}{3.52 × 10^{3} kg m^{-3}} \right) }
=10^{5} Pa + 1.51 × 10^{9} Pa =1.51 × 10^{4} bar
Fortunately for all those with diamond rings, although the conversion of diamond to graphite at 1 bar and 298 K is spontaneous, the rate of conversion is vanishingly small.