Chemical Thermdynamics Spntaneus Prcesses, Entrpy and the Secnd Law f Thermdynamics Review Reactin Rates, Energies, and Equilibrium Althugh a reactin may be energetically favrable (i.e. prducts have lwer energy than reactants), reactin rates are largely determined by the magnitude f the activatin energy barrier Fr example, mlecules require a minimum kinetic energy t cllide and react The higher the activatin energy barrier, the fewer the mlecules in the ppulatin with enugh energy t vercme the barrier At equilibrium, ppsing reactin rates (e.g. frward and reverse reactin rates) are equal in magnitude Reactin rates are related t the prduct f an intrinsic rate cnstant and the cncentratin f reactant(s) If equilibrium is related t reactin rates and reactin rates are related t energy, then equilibrium must be related t energy in sme way We have used the magnitude and sign f reactin enthalpies t predict whether a reactin will prceed r nt Reactins with negative enthalpies indicate heat release; the prducts are lwer in energy than the reactants and the reactin will prceed Reactins with psitive enthalpies indicate that heat must be input int the system fr the reactin t prceed. The prducts are higher in energy than the reactants and the reactin will nt usually prceed spntaneusly Hwever, we have seen that a cnsideratin f reactin enthalpies alne is nt enugh t predict whether a reactin (r physical prcess) will prceed NH 4 SO 3 (s) disslves in water spntaneusly even thugh the enthalpy f slvatin is psitive (it absrbs heat and is the basis f instant ice packs) Spntaneus Prcesses The first law f thermdynamics: In additin t enthalpy, we must cnsider the change in the randmness r disrder that accmpanies a reactin r physical prcess The Law f Cnservatin f Energy Any energy lst by a system must be gained by the surrundings, and vice versa. Energy is neither created nr destryed by a prcess (it is nly transfrmed frm ne frm t anther) Energy, E, can be assciated with wrk, w (i.e. frce * distance, in the displacement f an bject) r the transfer f heat, q ΔE = q w ΔE = the change in the internal energy f a system q = heat absrbed by the system frm the surrundings w = the wrk dne n the system by the surrundings We can keep track f the wrk dne n a system and the heat absrbed r emitted by a system, hwever, since the verall change in energy fr (the system surrundings) = 0, we cannt use the change in energy f a system t predict whether a reactin will ccur If the internal energy f a system is increased, then the energy f the surrundings must decrease, and vice versa 1
When the effects f the energy change f the surrundings are taken int accunt, the net energy change = 0 Sme prcesses are knwn t always ccur: Highly rdered systems becme mre disrdered Eggs have a tendency t break, yet I have never seen an intact egg spntaneusly reassemble frm a pile f egg bits A pile f leaves in my garden will eventually distribute themselves ver my entire lawn. Unfrtunately, I have never seen the leaves n my lawn spntaneusly assemble int a nice neat pile by themselves I have seen cean waves destry an elabrate sand castle, but have never seen a sand castle spntaneusly rise up and frm ut f the sandy surf. These spntaneus prcesses als have an intrinsic directin, even thugh there is n apparent net energy change By all accunts, in the absence f sme utside influence, the abve spntaneus prcesses are irreversible Althugh a prcess may be spntaneus, that des nt mean that it will necessarily be fast Thermdynamics prvides infrmatin abut the directin a reactin r prcess will prceed (i.e. which directin f a reactin is spntaneus, but nt abut hw fast it will g Spntaneity f a prcess depends nt nly upn the enthalpy change (e.g. exthermic) but als upn hw the disrder f the system changes Entrpy and the Secnd Law Cnsider the fllwing tw experiments: 1. Behavir f gas in a cntainer I have tw glass cntainers cnnect by a valve. In the lefthand side cntainer I have a sample f nitrgen gas. In the righthand side cntainer there is nthing (i.e. a vacuum): I then pen the valve cnnecting the tw cntainers. What happens? 2
There is a spntaneus prcess that invlves the flw f gas mlecules frm the left chamber t fill in the empty righthand side chamber There was n wrk perfrmed either n the system r n the surrundings (w = 0) There was n heat input r utput frm the system t the surrundings (q = 0) The spntaneus reactin results in pressure unifrmity Althugh the gases spntaneusly mved frm the left chamber t the right, it is highly unlikely (impssible) that the gases might spntaneusly mve back int the lefthand side chamber (t prduce the riginal cnditin). Thus, the prcess appears t be spntaneus and irreversible (in the absence f sme utside influence) 2. The behavir f bjects f different temperature I have tw cubes f metal. One cube I stick in the ven until it gets red ht. The ther I put in the freezer until it gets really cld. Then I put the tw blcks next t each ther: Over time what will happen t the temperature f the tw blcks? There is a spntaneus prcess that invlves the flw f heat energy frm the ht blck t the cld blck (the net heat energy is unchanged, hwever) There was n wrk perfrmed either n the system r n the surrundings (w = 0) There was n heat input r utput frm the system t the surrundings (q = 0) The spntaneus reactin results in temperature unifrmity Althugh the heat energy spntaneusly mved frm the ht blck t the cld blck, it is highly unlikely (impssible) that heat energy might spntaneusly mve back int the lefthand side blck (t prduce the riginal cnditin). Thus, the prcess appears t be spntaneus and irreversible (in the absence f sme utside influence) In bth f the abve cases, we had a situatin where the starting cnditin was highly rdered (gas n left, vacuum n right; ht blck n left, cld blck n right) The final cnditin was ne f unifrmity (f pressure r temperature) The spntaneity f the prcess appears t be assciated with a highly rdered system ging t a lessrdered, unifrm state The disrder is expressed by a thermdynamic quantity called entrpy (S) 3
The mre disrdered a state, the larger its entrpy (a large magnitude fr S means a lt f disrder) Entrpy is a state functin: ΔS = S final S initial In ther wrds, the pathway yu chse t get t S final is nt imprtant, what is imprtant is discussing the change in entrpy f a system is a cmparisn f the initial and final values f the entrpy A psitive value fr ΔS indicates an increase in disrder. A negative value fr ΔS indicates a decrease in disrder Hw is the change in the enthalpy f a system (ΔS) defined? (where T is sme cnstant temperature) Obviusly an enthalpy change is in sme way prprtinal t sme kind f energy term. If a system des nt d any mechanical wrk, then heat energy must be invlved. (Nte: heat, like wrk, is nt an entity but a methd f energy transfer) If the system is at cnstant T, then the nnmechanical energy flw is nt being used t change the temperature (it is assciated with the changing rder f the system) Why is ΔS inversely prprtinal t the abslute (K) temperature? At high temperatures what will the system be like? It will be a highly energetic gas and it will be difficult t get much mre disrder ut f it (thus at high temperatures, the change in disrder (ΔS) will be small fr a given amunt f nnmechanical energy transfer t the system. At lw temperatures the system will be mre rdered (highest rder is at abslute zer) and the change in disrder (ΔS) will be larger fr a given amunt f nnmechanical energy transfer t the system One f the classic examples f entrpy changes in respnse t nnmechanical energy transfer at a cnstant temperature is the melting f ice at 0 C (i.e. 273K) At 0K liquid water is in equilibrium with slid (i.e. ice) Nnmechanical energy can be transferred int the system withut raising the temperature (this is true as lng as there is sme ice and water present; if things get t the pint where all the ice is melted, then the temperature will raise as yu heat the water; if all the liquid water is frzen, then yu lwer the temperature f the ice as heat is remved; therwise, the effect f energy flw int and ut f the system results in either mre r less ice being present, but n temperature change) The amunt f heat transferred t the system during the fusin f ice is the heat f fusin, ΔH fus (6.01kJ/mle). Thus, ΔS = 6.01kJ/mle / 273K = 22 J/ml K ΔS is psitive, indicating that the liquid frm f water has greater disrder cmpared t the slid frm (ice) (Nte: this is an equilibrium situatin, nt an irreversible spntaneus reactin. Thus, input energy must increase the disrder f the system) We can als define the entrpy change in relatinship t expansin f a gas at cnstant T (i.e. isthermal expansin ΔS = nr ln (V final /V initial ) (fr an ideal gas at sme cnstant T) The Secnd Law f Thermdynamics 4
The law that expresses the idea that there is an inherent directin in which prcesses ccur is called the secnd law f thermdynamics We must cnsider the change in entrpy f the system and the surrundings Tgether, the system and the surrundings cnstitute the universe ΔS universe = ΔS system ΔS surrundings Fr an irreversible (spntaneus) prcess, we have an increase in entrpy. Since it is irreversible, the Universe has gained entrpy: ΔS universe = (ΔS system ΔS surrundings) > 0 Fr a reversible prcess (nt spntaneus, but manipulable by heat flw) there is n net ΔS as far as the Universe is cncerned (i.e. if we input heat energy int the system, then the heat energy f the surrundings decreases) ΔS universe = (ΔS system ΔS surrundings) = 0 Thus, there is n case where ΔS f the universe decreases ΔS universe is cnstantly increasing (the universe is mving twards greater and greater disrder) Chemical reactins fllw this same law Cnsider an exthermic reactin that is less S decreases; the disrder decreases) O 2 (g) 2H 2 (g) 2H 2 O(g) Let's perfrm this reactin in the fllwing way: We will keep the temperature cnstant. Since the reactin is exthermic, we d this by having the surrundings absrb the released heat. The prduct is tw mlecules f water. The reactants cmprise three mlecules (ne f O 2 and tw f H 2 ) Since temperature is cnstant, the verall change in entrpy f the system (reactants and prducts) is reduced (prducts are fewer gas mlecules due t atms being bnded tgether this reduces the entrpy f the atms because since they are bnded tgether they are restricted in their mvements) The released heat is taken up by the surrundings The surrundings are kept at same temperature. Since temperature is cnstant, the energy absrbed by surrundings is manifest as increased disrder. The surrundings becme mre disrdered q sys /T = q surr /T (where T is cnstant) in ther wrds the entrpy lst by the system = entrpy gained by surrundings If the entire prcess were reversible, ΔS universe = 0 If any small part is irreversible, ΔS > 0 Althugh the reactin results in a decrease in entrpy, the net entrpic change f the universe is either 0 r psitive 5
A Mlecular Interpretatin f Entrpy Many prcesses can lead t an increase in a system's entrpy Increasing the vlume that a gas can ccupy will increase the disrder f a gas Disslving a slute int a slutin will increase the entrpy f the slute typically resulting in an increase in the entrpy f the system. (Nte: the slvatin f a slute can smetimes result in a significant decrease in the slvent entrpy leading t a net decrease in entrpy f the system) Phase changes frm slid t liquid, r liquid t gas, lead t an increase in the entrpy f the system Sme prcesses can lead t a decrease in the entrpy f a system A gas mlecule disslved in a liquid is much mre cnfined by neighbring mlecules than when it is in the gaseus state. Thus, the entrpy f the gas mlecule will decrease when it is disslved in a liquid A phase change frm a liquid t a slid (i.e. freezing), r frm a gas t a liquid (i.e. cndensatin) results in an decrease in the disrder f the substance, and a decrease in the entrpy A chemical reactin between gas mlecules that results in a net decrease in the verall number f gas mlecules will decrease the disrder f the system, and result in a decrease in the entrpy 2NO(g) O 2 (g) 2NO 2 (g) ΔS < 0 What is the mlecular basis fr the abve bservatins fr the change in entrpy? Let's first cnsider the last example, the decrease in entrpy assciated with a decrease in the number f gas mlecules fr a chemical reactin 2NO(g) O 2 (g) 2NO 2 (g) The prduct (NO 2 ) invlves the frmatin f a new NO bnd. The O atms, riginally in a separate O 2 mlecule, are nw cnnected t the NO mlecule via a new NO bnd Since they are nw physically bnded t the ther mlecule (frming a new, larger, single mlecule) the O atms have less freedm t mve arund The reactin has resulted in a lss f the degrees f freedm f the atms (O atms) There is a reductin in the disrder f the system (i.e. due t the reductin in the degrees f freedm, the system is mre rdered after the reactin). ΔS < 0. Degrees f Freedm f mlecules Individual mlecules have degrees f freedm related t their mtins within a substance Translatinal mtin. The entire mlecule can mve in sme directin in three dimensins 6
Rtatinal mtin. The entire mlecule can rtate arund any axis, (even thugh it may nt actually change its psitin translatinally) Vibratinal mtin. The atms within a mlecule have certain freedm f mvement relative t each ther; this displacement can be peridic mtin like the vibratin f a tuning frk These frms f mtin are ways in which the mlecule can stre energy The greater the energy that is stred in these mtins, the greater the degrees f freedm, and the greater the entrpy The Third Law f Thermdynamics If we lwer the temperature f the system, the thermal energy decreases the energy stred in translatinal, rtatinal and vibratinal mtins decreases the entrpy f the system decreases The Third Law f Thermdynamics: The entrpy f a pure crystalline substance at abslute zer (i.e. 0 Kelvin) is 0. S = 0 crrespnds t perfect rder. The psitin f the atms r mlecules in the crystal wuld be perfectly defined As the temperature increases, the entrpy f the atms in the lattice increase Vibratinal mtins cause the atms and mlecules in the lattice t be less well rdered Cntinued heating f a slid lattice In a slid lattice, neighbring mlecules are cnstrained t a certain psitin in the lattice (that's what makes a slid a slid) In the liquid state, a mlecule is free t mve abut the entire vlume f the liquid Mlecules in a liquid have a higher degree f freedm than in a slid The liquid has a higher entrpy than a slid As we heat a slid, there is a gradual increase in the entrpy f the system At the melting pint there is a dramatic increase in the entrpy f the system as neighbring mlecules are free t mve past each ther After all the slid has melted, the system is in the liquid state As we heat the liquid, there is a graduate increase in the entrpy In the vapr state, a mlecule is free t mve abut the entire cntainer (ptentially the entire universe) 7
Mlecules in the vapr state have a higher degree f freedm than in the liquid state The vapr has a higher entrpy than the liquid At the biling pint there is a dramatic increase in the entrpy f the system After all the liquid has vaprized, the system is in a gaseus state As we heat the gas, there is a gradual increase in the entrpy f the system In general, the entrpy is expected t increase fr the fllwing types f prcesses: 1. The melting f a slid t frm a liquid 2. The vaprizatin f a liquid (r slid) t prduce a gas 3. Chemical reactins that invlve phase changes f slid liquid/gas, r liquid gas 4. Chemical reactins that result in an increase in the number f gaseus mlecules 5. Any time the temperature f a substance is increased Calculatin f Entrpy Changes We have seen that the energy given ff (r absrbed) by a reactin, and mnitred by nting the change in temperature f the surrundings, can be used t determine the enthalpy f a reactin (e.g. by using a calrimeter) Tragically, there is n cmparable easy way t measure the change in entrpy fr a reactin. Hwever, ne thing t think abut is the fllwing: suppse we knw that energy is ging int a system (r cming ut f it), and yet we d nt bserve any change in temperature. What is ging n in such a situatin? Changes in internal energy, that are nt accmpanied by a temperature change, might reflect changes in the entrpy f the system Fr example, cnsider water at 0C at 1atm pressure This is the temperature and pressure cnditin where liquid and slid phases f water are in equilibrium (als knwn as the melting pint f ice) H 2 O(s) H 2 O(l) At such a temperature and pressure we have a situatin (by definitin) where we have sme ice and sme liquid water If a small amunt f energy is input int the system the equilibrium will shift slightly t the right (i.e. in favr f the liquid state) 8
Likewise if a small amunt f energy is withdrawn frm the system, the equilibrium will shift t the left (mre ice) Hwever, in bth f the abve situatins, the energy change is nt accmpanied by a change in temperature (the temperature will nt change until we n lnger have an equilibrium cnditin; i.e. all the ice has melted r all the liquid has frzen) Since the quantitative term that relates the amunt f heat energy input vs. the rise in temperature is the heat capacity, it wuld seem that in sme way, infrmatin abut the heat capacity (and hw it changes with temperature) wuld allw us t determine the entrpy change in a system In fact, values fr the "standard mlar entrpy" f a substance have units f J/ml K. The same units fr mlar heat capacity. Standard Mlar Entrpy, S 0 The entrpy f a substance has an abslute value f 0 entrpy at 0K Standard mlar entrpies are listed fr a reference temperature (like 298K) and 1 atm pressure (i.e. the entrpy f a pure substance at 298K and 1 atm pressure). A table f standard mlar entrpies at 0K wuld be pretty useless because it wuld be 0 fr every substance (duh!) When cmparing standard mlar entrpies fr a substance that is either a slid, liquid r gas at 298K and 1atm pressure, the gas will have mre entrpy than the liquid, and the liquid will have mre entrpy than the slid Unlike enthalpies f frmatin, standard mlar entrpies f elements are nt 0. The entrpy change in a chemical reactin is given by the sum f the entrpies f the prducts minus the sum f the entrpies f the reactants As with ther calculatins related t balanced equatins, the cefficients f each cmpnent must be taken int accunt in the entrpy calculatin (the n, and m, terms belw are there t indicate that the cefficients must be accunted fr) ΔS 0 = Σ ns 0 (prducts) Σ ms 0 (reactants) Calculate the change in entrpy assciated with the Haber prcess fr the prductin f ammnia frm nitrgen and hydrgen gas. At 298K as a standard temperature: S 0 (NH 3 ) = 192.5 J/ml K S 0 (H 2 ) = 130.6 J/ml K S 0 (N 2 ) = 191.5 J/ml K N 2 (g) 3H 2 (g) 2NH 3 (g) Frm the balanced equatin we can write the equatin fr ΔS 0 (the change in the standard mlar entrpy fr the reactin): ΔS 0 = 2*S 0 (NH 3 ) [S 0 (N 2 ) (3*S 0 (H 2 ))] ΔS 0 = 2*192.5 [191.5 (3*130.6)] ΔS 0 = 198.3 J/ml K It wuld appear that the prcess results in a decrease in entrpy i.e. a decrease in disrder. This is expected because we are decreasing the number f gas mlecules. In ther wrds the N 2 (g) used t flat arund independently f the H 2 gas mlecules. After the reactin, the tw are bnded tgether and can't flat arund freely frm ne anther. (I guess yu can cnsider marriage as a negative entrpy prcess) 9
Gibbs Free Energy J. Willard Gibbs was the first persn t be awarded a Ph.D. in science frm an American University (Yale, 1863) Spntaneus reactins ften have: A negative enthalpy (release f heat energy, ΔH < 0). An increase in entrpy (increase in disrder, ΔS > 0) The spntaneity f a reactin appears t invlve tw thermdynamic prperties: enthalpy and entrpy Furthermre, spntaneus reactins are thse that g dwnhill in energetic terms. In ther wrds, the final state has a lwer energy cntent than the initial state Gibbs came up with an equatin, cmbining bth enthalpy and entrpy cntributins, that prvided a means t describe energy cntent and therefre a means t evaluate the spntaneity f a reactin when that energy cntent changes. The energy cntents f a substance was termed the Gibbs Free Energy and it was defined by the Gibbs Free Energy equatin: G = H T*S The free energy f a substance = stred heat energy inherent disrder at a reference temperature H is enthalpy, S is entrpy and T is the temperature in Kelvin If there is a lt f stred heat energy, then the substance has a lt f free energy The mre disrder a substance has, the less free energy it has Changes in a substance (as in a chemical reactin r physical phase change): ΔG = ΔH T*ΔS If the substance releases heat energy, then the prduct has a lwer value f stred heat energy and ΔH < 0. Such a change is dwnhill energetically (i.e. spntaneus) If the disrder (entrpy) increases, this is als a spntaneus prcess and ΔS > 0. Since ΔS > 0 this means that (T * ΔS) < 0. Therefre, bth the enthalpic (ΔH) and entrpic (T*ΔS) terms are negative fr prcesses that are spntaneus Hw t quantitate the cntributin f the entrpic term t the free energy? In "instant ice packs" a reactin ccurs that is spntaneus, and yet is endthermic (i.e. it is "cld" due t absrptin f heat energy). The absrptin f heat energy is unfavrable and therefre must be "driven" by an increase in entrpy (i.e. a large T*ΔS term). Therefre, the degree t which heat can be spntaneusly absrbed is actually prviding us with infrmatin regarding the magnitude f the entrpic increase (i.e. the entrpic increase is what is "driving" the unfavrable heat absrptin). Recall that entrpy was defined previusly as q/t (i.e. nnmechanical energy transferred at a cnstant temperature). This definitin is just a way f saying that if heat is absrbed but the temperature des nt change, then the heat energy is being sucked in due t an entrpic increase and that entrpic increases is energetically equal t the heat energy absrbed In such a case ΔS is psitive, entrpy increases, and (T*ΔS) is negative (favring spntaneus reactin) 10
Heat absrbed at a fixed temperature is essentially an infinite heat capacity, and this is just t pint ut that entrpy has units f J/ml K, like heat capacity values. A material with a high heat capacity has the ability t absrb a lt f heat energy with a small temperature change. It can d this because it is able t increase its internal disrder and "sak" up the heat energy and thus the temperature wn't rise much. If a material cannt increase its internal disrder, then it will increase temperature in respnse t even a small amunt f added heat energy. Fr a prcess ccurring at cnstant temperature, T: If ΔG = 0, it means that the enthalpy change assciated with a reactin is equal in magnitude (and ppsite in sign) t the entrpy change. Fr example, the prcess being cnsidered may result in a reductin in entrpy (i.e. mre rder) but it releases an amunt f heat energy that exactly cunteracts the effects f the TΔS term. Such a reactin is in equilibrium (i.e. n net reactin) If ΔG < 0. This wuld ccur fr nt nly an exthermic reactin (ΔH = negative) that verwhelms any unfavrable entrpic effect. But ptentially als fr an endthermic prcess that has a significant increase in disrder (i.e. ΔS is large, and thus, TΔS is large and negative). In either case, the prcess being cnsidered has a net driving frce (release f energy r increase in entrpy) that indicates the reactin is spntaneus If ΔG > 0. This will happen if the reactin is highly endthermic (ΔH psitive) and the entrpic term is nt s great. Or, if the reactin is exthermic (i.e. ΔH is negative), but the prcess results in significant increase in rder (i.e. the T*ΔS term ends up being a negative number, s that T*ΔS is psitive in magnitude). In any case, this is energetically an unfavrable prcess being cnsidered. T mve in the frward directin, energy must be supplied. If energy is nt supplied, then the reactin will prceed spntaneusly in the reverse directin. Free energy and chemical reactins at equilibrium. It wuld seem there are 4 pssible types f reactins r prcesses with regard t the enthalpic and entrpic cntributin t the free energy change: 1. ΔH = (), TΔS = (). Favrable enthalpic change (exthermic) and favrable entrpic change (disrder increases) 2. ΔH = (), TΔS = (). Unfavrable enthalpic change (endthermic) and unfavrable entrpic change (disrder decreases) 3. ΔH = (), TΔS = (). Favrable enthalpic change (exthermic) and unfavrable entrpic change (disrder decreases) 4. ΔH = (), TΔS = (). Unfavrable enthalpic change (endthermic) and favrable entrpic change (disrder increases) If yu lk at these fur types f energy changes, yu will ntice that 1) and 2) are cnsidering the same prcess, just frm different directins. Likewise, with 3) and 4) (i.e. an exthermic prcess in ne directin is endthermic in the ppsite directin). S, in principle, we just have t understand tw types f prcesses. Fr type 1) abve, bth enthalpy and entrpy favr the frward directin. There wuld appear t be n energetic term favring the reverse directin. Thus, this type f prcess wuld be expected t g t cmpletin (i.e. n equilibrium cnditin because n reverse reactin ccurs). Type 2) is the same situatin, but viewed frm the ppsite directin. Fr type 3) abve, enthalpy favrs the frward directin but the energy assciated with the entrpic change favrs the reverse directin. There are, therefre, energetic frces driving the prcess in ppsite directins, and it wuld seem likely that an equilibrium cnditin might exist. Type 4) is the same situatin, but viewed frm the ppsite directin. Cnsider ur ld friend, the Haber reactin: 11
If we start with nthing but NH 3 (g) in the sample, the reactin will prceed in the reverse directin t prduce H 2 (g) and N 2 (g) (i.e. ΔG fr the reactin will be a psitive value) N 2 (g) 3H 2 (g) 2NH 3 (g) Or, ΔG is negative fr the fllwing reactin (i.e. the fllwing reactin is spntaneus): 2NH 3 (g) N 2 (g) 3H 2 (g) If we start with nthing but H 2 (g) and N 2 (g) in the sample, the reactin will prceed in the frward directin t prduce NH 3 (g) (i.e. ΔG fr the reactin will be a negative value) N 2 (g) 3H 2 (g) 2NH 3 (g) If we start with cncentratins f all cmpnents such that Q = K c, then the reactin is at equilibrium and ΔG = 0. The reactin wants t be driven in a directin such that ΔG ges t 0 (i.e. equilibrium) At equilibrium the free energy f the system is at a minimum. T prduce either mre prduct, r mre reactants, requires an increase in free energy (i.e. sme mdificatin f heat r entrpy prperties) Standard FreeEnergy Changes The freeenergy term, G, is a state functin, thus values can be defined fr substances at specific cnditins f temperature and pressure knwn as the standard state. In this case, we will have ΔG values assciated with the frmatin f cmpunds frm their elemental cnstituents, knwn as the standard free energy f frmatin, ΔG f 0. Standard cnditins include: 1 atm pressure 1M cncentratin (slutins) Pure slid (if a slid) r pure liquid (if a liquid) Fr elements, the standard free energy f frmatin, ΔG 0 f f an element in its nrmal state is 0 There is n standard state fr temperature. G will vary with temperature. 298K (i.e. 25 C) is a cmmn temperature chsen fr standard reference values f G Standard freeenergy values can be used t calculate the standard free energy change assciated with a reactin: 12
ΔG 0 = Σ ΔG f 0 (prducts) Σ ΔG f 0 (reactants) What infrmatin will the calculatin f ΔG 0 prvide? ΔG 0 < 0, the reactin is spntaneus as written (i.e. ges t the right) ΔG 0 > 0, the reactin will prceed t the left as written ΔG 0 = 0, the reactin is at equilibrium Free Energy and Temperature Free Energy and Temperature Free Energy and the Equilibrium Cnstant Hw is the change in free energy (i.e. ΔG ) as we g frm state 1 t state 2 (r reactants t prducts) affected by temperature? The enthalpic term, ΔH ΔG = ΔH T*ΔS Exthermic reactins release heat t the surrundings and have a negative value fr ΔH If there were n entrpic cnsideratins, all exthermic reactins wuld result in negative values f ΔG, and wuld define spntaneus reactins The entrpic cntributin, T*ΔS The value f ΔS (the entrpic change) may be either psitive (prducts have a greater degree f disrder than reactants) r negative (prducts have a lesser degree f disrder than reactants) Fr thse reactins where ΔS is psitive, the TΔS term will be negative. This will cntribute t the verall negative value (i.e. spntaneity) f ΔG Fr thse reactins where ΔS is negative, the TΔS term will be psitive. This decrease in entrpy will ppse spntaneity. This will reduce the magnitude f a negative value f ΔG and may even cause ΔG t be psitive (indicating a nnspntaneus prcess) What is the effect f temperature n ΔH and ΔS values, and ΔG? Generally, ΔH and ΔS values d nt change much with temperature Althugh the ΔS term may nt change much with temperature, the magnitude f the TΔS term is bviusly dependent upn the value f T (temperature). Fr an equivalent change in disrder, greater energy is absrbed at higher temperatures. The greater the temperature, the greater the magnitude f the TΔS term If ΔS is negative (i.e. unfavrable entrpy change), the value f the TΔS term is psitive, and increasing the temperature will increase the magnitude f the psitive value f the TΔS term 13
At sme high temperature, the magnitude f the (psitive) TΔS term can verwhelm the magnitude f an exthermic (i.e. negative) ΔH term. And even thugh the reactin is exthermic it will nt ccur spntaneusly (i.e. ΔG = psitive in value) The ppsite type f situatin may ccur: a reactin may be endthermic (ΔH is psitive), and have a psitive value fr ΔS With a psitive value fr ΔS, the TΔS term favrs spntaneity. Hwever, at lw temperatures, its cntributin will be small (magnitude f TΔS at lw temperatures is small). In this case ΔH predminates and the reactin will be nnspntaneus (ΔG = psitive) At higher temperatures, the magnitude f the TΔS term increases and can verwhelm the (psitive) ΔH term. In this case the TΔS predminates and the reactin is spntaneus (ΔG = negative at high temp) 14
Summary: ΔH ΔS TΔS ΔG (exthermic) (prducts mre disrdered) (favrs spntaneity) (spntaneus at all T) (exthermic) (prducts less disrdered) (ppses spntaneity) (spntaneus) at lw T (nnspntaneus) at high T "Enthalpicallydriven prcess" (endthermic) (prducts mre disrdered) (favrs spntaneity) (nnspntaneus) at lw T (spntaneus) at high T "Entrpicallydriven prcess" (endthermic) (prducts less disrdered) (ppses spntaneity) (nnspntaneus at all T) 1
Cnsider the fllwing pssible states fr tw different types f mlecules with sme attractive frce: There wuld appear t be greater entrpy n the left (state 1) than n the right (state 2). Thus the entrpic change fr the reactin as written (i.e. ging t the right) wuld be () in magnitude, and the energetic cntributin t the free energy change wuld be () (i.e. unfavrable) fr the reactin as written. In ging t the right, there is an attractive frce and the mlecules adjacent t each ther is a lwer energy state (heat energy, q, is liberated). T g t the left, we have t vercme this attractive frce (input heat energy) and the left directin is unfavrable with regard t heat energy q. The change in enthalpy is () in ging t the right (q released), and this enthalpy change is negative () in ging t the right (and () in ging t the left). This reactin as written, is therefre, enthalpically favrable, and entrpically unfavrable. It is enthalpically driven. Frm the abve table, it wuld appear that we might be able t get the reactin t g t the right at lw temperatures (lw temperature wuld minimize the energetic cntributin f the entrpic change). Lking at the same prcess frm an ppsite directin: This reactin as written, is entrpically favrable, and enthalpically unfavrable. It is entrpically driven. Frm the abve table, it wuld appear that we might be able t get the reactin t g t the right at high temperatures (high temperature wuld increase the energetic cntributin f the entrpic change). Free Energy and the Equilibrium Cnstant Recall that standard freeenergy f frmatin values frm tables can be used t calculate the standard free energy change assciated with a reactin: ΔG 0 = Σ ΔG f 0 (prducts) Σ ΔG f 0 (reactants) These tabulated ΔG 0 f values are fr standard cnditins and a defined temperature Often we are interested in the value f ΔG 0 fr a reactin that is nt under standard cnditins (e.g. ther cncentratins than standard cnditins f 1 atm fr a gas, r 1M fr a slutin) 1
The general relatinship between the freeenergy change under standard cnditins (i.e. ΔG 0 ) and the freeenergy change under any ther cnditins, (i.e. ΔG), is defined as: ΔG (e.g. under nnstandard cnditins f cnc.) = ΔG 0 RT ln Q R is the gas cnstant, 8.314 J/ml K T is abslute temperature (K) Q is the calculated reactin qutient Q reflects the ratis f the varius cmpnents under starting cnditins. K reflects their ratis at equilibrium. If the starting system is at equilibrium, then Q = K At equilibrium ΔG = 0 (n spntaneus reactin in either directin) Therefre: at equilibrium becmes and ΔG = ΔG 0 RT ln Q ΔG = ΔG 0 RT ln K 0 = ΔG 0 RT ln K r ΔG 0 = RT ln K If K>1 it means the equilibrium favrs the prducts (i.e. spntaneity t the right). In this case lnk will be psitive and therefre ΔG 0 will be negative (again spntaneity t the right) If K<1 it means the equilibrium favrs the reactants (i.e. spntaneity t the left). In this case lnk will be negative and therefre ΔG 0 will be psitive If K = 1, lnk will be zer and s will ΔG 0. The reactin is at equilibrium The abve equatin als allws us t calculate K if we knw ΔG 0 : 2000 Dr. Michael Blaber 2