Question

The methods of this section can also be applied to reactions in which one set of solids converts to another. A geologically important example is the transformation of albite into jadeite + quartz:

${\mathrm{NaAlSi}}_3{\mathrm{O}}_8⟷{\mathrm{NaAlSi}}_2{\mathrm{O}}_6+{\mathrm{SiO}}_2$

Use the data at the back of this book to determine the temperatures and pressures under which a combination of jadeite and quartz is more stable than albite. Sketch the phase diagram of this system. For simplicity, neglect the temperature and pressure dependence of both $∆$S and $∆$V.

Short answer

The temperature is 100 K above room temperature, an extra 1.7 kbar of pressure is required to keep Jadeite+Quartz stable.

Step-by-step solution

Given information

The Gibbs free energy, Entropy and Molar Volumes for Albite, Jadeite, quartz is given as:

$$\begin{gathered} \text{Albite}{G}_a=-3711.5\mathrm{KJ}{S}_a=204.4\mathrm{J}/\mathrm{K}{V}_a=100.07{\mathrm{cm}}^3 \\ \text{Jadeite}{G}_j=-2852·1\mathrm{KJ}{S}_j=133·5\mathrm{J}/\mathrm{K}{V}_j=60·40{\mathrm{cm}}^3 \\ \text{Quartz}{G}_q=-856.64\mathrm{KJ}{S}_q=41·84\mathrm{J}/\mathrm{K}{V}_q=22·69{\mathrm{cm}}^3 \end{gathered}$$

$\text{Albite}\to \text{jadeite}+\text{quartz}$

For the above reaction

role="math" localid="1646935903662" $$\begin{gathered} \Delta G=G_{\text{final}}-G_{\text{initial}} \\ =G_j+G_q-G_a \\ =(-2852·1-856·64+3711·5)\mathrm{kJ} \\ =2·76\mathrm{kJ}>0 \end{gathered}$$

Explanation

As a result, under normal settings, the Albite is more stable.

At high pressures, the jadeite + quartz combination becomes more stable.

The change in volume is calculated as

$$\begin{gathered} \Delta V=V_a-V_j-V_q \\ =(100·07-60·40-22·69){\mathrm{cm}}^3 \\ =16·98{\mathrm{cm}}^3 \\ =1.698\times {10}^{-5}{\mathrm{m}}^3 \\ 1\frac{\mathrm{kJ}}{\mathrm{kbar}}={10}^{-5}{\mathrm{m}}^3 \\ \therefore \Delta V=1·698\frac{\mathrm{kJ}}{\mathrm{kbar}} \end{gathered}$$

Explanation

At standard temperatures, the pressure at which jadeite + quartz becomes stable is:

$$\begin{gathered} P=\frac{\Delta G}{\Delta V} \\ P=\frac{2·76}{1·698}\frac{\mathrm{KJ}}{\frac{\mathrm{KJ}}{\mathrm{Kbar}}} \\ P=1.6254\mathrm{Kbar} \end{gathered}$$

The coexistence line with the P-axis in the P-T diagram is thus of relevance.

The Clausius-Clapeyron equation's slope is

$$\begin{gathered} \frac{dP}{dT}=\frac{\Delta S}{\Delta V} \\ \Delta S=S_a-S_j-S_q \\ =(204·4-133·5-41·84)\mathrm{J}/\mathrm{K} \\ \mathrm{\Delta S}=29·06\mathrm{J}/\mathrm{K} \\ \text{Slope}=\frac{\mathrm{\Delta S}}{\mathrm{\Delta V}} \\ =\frac{29·06\mathrm{J}/\mathrm{K}}{1·698\mathrm{J}/\mathrm{bar}} \\ =17·11\frac{\mathrm{bar}}{\mathrm{K}} \end{gathered}$$

Conclusion

The phase diagram for Jadeite+Quartz and Albite is

Jadeite+Quartz is stable at pressures more than 1.6254kbar at room temperature. The Jadeite& Quartz-Albite phase border has a slope of 17.11 bar/K.

If the temperature is 100 K above room temperature, an extra 1.7 kbar of pressure is required to keep Jadeite+Quartz stable.