Question

Diamonds are formed from graphite under high pressure in coal mines. Calculate the equilibrium pressure (in atm) at which graphite is converted to diamonds at $25^{\circ} C$ (assumed constant) given densities of $\rho_{\text {graphite }}=2 g / c c \& \rho_{\text {diamond }}=3 g / c c \left(\Delta G_{f}^{o}\right)$ for diamonds is $3 kJm ^{-1}$ from graphite

Answer 1

Differential equation of free energy dG = VdP - SdT at constant temp. with change in allotropic modification there is a change in molar volume $\therefore$ $\mathrm{d}(\Delta \mathrm{G})=(\Delta \mathrm{V}) \mathrm{dP}$.
$\Delta \mathrm{V}=\frac{\mathrm{M}_{\text {Diamond }}}{\rho_{\text {Diamond }}}-\frac{\mathrm{M}_{\text {graphite }}}{\rho_{\text {graphite }}}=\frac{12}{3}-\frac{12}{2} $
$\Delta \mathrm{V}=-2 \mathrm{~cm}^3 / \mathrm{mole}=-2 \times 10^{-6} \mathrm{~m}^3 \mathrm{~mole}^{-1} $
$\Delta \mathrm{G}_{\mathrm{f}}^{\mathrm{o}} \text { at } 298 \mathrm{~K}, \mathrm{p} \text { atm }=0 $
$\int \mathrm{d}(\Delta \mathrm{G})^{=}=-2 \times 10^{-6} \mathrm{~m}^3 \mathrm{~mole}^{-1} \mathrm{P}=\mathrm{patm} $
$\Delta \mathrm{G}_{\mathrm{f}}^0 \text { at } 298 \mathrm{~K}, 1 \mathrm{~atm} $
$\mathrm{P}=1 \mathrm{~atm} $
Because at equilibrium when diamonds are formed both the phases are in equilibrium $\Delta \mathrm{G}_{\mathrm{f}(\text { Diamonds })}^{\mathrm{o}}$ at $298 \mathrm{~K}, \mathrm{P}$ atm $=0$
$\int^0 \mathrm{~d}(\Delta \mathrm{G})=-2 \times 10^{-6} \mathrm{~m}^3 \mathrm{~mole}^{-1} \times 10^5 \mathrm{Nm}^{-2} \int \mathrm{P}=\mathrm{patm} $
$3000 \mathrm{~J} \mathrm{~mole}^{-1} P=1 \mathrm{~atm} $
$\left(1 \mathrm{~atm}=10^5 \mathrm{Nm}^{-2}\right. \text { ) } $
$0-3000 \text { Joules mole }{ }^{-1}=-2 \times 10^{-1} \mathrm{~J} \mathrm{~mole}^{-1}(\mathrm{P}-1) $
$\frac{3000}{0.2}=15000=(\mathrm{P}-1) $
$\therefore P=15001 \text { atm. } $