Hydrothermal Solutions and Ore Deposits

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1 Hydrothermal Solutions and Ore Deposits Physical Chemistry of Minerals and Aqueous Solutions D.M. Sherman, University of Bristol Chalcophiles, Lithophiles, Siderophiles.. Lithophile = oxides, silicates Siderophile = Fe alloys Chalcophile = sulfides

2 Physical Chemistry of Minerals and Aqueous Solutions Hydrothermal Vein Deposits Hypothermal ( oc) Mesothermal ( oc) Sulfide Ore Minerals Molybdenite MoS2 Chalcopyrite, CuFeS2 Pyrrhotite Fe1-xS Bornite, Cu5FeS4 Chalcopyrite CuFeS2 Galena, PbS Sphalerite, ZnS Arsenopyrite, FeAsS Gangue Minerals Quartz Tourmaline Topaz Micas Quartz Carbonates Barite Epithermal ( oc) Cinnabar, HgS Stibnite, Sb2S3 Argentite, Ag2S Quartz Chalcedony Opal Calcite Chalcopyrite (CuFeS2) Primary copper mineral in porphyry-copper deposits: sulfides desseminated in felsic intrusive rocks. The most widespread copper mineral. Usually meso-to hypothermal deposits.

3 Physical Chemistry of Minerals and Aqueous Solutions Galena (PbS) Primary ore mineral of Pb. Primarily found in mesothermal Mississippi Valley Pb-Zn deposits. Simple rocksalt structure. Forms large cubic crystals. Sphalerite (ZnS) Primary ore mineral of Zn. Primarily found in mesothermal Mississippi Valley Pb-Zn deposits.

4 Molybdenite (MoS 2 ) Primary molybdenum ore. High-temperature deposits. Accessory in granites Fundamental Questions How are metals such as Cu, Zn, Au and Pb concentrated into ore deposits? What chemical signatures can we use to find ore deposits? Are there vast resources at depth that we haven t yet discovered?

5 K eq vs T Since lnk = -ΔG 0 /RT we find, pk = -(ln K)/2.303 = ΔG 0 /(2.303RT) = ΔH 0 /(2.303RT) - ΔS 0 /(2.303R) If we assume ΔH 0 and ΔS 0 are constant with T, then pk(t) = pk(298) + ΔH R 1 T Solubility of Sphalerite (ZnS) Under acidic conditions, we can express the dissolution of sphalerite as ZnS + 2H + = Zn +2 + H 2 S For this reaction, pk = 4.44 and ΔH 0 = 14.0 kj/mol at 298 K. pk = pzn + ph 2 S - 2pH

6 Solubility of Sphalerite (cont.) pk(t) = pk(298) + ΔH R 1 T 1 = pzn+ ph 2 S 2pH 298 Rearranging gives, pzn = R T 1 ph S + 2pH Solubility of Sphalerite (cont.) Elevated temperatures are not enough to account for the solubilities of sulfide minerals needed to from ore-deposits.

7 Nature of Ore-Forming Solutions Fluid inclusions in mineral grains preserve samples of hydrothermal solutions. Upon cooling, the hydrothermal brines separate into solid (usually NaCl, gas (CO 2 + CH 4 ) and aqueous phases. The temperature at which the fluid was trapped can be determined by heating the sample and measuring the temperature at which gas + liquid recombine. Cl Complexation of Zn Zn +2 + Cl- = ZnCl + pk = -0.2; ΔH = 43.3 kj/mol Zn Cl- = ZnCl 2 pk = -0.25; ΔH = 31.2 kj/mol Zn Cl - = ZnCl 3 - pk = 0.02; ΔH = 22.6 kj/mol Zn Cl - = ZnCl 4-2 pk = -0.86; ΔH = 5.0 kj/mol Zn(H 2 O) 6 + ncl = ZnCl n + 6H 2 O Complexation is driven by the entropy increase when solvation waters are released.

8 Cl Complexation of Zn We can combine the reaction ZnS + 2H + = Zn +2 + H 2 S (pk ZnS ; ΔH ZnS ) with each complexation reaction Zn +2 + ncl - = ZnCl 2-n n (pk n ; ΔH n ) to get the reactions ZnS + 2H + + ncl = ZnCl 2-n n + H 2 S with pk = pk ZnS + pk n and ΔH = ΔH ZnS + ΔH n pk(t) = pk(298) + Cl Complexation of Zn ΔH R 1 T = pzncl n 2 n + ph2 S 2pH - npcl To a close approximation, pcl = pcl tot. Rearranging gives 2 n ΔH pzncl 0 1 n = pk(298) R T 1 ph S + 2pH + npcl

9 Solubility of Sphalerite: Cl complexation Cl-complexation of Zn greatly enhances the solubility of ZnS at high temperature. Caution: we assumed that ΔH 0 was constant with T. Entropy and Complexation The complexation of metals at high temperature is driven by the increased translation entropy resulting from the breakdown of the metal hydration sphere: Zn(H 2 O) 6 + Cl - = ZnCl(H 2 O) H 2 O Zn(H 2 O) 6 + 2Cl - = ZnCl H 2 O (Hydration numbers are derived from molecular dynamics simulations.)

10 The Continuum Model of Aqueous Solutions Born (1920) theory of solvation free energy G: Where: ΔG = q2 e 2 2R 1 1 ε R = Born radius of cation with charge q ε = dielectric constant of the solvent Basis for HKF Equation of State used to predict stability constants of complexes at high P,T. Changes in Dielectric Constant of Water with P and T We expect decreased solvation of ions with increasing T. This will favor metal complexation by Cl -. Pressure should enhance solvation.

11 HKF Equation of State (cont.) The heat capacity and volume of a species depend on T and P as: C p0 (P,T) = c 1 + c 2 ω + ωtx + 2TY 2 (T θ) T P T 1 ε 1 2 ω T 2 P V p0 (P,T) = a 1 + a 2 P + ψ + a 3 T θ + a 4 P + ψ ( )( T θ) ωq + 1 ε 1 Where c 1, c 2, a 1, a 2, a 3 and a 4 are parameters for the particular solute species ω P T HKF Equation of State (cont.) ω is the Born coefficient of the ion, Y = 1 ε 2 ε T P, Q = 1 ε 2 ε P And, finally, θ and ψ are parameters for the solvent. T, X = 1 ε 2 2 ε T 2 P 2 ε ε P 2 P

12 Precipitation of Sulfides Given the general reaction ZnS + 2H + +ncl = ZnCl n 2-n + H 2 S ZnS will precipitate when H + is consumed: 2H + + CaCO 3 (calcite) = CO 2 + Ca +2 + H 2 O 3KAlSi 3 O 8 (feldspar) + 2H + = 6SiO 2 + 2K + + KAl 3 Si 3 O 10 (OH) 2 (muscovite) Volcanogenic Massive Sulfide Deposits

13 Convergent Plate Boundaries Porphyry Deposits Ore zone: CuCl 2 + FeCl 2 + 2H 2 S = CuFeS 2 + 4H Cl - Argillic: 2KAl 3 Si 3 O 10 (OH) 2 + 2H + + 3H 2 0 = 3Al 2 Si 2 O 5 (OH) 4 + 2K + Phyllic: 3KAlSi 3 O 8 + 2H + = KAl 3 Si 3 O 10 (OH) 2 + 6SiO 2 + 2K + Potassic

14 Summary Sulfide minerals are extremely insoluble. Hydrothermal solutions contain high concentrations of NaCl. Complexation of metals by Cl - (and possibly HS - ) greatly enhances the solubility of sulfides at high temperature Precipitation of sulfide minerals occurs either by cooling, boiling or by a drop in ph when fluids react with host rock (e.g., carbonates).

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