chemical kinetics rates of reaction：反應(yīng)的化學(xué)反應(yīng)動力學(xué)速率

Chemistry:TheMolecularScienceMoore,StanitskiandJursChapter13: ChemicalKinetics:RatesofReactionsInquiringMindsWanttoKnow….Willareactionhappen?(Ch.18)Howfastorslowdoesareactionhappen?(Ch.13)Whenwillthereaction“stop”orreachequilibrium?(Ch.14and17)Drugsinthebody.Ozoneintheatmosphere.Graphitetodiamond.ChemicalKineticsChemicalkineticsstudiestherateatwhichachemicalreactionoccursandthepathwaytaken.Rateisthechangeofsomethingperunittime. Examples:

distance/time $/time concentration/timeC(graphite) C(diamond) excruciatinglyslow

rate=-

[C(graphite)]

/

timeCH4

(g)+2O2

(g) CO2(g)+2H2O(g) rapid

rate=-

[CH4(g)]/timeHomogeneous

-reactants&productsinonephase.ex.moleculesmorelikelytocometogetheringasorliquid phase

Heterogeneous

-speciesinmultiplephases. ex.solidcatalystandair(catalyticconverter)ChemicalKineticsReactionRateFactorsaffectingthespeedofareaction:(Homogeneous)[X],MorPConcentrationofreactantsk,unitsincludes-1Properties(phase,structure…)TemperatureCatalystsRate=-?[Cv+]/?time=k[Cv+]n

laterReactionRateChangein[reactant](or[product])perunittime.rate= =changeinconcentrationofCv+elapsedtime-Δ[Cv+]ΔtCresolviolet(Cv+;adye)decomposesinNaOH(aq):Cv+(aq)+OH-(aq)

CvOH(aq)Foranygeneralreaction:

aA+bB cC+dDTheoverallrateofreactionis:ReactionRatesandStoichiometryRate = = = =?1aΔ[A]Δt?1bΔ[B]Δt+1cΔ[C]Δt+1dΔ[D]ΔtCv+(aq)+OH-(aq)

CvOH(aq)Lossof1Cv+→Gainof1CvOHRateofCv+loss==RateofCvOHgainRate=-?[Cv+]/?time=?[CvOH]/?timeReactionRatesandStoichiometryH2(g)+I2(g)→2HI(g)TherateoflossofI2is0.0040molL-1s-1.WhatistherateofformationofHI?Rate = = =?Δ[H2]Δt?Δ[I2]Δt+12Δ[HI]Δt ?(-0.0040)=+12Δ[HI]ΔtΔ[HI]ΔtSo=+0.0080molL-1s-1Time,t [Cv+]Averagerate

(s) (mol/L)(molL-1s-1)0.0 5.000x10-510.0 3.680x10-520.0 2.710x10-530.0 1.990x10-540.0 1.460x10-550.0 1.078x10-560.0 0.793x10-580.0 0.429x10-5100.0 0.232x10-513.2x10-79.70x10-77.20x10-75.30x10-73.82x10-72.85x10-71.82x10-7

0.99x10-7SamplecalculationAvg.rate=Δ[Cv+]

Δt=-(3.68x10-5-5.00x10-5)mol/L(10.0–0.0)s=1.32x10-6

molL-1s-1

(M/s)Ratesarealwayspositive.TherateoftheCv+reactioncanbecalculated:ReactionRatetrend:?Ratedependsonconc.Ratedecreasewithconc.React.AverageRateandInstantaneousRate5.0E-54.0E-53.0E-52.0E-51.0E-50[Cv+](mol/L)0 20 40 60 80 100t(s)Averagerate=slopeoftheblueorgreytriangle…buttheavg.ratedependsonintervalchosen.GraphicalviewofCv+reaction:Rate=-?[Cv+]/?time5.0E-54.0E-53.0E-52.0E-51.0E-50[Cv+](mol/L)0 20 40 60 80 100t(s)Instantaneousrate=slopeofalinetangenttothecurve.t=5sandt=75shavedifferentinstantaneousratesCannotpredict[Cv+]atatimetoofarawaysincetheratechanges.AverageRateandInstantaneousRateRate=-?[Cv+]/?timeRatemaychangewhen[reactant]changes.

t [Cv+] RateofCv+ Rate/[Cv+](s) (M) loss(M/s) (s-1)05.00x10-5 13.2x10-7 0.0264401.46x10-5

3.82x10-7 0.0262Cv+exampleshowsthis.ForCv+therateisproportionaltoconcentration.

rate=k[Cv+]nTheRateLawk(~10secondinterval)y=mx5.0E-54.0E-53.0E-52.0E-51.0E-50[Cv+](mol/L)0 20 40 60 80 100t(s)TheRateLawRate[Cv+](mol/L)

rate=k[Cv+]1Theratelaw.rateRateLawandOrderofReactionAgeneralreactionwillusuallyhavearatelaw: rate=k[A]m[B]n...where

k

rateconstant(T,properties,catalyst)

m,n

orderforA&B,respectively

m+n +… overallorderofthereactionTheordersareusuallyintegers(-2,-1,0,1,2…),butmayalsobefractions(?,?…)Theconcentrationofareactantwithazero-orderdependencehasnoeffectontherateofthereaction.(Therejustneedstobesomereactantaround.Thereactionis:secondorderinNOfirstorderinH2

thirdorderoverall.TheexperimentallydeterminedratelawisTheRateLaw–ReactionOrderTheRateLaw–ReactionOrder#22.2NO(g)+Br2

(g) 2NOBr(g)Experimentshowsthatthereactionisfirst-orderinBr2andsecond-orderinNO.Writetheratelawforthereaction.Rate=k[Br2][NO]2IftheconcentrationofBr2istripled,howwillthereactionratechange?

3XWhathappenstothereactionratewhentheconcentrationofNOisdoubled?4XDeterminingRateLawsfromInitialRatesInitialRateMethodTofindtheorderforareactant:Runtheexperimentwithknown[reactant]0.Measuretheinitialrateofreaction(slopeatt=0,<2%consumed.)Change[reactant]0of1reactant;keepallothersconstant.Re-measuretheinitialrate.Theratioofthetworatesgivestheorderforthechosenreactant.Ratelawsmustbemeasuredexperimentally.Theycannot

bepredictedfromreactionstoichiometry. Initialconcentration(M) Expt. [CH3COOCH3]0 [OH-]0 Initialrate(M/s)1 0.040 0.040 2.2x10-42 0.040 0.080 4.5x10-43 0.080

0.080 9.0x10-4Dataforthereactionofmethylacetatewithbase:CH3COOCH3+OH- CH3COO-+CH3OHStartwithgenericratelaw:

rate=k[CH3COOCH3]m[OH-]nDeterminingRateLawsfromInitialRateschangeconcentrationofeachreactantinturn>x2>x2Dividingthefirsttwodatasets([OH]changing): 4.5x10-4M/s=k(0.040M)m(0.080M)n 2.2x10-4M/s=k(0.040M)m(0.040M)nThus: 2.05=(2.00)n 2.05=(2.00)nand n=1 Initialconcentration(M) Expt. [CH3COOCH3] [OH-] Initialrate(M/s)1 0.040 0.040 2.2x10-42 0.040 0.080 4.5x10-43 0.080

0.080 9.0x10-4DeterminingRateLawsfromInitialRates

rate=k[CH3COOCH3]m

[OH-]1Useexperiments2&3tofindm(dependanceonCH3COOCH3]):So: 2.00=(2.00)m(1)n 2.00=(2.00)mand m=1Also1storderwithrespecttoCH3COOCH3.9.0x10-4M/s=k(0.080M)m(0.080M)n4.5x10-4M/s=k(0.040M)m(0.080M)nDeterminingRateLawsfromInitialRates

rate=k[CH3COOCH3]1

[OH-]1Theratelawis:

rate=k[CH3COOCH3][OH-]Overallorderforthereactionis:

m+n=1+1=2Thereactionis:2ndorderoverall.1storder

in

OH-1storder

in

CH3COOCH3DeterminingRateLawsfromInitialRatesIfaratelawisknown,kcanbedetermined: rate=k[CH3COOCH3][OH-]2.2x10-4M/s(0.040M)(0.040M)k=k=0.1375M-1s-1=0.1375Lmol-1s-1

rate[CH3COOCH3][OH-]k=Couldrepeatforeachrun,takeanaverage…Butagraphicalmethodisbetter.UsingExp.1:DeterminingRateLawsfromInitialRatesPractice2NO(g)+O2(g)2NO2Initialratesweremeasuredat25oCstartingwithvariousconcentrationsofreactants.InitialConcentrations(mol/L)InitialRate[mol/(L.s)]NOO2Exp.10.0200.0100.028Exp.20.0200.0200.057Exp.30.0400.0200.227Whatistheratelaw?(m,n)Calculatetherateconstant(k).Rate=k[NO]n[O2]mInitialConcentrations(mol/L)NOO2InitialRate[mol/(L.s)]Exp.10.0200.0100.028Exp.20.0200.0200.057Exp.30.0400.0200.227Rate=k[NO]n[O2]mWhycan’twecompareexperiment1and3???n=2ThereactionissecondorderwithrespecttoNO.Practice–Whatisn?2x4xInitialConcentrations(mol/L)NOO2InitialRate[mol/(L.s)]Exp.10.0200.0100.028Exp.20.0200.0200.057Exp.30.0400.0200.227m=1Thereactionisfirstorderwithrespecttooxygen.Practice–Whatism?Rate=k[NO]2[O2]m2x2xRate=k[NO]2[O2]

Practice–Whatisk?Units?...ratealwayshasunitsofM/s.UsingExp.1…Iodideionisoxidizedinacidicsolutiontothetriiodideion,I3-,byhydrogenperoxide.Aseriesoffourexperimentswererunatdifferentconcentrations,andtheinitialratesofI3-formationweredetermined.Fromthefollowingdata,obtainthereactionorderswithrespecttoH2O2,I-,andH+.Calculatethenumericalvalueoftherateconstant.Practice–Whatisk?28InitialConcentrations(mol/L)H2O2I-H+InitialRate[mol/(L.s)]Exp.10.0100.0100.000501.10x10-6Exp.20.0200.0100.000502.20x10-6Exp.30.0100.0300.000509.90x10-6Exp.40.0100.0100.001001.10x10-6Rate=k[H2O2]m[I-]n[H+]pPractice–Whatism,n,p?Rate=k[H2O2]1[I-]2[H+]02x2x3x9x2x1xWecannowcalculatetherateconstantbysubstitutingvaluesfromanyoftheexperiments.UsingExperiment1:Because[H]0=1,theratelawis:Rate=k[H2O2][I-]2ChangeofConcentrationwithTimeAratelawsimplytellsyouhowtherateofreactionchangesasreactantconcentrationschange.Theintegratedratelawshowshowareactantconcentrationchangesoveraperiodoftime.TheIntegratedRateLaw(Calculus)rate=– =k[A]Δ[A]

Δt =-k[A]d[A]dtIf

areactionis1st-order,

aplotofln[A]vs.twillbelinear.Considera1st-orderreaction:A productsIntegratesto: ln[A]t

=?kt

+ln[A]0

y =mx+ b (straightline)AbigchangeAlittlechange(differentialequation)32GraphingKineticDataInadditiontothemethodofinitialrates,ratelawsandkcanbededucedbygraphicalmethods.Thismeansifweplotln[A]tversust,wewillgetastraightlineforafirst-orderreaction.Forfirst-orderreaction:y=mx+bConcentration-TimeEquationsSecond-OrderRateLawYoucouldwritetheratelawintheformUsingcalculus,yougetthefollowingequation.Here[A]tistheconcentrationofreactantAattimet,and[A]oistheinitialconcentration.34GraphingKineticDataForsecond-orderreaction:Thismeansifweplot1/[A]tversustime,

wewillgetastraightlineforasecond-orderreaction.y=mx+bConcentration-TimeEquationsRearrangegetthefollowingequation.Here[A]tistheconcentrationofreactantAattimet,and[A]oistheinitialconcentration.Forazero-orderreactionwecouldwritetheratelawintheform36GraphingKineticDataForzero-orderreaction:Thismeansifweplot[A]tversustime,

wewillgetastraightlineforasecond-orderreaction.y=mx+bOrderRatelawIntegratedratelawSlopeTheIntegratedRateLaw0 rate=k [A]t=-kt+[A]0 -k1 rate=k[A]ln[A]t=-kt+ln[A]0 -kA products1[A]t2 rate=k[A]2=kt+ +k1[A]0Themostaccuratekisobtainedfromtheslopeofaplot.ln[A]timetFirst-orderreactionslope=-k1/[A]timetSecond-orderreactionslope=k[A]timetZeroth-orderreactionslope=-ky=mx+bTheIntegratedRateLawThereactionisfirstorder(theonlylinearplot)k=-1x(slope)ofthisplot.Ratedataforthedecompositionofcyclopentene

C5H8(g)

C5H6(g)+H2(g)weremeasuredat850°C.Determinetheorderofthereactionfromthefollowingplotsofthosedata:NOTaline line! NOTalineExample1ThedecompositionofN2O5toNO2andO2isfirstorderwitharateconstantof4.800x10-4s-1.IftheinitialconcentrationofN2O5is1.65x10-2mol/L,whatistheconcentrationofN2O5after825seconds?Thefirst-ordertime-concentrationequationforthisreactionwouldbe:ThedecompositionofN2O5toNO2andO2isfirstorderwitharateconstantof4.800x10-4s-1.IftheinitialconcentrationofN2O5is1.65x10-2mol/L,whatistheconcentrationofN2O5after825seconds?eln(x)=xPractice-usingintegratedratelawsExample2Considerthefollowingfirst-orderreaction: CH3CH2Cl(g)C2H4(g)+HCl(g)Theinitialconcentrationofethylchloridewas0.00100M.Afterheatingat500.0oCfor155.00stheconcentrationwasreducedto0.000670M.Whatwastheconcentrationofethylchlorideafteratotalof256.00s?Theinitialconcentrationofethylchloridewas0.00100M.Afterheatingat500.0oCfor155.00stheconcentrationwasreducedto0.000670M.Whatwastheconcentrationofethylchlorideafteratotalof256.00s?

=2.58×10-3s-1[C2H5Cl]0=0.00100M[C2H5Cl]t1=0.000670Mt1=155.00st2=256.00s,[C2H5Cl]t2=?Example1Theinitialconcentrationofethylchloridewas0.00100M.Afterheatingat500.0oCfor155.00stheconcentrationwasreducedto0.000670M.Whatwastheconcentrationofethylchlorideafteratotalof256.00s?Example1t1/2=

timefor[reactant]tofallto?[reactant]0.For1st-orderreactionst1/2

isindependentofthestartingconcentration.onlytruefor1storderreactions(not0th,2nd…)t1/2isconstantforagiven1st-orderreaction.Half-LifeFora1st-orderreaction: ln[A]t=-kt+ln[A]0Half-LifeWhent=t1/2,

[A]t=?[A]0Then:ln(?[A]0)=-kt1/2+ln[A]0ln(?[A]0/[A]0)=-kt1/2

{note:lnx–lny=ln(x/y)}ln(?)=-ln(2)=-kt1/2

{note:ln(1/y)=–lny}ln2k0.693kt1/2= =HalfLife0.0100.0080.0060.0040.0020[cisplatin](mol/L)0 400 800 1200 1600 2000t(min)Forcisplatin(achemotherapyagent):0.0100M0.0050M,after475min (t?=475minutes)t1/2ofa1st-orderreactioncanbeusedtofindk.k= =0.693475minln2t1/2=1.46x10-3min-1graphicallyt?t?t?Useanintegratedrateequationorhalf-lifeequation.ExampleIna1st-orderreaction,[reactant]0=0.500mol/Landt1/2=400.s.Calculate:[reactant],1600.safterinitiation.tfor[reactant]todropto1/16thofitsinitialvalue.tfor[reactant]todropto0.0500mol/L.Calculating[]ortln[A]t=-kt+ln[A]0Ina1st-orderreaction,[reactant]0=0.500mol/Landt1/2=400.s(a)Calculate[reactant],1600.safterinitiation.1storder:

k=ln2/t?=0.6931/(400.s)=1.733x10-3s-1

ln[A]t=-(0.001733s-1)(1600s)+ln(0.500) ln[A]t=-2.773+-0.693=-3.466

[A]t

=e-3.466=0.0312mol/LCalculating[]ortOR…1600s=4t1/2

so0.500→0.250→0.125→0.0625→0.0313M.Ina1st-orderreaction,[reactant]0=0.500mol/Landt1/2=400.s(b)Calculatetfor[reactant]todropto1/16thofitsinitialvalue. [reactant]0 [reactant]0

t1/2

12 [reactant]0 [reactant]0

t1/2

1412 [reactant]0 [reactant]0

t1/2

1814 [reactant]0 [reactant]0

t1/2

116184t1/2=4(400s)=1600sCalculating[]ortFrompart(a): k=1.733x10-3s-1

ln[A]t=-kt+ln[A]0then ln(0.0500)=-(0.001733s-1)t+ln(0.500) -2.996=-(0.001733s-1)t–0.693

t=1.33x103sIna1st-orderreaction,[reactant]0=0.500mol/Landt1/2=400.s(c)Calculatetfor[reactant]todropto0.0500mol/L?t= -2.303-0.001733s-1Calculating[]ortElementaryandComplexReactions

Onthenanoscalelevel,areactionmaybe:unimolecular–asingleparticle(atom,ion,molecule)rearrangesinto1or2differentparticles. (example:somedecomposition,geometricisomer)bimolecular–twoparticlescollideandrearrange (example:composition)Observedreactions,onthemacroscalelevel,maybe:elementary–directlyoccurbyoneofthesetwoprocesses,orcomplex–occurasaseriesofelementarystepsunimolecularbimolecularunimolecularunimolecularElementaryReactions

Example:Labeltheseelementaryreactionsasunimolecularorbimolecular:2-buteneisomerizationisunimolecular:cis-2-butene trans-2-buteneElementaryReactions-Unimolecular

FinalstateInitialstateReaction

Progress(angleoftwist)ΔE=-7x10-21JEa=435x10-21J5004003002001000Potentialenergy(10-21J)-30° 030°60°90°120°150°180°210°cis-transconversiontwiststheC=Cbond.Thisrequiresalotofenergy(Ea=4.35x10-19J/molecule=262kJ/mol)Evenmore(4.42x10-19J/molecule)toconvertback.FinalstateInitialstateReactionProgress(angleoftwist)ΔE=-7x10-21JEa=435x10-21J5004003002001000Potentialenergy(10-21J)-30° 030°60°90°120°150°180°210°transitionstateoractivatedcomplexEaistheactivationenergy,theminimumEtogooverthebarrier.ExothermicoverallUnimolecularReactionskinetics–iflowerEa,increasethespeedofthereactionthermodynamics–ifEproduct<Ereactant,thereactionisexothermicOccursatthetopoftheactivationbarrierExistsforveryshorttime(fewfs,1fs=10-15s).Fallsaparttoformproductsorreactants.themoleculesmusthaveenoughenergy–activationenergy–toformthetransitionstateandconvertitintoproducts.TransitionStateEndothermicreactionoverall“uphill”Eproducts>EreactantsEaExothermicreactionoverall“downhill”Eproducts>EreactantsBimolecularReactionsI-mustcollidewithenoughE

andwiththerightorientationtocausetheinversion.I-(aq)+CH3Br(aq) ICH3(aq)+Br-(aq)transitionstateForthereactiontotakeplacetwomoleculesmusteffectivelycollide.Oneoutoffourcollisionsissuccessful.(backsideoftetrahedron,oppositeBr).Thisgeometricconstraintonapproachiscalledastericfactor.BimolecularReactionsBimolecularReaction

BimolecularReaction

O3+NO O2+NO2ReactionEnergyDiagramsForwardreaction:Ea=10kJ/molExothermicBackwardreactionEa=?ExoorEndo?IncreasingTwillspeedupmostreactions.HigherT =higheraverageEk

forthereactants. =largerfractionofthemoleculescan overcometheactivationbarrier.25°CkineticenergynumberofmoleculesEa75°CManymoremoleculeshaveenoughEtoreactat75°C,sothereactiongoesmuchfaster.FactorsAffectingReactionRatesTemperatureandReactionRateRuleofthumb–approximatelydoubletherateforevery10K.250300350400T(K)0.000.100.200.30k(Lmol-1K-1)T(K)k(Lmol-1K-1)273 4.18x10-5290 2.00x10-4310 2.31x10-3330 1.39x10-2350 6.80x10-2370 2.81x10-1I-+CH3Br

Br-+CH3IRate=k[I-]m[CH3Br]nTheArrheniusequationshowshowkvarieswithT-Ea/RTk=A

eA

Frequencyfactor.Howoftenacollisionoccurswith thecorrectorientation.(1/4forI-+CH3Br)

e-Ea/RT

Fraction

ofthemoleculeswithenoughEa

tocross thebarrier.

T

Temperature MustbeinKelvin.R

Gaslawconstant 8.314JK-1mol-1.TemperatureandReactionRateexponentialrelationship250

300350400T(K)k(Lmol-1K-1

)TemperatureandReactionRaterate=k[I-]m[CH3Br]nrate=(Ae-Ea/RT)[I-]m[CH3Br]nCanrearrange….ln= –k1k21T21T1Ea

RkdependsonT,Ea(catalyst)kincreases,whenTincreaseskincreases,whenEadecreasesOWL13.5aOWLlinkPresenceofacatalyst.Acatalystisasubstancethatincreasestherateofareactionwithoutbeingconsumedintheoverallreaction.Thecatalystgenerallydoesnotappearintheoverall

balancedchemicalequation(althoughitspresencemaybeindicatedbywritingitsformulaoverthearrow).FactorsAffectingReactionRatesCatalystsAcatalystincreasestherateofreactionbyloweringtheactivationenergy,Ea.Toavoidbeingconsume