Answer:
(a). The Gibbs free energy change is 2.895 kJ and its positive.
(b). The Gibbs free energy change is 34.59 J/mole
(c). The pressure is 14924 atm.
(d). The Gibbs free energy of diamond relative to graphite is 4912 J.
Explanation:
Given that,
Temperature = 298 K
Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.
![\Delta H\ for\ diamond = 1.897 kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5C%20for%5C%20diamond%20%3D%201.897%20kJ%2Fmol)
![\Delta H\ for\ graphite = 0 kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5C%20for%5C%20graphite%20%3D%200%20kJ%2Fmol)
![\Delta S\ for\ diamond = 2.38 J/(K mol)](https://tex.z-dn.net/?f=%5CDelta%20S%5C%20for%5C%20diamond%20%3D%202.38%20J%2F%28K%20mol%29)
![\Delta S\ for\ graphite = 5.73 J/(K mol)](https://tex.z-dn.net/?f=%5CDelta%20S%5C%20for%5C%20graphite%20%3D%205.73%20J%2F%28K%20mol%29)
(a) We need to calculate the value of
for diamond
Using formula of Gibbs free energy change
![\Delta G=\Delta H-T\Delta S](https://tex.z-dn.net/?f=%5CDelta%20G%3D%5CDelta%20H-T%5CDelta%20S)
Put the value into the formula
![\Delta G= (1897-0)-298\times(2.38-5.73)](https://tex.z-dn.net/?f=%5CDelta%20G%3D%20%281897-0%29-298%5Ctimes%282.38-5.73%29)
![\Delta G=2895.3](https://tex.z-dn.net/?f=%5CDelta%20G%3D2895.3)
![\Delta G=2.895\ kJ](https://tex.z-dn.net/?f=%5CDelta%20G%3D2.895%5C%20kJ)
The Gibbs free energy change is positive.
(b). When it is compressed isothermally from 1 atm to 1000 atm
We need to calculate the change of Gibbs free energy of diamond
Using formula of gibbs free energy
![\Delta S=V\times\Delta P](https://tex.z-dn.net/?f=%5CDelta%20S%3DV%5Ctimes%5CDelta%20P)
![\Delta S=\dfrac{m}{\rho}\times\Delta P](https://tex.z-dn.net/?f=%5CDelta%20S%3D%5Cdfrac%7Bm%7D%7B%5Crho%7D%5Ctimes%5CDelta%20P)
Put the value into the formula
![\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130](https://tex.z-dn.net/?f=%5CDelta%20S%3D%5Cdfrac%7B12%5Ctimes10%5E%7B-6%7D%7D%7B3.51%7D%5Ctimes999%5Ctimes10130)
![\Delta S=34.59\ J/mole](https://tex.z-dn.net/?f=%5CDelta%20S%3D34.59%5C%20J%2Fmole)
(c). Assuming that graphite and diamond are incompressible
We need to calculate the pressure
Using formula of Gibbs free energy
![\beta= \Delta G_{g}+\Delta G+\Delta G_{d}](https://tex.z-dn.net/?f=%5Cbeta%3D%20%5CDelta%20G_%7Bg%7D%2B%5CDelta%20G%2B%5CDelta%20G_%7Bd%7D)
![\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}](https://tex.z-dn.net/?f=%5Cbeta%3DV%28-%5CDelta%20P_%7Bg%7D%29%2B%5CDelta%20G%2BV%5CDelta%20P_%7Bd%7D)
![\beta=\Delta P(V_{d}-V_{g})+\Delta G](https://tex.z-dn.net/?f=%5Cbeta%3D%5CDelta%20P%28V_%7Bd%7D-V_%7Bg%7D%29%2B%5CDelta%20G)
Put the value into the formula
![0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3](https://tex.z-dn.net/?f=0%3D%5CDelta%20P%28%5Cdfrac%7B12%5Ctimes10%5E%7B-6%7D%7D%7B3.51%7D-%5Cdfrac%7B12%5Ctimes10%5E%7B-6%7D%7D%7B2.25%7D%29%5Ctimes10130%2B2895.3)
![0=-0.0194\Delta P+2895.3](https://tex.z-dn.net/?f=0%3D-0.0194%5CDelta%20P%2B2895.3)
![\Delta P=\dfrac{2895.3}{0.0194}](https://tex.z-dn.net/?f=%5CDelta%20P%3D%5Cdfrac%7B2895.3%7D%7B0.0194%7D)
![\Delta P=14924\ atm](https://tex.z-dn.net/?f=%5CDelta%20P%3D14924%5C%20atm)
(d). Here, ![C_{p}=0](https://tex.z-dn.net/?f=C_%7Bp%7D%3D0)
So, The value of
and
at 900 k will be equal at 298 K
We need to calculate the Gibbs free energy of diamond relative to graphite
Using formula of Gibbs free energy
![\Delta G=\Delta H-T\Delta S](https://tex.z-dn.net/?f=%5CDelta%20G%3D%5CDelta%20H-T%5CDelta%20S)
Put the value into the formula
![\Delta G=(1897-0)-900\times(2.38-5.73)](https://tex.z-dn.net/?f=%5CDelta%20G%3D%281897-0%29-900%5Ctimes%282.38-5.73%29)
![\Delta G=4912\ J](https://tex.z-dn.net/?f=%5CDelta%20G%3D4912%5C%20J)
Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.
(b). The Gibbs free energy change is 34.59 J/mole
(c). The pressure is 14924 atm.
(d). The Gibbs free energy of diamond relative to graphite is 4912 J.