Question

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298K are
${\Delta }_f{G}^{\circ }\left[C\right(graphite\left)\right]=0kJmo{l}^{-1}$
${\Delta }_f{G}^{\circ }\left[C\right(diamond\left)\right]=2.9KJmo{l}^{-1}$
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. 58001 bar
• B. 1450 bar
• C. 14501 bar
• D. 29001 bar

Answer 1

The correct option is C 14501 bar

$G=H-TS=U+pV-TS$
$\Rightarrow dG=dU+pdV+Vdp-TdS-SdT=Vdp-SdT$
$[\because dU+pdV=dq=TdS]$
$\Rightarrow dG=Vdp$ if isothermal process (dT = 0)
$\Rightarrow \Delta G=v\Delta p$
Now taking initial state as standard state
$G_{gr}-G_{gr}^{\circ }=V_{gr}\Delta p$ . . . . .(i)
$G_d-G_d^{\circ }=V_d\Delta P$ . . . . . .(ii)
Now (iii) – (i) gives,
$(V_d-V_{gr})\Delta p=G_d-G_{gr}+(G_{gr}^{\circ }-g_d^{\circ })$
At equilibrium, $G_d-G_{gr}$
$\Rightarrow (V_{gr}-V_d)\Delta p=G_d^{\circ }-G_{gr}^{\circ }=2.9\times {10}^{3}J$
$\Rightarrow \Delta p=\frac{2.9\times {10}^3}{2\times {10}^{-6}}Pa=\frac{29}{2}\times {10}^{8}Pa=\frac{29000}{2}bar$
$p=p_0+\frac{29000}{2}=1+\frac{29000}{2}=14501bar$