chapter 14 kinetics 2 7 bv

7/30/2019 Chapter 14 Kinetics 2 7 BV

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Chapter14

ChemicalKinetics

bv,272009

I.Introduction

Gasolineandairinacarengineexplodeviolently,butleftuntouched,theywillnotreactforyearsata

time.Meatleftoutwillinvitebiochemicalreactionsthat,amongotherthing,generatebadsmelling

gases.Ifkeptatlowertemperatures,thesereactionstakeamuchlongertimetooccur.Enzymatic

reactionsoccurslowlyatlowtemperaturesandathightemperatures,butrapidlyatintermediate

temperatures.Whydosomeoccurquickly,whileothersslowly?Whydosomereactionsoccuratall

whileothersdonot?Inotherwords,whatcontrolschemicalreactivity?

Chemicalreactivityiscontrolledbytwobroadfactors:thermodynamicsandkinetics.Thermodynamics

considersthequestion:whichstateismorestable,reactantsorproducts.Thermodynamicsanswersthe

question:shouldthisreactionoccur?Kinetics thesubjectofthischapter considersthequestion:what

controlstherateofareaction?

Inorderforareactiontooccurinapracticalsense,itmustbeboththermodynamicallyandkinetically

favored.Thesearerelativeterms:thereactionmustbethermodynamicallyfavoredenoughtoformtheamountofproductdesired,anditmustbekineticallyfavoredenoughtobecompleteonthetimescale

required.

Thermodynamiccontrolofreactionsarisesfromenthalpy,entropyandthetemperature.Kineticcontrol

ofareactionarisesfromthemannerinwhichthereactiontakesplace itsmechanism,theenergy

barrierrequiredtobeovercome theactivationenergy,theconcentrationofreactants,andagainthe

temperature.Inthischapterweanalyze:


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howconcentrationcontrolsreactionrate themathematicsconnectingconcentration,rateandtime howactivationenergycontrolsreactionrate howtemperaturecontrolsreactionrate howeachoftheseisrelatedtothereactionsmechanism

II.CollisionTheory

Collisiontheoryrelateshowwethinkofreactionstothereactionratesweobserve.Theideaofcollision

theory,isthatmoleculesarecollidingallthetimeandsomefraction butnotall ofthosecollisionswill

leadtotransformationofthereactantstotheproducts.

1.Themoleculesmustcomeintocontact.Thisisacollision.

2.Theymustcollidewithenoughenergytoovercomeanenergybarriertoreactioncalledthe

activationenergy.3.Theymustcollideinanorientationthatallowsthenecessarybondbreakingandforming

neededtotransformthereactantstotheproducts.

Relationshipstoreactionrate:

1.Collisions:Thisissimple.Ifsomefractionofcollisionswillleadtocreationofproducts,thenthemore

collisionspersecond,thefasterthereactionwillproceed.Thiscollisionrequirementdoeshavealarge

effectonwhatmediaarechosentoperformchemicalreactions.Solidstendtobeveryslowreactors

becauseonlytheatomsonthesurfacecanhavecollisionswithotheratomsonothermolecules.Ever

noticehowslowironrusts?Mostreactionsaredoneeitherinsolutionorinthegasphasewherefreedomofmovementofthereactantmoleculesallowsthemtoeasilycomeintocontact.

Conclusion:Themorecollisions,thefasterthereaction.

2.OvercomingtheActivationBarrier:AnysampleofreactantswillhaveaBoltzmanndistributionof

molecularenergies.Somemoleculeswillhavehighenergy;somelow;manyintermediate.Onlythose

withenergiesgreaterthantheactivationenergywillbeabletoreact.Figure15.1ashowsBoltzmann

plotsforasetofreactantsattwodifferenttemperatures.Onlythosereactantswithenergygreaterthan

(totherightof,intheplot)theactivationenergywillbeabletoreact.Becauseagreaterfractionof

moleculesinthehightemperaturesampleexceedtheactivationenergy,thehightemperaturesamplewillhaveeffectivecollisionsandwillexperienceafasterreactionrate.


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Figure15.1a:Boltzmanndistributionsofmolecularenergiesatdifferenttemperaturesshow

thatmoremoleculesexceedtheactivationenergyathighertemperature.b:Atagiven

temperature,moremoleculesexceedaloweractivationenergythanahighone.

Figure15.1bshowsaBoltzmanndistributionofmolecularenergiesforasinglesample,butshowstwo

differentactivationenergies.Agreaterfractionofthesamplemoleculesexceedtheenergyofthelower

activation

energy

than

do

the

higher

activation

energy.

Therefore,

those

reactant

molecules

will

undergothereactionwiththeloweractivationbarriermorerapidlythanwouldwiththehigher

activationbarrier.

Conclusion:Thehigherthetemperatureandthelowertheactivationenergy,thefasterthe

reaction.

3.CollisionOrientation:ConsiderthereactionwhereachlorideionbondstoanelectrondeficientC

atomintheunstableC(CH3)3+carbocation.ThereactionintermsofLewisstructuresisshowninFigure

15.2a.

Figure15.2a

Thereactionismoreclearwhenviewedusing3dimensionalmolecularmodels.Theseareshownin

Figure15.2bandc.In(a)theCl ionapproachestheC(CH3)3+ionfromtheopenside,sotheCl

lonepair

ofelectronscanreachtheopensiteonthecentralCatomandformthebondtomaketheproduct.


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Figure15.2bandc

In(c)theClionapproachesfromtheendandisblockedbyoneoftheCH3groups.Itcollides,butnotin

therightplace.So,noreaction.Inthisreactionmanyofthecollisionswillbeintheproperorientation

butmanywillnot.Orientationeffectsslowthereaction,butnotbymuch.Somereactionsinvolvevery

largemoleculeswithveryspecificreactionsites,andinthesealowfractionofcollisionsoccurringlead

toproducts.

Conclusion:Themorespecificareactionsite,theslowerthereaction.

II.ExpressingReactionRate

Howfastisareaction?Weknowitwhenweseeit,buthowisitexpressedquantitatively?Wedothisby

writingaratioofchangeinconcentrationoverchangeintime.ConsiderFigure15.3,whichshowsthe

changesinconcentrationoveraperiodof8secondsforareaction,A2B.ReactantAstartsata

concentrationof0.50Manddropsto0Mover8seconds.ProductBstartsat0Mandincreasesto1.0

Mover8seconds.Therearethreecommonwayswemeasurerate.

Figure15.3ConcentrationTimeplotsduringthe

courseofthereactionofA2B


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AverageRateOverTime

Onewaywecouldexpresstherateisthechangeinconcentrationovertheperiodof8seconds.Therate

ofproductionofBwouldbe1.00M/8s,or0.125M/s.Thisisnotparticularlyusefulbecausethereaction

wasalreadydone.Betteristolookattherateofthereactionwhileitisoccurring.Ifwelookattheplot

inmoredetail,weseethatthechangeinconcentrationinthefirstsecond,is:

Inthefirstsecond,theconcentrationofBchangesfrom0Mto0.50M.Therateofchangeistherefore,

[ ] ( )0.50 0.00.50 /

1.0 0.0

B M MM s

t s s

= =

Likewise,inthefirstsecond,theconcentrationofAchangesfrom0.50Mto0.25M.Therateofchange

ofAistherefore,

[ ] ( )0.25 0.500.25 /

1.0 0.0

A M MM s

t s s

= =

Whythenegativesign?Ratesarealwaysconsideredtobepositive.Whenexpressingarateintermsofa

reactants(forwhichtheconcentrationdecreases)wechangethesigntomakesuretheratecomesout

positive.Rule:Ratesexpressedforproductsdontusethenegativesign;thoseforreactantsdo.


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Example1.

Forthedecompositionofhydrogenperoxideindilutesodiumhydroxideat20oC

2H2O2(aq) 2H2O(l)+O2(g)

thefollowingdatahavebeenobtained:

Time,minutes [H2O2],mol/L

0 9.12x102

434 5.66x102

868 3.51x102

1302 2.18x102

WhatistheaveragerateofdisappearanceofH2O2overthetimeperiodfrom0minto434min?

Solution:

Therateovertimeisgivenbythechangeinconcentrationoverthechangeintime.Fora

reactant,weaddaminussigntomakesuretheratecomesoutasapositivevalue.

[ ] ( )2 22 2 55.66 10 9.12 10H O 7.97 10 / mintime 434min 0 min

MRate M

= = =

WecanrelatetheratesofchangeofAandBusingthereactionequation.Because2molofBare

producedforeachmolofAreacting,thechangeinconcentrationsofAandBarerelatedbyafactorof

2.

[ ] [ ]2

B A

t t

=


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Example2.

Forthedecompositionofhydrogenperoxideindilutesodiumhydroxideat20oC

2H2O2(aq) 2H2O(l)+O2(g)

theaveragerateofdisappearanceofH2O2overthetimeperiodfromt=0tot=516minis

foundtobe8.08x105M/min.WhatistherateofappearanceofO2overthesametimeperiod?

Solution:

ThereactionequationshowsthatforeverytwomolesofH2O2thatreact,onemoleofO2is

formed.ThereforetherateofformationofO2ishalftherateofH2O2consumption.

[ ] [ ] 52 2 2 5O H O1 1 8.08 10 4.04 10 / min2 2 min

MM

t t

= = =

InstantaneousRate

Weareofteninterestedinhowfastareactionisgoing,rightnow.Thisiscalledtheinstantaneousrateandisequaltotheslopeoftheconcentrationtimecurve.Ifdrawtangentsonthecurveatthe2second

pointandmeasuretheirslopes,weseethattheinstantaneousrateat2secondsis+0.130M/sforthe

productionofBand 0.065M/sforconsumptionofA.

InitialRate

Theinitialrateistheinstantaneousraterightwhenthereactionstarts.Thisisofinterestexperimentally

becauseitistheeasiesttimeforustoknowtheexactconcentrationsofthedifferentspeciesinsolution.

Afterthereactiongoesawhile,weneedtomakeinstantaneousmeasurementstoknowwhatthe

concentrationsare.Sometimesthatseasy,butsometimesnot.Buttheconcentrationsatthestartof

thereactionarealwayseasytoknow.Youmadethesolutions,afterall.


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Theinstantaneousrateistheslopeoftheconcentrationtimecurveatthetime=0point.Again,though,

weseethattherateofproductionofBistwicetherateofconsumptionofA.

AFinalPoint

Youcanuseaconcentrationtimecurvetosayalotaboutareaction.Wevealreadynotedthatthe

concentrationtimecurveidentifiesAasareactantbecauseitdecreasesinconcentrationovertime,and

Basaproductbecauseitincreases.Wecanalsoseethat2molofBareformedforeachmolofAthat

reactsbecauseBgoesuptwiceasmuchasAgoesdown.

Butwecanalsoseeinthisplotthatthereactionslowsdownasitproceeds.Notallreactionsdo.WhatthistellsusisthatthereactionratedependsontheconcentrationofAavailabletoreact.Noticethat

when[A]islarge,theslopeofthecurveissteep.Thereactionisfastwhen[A]islarge.Laterinthe

reaction,[A]ismuchsmaller,andsoistheslope.ThereactionslowsdownasAisusedup.Thereisa

concentrationdependenceofrateonconcentration.Which,notcoincidentally,isthesubjectofthenext

section.


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III.ConcentrationEffects

Thefrequencyofcollisionswasthefirstconsiderationweexaminedwhenthinkingaboutwhatcontrols

reactionrates.Themorecollisions,thefastertherateshouldbe.Andthatsgenerallybutnotalways

true.Therearetwowaystoincreasethefrequencyofcollisionsbetweenmolecules:crowdmoreof

themtogetherinaspaceormakethemmovefaster.Movingfasterisaccomplishedbyraisingthe

temperature,andwelldealwiththatlater.Asforcrowdingmoremoleculestogether,weretalking

aboutconcentration.Thegreatertheconcentrationofacompound,themoremoleculeswillbepresent

inthesamevolume,andthemorecollisionswilloccur.Thisiswhycrowdeddancefloorsarepopular,

andemptyonesarenot.Inadditiontothatbeingtruebydefinition,ofcourse.

Thedependenceofarateontheconcentrationofthereactantsisexpressedinaratelaw(orsometimes

calledarateequation).ConsiderthedecompositionreactionofNH4NCO.

NH4NCO(aq)(NH2)2CO(aq)

Theratelawforthereactionis

Rate=k[NH4NCO]2

Theratelawhasthreeparts:

Whatdoesthismean,anyway?Well,forthisreaction,itmeansthattherateofthereactionis

proportionaltothesquareoftheconcentrationofthereactant.Youmightthinkthatifyoudoublethe

reactantconcentration,theratewoulddouble.Inthisreaction,nottrue.Itwillquadruple.

Moreexamples:Reaction(allgasphase) RateLaw OverallOrder

2N2O54NO2+O2 Rate=k[N2O5] firstorder

NO2+CONO+CO2 Rate=k[NO2]2 secondorderinNO2,zeroorderin

CO,secondorderoverall

2NO+O22NO2 Rate=k[NO2]2[O2] secondorderinNO,firstorderin

O2,thirdorderoverall


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Acouplethingstonoticeintheexamplesabove:

1.Theratelawsarenotderivedfromthereactionequations.Sometimestheymatch.

Sometimestheydonot.

2.Althoughtherateconstantisreallyanumberwithunits,whendescribingageneralratelaw,

weusuallyjustwritek.Thisisbecausetheratelawistrueatalltemperatures,butany

numericalkvalueistrueonlyatonetemperature.Weindicateanumberwhenwewanttouse

theratelawincalculations.

AnnoyingBoxAbouttheUnitsofk

Assumingtimeismeasuredinseconds,therateofareactionisalways

expressedintermsofM/s.Whataretheunitsofk?Well,itdependson

thereactionorder.

Zero

Order

Ifthereactioniszeroorder,theratelawis,

Rate=k

Theunitsofkmustbethesameasthatofrate,M/s.

FirstOrder

Inafirst

order

reaction,

the

rate

law

is,

Rate=k[A]

Intermsofunits,thisisM

k Ms

= .Theunitsofkmustbe1/s,ors1.

SecondOrder

Inasecondorderreaction,theratelawis,

Rate=k[A]2

Intermsofunits,thisis2M k M

s= .Theunitsofkmustbe1/Ms,or

M1s1.


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Example3:

Methylacetatereactswithhydroxideionaccordingtothefollowingequation,

CH3CO2CH3+OHCH3CO2

+CH3OH

withtheratelaw,

Ratek[CH3CO2CH3][OH]

Whatistheorderofreactionforeachreactant,andtheoverallreactionorder?

Solution:

Becausethetherearenosuperscriptstotherightofeachconcentrationintheratelaw,it

meanstheyareeachfirstorder.ThereactionisfirstorderinCH3CO2CH3andfirstorderinOH.It

issecondorderoverall,because1+1doesequal2.

Determining

Rate

Laws

Using

the

Method

of

Initial

Rates

Ratelawsarenotderivedfromthereactionequation.Nomatterhowmanytimeswesay,peoplekeep

thinkingitstrue.Butitsnot.

Ratelawsarederivedfromexperimentaldata.Therearetwomethods:themethodofinitialrates,and

thegraphicalmethod.Herewecoverthemethodofinitialratesinwhichareactionisrunmultipletimes

withdifferentinitialconcentrationsofeachreactant.Theinitialratesofeachexperimentarecompared

totheinitialconcentrationsofthereactantstodeterminetheorderofeach.Themethodismostoften

performedinthefollowingway:

1.For

each

reactant,

run

apair

of

reactions

in

which

that

reactants

concentration

isdoubled

while

all

otherreactantconcentrationsareheldconstant.Thisisolatestheeffectonratetothatsinglereactant.

2.Mostoftenoneofthreethingshappenstotheinitialratewhenthereactantconcentrationis

doubled:

a.Theratedoesnotchange:thismeanstherateisindependentofthatreactant.Thereactionis

zeroorderwithrespecttothatreactant.

b.Theratedoubles:thismeanstherateisproportionaltotheconcentrationofthatreactant.

Thereactionisfirstorderwithrespecttothatreactant.

a.Theratequadruples:thismeanstherateisproportionaltothesquareoftheconcentrationof

thatreactant.Thereactionissecondorderwithrespecttothatreactant.

3.Oncetheorderforeachreactantisfound,takeanyoftheexperimentsandpluginconcentration

valuesandthemeasuredratetodeterminethenumericalvalueofk.


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Example4.

Areactionisperformedbetweentworeactant,AandB,

A+BC

Thefollowingdatawasobtainedfortheinitialrateofthereactioninfourseparateexperiments.

Whatistheratelawforthereaction,andwhatisthenumericalvalueofk?

Solution:

Theideaistoidentifytwotrialsforwhichonereactantconcentrationisdoubledandtheotheris

heldconstant.

ForA,thepairtouseisTrial1andTrial2.Noticethe[B]isconstantat0.252M.When[A]is

doubled,therateincreasesfrom0.0204M/minto0.0817M/min.Thisisanincreaseof4times.

Thereactionratequadruples,soweknowthatthereactionissecondorderinA.

ForB,thepairoftrialstouseisTrial1andTrial3.[A]isheldconstantat0.213Mwhile[B]is

doubled.When[B]isdoubled,therateincreasesfrom0.0204M/minto0.409M/min.Therate

doubleswhen[B]isdoubled,sothereactionisfirstorderinB.

Theratelawisthen,Rate=k[A]2[B]

Finally,wefindthevalueofkbychoosinganyofthetrials,insertingthevaluesforrate,[A],and

[B],andsolvingfork.ChoosingTrial4(fornoparticularreason),wesee,

0.0901M/min=k[0.761M]2[0.630]

2 1

2

0.0901 / min 0.247 min[0.761] [0.630]

Mk M = =


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ConcentrationTimeRelationshipsforSingleReactantReactions

Forreactionsthatinvolveasinglereactant,wehavederivedmathematicalequationsthatrelatethe

concentrationatanytimeduringthecourseofthereactiontotheinitialconcentrationandtherate

constant.Thesearecalledintegratedrateequationsbecause,perhapswithoutsurprise,theyare

derivedbyintegrationoftherateequation.Becausetherateequationsdependonthereactionorder,

sodotheintegratedrateequations.Table13.1showsthethreeequationsintwoforms.Thefirstisthe

regularequation.Thesecondisaversionintheformofastraightlineequation.Wellusethatlaterto

makestraightlineplots,whicharealwaysusefulinscience.Ineachcase,[A]orepresentsthe

concentrationofAbeforethereactionstarts.[A]trepresentstheconcentrationofAaftertimetpasses.

Twoalgebraicversionsaregivenforthefirstordercase,becausebothareuseful.

Equation/Order ZeroOrder FirstOrder SecondOrder

RateLaw Rate=k Rate=k[A] Rate=k[A]2

IntegratedRate

Equation

[A]o[A]t=kt[ ]

[ ]t

o

kt

t o

Aln kt

A

[A] =[A] e

=

[ ] [ ]t o

1 1kt

A A =

LinearForm [A]t= kt+[A]o

y= mx + b

t o

ln[A] kt ln[A]= +

y = mx +b

[ ] [ ]t o

1 1kt +A A=

y = mx +b

StraightLinePlot y=[A]

x=time

slope= k

y=ln[A]

x=time

slope= k

y=1/[A]

x=time

slope=k

Theseequationsaregenerallyusefulinthreeways:

1.Topredicttheconcentrationatsomefuturetime,givent,kand[A]o.

2.Determinethetimerequiredfor[A]otodecreaseto[A]t.

3.Todeterminetherateconstant,k,given[A]oand[A]tattimet.


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Example5.

Thedecompositionofnitrousoxideat565oC

2N2O2N2+O2

issecondorderinN2Owitharateconstantof1.10x103M

1s1.Ifthereactionisinitiatedwith

[N2O]equalto0.108M,whatwillitsconcentrationbeafter1250shaveelapsed?

Solution

Becausethereactionisfirstorder,weusethefirstorderintegratedrateequation.

[ ] [ ]

1 1

t oktA A =

Weknow[N2O]o,k,andt.Wesolvefor[N2O]t.

[ ]( )3 1 1

2 t

1 1= 1.10 10 1250 s

N O 0.108M s

M

[ ]1 1

2 t

19.26 1.38

N OM M

= +

[N2O]t=0.940M


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Example6.Findingthetimerequiredfor[A]otoDecreaseto[A]t

Theisomerizationofmethylisonitriletoacetonitrileinthegasphaseat250oC

CH3NCCH3CN

isfirstorderwitharateconstantof3.00x103s

1.IftheinitialconcentrationofCH3NCis0.107

M,howmuchtimemustpassfortheconcentrationofCH3NCtodropto0.0142M?

Solution

Becausethereactionisfirstorder,weusethefirstorderintegratedrateequation.Theversion

bettersuitedtofindingtime(ork,forthatmatter)is,

[ ][ ]

ln t

o

Akt

A=

Weknow[CH3NC]t,[CH3NC]o,andk.Wesolvefort.

[ ][ ]

3 3 1t

3 o

3 1

3 1

CH NC 0.0142 Mln ln 3.00 10

CH NC 0.107 M

2.02 3.00 10

2.02 6733.00 10

s t

s t

t ss

= =

=

= =


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Example7.DeterminingtheValueofk

Thegasphasedecompositionofdinitrogenpentoxideat335K

2N2O54NO2+O2

isfirstorderinN2O5.Duringoneexperimentitwasfoundthataninitialconcentrationof0.249

Mdroppedto0.0496Min230s.Whatisthevalueoftherateconstant,k,ins1?

Solution:

Becausethereactionisfirstorder,weusethefirstorderintegratedrateequation.Theversion

bettersuitedtofindingkis,

[ ][ ]

Aln

A

t

o

kt=

Weknow[N2O5]t,[N2O5]o,andt.Wesolvefork.

[ ][ ]

2 5 1t

2 5 o

1

3 1

N O 0.0496 Mln ln 230

N O 0.249 M

1.61 230

1.61 7.02 10230

k s s

k s s

k ss

= =

=

= =

HalfLifeandRadioactiveDating

Theconceptofhalflifeisusefulfordescribingtheroughspeedofareaction.Thehalflifeofareactionis

thetimeittakesforofthereactantstoturnintoproducts.Considerthefirstorderdecompositionof

hydrogenperoxide.ThetablebelowshowstheconcentrationofH2O2inincrementsof654min.The

dataareplottedaswell.


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Noticethatinthefirst654min,theconcentrationdropsfrom0.020Mto0.010M.Thatis,itdropsin

half.Thehalflifeofthereactionistherefore654min.Then,whenanother654minpasses(fromt=654

tot=1308)theconcentrationdropsinhalfagain:from0.010Mto0.0050M.Thehalflifeisstill654

min.Successivehalflivescuttheoriginalconcentrationtofractionsof,

#halflives 1 2 3 4 5 6

fractionremaining1

2

1

4

1

8

1

16

1

32

1

64

Example8.

Thefollowingdataareforthedecompositionofsulfurylchlorideat383oC.

SO2Cl2SO2+Cl2

time,min 0 166 332

[SO2Cl2],M 5.18x103 2.59x10

3 1.30x10

3

Whatisthehalflifeofthereactionstartingattime0min,andattime166min?

Solution:

Theconcentrationdropsinhalffromto2.59x103overthefirst166minofreaction.Therefore

thehalflifestartingatt=0minis166min(bydefinition).

Startingat166minutes,theconcentrationdropsfrom2.59x103to1.30x10

3of166min(from

166minto332min).Thisisagaindroppinginhalfoverthatsametimeperiod.Thehalflife

remains166min.


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MathematicalRelationshipsforHalfLifeandk

Foranyfirstorderreaction,thehalflifecanberelatedtotherateconstantinthefollowingway:

Atthehalflife,theconcentrationforreactantAisofwhatitwasatthestartofthereaction.So,

[ ][ ]

[ ]

[ ]

ot

o o

1AA 2ln ln

A A

1ln

2kt

=

=

Therefore,

1/2

1/2

1/2

1/2

ln(0.5)

0.693

0.693 0.693and

kt

kt

t kk t

=

=

= =

RadioactiveDecay

Allradioactiveisotopesdecayviafirstorderreactions.Forthesecaseswewritetheratelawandtime

basedequationsintermsofnumbersofatomsoramountpresentinsteadofconcentration,wherethe

letterNisusedtodenotehowmuchoftheradioactiveisotopeispresentatanyparticulartime.

[ ][ ]

t

o

kt

t o

Nln ktN

[N] =[N] e

=

Example9.

Theisotope32Phasahalflifeof14.3days.Ifasamplecontains0.884gof32P,whatmassof32P

willremainafter22days?

Solution:

Thepathforthisproblemistousethehalflifetodeterminetherateconstantk.Thenusethe

firstorderconcentrationtimeequationtodeterminethefinalamount.

1

1/2

0.693 0.6930.0485 days

14.3 daysk

t

= = =

1(0.0485days )(22 days)0.884 0.304

kt

t oN N e ge g

= = =


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RadioactiveCarbonDating

Carbonexistsasmostly12C,withabout1%of13C,andaverysmallfractionofradioactive14C.Itisformed

intheupperatmospherebyreactionof14Nwithhighenergysolarradiation.Becauseitisconstantly

decayingandbeingreformedintheupperatmosphere,thereisarelativelyconstantconcentrationof14

CpresentasCO2intheatmosphere.ThatCO2,likeallCO2,canbesequesteredbyphotosynthesistoformplants.Thoseplantslivetheirplantlikelife,sometimesbeingeatenbyanimals,sometimesjust

dying.AsaplantoranimallivesitkeepsexchangingCwiththeatmosphere,andthefractionof14Cin

theplantoranimalstaysequaltothatintheatmosphere.

Oncethethingdies,however,itsCcontentislockedanditnolongerreceivesnew14C.Becausethe14C

isradioactive,thefractionofitpresentinthedeadthingstartstodecrease.Ifwecomparethefraction

presenttodaywiththefractionwepresumetobethesteadystate14Camount,wecanestimatethe

timesincedeath.ThisisshowninFigure14.5.

Theactualexperimentinvolvesmeasuringthe14Cradioactivityfromasample.Itisgenerallyreported

inunitsofcountsperminutepergramofC.Thehalflifeof14Cis5730years,andthistechniqueisgood

fordatingitemsasoldas50,000years.


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Example10.

AwoodenbowlisfoundinacaveinFranceisfoundtohavea14Cradioactivityof10.2countsper

minutepergramofcarbon.Livingwoodhasanactivityof13.6countsperminutepergramof

carbon.Howlongagodidthetreedie?

Solution.

Tomakethiscalculation,wemaketheassumptionthatthefractionof14Cintheatmospherehas

beenconstantovertime.Therefore,theactivityatt=0isassumedtohavebeen13.6

counts/ming.Becausetheradioactivityofasampleisdirectlyproportionaltotheamountof

radioisotopepresent,theratioof14Cpresentnowandthenisequaltotheratioofradioactivity

nowandthen.So,

4 1

4 1

4 1

ln

0.6931.21 10

5730

10.2ln 1.21 10

13.6

0.288 1.21 10

2380

t

o

Nkt

N

k y ty

y t

y t

t y

=

= =

=

=

=

DeterminingtheRateLaw:TheGraphicalMethod

Concentrationtimeplotscanbeusedtodetermingtheratelawforsinglereactantreactions.Thelinear

formsoftheconcentrationtimeequationsforzero,first,andsecondorderreactionspredictdifferent

formsofequationsthatwillyieldalinearplot.

Equation/Order ZeroOrder FirstOrder SecondOrder

LinearForm [A]t= kt+[A]o

y= mx + b

t oln[A] kt ln[A]= +

y = mx +b

[ ] [ ]t o

1 1kt +

A A=

y = mx +b

StraightLinePlot y=[A]

x=time

slope= k

y=ln[A]

x=time

slope= k

y=1/[A]

x=time

slope=k


21/41

Thegraphicalmethodfordeterminingratelawsinvolvesthesesteps:

1.Collectconcentrationtimedatawhilethereactionproceeds.

2.Makethreeplots:[A]vs.t,ln[A]vs.t,and1/[A]vs.t.

3.Determinethereactionorder:

if[A]vs.tislinear,thereactioniszeroorder

ifln[A]vs.tislinear,thereactionisfirstorder

if1/[A]vs.tislinear,thereactionissecondorder

4.Theslopeoftheplotthatgivesastraightlineistherateconstantforthereaction.


22/41

Example11.

Concentrationvs.timedataiscollectedforthedecompositionofH2O2.

2H2O2(aq)2H2O(l)+O2(g)

Usethesedatatodeterminetheratelaw,andthenumericalvalueoftherateconstant.

Solution

ThefirststepistoenterthedataintoagraphingprogramsuchasExcelandusethecalculation

functionstocreatecolumnsforln[H2O2]and1/[H2O2].

Next,createthreeplots:[H2O2]vs.t,ln[H2O2]vs.t,and1/[H2O2]vs.t.

Theln[H2O2]plotgivesastraightline.Thismeansthatthereactionisfirstorderandtheratelaw

is,

Rate=k[H2O2]

Finally,thenumericalvalueoftherateconstantistheabsolutevalueoftheslopeofthestraight

lineplot.TheslopecanbefoundinExcelbyrightclickingononeofthedatapoints,and

choosingAddTrendline/Linear/DisplayEquationonChart.

Inthisreaction,k=0.00106min1.


23/41

IV.ActivationEnergyandTemperature

Weallhavethefeelingthatreactionsgofasterathighertemperatures,andthatstrue.Givenequal

concentrations,allreactionswillproceedmorerapidlyastemperatureincreases.Why?Allreactions

haveanactivationbarrier,anenergytheymustovercomebeforeproceedingtoproducts.Consider

thecaseofthereactionofthisorganiciodidecompound.

The

first

step

involves

the

breaking

of

the

C

I

bond.

This

is

an

endothermic

process

and

requires

energy

tomakehappen.Eventhoughlaterstepsinvolvebondformationandareexothermic,andcanmakethe

overallreactionexothermic,energymustbeusedtogetthefirststeptooccur.

ReactionCoordinateDiagrams

Theenergeticchangesthatoccurduringtheprogressofareactionareoftendisplayedgraphicallyusing

areactioncoordinatediagram.Thisisaplotwithenergyofthechemicalsystemontheyaxisandthe

progressofthereaction fromreactantsonthelefttoproductsontheright alongthexaxis.

Figure14.5showsrepresentativereactioncoordinatediagramsfortwosimplereactions,one

exothermicand

the

other

endothermic.

Figure14.5Reactioncoordinatediagramsforanexothermicreactionandanendothermicreaction.


24/41

Importantfactsaboutreactioncoordinatediagrams:

Energymustbegainedbythereactantsinordertoreachtheactivatedcomplex(alsocalledthetransitionstate)

Theenergyrequiredisthereactionsactivationenergy,andthesymbolgivenisEa. Theactivationenergyisalwaystheenergydifferencebetweenthatofthereactantsandthatof

theactivatedcomplex.

o Activationenergyisalwayspositive. Theenergychangeforthereaction(E)istheenergydifferencebetweenthereactantsandthe

products.

o EcanbeeitherpositiveornegativeWhenintroducingcollisiontheory,wesawthatanysampleofreactantswillhaveaBoltzmann

distributionofmolecularenergies.Onlythosewithenergiesgreaterthantheactivationenergywillbe

Figure14.6aandb.Boltzmanndistributionsofmolecularenergiesatdifferenttemperatures

showthatmoremoleculesexceedtheactivationenergyathighertemperature.b:Atagiven

temperature,moremoleculesexceedaloweractivationenergythanahighone.

abletoreact.Figure14.6ashowsBoltzmannplotsforsamplesatlowandhightemperature.Agreater

fractionofmoleculesexceedtheactivationenergyinthehightemperaturesample.Thisleadstoa

greaterfrequencyofeffectivecollisionsandafasterreactionrate.Figure14.6bshowshowactivation

energyaffectsrate.Highactivationenergyleadstoasmallerfractionofmoleculesthatcanreact,anda

slowerrate.

Temperatureandactivationenergythereforeplayoffeachothertocontrolreactionrate.Greater

activationenergydecreasesthefractionofcollisionswithsufficientenergytoreact.Buthighactivation

energycanbeovercomewithhighertemperatures.

ConsidertheconcentrationtimecurvesinFigure14.7a.Iftworeactionsarerunatthesame

temperature,theonewiththegreateractivationenergywillbeslower.Figure14.7bshowsthatasingle

reactionwillproceedfasteratahighertemperature.


25/41

Figure14.7aTworeactionsrunatthesametemperature,butwithdifferentactivationenergies.

Thehighactivationenergycasegivestheslowerreaction.

Figure14.7bThesamereactionrunattwodifferenttemperatures.Thehightemperaturecase

givesthefasterreaction.

Remembertheratelaw?Itwassupposedtocontrolrate,andsoitstilldoes.Recallthatforasimplefirst

orderreaction,AB,theratelawis,

Rate=k[A]

Temperatureandactivationenergyeffectsarepartoftherateconstant,k,asdescribedbythe

Arrheniusequation.


26/41

TheArrheniusequationshowsusthat:

ahigherfractionofcollisionswithcorrectorientationleadstolargervalueofAandafasterrate(wegenerallydonotreportvaluesofAforreactions,buttheydomatter)

isthefractionofcollisionsexceedingtheactivationenergy asEaincreases, decreasesandtherateisslower asTincreases, increasesandtherateisfaster thefractionEa/Tcontrolsrate:thesmallerthisfraction,thefastertherate

UsingtheArrheniusEquation

ThetwopointversionoftheArrheniusequationis

2

1 2 1

1 1ln a

Ek

k R T T

=

andismostoftenusedintwoways:

1. predictingtherateofreactionatadifferenttemperatureifyouknowEaandtherateatanothertemperature

2. determiningtheactivationenergy


27/41

Example12.FindingkataNewTemperature

Theactivationenergyforthegasphasedecompositionoftbutyl

propionateis164kJ.

C2H5COOC(CH3)3 (CH3)2C=CH2+C2H5COOH

Therateconstantat528Kis3.80x104/s.Whatwillthe

rateconstantbeat569K?

Solution:

Inthiscase,weknow:

T1=528K

T2=569K

k1=3.80x104/s

k2=?

Ea=164kJ/molandR=8.314x103

kJ/Kmol

WeinserttheseintotheArrheniusequation,

2

3

1 2 1

2

4

2.692

4

1 1 164 / 1 1ln

8.314 10 / 569 528

ln 2.693.80 10 / s

14.83.80 10 / s

aEk kJ mol

k R T T kJ K mol K K

k

ke

= =

=

= =

i

k2=5.61x103/s

Whatofthiscanwecontrol?

Althoughwediscusshowchangesin

Eawillaffectreactionrate,the

activationenergyisinherenttoany

reaction.Wecancontrolthe

temperature,butnotEa.Forsome

reactions,Eacanbeloweredthrough

useofacatalyst,butnotina

controlledway.Ifyouwanttochange

reactionrate,youchangeeither

concentrationortemperature.


28/41

Example13.FindingEa

Forthegasphasedecompositionofethylchloroformate,

ClCOOC2H5 C2H5Cl+CO2

therateconstantat470Kis1.05x103/sandtherateconstantat508Kis1.11x102/s.Whatis

theactivationenergyforthisreaction?

Solution:

Inthiscase,weknow:

T1=470K

T2=508K

k1=1.05x103/s

k2=1.11x102/s

Ea=?

R=8.314x103kJ/Kmol

WeinserttheseintotheArrheniusequation,

2

3 3

4

3

1.11 10 / s 1 1ln

1.05 10 / s 8.314 10 / 508 470

2.36 1.59 10 / 8.314 10 /

a

a

E

kJ K mol K K

EK

kJ K mol

=

=

i

i

Ea=123kJ/mol

GraphicalDeterminationofEa

TheArrheniusequationcanbewrittenintheformofastraightlineequation.

1ln lna

Ek A

R T= +

y = m x + b

Inastraightlineplot,y=lnk,x=1/T,andtheslope=Ea/R.TodetermineEa,werunthereactionata

seriesoftemperaturesandmeasuretherateconstantateach.Thenweplotlnkvs.1/T.


29/41

Example14.

Solution:

Thefirststepistocreateaplotofthenaturallogofthemeasuredrateconstantsvs.theinverse

temperature.ThiscanbedoneinExceloranothergraphingprogram.Findtheslopeusingleast

squaresanalysis(inExcelthisiscalledatrendline).

Theactivationenergyisgivenby,

Ea= Rxslope= (8.314x103kJ/Kmol)(1.22x104)=102kJ/mol

V.ReactionMechanisms

Chemicalreactionsinvolverearrangingatoms.Atomsononemoleculeenduponanother;some

moleculesfallapart;othermoleculesform.Areactionmechanismisadetaileddescriptionofthebond

breakingandformingstepsinvolvedinthepathwayfromreactanttoproducts.Anoverallreactionis

composedofaseriesofindividualsteps.Eachofthesediscretestepsiscalledanelementarystep.


30/41

Reactionmechanismscannotbecalculatedorpredictedwithassuredness.Instead,allreaction

mechanismsmustbedeterminedexperimentally.InsectionsVfandVgwereviewsomeofthemethods

usedtoelucidatemechanisms.

Va.TypesofMechanismSteps

Molecularity

Chemicaleventscanbecategorizedsimplybythenumberofreactingspeciesinanelementarystep.

Almostalleventsareunimolecular,involvingasinglereactant,orbimolecular,involvingtworeactants.

Ofthemillionsofchemicaleventsobserved,onlyahandfulinvolvetermolecularsteps,inwhichthree

speciesreacttogetheratthesameinstant.Therefore,wewilldealwithunimolecularandbimolecular

steps,butnottermolecularsteps.

Fromamorechemicalperspective,anyreactionweobserveisthesumofaseriesofeventsthatoccurin

sequence.Therearethreekindsofchemicalevents:

bondbreaking bondforming concertedbondbreakingandforming

Abondbreakingstepinvolvesasingleentitybreakingupintotwopieces.Intheexamplebelow,aCO

bondbreaks.Asinglereactantformstwoproducts.Becauseonlyonereactantmoleculeisinvolved,this

stepisalsotermedunimolecular.

Abondforming

stepinvolvesbringingtwomolecularfragmentstogethertoformasingleproduct.In

theexamplebelow,aCObondisformed.Tworeactantsformasingleproduct.Becausetworeactant

moleculesareinvolved,thisstepisalsotermedbimolecular.


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Aconcertedbondbreakingandformingstepinvolvesonebondbreakingwhileanotherisforming.In

theexamplebelowaCFbondisformedwhileaCClbondbreaks.Tworeactantsformtwoproducts.

Thisstepisalsobimolecularbecausetworeactantmoleculesareinvolved.

Vb.ExamplesofSingleStepReactions

Manychemicalreactionsaresimpleeventsthatoccurinasinglestep.Manyoftheseyouhaveseen

before.Acidbasereactionsmostlyoccurinasinglestep.Theprotonationofammoniabyhydroniumion

involvesthetransferofanH+ionfromH3O

+toNH3.Thissinglestepisaconcertedbondbreakingand

formingprocess.

H3O+(aq)+NH3(aq)H2O(aq)+NH4

+(aq)

ThedecompositionofN2O4isasinglebondbreakingstep.

Vc.MultistepReactions,IntermediatesandCatalysts

Mostchemicalprocessesthatweobserveoccurinaseriesofelementarysteps.Herearesomerules:

1.Theoverallreactionisthesumofthereactionsteps.

2.Aspeciesthatisformedinonestepandthenusedupinalaterstepisanintermediate.

3.Aspeciesthatisusedinonestepandthenreformedinalaterstepisacatalyst.

4.Neitherintermediatesnorcatalystsareseenintheoverallreactionequation.

5.Acatalystwillnormallybeseeninthereactionsratelaw,butanintermediateisnot.


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Thedecompositionofozonetakesplaceintwosteps.

Step1. O3(g)O2(g)+O(g) unimolecular,bondbreaking

Step2. O3(g)+O(g)2O2(g) bimolecular,concertedbondformingandbreaking

WhenstudyingHesssLawinChapter5weintroducedtheideaofaddingaseriesofreactionstogivea

netreaction.Thesametechniqueisusedheretodeterminetheoverallreactionthattakesplaceina

seriesofsteps.Thisisperformedbyaddingallthereactantstotheleftofthereactionarrowandallthe

oftheproductstotherightofthereactionarrow,andcancellinglikeitemsthatarepresentonboth

sides.Inthiscase,theO(g)ispresentonbothsides.

Step1. O3(g)O2(g)+O(g)

Step2. O3(g)+O(g)2O2(g)

2O3(g)+O(g)3O2(g)+O(g)

Overallreaction: 2O3(g)3O2(g)

LetsnotjustyetcompletelywriteoffthatO(g)atom.Sure,itdoesntappearintheoverallreaction

equation,butthatdoesntmeanitdoesnotexistatleastforalittlewhile.Achemicalspeciesthatis

formedinonestepofamechanismandthenusedinalaterstepiscalledanintermediate.

Intermediatesneverappearintheoverallreactionequation.Sometimestheycanbeobservedwhilethe

reactionprogresses,butothertimestheyareformedandusedupsoquicklytheyarenotseen,nor

heard.

Thedecompositionofhydrogenperoxideinthepresenceofiodideionoccursintwosteps.

Step1. H2O2(aq)+I(aq)IO

(aq)+H2O(l) bimolecular,concerted

Step2. IO(aq)+H2O2(aq)I

(aq)+H2O(l)+O2(g) bimolecular,concerted

overallreaction 2H2O2(aq)2H2O(l)+O2(g)

Inthiscase,theIOionisformedinthefirststepandthenconsumedinthesecondstep.IO

isan

intermediateinthisreaction.Conversely,Iionisusedinthefirststepandthenreformedinthesecond

step.

I

is

a

catalyst.


33/41

Example15.

Chlorofluorocarbonsbreakdownintheupperatmospheretogivechlorineatoms.Theseare

involvedinthebreakdownofozoneviathefollowingmechanism.

Step1. Cl(g)+O3(g)ClO(g)+O2(g)

Step2. ClO(g)+ O3(g)Cl(g)+O2(g)

a.Whatisthemolecularityofeachstep?

b.Whatistheoverallreactionequation?

c.Identifyanyintemediatesand/orcatalysts.

Solution:

a.Bothstepsinvovletworeactants.Theyarebothbimolecular.

b.Theoverallreactionisfoundbeaddingthetwostepsandcancellinglikespeciespresenton

bothsides.Inthiscase,bothClOandClarecancelled.

Step1. Cl(g)+O3(g)ClO(g)+O2(g)

Step2. ClO(g)+ O3(g)Cl(g)+O2(g)

OverallReactionbeforecancelling: Cl(g)+ClO(g)+2O3(g)Cl(g)+ClO(g)+2O3(g)

OverallReactionaftercancelling: 2O3(g)3O2(g)

c.Cl(g)isareactantintheStep1,butisregeneratedinStep2.Clisthereforeacatalyst.ClOis

generatedinStep1,butisconsumedinStep2.ClOisthereforeanintermediate.

ReactionCoordinateDiagramsandMechanisms

Reactioncoordinatediagramsareusedtorelatetheprogressofareactiontotheenergyofthesystem

atanyparticularstep.Figure14.8showsreactioncoordinatediagramsfora1stepanda2step

reaction.

Figure14.8ReactionCoordinatediagramsfor1step(RP)and2step(RInP)reactions.


34/41

Intheonestepreaction,theenergyofthereactants(R)mustincreasetothatofthetransitionstate

beforeprocedingtoformproducts(P).Inthetwostepreaction,atransitionstatemustfirstbereached

toallowthereactants(R)toundergoesthefirststeptoformtheintermediate(In).Theintermediatelies

inanenergywell.Asecondactivationbarriermustbeovercomefortheintermediatetoreactfurtherto

formthefinalproducts(P).Anyreactionsoverallactivationenergyistheenergydifferencebetweenthe

reactantsandthetransitionstateofhighestenergy.

Vd.ComplexReactionMechanismsandConcentrationTimeCurves

Consideratwostepreaction.

Step1. AB

Step2. BC

OverallReaction: AC

HowwilltheconcentrationsofA,B,andofCchangeovertime.Clearlytheconcentrationofthereactant

Awilldecreaseasthereactionproceeds,andtheconcentrationoftheproductCwillincrease.Butwhat

abouttheconcentrationoftheintermediate,B?Itisformedinonestepandconsumedinanother.Its

concentrationmustbezeroatthestartofthereaction,andmustbezerowhenthereactioniscomplete,

butnonzeroinbetween.Thisisbestobservedbyexaminingconcentrationtimecurves.Figure14.9

showsconcentrationtimecurvesforA(red),B(yellow),andC(blue)forthreecases.

a. b. c.

Figure14.9a.Step1fasterthanStep2.b.Step2alittlefasterthanStep1.c.Step2mustfasterthan

Step1.

In(a)thefirststepissomewhatfasterthanthesecondstep.Initially,asABoccurs,therateof

formationofBisgreaterthanitsrateofconsumptioninStep2.So,theconcentrationofBincreases

overtime.Then,asreactantAisdepletedandtheconcentrationofBincreases,therateofformationof

Bdecreasesanditsrateofconsumptionincreases.Atthispoint,theconcentrationofBstartsto

decrease.EventuallytheconcenrationsofbothAandBdroptonearzeroasthereactionnears

completion.

Part(b)showsacasewhereStep2issomewhatfasterthanStep1.Inthiscase,Bisconsumedmore

quicklyanditsconcentrationdoesnotbuildupasmuchasinPart(a).


35/41

Part(c)showsacasewhereStep2ismuchfasterthanStep1.Inthiscase,Bisconsumedalmost

instantlyasitisformed.TheconcentrationofBneverrisestoanappreciablelevelandBmayneverbe

concentratedenoughtoobserveinanexperiment.

Youmightwonder,ifacaselikePart(c)occurs,howwouldweeverknowtheintermediateeverexisted?

Insomecasesweneverdo.Inothercases,wedesignexperimentstoshowthattheintermediatewaspresent,evenifnotdirectlyobserved.Oneprimemethodforthisisatrappingexperiment.Ifwe

suspectedsuchanintermediatewaspresent,weaddanotherreagentthatwepredictwillreactwiththe

intermediate.Ifweseetheproductofthatexpectedreaction,thenwehaveevidencethatthe

intermediatewaspresent.Inoursimpleexample,

AB

B+DE

weaddreagentD,whichinterceptsintermediateBbeforeiscanformproductC.Itinsteadforms

productE,whichwecanobserveasitbuildsupinsolution.

Ve.ReactionsTooComplextoDescribeWithMechanisms

Reactionsthatinvolvemorethanonephaseareinherentntlycomplexandaregenerallynotwrittenas

mechanisms.Takeforexampletheformationofsolidesilverchloridethatformswhensolutionsofsilver

nitrateandsodiumchloridearemixed.Theoverallreactionandnetionicequationsarebothsimple:

AgNO3(aq)+NaCl(aq)AgCl(s)+NaNO3(aq)

Ag+(aq)+Cl(aq)AgCl(s)

Thisreactionseemsprettysimple.Twoionscometogetherandformasolid.Butitisthesolidthat

makesitcomplex.Thesolidthatformsishugeonthemolecularscale,containingbillionsofions.Each

crystalformsbyaddingmoreandmoreAg+andClions.

Intheimagehere,thecrystalgrowsbyafewaddedions.Thereisnoclearwaytowriteamechanismfor

thisprocess.Itinvolvesbillionsofstepsandalthoughthegeneralprocessisalwaysthesame,the

intemediatecrystalsinvolvedaredifferenteachtimethereactionisrun.


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Vf.RelatingMechanismstoaReactionsRateLaw

Thepathareactiontakeshasaneffectonhowrapidlythereactiontakesplace.Therateofanoverall

reactionisequaltotherateofthatreactionssloweststep,whichiscalledtheratedeterminingstep(or

sometimestheratelimitingstep).Acommonmisperceptionisthattheratelawisdirectlyderivedfrom

thereactionequation.Thisisnottrue removethatthoughtfromyourmind.Theratelawisinstead

deriveddirectlyfromthechemicalequationoftheratedeterminingstep.Thereareonlythreetypesof

steps:

unimolecular AD Rate=k[A]

bimolecular,onereactant 2AD Rate=k[A]2

bimolecular,tworeactants A+BD Rate=k[A][B]

Theexperimentallydeterminedratelawisoneofthemostusefultoolsintryingtodeterminea

mechanism.Theprocessinvolvesthreesteps:

1.Proposeamechanism,includingwhichstepwillberatedetermining.

2.Predicttheratelawusingtheratelawoftheslowstep.

3.Measureandcomparetheexperimentalratelawtothatpredictedforthemechanism.Ifthey

donotmatch,thentheproposedmechanismiswrong.Iftheydomatch,thentheproposed

mechanismmightbecorrect.Butitmightnotbecausemultiplemechanismscanleadtothesamepredictedratelaw.


37/41

Example16.

AnoxygenatomtransferreactionoccursbetweenNO2andCO.

NO2(g)+CO(g)NO(g)+CO2(g)

Therearetwoproposedmechanisms:

MechanismA.

Step1. NO2(g)+CO(g)NO(g)+CO2(g) ratedeterminingslowstep

MechanismB.

Step1. 2NO2(g)NO3(g)+NO(g) ratedeterminingslowstep

Step2. NO3(g)+CO(g)NO2(g)+CO2(g) fastsecondstep

Theexperimentallydeterminedratelawisrate=k[NO2]2.

Whichofthesetwomechanismsissuppportedbytheexperimentalevidence?

Solution

TheratelawpredictedforMechanismAisthatforthesinglestep.Becausethatstephasboth

NO2andCOasreactants,thepredictedratelawforthatstepandtheoverallreactionis:

rate=k[NO2][CO]

Thisdoesnotmatchtheexperimentalratelaw,sothismechanismiswrong.

TheratelawpredictedforMechanismBisthatfortheslow,firststep.Thisstepinvolvestwo

moleculesofNO2,sothepredictedratelawforthatstepandtheoverallreactionis:

rate=k[NO2]2

Thisdoesagreewiththeexperimentalratelaw,sothismechanismissupportedbutnotproved

correct.

MechanismswithReversibleSteps

Somereactionstepscanreactinboththeforwardandreversedirections.Thatis,reactantscangotoformproducts,butthoseproductscanbackreacttoreformthereactants.Ifbothstepsarefast,they

quicklyformanequilibriumstate,wheretheratesoftheforwardandreversereactionsproceedatthe

samerate.

Bromineandhydrogenreacttoformhydrogenbromide.Theoverallreactionis

Br2(g)+H2(g)2HBr(g).


38/41

Theproposedmechanismis:

1

1

2

3

2

2

2

Step 1. Br (g) 2 Br(g) fast in both directions

Step 2. Br(g) + H (g) HBr(g) + H(g) slow

Step 3. H(g) + Br (g) HBr(g) + Br(g)

k

k

k

k

fast

Inthismechanismdepiction,wewritetherateconstantforeachstepalongwithitsreactionarrow.The

stepnumberiswrittenasasubscriptforeachstep.Thereversiblefirststepiswrittenwithadouble

arrow,withtherateconstantforthereversestepbeingdenotedwithaminissign.Notethatk1isnot

equaltok1.The 1justmeansitstherateconstantforthereverseofstep1.

Theratelawfortheoverallreactionistheratelawfortheslowstep:

Rate=k2[Br][H2]

However,Brisnotareactantandweneverknowitsconcentration.InordertoobtainthepredictedratelawfortheoverallreactionweneedtoexpresstheconcentrationofBrintermsofspecieswedo

know.TodothiswesolvetheratelawforStep1for[Br].ThekeytothisisthatthereversibleStep1is

atequilibrium,astatewheretheforwardandreversereactionratesareequal(nottherate

constants).So,

2

1 2 1

1/2

1/21 2 12

1 1

[Br ] [Br]

[Br ][Br] [Br ]

k k

k k

k k

=

= =

Wetheninsertthissolutionfor[Br]intotherateequationfortheslowstep2.

1/2

1/212 2 2

1

Rate [Br ] [H ]k

kk

=

Experimentally,wecouldnotdifferentiatebetweenthecollectionofkvaluesandtheywouldcollapse

intoasingle,measuredrateconstant.

Rate=k[Br2]1/2

[H2]

Theoverallreactionorderis3/2.


39/41

Vg.AdvancedMethodsforDeterminingMechanisms

Alargepartofthechemistsworkisdeterminingthepathwaybywhichareactionoccurs.Ifyoure

lucky andyoureusuallynot someeasyexperimentscanhelpdetermineamechanism.Ifnot,more

advancedmethodsareemployed.Somecommonmethodsaredescribedbelow.

1.Reactiontotheexperimentalratelaw.Thiswasdescribedintheprevioussection.

2.Detectionofanintermediate.ThiswasdescribedinSectionVd.Detectionisusuallydone

spectroscopically,byobservingabsorptionofradiationintheUV,VIS,orIRregions.SometimesNMR

spectroscopyisused.Themainobstructiontodetectingintermediatesisthattheyareunstableand

tendtoreactquickly.Thatswhytheyareintermediates.But,sometimestheystickaroundawhileand

youcanseethem.

3.Trappingexperiments.Theseexperimentsareusedtochemicallydetectintermediatesbyhavingthe

intermediatereactwithasecondaryreagenttoproduceanidentifyableproduct.Thatis,youdesignan

experimenttohighjackanintermediateandcutofftherestofthemechanismunderstudy.

Inthereactiondiagramsabove,theproposedmechanisminvolvesthreesteps,withformationoftwo

intermediates,Band

C.

In

the

trapped

mechanism,

areagent

Risadded

to

siphon

off

intermediate

C.

If

additionofRleadstoformationofEandoflessD,thatsupportstheproposedmechanism.IfnoEis

observedandDisstillformed,thatmeanstheproposedmechanismiswrong.Thisallassumesyouare

surethetrappingreaction(C+RE)bothoccursandoccursfasterthanconversionofCD.Butyou

canknowthatwithreasonablecertainty,soitsok.

4.Isotopiclabeling.Inthistypeofexperiment,oneofthereactantsissynthesizedcontainingaparticular

minorisotopeofoneofthereactingatoms.Youcanthentrackwherethatisotopicallylabeledatom

endsupintheproducts.Inoxidationreactionsinvolvingmetaloxides,itisunclearwhethertheoxygen

thatisaddedtothespeciesoxidizedcomesfromthemetaloxide,orjustfromwaterinsolution.

Todeterminethis,themetaloxidecanbesynthesizedusingtherareisotope18O,inasolutionwherethe

watercontainsalmostall16O.Iftheproductcontains

18O,youknowitcomesfromthemetaloxide.Ifit

contains16O,youknowitcomesfromthewater.Tobesurethisexperimentisvalid,youneedtomake

surethemetaloxideandwaterdonotexchangeOatoms.


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Vh.Catalysis

Catalysisisthetechniqueofusingacatalysttoinfluencetherateofareaction.Catalystsserveoneof

twofuncitons:makingreactionsfasterormakingthemmoreselective.Ineithercase,thecatalyst

providesanalternativemechanisticpathwayforthereaction,whichchangestheactivationenergyand

changestherate.

IncreasingReactionRate.

Earlier,wesawthemechanismoftheI ioncatalyzeddecompositionofH2O2.Thereactioncanbe

observedtooccuroverashortperiodonceaniodideisadded.However,H2O2canalsodecomposeall

onitsown,thankyouverymuch.Ifyouhaveanunopenedbottleofhydrogenperoxidethathasbeenon

theshelfforafewmonths,payattentionwhenyoufirstopenittherewilloftenbeaswooshofair

escaping.ThisisduetothebuildupofO2gasthathasformedinthedecompositionovertime.

Figure14.10showsareactioncoordinatediagramforthisreaction,bothwithacatalystpresent,and

withoutone.Theloweractivationenergyinthecatalyzedcaseleadestoafasterreaction.

Figure14.10

Reaction

Coordinate

diagrams

for

the

uncatalyzed

and

thecatalyzeddecompositionofhydrogenperoxide.

IncreasingSelectivity

Welietoyou.Youvesuspectedit.Wetendtowritechemicalreactionsasiftheywilloccuraswritten

andgoallthewaytocompletion.Oftentimes,though,reactionsaremessy.Oneformofthemessinessis

thatasetofreactantscanoftenundergodifferentreactionstoformdifferentsetsofproducts.Thats

bad.Youalwayswantonesetofproductsbecauseitisdifficultandexpensivetopurifytheproductsofa

mixedreaction.Thesemultiplereactionsoccurwhenbothmechanisticpathwayshavesimilaractivation

energies.Selectivityisameasureofhowmuchthereactioncreatesonesetofproductsoveranother.(Uptillnowwehavebeenassumingallreactionsareperfectlyselective.)

Acatalystcanincreaseselectivitybychangingthepathwayofonereactionpathandnottheother.

Whenthishappens,thereactiontoformonesetofproductswillincreasewhiletheotherstaysslow.

Thereactionprefersthefasterformedproductsandismoreselective.Figure14.11showsreaction

coordinatediagramsrepresentinghowthisworks.


41/41

Figure14.11ReactionCoordinatediagramsshowingtheprogressfortwosimultaneousreactionsthat

occurfromasinglesetofreactants.Thetwosetsofproducts(AandB)areshowntoeithersideofthe

reactants,whichareinthecenterofthediagram.

Intheuncatalyzedcase,thetworeactions havesimilaractivationenergiesandproceedatclosetothe

samerate,formingsignificantquanitiesofproductsforbothcases.Additionofthecatalystdecreases

theactivationenergyforformationofproductsetB,butnotproductsetA.Inthiscase,formationofBis

muchfasterthanformationofA,sothereactionisselectiveforformationofB.

TypesofCatalysts

Catalystsaregenerallyhomogeneousorheterogeneous.Homeogeneouscatalystsaresoluble

compoundsthatareaddedtosolutionsalongwiththereactants.Thesehavetheadvantageofbeing

chemicallymodifiedtoagreatdegree.Chemistsareverygoodatsynthesizingsmallmoleculesandcan

generallycreatesmall,solublecompoundsthatactinveryspecificways.Homogeneouscatalystsare

goodatmodifyingreactionsinahighlyselectiveway.Themaindisadvantageofhomogeneouscatalysts

isthattheymustberemovedfromsolutionwhenthereactioniscomplete.Thatis,oncetheproducts

areformed,theyarestillmixedwiththecatalystandthetwomustbeseparated.Thisisoften

expensive.

Heterogeneous

catalysts

are

solids

that

serve

as

a

reactive

surface

upon

which

reactants

in

either

the

gasorsolutionphasecanreact.Thesehaveadvantagesanddisadvantagesthatareoppositefrom

homogeneouscatalysts.Theyareeasytoseparatefromthereactionproducts:justfilteroffthesolid.

Theyare,however,muchmoredifficulttomodify.Becauseofthesedifferences,homogeneouscatalysts

aremoreoftenusedforreactionsinwhichgreatselectivityisneeded;heterogeneouscatalystsare

neededwhereselectivityisnotanissue.

chapter 14 kinetics 2 7 bv

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