Thermodynamics and Thermochemistry - Result Question 22
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.)
[Useful information : $1 J=1$ $ kg$ $ m^{2}$ $ s^{-2}$, $1 Pa=1$ $ kg$ $ m^{-1} $ $s^{-2} ; 1 bar=10^{5} Pa]$
(a) $58001$ bar
(b) $1450$ bar
(c) $14501$ bar
(d) $29001$ bar
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Answer:
Correct Answer: 22. (c)
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 $
$ [\because d U+p d V=d q=T d S]$
$ \Rightarrow d G=V d p \text { if isothermal process }( d T=0) $
$ \Rightarrow \Delta G=V \Delta p$
Now taking initial state as standard state
$G_{g r}-G_{g r}{ }^{\circ} =V_{g r} \Delta p \quad$ ….(i)
$G_d-G_d{ }^{\circ} =V_d \Delta p \quad$ ….(ii)
Now (ii)-(i) gives,
$\left(V_d-V_{g r}\right) \Delta p=G_d-G_{g r}+\left(G_{g r}^{\circ}-G_d^{\circ}\right)$
At equilibrium, $G_d=G_{g r}$
$\begin{gathered} \Rightarrow \quad\left(V_{g r}-V_d\right) \Delta p=G_d{ }^{\circ}-G_{g r}{ }^{\circ}=2.9 \times 10^3 \mathrm{~J} \\ \Rightarrow \quad \Delta p=\frac{2.9 \times 10^3}{2 \times 10^{-6}} \mathrm{~Pa}=\frac{29}{2} \times 10^8 \mathrm{~Pa}=\frac{29000}{2} \text { bar } \\ p=p_0+\frac{29000}{2}=1+\frac{29000}{2}=14501 \mathrm{bar} \end{gathered}$
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