Geological Sciences 4550: Geochemistry

Transcription

1 1. Consider the following hypothetical gaseous solution: gases 1 and form an ideal binary solution; at 1000 K, the free energies of formation from the elements are - 50kJ/mol for component 1 and - 40kJ/mol for component. a.) Calculate Gmixing for the solution at 0.1 increments of X. Plot your results. b.) Calculate G for ideal solution at 0.1 increments of X. Plot your results. c.) Using the method of intercepts, find µμ1 and µμ in the solution at X = 0. This problem also involves repeated calculations of the same formulae, so it is ideal for a spreadsheet (below). For part c, we take the derivative dg/dx to get the slope, which is µ µ 1 RT ln(x /X 1 ). (We need to make sure the units are right as the G f s are in kj, but R is in J). The chemical potentials are calculated as µ 1 = G ideal (X ) slope*x and µ = G ideal (X ) slope (1-X ). G id. Mix. G ideal X1 X ln(x1) ln(x) KJ KJ µ1 slope µ1 µ G id mix = RT S (xlnx) G ideal = S(xµ) G id mix slope = dg/dx = µ- µ1rtln(x/x1) R J/K- mol T 1000 mu1-50 kj/mol mu - 40 kj/mol 1

2 Gbar Gbar mixing G ideal (kj) X. The fol lowing is an analysis of Acqua di Nepi, bottled water from the Italian province of Viterbo: HCO ppm SO ppm F 1.3 ppm Cl - 0 ppm NO ppm Ca 8 ppm Mg 7 ppm Na 8 ppm K 50 ppm a. Calculate the ionic strength of this water. (Recall that concentrations in ppm are equal to con- centrations in mmol kg - 1 multiplied by formula weight.) b. Using the Debye- Hückel equation and the data in Table 3., calculate the practical activity co- efficients for each of these species at 5 C. See the spreadsheet below.

3 z conc ppm mol. wt. conc mmol mmol*z^ å log gamma gamma HCO Cl Ca Na SO NO Mg K F I, mol A root I, mol B Consider the following analysis of a pyroxene: Site Ion Ions per site Tetrahedral Si 1.96 Al 0.04 Octahedral M1 Al 0.1 Mg 0.88 Octahedral M Fe 0.06 Ca 0.8 Na 0.1 Use the mixing on site model of ideal activities to calculate the activity of jadeite (NaAlSiO6) and diopside (CaMgSiO6) in this mineral. The relevant equation is To calculate mole fraction, we need to divide the ions per site, by the stoichiometric coefficient, which is 1 for the octahedral sites, but for the tetrahedral one (the one with Si). So the mole of Si is 1.96/ and its contribution to activity of both phase components is (1.96/). Activities are as shown below. 3

4 atoms/site Coeff mole frac. Si Al M1 Al Mg M Fe Ca Na jadeite act diopside act Notice that the Al in the tetrahedral site as well as iron are effectively dilutants and don t contribute to the activity of either jadeite or diopside. Also note that to do this properly, we need some knowledge of the lattice structure of the mineral. 4. Write the equilibrium constant expression for the reaction: CaCO 3(s) H (aq) assuming the solids are pure crystalline phases and that the gas is ideal. A general form would be: K = SO 4 H O (liq)! CaSO 4 H O CO (g) a gypsum a CO ( g ) a calcite a SO4 a H O Assuming the solids are pure phases means their activities are 1. If CO is ideal, then its activity may be replaced by its partial pressure, hence: K = p CO ( g ) a SO4 a H O Provided the solution is reasonably dilute, we would commonly also assume the activity of water was one, so that: K = p CO ( g ) a SO4 5. The equilibrium constant for the dissolution of galena: PbS solid H Pb aq H S aq is at 80 C. Using the γ Pb = 0.11 and γ H S = 1.77, calculate the equilibrium concentration of Pb in aqueous solution at this temperature and at ph s of 6, 5 and 4. Assume the dissolution of galena is 4

5 the only source of Pb and HS in the solution and that there is no significant dissociation of HS. Hint: mass balance requires that [HS] = [Pb ]. Since galena is a solid, we assume an activity of 1, so the equilibrium constant expression is 1: K = a Pb a H S a H a PbSsolid = a a Pb H S a H Expressed in log form, the equilibrium constant expression for this reaction is: log K = ph log a Pb log a H S = ph log[pb ] logγ Pb log[h S] logγ H S (assuming PbS is a pure phase). Since [H S] = [Pb ], we may write: Rearranging: log K = ph log[pb ] logγ Pb logγ H S log[pb ] = log K logγ logγ Pb H S We obtain the following answers: ph [Pb ] M ph 5