joelbrugger Posted May 20, 2009 Share Posted May 20, 2009 Dear group, I'm trying to model the kinetics of replacement of calaverite (AuTe2) by gold (Au), in order to fit experimental data. A simple model with a simple kinetics low for calaverite works rather well if I suppress Au(OH)(aq) from the run. If I use Au(OH)(aq), then the model predicts that my experimental solution is always undersaturated with respect to native Au, and hence calaverite dissolves, but no gold is formed. There is little justification for suppressing Au(OH)(aq) - it is well characterized at 300C and Steffanson and Seward (2003) suggest that the properties are good to ~200C (I'm working at 220C). It seems that my problem could be solved if I could 1) force gold to precipitate, even in an undersaturated solution (Q/K = 1e-5), and 2) prevent the formed Au to dissolve once it has formed. I've tried a negative reaction rate for Au (negative is precipitation, right?), in conjuction with the nucleus keyword, but had no luck - always zero mole of Au in the system. Two example runs are visible at http://gallery.me.com/minmetsol/100108. I'm using react v. 7.0.3, and the script with kinetic laws for both Au and Calaverite is below. Let me know if you want the database and conf file to have a play. Thanks for your help, JOEL React> show Reaction runs from 0 years to 2.5 days Temperature is 220 C Thermo dataset: ..\thermo_MMS_11May18.dat Working directory: z:\gwb_work\tellurium Options: Debye-Huckel Basis is: H2O .015 free kg Na+ .1054 molal Cl- .01 molal H2TeO3(aq) 1e-8 molal O2(aq) 8.45e-5 mol Au+ 1e-8 molal HPO4-- .1998 molal H+ charge balance Reactants: 10 mg of Calaverite by kinetic rate law Rate constant = 1.8e-9 mol/cm2 sec Specific surface area = 51.06 cm2/g React 0 mol of O2(aq) Au by kinetic rate law Rate constant = -3e-9 mol/cm2 sec Specific surface area = 100 cm2/g Nucleus density = 200 cm2/cm3 Critical saturation index = 1e-8 Other "show" options: type "show show" Quote Link to comment Share on other sites More sharing options...
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.