Aluminum silicate – stability of three crystal structures

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.29.

We can apply a similar analysis to that for graphite and diamond to the more complex case of aluminum silicate {\mbox{Al}_{2}\mbox{SiO}_{3}}, which has three crystalline forms: kyanite, andalusite and sillimanite. Each is stable under some range of temperature and pressure.

Using the data in Schroeder’s book, we can get some insight into the stability of these three forms by considering the Gibbs free energy. The thermodynamic potential for the Gibbs free energy is

\displaystyle dG=-S\;dT+V\;dP+\mu\;dN \ \ \ \ \ (1)

 

First, if {T} and {N} are constant, we have

\displaystyle \left(\frac{\partial G}{\partial P}\right)_{T,N}=V \ \ \ \ \ (2)

 

Ignoring compressibility, we can assume that {V} is constant as the pressure is increased. The molar volumes for the three forms are {V_{k}=44.09\times10^{-6}\mbox{ m}^{3}} for kyanite, {V_{a}=51.53\times10^{-6}\mbox{ m}^{3}} for andalusite and {V_{s}=49.90\times10^{-6}\mbox{ m}^{3}} for sillimanite. Therefore the Gibbs energies are

\displaystyle G_{k} \displaystyle = \displaystyle G_{k0}+44.09\times10^{-6}P\ \ \ \ \ (3)
\displaystyle G_{a} \displaystyle = \displaystyle G_{a0}+51.53\times10^{-6}P\ \ \ \ \ (4)
\displaystyle G_{s} \displaystyle = \displaystyle G_{s0}+49.90\times10^{-6}P \ \ \ \ \ (5)

The constants of integration {G_{k0}}, {G_{a0}} and {G_{s0}} can be found by requiring the energies to be those quoted in Schroeder for {P=1\mbox{ bar}=10^{5}\mbox{ Pa}}. As with our calculations for calcium carbonate, the difference in Gibbs energy between a pressure of zero and 1 bar is negligible so we can just set the constants of integration to the values quoted in Schroeder for {P=1\mbox{ bar}}. We get

\displaystyle G_{k} \displaystyle = \displaystyle -2443.88\times10^{3}+44.09\times10^{-6}P\ \ \ \ \ (6)
\displaystyle G_{a} \displaystyle = \displaystyle -2442.66\times10^{3}+51.53\times10^{-6}P\ \ \ \ \ (7)
\displaystyle G_{s} \displaystyle = \displaystyle -2440.99\times10^{3}+49.90\times10^{-6}P \ \ \ \ \ (8)

At {P=0}, {G_{k}} is the lowest and since the slope of the line in the plot of {G_{k}} versus {P} is also the smallest, {G_{k}} will be less than both {G_{a}} and {G_{s}} at all pressures, so at a temperature of 298 K, kyanite is always the most stable form. The graph looks like this:

The red line is kyanite, green is andalusite and yellow is sillimanite. The latter two lines do intersect at a pressure of around 9 kbar, but kyanite is always the most stable.

At constant pressure, we can use 1 to investigate the stability of the three forms as temperature varies. We have

\displaystyle \left(\frac{\partial G}{\partial T}\right)_{P,N}=-S \ \ \ \ \ (9)

By subtracting the two forms of this equation for two crystal types, we can get a relation for the difference in Gibbs energy between two types:

\displaystyle \left(\frac{\partial\Delta G}{\partial T}\right)_{P,N}=-\Delta S \ \ \ \ \ (10)

 

The entropy of a substance can be written as

\displaystyle dS=\frac{dQ}{T}=C_{P}\frac{dT}{T} \ \ \ \ \ (11)

 

so the total entropy as a function of temperature depends on the heat capacity {C_{P}}. If we take Schroeder’s word for it that {C_{P}} at high temperatures depends only on {N}, then even though the raw value of {S} for a sample of solid increases with {T}, the difference between the entropy values for two substances should remain relatively constant as we increase {T}. We can therefore integrate 10 to get

\displaystyle \Delta G\left(T_{2}\right) \displaystyle = \displaystyle \Delta G\left(T_{1}\right)-\int_{T_{1}}^{T_{2}}\Delta S\left(T\right)\;dT\ \ \ \ \ (12)
\displaystyle \displaystyle = \displaystyle \Delta G\left(T_{1}\right)-\Delta S\left(T_{2}-T_{1}\right) \ \ \ \ \ (13)

The entropy values quoted in Schroeder’s book for 1 mole at {T=298\mbox{ K}} and {P=1\mbox{ bar}} are {S_{k}=83.81\mbox{ J K}^{-1}}, {S_{a}=93.22\mbox{ J K}^{-1}} and {S_{s}=96.11\mbox{ J K}^{-1}} so using the Gibbs energy values from above, we get (with energies in kJ):

\displaystyle \Delta G_{ka} \displaystyle = \displaystyle G_{k}-G_{a}-\left(S_{k}-S_{a}\right)\left(T-298\right)\ \ \ \ \ (14)
\displaystyle \displaystyle = \displaystyle -1.22+9.41\times10^{-3}\left(T-298\right)\ \ \ \ \ (15)
\displaystyle \Delta G_{ks} \displaystyle = \displaystyle -2.89+12.30\times10^{-3}\left(T-298\right)\ \ \ \ \ (16)
\displaystyle \Delta G_{as} \displaystyle = \displaystyle \Delta G_{ks}-\Delta G_{ka}\ \ \ \ \ (17)
\displaystyle \displaystyle = \displaystyle -1.67+2.89\times10^{-3}\left(T-298\right) \ \ \ \ \ (18)

If we plot these three lines, we get

Here, {\Delta G_{ka}} is red, {\Delta G_{ks}} is green and {\Delta G_{as}} is yellow. To use this graph to determine the temperature range for which each structure is the most stable, we observe that the most stable form has the lowest value of {G} at that temperature. For kyanite to be the most stable, for example, we must have {G_{k}<G_{a}} and {G_{k}<G_{s}}. Therefore, {\Delta G_{ka}=G_{k}-G_{a}<0} and {\Delta G_{ks}=G_{k}-G_{s}<0}. Looking at the graph, these conditions are true if both the red and green lines are below the {T} axis, which occurs when {T} is less than the point where the red line crosses the {T} axis, which is where {\Delta G_{ka}=0}. From 15 this occurs at {T=427.6\mbox{ K}}.

For andalusite to be stable, we must have {G_{a}<G_{k}} and {G_{a}<G_{s}}, which means {\Delta G_{ka}>0} and {\Delta G_{as}<0}. Thus the red line must be above the {T} axis and the yellow line must be below the {T} axis. The minimum {T} satisfying these conditions is {T=427.6\mbox{ K}} that we found above. The maximum {T} is where the yellow line crosses the {T} axis, which occurs when {\Delta G_{as}=0}, at {T=875.9\mbox{ K}}.

For sillimanite to be stable, we must have {G_{s}<G_{k}} and {G_{s}<G_{a}}, which means {\Delta G_{ks}>0} and {\Delta G_{as}>0}. Thus both the green and yellow lines must be above the {T} axis, which is true for {T>875.9\mbox{ K}}. To summarize: kyanite is stable for {T<427.6\mbox{ K}}, andalusite is stable for {427.6\mbox{ K}<T<875.9\mbox{ K}} and sillimanite is stable for {T>875.9\mbox{ K}}.

How accurate is the assumption that {\Delta S} (the difference in entropy between two states of the substance) is actually a constant over a wide range of temperature? From 11 for one state, we can integrate:

\displaystyle S\left(T_{2}\right) \displaystyle = \displaystyle S\left(T_{1}\right)+C_{P}\int_{T_{1}}^{T_{2}}\frac{dT}{T}\ \ \ \ \ (19)
\displaystyle \displaystyle = \displaystyle S\left(T_{1}\right)+C_{P}\ln\frac{T_{2}}{T_{1}} \ \ \ \ \ (20)

Thus {\Delta S} between two states, {i} and {j} is obtained by subtracting one equation of this type from the other:

\displaystyle \Delta S_{ij}\left(T_{2}\right)=\Delta S_{ij}\left(T_{1}\right)+\left(C_{Pi}-C_{Pj}\right)\ln\frac{T_{2}}{T_{1}} \ \ \ \ \ (21)

 

{\Delta S_{ij}} will be constant if the two states have equal heat capacities. From the data in Schroeder’s book, {C_{Pk}=121.71\mbox{ J K}^{-1}}, {C_{Pa}=122.72\mbox{ J K}^{-1}} and {C_{Ps}=124.52\mbox{ J K}^{-1}}. Using a reference temperature of {T_{1}=298\mbox{ K}}, the log factor over the range of temperatures over which the state changes occur varies from {\ln\frac{427.6}{298}=0.361} to {\ln\frac{875.9}{298}=1.08}. The largest difference in heat capacity is {C_{Ps}-C_{Pk}=2.81}, so we can expect the last term in 21 to be in the vicinity of {1\mbox{ J K}^{-1}}. The entropy differences used in 15 and following equations are between around 2 and 12 {\mbox{J K}^{-1}} so actual value of {\Delta S} could be in error by a fair bit. Of course, we’ve also used the heat capacities for {T=298\mbox{ K}} and assumed that they don’t vary over the temperature range, so we’d really need to correct that error as well.

2 thoughts on “Aluminum silicate – stability of three crystal structures

  1. Pingback: Aluminum silicate phase diagram | Physics pages

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