Question

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at $T=298K$ are
${\Delta }_f{G}^0[C(\text{ graphite })]=0kJmo{l}^{-1}$
${\Delta }_f{G}^0[C(\text{ diamond })]=2.9kJmo{l}^{-1}$
The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by $2\times 10^{-6}{m}^{3}mo{l}^{-1}$. If C(graphite) is converted to $C$ (diamond) isothermally at $T=298K$, the pressure at which C(graphite) is in equilibrium with $C$ (diamond), is [Useful information:
$1J=1kg{m}^2{s}^{-2};1Pa=1kg{m}^{-1}{s}^{-2};1bar=10^{5}Pa]$

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

Answer 1

Answer: (A)

$\Delta G=PdV$
$[{\Delta }_f{G}^{\circ }$ (diamond) $-{\Delta }_f{G}^{\circ }$ (graphite) $]=PdV$
$2.9\times 10^{3}Jmo{l}^{-1}=P\times 2\times 10^{-6}{m}^{3}mo{l}^{-1}$
$P=1.45\times 10^{9}Pa$
$P=1.45\times 10^9\times 10^{-5}bar$
$P=1.45\times 10^{4}bar$
$P=14500bar$