Topic 3 : Thermodynamics

Transcription

1 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 1/20 GEOL360 Topc 3 : Thermodynamcs 3.1 Introducton and vocabulary Thermodynamcs deals wth the physcal and chemcal changes of matter due to work and hear flow (thermo = heat, dynamcs = movement) Real rocks are end products of long complex processes, cannot be exactly duplcated n the lab, but we can use thermodynamc measurements and reasonng to nvestgate nature of these processes (.e. t s a model of real lfe) System = materal of nterest (glasses, lquds, solds, any combnaton) Surroundngs = everythng else (especally the materal around the system) An solated system (or a thermally solated system) can exchange nether matter nor energy wth the surroundngs. A closed system can exchange energy (n the form of heat) wth the surroundngs, and can also do work on the surroundngs, but ts composton remans fxed. An adabatc system s of fxed composton and cannot exchange heat wth the surroundngs, but t can do work An open system can exchange both heat and materal wth ts surroundngs (e.g. rocks that have fluds passng through them). Phase = physcally homogeneous porton of a system wth a defnte boundary (e.g. a quartz crystal, or a pool of water) Component = chemcal consttuent : choose the smallest number possble (e.g. for the system: gypsum CaSO 4.2H 2 O anhydrte CaSO 4 water H 2 O, we need just two components, CaSO 4 and H 2 O) To study a system we need to know the number and nature of phases present, and ther chemcal composton. A system can be descrbed completely by a set of physcal and chemcal propertes, known as varables. A change n one or more propertes s a change of state (e.g. ncreased pressure, change n composton, etc. ). Intensve varables are ndependent of the amount of materal beng consdered, e.g. pressure (P), temperature (T), densty (ρ), chemcal potental (µ ), actvty (a ), fugacty (f ), Extensve varables depend on the amount of materal beng consdered, e.g. volume (V), mass (M), nternal energy (U), chemcal composton, An equaton of state relates certan propertes of the system. A well-known example s the deal gas equaton: PV = nrt

2 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 2/20 (pressure x volume = number of moles x gas constant x temperature) Note that only three of the four varables (P, V, n, T) needs to be known to calculate the fourth, so they do not all need to be defned ndependently. Equlbrum may be physcal, chemcal, or both. A system n equlbrum does not change wth tme. e.g. f a system ncludng the reacton A + B = C + D s n equlbrum, the reacton wll occur n both drectons, but there wll be no net change n the amount of A, B, C and D wth tme (and no change n P, T, etc. ether) In mechancs, energy s the capacty to do work (e.g. do work to lft a body, then ths body has ncreased gravtatonal potental energy) In thermodynamcs, we are concerned wth the nternal energy of the system (U): Work can be done on the system by the surroundngs (e.g. compresson of a gas), to ncrease the nternal energy of the system. Or work can be done by the system on the surroundngs (e.g. expanson of a gas) whch decreases the nternal energy of the system. Temperature s a measure of the ntensty of moton of atoms and molecules n a system. If two systems are n thermal equlbrum, there s no net heat flow between them, and they have the same temperature. Zero th Law of Thermodynamcs: If two systems are each n thermal equlbrum wth a thrd system, they are also n thermal equlbrum wth each other. Temperature s measured usng the Kelvn temperature scale, whch s based on the trple pont of water. The trple pont s the unque pressure and temperature at whch sold, lqud and gas phases of a substance all coexst n equlbrum. The trple pont of pure water s at K and atmospheres pressure. 1 Kelvn (1 K) s defned as 1/ x T trple pont. The freezng pont (ce-water equlbrum) at atmospherc pressure s K. Thermodynamcs can say what the fnal state of the system should be for a gven change n varables affectng the system: t does not specfy the transformaton mechansm or pathway or rate. These are the subject of knetcs. Expermental studes ndcate that the rate of reacton depends on the temperature, and on the concentraton of reactng molecules. Temperature measures the amount of heat energy, whch s requred to overcome the actvaton energy of a reacton. Reactons wth smaller actvaton energes go qucker. Reactons often consst of a seres of several steps; the slowest s the rate-determnng step. e.g. the stable form of carbon n ths room s graphte, but damonds exst metastably (luckly for jewellers). A metastable substance s n a form that s not the most stable under prevalng condtons, but does not change spontaneously to a more stable form (hgh E A slow reacton rate).

3 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 3/ Frst Law of Thermodynamcs and Enthalpy Frst Law of Thermodynamcs: If a thermally solated system s brought from one equlbrum state to another, the work necessary to acheve ths change s ndependent of the process used. Ths s equvalent to the law of conservaton of energy: f we could acheve the same change n a system by a dfferent route requrng less work, we could effectvely create energy. Unfortunately for Calforna, ths cannot happen. A system n a gven state possesses a certan amount of nternal energy (U); however, there s no absolute reference avalable, and we can only measure dfferences n nternal energy. An adabatc system has no heat flow n or out of the system, so amount of work done on the system = change n nternal energy of the system (recall work done on system +ve change n nternal energy) We can wrte ths as: du = -dw where du s change n nternal energy of the system and dw s work done by the system (dw s -ve f work done on the system) For a non-adabatc system, the same amount of work s found to produce a dfferent change n temperature (and hence U); dfference attrbuted to transfer of energy n the form of heat, so du = dq -dw where Q s heat absorbed by the system (so t s +ve). Ths s another way of wrtng the frst law of thermodynamcs. Note that work = force x dstance, so dw = P x dv. So another way of wrtng ths equaton s: du = dq - P dv e.g. aragonte calcte at 25 C, 1 atm (1 atm = Pa). Heat absorbed by system = 59 cal mol -1, or +247 J mol -1 (1 cal = J). The molar volumes are cm 3 mol -1 for aragonte, cm 3 mol -1 for calcte, so V = cm 3 mol -1. Ths reacton s endothermc. U = Q - P V = 247 J mol -1 ( Nm -2 x 2.78 x 10-6 m 3 mol -1 ) = 247 J mol Nm mol -1 = J mol -1 We can see that the ncrease n nternal energy s almost entrely due to heat absorbed by the system.

4 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 4/20 In geology we are usually lookng at systems at constant pressure, rather than at constant volume. We defne a new varable, the enthalpy (H): H = U + PV Agan we can only measure changes n enthalpy, snce we can only measure changes n the nternal energy. The change n enthalpy s then: dh = du + PdV + VdP rememberng that du = dq PdV, we can wrte: dh = dq + VdP now, at constant pressure, dh = dq and enthalpy change = heat transferred There are some specfc names for H of certan processes, all at constant P: enthalpy of fuson sold lqud enthalpy of vaporzaton lqud gas enthalpy of reacton heat absorbed (+ve) or released (-ve) by reacton The enthalpy of formaton of a substance, H f 0 s the enthalpy change produced n the reacton of the necessary elements to form that compound under standard condtons (stp: T = 25 C = K; P = 1 atm = Pa). Note that the enthalpy of formaton of the stable form of an element at stp s arbtrarly set to zero. For example, H f 0 of H 2 gas s zero, but the H f 0 of H 2 O lqud s not (must form H 2 O gas from H 2 and O 2 gases, then condense to form H 2 O lqud). Two equatons for two chemcal reactons can be combned to get a thrd equaton, whch s very useful. For example, the converson of graphte to damond cannot be measured drectly at stp. So we combne two reactons whch can be measured: () C(graphte) + O 2 (gas) CO 2 (gas) () C(damond) + O 2 (gas) CO 2 (gas) H = J mol -1 at s.t.p. H = J mol -1 at s.t.p. for reacton (), H = J mol -1, so t s an exothermc reacton (.e. t releases heat energy). Ths s also H f 0 of CO 2 (gas). for the reacton C(graphte) C(damond), H = H = J mol -1 Ths H s postve, and so ths s an endothermc reacton (energy nput s requred to make ths reacton go). Ths s also H 0 f of C(damond). [Remember that H 0 f of C(graphte) s zero, snce t s the stable form at s.t.p.] See handout for tables of: () Thermodynamc unts and converson factors (handout) () Standard enthalpes, entropes, and free energes of formaton at s.t.p.

5 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 5/ Entropy: the Second and Thrd Laws of Thermodynamcs H s not the only factor drvng chemcal reactons, e.g. halte dssolvng n water, endothermc reacton ( H = J mol -1 ) Although ths reacton absorbs heat, t vastly ncreases the dsorder of NaCl. The entropy (S) of a substance n a gven state can be consdered a measure of the probablty that the substance wll occur n that state. It s also a measure of the dsorder of a system (hgher entropy = greater dsorder = greater probablty of occurrng n that state, e.g. mxture of two gases n a box) (note unts of entropy are J K -1, usually expressed per mole,.e. J mol -1 K -1 ) Competng effects: favoured reactons have negatve H (release heat) and postve S (change to a more probable, more dsordered state). Examples: exothermc ncrease entropy always occur e.g. burnng coal, produces CO 2 gas exothermc decrease entropy sometmes occur* e.g. freezng water produces ce endothermc ncrease entropy sometmes occur* e.g. bolng water produces steam endothermc decrease entropy never occur e.g. heat lqud water ce * we shall see what determnes ths n the next secton (Gbbs Free Energy) A reversble process s conducted n nfntesmal steps so that the system s always n equlbrum, and can be reversed at any tme by nfntesmal opposte changes n the surroundngs. For a closed system (exchange of heat but not materal wth surroundngs), and a reversble process, ds = dq r /T, where Q r s heat absorbed from the surroundngs (reversbly) S s change n entropy at constant temperature (T) For an adabatc system (no net heat transfer between system & surroundngs), Q = 0, so a change can only occur f t ncreases entropy. Lke enthalpy, entropy depends only on the state of the system, and not how t got there (.e. we can measure S, but cannot nfer any partcular pathway). Second Law of Thermodynamcs: Heat cannot be converted nto work wth 100% effcency Alternatve statements of ths law nclude: () In nature, all processes are rreversble () ds > dq/t for real processes () You can t shovel manure nto the rear end of a horse and expect to get hay out of ts mouth (Nordstrom & Munoz, 1985)

6 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 6/20 Thrd Law of Thermodynamcs: Every perfectly ordered pure crystallne substance has the same entropy at 0K (Ths value s arbtrarly set to zero n all cases). Another verson of ths law s: It s mpossble to reach absolute zero usng a fnte number of processes We can calculate the entropy of a substance at dfferent temperatures usng the heat capacty (C), defned as C = dq/dt S2 e.g. for reversble heatng, ds = dq r /T, then ds = S S C T T S 2 1= [ ln 2 ln 1] 1 {Note that C usually vares wth temperature, and s ether measured at constant volume, C V, or at constant pressure C P. The two are dfferent!] Note that although we cannot perform a reversble process n the lab, we can acheve the same change by a seres of rreversble processes to measure changes n state functons such as P, V, T, G, U, S, H In general, entropy depends on volume, chemcal composton, crystal structure, degree of sold soluton, type of bondng, etc. e.g. consder hgh pressure phases. Graphte has a hgher molar volume and hgher entropy than damond at standard state. Lkewse, standard state entropes decrease from trdymte (hgh P, hgh S) to crstobalte to quartz. As a rule of thumb S(gas, low P) > S(gas, hgh P) > S(lqud) > S (sold). 3.4 The Gbbs Functon State functons depend only on the state of the system, not on how the system got there. These nclude U, H, G, S, etc For a reversble change n state, the Frst Law states that U = Q r W r. Usng the Second Law (ds = dq r /T) and W r = P V, we get: U = T S r - P V r (useful where S and V are varables) (Although ths equaton s only vald for reversble processes, the value of U has to be the same f the same change s acheved by an rreversble process) Lkewse, snce H = U + (PV), we can show that H r = T S r + V P r (useful where S and P are varables) In geology we are usually most nterested n varyng P and T, e.g. to predct stable phases n the mantle We defne the Gbbs functon (a.k.a. Gbbs free Energy) as G = U + PV TS then G = V P - S T = H - (TS), and at constant T, G = H - T S Free energy of reacton = Σ (free energy of products) Σ (free energy of reactants)

7 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 7/20 Standard free energy of formaton of a compound, G f 0, s defned as the free energy change resultng from the formaton of that compound from ts consttuent elements at s.t.p. (analogous to H f 0 ). e.g. CaCO 3 (calcte) + SO 2 (quartz) = CaSO 3 (wollastonte) + CO 2 (gas) Phase dagram (Fg. 3-1, overhead): G 0 r = H - T S = H 0 f (woll) + H 0 f (CO 2 ) { H 0 f (cc) + H 0 f (qtz)} x{s 0 (woll)+s 0 (CO 2 )} x {S 0 (cc)+s 0 (qtz)} = J mol -1 (Ths s an ncrease n free energy, therefore the reacton does not occur at stp) Now consder the same reacton at 650K and 1 atm. Recall dh = dq + VdP So at constant pressure, dh = dq = C P dt 650 () For each compound, H 650K H K = CdT P, (equaton ) so can calculate H 650K for all the reactants and products, and hence H 650K for the reacton. 650 CP () Also, for each compound, S 650K - S K = T dt and the entropy change for the reacton at 650 K s gven by (equaton ) S 650K = 650 CP T dt + S K, where C P = Σ C P (products) Σ C P (reactants)

8 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 8/20 Fnally, to get G 650K for the reacton, determne H and S, and use G = H - T S. In fact we fnd that G 650K s approxmately zero, so ths reacton s very close to equlbrum at 650 K and 1 atm (f G s close to zero there s lttle push n ether drecton; at equlbrum, G=0 as we see later). Now consder the same reacton at hgher pressures: Recall G = V P S T, so at constant temperature G = V P, and change n G of a reacton due to a change n pressure (at constant temperature) s G = V P and the free energy of reacton at selected P and T s: T P G( T, P) = G( T, 1atm) + VdP 1 (equaton ) where V = Σ V (products) Σ V (reactants) As long as P and T are moderate (reasonable crustal values, so T<1000 C and P<10 kbar), can often assume that V s approxmately the same as t s at s.t.p. (only f the reacton nvolves only sold phases). If a gas phase s nvolved, can assume: () that V of reacton s due entrely to the gas phase () that the gas obeys the deal gas equaton, PV=nRT (compare s.t.p. molar volumes of quartz, cm 3 /mol and CO 2 gas, cm 3 /mol). R s the gas constant, J mol -1 K -1. Usng these assumptons, G( T, P) = G( T, 1atm) + 1 P RT P dp e.g. at 650K and 2000 atm, G for our reacton s 0 + (8.31 x 650) x {ln(2000) ln(0)} = J mol -1 (.e. stll postve so reacton does not occur for closed system) At very hgh pressures and/or temperatures, we must take account of the thermal expansvty and compressblty of sold phases. (We can assume these effects more or less cancel out over crustal values of P and T, snce ncreased temperatures lead to hgher molar volumes, and ncreased pressures lead to lower molar volumes) Thermal expansvty, α = 1 V V T P 1 Compressblty, β = V V P T

9 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 9/20 So the molar volume V at any pressure and temperature s then [ ( )] V( P, T ) = V(. K, atm ) + ( α T ) β P 1 Fnally we can combne everythng we have just learned (equatons, and ) nto a sngle equaton for G of any reacton at any P and T: T T C P P GTP (, ) = H( K, 1atm) + CdT P T S K atm T dt VdP. (., ) T where C P = ΣC P (products) ΣC P (reactants); V = ΣV(products) ΣV(reactants) In geology we are prmarly concerned wth workng out where reactons wll occur n P-T space. So we want to know the slope (dp/dt) of equlbra (where G = 0). Skppng over the math (whch you can fnd n textbooks), we obtan the Clapeyron (or Clausus-Clapeyron) equaton: dp dt S = V Note that f V and S are both postve, the slope dp/dt wll be postve. If V s negatve (volume reducton) and S s postve, then the slope wll be negatve. As long as S and V only change slghtly over the P-T range of nterest, they wll plot approxmately as straght lnes, and can also be approxmated usng volumes and entropes tabulated for 25 C and 1 bar. As an example, consder the reacton jadete + quartz = albte NaAlS 2 O 6 + SO 2 = NaAlS 3 O 8 From thermodynamc data tables we get the followng molar standard state volumes and entropes: S 0 (J mol -1 K -1 ) V 0 (cm 3 mol -1 ) Then Jadete Quartz Albte S 0 V 0 = S 0 (albte) - S 0 (jadete) - S 0 (quartz) = = J mol -1 K -1 = V 0 (albte) - V 0 (jadete) - V 0 (quartz) = = cm 3 mol -1

10 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 10/20 and from the Clapeyron equaton, dp/dt = S/ V S 0 / V J mol -1 K -1 / cm 3 mol J cm -3 K bar K -1 (notng that 1 J = 10 cm 3 bar) or an ncrease of about 1.9 kbar n pressure for every 100 K n temperature: In ths fgure you can also see the effect of varable S and V over the P-T range of nterest, exemplfed by the breakdown of tremolte (an amphbole) to dopsde (a clnopyroxene) + enstatte (an orthopyrxene) + quartz + water. dp/dt s postve where both S and V are postve, but at hgh pressures the volume change of the reacton becomes negatve and so does the P-T slope. Ths s qute common where a reacton has all sold phases on one sde and a flud phase (lqud or gas) on the other sde Recap: If you only remember three thngs from ths secton, remember that () G = H - T S () reactons only occur f G s negatve at the specfed P, T condtons; at equlbrum, G = 0 () the slope of a reacton s gven by dp/dt = S/ V At equlbrum, both reactants and products coexst. We can quantfy the amounts of reactants and products usng the equlbrum constant (secton after next).

11 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 11/ Chemcal potental, fugacty, and actvty Rememberng that a change n the nternal energy of a closed system s (du=tds - PdV), we consdered changng U ether by dong work on the system (PdV), or by changng the nternal order of the system (TdS). For open systems, whch can exchange materal wth the surroundngs, we must also consder changng the nternal energy of the system by changng ts chemcal composton. The chemcal potental, µ, of component n a system s defned as: U µ = n SV,, nj.e. the chemcal potental of component s equal to the change n nternal energy (U) of a system, when the number of moles of component (n ) s changed at constant entropy, volume, and amounts of other components present (n j ). The total change n nternal energy for an open system s then: where du = TdS PdV + µ dn µ dn s the sum of (change n amount of component x chemcal potental of that component) for all components n the system. Lkewse, chemcal potental s also the change n Gbbs Free Energy of a system when the number of moles of component s changed at constant P, T, n j (.e. the partal molar Gbbs Free Energy): Therefore at equlbrum, G µ = n PT,, nj dg = VdP SdT + µ dn (Gbbs-Duhem equaton) VdP SdT + µ dn = 0

12 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 12/20 (Asde: the total Gbbs functon of a system consstng of a sngle component n a sngle phase s equal to the Gbbs functon per mole of the pure component multpled by the number of moles present. When two or more components are mxed together n a system, the Gbbs functon of the system cannot be calculated by summng the respectve Gbbs functon per mole values of the pure components. The chemcal potental s the effectve Gbbs functon per mole of each component). At equlbrum, the chemcal potental of every component n a phase s equal to the chemcal potental of that component n every other phase n whch that component s present. (Asde: a dfference n the chemcal potental of a component between two dfferent phases causes some of that component to transfer from the phase where t has hgher potental to the phase where t has lower potental: the system readjusts towards mnmzng the chemcal potentals of all components n all phases). Now we need a way of relatng the chemcal composton of a phase to the chemcal potentals of components wthn that phase. Ths s where actvty comes nto t but frst of all let s look at deal gases as a smple startng pont. For an deal gas (no forces between atoms), PV=nRT. For a mxture of several deal gases (each of whch s a separate component), µ = µ * +RT ln P (deal gases) where µ * s the chemcal potental of pure component at unt pressure, and P s the partal pressure of component (.e. P = mole fracton of component x total gas pressure) Real gases are not deal, unfortunately. We get round ths by defnng the fugacty (f ) as the pressure needed (at a gven T) for the propertes of a real gas to satsfy the deal gas equaton: f = γ P where γ s the fugacty coeffcent, and P s the partal pressure of the gas. For deal gases, γ =1. For real gases, γ I 1 as total pressure 0. Now we can defne µ for real gases: µ = µ * +RT ln f (real gases) In geology; the fugacty of gases such as H 2 O and CO 2 can be as mportant as P and T n determnng gneous & metamorphc mneral assemblages. e.g. the fayalte-magnette-quartz (FMQ) oxygen buffer reacton s: 3Fe 2 SO 4 (fayalte) + O 2 = 2Fe 3 O 4 (magnette) + 3SO 2 (quartz) As long as fayalte, magnette and quartz coexst, the fugacty of oxygen s fxed for a gven P and T.

13 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 13/20 (Fg. 3-2, overhead) Now let s move from gases and apply smlar logc to real sold and lqud solutons. For deal solutons, µ = µ * +RT ln N (deal solutons) where N s the mole fracton of component (.e. the concentraton of component ), and µ * s the chemcal potental of pure (when N = 1) at the temperature of nterest. (Remember ths, t s mportant). Agan, real solutons are more complcated than deal solutons so we defne the actvty, a, of a component n soluton as: a = γ N where γ s the actvty coeffcent, and N s the concentraton of the component. For aqueous solutons, a = γ m, where m s the molalty (mols/kg of solvent). For deal solutons, γ =1. For real solutons, γ I 1 as concentraton 0. Now we can defne µ for real solutons: µ = µ * +RT ln a (real solutons) For real (non-deal) solutons, a s the effectve concentraton of a component (.e. the soluton behaves as though the amount of present s a, even though t s actually N or m ) (Asde: actvty coeffcents can be calculated usng a number of dfferent methods, ncludng the Debye-Huckel equaton for dlute electrolyte solutons: logγ 2 = Az I where A s a constant (functon of P and T), z s the charge on on, and I s the onc strength of the soluton. Ionc strength s defned as I = 1 2 where m s the concentraton of component n molal unts (mol kg-1 of solvent)). The reason we had to go through ths s that when dealng wth mneralmneral, mneral-soluton, or soluton-soluton reactons, what geochemsts actually measure s the concentratons of the components, but the thermodynamc relatonshps used to model these solutons nvolve actvtes. The good news s that for very dlute solutons (<50 mg of dssolved ons / kg of solvent), γ s >0.95 for most ons, and actvtes are smlar to measured concentratons. (.e. the behavour of very dlute solutons approaches that of deal solutons). mz 2

14 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 14/20 For example, stream water usually has low concentratons of dssolved ons and concentratons can be used n place of actvtes wth only small errors. In contrast, seawater cannot be treated as an deal soluton. For example, the concentraton of carbonate on (CO 3 2- ) n seawater s about mol kg -1, but t has a low actvty coeffcent of 0.20, and about 90% of (CO 3 2- ) occurs n other forms (not as free carbonate on), so the actvty of the free (CO 3 2- ) on s only (about 2% of ts measured concentraton). Trace elements n mnerals can be consdered as dlute solutons, and behave approxmately deally, so we can apply thermodynamcs to trace element dstrbuton between mnerals, and between coexstng mnerals and melts. e.g. cobalt (Co) dstrbuton between botte and hornblende: at equlbrum, the chemcal potental of Co must be the same n both mnerals. Therefore: bt * bt bt hbl * hbl µ = µ + RT lna = µ = µ + RT lna Co Co Co Co Co * bt * hbl where µ Co and µ Co are constants (recall that µ * s the chemcal potental when N = 1,.e. pure component, a slghtly bzarre concept n ths case ) From ths we can deduce that a bt aco s a constant, whch we shall call K hbl D. Ths s Co also known as the dstrbuton coeffcent or partton coeffcent, snce t s a measure of how Co s dstrbuted between two mnerals (botte and hornblende n ths example). As long as we are lookng at only trace quanttes of Co n these mnerals (whch we are), we can reasonably assume deal behavour, and use concentratons n place of actvtes, thus n ths example: K D = hbl Co molar fracton of Co n botte molar fracton of Co n hornblende (NB we measure concentraton by molar fracton,.e. number of atoms, NOT wt. %!) (Fg. 3-3, overhead) We can also use composton ratos to express K D, rather than concentratons. Ths avods the problem of convertng compostons to actvtes. For example, a well-known geothermometer uses the dstrbuton of Fe 2+ and Mg between coexstng garnet and botte to determne the temperature at whch they last equlbrated. The reacton looks somethng lke ths: Fe-botte + Mg-garnet = Mg-botte + Fe-garnet

15 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 15/20 ( Fe Mg ) Then we can wrte K D = ( Fe Mg) garnet botte and use measured botte and garnet compostons to calculate temperatures of equlbraton, usng expermental relatons between K D and T. (Fg. 3-4, overhead) We wll dscuss the actvtes of components n sold solutons n topc 5. Recap: If you only remember one thng from ths secton, remember that the Gbbs free energy of an open system can be changed not only by changng the pressure and temperature, but also by changng ts chemcal composton: dg = VdP SdT + µ dn µ s the chemcal potental: µ = µ * +RT lna a s the actvty: a = γ N where N s the concentraton of component n the system, µ * s the chemcal potental when N = 1 (.e. pure component ), and γ I s the actvty coeffcent. Actvtes are used n place of concentratons wherever we are dealng wth a non-deal soluton (usually whenever there s more than a trace amount f component n the system). The actvty of a substance s ts effectve concentraton.

16 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 16/ Equlbrum and the equlbrum constant For a closed system*, at fxed P and T, trade-off between enthalpy and entropy s represented by change n Gbbs Functon, G = H - T S. exothermc reactons (-ve H) whch ncrease dsorder (+ve S) must have negatve G, and wll always occur endothermc reactons (+ve H) whch decrease dsorder (-ve S) must have postve G, and wll never occur f we know H and S at a partcular P and T, we can calculate G and predct whether or not the reacton wll occur (at that partcular P and T) (If we know varatons n C P wth P and T, we can predct G for other P-T condtons as well). (*recall: closed system = exchange of heat but not materal wth surroundngs) e.g. recall our prevous example reacton, cc + qtz = wo + CO 2 at 298K, 1 atm, G = +42 kj mol -1 (no reacton) at 650K, 1 atm, G 0 ( equlbrum) at 650 K, 2000 atm, G = +41 kj mol -1 (no reacton) at T>650K, 1 atm, G <0 (reacton proceeds) We can calculate G to plot the stablty felds of calcte + quartz, and of wollastonte + CO 2, whch are separated by an equlbrum curve where G 0 (note; open system has dfferent free energy relatonshps, so dfferent stablty felds) Le Châteler s prncple descrbes qualtatvely how a system n equlbrum wll respond to a change n P or T (or volume, etc),: When a reacton at equlbrum s dsturbed, t re-establshes equlbrum by counteractng the dsturbance For example, exothermc reactons are favoured by a decrease n temperature (the reacton releases some heat energy to compensate for the decrease n T), and endothermc reactons are favoured by an ncrease n T (the reacton absorbs some heat to compensate for the ncrease n T) The magntude of G at a gven P and T s a measure of the dstance from equlbrum. For any reacton: aa + bb = cc + dd at a gven T, the equlbrum constant for an deal system s: c C D K = ( ) ( ) a ( A) ( B) d b

17 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 17/20 where (A) = concentraton of A, etc. For non-deal systems: c C D K = [ ] [ ] a [ A] [ B] where [A] = actvty of A, etc. The value of K vares wth P and T. (When the system s not at equlbrum, ths rato s called the reacton quotent) How can we determne K? At ths pont we ntroduce a very useful equaton relatng G r to K: G = G + r RT lnk or, more generally (not necessarly at equlbrum): c C D Gr = G + RTln [ ] [ ] a [ A] [ B] where G r s the free energy change for the reacton at a gven P and T; G s the free energy change when the reactants and products are n a standard state of unt actvty From these equatons, t s clear that at equlbrum (when G r must = 0): G = RT ln K or G = RT log K (note that s very mportant to wrte G and not just G!) Because we can work out G from tables of thermodynamc data, we can work out the equlbrum constant at the standard temperature for dfferent reactons (even though most of these reactons wll not be at equlbrum at s.t.p. later we wll see how to work out K at other T and P). For example, NaAlS 2 O 6 (jadete) + SO 2 (quartz) = NaAlS 3 O 8 (albte) d b H f 0 (kj mol -1 ) G f 0 (kj mol -1 ) d b Jadete Quartz Albte G r = G f(naals 3 O 8 ) G f(naals 2 O 6 ) G f(so 2 ) = ( ) (-856.6) = -2.8 kj mol -1, or J mol -1

18 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 18/20 Now G r = -RT ln K 0, so ln K 0 = - G r /RT log K 0 = -(-2800)/(2.303 x x ) = 0.49 and K 0 = 3.1 at K and 1 bar { remember log q = (ln q/ ln 10) where q s a number, and ln } To return to our orgnal example, CaCO 3 (s) + SO 2 (s) = CaSO 3 (s) + CO 2 (g) G r = G f(caso 3 ) + G f(co 2 ) G f(caco 3 ) G f(so 2 ) = ( ) ( ) ( ) = J mol -1 G r = -RT ln K 0, so ln K 0 = - G r /RT log K 0 = /(2.303 x x ) = K 0 = , so a a a CaSO3 CO2 a CaCO3 SO2 = at a temperature of K The standard state of a substance refers to a pure phase, n whch case ts actvty s 1. So f CaCO 3, SO 2 and CaSO 3 are all pure phases, ther actvtes must be 1 (by defnton). The only varable left n the equaton s the actvty of CO 2, and n ths case f wollastonte, calcte, quartz and CO 2 were all n equlbrum at 25 C then the actvty of CO 2 would be a CO2 = Remember that the actvty of a gas s called ts fugacty (t s the same thng, only gases get a specal name for ther actvtes) so we have just calculated the fugacty of CO 2, whch would be Ths also means that f wollastonte, calcte, quartz and a gas phase contanng CO 2 were all n equlbrum, the partal pressure of CO 2 n the gas phase would be Sometmes we want to nvestgate reactons at T other than standard state (25 C, 1 atm). So how does K vary wth T? Agan, we wll skp over the math and go straght to the answers. Varaton of K wth T (at constant P) s descrbed by the van t Hoff equaton: lnk H = 0 2 T RT P so that ln K K T T 0 0 H 1 1 = 0 R T T

19 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 19/20 Note that ths equaton assumes that H r and S r are constant over the P-T range of nterest although ths s never strctly true, t s a useful assumpton over lmted P-T ranges. We can see that f H s postve (endothermc reacton), an ncrease n temperature wll lead to an ncrease n the equlbrum constant (the reacton wll move to the rght, absorbng heat energy). Ths s an example of Le Châteler s prncple. Now f H and K and known for one temperature, they can be calculated at another temperature. Lkewse, f we know K for the reacton at several dfferent temperatures we can calculate H for the reacton. We return to the example of jadete + quartz = albte, where we have already determned log K at K to be 0.49 (K = = 3.1). If we wsh to calculate K at 500K, we use the equaton ln K 0 T H 1 1 = 0 K R T T T 0 Frst we need to calculate H : H r = H f(naals 3 O 8 ) H f(naals 2 O 6 ) H f(so 2 ) = ( ) (-910.9) = +6.7 kj mol -1, or 6700 J mol -1 So ln K 500 = ln K /8.31 x (1/500 1/298.15) = x ( ) = (-1.09) = 2.22 K 500 = 9.21 At ths pont t s worth rememberng that f albte, quartz and jadete are all present as pure phases, ther actvtes are all equal to 1 by defnton, so the rato a albte /(a jadete x a quartz ) s fxed at 1 and can never be equal to 9.21,.e. the three pure phases wll never be n equlbrum at 500 K and 1 bar (or at K and 1 bar). If we wsh to calculate K at 210K, we use the same equaton: So ln K 210 = ln K /8.31 x (1/210 1/298.15) = x ( ) = = 0 K 210 = 1 Therefore the reacton s at equlbrum at 210 K and 1 bar; ths can be verfed by settng G = H - T S equal to zero. (Agan we have assumed that H r and S r are the same at 210 K as at 298 K)

20 GEOL360 LECTURE NOTES: T3 : THERMODYNAMICS 20/20 Varaton of K wth P (at constant T) s descrbed by: lnk V = 0 P RT T so that ln K K P P 0 0 V = RT P P 0 ( ) Ths tme we can see that f V s postve, an ncrease n pressure wll lead to a decrease n the equlbrum constant (the reacton wll move to the left, reducng the amount of hgher-volume products and ncreasng the amount of lower-volume reactants). Ths s another example of Le Châteler s prncple. Recap: If you only remember three thngs from ths secton, remember that () the Gbbs free energy of a reacton s related to the equlbrum constant: c C D Gr = G + RTln [ ] [ ] a [ A] [ B] d b so at equlbrum, G = RTlnK where K s the equlbrum constant at standard temperature, and G s calculated for the standard state (equlbrum and standard state are usually NOT the same thng, so you cannot calculate K at any T from G =-RT lnk, only K at T 0 )! () We can calculate K at other temperatures usng the relaton: ln K K T T 0 0 H 1 1 = 0 R T T assumng that H r and S r are constant at dfferent temperatures A fnal note: For the purposes of ths course you should understand and be able to use the equatons mentoned n the recap sectons. You do not have to know how to derve thermodynamc relatons, but you should be able to use them. We wll return to them n the next two topcs, water chemstry and crystal chemstry.