Thermochemistry of Geological Processes
Thermochemistry of Geological Processes GEOL 555
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I TABULATED TOPIC 2 THERMODYNAMIC PROPERTIES THERMODYNAMICS or SIMPLE SYSTEMS FORMATION FROM THE A PROBLEM ELEMENTS Thermodynamic parameters such as U H G A etc are of clear interest to us We need to know how these are measured and where to get values for them 0 However we face the problem that we cannot determine absolute values of these parameters we can only determine differences For every pure compound We tabulate the difference between the absolute G or H values of the compound and the sum of the absolute G or H values of its constituent elements In reality We determine AG orAH for the reaction in which the compound is formed from its elements in their stable states These differences can be determined experimentally even though the absolute values cannot GIBBS FREE ENERGY OF FORMATION OF WATER FROM THE ELEMENTS The Gibbs free energy of formation of liquid Water from the elements is given by a 7 a a 1 a H200 7 6112007 GH 73602 which is just the free energy of reaction A39GD for the reaction HZ 12OZ lt gt HZO1 The subscript f denotes formation from the elements and the superscript denotes standard conditions of some kind usually 1 bar and pure substances at any speci ed tempemture For the free energy of formation of ice from the elements ie HZ 1ZOZ lt gt HZOs We Write a 7 a a a AGHpltgt GHzoc T GH T zLGoz Note that if We Write a balanced chemical reaction then all the G terms for the elements cancel For example for the melting of ice HZOs lt gt HZO1 We Would Write a 7 a a ArG AGHzo T AGH20 G22017 G212 TTng T Gilligan T G22 TiTGa a a T GH201 T GH205 EFFECT OF PHASE CHANGES Free energies and enthalpies of formation from the elements are based on the elements in their most stable forms but this form may change as a function of tempemture and pressure For example consider the free energy and enthalpy of formation of NaCls from the elements These refer to the reaction Na12ClZ lt gt NaCl From 298 K to 1700 K Clzg is the stable form but Na melts at 37098 and boils at 1175 Thus the most stable reference phase for Na changes Moreover NaCl melts at 10738 K 7 ENTHALPY OF FORMATION OF NaCl VS TEMPERAT 736 New kJmol ronn run 2n tun EDD Bun mun nun Mun tun TM 3 GIBBS FREE ENERGY OF FORMATION OF NaCl VS TEMPERATURE b p Na fa m p NaCl Ave Nam KJmol jzu mm mm sun auu mun 12m Nun 15m TM TABULATION OF ENTROPY Entropy is different from enthalpy and free energy absolute as opposed to relative values can be tabulated This is a result of the thde law As shown above entropy at 0 K is zero and entropy at a higher temperature is given by T s I dT u Entropies so computed are called ThirdLavv absolute or conventional entropies Entropies of formation AIS can also be calculated and are used to calculate AG from enthalpy and entropy data but are usually not tabulated m II TEMPERATURE AND PRESSURE DEPENDENCE AG AND AH AT ELEVATED T AND P Standard free energies and enthalpies of formation Consider the compound AZB formed from the elements according to 2A B gt AZB The standard free energy and enthalpy at elevated T and P are de ned as follows AGquot G 726 0 f A181 14151 ALP 65 12 AHZETF H315 ZHXw HEN AFFARENT FREE ENERGIES AND ENTHALFIES AT T A Incdzuh rlglmvdnd ee ust ma mums Jams expmdzdmwmbmgk hug AndamlyhxenMPyve 1m A inraznrwm 7H1 mAa human m me my 50mm m Hump kseynpmsmekvdede a anywmmm Him mi Am smqnni m m GM 5 emYENEEstIklgesm a me am seenby mwn rvgk Miami r mm M GIG39AG 72 GHGG AGL r AG quot716376 Hanks m nmmm Dru1mm Mm 9mm Emmy Hlkum Whemth Ms rm was h zypmmehbs am 9192 Emma hammnamm Wm A nmm rm 1 v gquot m Wmm dbecmekycmdm Alarm5mmgiwmn b gamma F x mm a mm quwa a an hummer mm mum mmmmmm Am Murmur r Hymn Melmcomn mmmkmm W m hquot th y mammame rm mmm A 39 W mmem mcmumdwm manmaquotEmurm odmwin mint 4 mwm melmvm39 TWO U SEFUL FUNC TIONS mummmn 1k thbsuwzv lmcmmmw Mmr to rr gamer ng mm 3911 mmmmm m hmnfmmk mm man mwa AH VAH mm 7 Ann Man quotm bekmwnmmeduse Mom mym w yuhxdmd fut durum 45o ma cmbe used m comm 59m 5mm m Aypmmympmus CONVERSION FROM STANDARD TO AFFARENT GIBBS FREE ENERGY FORMATION FROM THE OXIDES CALORIMETRY Wm M km E Monmzm mmi39 o dim Wm DH 1 mm MOMMA m mm 3911 m away Ca mm mm ME 1m m Momma mmmm Mm ukthymc 11an mm Ema rm gamma m 3911 Exam 41 6 cmzr Tine mm ade39ze ngemn ysmallu AG m7 A memswmwmrmnmm Mm dunquot m commm SOLU TION CALORIMETRY MAJOR TYPES OF CALORIMETRY Owen m k runhexde m mm d solves 2nd mm mm 3911 m 39 5mm Edmmeky mammth cornymmka m Dudes Hrmmm w d solvewgweksmsohmmcunwosmnn Mummeme mad mu chm Dmp maan mm mm schemf m cmgm mummy 7cumyuxun durum suhmnn39a gummy m1 avmam cmnyummelzmms 9mm AHA AH AHvAHfAr HF SOLUTION CALORIMETRY Rarely as simple as the tworeaction scheme outlined At least some of the component elements cannot be placed directly in calorimeter r en must be added in form of oxides 7 Na and K are volatile so must be added as KCl and NaCl Heats of reaction involving gaseous components must be done in abomb calorimeter Chloride salts ionize KCllt gt K Cl39 Determination ofAH for a single silicate may require adding together many AH s for many different re actions HIGHTEMPERATURE OXIDE CAL ORIMETR Y I Usedin the tempemture mnge 600900 C I Involves using fused salts such as 2PbOBZO3 or 3NazO4MoO3 as solvents I Readily dissolve refractory oxides and silicates I Heats of formation from the oxides canbe determined directly I Heats of formation from elements rarely measured by this method I Simpler but less precise than HF calorimetry DROP CALORIMETRY I Amount of heat required to change the temperature of a substance from Tr to T is simply I Determined by cooling the substance from T to Tr in a calorimeter and measuring the heat evolved I Calorimeter is placed directly under a lich and sample is dropped from furnace at T into calorimeter at Tr STANDARD HEAT OF FORMATION OF HEMATITE AT ELEVATED T USING HEAT CONTENTS For any substance AfH AfH AfH 7H For hematite 2Fes 3202g lt gt FeZOKs AHi Hint Hi HZMQ 2Hi HSJFE H2 H2 0 AfHi Hing HMO ZHi ZHioz H2121Q T 2111212 yzHig AfH AfH AfH7HyFEZOZ N 112203 T 12203 DETERMINATION OF HEAT CAPACITY HT Hr is determined as a function of tempemture and then t to the equation HT HTr ABT CTZ DTquot The heat capacity is derived according to aHhH cpz T T B2c39erT392 3139 T P this is usually expressed as the MaierKelley equation CF abT cT39Z STANDARD HEAT OF FORMATION OF HEMATITE AT ELEVATED T USING CP Consider the formation of hematite from its elements 2Fes 3202g lt gt FeZOKs AHha H2011 2H2 XHQ CpU EzOz and bFame Ema r72 CPFe ah bF39T 7 CF39T392 cp02 aq may gain Acp AaAbT rAci39z AFFARENT HEAT OF FORMATION Mam n ma AT ELEVATED TEMPERATURES w AHM HMFZHn TH39VA wan Immrurar mmmayngwqm AH rm AH m HMO 4 v m 1 i y I m admmge pr PWEMS vs sunkd Farms 5 Am rm Am 7 7 I cmmamh mmm gnu myan39y mm mm range acumen m Agni Dru x Megn nnllmmegmdlh m ms luseq nudnsumlmT m r 7 39o n 5 muduedwxrgmdzcm Tum MVIWI2M Mmmmm mcpmma T mdzum5mzwmmazrdexmwhmd FO A TA EEJEAW mm m mm w v nduenhvpyxscdmhmdfm IdAH IACdoAVHoIAC HY Xr CI 9 om mm 1115quot ENTROFIES AB ovE 298 K mm m mth cm s mam VH dzsmhedbefme gxmssmms W Amway w u ENTROPIES ABOVE 298 K FROM CP This method is genemlly preferred by Earth scientists It is more amenable to computer use as dT cgT guysquot AycgTW W T 7 My 7 jdr Acp AaAbT 7mm 7 o o T Ava AVE 45745 JTAyb7dT o o T A 1 1 AST7AST AalnT YAbT7TTF7T Y2 7 APPARENT ENTROPY OF FORMATION The apparent entropy of formation may be calculated om T C Ans 7 As Tj Tidr For the hematite example this becomes Ansipeaz ASF903 aw m0 y CF20 l I TiT lt 2 T2 CALCULATION OF GIBBS FREE ENERGIES OF FORMATION FROM THE GIAUQUE FUNCTION Standard free energies of formation from the elements at the reference temperature are given by A G A HaiTA S and at higher G7HI f AfG AfH 7TAfL EJ CALCULATION OF APPAREN T GIBBS FREE ENERGIES OF FORMATION FROM CP For any compound 139 We may write 7 86quot Ag AGga 8T 7 T 7L6 j7sdr T The integration in the last term is performed by parts 7 Co judv7uv7jvdu where u dT and vT r This results in T T AaG Afagt 75T7T joginf l L Next We substitute ffcquot dTaT7T b T17TZc i7i T P t v 2 y T TY ic Z dr a1nfmbf 7 T C L Li r n 1 K VUIIBLLIUII Ul Lne term ieau Lo 0 o o T AaGn AGn st1quot 1 a T Ty Th F y 2 Z 2ITYAT17TZC39 T T 22 77739 This can of course be adapted as follows for the LdlLUJdLlUIl UlurU JUJ an chLUUJI AG AGE 7 Asia 7 Ty Aya T7 Ty 7 Th1 y 2 2 2miTziTyzA CT T272TT o 077 APPROXIMATIONS FOR CALCULATING TDEPENDENCE ACP constant AH AH A013 T 7 I As As Acgvz 111 These expressions are rarely used More common is the assumption that ACF m 0 which results in LG AH 7 TA Z This approximation is o en good if only solids are involved but usually fails misembly if gases liquids or aqueous species appear in the reaction a PROPERTIES OF SOLIDS AT ELEVATED TEMPERATURE AND PRESSURE For conditions of elevated temperature and pressure additional terms must be added to calculate the apparent Gibbs free energy of a compound 7 r o 0 BGquot aG A Gm LG BT dT 81 7 pp p r r The last inteng can only conveniently be evaluated for solids Gases and liquids are separately as We shall see later 011 THERMAL EWANSIVITY AND COMPRESSIBILITY The derivative ofG With respect to P is V ie e 6P 7 To integrate this expression We need to know how V varies as a function ofP and T Isobaric thermal expansivity a 1 r Isothermal compressibility 3 2 7 r EVALUATION OF PRESSURE L For solids values of on and are relatively small and it can be assumed that V is independent of T and P over modemte mnges F we dPGrGF 47071 The equation for the apparent Gibbs free energy then hPr nmP AEG AG 73 Ti 7aTi T iTlniii 39 39 39 Ty 2 2 2T17T27T3C T M 2m VT ltP71gt Z ZTT m WHAT IF V IS NOT INDEPENDENT OF TEMPERATURE AND PRESSURE Berman 1988 has t volume data for rockforming minerals to the expression r m Using this expression the pressure integral then becomes 1 WP IV 1mm 12 v2P7P2 VT z V T TyfldP n V Matamamav3ltT7rgtvtltT7rgt2 III HEAT CAPACITY THEORY OF HEAT CAPACITY As we 1 5m duedexrnmzmal mmnmm 5 3914 Pmsme arm czpnmzs x1 and am m 120mb Hutczyx y ms Banning aha yardde enyhmccmdwmmzlmazvas bemoan mm In m c m m Mm an m mum no my m Alwekayexzms RULE OF DULONG AND FETIT 1819 mmwmmm m mm gum Emma yfmrd mbezywuxmmly m mmmmmm mm He mm m m m mm mm T as Mm mum x E m whan eagle urinednmslrmdd mm x mz w sukwmlgzrdkmcmm 514mme 3RS 2K mol SOLID ELEMENTS AT LOW DEE YE EQUA TION 71 Debye EmmdvunKanmnslvued m away Majkamuymgwlwwmm Y H a mC a k Futon my mums c mink 3R 5mm my dues WNW daperdeme as n u AsT T x u e iu c z 7 D X My 7 ALB x0 WHY More mammm m mmsmgmmsmwmw bmmxymmmmm DEE YE EQUATION r u A mm T 3911 mm lm nnslmyh es m 1275 T A A c 76mm m exw sm 5 used m emzwlm 6mm am m T MW 3914 exyenmnh y mm Fig mama Mlzw mum mm mums WWW Mm mums maxi Dunmwmnmm 16 Had y Dumas mammgm m m 11 121mm cK Emmdsdmkmg mum given THE HEAT CAPACITY cal K391 molquot AND THE DEB YE 0D FOR SEVERAL ELEMENTS Pb Au Ag Cu Fe Al Diamond Eb 90 3 169 213 313 417 389 1890 CP25K 336 123 075 023 010 011 000 C 100 K 5 84 5 10 482 385 2 88 3 12 006 CP298K 641 6 609 585 597 582 145 COMPLEX SOLIDS I I Kopp sRule 1964 The heat capacity ofa solid compound is equal to the sum of the heat capacities of its constituent elements quite reliable at T s Where Dulong amp Petit rule applies Example Estimate CFVN ZS from CPS and VN CFVS 570 102x10 3T 106x105T39Z CFVN 590 205x10393T CFNizsi 2CFNi Casi CP N S 175 512x10393T 106x105T39Z v z The accuracy ofKopp s rule may be improved by using similar substances same formula andor structural class instead of the elements Example 1 We Would like CFRbZS 04 but We only know cKZ s04 CFRbC104 and CFKC104 CFRbZSOA m CPKZSOA 2CFRbC104 2CFKC104 3108 22466 22396 3248 cal Kquot molquot Example 2 We Want vaakermamte but We know vagemmne CaZMgSiZO7 oLAlZO3 gt CaZ AlZSiO7 MgO SiOZ akermanite corundum gehlenite periclase CEaker mamte CF gehlemte CFquartz cppenclase 39 cpcumndum 57 Example 3 Estimate CF of muscovite KAlKSiKOmOHZ at 25 C and lbar By Kopp s rule using oxides CF mus 12cpv K20 32cvA1203 3cmOZ CFHzo CF mus 122010 321890 31063 957 CE mus 7986 cal Kquot molquot By assumingArCP 0 for the following reaction muscovite quartz lt gt pyrophyllite 12AlZO3 12190 CF mm cFpyr12cFK o 12CFA1203 39 CES 02 CF mus 7023 122010 121890 1063 CPV mus 791 cal Kquot molquot The actual experimentally determined value is CH mus 779 cal Kquot molquot Kopp s rule using oxides results in CFV mus too high by 25 The second approach is too high by only 15 I Electronic contribution Quantum theory predicts a contribution by electrons to the heat capacity Cdmmm YT Becomes important at both loW and high T ELECTRONIC CONTRIBUTION TO HEAT CAPACITY cal molquot Kquot At 2 K At 30 K 7x104 C21 Cm C31 Cm Cu 180 000036 000045 0005 0412 A1 346 000070 000075 0010 0203 COMPLEX SOLIDS V II Mum cmmbncnxs m Mayach my mmm m mx gws makmk nmnmvfmlgmmmm DWIm rummla Aledluws 4 mm Tmm Ad ma mm do m A mm cmkkaby n ixms mm M a y mi mm hum mm m WWII my wry ESTIMATION OF HEAT CAPACITIES S AND LIQUIDS M mMgms mumnmmamImmmm gawmymmkmgmgmm cv m a c m cpzcm INN 4m mm maddmmd anW m works A mI cm thmmmmp Dam 44 gases madan the dean m Inmme AI Mann wa suchdrul 3914 whammy m Myexcmdwegu CV vamp m m m sun mum I Vibucnns dB mu39ollml Hank39s szexidly Admme J Pulyzmmc 44 gases mysmmwmmwaemm mammmxmmgwmrsmm mmwaemmnmammmhm mm c 74o jaw c xz 24w Lmymk MIAMI me pnlyummm C ZR 9 sz Nanhnearpnlyamxmc CV7 3R 3mm Fax ml gases nlhghpusslxe anle umpenlu39e we mug rely an Exp1xan measlxemzn39s HEAT CAPACITIES OF LIQUIDS Very Mmhmcdmhmdmdy Usua yxns W q ltc gtc m1 CAPACHB um mx 7 m mums cumm WHHVALUB m maxmm Mme I sun a a s n l a m 2 s E s 7 cu 7 a an 7 y m 7 y 7 6 HEAT CAPACITIES I K m1 FOR LIQUIDS COMPARED WITH VALUES FOR THE soLID Meter HEAT CAPACITY OF voLA TILE m oImLm nmzlly lgdwnhk mausea Plus hndso In mm mm aft MAM 5 c xdzss k VAPDIPxessu 5 gemme m dam c cl Wham k exadxehmzmshlybeween c ma c m m A A 39 siiJ E c a a a 51 1511 WI WA 3 a Substituting We obtain C ac V W V 6P m P or P or m Cr crntT 2 0T P 8T m ct 7cm mg For Water at 373 K CF Csat 00046 cal K39l mol39l or 003 of CF At low temperature the last term in the above equation is negligible but it increases mpidly With increasing tempemture mi 73 IV MANIPULATING THERMODYNAMIC EQUATIONS SOME PRELIMINARY WARMUPS We have already derived the master equation for ermodynamics to be dU TdS PdV We may express any given exact differential dz as function of two other variables x and y such 39 dz dx dy 3quot y 3y x We may thus Write for dU dU j ds j W as V aV Thus Urnay be consideredto be a simple function ofS and V a1one Moreover comparison of the above equation with the Master Equation s owst at BU T and 7 a V P BS V S This shows that both Tand Pcan be considered gradients or potentials Now if we divide the Master Equation through by dV at constant Twe obtain 6V T 6S V W T W S 6V do EULER S THEOREM For any exact differential We can Write Bz Bz dz 5de a or dz ax ydx bg ydy It must be true that 1 1 Kax y Kayak which can also be 39 MAXWELL S RELATIONS I Applying Euler s theorem to the relation dU TdS PdV We obtain the rst A V Applying Euler s theorem to the relation dH TdS V MAXWELL S RELATIONS II Applying Euler s theorem to the relation dA SdT PdV we obtain the third Maxwell relation Applying Euler s theorem to the relation dG SdT VdP we obtain the fourth Maxwell rel 39on A THERMODYNAMIC EQUATION OF STATE Starting with the equation we already derive d 11 T j 7 P 8V 7 8V T and substituting the third Maxwell relation BS Ef y we obtain athermodynamic equation of state tenet APPLICATION DEMONSTRATE THAT U IS NOT AFUNCTION OF VOLUME FOR AN lDEALGAS inRT 7 6 ll a T 6T V E an a l V lT7PP7PO APPLICATION WHAT IS THE GENERAL RELATIONSHIP BETWEEN CP AND CV CT a T M E in i 8T T 8T T 8T T 8T T dU j drafts Z d7CVdT W or T 6V T 6V T New divide through by dT at constant pressure to get at ic 6U 617 or T V 617 T or T E cTTP 8T T or T r C C 6U 617 an F V 6V T or T or T C C i 6 WI P V 6V T or T But we have previously derived the expression BU aP T 71 0V T 8T T Substituting we obtain a in rtf JJTJIZ fl T lgl r According to the chain rule for partial derivatives we have 617 or 617 5 l5le 8P BP 017 a a a rip 0V or iota We now de ne the following 1 617 1 617 u2 27 V or T V a 2 2 2 7CTTTI7 1W 13 r APPLICATION WHAT IS THE RELATIONSHIP BETWEEN CP AND CVFOR AN IDEAL GAS CP7CVTI7 thaw 4PM 1 17 aTP I7 aT P PV T 71617 771 BQITP fixr p I 7al R an 7 aP T P 6P 2 7i 7 7 an 71 122 1 TVP CP7CVTV TV p l T2 P P R T Note that We used this expression earlier in our calculation of the Work involved in the reversible adiabatic expansion of an ideal gas APPLICATION DERIVE AN EXPRESSION FOR THE JOULETHOMSON COEFFICIENT IN TERMS OF MORE READILY AVAILABLE THERMODYNAMIC VARIABLES aT Mn aPH dH dP E or a T aT P aH dH dP CPdT 0 Contant enthalpy 3 T process Divide through by dPto get 8H 8T Cp 0 BF T aP H 7 l aH Mn C a Now we must calculate the partial differential on the right starting with H TdS VdP Dividing through by dP at constant Twe get 6H BS T V an 7 an 7 It seems like we substituted on partial for another but we can now apply the Maxwell relation IEHQ rr Now what is the JouleThomson coef cient for an ideal gas We already derived for an ideal gas L T 70706017 71iTI7 17117 0 CF CF CF So an ideal gas should exhibit no JouleThompson effect a Mn APPLICATION WHAT IS THE TEMPERATURE CHANGE DURING ISENTROPIC DECOMPRESSION Isentmpic adiabatic reversible The relevant partial differential is 5 r as as Manatee Divide through by dPto obtain as aT as 0 ar P an 5 an 7 as ar 7 as aT P a A aP T in an A 6P 7 as F We have already shown that 3 5 7 BP T Also we lmow that as CFTMT so as C GEOLOGIC CALCULATIONS How Will tempemture vary as a function of pressure for isenthalpic vs isentropic expansion of solid mantle Assume pure forsterite on 2x105 Kquot I7 029 cng cF 03 cal g391K39l At T 300 K E 015KKb 723KKb a A a H At T 1300 K 6T 6T 5106KKb 5 725KKb Meltsatl300K E 22KK1 732KKb 6P 5 8P H WateratZKb 600K 6T 100KKb ZSKKb 6P 5 BP H APPLICATION WHAT IS THE SLOPE OF AN EQUlLlBRIUM BOUNDARY ON A PRESSURE TEMPERATURE PHASE DIAGRAM dG VdP SdT At an equilibrium boundary it must be true that 0G VdP SdT 0 VdP SdT 7 i Clapeyron if G T 7 Equation Very use Jl expression relating the slope of a boundary on a phase diagram With volume and entropy changes of the reaction IACOBIAN ALGEBRA Jacobian A 2x2 determinant Whose elements are panial derivatives 1 11 amp y l l M i i l 1414 Rulel Jkxyyro Rule 2 Interchanging the order of any two variables changes the sign of the Jacobian u Rule 3 Any partial derivative can be expressed as a Jacobian JKyyy ax z xz Rule4 Tia31 Jyz Bx xz JUJ JK AW Rule 5 Jx ydzJyzdx Jzxdy 0 Rule 6 JxyJzw JyzJxwJzxJyw0 n USING THE SHAW TABLES 6G I Find the following in terms of TV V aG 7JG7Sa7b if V JTV yiupf 7 JP17 7 BF a JVr11rV l 11 Find u as a function of V and T using the Shaw tables ar 1011 r1471 1 171 M H 5 JPH Tc 5 Tc 6TH JTP 1 71 17 unit CF 7m Vin I7 quot751 c 1 c JST 71 J l 7 CH RT CH Mn 6V Using one ofMaxwell s T V relations we get P 111 Find the following in terms ofP and T using the Shaw Tables 6quot c H M H 839q JqH 4175 if H JTH TbrVZ Easiertowork ilrrlbi Withtheinverse qu I71 Tc T1171 CH 7Tb TbiV 611 1ch CH C VCH VI 1 T E H7 8739 8P B39q H 17 v g Q 6V 7 17 1 7 5T 6394 H 17C ar rTocV7 170700 a39q H CP 7GP IV Find the following in terms of P and T using the Shaw Tables 6P 7 EU JUT 7 7TbrPa77 b a r Tp ap JPT 1 1 1 all 77 710413 JVT Elf 71mm Jinan CALCULATION OF A SIMPLE PHASE DIAGRAM I The simplest phase diagram involves a single component that can exist in more than one crystalline orm eg a mineml having two or more polymorphs One example of this is CaCOK which can exist as calcite or amgonite Consider the reaction CaCOKaragonite lt gt CaCOKcalcite I These two phases are in equilibrium with each other when ArG 0 I An equilibrium phase boundary in say PT space is the locus of all points where ArG 0 m APPL YING THE CLAPEYRON EQUATION We derivedthe Clapeyron equation earlier to be dP 7 AS dT 417 To apply this we need to know the following information V calm 36934 cm3 molquot V magmas 34150 cm3 molquot S 298V1calcite 2215 cal K39l molquot S 298V1amgonite 2156 cal Kl molquot To make the units work out we have to convert volume from units of cm3 mol39l to units of cal bar391 molquot The conversion factor is 1 cal barl mol39l 41839 cm3 molquot For the reaction CaCOKamgonite lt gt CaCOKcalcite AIS 2215 2156 059 cal K391mol391 ArV 36934 341504l839 00665 cal bar1 mol 1 dPdT 059 calK39l mol39100665 cal barl molquot 8867 bar K39l So this yields apositive slope of the equilibrium boundary on aPT phase diagram at tempemtures and pressures near STP Calcite is the lowpressure phase because it occupies the largest molar volume Calcite is also the 39g temperature phase because it is the higher entropy e So schematically the PT diagram looks like ms AG 0 aragonite AG gt o calcite AG lt o T constant G 1m calcite HOW DO WE ACTUALLY CALCULATE THE BOUNDARY We have to determine a suf cient number of PT conditions Where AG 0 that We can de ne the boundary We must make use of the equation 7 o r AG 7 Ami J MEAT dT 340 81 011 7 F7 p 777 AGT7AGT ASTT TA T T Tlnr Ab AcT2T272TT 42TT7T27T2 39 39 39 AV P71 2 m ltmgt But because at the boundary the tWo phases are in equilibrium the following 1s true 0 A3 7 A3 T7 T AaT7 T 7T1nT1H Ab AcT2T272TT 42TT7T27T2 39 39 39 AV P71 2 m lt gt T 7AV P71AG 7As T7T A T4 7T1 a nKTJ 2 2 2TT 7T2 7T1J HC T T39 HT39 y m 2 v ArG 298 AtG 298ca1me 39 AtG 298 aragumte 270100 269875 225 Cal 1011 AS m 2215 2156 059 cal Kquot molquot ArV m 36934 3415041839 00665 cal barquot rnolquot A a aalum amg mtg 2498 2013 485 cal K39l rnol39l b bcme 713mm 524x103 71024x103 501x10 3 calK39Z rnol391 Arc ccame cmg mtg 620x10S 334x105 286x105 calK391mo139l 7 00665P71 7225 7 059T7 29815 485T7298157Tln T i L 29815 j 7 73 5012gtlt10 2T298157T27298132 86 x105 T2 298152 72T298 15 2T298152 Temperature c 100 00 1561315 16000 12000 aragonite Pbars 8000 calcite 4000 T C TOPIC 7 GEOTI IERMOMETRY GEOBAROMETRY BASIC PRINCIPLE OF GEOTHERMOBAROMETRY Consider the reaction 2CaMgSiZO cpx FeZSiOAo1 lt gt 2CaFeSiZO cpx Mg2 SiOAol We can calculate ArG ZAEG CaFeSlzo AfG MngiO4 39ZAtG CaMgSizo AfG FeZSi04 cpxz 11 A Go iRT an iRT 1n 6152506 39aMg2amp04 5px a1 aCaMngzo Fgmu 2 A39G T 1bar P 7 my ART In K This equation tells us that In K is a mction of pressure and tempemture Thus if We had a rock With clinopyroxene and olivine which We can assume Were in equilibrium We could calculate In K from a microprobe analysis and an activity model Given anindependent estimate of P We could calculate T or vice versa If the above equation Was relatively independent of pressure We could use it as a geo ermometer even if We did not have an independent estimate of pressure Similarly if the reaction is relatively T independent We might have a geobarometer CRITERIA FOR GEOTHERMOMETERS AND GEOBAROMETERS The ideal geothermometerwould depend stroneg on T and Weakly on P The ideal geobarometer Would depend strongly on P and We The temperature dependence is given by 8011 K 7 AH M F 7 W and the pressure dependence is given by aln K 7 AVquot T RT T 6 Thus a good geothermometer is one in which the absolute value ofArH is relatively high and AV is near zero Conversely a good geobarometer is one in which All is nearly zero and the absolute value of AV is relatively high Potential geothamometa Potential geobarometer K 10 5 2 K 1C P p K 5 K 2 OTHER CRITERIA 1 Reasonably accurate standardstate thermodynamic data are either known or can be estimated from simplesystem experiments 2 The relationships between activities of the relevant components in complex phases and the compositions ofthe phases should be Wellde ned In pmctice reactions involving major constituents are genemlly as concentrationincreases y gt 0 EXCHANGE REAC TIONS Coxsldudrlz xexcnnwewmk 112le stkolcm 173019 9 2c ex10cpxo Max010 THE GARNETVEIOTITE Essng EXCHANGE THERMOMETER m 5 ma until Mom Fawxso xm 39 msswxpmtomxba ohm cmbewn anmtm AAMum mymm x eM Bathing mm manym sand Semirme nuns m Wyn th SMAV mm 3911 mm 5 bmanamm exam damages Mmng mm m coma Mg mm c m mm Emma a Mg m mm m W mm m mm mm mm v r 39 Hm FEW c1 L 39 MN an F ax gt39F M Tinn lzmudibmlmcmhmcmyewmm 1m Wnagmmemm39mmu cmmemmmcmm mmm mm m K 1 x mm 39 MgFl 39 Emsde mmmvfgmbmhu msud mn mmg mm mmm mamammm ufFamMgbmmmymsgdq mi mmxnm smmmmmm mq u mamvn FERRY at SFEAR 1978 ALIERA 139on Ema k gammy mgr amm u ymn Hmm ms 45o mm m dznve memsm may WWW 1mm quotwe m hammmh ACgt mdA 5 mmm ufynssm ma may we an m AlivaK1burn 2vxKIbxoAJPrnoquotmt n 171mm mtxwmeq mmdm m m P2n7ym mymmmmmm wrm mmmwmw my mu m D W m ammmmup El mmmumm czrcelhcnn wad Acpwmmmemem marl ng my be a Idnmmrgc ybmmawh n Mm mmthms 1 mmth gum m vhmmmaahm mm mtg mm bewam gm ma mm Fgrmemmbeweengmmddm he a m OTHER EXCHANGE GEOTHERMOMETERS soLst GEOTHERMOMETERS mm mm Mm ohm c ma Mneman mg m szmMngemm mad mm m a Mwmmum ra mm DfK msz beweenmlscmk ma Pagan NET TRANSFER REACTIONS Many geobarometers are based on nettmnsfer reactions ie reactions that cause the production and consumption of phases Such reactions o en result in large volume changes so the equilibrium constant is pressuresensi 39ve An example garnetplagioclaseolivine 3Fe7 8104 3CaAlZSiZOB lt gt CajAlZSiKOlZ 2 FezAlZSiKOlz THE GASP GEOBAROMETER GASP GarnetAlumino SilicatePlagioclasequartz This is a net tmnsfer geobarometer based on the reaction 3CaAlzsizo8 lt gt CajAlZSiKOIZ 2Alsizo5 8102 Solid solution of grossular with other garnet components mainly almandine and pyrope loWers the pressure at Which the righthand side assemblage is stable Solid solution of anorthite With albite in plagioclase tends to stabilize the le hand side to higher pressures Because grossular o en quite dilute the pressure loWering effect prevails Garnet plagioclase AlZ SiO5 quartz is a common assemblage inthe crust 2n THERMODYNAMICS OF GASP Newton amp Haselton 1981 This geothermometer is based on the thermodynamic eq 39 I gr 3 AyG TP71AVRTln 2 0 Experimental determinations of the PT curve for the endmember reaction with kyanite have been expressed in the form P a bT and the error is expected to be on the order of i1 kbar The analogous expression for the GASP reaction With sillimanit 39 39 Once you have P T then AG T P TAYV Where P T is the breakdown pressure of anorthite at a given temperature for the endmember reaction and AV 662 cm3 mol391 at 298 K and 1 bar the molar volume change for the endmember reaction 22 KOZIOL amp NEWTON 1988 CALIBRATION This calibmtion can be Written in the alternative form 0 48357 15066 TK Pl6608 RT In K Thus for this calibmtion We get AH 298 K1 bar 48357 Jmolquot AS 298 Kl bar 15066 J K391 mol 1 Av 298 K1 bar 6608 J bar molquot and it is assumedthat AV is constant and App 0 Margules Parameters For Game Hensen et al 1975 obtained the expression Wm 7460 43TK in calories per MAlZKSiOA Which goes into the equation RT 1 Var453310 CmMgXAZIgAIZBSo Note the strong tempemture dependence of non ideality The above expression assumes a regular or symmetrical solution model Which is usually only strictly valid over a narrow range of composition In the Work of Hensen et al 1975 the mole fmction of grossular only ranged from 01 to 022 z Mama 1977 anhm 33m x J Cnsseyetzl mama CrFemmrlgmgzm Thyme u WM cmbe mm m be m an 5mm mm gm MW 5 Addmmdexyenmmal dzususgest 43 WW me mglactnd have w W 4 Expmmmb d hinges m Wm 5 yummy m mshdhlz so 3914 somannxmdzl errwhyed 14 5mm mlbe emFl yEd rm wrangling rommgmmmmm 3914 mmk y m v15 Tiny32m amm my my ned anth E Damn39s 19751 A1 mmgvvmm u mud mm m 4 mm mm mum u x PARTIAL MOLAR VOLUMES 0F GROSSULAR x K wave 117uw7 x s z r zrdAlZSZSErHZEBC u512D4MIE 1 9w DDEEkagmsquzrdmrdn mm mm an PnuAlmanahme m a cwswumm gum mm mu m 5 mmde mmuu um V quotx um um m mum 11 no ma ma Val rum 0215an swam mmbak mama quotms mimdmmk mm mm cmmwammwmaywumm mm pmm mammm mm smmmm SOME OTHER NETVTRANSFER OTHER GEOEAROMETERS EQUILIBRIA F h yum coma Wm ems mrmmtnmmurmgmmmm m chimkom zrguiskon 511019 am hmn h yrhcomm h xmgkoxosrmoi moi comm wmmu ma mm Mammnmr ma mm cm cmmmrqumrmur mm mm 31304 01 e mic mmhmr am until m comm mam mcemn we mugs SOURCES OF ERROR m5 1me m mmm Nu m mm WRMSMWMS Wmmmmmw mm DAWN mm mm mm 2 memmmndAV ma ydzkmd erdzm nyUD man 3 mm n awn Maugham v hmabmm hmr m smPhSE 16mmCaulmuvxmwmgumxsknmuf mmmw 12 m n him 3 2 u mu 5m 6m m m s Tc quot TOPIC 3 IDEAL SOLUTIONS FUGACITY ACTIVITY AND STANDARD STATES I PARTIAL AND APPARENT MOLAR PROPERTIES MOLAR VS PARTIAL MOLAR QUANTITIES Molar values ofstate functions are de ned as follows Z7Un Sn I7Vn etc These are use Jl only in the case of single component systems and dependent only on pressure and temperature not composition Partial molar quantities are de ned according to aU W U 6 S 1 l K a 8 9 KW 19 1 mm These are dependent on T P and composition PHYSICAL INTERPRETATION OF PARTIAL MOLAR VOLUIWES The partial molar volume of componenti in a system is equal to the in nitesimal increase or decrease in the volume divided by the in nitesimal number of moles of the substance Which is added While maintaining T P and quantities of all other components constant Another Way to visualize this is to think of the change in volume on adding a small amount of componenti to an in nite total volume of the system Note partial molar quantities can be positive or negative SUMNIING PARTIAL MOLAR QUANTITIE S The total value for a state function of a system is obtained by summing the partial molar volumes of its components according to n U r471 7sz 1133 1137A Vlan 11 v1 v1 11 H Z ref 1 V Z quot7 11 11 We can manipulate partial molar quantities in a manner identical to the Way We manipulate total quantities As With total state functions We cannot know absolute values only differences except for V and S s We can also express the summations in terms of molar state functions and mole fractions 11 v1 11 I7 2m 72217 72fo 11 11 11 n X 1 Z quot1 11 In the case of the volume of atWocomponent system egNaClHZO We can Write 7 XNaCVNacI XHZOVHZO Schematic plot ofthe molar volume ofaqueous Nacl solutions as a function of mole fraction ofNaC l V solution HOW TO DETERMINE PARTIAL MOLAR VOLUME Refer to the previous diagram 39llrianglesA andB are similar so it is true that erad 75er x X1120 y Zorro XNnCIy XNacJZoum anirdm anox Xap Xrtahiam Xnay Xpr but Xian ano 1 so may anot comparison with 7 XM C17MICIXWVW shows that 72W y ng x So the partial molar volumes can be determined from the intercepts ofa line tangent to the plot ofvolume vs mole fraction r PARTIAL MOLAR FREE ENERGY THE CHEMICAL POTENTIAL as G l 2 it chemical potential We Area The previous relationships also apply G Z W 5 21 H 11 It can also be shown that 3U u 4 kenW laxW kenrs Schematic plot of chemical potential vs mole fmetion for a binary system gt39 a 9 e p quot Lu 12 g 13 EM aquot mm Q Va 9 I L6 o E a 3 8 VA 9 lt I I I 2 El 4 El 6 I E 1 El COMPOSITIONAL CHANGES The Master equations that We developed previously for onecomponent systems can now be Written as dU TdsiPdVJr Zmdn rl dH TdS VdPZ udn rl dG rSdT VdPZ Mdquot rl dA 7SdT7PdVZMjah a AN ADDITIONAL REQUIREMENT FOR EQUILIBRIUIVI Consider a system with components i j k l distributed among phases 0t 3 y 5 At equilibrium it must be true that W Wr wr HP H HBWH5 Hie pk wpif Mr M W H etc Chemical potentials represent the slope of the Gibbs free energy surface in compositional space Thus has has a low chemical potential until its chemical potential in all phases is the same Speci c example Consider a silicate melt in equi 39 39um with forsteritelVIgZSiOA and enstatite MgZSiZO At equilibrian the following must be Mgmelt HMgFu Mgfn tame 115 115 Home HOFU Olin GIBB SDUHEM EQUATION For a homogenous phase of two components A and B the Master Equation becomes dG rSdT VdPMAahA M5dn5 If we now specify equilibrium at constant T and P dG uAdnA uEdnE 0 Now we have shown above that G mm Hang Differentiating this we obtain dG Mm Madng quotAdMA quotMtg GIBB SDUHEM EQUATION C ONTINUED At equilibrium dG mm Mama quotJim quotEdME 0 Substituting the previous expression dG uAdnA uEdnE 0 we obtain nd n In the geneml case we get the GibbsDuhem equation GIBB SDUHEM EQUATION CONTINUED Starting with the expression quotAdm Manila 0 If we divide through both sides by dXA we get BXA TPII BXA TPn And now dividing by nA nB we get X A 1 X A 0 aXA PM aXA TPM mum APPARENT MOLAR QUANTITIES Although in principle partial molar quantities can be they are not determined this way in pmctice a state function like volume the apparent molar volume LpV is given by VV ntV1 Vnz solu te respectively and V1 is the molar volum f pure so ven meas e from intercepts of lines tangent to a plot of state functions vs mole fmction as outlined previously In practice apparent molar quantities are determined For wher n1 and n2 are the number of moles of solvent and 17 Total volume of a solution as a function of solute concentration Volume 3 cm V2 Volume attrtbuted El solute pure solvent 2 a Volume attrtbuted quot1V1 msutvent 01 n2 gt Moles of solute n2 The apparent molar volume is the volume that would be attributed to one mole of solute in solution if it is assumed that the solvent contributes the same volume it has in the pure state Starting with the de nition of apparent molar volume VV n1V1 Vnz we can rearrange to get V 1 71 H2 Pv and dividing by n1 n2 V X1 V1 X2 LPV Thus the volume of solution can be calculated knowing va instead ofthe partial molar volume w Comparison ofapparent molar and partial molar volumes V2 V1 V1 V V2 W n u u 2 on as u E in PARTIAL MOLAR VOLUMES FROM APPARENT MOLAR VOLUMES V m mm Kan 7 K an rpm a 8 If tbv measurements are t by an equation of the type Vabmcm then we have V23 mb 2cm abm cmZ or VZ a 2bm 3cmZ II IDEAL SOLUTIONS THERMOD YNAMICS OF IDEAL S OLUTIONS An ideal solution is one that satis es the following equation H p RT lnX where p is the chemical potential of some component i in a solution and p is the chemical potential of that component in the pure form Recall that G 2X M substituting we get G xdealml n ZXufRT1nX thmkol n 2X44 RTZXlnX vazch mx XIX10 Gxdealml n Gmech mxx RTZX11HXI AGIdaalmx Idealsol n Gmach m RTZX IIIXI These equations tell us that the free energy of an ideal solution is the sum of two terms the free energy of amechanical mixture and a free energy of ideal z ENTHALPY AND VOLUME OF AN IDEAL SOLUTION o o thtzalsol n Hmtzchvmx ZXH ZXH I I o o Vague vazch mx XXIV XXIV I I There is no volume or enihalpy change upon ideal In other Words A V 11211 mxx Lfi is a pure substance Adeealmxx 0 However there is a change in enLropy upon ideal mixing Because the soluLion is more disordered enLropy increases ENTROPY OF MIXING Minimum RT ZXJHX bul We have AGmeal mm IAHmeal mm TASmeal mm and AHmeal mm 0 so AQdEalmxx TASxdgazm RTZXx 1 Xx Asxdealmxx RZXx 1 Xx Thus the only contribution to AGmeal mm is an enLropy conLribuLion n a quot u a gtltML quot E Ew quot a J HA quot g a VB 3 g W mVi t E Vs L i guso E ememe E we 396 EOVAa f 2 E 00 02 04 06 08 10 00 02 04 06 08 10 XE XE E Ha m E gmii i III FUGACITY AND 9 a 39 n E xx MSW ACTIVITY EHA 00 02 04 06 08 10 FUGACITY a I 0 anacxlyxrlzybe mm Dfas I mm Pussma ms I arms MEASUREMENT OF FUGACITY as unwammbemmm aw V 3p E Bum W Mn IL comnssfhd yfame z E m H amequm Mm af ueka Newmangme rmwnm We unsubme mm 3911 mm LUP cdmhtad 59m CALCULATION OF EQUILIBRIUM BOUNDARIES INVOLVING FLUIDS am In 9 LI P man s magmaI10 DICO eg MIASIAOIAOHL e mp A1103 H10 Czcoi Ioiecmoi coi 9 mm Ms e 55 o Cmd AVV Vlt39VAV VML A 7 AV o m J39 AYVdP J39Aydm J39 V wdp The pressure integral for the solids is then evaluated using the constant ASV approximation and that for the uid is evaluated using lgacities f AdeP AVP7 1 Rrh1fr M f H For the muscovite breakdown reaction above we can 39 the equation39 AYGVFAYGVF 4346 ZT dBKBAGzPleP rr Aqu 7 34min iAV dP T p p p p IAV HP IASV dPIV1320dP ASV P71RTln p p p Arv V Ksparv curv H20 39V mus AYV V Km V curV mm V HZO Ay ASV v H20 T 0 LG 7AST7EA11T77 7TIHEFH fem sza my J Ayb 2 At T2 T 7277 2 2TT7T 77 2T ASV P71 RT In szW function ofpressure szD My nd temerature FUGACITIES IN GASEOUS SOLUTIONS Starting with the following in terms of partial molar volumes a 111 f j aP T RT We obtain the expression for the fugacity coef cient of a compone 39 39 ALL CONSTITUENTS HAVE A FUGACITY The expression dG RT dln f may be integrated between two states 1 and 2 to give G2 G1 RTl MA This equation applies to apure onecomponent system For a solution we must use chemical potentials and we write Hz M RTh1f f This equation makes no stipulation as to the state of component i and can therefore refer to solid liquid or gas m I Solids and liquids therefore are also associated with a fugacity In some cases this fugacity can be thought of as a vapor pressure Fugacity can also be thought of as an escaping tendency I However in some cases avapor phase may not exist but a lgacity always exists One must reallze that the fugacity is athermodynamic model parameter not always an approximation to a real pressure I Fugacities of solid phases or individual components of solid solutions are not genemlly known I Fugacities are absolute physical properties ACTIVITIES The absolute values of the fugacities of solids and liquids cannot always be determined but their mtios can be Consider H M RT1nf fj If we let one of these states be a reference state this can be rewritten u Hf er WI gt We now de ne the activity of constituent i to be a ff Thus y Hf RTln q DALTON S LAW Dalton 1 81 1 discovered that at low total pressures a mixture o gases exerts apressure equal to the sum of the pressures that each constituent gas would exert if each alone occupied the same volume Strictly true only for ideal gases but is approximately true at low total pressure where real gases approach ideality For each gas we 39 Plv anT sz nZRT etc For the gas mixture we have Romy Z ntRT u If we divide the expression for each constituent by the expression for the mixture we obtain39 1 quott 2 12 i y i y Z i Z 2 etc or 131139Pmtal P2 Xz 39Pmtat etc Pm 21 Pl PZ P3 P1PZ etc are called the partial pressures HENRY S LAW Henry 1803 was studying the solubility of gases in liquids He found that the amount of gas dissolved in a liquid in contact with it was directly proportional to th the gas ie Km is a constant called the Henry s Law constant In pmctice this law holds only at re1ative1y1ow RAOULT S LAW Raoult 1887 studied vaporliquid systems in which two or more liquid components were mixed in known proportions and the liquid was equilibrated with its own vapor The composition of the vapor was then determined The total vapor pressure of the system was low so the vapor behaved ideally and conformed to Dalton s law In such systems the partial pressures ofthe gaseous components were found to be a linear function of the their mole fraction in the liquid Thus for a binary system AB he obtained PA XAPA and PB XBPB where P A and PB are the vapor pressures of pure components A and B respectively The only way that such a simple relationship as R oult s law can hold is if the intermolecular forces between AA BB and AB are identical Solutions in which this is the case are called idea solutiom The most geneml wamg Raoult s Law is t t 1 Very few systems follow Raoult sLaw over the entire mnge of composition from X 0 to X 1 However Raoult s Law o en applies to the emin dilute solutions whereas the solute in dilute solutions follows Henry s Law THE GIEE SVDUHEM EQUATION REV ISITED fut zbmsuhmnn a m mmst hm dupes Drum n ms Drummdmm vs mngan My Chunkleka n smmmm Mm m H DUHEMVMARGULES EQUA TION Smngwnhk ommkmewm 17x x mmwmmnxxwwmg gsmmvmmym P smsfm lakusm mude Wm m Mum mmmmmwwmmm mmgmammmmwammm mummdym 1 mum V E i n u 2 u u n m n14 M m u mm H N THE LEWIS FUGACITY RULE DEAL MMNG AND Acmm Mnmmm quotwe mm m dz mcnnm39du mm Waughth mmmmmn z ma Axemzrglmzm aft Lew mum gtlt i we see remnants am My Lm anamly m x We cmdsu W W A mm 5 mxmdued m x mm hmm39bmlh39s sz Activity lemons for an ideal binary system Activity It turns out that these relations hold not only for liquid and ans gaseous solutions but also for solid solut39 M NONIDEAL MIXING I As already discussed most real solutions do not conform to Raoult s Law over the entire compositional range I However whether solid liquid or gas in many solutions the component in excess solvent follo Raoult s Law and the minor component solute follows Henry s Law over a limited mnge at low mole fractions Positive deviation fromRaoult s Law 3A I Raoultian activity s x ANOTHER NIFTY APPLICATION OF THE GIBB SDUHEM EQUATION Task Prove that if the solute in a binary solution obeys Henry s Law then the solvent obeys Raoult s Law Starting with the Gibbs Duhem equation for a binary system quotAdm Mam 0 and dividing through both sides by quotA quotB we get X Adm Xdes 0 Xian Xad a At low pressures we have do RT dln P So we cannow write XAdlnPA rXEdanT Now if component A is the solute and obeys Henry s Law we have PA Kh A Taking the natuml logarithm ofboth sides we have lnPA1nKthlnXA Now differentiating we get 0 ln P A 0 ln X A SOHOW XAdlnXA 7X5dlnPE dX XAT 7X5dln117 dXA rXEd 1n 1 dXB 01XA so 74X rXBdlnfg Now dX JdlnPs dlnXEdlnI XE Nowwe integrate fromXB 1 to XB XB ln ln where P B is the partial pressureEofB when XB 1 ie the partial pressure ofpure B mquot L P lXlE X 11511 E a X P Raoult sLaw ACTIVITY COEFFICIENTS The ideal solution is use Jl as a model with which real solutions are compared This comparison is effected by taking the ratio of the activity of the real solution relative to that of the ideal solution This ratio is called the activity coef cient Deviations from Raoult s Law are expressed by the Raoultian activity coef cient 7R q 7 122 Deviations from Henry s Law are expressed by the Henryian activity coef cientyH q MK Activity coef cients because they are ratios of activities are uni ess A major difference between the two types of activity coef cients is t 7R lasXgt lbut 7H gt l as X gt 0 Thus 7H is usually more useful for solutes in dilute solutions IV STANDARD STATES STANDARD STATES Because the activity is the ratio of two fugacities ie f f i the value of the activity depends on the reference state chosen for f This state we usually refer to as the standard state The choice of the standard state is completely The standard state need not be a real state It is only necessary that we be able to calculate or measure the ratio of the fugacity of the constituent in the real state to that in the standard state m A STANDARD STATE HAS FOUR ATTRlBUTES 0 Temperature 0 Pressure 0 Composition 0 A particular welldefined state eg ideal gas ideal solution solid liquid etc If desirable we can permit T or P to be variable ie on a sliding scale STANDARD STATES FOR GASES Single Ideal Gas Starting with the relationship p2 ul RT 111 PlPl we can assign our standard state to be the ideal gas at 1 bar and any temperature In this case we can write it u R and u u is the difference in chemical potential between an ideal gas at T and P and an ideal gas at T and 1 bar Ideal mixture ofideal gases For such a mixture we can write P1 11 RT h1X1PX1P l Ifwe choose our standard state to be the pure ideal gas 1 at any temperature and 1 bar then Xl l andP 1 so ulul RTlnX1PRTlnP1 Noniideal gases For nonideal gases we would write 1 1 RT 111 flf1 but recall that lim jP Pan 1 So if we chose our standard state to be the pure ideal gas at any temperature and P 1 bar we get 73 Ht 1 RT Inf This equation is frequently written but mrely understood It only has meaning if the standard state is speci ed to be the pure ideal gas at any tempemture and 1 bar This is the most commonly chosen standard state for gases and supercritical uids However there is no reason why this particular standard state has to be chosen We couldjust as easily choose 1 the pure ideal gas at any T anle bars 2 apure real gas at 25 C and 1 bar 3 a speci c mixture of gases at any T and rt bars or 4 any other well de ned standard state LIQUIDS AND SOLlDS The following equation applies to liquids and solids as well A u RT 111 Wf Fixed pressure standard state The standard state is chosen to be the pure phase at the tempemture of interest and 1 bar Then j 1 so a j In this case it is necessary to know at each and ever set of PT conditions of interest Variable pressure standard state The standard state is the pure phase at the pressure and tempemture of interest 75 Under these conditionsj so a l The only way the activity of a solid deviates from unity under this standard state is when the solid is not pure but is a solid solution It may seem that the second standard state is easier to deal with in terms of pressure corrections However with the rst standard state the pressure correction is applied to j whereas in the second standard state the correction is applied to u In either case volume data for the constituent are required to make the correction AQUEOUS SOLUTIONS The activities of solutes in dilute solutions are more closely approximated with Henry s Law than Raoult sLaw Thus a somewhat different standard state is applied We start with the equation expressing the difference in chemical potentials between two solutions with different molalities H u RT h1vH m 7H m If we let one solution be the standard state we can H 11 RT 111 mmmm We then de ne the standard state to be the hypothetically ideal onemolal solution at the ure and pressure of interest Under these conditionsyH l and m 1 so yHm l andwe H 11 RT h1va This somewhat strange standard state is necessary because if we let the standard state be the in nitely dilute solution we would have H 1 an m 0 so yHm 0 which would result in an unde ned value of H 11 RT 111 YHMV YHM