construction for the System K 2 OAl 2 O 3 SiO 2 amorphous silica quartz gibbsite Where is Si Where is Al Which should be more soluble By how many orders of magnitude at ID: 1022880
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1. Korjinski (activity/activity) Diagram construction for the System K2O-Al2O3-SiO2
2. amorphous silicaquartzgibbsiteWhere is Si?Where is Al?Which should be more soluble?By how many orders of magnitude at a typical pH range for Earth’ssurface?4Si speciesChoose your inert component, which in this case is Al2O3. Review the locations of Al3+, K+, and Si4+ on the Ionic Potential graph and think about the relative solubilities of Si and Al species in solution to remind yourself why we choose Al2O3 as inert.
3. (not actually a mineral)If the denominator in your ratio is 0, use “∞” for the value of the ratio. If the numerator in your ratio is 0, use “0” for the value of the ratio. 2) Calculate the mole ratios of the mobile components to the inert component for all the solids which appear on the triangular diagram. The table of moles of oxides in each phase (calculated for the triangular diagram lab) and valuesof ∆Gf are on the next page.MineralK2O / Al2O3SiO2 / Al2O3 Quartz Microcline Kaliophilite Muscovite Kaolinite Gibbsite KOH 000000.333622211∞∞
4. ∆Gf∘ kJ/mol# of moles and mol percentFormulaCompoundK2OAl2O3SiO2-322.11 100%0 0%0 0%K2OK-oxide-379.11 100%0 0%0 0%KOHK-hydroxide-856.60 0%0 0%1 100%SiO2Quartz-1582.30 0%1 100%0 0%Al2O3Corundum-1155.10 0%0.5 100%0 0%Al (OH)3Gibbsite-915.80 0%0.5 100%0 0%Al O (OH)Boehmite-3799.70 0%1 33%2 67%Al2 Si2 O5 (OH)4Kaolinite-5266.10 0%1 20 %4 80%Al2 Si4 O10 (OH)2Pyrophyllite-2005.30.5 25%0.5 25%1 50 %KAlSiO4Kaliophilite-2871.40.5 17%0.5 17%2 67%KAlSi2O6Leucite-3742.90.5 12.5%0.5 12.5%3 75%KAlSi3O8Microcline-5608.40.5 10%1.5 30%3 60%KAl3Si3O10(OH)2Muscovite-2443.90 0%1 50 %1 50 %Al2SiO5Kyanite-237.1 waterH2O-1308.1H4SiO4Dissolved silica-282.5K+Dissolved K+ ion
5. 3) Plot the minerals on the oxide ratio diagram.KOH will plot at infinity on the y-axis andquartz will plot at infinity on the x-axis.
6. 4) Draw in all stable tie-lines from your triangular diagram.KOHSiO2The lines shown as dashed representintersections of mineral stability fieldswith the saturation lines of quartz andKOH. A later slide will show you how tocalculate the saturation lines for these two solids.(not actually a mineral)
7. 5) Draw a line perpendicular to each tie-line a) These perpendiculars represent orientations of the mineral phase boundaries which will appear on your final activity diagram. b) This equivalency can be derived by laws of thermodynamics but we will accept it without this proof.
8.
9. 6) Transfer these perpendiculars to the qualitative activity diagram with the axes indicated here.●Just add each perpendicular to the graph approximating its orientation from the previous oxide ratio diagram.● Start in the lower, left-hand corner where gibbsite is. Sizes of fields are not important here.Use the blank diagram and try plotting the lines yourself before you see them here by clicking.The intersection of the gibbsite field with the KOH saturation line occurs only when no H4SiO4 is present in solution so it will not appear on our diagram be-cause we assume the presence of some H4SiO4 in solution.microclinegibbsitequartz saturation boundarymuscovitekaliophilitekaoliniteKOH saturation boundaryWe won’t bother to calculate the quartz/KOH boundary because the required solution chemistry is notfound (or is exceedingly uncommon) in nature.(not actually a mineral)
10. 7) Now use free energy data to quantify the diagram by writing balanced reactions representing conversion of one mineral to another using only water, K+, H+ and H4SiO4 to balance your reaction. Remember to conserve Al2O3 in the solids. Remember, for units of kJ/mol; GR = -5.709 log K (in kjoules/mol)It's usually easiest to start with reactions which involve only one variable (either the log aK+/aH+ ratio or the log aH4SiO4. These are the horizontal or vertical boundaries. Use the thermodynamic data from the triangular-diagram tutorial in your calculations.To ensure you can do all calculations for your problem set, remember to generate many of these chemical reactions and equilibrium-constant expressions yourself before clicking through the following slides.
11. -58.810.3010.30-5.15
12. 12.23-69.8= -6.11
13. 13.8 -157.527.6
14. ∆Gro = 36 kJ/mol, therefore,-6.33.2
15. -5.129.15.1
16.
17. Finally: To simplify this problem set, amorphous silica is not included so instead calculate the quartz saturation line.Then calculate the KOH saturation line.-140.524.6 24.6 log [H4SiO4(aq)] = -4.01 -3.98K = [H4SiO4(aq)] = 10-4.01 -3.982 H2O(l) + SiO2(s) ↔ H4SiO4(aq) K= 10-4.01 -3.98
18. Don’ forget to plot the saturation boundaries for quartz and KOH.
19. 181614121086420-2 aK+log ------ aH+log aH4SiO4-10 -9 -8 -7 -6 -5 -4 -3 GIBBSITEMUSCOVITEKALIOPHILITEMICROCLINEKAOLINITEQuartz Saturation LineThe KOH Saturation Line plotsup off thegraph at aK+log ----- = 24.6 aH+
20. For a few of the equations in your CaO/MgO/SiO2 problem set you will have to add H+ to both sides of your reaction in order to get the proper number of hydrogen ions to yield the variables for your graph. Remember that your variables are the logs of ion ratios: log a(Mg2+)/a(H+)2 and log a(Ca2+)/a(H+)2 . For the chemical reactions to yield the appropriate ratios for the axis labels your reactions must include twice as many H+ ions as they do Ca2+ and Mg 2+ ions. Below is an example of how to accomplish this. One of the minerals in this reaction, diopside (CaMgSi2O6), is not one that you have to consider for your exercise because it does not form at 25oC under earth-surface conditions, so you will not be writing this reaction. 2CaSiO3(s) + Mg 2+ (aq) = CaMgSi2O6(s) + Ca 2+ (aq) Clearly, this reaction can be mass balanced without the addition of any water or H+. Therefore, to generate the proper variables you must add H+ to both sides so that there are twice as many H+ as Mg 2+ on opposite sides and twice as many H+ as Ca 2+ on opposite sides. The H+ must be on opposite sides of the reactions from the Ca 2+ and Mg 2+ to yield the proper ratios. It just happens that in order to balance this reaction you will end up with the same number of H+ on both sides. For some of the actual reactions in your problem set this will not be the case. However, you will always be adding the same number of H+ to both sides, because otherwise you would disturb the mass balance of the reaction. 2CaSiO3(s) + Mg 2+ (aq) + 2 H + (aq) = CaMgSi2O6(s) + Ca 2+ (aq) + 2 H + (aq) The chemical reaction yields this “Keq” expression and algebraic equation for the boundary of wollastonite and diopside. (a(Ca2+)) (a(H+))2 Keq = ------------------------- (a(Mg2+)) (a(H+))2 log a(Mg2+)/a(H+)2 = log a(Ca2+)/a(H+)2 - log K