1 / 33

Today's lecture

Today's lecture. More Aqueous Geochemistry First a look back at calcite solubility as a function of pH and Temperature. SOLUBILITY OF CALCITE IN AN OPEN SYSTEM.

hateya
Télécharger la présentation

Today's lecture

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Today's lecture • More Aqueous Geochemistry • First a look back at calcite solubility as a function of pH and Temperature

  2. SOLUBILITY OF CALCITE IN AN OPEN SYSTEM

  3. The pH of pure water in equilibrium with calcite at 25°C as a function of the partial pressure of CO2. Note that pH decreases linearly with increasing CO2 partial pressure.

  4. Changes to Ocean pH as a function of PCO2

  5. Why do we care about mineral precipitation in CO2 sequestration?

  6. Solubility Product • CaCO3(s)   <==>    Ca2+(aq)  +   CO3-2(aq) • In writing the equilibrium constant expression for a heterogeneous equilibria, we ignore the concentrations of pure liquids and solids.  So the equilibrium constant expression for the equilibria above is: • Ksp = [Ca2+][CO3-2]  • This equilibrium constant is called a solubility-product constant. Even though [CaCO3] is excluded from the equilibrium constant expression, some undissolved CaCO3(s) must be present in order for the system to be at equilibrium.  In general, the solubility product constant (Ksp) is the equilibrium constant for the equilibrium that exists between a solid ionic solute and its ions in a saturated aqueous solution.  The rules for writing the solubility-product expression are the same as those for writing any other equilibrium constant expression: The solubility product is equal to the product of the concentrations of the ions involved in the equilibrium, each raised to the power of its coefficient in the equilibrium equation.

  7. The Common-Ion Effect • The solubility of a substance is affected not only by temperature but also by the presence of other solutes.  We now consider the case in which the other solute has an ion in common with the substance involved in the equilibrium, in other words we are going to look at the common-ion effect in solubility.  Just as in the case of the common-ion effect in acid-base equilibria, the common-ion effect in solubility is just a special case of LeChattelier's principle.  Consider the equilibrium of CaCO3(s) with its saturated solution, if we add a solute containing Ca2+ or a solute containing CO3-2 to the saturated solution, the equilibrium will shift to the left making CaCO3 less soluble. • CaCO3(s)   <==>    Ca2+(aq)  +   CO3-2(aq) • <----Addition of Ca2+ or CO3-2 shifts equilibrium left decreasing solubility--------<< • Example: What if a solution undersaturated with respect to calcite passes through an evaporite bed with anhydrite (Ca SO4). Anhydrite will dissolve putting CA++ and SO4– into solution. The sulfate will not affect the solubility of calcite but the added Calcium might raise the Ca++ concentration enough to exceed the Solubility Product of Calcite.

  8. Carbonates( such as calcite) are one of the mineral groups that may form to sequester CO2. But remember that in the last class we say that adding CO2 to the aqueous solution caused a pH decrease. How does that affect the precipitation of calcite? http://www.ldeo.columbia.edu/~sanpisa/OceanSed%20project/class%20project/caco3phsol.JPG

  9. Hydrolysis reactions neutralize acid These are very common reactions in weathering and hydrothermal alteration CO2 + H2O = H2CO3 H2CO3 = H+ + HCO3- HCO3- =H+ + CO3-2

  10. Feldspar compositional variation • Or - Orthoclase KAlSi3O8 • Ab - AlbiteNaAlSi3O8 • An - Anorthite CaAl2Si2O8 Feldspars exhibit two solid solutions One from Or to Ab and the other from Ab to An. Labradorite: An50 to An70 Feldspars are the most common mineral in the earths crust !!

  11. Solid Solution Solid solution is not seen just in Feldspar. One other example is Carbonate which is important for this class

  12. Feldspar compositional variation • Or - Orthoclase KAlSi3O8 • Ab - AlbiteNaAlSi3O8 • An - Anorthite CaAl2Si2O8 Feldspars exhibit two solid solutions One from Or to Ab and the other from Ab to An. Labradorite: An50 to An70 Feldspars are the most common mineral in the earths crust !!

  13. Where do we find Feldspar? • Feldspar rich sandstone (Arkose)-- Reservoir • Shales—Caprock • Basalts

  14. I. Gaus et al. / Chemical Geology 217 (2005) 319–337

  15. I. Gaus et al. / Chemical Geology 217 (2005) 319–337

  16. If you have kinetic rate constants you can calculate how long these reactions will take.

  17. Feldspar is not the only mineral you need to be concerned with in a sandstone reservoir. • The amount of initial carbonate (probably as cement) is also important

  18. Effects of Carbonate buffers

  19. Siliciclasticvs carbonate buffered reservoirs.

  20. Ca-rich plagioclase feldspar is an essential mineral in basalt. In felsic rocks both plagioclase and Alkali feldspars are important

  21. Reaction of Ca-rich feldspar to calcite Volcanic glass may be important too. It is easily reacted.

  22. Another important mineral in basalt is OLIVINE

  23. Olivine reactions

  24. Mineral Carbonation for Sequestration http://www.esc.ethz.ch/research/efficientconversion/cotwo/mineralization.jpg

More Related