Question

The standard state Gibbs free energies of formation of $C_{\left(graphite\right)}$ and $C_{\left(diamond\right)}$ at $T=298K$ are

${\Delta }_f{G}^{\circ }{C}_{\left(graphite\right)}=0kJmo{l}^{-1}$

${\Delta }_f{G}^{\circ }{C}_{\left(diamond\right)}=2.9kJmo{l}^{-1}$

The standard state means that the pressure should be $1bar$, and substance should be pure at a given temperature. The conversion of graphite

$C_{\left(graphite\right)}$ to diamond $C_{\left(diamond\right)}$ reduces its volume by $2\times {10}^{-6}{m}^{3}mo{l}^{-1}$. If $C_{\left(graphite\right)}$ is converted to $C_{\left(diamond\right)}$ isothermally at $T=298K$, the pressure at which $C_{\left(graphite\right)}$ is in equilibrium with $C_{\left(diamond\right)}$, is?

[Useful information: $1J=1kg{m}^2{s}^{-2};1Pa=1kg{m}^{-1}{s}^{-2};1bar={10}^{5}Pa]$

• A. $14501bar$
• B. $29001bar$
• C. $1450bar$
• D. $58001bar$

Answer 1

Answer: (A)

$14501bar$

$C_{graphite}\to C_{diamand}$

$\Delta G_r^o=\Delta G_{diamond}^o-\Delta G_{graphite}^o$

$2.9KJmo{l}^{-1}-0KJmo{l}^{-1}=2.9KJmo{l}^{-1}$

$\Delta G_r^o=\Delta H_r^o-T\Delta S_r^o$

$\Delta S_r^o\approx 0$

$T\Delta S_r^o\approx 0$

$\Delta H_r^o=\Delta E_r^o+\Delta \left(PV\right)$

${\Delta }_r^o=0$ because of isothermal process $\Delta T=0$

$\Delta H_r^o=2.9\times {10}^3=\Delta \left(PV\right)$

At equilibrium pressure is constant

$2.9\times {10}^{3}Jmo{l}^{-1}=P(V_2-V_1)$

$P=\frac{2.9\times {10}^{3}Jmo{l}^{-1}}{2\times {10}^{-6}{m}^{3}mo{l}^{-1}}$

$=1450\times {10}^6$ pascal

$=14500bar$

Total pressure $=$ equilibrium pressure $+$ initial pressure

$=14501bar$

Hence, the correct option is $A$.