Thermodynamics and Thermochemistry 1 Question 19
22. The standard state Gibbs free energies of formation of $\mathrm{C}$ (graphite) and $\mathrm{C}$ (diamond) at $T=298 \mathrm{~K}$ are
$\Delta_{f} G^{\circ}[\mathrm{C}($ graphite $)]=0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\Delta_{f} G^{\circ}[\mathrm{C}($ diamond $)]=2.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$
The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite $[\mathrm{C}$ (graphite) $]$ to diamond $[\mathrm{C}$ (diamond) $]$ reduces its volume by $2 \times 10^{-6} \mathrm{~m}^{3} \mathrm{~mol}^{-1}$. If $\mathrm{C}$ (graphite) is converted to $\mathrm{C}$ (diamond) isothermally at $T=298 \mathrm{~K}$, the pressure at which C(graphite) is in equilibrium with $\mathrm{C}$ (diamond), is
(2017 Adv.)
Thermodynamics and Thermochemistry 105
[Useful information : $1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}$,
$\left.1 \mathrm{~Pa}=1 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-2} ; 1 \mathrm{bar}=10^{5} \mathrm{~Pa}\right]$
(a) 58001 bar
(b) 1450 bar
(c) 14501 bar
(d) 29001 bar
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Solution:
- $G=H-T S=U+p V-T S$
$\Rightarrow d G=d U+p d V+V d p-T d S-S d T=V d p-S d T$
$\Rightarrow d G=V d p$ if isothermal process $(d T=0)$
$$ [\because d U+p d V=d q=T d S] $$
$\Rightarrow \Delta G=V \Delta p$
Now taking initial state as standard state
Now (ii)-(i) gives,
$$ \begin{aligned} G_{g r}-G_{g r}{ }^{\circ} & =V_{g r} \Delta p \ G_{d}-G_{d}{ }^{\circ} & =V_{d} \Delta p \end{aligned} $$
$$ \left(V_{d}-V_{g r}\right) \Delta p=G_{d}-G_{g r}+\left(G_{g r}^{\circ}-G_{d}^{\circ}\right. $$
At equilibrium, $G_{d}=G_{g r}$
$$ \begin{gathered} \Rightarrow \quad\left(V_{g r}-V_{d}\right) \Delta p=G_{d}{ }^{\circ}-G_{g r}{ }^{\circ}=2.9 \times 10^{3} \mathrm{~J} \ \Rightarrow \quad \Delta p=\frac{2.9 \times 10^{3}}{2 \times 10^{-6}} \mathrm{~Pa}=\frac{29}{2} \times 10^{8} \mathrm{~Pa}=\frac{29000}{2} \mathrm{bar} \ p=p_{0}+\frac{29000}{2}=1+\frac{29000}{2}=14501 \mathrm{bar} \end{gathered} $$
