Chemical Thermdynamics Objectives 1. Be capable f stating the First, Secnd, and Third Laws f Thermdynamics and als be capable f applying them t slve prblems. 2. Understand what the parameter entrpy means. 3. Recgnize that Gibbs Free Energy change is related t the entrpy change f the Universe fr a physical r chemical change and als represents the maximum amunt f wrk that can be achieved frm the prcess. 4. Define standard cnditins as this term applies t chemical thermdynamics. 5. Be able t calculate ΔHº, ΔSº, and ΔGº frm values tabulated at 298 K. Understand that these changes apply t the physical r chemical change ccurring at standard cnditins. 6. Find ΔGº at temperatures ther than 298 K using the equatin: ΔGº = ΔHº - TΔSº. 7. Be capable f interpreting the equatin in the previus item t explain the fur pssible types f temperature dependence fr reactin spntaneity. 8. Find ΔG frm ΔGº at nnstandard cnditins with: ΔG = ΔGº + RTlnQ and understand that when Q = K, ΔG = 0 meaning the reactin is at equilibrium (Nte: there are cases when ΔGº = 0 which means that the prcess is at equilibrium at standard cnditins). 9. The result f the previus statement is that ΔGº = -RTlnK (always). That means that ΔGº and K bth prvide the same type f inf abut a reactin (i.e. which are favred at equilibrium: prducts r reactants). Als, nte that fr the case when ΔGº = 0, K = 1. 10. We may cmbine the equatin frm number 9 with the equatin frm number 6 t find the van t Hff equatin that we intrduced in Chapter 14 (which is, if yu recall, a mre general versin f the Clausius-Clapyern equatin frm Chapter 12). 11. Be capable f sketching reactin crdinate diagrams t illustrate the relative Gibbs Free Energy befre and after a reactin and als where equilibrium ccurs. 12. Be capable f demnstrating, via calculatins, the way in which nnspntaneus reactins can be made spntaneus thrugh the cupling f nnspntaneus and spntaneus prcesses. This may be dne by taking prducts f the equilibrium cnstants f the added reactins OR by adding the ΔG terms f the added reactins. 13. Take time t really think abut all f the abve... what we have really dne is prvided a cnceptual framewrk f thery t explain everything we have been ding with equilibria fr the past several chapters. Chemical Thermdynamics Chemical Thermdynamics is the area f chemistry that answers questins abut why certain reactins ccur (under a certain set f cnditins) while ther reactins d nt ccur. At atmspheric pressure, water will freeze t frm ice: but nly when the temperature falls t (r belw) 0ºC. Water will als bil at 1 atmsphere pressure, but nly when heated t 100ºC. Irn will slwly rust t frm irn xides (r irn xyhydrxides in the presence f water) at rm temperature and xygen and hydrgen will cmbine rapidly t frm water (if a spark initiates the reactin). S what it is that cntrls whether a prcess will ccur and at what temperatures? Why 141
are sme reactins fast and thers are slw was addressed in an earlier chapter n chemical kinetics. Nw we will get the rest f the stry. As we learned when we studied the First Law f Thermdynamics: energy is cnserved but may be cnverted frm ne frm t anther. In an even earlier chapter, we had learned that matter is cnserved in nrmal physical r chemical changes. In rder t explain why sme reactins are thermdynamic favrable, and under what cnditins, we need t intrduce a new cncept - r if yu prefer parameter. This parameter is given the name entrpy (given the symbl S and expressed in energy units per quantity per K, ex. J / mlek ) and it is a rather subtle subject t fully understand. If yu think yu fully understand entrpy after cmpleting this chapter, yu have almst certainly versimplified the situatin! Smetimes freshman texts will discuss entrpy in terms f the amunt f energy stred in a substance as it is heated frm abslute zer t sme higher temperature. In fact this descriptin is useful and will be discussed during class. Other times, the entrpy issue will be discussed as a measure f randmness r disrder present. Many spntaneus prcesses d increase the randmness r disrder present, hwever, this still des nt quite capture the cmplete nature f entrpy. T better appreciate what is meant by entrpy, we are ging t have t d a bit f thinking abut bth f these descriptins. First let us cnsider stred energy. A substance is heated when energy is depsited int that substance. The temperature f a substance prvides a measure f the amunt f heat stred and when tw bjects the differ in temperature are placed in cntact with ne anther, energy flws frm the htter t the cler bject. But what des this really mean is happening at the mlecular level? Cnsider first a perfectly rdered perfectly pure piece f metal (maybe it is a nice chunk f gld). As energy flws int the piece f gld it des nt simply vanish... instead it is stred as vibratins f individual gld atms and cllectins f gld atms. Nw if this gld blck is brught int cntact with a cup f cl water, the vibratins f the gld atms are diminished as energy is passed alng t water mlecules in the frm f bth vibratins f the O-H bnds in each water, rtatins f the water mlecules, and translatin f the water mlecules. S, all substances stre energy in the frm f atmic r mlecular mtins. In rder t figure ut hw much energy is stred, we can d sme simple calculatins yu may recall frm ur calrimetry calculatins. 142
Nw, the Third Law f Thermdynamics states the fllwing (yes I knw that I skipped the Secnd Law... but bear with me): all perfectly pure perfectly crystalline materials cntain the same entrpy at abslute zer. Fr cnvenient bkkeeping, we have decide t state that this value is called entrpy = S = zer. After we have decided this, it is easy enugh t find the entrpy f any pure substance at any temperature if we can track hw much heat energy is required t raise the substance t that temperature. It is imprtant t understand that the standard entrpy, Sº, f a substance at any temperature (mst surces tabulated the values at 25ºC, 298 K) is really the integratin f the amunt f heat energy required t heat the H e a t A d d e d Figure 1. Temp. in K substance frm 0 K t that temperature. A general idea f what this might lk like in graphical frm is shwn in figure 1. The first slped line segment in the graph represents heating the slid, the first vertical segment represents melting the slid (nte that T is cnstant during that prcess), the secnd slped line segment represents heating the liquid, the secnd vertical segment crrespnds t biling (vaprizatin), and finally the last slped segment t the heating f the vapr. Energy that is used t heat substances can be stred as mtin f the atms and mlecules in that substance. Three types f mtin (rtatin, vibratin, and translatin) are pssible fr mlecules in the gas phase and translatin is nearly eliminated fr slids (because atms and mlecules cannt easily translate thrugh a slid medium). Hwever, atms in the gas phase can nly stre energy as translatin (since they are islated atms they cannt have vibrating bnds and rtatin f a spherical atm is nt energetically imprtant). It may be illustrative t lk at sme tabulated entrpy values. Table 1. Standard mlar entrpies f sme substances at 298 K. SOLIDS LIQUIDS GASES Substance Sº (J/Kmle) Substance Sº (J/Kmle) Substance Sº (J/Kmle) Al 28.3 Hg 76.0 H 2 130.6 AgCl 96.2 Br 2 152.2 F 2 202.7 KCl 82.6 H 2 O 69.9 Cl 2 223.0 KClO 3 143.1 CH 3 OH 126.8 HCl 186.8 CH 3 CH 2 OH 160.7 I 2 260.6 NO 210.7 NO 2 240 Sme general trends shuld be pinted ut in this table. First, a mle f AgCl has twice the number f atms as a mle f Al. With mre atms in a mle f substance, mre energy can be stred in vibratins f the atms in the slid. Secnd, the mass f the atms in the cmpund als cntributes t the stred energy (cmpare the hmnuclear diatmic gases). Finally, the bnd strengths can affect the amunt f vibratinal energy that can be stred. 143
The case in which entrpy changes are described as increases in randmness will be discussed further belw. Enugh Already... What makes prcesses happen? As it turns ut, the prcesses that are thermdynamically favrable (i.e. spntaneus) are the nes that generate an increase (r n change) in the entrpy f the Universe. This is the Secnd Law f Thermdynamics which can be written as: a prcess is spntaneus when ΔS univ 0. The prcesses that cause n change in the entrpy f the Universe are really ideal cases and d nt really affect ur discussin. S we can really state, a prcess is spntaneus when ΔS univ > 0. There are primarily tw factrs that cntribute t raising the entrpy f the Universe: (a) first, is the prcess endthermic r exthermic and (b) whether r nt the prcess increases the entrpy f the system. In the traditin f energy cnservatin, let s divide the Universe int tw parts (the system and the surrundings). Let us cnsider the virtues f endthermic vs. exthermic reactins. Fr an exthermic reactin, heat is released and generates an increase in the entrpy f the part f the Universe utside the system (the surrundings) by creating mre mtin in the atms and mlecules in the surrundings. An endthermic reactin (r prcess) actually absrbs heat energy frm the surrundings cntributing t a decrease in the entrpy f the Universe. S... frm the surrundings pint f view, EXOTHERMIC reactins prmte increase the entrpy f the Universe by raising the entrpy f the surrundings. But what f the entrpy f the system? Nw we need t discuss the ther mdel f entrpy... the randmness issue. When we mix tw ideal gases at very lw pressures the prcess is neither endthermic r exthermic (i.e. ΔH = 0 fr the prcess f mixing). S, the entrpy change in the surrundings must be zer. Hwever, I knw that yu knw that gases will mix when given the chance. IF the Secnd Law f Therm is crrect then ΔS univ > 0 fr the prcess f gases mixing. Since the Universe can be divided int t prtins (the system and the surrundings) we can rewrite this expressin as: ΔS univ = ΔS surrund + ΔS system > 0 fr spntaneus prcesses (1) Nw, fr the case under discussin (mixing f ideal gases) I have explained that ΔS surrund = 0. This leads us t the cnclusin that fr this mixing prcess ΔS system > 0 unless we want t insist the Secnd Law is wrng. But why is mixing tw gases a prcess that increases entrpy? The true prf f this is a rather cnvluted prcess. S let s just cnvince urselves that mixing must prduce entrpy by cnsidering the reverse prcess. In rder t separate a mixture f argn and nen (tw ideal gases), we culd nt just rely n the gases spntaneusly segregating themselves t ppsite sides f a cntainer. Instead we must supply an input f energy (in thermdynamic ling we say it requires wrk) t frce the separatin t ccur. Nw, in the thermdynamic sense, prcesses that require an input f wrk are NOT spntaneus (and spntaneus prcesses can be used t d wrk as we shall see shrtly). I knw that this might nt be the mst satisfying answer, but I ll try t explain a bit mre during class. 144
Nw let us see if we can make use f the Secnd Law t make sme useful predictins abut equilibria. Abve we stated that: Earlier we stated that an exthermic prcess (with a negative ΔH sys ) causes a rise in the entrpy f the surrundings. As it turns ut, ΔS surrund = -ΔH system /T where ΔH system is the enthalpy change in the system. We can nw rewrite the relatinship: ΔS univ = ΔS surrund + ΔS system > 0 using this new inf t find ΔS univ = -ΔH system /T + ΔS system > 0 fr spntaneus prcesses (2) Multiplying thrugh by -T we find ΤΔS univ = ΔH system - TΔS system < 0 and we will nw state ΤΔS univ = ΔG giving ΔG = ΔH system - TΔS system < 0 (3) where ΔG is the change in Gibbs free energy. Tw pints: First we multiplied by a negative T. Therefre, ΔG has the ppsite sign f the entrpy change f the Universe (i.e. fr the entrpy f the Universe t increase ΔG must be negative). Secnd, we nw have an equatin that invlves knwing nly ΔH and ΔS fr the system in rder t find a parameter (ΔG) that is related t the entrpy change f the Universe. I think yu will agree that it shuld be easier t experimentally measure changes in the system vs. the Universe! Nuts and Blts We (perhaps) understand what is meant by the term entrpy and we can imagine entrpy at the mlecular level. Knwing the laws f thermdynamics (First, Secnd, and Third), we can nw apply the fllwing equatins t determine when prcesses shuld be spntaneus. First understand that the entrpies f pure substances (and f species in slutin) have been determined and are tabulated. The values tabulated are fr standard state cnditins: fr slids and liquids the substances are pure and fr gases the substance and pure and the pressure is 1 atm. Fr slutins the cncentratin is 1 M. Mst tables tabulate these values at 298 K (but that need nt be the case). Therefre, fr any given reactin we may find the entrpy change IN that reactin using the fllwing equatin: ΔS = ns ( prducts ) ms ( reac tan ts) (4) This equatin states that the entrpy change (at standard cnditins, dented by the º) can be fund by taking the difference between the entrpy f the prducts and the entrpy f the 145
reactants. The n and m merely accunt fr the stichimetric cefficients since the entrpies are nrmally tabulated per mle. Nw the enthalpy change in the system can be fund (as we learned in ur previus curse) using a similar equatin. This equatin states that the enthalpy change (at standard cnditins, dented by the º) can be fund by taking the difference between the enthalpy f the prducts and the enthalpy f the reactants. The n and m merely accunt fr the stichimetric cefficients since the enthalpies are nrmally tabulated per mle. Using these values we culd find ΔGº using the relatinship: ΔGº = ΔHº - TΔSº (6) and the ΔHº and ΔSº values fund abve. Nte that ΔGº may als be fund frm tabulated values f ΔGº f tabulated at standard cnditins using an equatin similar t the tw used abve (equatins 4 and 5) fr ΔSº and ΔHº. (5) This equatin states that the change in Gibbs free energy (at standard cnditins, dented by the º) can be fund by taking the difference between the Gibbs free energy f the prducts and the Gibbs free energy f the reactants. The n and m merely accunt fr the stichimetric cefficients since the Gibbs free energy are nrmally tabulated per mle. Equatin (7) is useful, but nrmally it will be mre infrmative t knw ΔHº and ΔSº and use equatin (6) t find ΔGº. It is als knwn that the changes in enthalpy and entrpy are nearly cnstant ver reasnable temperature ranges (with the prvisin that the physical states are nt changed by raising r lwering temperature). T find the change in the free energy at temperatures ther than 298 K we may use equatin (6) with a value f T ther than 298 K. Equatin (6) can als be used t evaluate the temperature ver which a reactin (r prcess) shuld be spntaneus (slve fr the temperatures ver which. Yu will find that there are fur pssible types f temperature behavir (summarized belw in Table 2). Table 2. Temperature dependence f reactin spntaneity. ΔH ΔS ΔG Reactin Characteristics Example (-) (+) Always negative Spntaneus at all temperatures due t bth ΔH and ΔS 2 O 3 (g) à 3 O 2 (g) (+) (+) Negative at Spntaneus fr temperatures CaO (s) + CO 2 (g) à CaCO 3(s) (7) 146
high T when -TΔS > ΔH (Entrpy driven) (-) (-) Negative at lw T Spntaneus fr temperatures when ΔH > -TΔS (Enthalpy driven) (+) (-) Never negative Nn-spntaneus at all temperatures due t bth ΔH and ΔS CaCO 3(s) à CaO (s) + CO 2(g) 3 O 2 (g) à 2 O 3 (g) Nn-Standard Cnditins When a prcess is carried ut with reactants and prducts at nn-standard cnditins, use the fllwing equatin t predict the directin f reactin: rxn Δ G = ΔG + RT lnq Q is the reactin qutient. (8) rxn When > 0 the prcess prceeds t the left, when < 0 the prcess prceeds t the right, and when = 0 the prcess is at equilibrium and Q = K. This means that Δ = RT ln K. (9) G rxn When Δ G rxn > 0, the K is < 1, when ΔG rxn < 0 then K > 1, and when ΔGrxn at equilibrium at standard cnditins and K = 1. = 0 then prcess is Cupled Reactins All reactins that take place are nt (as islated prcesses) spntaneus (fr example, steel cmpanies can extract irn metal frm xide res f irn... but this is dne by cupling reactins tgether alng with an input f heat energy t create an verall prcess that is spntaneus: in the thermdynamic sense). Fr example the prductin f steel requires irn (Fe), but in nature irn is fund in res as Fe 2 O 3 r Fe 3 O 4. The decmpsitin f bth either f these cmpunds cannt be accmplished under reasnable cnditins (even at high T) withut the additin f a reducing agent (a reactant that is spntaneusly xidized). Fr example, hydrgen gas r carbn can be used. Fe 2 O 3 (s) ß à 2 Fe (l) + 3/2 O 2 (g) (10) 3/2 O 2 (g) + 3 H 2 (g) ß à 3 H 2 O (g) (11) Fe 2 O 3 (s) + 3 H 2 (g) ß à 2 Fe (l) + 3 H 2 O (g) (12) Using the tables at the back f yur text, find ΔHº, ΔSº, and ΔGº at 298 K and cmment n the signs and magnitude f each fr reactins 10, 11, and 12. Then fr reactin 12, shw that yu 147
can find ΔHº, ΔSº, and ΔGº by simply adding tgether the values f these parameters fund fr reactins 10 and 11. Als, determine the values f K eq (at 298 K) fr all three reactins and demnstrate that K 12 = (K 10 )x( K 11 ). Nw determine if reactin 12 is favrable under standard cnditins at the temperatures used in blast furnaces at steel mills (abut 900ºC). Aside frm altering the temperature, what else can be changed t make the frward reactin in reactin number 12 mre favrable? Finally, is the frward reactin in reactin 12 entrpy driven r enthalpy driven? Warning As yu lk back ver the past chapters, yu will see many reactins with different values f K (sme large, thers small). It is imprtant that yu d nt misinterpret what yu have learned in this chapter. Thermdynamic spntaneity shuld NOT be cnfused with ther reactin terms. Fr example, d nt cnfuse thermdynamic favrability with chemical kinetics. A reactin that is thermdynamically favrable may nt ccur at a useful, r even an bservable, rate. The frmatin f ammnia frm hydrgen gas and nitrgen gas has a large K (negative ΔGº) at rm temperature and yet a mixture f nitrgen and hydrgen gas wuld nt prduce measurable amunts f ammnia after weeks at rm temperature because the reactin is kinetically hindered (that is, it has a large activatin energy, E a ). N 2 (g) + 3 H 2 (g) ß à 2 NH 3 (g) (13) Refer back t Chapter 14 fr a discussin f the Haber prcess fr ammnia prductin and see hw the reactin is run at elevated temperature ver a catalyst t increase reactin rate. Since ΔHº fr the reactin is negative (abut -92 kj as written), the equilibrium cnstant becmes smaller than 1 (K<1) and the ΔGº becmes large (ΔGº>0) when the temperature is raised. In rder t make the reactin useful, the pressure is raised by pumping in a large number f mles f nitrgen and hydrgen (this favrs the side f the reactin with fewer mles f gas, the prduct side). Simultaneusly the reactin is cycled thrugh a heat exchanger t remve the ammnia frm the reactin mixture as a liquid. This keeps the reactin frm reaching equilibrium (i.e. ΔG is always < 0 and Q is always < K). Als, reactins that, at first glance, seem t be thermdynamically impssible may ccur quickly. Cnsider the acid inizatin f acetic acid (K a = 1.8x10-5... yu will find ΔGº>0). Hwever, yu knw frm experience that acetic acid des inize! What is missing frm this picture? Recall that ΔGº refers nly t a reactin ccurring at standard cnditins. That is fr the inizatin f acetic acid we wuld be cnsidering the fllwing starting cnditins fr the reactin: CH 3 COOH (aq) + H 2 O (l) <===> H 3 O + (aq) + CH 3 COO - (aq) initial 1.0 M ---- 1.0 M 1.0 M S, fr these starting cnditins Q = 1 and that is NOT at equilibrium and Q > K therefre the reverse reactin wuld ccur. Hwever, when we did mst acid dissciatin prblems we started ut at NONSTANDARD cnditins like these... 148
CH 3 COOH (aq) + H 2 O (l) <===> H 3 O + (aq) + CH 3 COO - (aq) initial 0.10 M ---- ~0.0 M 0.0 M Nw in this case, Q is < K and the frward reactin must ccur. It might be f interest fr yu t find ΔG fr this set f starting cnditins and cnvince yurself that the frward reactin is the predicted directin. We culd d the same srt f arguments with slubility (K sp ) type reactins, base inizatin (K b ) type reactins, cmplex frmatin (K f ) type reactins, with the autprtlysis (self-inizatin reactin (K w ) fr water, and even with vaprizatin, sublimatin, melting, and slutin frmatin prcesses. Maximum Wrk The First Law f Thermdynamics tells us that chemical prcesses cannt create r destry energy (energy is cnserved s the best we can hpe t is t break even when cnverting energy frm ne frm t anther). The Secnd Law f Thermdynamics is mre restrictive: it indicates that we cannt hpe t break even. Althugh it may nt be apparent frm the prir discussin, ΔG and ΔGº type calculatins will give us infrmatin abut the maximum amunt f useful wrk we can hpe t extract frm any chemical r physical change. We will fllw up n this t sme extent in the chapter n electrchemistry. Cnclusin Every single equilibrium prblem that yu have wrked with Q s and K s can als be dne with ΔG and ΔGº type calculatins. When yu fund a K eq at a new temperature using the van t Hff equatin, yu were really finding a new ΔGº (at that same new temperature). S... what have we gained fr all f ur trubles? Well... we shuld appreciate why sme reactins ccur at all temperatures while thers becme spntaneus at lw OR at high temperatures. Recgnizing whether a reactin is driven by enthalpy change r entrpy change is useful infrmatin. We nw have a better understanding f why endthermic reactins are favred at higher temperatures (after all we have seen the equatin frm which the van t Hff equatin is derived). We als shuld better appreciate the subtle cnnectins between kinetics and equilibrium. Als, nw that yu understand thermdynamics, yu shuld knw that any prcess that des ccur must be spntaneus in the thermdynamic sense (even if that spntaneity is the result f cupling the reactin t a spntaneus prcess). This is particularly imprtant in bilgical system. Mst f the prcesses that keep yu alive are nt spntaneus... until yur bdy cuples them t the pathways that metablize yur fd. The energy needed t make yur heart beat, t create prteins, the functining f yur brain ALL rely n being cupled with the cnversin f fd int mlecules that are thermdynamically mre stable than the fd itself. Fr example, a triglyceride (fat) mlecule is a high energy mlecule. Cnverting it int carbn 149
dixide and water releases energy just as burning whale il in a lamp des (yur bdy simply rchestrates the energy release in a manner which allws yur bdy t use the energy - instead f burning it in an pen flame). Hwever, bth cnversins must bey the Secnd Law f Thermdynamics and as such invlve waste energy. It is useful t nte that the sun prvides the energy required t manufacture the fd we eat n this planet. Withut the cnstant input f energy frm the NUCLEAR reactr that we call the SUN, life as we knw it (and perhaps n life at all) wuld exist n this planet. We will discuss NUCLEAR energy PRODUCTION (nte that I did nt say cnversin) later in this semester. Keywrds Clausius-Clapyern Enthalpy Enthalpy driven Entrpy Entrpy driven Equilibrium Equilibrium cnstant Gibbs Free Energy Nnspntaneus Reactin Crdinate Reactin Qutient Spntaneus Van t Hff Equatin Wrk 150