Figure 2.06. Spherical Nucleus Gibbs Energy vs. Radius. Click on the image to see a larger version with sliders to change the calculation parameters along with a more detailed explanation.
As you increase the |ΔT| with its slider in Figure 2.06, you will see the ΔGV increase and the calculated critical radius decrease. Although the interfacial Gibbs energy does change with temperature, typically decreasing with increasing T and increasing with decreasing T, these changes are poorly known and believed to be small so they are not included in this model.
As an example, consider the kyanite = sillimanite reaction at a pressure of 0.6 GPa. You can see Gibbs energy data for this reaction here. The value of dΔGrxn/dΔT for this reaction is about -12 J/K/mole of Al2SiO5. The molar volume for sillimanite is about 50 cm3/mole of Al2SiO5. Using these values and an interfacial Gibbs energy of 0.015 J/m2, what is the ΔT required to have a critical nucleus with a radius of 1 nm? Use Figure 2.06 to answer the question, type your answer in the box.
Press "Enter" after you type in the number.
Your answer is incorrect. Open Figure 2.06 and use the sliders to set the dΔGrxn/dΔT to -12.0 J/mole/K, the molar volume to 50 cm3, and the interfacial energy to 0.015 J/m2. Then change the ΔT until the critical radius drops to 1 nm. Please try again.
Press "Enter" after you type in the number.
No. You still have not given the correct answer. According to the model, a ΔT of 120 to 130 °C beyond the equilibrium reaction temperature is required to have a (spherical) critical nucleus with a radius of 1 nm for the spontaneous growth of sillimanite from kyanite.
Yes! Using the specified parameters, the model yields a ΔT of 120 to 130 °C beyond the equilibrium reaction temperature to have a (spherical) critical nucleus with a radius of 1 nm for the spontaneous growth of sillimanite from kyanite.
Use the data of Figure 2.06 to help answer this question. Press "Enter" after you type in the number.
Your answer is incorrect. Use Figure 2.06 with the same parameters as for the last question. See how many forumla units there are in a 1 nm radius sphere of sillimanite and use the sillimanite formula to convert formula units to atoms. Please try again.
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No. You still have not given the correct answer. The total number of formula units of sillimanite in a 4.2 nm sphere is given in Figure 2.06 as about 50. There are 8 atoms per sillimanite formula (Al2SiO5). Therefore, there should be about 400 atoms in a 1 nm sphere of sillimanite.
Yes. The total number of formula units of sillimanite in a 4.2 nm sphere is given in Figure 2.06 as about 50. There are 8 atoms per sillimanite formula (Al2SiO5). Therefore, there should be about 400 atoms in a 1 nm sphere of sillimanite.
The thermodynamic (macroscopic) model of Figure 2.06 indicates that even if a large degree of reaction overstepping occurs so that the critical radius for spontaneous growth of a spherical sillimanite nucleus is reduced to 1 nm, 400 atoms must be organized by the semi-random motions into the sillimanite structure to make the nucleus. This seems like a large number atoms to organize. This number can be reduced further by a larger ΔT. It also can be reduced by nucleation on another phase with a lower interfacial energy.
