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 ${\rho }_{\text{graphite }}=2g/cc\&{\rho }_{\text{diamond }}=3g/cc\left(\Delta G_f^o\right)$ for diamonds is $3kJ{m}^{-1}$ from graphite

Answer 1

Answer: 15001

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 $=0$
${\int }^0\mathrm{d}(\Delta \mathrm{G})=-2\times 10^{-6}{\mathrm{m}}^3{\mathrm{mole}}^{-1}\times 10^5{\mathrm{Nm}}^{-2}\int \mathrm{P}=\mathrm{patm}$
$3000\mathrm{J}{\mathrm{mole}}^{-1}P=1\mathrm{atm}$
$(1\mathrm{atm}=10^5{\mathrm{Nm}}^{-2}\text{ ) }$
$0-3000\text{ Joules mole }{}^{-1}=-2\times 10^{-1}\mathrm{J}{\mathrm{mole}}^{-1}(\mathrm{P}-1)$
$\frac{3000}{0.2}=15000=(\mathrm{P}-1)$
$\therefore P=15001\text{ atm. }$