Question

The standard state Gibbs free energies of formation of $C_{(graphite)}\text{ and }{C}_{(diamond)}$ at T=298K are $\begin{array}{rl}& {\mathrm{\Delta }}_f{G}^{\circ }{C}_{(graphite)}=0KJ/mol\\ & {\mathrm{\Delta }}_f{G}^{\circ }{C}_{(diamond)}=2.9KJ/mol\end{array}$
The standard state means that the pressure should be 1 bar and the substance should be pure at the 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: 1 J = $1Kg{m}^2{s}^{-2}$; 1 Pa = $1Kg{m}^1{s}^{-2}$; 1 bar = ${10}^{5}Pa$ ]
[A] 14501 bar
[B] 29001 bar
[C] 1450 bar
[D] 58001 bar

Answer 1

HINT: We can also solve this question by using the relation between pressure and volume and Gibbs free energy at constant temperature. The relation is - $\mathrm{\Delta }G={\displaystyle \frac{P}{\mathrm{\Delta }V}}$ . By simply putting the values of change in Gibbs free energy and change in volume, we can easily find out the pressure. However, we have to be careful with the units and convert them in terms of bar from Pascal.

COMPLETE STEP BY STEP SOLUTION: We know that in thermodynamics, we can write Gibbs free energy at standard state in terms of standard enthalpy, entropy and temperature as-
$$\mathrm{\Delta }{G_r}^{\circ }=\mathrm{\Delta }{H_r}^{\circ }-T\mathrm{\Delta }{S_r}^{\circ }$$
In the question, it is mentioned that the temperature is isothermal.
Therefore, we can write that T = 0. When, T = 0, $$T\mathrm{\Delta }{S_r}^{\circ }$$ will also be zero.
So, we can write Gibbs free energy in terms of enthalpy i.e. $$\mathrm{\Delta }{G_r}^{\circ }=\mathrm{\Delta }{H_r}^{\circ }$$
Now, we know that $\mathrm{\Delta }{H_r}^{\circ }=\mathrm{\Delta }{U_r}^{\circ }+P\mathrm{\Delta }V$
We know that for constant temperature, internal energy is zero.
Therefore, we can write that $\mathrm{\Delta }{H_r}^{\circ }=P\mathrm{\Delta }V$
Now, if we put this in the equation of Gibbs free energy, we can write that-
$$\mathrm{\Delta }{G_r}^{\circ }=P\mathrm{\Delta }V$$
Here, $C_{(graphite)}$ is converted to $C_{(diamond)}$,
Therefore, $$\mathrm{\Delta }{G_r}^{\circ }$$ = ${\mathrm{\Delta }}_f{G}^{\circ }{C}_{(diamond)}-{\mathrm{\Delta }}_f{G}^{\circ }{C}_{(graphite)}$ = 2.9 KJ/mol – 0KJ/mol = 2.9 KJ/mol
Also, the change in volume is given to us as $2\times {10}^{-6}{m}^{3}mo{l}^{-1}$.
Now, if we put these values in the above equation and solve for pressure, we will get that-
$$\begin{array}{rl}& {\displaystyle \frac{\mathrm{\Delta }{G_r}^{\circ }}{\mathrm{\Delta }V}}=P\\ & Or,{\displaystyle \frac{2.9J/mol\times {10}^3}{2\times {10}^{-6}{m}^{3}mo{l}^{-1}}}=P\end{array}$$
Here we have converted kilojoule to joule by multiplying it by 100.
Therefore, P = 1450000000 Pa = 14500 bar.
The initial pressure is 1 bar so the total pressure will be 14500 + 1 i.e. 14501 bar.

So, the correct answer is option [A] 14501 bar

NOTE: In thermodynamics, at constant temperature and pressure the change in free energy of a reaction can be written in terms of change in entropy, temperature and enthalpy. It is given as-
$$\mathrm{\Delta }G=\mathrm{\Delta }H-T\mathrm{\Delta }S$$
Where $\mathrm{\Delta }$G is the free energy,
$\mathrm{\Delta }$S is the entropy and T is the temperature, on which entropy depends,
$\mathrm{\Delta }$H is the enthalpy.
$\mathrm{\Delta }$G can be either positive or negative. But, for a reaction to be spontaneous, the change in free energy should be negative which means there should be release of free energy.