Thermodynamics and Thermochemistry - Result Question 97
####22. The standard state Gibbs free energies of formation of $C$ (graphite) and $C$ (diamond) at $T=298 K$ are
$\Delta _f G^{\circ}[C($ graphite $)]=0 kJ mol^{-1}$
$\Delta _f G^{\circ}[C($ diamond $)]=2.9 kJ mol^{-1}$
The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite $[C$ (graphite) $]$ to diamond $[C$ (diamond) $]$ reduces its volume by $2 \times 10^{-6} m^{3} mol^{-1}$. If $C$ (graphite) is converted to $C$ (diamond) isothermally at $T=298 K$, the pressure at which C(graphite) is in equilibrium with $C$ (diamond), is
(2017 Adv.)
Thermodynamics and Thermochemistry 105
[Useful information : $1 J=1 kg m^{2} s^{-2}$,
$\left.1 Pa=1 kg m^{-1} s^{-2} ; 1 bar=10^{5} Pa\right]$
(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, $G _d=G _{g r}$
$$ \begin{gathered} \Rightarrow \quad\left(V _{g r}-V _d\right) \Delta p=G _d{ }^{\circ}-G _{g r}{ }^{\circ}=2.9 \times 10^{3} J \ \Rightarrow \quad \Delta p=\frac{2.9 \times 10^{3}}{2 \times 10^{-6}} Pa=\frac{29}{2} \times 10^{8} Pa=\frac{29000}{2} bar \ p=p _0+\frac{29000}{2}=1+\frac{29000}{2}=14501 bar \end{gathered} $$
