AChemist’sViewoftheUniverse
• System – whatwecareabout
• Surroundings – everythingelse
• Boundary – separatessystemfromsurroundings
Somebasicdefinitions
• Work– motionagainstanopposing(external)force
• Heat– energychangeassociatedwithachangeintemperature
– Exothermic – releasesheattothesurroundings– Endothermic – absorbsheatfromthesurroundings
InternalEnergy(UorE)
• Totalenergyofthesystem(kineticandpotential)
• Internalenergyisastatefunction,whichmeanswecandefine
achangeinitasDU=Uf – Ui.
Enthalpy(H)
• Totalpotentialenergyofthesystem
• Enthalpyisastatefunction,whichmeanswecandefinea
changeinitasDH=Hf – Hi.
1st LawofThermodynamics
• Theinternalenergyofanisolated systemisconstant.
• Theonlywaystochangetheinternalenergyofasystemareheatand
work
– DU=Q+W• Thechangeininternalenergyforasystemisequaland
oppositetothechangeininternalenergyforthesurroundings
– DUsys +DUsurr =DUtot =0
1st LawofThermodynamics
• “Greedy”convention(Ch andChE’s)
– Heatabsorbed bysystemà Q>0
– Heatreleased bysystemà Q<0
–WorkdonetosystemàW>0
–Workdoneby systemàW<0
Hess’sLaw
• Directapplicationofthepropertiesofstatefunctions.
• Thestandardenthalpyofanoverallreactionisthesumofthestandard
enthalpiesoftheindividualreactionsintowhichareactionmaybe
divided.(whethertheyarerealornot)
• ThiscanbeextendedtoANYthermodynamicvariable
PhaseChanges(Transitions)
• Anychangeofphasewillhaveacorrespondingchangeinenthalpy
• Becauseenthalpyisastatefunction,severalusefulpropertiesemerge.
• Considerachangeofphase(orstate)fromAtoBwithachangeofenthalpy=DH.Forthereverseprocess(goingfromBtoA),thechangeinenthalpywillbe–DH.
• ConsiderachangefromAtoC.Wecanconsiderthisashappeningintwosteps:first
fromAtoBandthenfromBtoC.ThusDHAàC=DHAàB+DHBàC
Enthalpiesofreaction
• Asimilartypeofanalysiscanbeperformedforachemicalreaction.We
candefinethestandardreactionenthalpyasthechangeinenthalpy
betweentheproductsandthereactants,inthestandardstate:
whereni referstothestoichiometriccoefficientofspeciesi,andHm,i
referstothemolarenthalpyofspeciesi(akaenthalpyofformation).
Theo indicatesstandardstate.
The2nd LawofThermodynamics
• Usedtopredictspontaneity(tendencyforaprocessto
happennaturally)
• The1st Lawonlytalksaboutconservationofenergy,it
saysnothing aboutthedirection thataprocesswilltendtogoin!
– Whydon’tballsleavethegroundandbounceup?
– Whydoesn’tshatteredglassreform?
– Whydoesn’tgreenpigmentseparateintoblueandyellow
pigments?
The2nd LawofThermodynamics
• The2nd Lawdescribeshowspontaneityisrelatedtothedistribution ofenergy,not tothetotal energy.
• Energytendstoflowinadirectionwhereitwillbemore“spreadout”,
ordispersed.
SoWhat’sEntropy?
• Thisleadstoanotherviewofthe2nd law:“Theentropyofanisolated
systemincreasesinthecourseofaspontaneouschange”
– DStot>0
• Relatedtochaos,randomness,disorder
• NoticethatQisnotastatefunctionbutSis!
Phasetransitions
• Previouslywesawthatforaphasetransitionoccurringataconstant
pressure,Q=DHtr
• ThismeansthatwecanalsocalculateDStr:whereTtr isthetemperatureatwhichthetransitionoccurs
Thusexothermicprocesses(freezing,condensing)have(-)changesin
entropy,whileendothermicprocesses(melting,boiling)have(+)
changesinentropy
Howisentropymeasured?
• Inananalogousfashiontoenthalpy,wecandefinethe
reactionentropychangeas:
• NotethatunlikeHmo,whichcan=0forsubstancesintheir
standardstate,Smo is≠0(unlessT=0)
Gibbsfreeenergy
• let’sdefinetheGibbsfreeenergyasG=H-TS.
• Foramacroscopicchange,DG=DH– TDS• Thisalsoleadstothefamiliarconclusionthatfora
spontaneousprocessDG≤0• DGalsorepresentsthemaximumnon-PVwork thatcanbedonebyasystem
HowcanDGbemeasured?
• Asfortheotherthermodynamicquantitieswehaveencountered,wecandefinethestandardfreeenergychangeforareactionas:
• Aswithenthalpyofformation,DGfo foranelement(ora
naturallyoccurringdiatomicmolecule)=0.
• ExperimentallyDGisoftenobtainedbydeterminingDHandDSseparately.
Gibbsenergyandtheequilibriumconstant
• WhatdoesDGrxno represent?Itisthedifferenceinthe(molar)Gibbs
energiesofproductsandreactants,intheirstandardstate.
isthecruciallinkbetweenthermodynamics(energy)andchemicalequilibrium
• NoticethatifK>1thentheproductsarefavoredatequilibrium,whileifK<1thenthereactantsarefavored.
• Ingeneral, whereQisthereactionquotient
Calculate ΔH°298 for the process Sb(s) + 5/2 Cl2(g) ⟶SbCl5(g) from the following information:Sb(s) + 3/2Cl2(g)⟶SbCl3(g) ΔH°298 = −314 kJSbCl3(s) + Cl2(g)⟶SbCl5(g) ΔH°298 = −80 kJ
Calorimetry– it’sdabomb!• Experimentsmaybedoneatconstantvolume(bomb)orconstantpressure(coffee-cup)
• Entiresystemisadiabatic – thereisnoheatlostbetweenthesystem(sample)andthesurroundings(waterbathandmetalcasing)
• Thechangeintemperatureofthecalorimeterwillbeproportionaltotheheatthatitabsorbs:QαΔT,orQ=CΔT,whereCistheheatcapacityofthecalorimeter.
Example
• Inapreliminaryexperiment,theheatcapacityofabombcalorimeterassemblyisfoundtobe5.15kJ/oC.Inasecondexperiment,a0.480gsampleofgraphite(carbon)isplacedinthebombwithanexcessofoxygen.Thewater,bomb,andothercontentsofthecalorimeterareinthermalequilibriumat25.00oC.Thegraphiteisignitedandburned,andthewatertemperaturerisesto28.05oC.CalculateΔHforthereaction:C(graphite)+O2(g)à CO2(g)
Solution
• Thekeytothisproblemistorealizethatalltheheatabsorbedbythecalorimetermusthavecomefromthecombustionreaction.
• FirstcalculatetheheatabsorbedbythecalorimeterusingQ=CΔT:
• Theheatgivenupbythereactionmustbeequalandoppositetothis:Qrxn =-15.7kJ
• WecanthenequatethistoΔUforthecombustionof1molofgraphite:
• Finally,recallthatΔH=ΔU+RTΔnforanisothermalprocess.Sincethetemperaturechangeisprettysmall(3.05oC)wecanassumethatitisisothermal.Δn=(1-1)=0soΔH≈ΔU.ThusΔH=-393kJ/mol
Calorimetryrevisited– that’sonefancycoffeecup!
• Forisobaricmeasurements,useathermallyinsulatedvesselthatisopentotheatmosphere
• Moresophisticatedcalorimeterscanbeused–Adiabaticflamecombustion–Differentialscanning–Isothermaltitration• Forsolidsandliquids,ΔH≈ ΔUsincetheirvolumeisnegligible(atleastcomparedtogases)
Example
• A15.5gsampleofametalalloyisheatedto98.9oCandthendroppedinto25.0gofwaterinacalorimeter.Thetemperatureofthewaterrisesfrom22.5to25.7oC.Calculatethespecificheatofthealloy.
Solution• Thekeytosolvingthisproblemistorealizethatalltheheatlostbythehotsolidmustbegainedbythewaterinthecup.
• Firstwewillfindtheheatabsorbedbythewater:QH2O =mcΔTso
• Thismustbeequalandoppositetotheheatlostbythealloy(rememberthesignconvention!)soQalloy =-334J
• Finallycalculatethespecificheatofthealloy:
Example
• A50.0mLsampleof0.250MHClat19.50oCisaddedto50.0mLof0.250MNaOH,alsoat19.50oC,inacalorimeter.Aftermixing,thesolutiontemperaturerisesto21.21oC.Calculatetheheatofthisreaction.
Solution• Firstrecognizethereactionthatistakingplace:HCl(aq)+NaOH(aq)à NaCl(aq)+H2O
• Nowlet’smakeafewassumptions/simplifications:–Takesolutionvolumestobeadditivesothetotalvolumeofsolutionis50.0+50.0=100.0mL–ConsidertheNaCl(aq)solutiontobesufficientlydilutethatthedensityandspecificheatarethesameasthoseforpurewater(1.00g/mLand4.184J/goC)–Thesystemisperfectlyinsulated,sonoheatescapesfromthecalorimeter–Theheatrequiredtowarmanypartofthecalorimeter(otherthantheNaClsolution)isnegligible)
• Findtheheatretainedinthecalorimeter:
• Finallyfindtheheatofreaction:Qrxn =Qp =-Qcal =-715J
ChemicalKinetics
• Studyoftheratesofchemicalreactions– Howquicklyaprocesscantakeplace
• Understandingofthemechanismofareaction– How(onamolecularlevel)aprocesscantakeplace
• Becausethesemeasurementsarechangingwithrespecttotimeandaresensitivetomanyvariables,theyarenotoriouslydifficultexperimentstocarryout!
Factorsaffectingtherateofachemicalreaction
• Concentration• Pressure(gasesonly)• Temperature• Presenceofacatalyst• Understandingthedependenceofareactiononthesefactorscanaidouroptimizationofachemicalprocess
Measuringtherateofareaction
• Wecandefinetherateintermsofthelossofareactantortheformationofaproduct:
– Noticethatthismeanstherateisrelatedtotheslope(tangentline)tothecurve
• Howeverthisdoesn’ttakeintoaccountthestoichiometryofthereaction(i.e.ifthereactionisRà2Ptherateofformationofaproductwillbetwiceasgreatasthelossofreactant).
• Intermsofagivencomponenti, whereni isthestoichiometric number.
Inpictures
http://textbook.s-anand.net/ncert/class-xii/chemistry/4-chemical-kinetics
Ratelaws• Itisfoundexperimentallythattherateofareactionisusuallyproportionaltothe
concentrationofeachreactant,raisedtoacertainpower:wherexandyaretheorders ofthereactionwithrespecttoAandB,respectively.Theordersmaybeanyrealnumber(including0andfractions).Theoverallorderisgivenbyx+y.
• IngeneraltheordersmustbedeterminedexperimentallyandareNOTnecessarilythestoichiometriccoefficientsofthereaction(unlessthereactioniselementary).
Determinationoftheratelawforareaction
• Isolationmethod– systematicallyvarytheconcentrationsofthereactantssothatallareinalargeexcessexceptforone.Thisallowsthedeterminationoftheorderofthatonespecies.
• Forexample,ifthegeneralratelawisandBispresentinalargeexcess,thenitsconcentrationcanbeassumedtobeconstantasthereactionproceeds,whichmeansthatd[B]/dtà0.Thuswecanwritetheratelawas whereandwecanfindthevaluesofk’andxbycurve-fitting.
Determinationoftheratelawforareaction
• Typicallytheinitial ratesaremeasured.ForexampleifBisinexcessthentheinitialratecanbewrittenas
• Thuswecanplotvo vs.[A]o andgetk’andxbyfittingthedatanonlinearly(powerlaw),orwecanlinearizetheequationbytakingthelogofbothsides:
• WecanrepeatthisprocesswhereAisheldinexcess,anddetermineanewpseudo-rateconstantk’’( )andy,andthereforealsogettheoriginalrateconstantk.
Example
• Therecombinationofiodineatomsinthegasphaseinthepresenceofargonwasinvestigatedandtheorderofthereactionwasdeterminedbythemethodofinitialrates.Theinitialratesofthereaction2I(g)+Ar(g)àI2(g)+Ar(g)wereasfollows:
[I]o (10-5 mol/L) 1.0 2.0 4.0 6.0vo (mol/L*s) a)8.70X10-4 3.48X10-3 1.39X10-2 3.13X10-2
b)4.35X10-3 1.74X10-2 6.96X10-2 1.57X10-1c)8.69X10-3 3.47X10-2 1.38X10-1 3.13X10-1
Example
• TheArconcentrationsarea)1.0mmol/L,b)5.0mmol/Landc)10.0mmol/L.DeterminetheordersofreactionwithrespecttotheIandAratomconcentrationsandtherateconstant.
Solution• Sincetheratelawwillbeoftheform
,weneedtoplotthedependenceoftherateonboth[I]o and[Ar]o.Alog-logplotwillbehelpfulastheslopewillgiveustheorderwithrespecttothatsubstance.
Solution• Fromthefirstgraph,theslopeis2whichmeansthatthereactionis2nd orderwithrespecttoI.Fromthesecondgraph,theslopeis1whichmeansthatthereactionis1st orderwithrespecttoAr.Thustheratelawis
• Noticethatthisistheratelawonlyfortheinitialrate– itispossiblethatthereactionhasadifferentratelawasthereactionproceeds.
Solution• Wecangettherateconstantfromtheinterceptsofeithersetoflines.Inthefirstexperiment,k’=k[Ar]0.Inthesecondexperiment,k’’=k[I]02.Ineithercase,k≈8.6*109 L2/mol2s.
logk' k' [Ar]0 k6.9365 8639727 0.001 86397266077.6439 44045343 0.005 88090686747.9326 85624885 0.01 8562488476
logk‘’ k‘’ [I]0 k-0.0616 0.867761 0.00001 86776074500.5387 3.457005 0.00002 86425123471.1363 13.68674 0.00004 85542121601.4969 31.39786 0.00006 8721626801
How will each of the following affect the rate of the reaction: CO(g) + NO2(g) ⟶CO2(g) + NO(g) if therate law for the reaction is rate = k [NO2] [CO] ?
(a) Increasing the pressure of NO2 from 0.1 atm to 0.3 atm(b) Increasing the concentration of CO from 0.02 M to 0.06 M.
TheArrheniusequation
• Aplotoflnkvs1/Tshouldgiveastraightline,withaslopeofEa/RandaninterceptoflnA:
• AsTà∞,kàko (orA),whichiscalledthefrequencyfactor.Itisalsocalledthepre-exponentialfactorsincetheaboveequationcanbewrittenas
http://chemistry.tutorvista.com/inorganic-chemistry/arrhenius-equation.html
MoreontheArrheniusequation• Thepre-exponentialfactorA representsthefastestpossiblerateforareaction,
whichwouldonlybelimitedbydiffusion.Thiscanbeinterpretedasbeingrelatedtotherateofsuccessfulcollisionsbetweenreactantmoleculestoyieldproductmolecules.
• NoticethatthefractionofmoleculeswithanenergygreaterthanEa isgivenbyanexponentialdecay,knownasaBoltzmanndistribution.Onlythosemoleculeswithanenergythatexceedstheactivationenergywillbeabletoreacttoformproducts.
• Thehighertheactivationenergy,themoretherateconstantwilldependonT.
Temperaturedependenceofthereactionrate
Arrheniusequation
Ea istheactivationenergyofthereaction.
http://www.wiley.com/college/boyer/0470003790/reviews/kinetics/kinetics_stability.htm
Roleofacatalyst• Acceleratetherateofareaction– Loweractivationenergy– Providealternatepath• Doesnotchangethethermodynamicsorequilibriumofareaction– ΔG,ΔH,ΔSandKareallthesameasfortheuncatalyzedreaction
Describe how graphical methods can be used to determine the activation energy of a reaction from a series of data that includes the rate of reaction at varying temperatures.
Theoreticaldescriptionoftheratelaws
• Althoughtheparametersforaratelawareexperimentallydetermined,itis
sometimespossibletocalculate(orpredict)themfromfirstprinciples.Thisrequires
aknowledgeofthewayinwhichthereactiontakesplace,alsoknownasthe
mechanism.
• Mostreactionsarethoughttooccurinaseriesof(relatively)wellunderstoodsteps,
eachofwhichisknownasanelementaryreaction.
• Inanelementaryreaction,typicallyasmallnumber(1-3)atoms,moleculesorions
collidewitheachotherandformaproduct.
Theoreticaldescriptionoftheratelaws
• Molecularity – thenumberofmolecules(oratomsorions)
thatcometogetherinanelementaryreaction
– Unimolecular- oneparticlebreaksapartorrearranges,i.e.dissociationorisomerization
– Bimolecular- twoparticlescollidewitheachother– Termolecular (rare)-threeparticlessimultaneously collidewitheachother
Theoreticaldescriptionoftheratelaws
• Becauseelementaryreactionsareofaknownmolecularity,wecan
writedowntheratelawsforthemsimplybylookingathowmany
speciesarepresent:
– AàP v=k[A]
– A+Bà P v=k[A][B]
• Amechanism isacombinationofelementaryreactionsthattriesto
explainthesequenceofevents(steps)ofachemicalreaction
• Therate-determiningstepistheslowest
“relevant”stepinachemicalreaction,and
determinestheoverallrateofthereaction.It
alsodictateshowmuchproductcanbe
formed.
• Typicallythesloweststepinachemical
reactionhasthehighestactivationenergy
since
• Inthediagramontheright,thelocalminima
correspondtointermediates – thismeansthat
theycanbe(theoretically)isolated.
• Themaximacorrespondtotransitionstates–theyareveryunstableandnotisolable
Rate-determiningstep
http://chemwiki.ucdavis.edu/Wikitexts/UC_Davis/UCD_Chem_2C%3A_Larsen/Chem_2C%3A_H
omework/Team_1/Chapter_24%3A_Kinetics