New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here

Principles of Geochemistry

by: June Gottlieb

Principles of Geochemistry GEOL 423

Marketplace > University of Idaho > Geology > GEOL 423 > Principles of Geochemistry
June Gottlieb
GPA 3.52


Almost Ready


These notes were just uploaded, and will be ready to view shortly.

Purchase these notes here, or revisit this page.

Either way, we'll remind you when they're ready :)

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Class Notes
25 ?




Popular in Course

Popular in Geology

This 91 page Class Notes was uploaded by June Gottlieb on Friday October 23, 2015. The Class Notes belongs to GEOL 423 at University of Idaho taught by Staff in Fall. Since its upload, it has received 49 views. For similar materials see /class/227868/geol-423-university-of-idaho in Geology at University of Idaho.


Reviews for Principles of Geochemistry


Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/23/15
TOPIC 3 THE TERRES TRIAL ENVIRONMENT Required reading Chapter 3 in Andrews et al 1996 Processes at the Earth s surface are largely the readjustment of unstable highpressure and temperature minerals to oxidizing acidic lowtemperature conditions ie weathering To fully understand the process of weathering we need to understand bonding and structure of silicate minerals CRYSTAL CHEMISTRY See Chapter 7 in Faure Boxes 33 and 34 in Andrews et al Cations elements that give up their valence electrons completely to form positively charged ions Anions elements that acquire a complete valence shell of electrons to become negatively charged ions IONIC BONDS Ionic bonds bonds formed via the electrostatic attraction between oppositely charged ions I Coulomb s law F ZtZ 47280r2 so permittivity of free space 884X10399 farads m391 farad coulomb volt391 Most minerals are held together via ionic bonds COVALENT BONDS Covalent bonds bonds formed by sharing of electrons Many gaseous species and organic compounds contain covalent bonds eg N2 A few minerals such as diamond and graphite contain covalent bonds Sharing may not be equal That is covalent bonds may have some ionic character and vice versa MEASURE OF IONICCOVALENT CHARACTER Electronegativitv x measures the ability of an atom in a compound to attract electrons to itself Metals have low x values and nonmetals have x values We use differences in electronegativity to determine degree of ionic character of a bond Example CsF XCS 0739 XF 40 XF XCS 33 gt 92 ionic39 lt 8 covalent i 7 m1 6 Pemnl lonn Characch ah Smgl cmm am mm xygen nilum m 1m MM M M u Arm am my quot u Mumv mmmmw OTHER EXAMPLES ZnS xz 15 k 25 35 xz 25 15 9 19ionic 81 covalent H2 XE 21 0 ionic 100 covalent CClA x0 25 Kc 30 Xm xci 025 05 5 ionic 94 covalent WHAT DIFFERENCE DOES THE TYPE OF BONDING MAKE All physical and chemical properties of a compound depend on the character of the bonds Example Solubility in water Ionic bonded compounds have high aqueous solubilities NaCl XCIXN30 09 21 67 ionic very soluble Cdiamond xc xc 0 0 ionic insoluble Example Physical properties Ionic bonded solids tend to be isotopic covalent bonds anisotropic 11 IONIC CRYSTAL STRUCTURES Ions form a crystal such that they are closest packed This is a consequence of Coulomb s law Ions arrange themselves such that interionic distances are minimized If we assume ions are hard incompressible spheres like billiard balls we can use the concept of radius ratio as a key to explaining crystal structures COORDINATION NUMBER Coordination number Number of anions surrounding a given cation The coordination number is well predicted by the ratio of the radius ofthe cation to the radius ofthe anion RLRa As the radius ratio increases more anions can be t exac y around a cation Coordination numbers ofs 7 910 and11 are very rare in mineral structures Derivation of the radius ratio for three fold coordination 660o sine R RCRa sin 6 0866 R 0866 R R3 0866Rd 086611 0134Ra 0866Rc 1mg 01340866 0155 15 Derivation of the radius ratio for fourfold coordination square planar ampamp Ra 0707Rd 070711 0293Ra 0707Rc 1mg 02930707 0414 IONIC RADII Determined from Xray studies ofinteratomic distances between ions in crystals Ions are not really hard spheres They can be formed by electrical charges surrounding them Thus ionic radius is dependent on CN Example Na CN 4 o 8 9 radiusA 107 110 124 140 Ionic radii are important for radius ratio arguments Also ions with similarradii can 39nera1s substitute for one another in mi NW straw TRENDS IN IONIC RADII The ionic radii of isoelectronic series decrease with increasing atomic number for both cations and anions The radii of ions with the same charge in a group increase with increasing atomic number ie downward in the periodic table The radii of ions of the same element decrease with increasing positive charge and increase with increasing negative charge The radii of ions with charges of 2 and 3 among the transition metals of the fourth period decrease with increasing atomic number implying a contraction of the electron cloud as the 3d orbitals are lled Ions of different elements may have similar radii Na Ca2 Cdz Y T1 REE Th4 N 100 A CN 6 Hf Zr4 N 080 A CN 8 Fe Co Ni3 065 A CN 6 CRYSTAL STRUCTURES 0 We can describe ionic crystal structures as cation coordination polyhedra outlined by the location of the anions around the cations The nature of this polyhedron can be determined by the radius ratio rule STRUCTURE OF HALITE NaCl PAULING S RULES These rules rationalize crystal structures of minerals 1 A coordination polyhedron of anions is formed about each cation the cationanion distance being determined by the sum ofthe two radii and the coordination number of the cation being determined by the radius ratio 2 In a stable ionic structure the Valence of each anion with changed sign is exactly or nearly equal to the sum of the strengths of the electrostatic bonds to it from adjacent cations This rule states that crystals should be electrostatically neutral over Very short distances 2 3 PAULING s RULES CONTINUED The presence of shared edges and especially of shared faces in a coordinated structure decreases its stability this effect is large for cations with high valency and small coordination number Essentially this says that any two cations tend to be as far from each other as possible When polyhedra do share edges or faces the cations have to be fairly close together and the polyhedra are thus distorted due to the mutual repulsion of positively charged cations 4 L11 PAULING s RULES CONTINUED In a crystal containing different cations those with large valence and small coordination number tend not to share polyhedron comers edges or faces with each other This also reflects the dislike of cations for each other The number of essentially different kinds of constituents in a crystal tends to be small This rule can be restated as follows The structural environment of a given cation or anion tends to be the same throughout a crystal39 ie only one type of coordination is likely for a given element APPLICATION OF PAULING S RULE TO HALITE Each Na ion is surrounded by 6 Cl39 so bond strength ofNaCl bond is 16 Each Cl39 ion in turn has 6 NaCl reaching it so 6 x 16 1l1i So the crystal is electrostatically neutral because the sum ofbo of all NaCl bonds reaching a single Cl39 ion is equal to the absolute Value of the charge of that ion bonds 25 PEROVSKITE STRUCTURE on ForCaz cN12 39 fem cN 6 Do APPLICATION OF PAULING S RULE 2 TO PEROVSKITE TiO bond strength 46 CaO bond strength 212 16 Total bond strength reaching each oxygen ion is 46X2l6X4 1262 l 2l from Ti from Ca Thus Pauling s 2nd rule is satis ed charge on O ARE THERE ANY PRACTICAL APPLICATIONS OF PAULING S RULES Of course Bond strength calculations useful in determining cleavage planes in minerals Minerals will break along planes of weakness low bond strength We can also use Pauling s rules to predict likely structures for new minerals COVALENT BONDING IN MINERAL S Sul des are me most important group of eowdendy bonded minerals Pyrite FeSz similar structure to halice wim Fe on me Na sites and 52 on c1 sites see gure Bond distance between s awms is 210 A which is less than am 312 A for 52 This implies strong covalentbonding be Xs25 xii1507 12 ionic 88 covalent STRUCTURE OF PYRITE OF Os HYBRIDIZATION OF ORBITALS FeS2 bonds are a good example of hybridization Fe is octahedrally coordinated by 82 buts p and d orbitals by themselves do not allow octahedral coordination However if we combine two 0 one s and three p orbitals mathematically we can get 6 new hybrid dzsp3 orbitals which point to the comers of an octahedron ORGANIC COlVlPOUNDS Mostly held together by covalent bonds XC25XH21 xCXHO4 4 ionic 96 covalent Solid organic compounds consist of molecules strongly bound intramolecularly by covalent bonds the molecules are weakly bound to one another via van der Waals forces Thus organic solids tend to melt at lower temperatures than ionic solids MANY SOLIDS CONTAIN BOTH IONIC AND COVALENT BONDS Calcite CaCO3 CO bonds form a trigonal planar array in the CO3239 ion x0 XC 3525 10 22 ionic 78 covalent O 39 O CO bonds form from sp2 hybrids o CO3239 ions are distinct within the calcite structure CO3239 and Ca2 are in turn bonded ionically Calcite dissolution CaCO3 lt gt Ca2 CO3239 SILICATE MINERALS SiO bonds x0 xSi 3518 17 51 ionic 49 covalent Si tends to form tetrahedral SiO4439 units Tetrahedral bonding allowed by sp3 hybridization Tetrahedral units then link by sharing comers to form all the silicate structures known Only a limited number of ways to link these tetrahedra so we only have only a few fundamentally different silicate structure types SILICATE CLASSIFICATION Class Arrangement of Shared corners Repeat unit SiO Example te ahredra Nesosilicates Independent 0 SiOAA39 14 Olivine te Sorosilicates Pair of 1 SizO7 39 135 Hemimorphite tetrahedra sharing comer Cyclosilicates Closed rings of 2 Sing39 13 Tourmaline tetrahedra Inosilicates In nite single 2 SiOzz39 13 Pyroxenes ain 0 In nite double 25 Si4011639 1275 Amphiboles chains of tetrahedra Phyllosilicates In nite sheets 3 SizOsz39 125 Micas of tetrahedra Tektosilicates Unbounded 4 SiOz 12 u framework of feldspars tetrahedra 35 SILICATE STRUCTURAL UNITS I Solmilicin Nesasiliclll cvclmilicl Incliliult 5309 Chain lnosilicall double chain SILICATE STRUCTURAL UNITS H v A A quotOquot 0quot VA A V V V humiliat IONKISUBSTHIHIONIN CRYSTALS Read Chapter 8 in Faure 1998 SOLID SOLUTION Occurs when in a crystalline solid one element substitutes for another 0 For example a garnet may have the composition Mgl7Fe09Mn02Ca02A12Si3012 The garnet is a solid solution of the following end member components Pyrope Mg3AIZSi3012 Spessartine Mn3AIZSi3012 Almandine Fe3A128i3012 and Grossular Ca3AlZSi3012 GOLDSCHIVIIDT S RULES 1 The ions of one element can extensively replace those of another in ionic crystals if their radii differ by less than approximately 15 N Ions whose charges differ by one unit substitute readily for one another provided electrical neutrality of the crystal is maintained If the charges differ by more than one unit substitution is generally slight DJ When two different ions can occupy a particular position in a crystal lattice the ion with the higher ionic potential forms a stronger bond with the anions surrounding the site 40 RINGWOOD S MODIFICATION OF GOLDSCHIVIIDT S RULES 4 Substitutions may be limited even when the size and charge criteria are satisfied when the competing ions have different electronegatiVities and form bonds of different ionic character This rule was proposed in 1955 to explain discrepancies with respect to the first three Goldschmidt rules For example Na and Cut have the same radius and charge but do not substitute for one another COUPLED SUB STITUTIONS Can Th4 substitute for Ce3 in monazite CePO4 Rule 1 When CN 9 rTh4 117 A rCe3 123A OK Rule 2 Only 1 charge unit difference OK Rule 3 Ionic potential Th4t 4 1 17 34239 ionic potential C t 3 1 23 244 so Th4 is preferred Rule 4 xTh 1339 xCe 11 OK But we must have a coupled substitution to maintain neutrality Th4 Si4 lt gt Ce3 P5 21 But can Si4 substitute for P5 according to Goldschmidt s rules Rule 1 When CN 4 rSi4 034 A rP5 025 A 0K Rule 2 Only 1 charge unit difference 0K Rule 3 Ionic potential Si4t 4O34 1176 ionic potential P5f 5O25 20 so P5 is preferred Rule 4 xSi18xP 21 0K OTHER EXAMPLES OF COUPLED SUB STITUTION Plagioclase NaAlSi3O8 CaAlZSiZO8 Na Si4 lt gt Ca2 Al3 Gold and arsenic in pyrite F e82 Au As3 lt gt 2Fe2 REE and Na in apatite Ca5PO43F REE3 Na lt gt 2Ca2 22 INCOMPATIBLE VS COMPATIBLE TRACE ELEMENTS Incompatible elements Elements that are too large andor too highly charged to t easily into common rockforming minerals that crystallize from melts These elements become concentrated in melts Largeion lithophile elements IL s Incompatible owing to large size eg Rb Cs Sr Ba K High eld strength elemenm LIFSE S Incompatible owing to high charge eg Zr Hf Ta Nb Th U Mo W etc Compatible elements Elements that t easily into rock forming minerals and may in fact be preferred e g Cr V Ni Co Ti etc 45 Changes in element concentration in the magma during crystal fractionation of the Skaergaard intrusion Divalent cations tag 57 Percemuun mm Changes in element concentration in the magma during crystal fractionation of the Skaergaard intrusion Trivalent cations THREE TYPES OF TRACE ELEIVIENT SUBSTITUTION 1 CAMOUFLAGE 2 CAPTURE 3 ADIVHSSION CAMOUFLAGE Occurs when the minor element has the same charge and similar ionic radius as the major element same ionic potential no preference Zr4 080 A Hf4 079 A Hf usually does not form its own mineral it is camou aged in zircon ZrSiO4 CAPTURE Occurs when a minor element enters a crystal preferentially to the major element because it has a higher ionic potential than the major element For example Kfeldspar captures Ba2 144 A Zr 139 or sr2 121 A Zr 165 in place of K 146 A Zr 068 Requires coupled substitution to balance charge K Si4 lt gt Sr2 Ba2 Al3 25 ADMISSION Involves entry of a foreign ion with an ionic potential less than that of the major ion Example Rb 157 A Zr 0637 for K 146 A Zr 068 in Kfeldspar The major ion is preferred WEATHERING PROCESSES 26 MECHANISMS OF WEATHERING Dissolution Oxidation Acid Hydrolysis DISSOLUTION Read Chapter 10 in Faure 1998 27 DISSOLUTION The simplest weathering reaction is dissolution of soluble salts eg CaSO4anhydrite lt gt Ca2 804239 0 olubili The total amount of a substance that will dissolve in a solution at equilibrium SOLUBILITY PRODUCT The solubility of a mineral is governed by the solubility product the equilibrium constant for a reaction such as CaSO4anhydrite lt gt Ca2 SO 4239 The solubility product is given by K aCaztasoj SP aCaSQ If anhydrite is a pure solid then aCaSO4 l and in dilute solutions aCa2z Ca2 and aSO42 m 804239 28 So we may write KSP 10 45 g Ca2so42 What is the solubility of anhydrite in pure water If anhydrite dissolution is the only source of both Ca2 and SO 4239 then by stoichiometry Ca2 5042 X X2 10 45 X 10225 562X10393 mol L391 MWanhydnte 13614 g rnol391 solubility 562X10393 mol L391l36 14 g mol391 0765 g L1 A MORE CONIPLICATED SALT A12SO43s lt gt 2Al3 3804239 a2 a3 K 2 A1 303 SP aAll so4 3s Assume aAlzltSOBgt3 l aA13 m Al3 and aSO42 z 804239 Then KSP 6919 m Al32SO42393 Let X the total number of moles of A12SO43s dissolved Then by stoichiornetry Al3t 2X and 804239 3X 29 6919 2X23X3 6919 108x5 X 09147 mol L391 A13 2X 1829 1 1 101L391 8042 3X 2744 mol L391 x 09147 mol L1342 g mol39l 3128 g L 1 A12SO43s SATURATION INDEX In a natural solution it is not likely that Ca2 804239 for example because there will be more than one source of each of these ions In this case we use saturation indices to determine if the water is saturated with respect to anhydrite KSP 1045 m ca2 eqSO42eq IAP ca2 act S042 act Saturation Q 2 m IAP index K KSP SP Suppose a groundwater is analyzed to contain 5X102 mol L391 Ca2 and 7XlO393 mol L391 804239 Is this water saturated with respect to anhydrite KSP 1045 mol2 L392 IAP 5X103927X10393 35X104 10345 mol2 L392 Q 10 34510 45 10105 1122 Q gt 1 ie IAP gt KSP so the solution is supersaturated and anhydrite should precipitate If Q l ie IAP KSP the solution would be saturated If Q lt l ie IAP lt KSP the solution would be undersaturated the mineral should dissolve ANOTHER EXAlVIPLE Suppose the drainage from a tailings pile contains 5X10393 mol L391 804239 10394 mol L391 Fe3 10393 mol L391 Ki and has pH 3 Is this water saturated with respect to the mineral jarosite KFe3SO42OH6 KSP 2 10965 m0112 L 12 IAP KW Fe3l3 S04239 2OH 6 OH39 K H 10391410393 103911 mol L1 IAP l031039435X1039321039116 108560 mol12 L12 Q 10856010965 2 1010898 2 791X1010 So the solution would be highly supersaturated HOW MUCH SALT SHOULD PRECIPITATE Returning to the previous example ie the groundwater with 5XlO392 mol L391 Ca2 and 7XlO393 molL391 804239 how much anhydrite should precipitate at equilibrium If X mol L391 of anhydrite precipitate then at equilibrium Ca2 5XlO393 X39 804239 7XlO393 X and Ca2SO4239 1045 5XlO393 X7X10393 X 103945 X2 57X10392X 3249X104 0 X1 505X10392 mol L391 39 X2 645X10393 mol L391 The rst root is unreal because it would cause both final concentrations to be negative So 645X10393 mol L391 of anhydrite precipitates or 645X10393 mol L391l36l g mol391 0878 g L391 and Ca2 5XlO392 645X10393 436X10392 mol L1 so421 7x103 645X10393 55x104 mol L1 THE COMlVlON ION EFFECT Natural waters are very complex and we may have saturation with respect to several phases simultaneously Example What are the concentrations of all species in a solution in equilibrium with both barite and gypsum 1 Species Ca2 Baz 804239 Ht OH 2 Mass action expressions CaSO42HZO lt gt Ca2 SO42 ZHZO KSP Ca2so421 1046 BaSO4lt gt Ba2 SO42 KSP Ba2 5042 2 10 100 H20 lt gt H OH KW H OH 1014 3 Massbalance Ba2t Ca2 804239 4 Chargebalance 2Ba2 2Ca2 Ht 2SO4239 OH39 Ca2107rso1 Ba l10o 1039so1l 10 1so jL905 1046 10100 2 80422 8042 104612 1023 mol Ll Ca2 1046103923 1023 mol L1 Ba2 10 10010 23 10 77 molLl The least soluble salt barite contributes a negligible amount of sulfate to the solution The more soluble salt supresses the solubility of the less soluble salt commonion effect 67 The solubility of gypsum is hardly affected by the presence of barite Solubility of barite alone Ba2so42 2 10 100 Ba22 2 10 100 Ba2t 1050mol L391 Solubility of gypsum alone Ca2so42 1046 Ca22 10 23 Ca2 1023mol L391 REPLACEMENT REACTIONS We can also calculate Ba2Ca2 in equilibrium with both barite and gypsum 65121404750271 Bazldo lylso Ca2 10100 0754 Ba2 10746 What would happen if a solution with Ba2Ca2 10393 came into contact with a gypsumbearing rock Barite will precipitate and gypsum will dissolve until Ba2Ca2 1054 69 REDOX REACTIONS Read Chapter 14 in Faure 1998 REDOX REACTIONS Oxidation a process involving loss of electrons Reduction process involving gain of electrons Reductant a species that loses electrons Oxidant a species that gains electrons RULES FOR ASSIGNMENT OF OXIDATION STATES l The oxidation state of all pure elements is zero 2 The oxidation state of H is 1 except in hydrides eg LiH PdHZ Where it is l 3 The oxidation state of O is 2 except in peroxides eg H202 Where it is l 4 The algebraic sum of oxidation state must equal zero for a neutral molecule or the charge on a complex ion VARIABLE VALENCE ELEMENTS Sulfur SO42396 SO32394 80 FeS2l HZS2 0 Carbon C024 CO CH44 Nitrogen NO3395 N02393 NO2 N20l N20 NH33 Iron Fe2033 FeO2 Fe0 Manganese MnO4397 Mn024 Mn2033 MnO2 Mn0 Copper CuO2 Cu20l Cu0 Tin Sn024 Sn22 Sn0 Uranium U0226 U024 U0 Arsenic H3As0405 H3As0303 As0 AsH3l Chromium CrO42396 Cr2033 Cr0 0 Gold AuCl4393 AuCN239l Au0 BALANCING OVERALL REDOX REACTIONS Example balance the redox reaction below Fe Cl2 lt gt Fe3 Cl39 Step 1 Assign valences Fe0 C12O lt gt Fe3 Cl Step 2 Determine number of electrons lost or gained by reactants Fe0 C12O lt gt Fe3 Cl i T 3e39 2e Step 3 Cross multiply 2Fe 3C12O lt gt 2Fe3 6Cl39 74 HALFCELL REACTIONS The overall reaction 2Fe 3C12O lt gt 2Fe3 6Cl39 may be written as the sum of two halfcell reactions 2Fe lt gt 2Fe3 6e39 oxidation 3C12O 6e39 lt gt 6Cl39 reduction All overall redox reactions can be expressed as the sum of two halfcell reactions one a reduction and one an oxidation Another example C2H6 NO339 lt gt HCO339 NH 4 C3932H6 N5O339 lt gt 2HC4O339 N393H4 i T l4e39 8e 8C2H6 14NO339 lt gt 2HCO339 NH4 8C2H6 14NO339 lt gt 16HCO339 14NH4 8C2H6 14NO339 12H 6H20lt gt 16HCO339 14NH4 Final example balance the redox reaction FeS2 02 lt gt FeOH3 SO42 Fet282391 020 lt gt Fe3OH3 296042 i T 15e39 4e39 4FeS2 1502 lt gt FeOH3 2804239 4FeS2 1502 lt gt 4FeOH3 88042 4FeS2 1502 14H20 lt gt 4FeOH3 8804239 16H This reaction is the main cause of acid generation in drainage from sulfide ore deposits Note that we get 4 moles of H for every mole of pyrite oxidized ACIDMINE DRAINAGE 4Fes2 1502 14H20 lt gt 4FeOH3 88042 16H Hardrock and coal mining can expose pyrite to oxidation and can lead to acid mine drainage AMD In acid mine drainage the pH can be as low as 02 or even negative Acidity causes increased solubility of Al and other toxic heavy metals WHAT SULFIDES OTHER THAN PYRITE CAN CAUSE AlD Pyrrhotite FeS 94O2 SZHZO lt gt FeOH3 2H SO42 Yes oxidation of pyrrhotite can form acid but not as much as pyrite Galena PbS 202 lt gt Pb2 SO42 No oxidation of galena does not produce acid PYRITE OXIDATION A TWO STAGE PROCESS Pyrite oxidation can be modeled as a twostep process 1 Oxidation of pyrite sulfur to sulfate 4FeS2 4HZO 1502 lt gt 8H 8SO4239 4Fe2 2 Oxidation of ferrous to ferric iron 4Fe2 O2 lOHZO lt gt 4FeOH3 8H Equal amounts of acid are formed in each step 40 ROLE OF BACTERIA Inorganic oxidation is very slow at low pH At pH lt 35 Fe oxidation is catalyzed by the bacterium T hiobacillus thiooxidans At pH 35 45 oxidation is catalyzed by Metallogenium The bacteria utilize Fe oxidation as a source of energy This is not an efficient process 220 g Fe2 must be oxidized to produce 1 g of cellular carbon This leads to large deposits of FeIII oxide Presence of H20 also catalyzes FeII oxidation ORGANIC MATTER Oxidation of reduced organic matter in soils is also catalyzed by microorganisms This is important because CO2 is produced increasing soil acidity In biologically active soils PC02 may be 10 100 times atmospheric 41 HALFCELL REACTIONS AND THE SHE Halfcell reactions involve electron transfer and so give rise to electrode potentials The reaction H2g lt gt 2H 2e39 is arbitrarily assigned a standard electrode potential ofE 000 V ie the ElVIF ofthis electrode is 000 V when pH2 1 atm and Ht 1 mol L39l The above electrode is called the standard hydrogen electrode SHE All other electrode potentials are measured and reported relative to SHE AN IMPORTANT RELATIONSHIP 0 Cquot 0 AGF mg Faraday constant number of electrons transferred 233906 kcal V1 And by convention AGfe39 AGfH 0 42 THE NERNST EQUATION Consider the halfcell reaction Fe 2 lt gt Fe3 e39 The following equation applies under all conditions AGr AGr RT 1n IAP IAP F Fe2 AGr AGr RT ln Fe3Fe2 AGr nSE AGr nSE S 2306 kcal V1 NemstEquation E E 0 logU1P no At 25 C E Equot wlogUIP n At equilibrium AGr 0 so E 0 Also at equilibrium IAP K E Z 0059210gK n Example Zn Cu2 lt gt Zn2 Cu EE wlogUIP n 2 2 110 00296log 2 21 Cu l If we fix E we fix the ratio Zn2Cu2 and Vice versa If Zn2f Cu2 10 mol L39l then E E0 l 10 V 86 43 HOW DOES THIS APPLY TO NATURAL WATERS We cannot measure an EMF unless we define a reference halfcell or electrode Define Eh as the EW generated between an electrode in any state and the H2 electrode in the standard state SHE Fe2 lt gt Fe3 e39 H e39 lt gt l2H2 Fe2 H lt gt Fe3 l2H2 For this reaction the Nernst Equation is Fe3 Eh 2E0 0059210 2 11 1 Fe H 3 But by de nition E 0059210g Fe 0ftheSHEpH 2 2 F9 l 1 atm and Ht F 3 1 mol L391 0769005921o 82 Fe ll Eh is an environmental parameter just like pH or temperature and is a characteristic of a given natural water High Eh implies oxidizing conditions Low Eh implies reducing conditions 44 In the above example if we can measure Fe3tFe2 then we can calculate Eh and Vice versa But what controls what Eh is one of the most difficult parameters to measure Redox reactions may not attain equilibrium in natural waters at low temperatures Consider a second redox reaction H3AsO3O H20 lt gt H3AsO4O 2e39 2H 0059 H3As0ffH2 2 H3As030 We don t always get agreement among the Eh values calculated from different redox couples EhE log STABILITY LIMITS OF WATER IN EhpH SPACE Upper limit H200 lt gt 2H 1202g 2e Eh E0 00295 log p0212H2 E0 AGrOnS 5668722306 123 V Eh 123 00148 log pO2 00592 pH At the Earth s surface p02 can be no greater than 1 bar so Eh 123 00592 pH 45 Lower limit 12H2g lt gt H e Eh 0 0059 10g HpH212 Again let pH2 1 bar Eh 0059 pH T 25 C pHZ 1 bar EhpH dlagram P 1 bar de ictin the 1 g 02 p g 20 11m1ts of stablhty of liquid water 1 E LU 0 go He 1 0 2 4 6 8 10 12 14 pH 92 46 Range of ElipH conditions in 0 natural 6 environments 39 based on data of 0 4 BaasBecking et al 39 1960 Jour Geol gt r 68 243284 45 111 A m o oz HOW IS Eh MEASURED Measure the voltage between a Pt indicator electrode and a reference electrode A SHE is not convenient in the eld so we often use a secondary reference called a saturated calomel electrode SCE 2Hg 2Cl39 lt gt HgZCl2 2e39 Then to correct to SHE we need to add 0244 V to the reading obtained with the SCE Dif cult to measure Eh in the eld Electrode itself disturbs equilibrium of system Nonequilibrium among different redox couples EhpH DIAGRAMS 0 Show both redox and acidbase reactions Depict mineral stabilities and solubilities Depict the predominant aqueous species 0 Important for understanding processes in environmental geochemistry eXploration geochemistry corrosion science hydrometallurgy and a variety of other fields THE SYSTEM FeHO Species AGf kcalmol391 Species AGf kcalmol391 H200 56687 Fe3 l12 Fe3O4 24260 Fe2 l885 Fe203 l776 THE SYSTEM FeHO FeFe304 boundary 4H20l 3Fe lt gt Fe3O4 8e39 8H 00592 Eh E IogH8 The activities of water all pure solids and the electron are unity by convention AGr 24260 456687 1585 kcal E AGfnS 1585 kcal82306 kcal V39l 00086 V Eh 00086 00592pH This plots below the water stability line so Fe not stable in the presence of water 97 Fe3O4Fe203 boundary 2Fe3O4 H20 lt gt 3Fe203 2H 2e39 00592 2 Eh E IogH2 AGr 31776 224260 56687 9087 kcal E AGfnS 9087 kcal2 2306 kcal W 020 V 00259210gH2 Eh 02 Eh 02 00592pH 49 Fe3O4Fe2 boundary 3Fe2 4H20 lt gt Fe3O4 8H 2e39 8 Eh E00059210g H2 3 2 Fe A r0 24260 31885 456687 40698 kcal E0 ACTS1 13 40698 kcal22306 kcal V39l 088 V Need to x Fe2 00592 H 8 arbitrarily We Eh 088 Tlog Fe23 choose 10395 mol L39l39 Eh 088 0237pH 0089logEe2 Eh 088 0237pH 0089 51325 0237pH 99 Fe203Fe2 boundary 2Fe2 3H20 lt gt FeZO3 6H 2e 00592 HT 1 2 0g Fe 2 2 AGr 1776 21885 3 56687 3016 kcal EhE E AGfI IS 30161 kca1 3923 06 kcal V39IW 065 V Need to x Fe2 at 00592 H15 10 ie 10395m011391 same value as before Eh 065 2 gFe22 I V Eh 065 0177pH 0059210gEe2 Eh 065 0177pH 00592 5 0946 0177pH 100 Fe203Fe3 boundary F6203 6H lt gt 2Fe3 3H20 AGro 2112 356687 1776 5299 kcal kng AGf 52979ca17 3 8 23025RT 230251987ca1 K 1m01129815K K 2107388 2 F6312 H 16 210g Fey 610g H 388 log Fe3 3pH 194 Need to X Fey at 5 3pH 1 3994 same value as before pH 102 ie 10395 mol L39l Fe3 Fe2 boundary Fe2 lt gt Fe3 e 00592 0gFf 1 F82 AGr 112 1885 1773 kcal E AGfnS 1773 kcal2306 kcal V4 077 V Eh 077 wlog 173 1 Fe By definition this boundary occurs where Fe2 Fe3 so Eh 077 V EhE T 25 C pH 1 bar p02 1 bar 2Fe10395 EhpH diagram for the system FeOH showing the stability and solubility of hematite and M magnetite 706 0 4 6 8 10 12 M 103 THE SYSTEM SHO Species AGf kcalmol391 Species AGf kcalmol391 H201 56687 HSO439 18069 H280 666 804239 17775 HS39 293 8239 2051 SHO SYSTEM HSO439ISO4239 boundary 804239 H lt gt HSO439 AGr 18069 17775 294 cal mol391 AG 29400a1 10gK 7 7 216 23025RT 230251987ca1K 1m01129815K K 210216 HSOH H1115 42 By definition this boundary occurs where SO 4239 HSO439 so pH 216 H2S0HS39 boundary H280 lt gt HS39 H AGr 293 666 959 cal mol391 AG 95900a1 10gK 7 7 7 03 23025RT 230251987ca1 K 1m01129815K K 210403 2 HS If H 25 01 By definition this boundary occurs where H280 HS39 so pH 703 HS39ISZ39 boundary HS39 lt gt 8239 H AGr 2051 293 1758 calmol391 AGf 175800a1 10gK 7 7 1289 23025RT 230251987ca1 K 1m01129815K K 2 1 SZ HHW H57 By definition this boundary occurs where HS39 8239 so pH 1289 HZSOHSO439 boundary H280 4H20 lt gt HSO439 9H 8639 AGr 18069 666 456687 52718 cal mol391 9 r Eh 2E0 0059210g H H504 8 H25 1 E AGfnS 52718 kcal82306 kcal V39l 0286 V 9 r Eh 0286 039059210g H H0504 8 H25 1 Eh 0286 00665 pH H2S0S04239 boundary H280 4H20 lt gt 804239 10H 8639 AGr 17775 666 456687 55658 cal mol391 Eh 2 E0 wlog H1 S00 8 H25 E AGfnS 55658 kcal82306 kcal VI 0302 V wlog H1 HOSO 8 H25 Eh 0302 00739 pH Eh 0302 HS39ISO4239 boundary HS39 4H20lt gt SO42 9H 8639 AGr 17775 293 456687 46068 cal mol391 9 27 E E0 wlogw 8 HS E AGfnS 46068 kcal82306 kcal V39l 0249 V 0059210 H9HSO H57 Eh 0249 00665 pH Eh 0249 S239S04239 boundary 82 4H20lt gt SO42 8H 8e AGr 17775 2051 456687 28488 cal mol 1 wk 150145037 52 E AGfnS 28488 kcal82306 kcal W 0154 V Eh 01544 10g HIf04 Eh 0154 00592 pH EhE H2S0Ss boundary H280 lt gt 85 2H 2639 AGr 666 cal 1 1 101391 wk H112SS1 2 H250 E AGfnS 666 kcal22306 kcal VI 0144 V 2 Eh 0144 0059210g H 1 Q 2 H250 Eh 0144 00592 pH 0029610g H280 set ZS H280 01 mol L391 Eh 0174 00592 pH Eh E 112 HS Ss boundary HS39lt gt 85 H 2639 A r 293 cal mol1 wlo HSS 2 HS E AGfnS 293 kcal22306 kcal VI 00635 V Eh 00635 wlog Ht 2 HS Eh 00635 00296 pH 0029610g HS39 set ZS HS39 01 mol L391 Eh 0034 00296 pH EhE HSO439Ss boundary 85 4H201 lt gt HSO439 7H 6639 AGr 18069 456687 46058 cal mol1 Eh wlogLWW 6 55 E AGfnS 46058 kcal62306 kcal VI 0333 V Eh 0333 logH7HSO l swj I Eh 0333 00691 pH 000987 10g HSO439 set ZS HSO439 01 mol L391 Eh 0323 00691 pH 114 S04239Ss boundary 85 4H201 lt gt SO42 8H 6e A r 17775 456687 48998 cal moi1 8 27 Eh E0 0059210 H SO4 6 55 E AGfnS 48998 kcal62306 kcal W 0354 V o39o sgzlogtHTtSOf a 10 Eh 0354 00789 pH 000987 log 804239 set ZS 804239 01 mol L391 Eh 0344 00789 pH Eh 0354 EhpH diagram for the system SOH showing the solubility of native sulfur THE SYSTEM CrHO Species AGf kcalmol391 Species AGf kcalmol391 H200 56687 CrOH2 1030 HCrO439 18277 CrZO3 s 25289 CrO4239 17394 CrOZ39 1280 CrOH SYSTEM HCrO439Cr04239 HCrO439 lt gt CrO4239 H AGr 17394 18277 883 kcal mol1 lOgK AGf 88300a1 6 23025RT 230251987ca1 K 1 mol 129815K K 107647 C70 H H C r0 By definition this boundary occurs where HCrO439 CrO4239 so pH 647 Cr0H2Cr203s 12Cr203s 2H lt gt CrOH2 12H201 A r 1030 12566871225289 490 kcal 1 1 101391 AGf 49000a1 359 23025RT 230251987ca1K 1m01 129815 K CV0H2 H 2 359 10gCrO 2 2pH CrOH2 ZCr 10396 mol L391 359 6 2pH pH 480 logK K 210359 CrOz39lCr203s 12Cr203s 12H201 lt gt CrOZ39 H AGr 1280 1256687 1225289 2679 kcal 1 1 101391 AG 26790ca1 23025RT 230251987ca1K 1m01 129815K K 2104964 002 H logK 1964 1964 10g CrOZ39 pH CrOZ39 ZCr 10396 mol L391 1964 6 pH pH 1364 60 Cr0H2HCr0439 CrOH2 3H201 lt gt HCrO439 6H 36 A r 18277 1030 356687 90291 kcal mol1 Eh 2E0 00592 10 H HCr0 CVOH2 E AGfnS 90291 kcal32306 kcal W 13052 V Eh 13052 003592 6 7 log H 1 H004 CVOH2 Eh 13052 01184 pH Cr203sHCrO439 12Cr203s 52H201 lt gt HC104 4H 36 AGr 18277 1225289 5256687 85393 kcal mol1 Eh E0 0390359210gH4HCr0 E AGfnS 85393 kcal32306 kcal VI 1234 V Eh 1234 03903 logH4 HCrO4 Eh 1234 00789 pH 0019710gHCrO439 Eh 1116 00789 pH 122 61 Cr203sCr04239 12Cr203s 52H201 lt gt CrO4239 5H 3e A r 17394 1225289 5256687 9422 kcal 1 1 101391 Eh E0 glogmj crof E AGfnS 9422 kcal3 2306 kcal W 1362 V Eh 1362 0390359216gH5Cr0f Eh 1362 00987 pH 0019710g CrO42 Eh 1244 00987 pH 123 CrOz39lCrO4239 CrOZ39 2H201 lt gt C1042 4H 36 AGr 17394 1280 256687 67434 kcal mol1 Eh wlogwr 0 3 C702 E AGfnS 67434 kcal32306 kcal W 0975 V Eh 0975 wlo H4Crof 3 Eh 0975 00789 pH 62 12 T 2500 EhpH diagram for the system Cr OH showing the solubility of Cr203s ACID HYDROLYSIS Optional reading Chapter 12 in Faure 1998 63 ACID HYDROLYSIS Acid hydrolysis Reaction between a mineral and acid weathering agents Example CaCO3 C02g H200 lt gt Ca2 2HCO339 such a reaction neutralizes acid in water Increased PCO2 causes increased calcite solubility Decreased PCO2 causes precipitation This equation is the basis of cave formations e g stalagmites and stalactites HYDROLYSIS OF SILICATES Congruent dissolution no solid products MgZSiO4 forsterite 4H2CO30 lt gt 2Mg2 4HCO339 H4 SiO4O lncongruent dissolution leaves alteration products CaAl2 Si208anorthite 2H2CO3O H200 lt gt Ca2 2HCO339 AlZSi205OH 4kaolinite Thus weathering reactions tend to lead to surface waters with nearneutral pH HCO339 as the major anion and Ca2 and Mg2 as major cations 64 A SINK FOR C02 By itself the precipitation of carbonates is not a sink for C02 CaCO3 C02g H201 lt gt Ca2 2HCO339 However acid hydrolysis reactions of silicates eg MgZSiO4 4C02g 4H201 lt gt 2Mg2 4HCO339 H4Si040 CaAIZSiZOS 2C02g 3H201 lt gt Ca2 2HCO339 A123i2050H4 2NaAlSi3O8 2C02g 11H20l lt gt 2Na 2HCO339 AlZSi205OH4 4H4Si040 Combined with the precipitation of carbonates Ca2 2HCO339 lt gt CaCO3 C02g H201 Ca2 Mg2 4HCO339 lt gt CaMgCO32 2C02g 2H201 result in the net loss of CO2 into the solid phase 129 WEATHERING RATES Read Chapter 15 pp 253256 in Faure 65 CONTROLS ON MINERAL WEATHERTNG RATES 1 Temperature consider the following chemical reaction A B lt gt C The rate of this reaction is given by the equation rate lltAI11 B 2 Arrhenius equation k k0 6XpEaRT Higher temperatures lead to faster chemical reactions A rule of thumb an increase of 10 C in temperature generally leads to a doubling of reaction rate tropical regions Tmeanannual m 20 C 12 C So there will be about a factor of 2 faster weathering rate in tropical as opposed to temperate regions 2 Flow rate High ow rates of water also lead to greater weathering rates because ions are more efficiently flushed from the system temperature regions T N N mean annual dissolution rates 0c 1 Q Thus mineral dissolution rates are much faster when solutions are undersaturated The weathering rate of more soluble minerals is especially dependent on flow rate 132 66 3 Parent bedrock Weathering rates follow a series rough the erals y of original crystallization for 6 g g ohvine E E w E pyroxene a o E m 9 U amphiboie 07 U c g E biome 9 E u w o 0 E Bowen s Reaction Series Ca4e dspar Na4e dspar Meidspar m usfovite quartz nCreasH39vq rate of weatherin m 4 Mechanical factors 39n 39ng and fracturing by mechanical processes leads to increased surface area andincreased reacti on rate 5 Plants and bacteria The in uence ofplants and bacteria can enhance weathering rates by factors of 1001000 times a Increased co2 partial pressure b Organic acids oxalic citric etc c polysaccharides andbio lms SOLID PRODUCTS OF WEATHERING Read Chapter 13 in Faure 1998 SOLID PRODUCTS OF WEATHERING The final stable products of weathering consist of quartz and clay minerals Clay minerals Hydrous sheet silicates phyllosilicates with a grain size lt 4 um Clays are constructed of two major structural components 1 Sheets of SiO4439 tetrahedra sharing three oxygens with neighbors 2 Sheets of Al Fe andor Mg in octahedral coordination with 0239 andor OH39 69 DIOCTAHEDRAL VS TRIOCTAHEDRAL Dioctahedral Only two out of three octahedral sites are occupied by trivalent ions Trioctahedral All three out of three octahedral sites occupied by a divalent ion 11 CLAY MINERALS Serpentine Kaolin Group Kaolinite AlZSi205OH4 I hydrogen bonds tetrahedral sheet 1 Cations cannot get between layers 2 Solid solution is limited 70 21 CLAY MINERALS micas illite smectite chlorite tetrahedral sheet octahedral sheet tetrahedral sheet solid solution is quite common in the 21 clays ILLITE Illite A general term to describe claysize mica type minerals Generally the composition is similar to muscovite One out of four Si4 ions are replaced by Al3 in the tetrahedral sheet This leads to a strong net negative charge Some octahedral Al3 may be replaced by Fe2 and Mg2 which also leads to net negative charge The charge is neutralized by large cations usually Kt in the interlayer spaces 7l ILLITE STRUCTURE Interlayer sites lled with K Strongly 4 bonded so cations cannot easily exchange with K SMECTITE Smectite similar structurally to illite However the 21 units are not as tightly bound Water can penetrate the interlayer sites causing them to swell Cations such as Ht Nat Ca2 and Mg2 also can enter the interlayer sites Thus the weak interlayer bonding makes smectites prone to replacement by other cations This leads to a high cation exchange capacity CEC 72 THE CHEMICAL INDEX OF ALTERATION It is predominantly feldspars that weather to clays We can thus base a measure of the degree of weathering on how far the composition is from that of an ideal feldspar During weathering Al and Fe are insoluble as oxides or oxyhydroxides Other cations and Si are quite soluble CIA J x 100 A1203 Ca0 NaZO K20 The concentrations are in molecular proportions CaO is CaO in silicates excluding that in carbonates and phosphates 0 CIA values of z 100 are typical of heavily leached materials such as topical laterites and bauxites Kaolinite and gibbsite occur in well drained heavily leached soils 0 Smectites form in poorly drained soils 73 Percentage of soil Smeait Kaohmie Gibbsite 2000 Rainfall mm saprolites representing increasing intensity of chemical Weathering of granitic gneisses from Minnesota saprolites shown above From Faure 1997 Figure 191 A Variation of the chemical composition of B Variation of the measured abundances of minerals in the ION EXCHANGE ClayOH K lt gt ClayOK H Clays smectites can hold ions both on their surfaces on their edges and in interlayer sites Clays can be used as adsorbents eg as backfill in nuclear waste repositories Natural clays in groundwater aquifers retard the migration of pollutants by adsorption Clay surfaces may act as catalysts 149 Figure 132 Stability of kaolinite Kmontmorillonite and Na montmorillonite in the presence of amorphous silica From Faure 1998 16 i Kmont 3 quotIquot E 14 T 1 96 Do 124 K Namont 10 l I I f l I I 10 12 14 3916 150 3908 NEHHf 75 Figure 133 Stability of selected minerals in the system KZOMgO AlZO3SiOZHZOHC1 at 25 C and 1 atm in the presence of amorphous silica From Faure 1998 7 8 Mi 10 I 391 microcline lt Ph 5 6 l 8 phlogopite To lr1slivite 2 Mu 2 I illite 7 4 l kaolinite I K Mgmont i G 5 111011 2 I I 39 I j I 1 4 6 8 10 12 10 M82l W12 151 CLAYS IN ENVIRONMENTAL RElVlEDIATION 2378tetrachlorodibenzopdioxin or more simply dioxin Gnome Cl 0 Cl one of the most toxic priority pollutants of EPA nerve poison no safe lower limit of enVironmental contamination Use clays to absorb dioxin then heat to destroy the dioxin Problem smectite clays have high capacity to adsorb dioxin but heating to T gt 200 C causes dehydration of interlayer sites and conversion of smectite to illite 152 76 Need to build pillars in the interlayers to keep the layers from collapsing A pillar is a thermally stable large cation such as A11304OH28339 Use of pillars is advantageous because they I prevent interlayer collapse 2 increase internal surface area of interlayer sites 3 height of pillar can be engineered Also need a surfactant to make clay more favorable for absorbing organic molecules Box 311 Sodium lauryl sulfate polar heart hydrophlhc hydrocarbon tail hydrophobic CH3CH211 O O Na 153 O CHEMISTRY OF CONTINENTAL WATERS Optional reading Chapter 20 in Faure 1998 CHEMISTRY OF CONTINENTAL WATERS The 20 largest rivers on Earth carry 40 of total continental runoff the Amazon alone accounts for 15 of the total These 20 rivers give a good indication of global average river composition There are three major characteristics 7 Four metals dominate all freshwater ie Ca2 Na Kt Mg All are present as simple ions 7 Low overall abundance of ions 7 The composition of freshwater is markedly different from continental crust The relative proportions of Al and Fe to other metals are particularly different Al3 3H20l lt7 AlOH3s Fe3 3H20l lt7 FeOH3s Ions with low I Z I r dissolve as simple ions e g Cl39 Naf Ca2 etc Ions with intermediate I Z I r precipitate as hydroxides or oxides Ions with I Z I r form oxyanions and are soluble e g Mo6 4H201 lt gt M0042 2H 78 river paniculatemncemra on mg gIxma a 9 mar dissolved Concentration Ratio of average elemental riverine particulate to dissolved 39 vs ratio of charge to ionic radius mg H 2 0 06 3 Loglzlll VARIATIONS IN IONIC COMPOSITION OF CONTINENTAL WATER OCCUR FROM PLACE TO PLACE Dissolved ion composition depends on 1 The composition of rainfall and dry deposition 2 Modi cation of atmospheric inputs by evapotranspiration 3 Varying inputs from weathering and organic matter decomposition in soils and rocks 4 Differential uptake in biological processes When weathering inputs are low sea spray dust and anthropogenic gases control freshwater chemistry The Na and Cl contents are a measure of seasalt input Where rainwater input is dommant Na will be the dominan cation Where weathering processes are important the dominant cations will be controlled by rocks but usually Ca2will be dommant Evaporation causes Ca to be removed as CaCOZ so Ca decreasesrelative to Na Bicarbonate HCOZ39 is the dominant anion owing to hydrolysis reactions is ation in the weight ratio NaNa Ca as a function oftotal dissolved so ids andlom strength for m v weathering vs all p atio eontributions to freshwater eomposiaons IONIC STRENGTH I ZciZi2 Ion mmol L391 Ion mmol L391 Nat 10 Li 001 1 HC0339 12 Call 2 304239 15 Mg 05 CI39 1 I l2cNal2 cKl2 Cca23922 cMg23922 cLi12 choj39 Dz cso42393922 cc1393912l 112101ll 24 054 001l 121 154 11 I 2005 mmol L391 ALKALINITY Alkalinig The amount of acid millieq L39l required to neutralize all bases in a water sample usually the amount of acid required to titrate to a pH lt 4 In most waters this can be calculated as a1k1 1HCO31 21CO321 This is an important quantity The higher the alkalinity of a natural water the greater capacity it will have to buffer the pH upon addition of acid Via e g acid rain or acid mine drainage 81 PROBLEMS ENGENDERED BY ACIDIFICATION Increased solubility of toxic metals For example Al is insoluble in most natural waters but solubility increases at extremes of low and pH Al is toxic to fish and other organisms Loss of forests owing to acid leaching of nutrients from soil Acidification a significant problem in poorly buffered freshwaters in the Adirondacks Scandinavia and Scotland For example in Adirondacks in 1930 the median pH of lakes was 67 In 1975 it fell to 51 163 Solubility of ZnO as a function of pH Zr12 Oversaturated ZnOH4239 log EZn J Undersaturated 82 Solubility of FeOH3 as a function of pH 1 c 3 oversaturated FeOH3s E 5 D O C O o 7 U 4 Fe2OH2 FeOH2 FeOH4 9 11 FeOH2 3 2 4 Fe 6 8 10 12 14 pH Solubility of AlOH3 as a function of pH 1 oversaturated 1 AIOH3s C Q E E 8 c 5 O O U o 0 7 AOH3 undersaturated undersaturated 2 4 6 8 10 12 14 83 TOPIC 2 THE ATMOSPHERE AND ITS RELATION TO THE CRUST AND HYDROSPHERE ACCRETION OF EARTH As accretion built up the Earth to its present mass it heated up owm o e radloactlve decay ofunstable isotopes a kinetic energy rom impa s Heating melted Fe and Ni which sank to the center forming the core Subsequent cooling permitted solidi cation of remainderto form mantle ofroughly MgFesilicate composition Crust hydrosphere and atmosphere formed from upper mantle during the early history ofEarth o Formation of the Solar System planets ofsolar system remnants ofsupemoyas ie discshaped clouds ofho gases ACCRETION OF EARTH nllhnmblllon gt aquot Condensing Wpors formed solids that coalesced to form planetesimals small bodies 5 Formation of the Solar System THE CRUST Accretion ofplanetesimals lead to formation of inner planets MercuryV enus EarthMars Large outer planets condensed from gases at muc ower temperatures 39 Crust is shell of volume lt00001 of total th volume a small but important part to 39 Crust evolved through time as incompatible elements Were removed from mantle by partial melting 39 Relative elemental abundances O gt Si gt A1 gt gt Ca gt Na gt M nm lm 7 m n u ammaumamawwm ramme m lmmlmm unmalalammmuawmncw armmmmmwk 7 eat raw wwwmmluwwmmuxmm n wk Camus1M ORIGIN OF LIFE THE ATMOSPHERE SynLhesls ofblologlcally lmportant molecules Volatile elements escaped fmm mantle probably took place m restncted speclallzed envrronments suc as surfaces ofclay mmerals dunng crust formauon e 3 vol Camc or near mbmmne hydro emal vents degassing Some were retained to form s here Llfe probablybegan m oceans 4 273 8B 1 ago butno fosslls oust Earllest fossll bactena N 3 5 Primitive atmosphere probably co2 N2 B Y old with minor H2 and H20vap0r Modern Earhestreactlons to synLheslze organlc matenal atmosphere had to await evolution of life probablybased on s from hydrothermal vean C02g ZHZSg gt CHZO s 235 H200 THE HYDROSPHERE PHOTOSYNTHESIS 35 BY Bulk ofwata at Earih s surface ls m oceans mm and m polar lcercaps and glaclers 01 cozg a CHZOs 02g 39 lt1 n ls Contlnental esh Wata most stored as groundwater Thlsreacuonhad greateffec l 02 ons edb mg surface compounds Source ofwaternotwelllmown andmmaals Event BUBrm r meP1 7 Water buund as OH m meteunte e waternch cumets7 Llfe adapted to use 02 produced whlch IS oLherWlse Loxlc 39 As Earth s surface Cooled w 100 C WHEY Could sedlmentary rocks 0 und 8B 1 condense From ailstence ofold 2 erwentreactlonsto form 03 ozone whlch w now the ocean exlshed by 3 protects the surface fromUV radlatlon THE ATMOSPHERE The smallest of Earth s geological reservoirs particularly susceptible to pollution Mixing time very short so contamination ry quickly spreads throughout the globe but is also dilutedrapidly Bulk composition of atmosphere very similar world wide ATMOSPHERE CHARACTERI STT CS COMPOSITION OF THE ATMOSPHERE Water and co2 are somewhat variable in abundance Most gases are relatively constant in abundance 39 Most attention is focused on reactive trace gases Lower atmosphere troposphere is well mixed by convection AL 15725 loan the atmosphere is heated by absorption ofUV by 02 and 03 stabilizesupper atmosphere stratosphere against mixing This is where ozone layer occurs bove N lzo lon mixing is so weak that gas molecules can separate gavltatlonally 7 H2 and He are enriched at the top 02 and N2 atbottom Definitions 7 Hetemsphere where gavitaannal sepamtlun aeeurs e Humusphere Wellrmlxedpmu atmusphere 7T rbapaise sepamtes heterasphere and hnmnsphere 1m u mm mm mm gun eflhz m From um um Ruiama Lima Coman ppm Cuban aimu you 60 mo cm mud 0391 yet 107 Mm L6 ym 1 6w Fannie Icid 10 any Nnmu am maa m 300 N inc 0 rd I whys oI Nitrogen mom 4 my 01 Amman 2 a I Sulphur dioxide J Tdm norN Hyman unpaid l day was mm mm my m cm su pm I ya u Dimmyl man I 9 am Medin cum 0 m 07 Methyl mm s m nooz Hydrogcn chkrid any um PARTIAL PRESSURE Box 2 3 or3 1 Andrews The total pressure of a mixture ofgases is the sum of the pressures of the individual componenm For example in the atmosphere Ptotal P02 PN2 PCO2 PH2 Pm Dalton s Law Gas chemistry in the Atmosphere BONDING Box 22 or 33 in Andrews et al Lewis dot structures Ar occurs as a monoatomic substance because its valence shell 1s and 3 p orbitals is completely full Ar An oxygen atom has an unhappy electron con guration 39 IO At constant temperature and volume GasA GAsE GASC Pam ram PD5Mm 5 E lt m E 8 Pofnuxlule I um2 mums mas sun o u o QQ Covalent double bond sharing two pairs of electrons O O N N a szzN Covalent triple bond sharing three pairs of electrons Atmospheric gases H20 and organic compounds are typically formed via covalent bonds IDEAL GAS LAW PV nRT P pressure V volume n moles ofgas T absolute temperature K R gas constant 8314 J mol391 K391 1987 cal mol391 K391 00820575 atrn L mol391 K391 I An ideal gas is one whose component molecules occupy no measurable volume and exert no forces of attraction or repulsion on one another I Real gases approach ideality at low pressure and high temperatures GASEOUS MIXTURES PlV anT sz nZRT sz nKRT 131 PZ P3V PTV nl nZ n3RT so P1 X1PT P2 XzPT P3 XKPT The parti pressure ofa gas is equal to the total pressure multiplied by the mole fmction of that gas LAW OF MASS ACTION C C K C D constant CA Le Chatlier s Principle If a system at equilibrium is perturbed the system Will react in such a Way to minimize the change Example Hydrolysis of urea NHZCONHZaq HZOl lt gt 2NH3g COZg a12VH CO amx comx YNH comx CNH comx K 4 z z z z z z NHZCONHzaHZO a 20 m 1 115mm 22 BAROMETRIC LAW PZ PnexpzH PZ pressure at altitude 2 PD pressure at ground level H scale height m 84 km The barometric law gives the pressure of a gas as a mction of altitude Dalton s law implies n2 n expltzHgt partial pressures and number of moles are related 25 For gases We have aNH3 1quotNi13Pm13 acoZ 1quotcozpcoz In dilute solutions 7 m l and at loW pressures Where gases behave ideally lquot m 1 so 2 PNHZPCOZ CNHZCONHZ K A more geneml form aA bB gt CC dD CHEMICAL EQUILIBRIUIVI Box 24 or 32 Andrews Sections 91 and 92 Faure Consider the reaction A B gt C D At equilibrium it must be true that ac a constant A 15 Where aA yACA aB yBCB etc In geneml a y C In dilute solutions 7 m 1 so a m c K Steady State or Equilibrium Steady state in ux out ux Equilibrium rate of forward reaction equals rate of reverse reaction STEADY STATE OR EQUILIBRIUM HOW MUCH METHANE SHOULD BE N E ATMOSPHERE How much cm should be present In the atmosphere lthls trace gas ls In equlllbrlum wlth the more abundant gases7 CH4g 202 lt gt C02g 2H20g P P2 a STEADY STATE CALCULATIONS FOR METHANE Fmc14 500 Tgyr noteTera 10 26 0el4 gyr Total mass of atmosphere 5 2x1018 kg K 7 01 a1 7 10m Total mass ofCH417 mollo6 mol5 2x102 lo29 Pm a 4 g A Thls suggeststhat cm pressure should be very low 7 7 43x1015g 79 75 but exactly how low7 7E750X10nW W3 a o 21 atm PCO2 a 3 6x104 atm PHp a o 01 atm WHAT IS THE MEANING OF RESIDENCE TIME I x104 012 PM 0212 mm 3 PCHA 8x1 0 47 atm But the pamal pressure ofmethane actually measured 15 1 7x106 atm much larger than that expected at equlllbnum What s Wrong7 clearly trace gases m the atmosphere may not be m emlcal equlllbnum wth other components m the osphere The resldence tlme IS a measure othe average length oftlme an lndlvldual molecule stays h the atmosphere assumlng atmosp ere lswell mlxed Very lmportantcharacterlstlc descrlblng steady state systems In general compounds wth hlgn I tend to bulld o s E eyen th u wlth e raprdly ths hlgh reactryrty may ltselfbe a cause for concern 0 no r E e o 1 amp STEADY STATE A steady state 15 a srtuatron ln whlch the sources and Slnks ofa component In the atmosphere arejust In s Approprlate analogy ls a leaky bucket A F F m our 139 F ux ofcomponent ln or outmass lee 9 unt of component In atmosphere I 2 resldence trme m the atmosphere Alsuluwtlmplles lnsuf nent ame furmlxmgsu e memes WthluWI shnuldhe qulte yanahle u I II 39 v A I a a 39 LI I quotan Raiders rim Curm iun ppb Carbon dioxide 4 ya 350000 Cum mm 01 m m Madame 36 y s 6m Fermi acid in days Nitrous oxide 21130 year 300 Nitric oxide I dun DJ Nitrogen dioxide 6 dlys 03 Alumni 2 d1 1 Sulphur dioxide 37 days Mil0 Hydrogen sulphide 1 my ans WWW 439 m 11 Ablation fro Cuhonyl sulphide 1 ya 05 Dimclhyl sulphide 1a mm Important for m t s Sm Medulcblo d may 07 important in lowdensity upper Muhyl iodide s dqs mm quotidiom a a mu atmosphere ATMOSPHERIC BOX MODEL Reservoir S nk gt gt 1 atmosphere I sunhigh GasesfS02 602 HO Acid39raih a serious problem on the The mam goal IS to 1dent1fy sources and smks of Of39some volcanoes in Hawaii components 1n the atmosphere as well as reactlons among components in the atmosphere Volcanlc sources are very dlscontlnuous BOX 2 5 ANDREWS ET AL 1996 SECTION 9 3 FAURE 1998 I Windblown dusts and sea sprays proba Bronsted de nition 1m Omar 7 Acid any substance that can donate a proton 7 Base any substance that can accept a proton HClaq lt gt H4r Cl39 NaOHaq H4r lt gt NaJr H200 0 Lewis de nition 7 Acid any substance that can accept electrons 7 Base any substance that can donate electrons H3BO3aq OH39 gt BOH439 H200 lt gt H4r OH39 H3BO3aq H200 lt gt BOH439 H4r 42 ACIDS AND BASES CONTINUED I Acids and bases react together to form salts and Water 2KOHaq HZSOAaq lt gt KZSOAacJ HZO1 I The strength of an acid or base is de ned as the degree to which it dissociates Strong acids and bases HC1aq gt H Cl39 NaOHaq lt gt Na OH39 C C C C KA H 0 gtgt1 K N 0 gtgt1 swung cNaOHUzq u ACIDS AND BASES CONTINUED Weak acids and bases F 9 H F39 BOH439 lt gt BOH3aq OH39 K CH CF 1072 2 KB 850H3aqco 1074 8 Emma 55mg It is sometimes convenient to use the notation pKX pKX Jog Ky pKA 10g K 13KB 40 KB A small pKA gt strong acid large pKA gt Weak acid N


Buy Material

Are you sure you want to buy this material for

25 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Bentley McCaw University of Florida

"I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Amaris Trozzo George Washington University

"I made $350 in just two days after posting my first study guide."

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."


"Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.