Question

Question: Silica$\left({\mathrm{SiO}}_2\right)$exists in several forms, including quartz (molar volume 22.69 cm³ mol^-1) and cristobalite (molar volume 25.74 cm³ mol^-1).

(a)Use data from Appendix D to calculate$\mathrm{\Delta H}°,\mathrm{\Delta S}°,\mathrm{and}\mathrm{\Delta G}\mathrm{at}25°\mathrm{C}$.

(b)Which form is thermodynamically stable at 25°C?

(c)Which form is stable at very high temperatures, provided that melting does not occur first?

Short answer

a)$\mathrm{\Delta H}°=-910.94,\mathrm{\Delta S}°=182.02,\mathrm{and}\mathrm{\Delta G}=856.67\mathrm{at}25°\mathrm{C}$

b)Quartz form of silica is thermodynamically stable at 25°C

c)The beta-Quartz is stable at very high temperatures.

Step-by-step solution

(a) Calculation of ΔH°, ΔS°, and ΔG at 25°C

The reaction of silica is as follows,

$\mathrm{Si}+{\mathrm{O}}_2\to {\mathrm{SiO}}_2$

So the change in enthalpy is as follow,

$∆\mathrm{H}°={∆\mathrm{H}°}_{\mathrm{f}\left(\mathrm{Product}\right)}-{∆\mathrm{H}°}_{\mathrm{f}\left(\mathrm{Reactant}\right)}$

According to Appendix D we can find the valus of ${∆\mathrm{H}°}_{\mathrm{f}\left(\mathrm{Product}\right)}$and ${∆\mathrm{H}°}_{\mathrm{f}\left(\mathrm{Reactant}\right)}$

$$\begin{gathered} \mathrm{\Delta H}°=-910.94-\left(0+0\right) \\ =-910.94 \end{gathered}$$

The change in entropy of the reaction is as follow,

$∆S°={∆S°}_{\mathrm{f}\left(\mathrm{Product}\right)}-{∆S°}_{\mathrm{f}\left(\mathrm{Reactant}\right)}$

According to Appendix D we can find the valus of ${∆S°}_{\mathrm{f}\left(\mathrm{Product}\right)}$and ${∆S°}_{\mathrm{f}\left(\mathrm{Reactant}\right)}$

$$\begin{gathered} \mathrm{\Delta S}°=41.84-\left(18.83+205.03\right) \\ =-182.02 \end{gathered}$$

The change in Gibbs free energy of the reaction is as follow,

$∆\mathrm{G}°={∆\mathrm{G}°}_{\mathrm{f}\left(\mathrm{Product}\right)}-{∆\mathrm{G}°}_{\mathrm{f}\left(\mathrm{Reactant}\right)}$

According to Appendix D we can find the valus of ${∆\mathrm{G}°}_{\mathrm{f}\left(\mathrm{Product}\right)}$and${∆\mathrm{G}°}_{\mathrm{f}\left(\mathrm{Reactant}\right)}$

$$\begin{gathered} \mathrm{\Delta G}°=-856.67-\left(0+0\right) \\ =-856.67 \end{gathered}$$

(b)

Quartz form of silica is thermodynamically stable at 25°C because it have small deviations in the bond angle.

(c)

The beta form of Quartz is stable at very high temperatures, provided that melting does not occur first and the alpha form of Quartz is nomenclature for lower temperatures.