化學(xué)原理Chemistry課件-post+9+Thermodynamics

Chapter18ThermodynamicsInthisChapter,wewilladdressseveralaspectsasfollows:I)Howto

predict

whetherareactionunderagivenconditionscanoccurornot?II)Howto

makethereactionhappen,ifitdoesnotproceedunderprimaryconditions?

SpontaneityandEntropy(S).TheSecondandThirdLawsofthermodynamics.FreeEnergy(G)andEquilibrium(K).Chapter18Thermodynamics1Spontaneous

process:

spontaneousnon-spontaneous18.2Goaheadbyitself,withouthelpfromsurroundings.spontaneousNon-spontaneousSpontaneousprocess:2Whatarethefactors

controllingthespontaneity?CH4

(g)+2O2

(g)CO2

(g)+2H2O(l)

DH0=-890.4kJH+

(aq)+OH-

(aq)H2O(l)

DH0=-56.2kJH2O(s)H2O(l)

DH0=6.01kJNH4NO3

(s)NH4+(aq)+NO3-

(aq)

DH0=25kJH2O

Energy

Freedom

Whatarethefactorscontrolli3Entropy(S):ameasureofsystemdisorder.orderSdisorderSDS=Sf-SiDS>0Foragivensubstance:Ssolid<Sliquid<<SgasH2O(s)H2O(l)DS>018.2Heatinghydrogengasfrom600Cto800CCrystallizationfromasupersaturatedsolutionDS<0Foraprocess:Entropy(S):ameasureofsyst4EntropyisalsoaStatefunction.

E1H1S1E2H2S2n1

P1V1T1n2

P2V2T2State1State2DE=E2–E1DH=H2–H1DS=S2–S1ThevalueofSisonlydeterminedbythestate,regardlessofhowthestateisreached.EntropyisalsoaStatefuncti5Howtoquantify

theEntropy?MicrostatesW=1W=4W=6SIn1868,BoltzmannproposedS=klnWDS=Sf-SiForaprocess:DS=klnWfWiBoltzmannconstantk=1.38×10-23J/KForCH3OHandC2H5OH:Smethanol<SethanolHowtoquantifytheEntropy?Mi6TheSecondLawofThermodynamicsTheentropyofthe

universe

increasesinaspontaneousprocess,andremainsunchangedinanequilibriumprocess.DSuniv=DSsys+DSsurr>0(1)Spontaneousprocess:DSuniv=DSsys+DSsurr=0(2)Equilibriumprocess:Inordertopredictthedirectionofachemicalreaction,however,bothDSsysandDSsurrneedtobecalculated,whichisaquestionremainingtobesolved.TheSecondLawofThermodynami7Howtocalculate

DSsurr,theentropychangeinthesurroundings?EndothermicDHsys>0DSsurr

DHsysDSsurr>0Foraconstant-pressureprocess,itisknownthatQp=DH.ExothermicDHsys<0DSsurr<0HowtocalculateDSsurr,thee8DSsurr=

DHsysTsurrSsurr

TsurrAthigh

Tsurr,thesurroundingsmoleculesarealreadyveryenergetic.Then,theabsorptionofheatfromthesystemwillnotchangethemolecularmotionverymuch.Thatis,theentropychangeinthesurroundingswillbesmall.Atlow

Tsurr,however,thesameamountofsuchheatwillcausearelativelylargechangeinthemolecularmotion,andthentheentropychangeofthesurroundingswillbelarge.How?DSsurralsodependsonTsurr,thesurroundingstemperature.Therefore,foraconstant-pressureprocess,DSsurr=DHsys9DSuniv=DSsys+DSsurr>0foraspontaneousprocess:DSuniv=DSsys

->0TsurrDSuniv=Tsurr

DSsys-DHsys>0-Tsurr

DSuniv=DHsys-Tsurr

DSsys<0DG<0spontaneous.DG>0non-spontaneousDG=0atequilibrium.Definition:

G=H-TS

GiscalledGibbsfreeenergydG=dH–TdS–SdTatconstantPandT:DHsys-Tsys

DSsys<0ForaprocessatconstantPandTDGsys=DHsys–TsysDSsysDSuniv=DSsys+DSsurr>0for10NotethatShasatrueandabsolutevalue,whichcanbedeterminedbyexperiment,orbystatistics.Thisisquitedifferentfrom

E

andH,

thevaluesofwhicharenotavailablebyexperiments.Foranyperfectcrystallinesubstanceat0K,theentropy

S=0.TheThirdLawofThermodynamicstheparticles(atom,moleculeorion)arrangedinspaceinamostlyoptimizedway;Nodefectsitesbothinbulkandonthesurface;Freeofimpurities.Undersuchconditions,onlyonewaytopacktheparticles.Thatis:W=1S=klnW

S=0NotethatShasatrueandabs11and1atmand1atm12Calculatethestandardentropychangeforthefollowingreactionat250C?2CO(g)+O2

(g)2CO2

(g)DS0rxn=2S0(CO2,g)–[2S0(CO,g)+S0(O2,g)]=427.2–[395.8+205.0]=-173.6J/K=2x(-393.5)

–[2x(-110.5)+0]=-566.0kJ=2(CO2,g)–[2(CO2,g)+(O2,g)]DH0fDH0fDH0fDH0rxnIsthereactionspontaneous?DGsys=DHsys–TsysDSsysDGorxn=(-566.0kJ)–(298K)x(-173.6J/K)Yes,theabovereactionisspontaneousat250C.=-514.3kJ<0Calculatethestandardentropy13SinceG=H-TS,

Gibss

freeenergyGisalsoastatefunction.Buttheabsolutevaluecannotbedeterminedbyexperiment,similartoEandH.E1,H1S1,G1n1

P1V1T1State1E2,H2S2,G2n2

P2V2T2State2DE=E2–E1DH=H2–H1DS=S2–S1DG=G2–G1SinceG=H-TS,Gibssfreeene14Standardfree-energyofformationDGf0

DGf0

isthefree-energychangethatoccurswhen1moleofthesubstanceisformedfromitselementsinmoststableformatastandardstate.

Standardfree-energyofformat15DG0rxnnDG0(products)f=SmDG0(reactants)fS-CalculatetheGibssfreeenergychangeforthefollowingreactionat250C?2CO(g)+O2(g)2CO2(g)DG=2x(-394.4)

–[2x(-137.3)+0]=-514.2kJ0rxnIsthereactionspontaneous?Yes,itisspontaneousat250C=2(CO2,g)–[2(CO2,g)+(O2,g)]DG0fDG0fDG0fDG0rxn

DG=DH-TDS0rxn0rxn0rxn=(-566.0kJ)–(298K)x(-173.6J/K)=-514.3kJRecallthatDG0rxnnDG0(products)f=SmDG0(16CaCO3

(s)CaO(s)+CO2

(g)Canthisreactionoccuratroomtemperature?Ifnot,howtomakethestonedecomposed?thisreactioncannothappenat25oC.=[(-604.2)

+(-394.4)]–(-1128.8)=130.2kJSinceDG>0,fortheproductionof1atmCO2.CaCO3(s)CaO(s)+C17

(2)Ifthereactionproceeds,DGshouldbelessthanzero.SinceDG=DH–TDS

DH–TDS<0ItistruethattemperatureThasgreateffectonH,SandG.ButThasaverysmalleffectonDHandDS,exceptDG.Assume:

DH0(T)DH0(298.15K),

DS0

(T)

DS0(298.15K)thenT

trans>thenT

trans>T

trans≈≈1108K(835oC)(2)Ifthereactionproceeds18CaCO3

(s)CaO(s)+CO2

(g)CaCO3(s)CaO(s)+C19Estimatetheboilingpointofwater?H2O(l)H2O(g)DG=DG(H2O,g)–DG(H2O,l)

0rxn0f0f(1)=-228.6–(-237.2)=8.6kJWatercannotevaporateintovaporat25oCwith1atmpressure.(2)DG=DH–TDST

boil>Waterwillboilat97oCat1atm.Estimatetheboilingpointof20ForachemicalreactionunderconstantPandconstantTaA+bBcC+dD(1)Atastandardstate(Pi=1atm,Ci=1M)andT=298K.DHrxn(T)DHrxn(298K),andDSrxn(T)DSrxn(298K)ooooAssume:Thereactiontemperaturecanbeestimatedfirstby:(2)Atastandardstate(Pi=1atm,Ci=1M)andT=aK.Forachemicalreactionunder21(3)Atanyconditions(T298K,

Pi

1atmorCi

1M)HereRisgasconstant(8.314JK-1mol-1)DGrxn(T,P)=DGrxn(T)+RTlnQ

oP(pa),andPo=101325paC(mol/L),andCo=1mol/LaA+bBcC+dD(3)Atanyconditions(T29822DoesNH4HCO3decomposeinair?

DGrxn=[DGf(NH3)+DGf(H2O,g)+DGf

(CO2,g)]-DGf

(NH4HCO3,s)ooooO=31.1kJNoreactionat25oCwithPi=1atmIfinaireachoccupies10%,Qp=(PNH3/P0)(PCO2/P0)(PH2O/P0)=0.001=31.1+8.314x0.298ln(0.001)=-3.1kJIngeneral,IfDGorxn<|±40|kJ,Itispossibletoreversethereactionbychangingthereactionconditions.DGrxn(T,P)=DGrxn(T)+RTlnQoNH4HCO3(s)NH3(g)+H2O(g)+CO2(g)Itwilloccurat25oCwithPi=0.1atmDoesNH4HCO3decomposeinair?23DG=DG0+RTlnQAtanequilibriumDG=0Q=K0=DG0+RTlnKDG0=-

RTlnKaA+bBcC+dDequilibriumconstantDG=DG0+RTlnQAtanequilib24DG0=-

RTlnKKcanbecalculatedfromDG0andviceversa;

Butitdoesnotmeansthatthereactionoccursatastandardstate.Kisonlyafunctionoftemperature.DGrxn(T)=DHrxn(T)–TDSrxn(T)ooo-RTlnK=DHrxn(T)–TDSrxn(T)ooDHrxn<00DHrxn>00T,KT,KDHrxn(T)DHrxn(298K),andDSrxn(T)DSrxn(298K)ooooAssume:DG0=-RTlnKKcanbecalcul25Summary:aA+bBcC+dDDS0rxndS0(D)cS0(C)=[+]-bS0(B)aS0(A)[+]DH0rxndDH0(D)fcDH0(C)f=[+]-bDH0(B)faDH0(A)f[+](1)Thestandardenthalpy(2)ThestandardentropyDG0rxndDG0(D)fcDG0(C)f=[+]-bDG0(B)faDG0(A)f[+](3)ThestandardGibbsfreeenergy(4)DGrxn=DHrxn–TDSrxnDG=DG0+RTlnQT

rxn>DG0=-

RTlnKSummary:aA+bBcC+26PracticeExercises:18.11;18.1818.2918.5018.7818.80PracticeExercises:18.11;27Chapter18ThermodynamicsInthisChapter,wewilladdressseveralaspectsasfollows:I)Howto

predict

whetherareactionunderagivenconditionscanoccurornot?II)Howto

makethereactionhappen,ifitdoesnotproceedunderprimaryconditions?

SpontaneityandEntropy(S).TheSecondandThirdLawsofthermodynamics.FreeEnergy(G)andEquilibrium(K).Chapter18Thermodynamics28Spontaneous

process:

spontaneousnon-spontaneous18.2Goaheadbyitself,withouthelpfromsurroundings.spontaneousNon-spontaneousSpontaneousprocess:29Whatarethefactors

controllingthespontaneity?CH4

(g)+2O2

(g)CO2

(g)+2H2O(l)

DH0=-890.4kJH+

(aq)+OH-

(aq)H2O(l)

DH0=-56.2kJH2O(s)H2O(l)

DH0=6.01kJNH4NO3

(s)NH4+(aq)+NO3-

(aq)

DH0=25kJH2O

Energy

Freedom

Whatarethefactorscontrolli30Entropy(S):ameasureofsystemdisorder.orderSdisorderSDS=Sf-SiDS>0Foragivensubstance:Ssolid<Sliquid<<SgasH2O(s)H2O(l)DS>018.2Heatinghydrogengasfrom600Cto800CCrystallizationfromasupersaturatedsolutionDS<0Foraprocess:Entropy(S):ameasureofsyst31EntropyisalsoaStatefunction.

E1H1S1E2H2S2n1

P1V1T1n2

P2V2T2State1State2DE=E2–E1DH=H2–H1DS=S2–S1ThevalueofSisonlydeterminedbythestate,regardlessofhowthestateisreached.EntropyisalsoaStatefuncti32Howtoquantify

theEntropy?MicrostatesW=1W=4W=6SIn1868,BoltzmannproposedS=klnWDS=Sf-SiForaprocess:DS=klnWfWiBoltzmannconstantk=1.38×10-23J/KForCH3OHandC2H5OH:Smethanol<SethanolHowtoquantifytheEntropy?Mi33TheSecondLawofThermodynamicsTheentropyofthe

universe

increasesinaspontaneousprocess,andremainsunchangedinanequilibriumprocess.DSuniv=DSsys+DSsurr>0(1)Spontaneousprocess:DSuniv=DSsys+DSsurr=0(2)Equilibriumprocess:Inordertopredictthedirectionofachemicalreaction,however,bothDSsysandDSsurrneedtobecalculated,whichisaquestionremainingtobesolved.TheSecondLawofThermodynami34Howtocalculate

DSsurr,theentropychangeinthesurroundings?EndothermicDHsys>0DSsurr

DHsysDSsurr>0Foraconstant-pressureprocess,itisknownthatQp=DH.ExothermicDHsys<0DSsurr<0HowtocalculateDSsurr,thee35DSsurr=

DHsysTsurrSsurr

TsurrAthigh

Tsurr,thesurroundingsmoleculesarealreadyveryenergetic.Then,theabsorptionofheatfromthesystemwillnotchangethemolecularmotionverymuch.Thatis,theentropychangeinthesurroundingswillbesmall.Atlow

Tsurr,however,thesameamountofsuchheatwillcausearelativelylargechangeinthemolecularmotion,andthentheentropychangeofthesurroundingswillbelarge.How?DSsurralsodependsonTsurr,thesurroundingstemperature.Therefore,foraconstant-pressureprocess,DSsurr=DHsys36DSuniv=DSsys+DSsurr>0foraspontaneousprocess:DSuniv=DSsys

->0TsurrDSuniv=Tsurr

DSsys-DHsys>0-Tsurr

DSuniv=DHsys-Tsurr

DSsys<0DG<0spontaneous.DG>0non-spontaneousDG=0atequilibrium.Definition:

G=H-TS

GiscalledGibbsfreeenergydG=dH–TdS–SdTatconstantPandT:DHsys-Tsys

DSsys<0ForaprocessatconstantPandTDGsys=DHsys–TsysDSsysDSuniv=DSsys+DSsurr>0for37NotethatShasatrueandabsolutevalue,whichcanbedeterminedbyexperiment,orbystatistics.Thisisquitedifferentfrom

E

andH,

thevaluesofwhicharenotavailablebyexperiments.Foranyperfectcrystallinesubstanceat0K,theentropy

S=0.TheThirdLawofThermodynamicstheparticles(atom,moleculeorion)arrangedinspaceinamostlyoptimizedway;Nodefectsitesbothinbulkandonthesurface;Freeofimpurities.Undersuchconditions,onlyonewaytopacktheparticles.Thatis:W=1S=klnW

S=0NotethatShasatrueandabs38and1atmand1atm39Calculatethestandardentropychangeforthefollowingreactionat250C?2CO(g)+O2

(g)2CO2

(g)DS0rxn=2S0(CO2,g)–[2S0(CO,g)+S0(O2,g)]=427.2–[395.8+205.0]=-173.6J/K=2x(-393.5)

–[2x(-110.5)+0]=-566.0kJ=2(CO2,g)–[2(CO2,g)+(O2,g)]DH0fDH0fDH0fDH0rxnIsthereactionspontaneous?DGsys=DHsys–TsysDSsysDGorxn=(-566.0kJ)–(298K)x(-173.6J/K)Yes,theabovereactionisspontaneousat250C.=-514.3kJ<0Calculatethestandardentropy40SinceG=H-TS,

Gibss

freeenergyGisalsoastatefunction.Buttheabsolutevaluecannotbedeterminedbyexperiment,similartoEandH.E1,H1S1,G1n1

P1V1T1State1E2,H2S2,G2n2

P2V2T2State2DE=E2–E1DH=H2–H1DS=S2–S1DG=G2–G1SinceG=H-TS,Gibssfreeene41Standardfree-energyofformationDGf0

DGf0

isthefree-energychangethatoccurswhen1moleofthesubstanceisformedfromitselementsinmoststableformatastandardstate.

Standardfree-energyofformat42DG0rxnnDG0(products)f=SmDG0(reactants)fS-CalculatetheGibssfreeenergychangeforthefollowingreactionat250C?2CO(g)+O2(g)2CO2(g)DG=2x(-394.4)

–[2x(-137.3)+0]=-514.2kJ0rxnIsthereactionspontaneous?Yes,itisspontaneousat250C=2(CO2,g)–[2(CO2,g)+(O2,g)]DG0fDG0fDG0fDG0rxn

DG=DH-TDS0rxn0rxn0rxn=(-566.0kJ)–(298K)x(-173.6J/K)=-514.3kJRecallthatDG0rxnnDG0(products)f=SmDG0(43CaCO3

(s)CaO(s)+CO2

(g)Canthisreactionoccuratroomtemperature?Ifnot,howtomakethestonedecomposed?thisreactioncannothappenat25oC.=[(-604.2)

+(-394.4)]–(-1128.8)=130.2kJSinceDG>0,fortheproductionof1atmCO2.CaCO3(s)CaO(s)+C44

(2)Ifthereactionproceeds,DGshouldbelessthanzero.SinceDG=DH–TDS

DH–TDS<0ItistruethattemperatureThasgreateffectonH,SandG.ButThasaverysmalleffectonDHandDS,exceptDG.Assume:

DH0(T)DH0(298.15K),

DS0

(T)

DS0(298.15K)thenT

trans>thenT

trans>T

trans≈≈1108K(835oC)(2)Ifthereactionproceeds45CaCO3

(s)CaO(s)+CO2

(g)CaCO3(s)CaO(s)+C46Estimatetheboilingpointofwater?H2O(l)H2O(g)DG=DG(H2O,g)–DG(H2O,l)

0rxn0f0f(1)=-228.6–(-237.2)=8.6kJWatercannotevaporateintovaporat25oCwith1atmpressure.(2)DG=DH–TDST

boil>Waterwillboilat97oCat1atm.Estimatetheboilingpointof47ForachemicalreactionunderconstantPandconstantTaA+bBcC+dD(1)Atastandardstate(Pi=1atm,Ci=1M)andT=298K.DHrxn(T)DHrxn(298K),andDSrxn(T)DSrxn(298K)ooooAssume:Thereactiontemperaturecanbeestimatedfirstby:(2)Atastandardstate(Pi=1atm,Ci=1M)andT=aK.Forachemicalreactionunder48(3)Atanyconditions(T298K,

Pi

1atmorCi

1M)HereRisgasconstant(8.314JK-1mol-1)DGrxn(T,P)=DG