1

EXPERIMENTB3:CHEMICALKINETICS

LearningOutcomesUponcompletionofthislab,thestudentwillbeableto:

1) Measuretherateofachemicalreactionanddeterminetheratelaw.2) Evaluatetheeffectoftemperatureontherateofachemicalreactionand

determinetheactivationenergy.3) Assesstheroleofacatalystontherateofachemicalreaction.

IntroductionChemicalkineticsisthestudyofspeedofchemicalreactions.Theratelawisamathematicalexpressionthatindicatestherelationshipbetweentheconcentrationsofthereactantsandtherateofthereaction.Therateofareactionmaybeexpressedintwoforms:1)thechangeinconcentrationofproductsovertimeor2)thechangeinconcentrationofreactantsovertime.Considerthefollowingchemicalreaction:

aA(aq)+bB(aq)!cC(aq)+dD(aq) Equation1

Intheaboveequation,a,b,c,anddarestoichiometriccoefficientsofthereactantsandproducts.Therateofthereactioncanbeexpressedasfollows:

€

Reaction Rate = − 1aΔ[A]Δt

= −1bΔ[B]Δt

= +1cΔ[C]Δt

= +1dΔ[D]Δt

Equation2

IntheexpressionforReactionRate,Δtisaunitoftimeduringwhichthechangesinconcentrationsofthereactantsorproductsaremeasured.TheunitforReactionRateisthereforemolaritypertimeandisMs-1orMmin-1.Theratelawforthisreactioniswrittenas:

RateLaw:ReactionRate=

€

k × [A]x × [B]y Equation3

IntheRateLawexpression,“k”isreferredtoastherateconstant,aproportionalityconstant.“x”isreferredtoasthe“order”ofthereactionwithrespecttothereactantAand“y”isreferredtoasthe“order”ofthereactionwithrespecttothereactantB.Theoverallorderofthereactionisgivenbyx+y.Mostchemicalreactionsarezeroorder,firstorder,orsecondorder.Theorderofachemicalreactionandtheorderwithrespecttoeachindividualreactant,canonlybedeterminedexperimentallyandisunrelatedtothestoichiometriccoefficientsofthebalancedchemicalequation.

2

Therefore,forthereactioninEquation1,“x”and“y”areunrelatedto“a”and“b”and“x”and“y”canonlybedeterminedexperimentally.Ifthevaluesof“x”and/or“y”happentobethesameas“a”and/or“b”,itispurelyacoincidence.Theorderofareactionisalwaysanintegerorhalf-integer.Valuesofreactionorderlargerthanthreearealmostimpossibleduetotheextremelylowprobabilitycollisionsthatwouldbenecessarytoresultinsuchanorder.Iftheexperimentalvalueof“x”inEquation3is0,thenitimpliesthattherateofthereactionisindependentoftheconcentrationofA.Iftheexperimentalvalueof“x”inEquation3is1,thenitimpliesthattherateofthereactionisproportionaltothefirstpoweroftheconcentrationofA.AnimportantpieceofinformationobtainedfromtheRateLawistheunitoftherateconstant“k”.Theunitoftherateconstantdependsontheorderofthereaction.Forinstance,supposeforaparticularreactionx=1andy=2,theunitof“k”maybederivedbydimensionalanalysisasfollows:

€

Reaction Rate = k × [A]1 × [B]2

k =Reaction Rate

[A]1 × [B]2

Unit of k = Ms−1

M1M 2 = M −2s−1

ExperimentalDeterminationofOrderofaReactionInordertodeterminetheorderofareactionwithrespecttoeachreagent,thereactionratemustbedeterminedusingvariousconcentrationsofthereactants,AandB.TheexperimentmustbedesignedsuchthattherateofthereactionismeasuredbyvaryingtheconcentrationofAwhilekeepingtheconcentrationBaconstant,andviceversa.Considerthefollowingdata:

Experiment [A],M [B],M Rate,Ms-11 0.1 0.1 0.12 0.2 0.1 0.23 0.1 0.2 0.4

Equation3givestheratelawexpression:ReactionRate=

€

k × [A]x × [B]y .Substitutingtheexperimentaldataintotheexpressionfortheratelawresultsinthefollowingthreeequations:

3

0.1Ms-1=k[0.1]x[0.1]y Equation4 0.2Ms-1=k[0.2]x[0.1]y Equation5 0.4Ms-1=k[0.1]x[0.2]y Equation6Tosolvefor“x”,divideEquation5byEquation4:

€

0.2Ms−1

0.1Ms−1=k[0.2]x[0.1]y

k[0.1]x[0.1]y

2 = 2x

x =1

Tosolvefor“y”divideEquation6byEquation4:

€

0.4Ms−1

0.1Ms−1=k[0.1]x[0.2]y

k[0.1]x[0.1]y

4 = 2y

y = 2

Therefore,basedontheexperimentaldata,theorderofthereactionwithrespecttoAisoneandtheorderofthereactionwithrespecttoBistwo.NOTE1:Usingthedataprovidedabove,thevaluesof“x”and“y”determinedwereexactintegers.However,whenusinglaboratoryexperimentaldatathiswilllikelynotbethecase.Sincereactionordersarenecessarilyintegerorhalf-integervalues,thevaluesof“x”and“y”mustberoundedappropriatelybeforeproceedingfurtherwiththecalculations.NOTE2:Theabovedatasetwasobtainedfromtheminimumnumberofexperimentsneedtosolvetheratelawexpressionmathematically.However,inthelaboratorymorenumberofexperimentsmustbeconductedtoaccountforexperimentalerrors.DeterminationofRateConstantandRateLawInordertodeterminetheRateLaw,theexactvalueoftherateconstantisalsorequired.Sincetheorderofthereactionisknown,therateconstantcanbedeterminedbysubstitutingthevaluesof“x”and“y”inEquations4,5,and6andsolvingfor“k”fromeachequationandthenobtainingtheaveragevalueof“k”.

4

Equation4isrewrittenas: 0.1Ms-1=k[0.1]1[0.1]2 Thereforek=100M-2s-1Equation5isrewrittenas: 0.2Ms-1=k[0.2]1[0.1]2 Thereforek=100M-2s-1Equation6isrewrittenas: 0.4Ms-1=k[0.1]1[0.2]2 Thereforek=100M-2s-1Theaveragevalueoftherateconstantk=100M-2s-1ThereforetheRateLawis:ReactionRate=100M-2s-1[A]1[B]2EffectofTemperatureontheRateofaReactionWhenreactionsareconductedatelevatedtemperatures,thenumberofcollisionsofthereactingmoleculesincreasesduetothefactthatthekineticenergyofthereactingmoleculeshasincreased.Asaresult,therateofthereactionincreaseswhenthetemperatureisincreased.TheeffectoftemperatureontherateofareactionisgivenbytheArrheniusequation.

€

k = Ae−EaRT Equation7

InEquation7: kistherateconstant AistheArrheniusconstant Eaistheactivationenergyofthereaction Ristheuniversalgasconstant(8.314J/molK) TistheabsolutetemperatureTheactivationenergy,Ea,canbedeterminedgraphicallybymeasuringtherateconstant,k,anddifferenttemperatures.ThemathematicalmanipulationofEquation7leadingtothedeterminationoftheactivationenergyisshownbelow.

€

k = Ae−EaRT

lnk = lnA − Ea

RT

lnk = −Ea

R⎛

⎝ ⎜

⎞

⎠ ⎟

1T

+ lnA

Therefore,aplotofthenaturallogarithmoftherateconstant(lnkonthey-axis)vs.reciprocaloftheabsolutetemperature(1/T)shouldyieldastraightlinewhose

5

slopewillbe

€

−Ea

R.Since,Ristheuniversalgasconstantwhosevalueisknown

(8.314J/mol.K),theactivationenergyEacanbecalculated.EffectofCatalystontheRateofaReactionCatalystsarechemicalsubstancesthatincreasetherateofachemicalreaction.Acatalystisnotchemicallymodifiedasaresultofthereactionandisrecoveredcompletelyattheendofthereaction.Onlyasmallamountofthecatalystisrequiredfortherateenhancement.Metalsandconcentratedacidsareexamplesofsomecommoncatalysts.Therateofareactiondependsontheactivationenergyofthereaction.Theactivationenergymaybethoughtofasanenergybarrierthatreactantsmustcrosspriortobeingtransformedintoproducts.ConsidertheexothermictransformationofreactantAtoproductB,inasinglemechanisticstep,whoseenergydiagramisshowninFigure1.

FIGURE1

AsshowninFigure1,thetransformationofAtoBisaone-stepmechanismthatpassesthroughasingletransitionstatewithanactivationenergyofEa.Thecatalystaltersthereactionmechanism.Themechanismofthereactioninthepresenceofacatalyst,sayX,willhavemorethanasinglestep,butwithloweractivationenergy.Apossiblereactionmechanismisshownbelow.

Energy

ReactionProgress

Reactant,A

Product,B

TransitionState

ActivationEnergy,Ea

6

€

A +C⇔ ACAC→B +C

Inthereactionmechanismshownabove,thereactantA,firstcombineswiththecatalystC,toformacomplexAC.ThecomplexACinthesecondstepformstheproductB,andregeneratesthecatalystinitsoriginalform.Inthisscheme,ACisknownasaReactionIntermediate.NOTE:Theaboveschemeisjustonepossiblemechanisticschemeinthepresenceofacatalyst.Differentcatalystswillworkdifferentlyandresultinavarietyofpossiblemechanisms.Theonlyconstantfeatureofallthesepossibilitiesisthattherewillbemorenumberofmechanisticstepscomparedtowhennocatalystispresent.TheenergydiagramforthetransformationofAtoBinthepresenceofthecatalystundertheschemediscussedaboveisshowninFigure2(dashedlines).TheplotforthetransformationofAtoBintheabsenceofthecatalystisalsoincludedinFigure2forcomparison.

FIGURE2

Duetothefactthatthecatalysthasalteredthereactionmechanismsuchthattherearenowtwostepsinthemechanism,therearenowtwotransitionstatesaswell(TransitionState1andTransitionState2).TheenergydifferencebetweenTransitionState1andthereactantAistheactivationenergy,Eac,inthepresenceofthecatalyst.

Product,B

Energy

ReactionProgress

Reactant,A

TransitionState

ActivationEnergy,Ea

ComplexAC

TransitionState2TransitionState1

ActivationEnergy,Eac(withcatalyst)

7

Byprovidinganalternatepathforthereaction,thecatalyst,likehightemperature,alsoincreasesthenumberofcollisionsofthereactingmolecules.Asaresulttherateconstantk,forthereactionisalsoalteredinthepresenceofthecatalyst.Theeffectofthecatalystontherateconstantofareactionwillbeexploredinthisexperiment.

8

ExperimentalDesignThekineticsofthereactionbetweeniodide(I−)andperoxydisulfate(S2O82-)willbeexploredinthisexperiment.Thereactionisalsoreferredtoasthe“IodineClockReaction”.ThechemicalreactionbetweenthesetwosubstancesisshowninEquation8below. 3I−(aq)+S2O82-(aq)

€

⇔I3−(aq)+2SO42-(aq) Equation8BoththereactantsinEquation8arecolorlessaqueoussolutions.Inordertomeasuretherateofthereaction,furthermanipulationsarenecessary.Afixedamountofthiosulfate,S2O32-andstarchareaddedtothereactionmixture.Thethiosulfate,reactswiththetriiodide(I3−)producedinEquation8accordingtothefollowingchemicalreaction. I3−(aq)+2S2O32-(aq)!3I−(aq)+S4O62-(aq) Equation9Onceallthethiosulfate,S2O32-,isconsumed,thechemicalreactionshowninEquation9willstop.Atthispoint,thetriiodide,I3−willcombinewiththestarchandformacomplexthatisdarkblue/purpleincolor. I3−(aq)+starch!I3−-starch Equation10Therateofareactionmaybemeasuredbytwodifferentmethods:

1. Theamountoftimeittakesforacertainamountofreactanttobeconsumed.2. Theamountoftimeittakesforacertainamountofproducttobeformed.

Inthisexperiment,theamountoftimetakenforafixedamountofoneoftheproducts,I3−,toformismeasuredindirectly.Duetothefactthataconstantamountofthiosulfate,S2O32-,isaddedtoallthereactions,eachreactionisthencarriedoutuntilaconstantamountofI3−isformed.AssoonasthisamountofI3−isformed,allthethiosulfate,S2O32-,isconsumed(asperEquation9),andtheI3−formedthereafterformsacomplexwiththestarch(asperEquation10)andgivesthereactionmixturethedarkblue/purplecolor.Therefore,bymeasuringthetimeittakesfortheappearanceofthedarkblue/purplecolor,thetimeittakesforafixedamountofI3−toformisbeingmeasured.Themolarityofthethiosulfateandthevolumeofthiosulfateusedinthisexperimentarerespectively,0.012Mand0.20mL.Thetotalvolumeofthereactionmixtureis1.9mL.AccordingtoEquation9,1moleofI3−reactswith2molesofS2O32-.

9

ThereforethemolarityofI3−thatwillbeformedtodeterminethereactionrateisgivenby:

1. Molesofthiosulfate =MolarityofS2O32-×VolumeofS2O32-=0.012M×0.20×10-3L=0.0000024

2. MolesofI3−

€

= 0.0000024 moles S2O32− ×

1 mole I3−

2 moles S2O32−

= 0.0000012 moles I3−

3. MolarityofI3− =

€

0.0000012moles0.0019L

= 0.00063molesL

Therateofthereactioncanthenbecalculatedas:

€

Δ[I3−]

ΔtwhereΔtisthetimeittakes

fortheappearanceofthedarkblue/purplecolor.Part1:DeterminationoforderofthereactionwithrespecttoI−:TheconcentrationofI−inthereactionmixturewillbevariedwhilekeepingallotherreagentsataconstantamount.Therateofthereactionwillbemeasuredaccordingtomethodsdescribedabove.Part2:DeterminationoforderofthereactionwithrespecttoS2O82−:TheconcentrationofS2O82−inthereactionmixturewillbevariedwhilekeepingallotherreagentsataconstantamount.Therateofthereactionwillbemeasuredaccordingtomethodsdescribedabove.Part3:Determinationofactivationenergyofthereaction:Thereactionwillbeconductedatfourdifferenttemperatures.Therateofthereactionwillbemeasuredaccordingtomethodsdescribedaboveateachtemperature.Theactivationenergywillbedeterminedgraphically.Part4:Studytheeffectofacatalystontherateofthereaction:Thereactionwillbecarriedoutinthepresenceandabsenceofacatalyst.Therateofthereactionwillbemeasuredaccordingtomethodsdescribedundereachcondition.Thecatalysttobeusedforthisreactioniscopper(II)nitrate.

10

ReagentsandSupplies0.2%starch,0.012MNa2S2O3,0.20MKI,0.20MKNO3,0.20M(NH4)2S2O8,0.20M(NH4)2SO4,0.0020MCu(NO3)2Hotplate,thermometer,stopwatch(SeepostedMaterialSafetyDataSheets)NOTE:Eventhoughthereactionmixtureonlyconsistsof0.2%starch,0.012MNa2S2O3,0.20MKI,and0.20M(NH4)2S2O8,thefollowingtworeagents-0.20M(NH4)2SO4and0.20MKNO3arealsoneededinordertomaintainaconstantionicstrengthinallthereactionmixtures.

11

ProcedurePART1:DETERMINATIONOFORDEROFTHEREACTIONWITHRESPECTTOI−1. Obtainfourtesttubesandlabelthemas1A,2A,3A,and4A.2. Obtainfourtesttubesandlabelthemas1B,2B,3B,and4B.3. Addthefollowingvolumesoftheindicatedreagentstotesttubeslabeled1A,2A,

3A,and4A(AllvolumesareinmL).

Reaction TestTube

0.2%Starch

0.012MNa2S2O3

0.20MKI

0.20MKNO3

1 1A 0.10 0.20 0.80 0.002 2A 0.10 0.20 0.40 0.403 3A 0.10 0.20 0.20 0.604 4A 0.10 0.20 0.10 0.70

4. Addthefollowingvolumesoftheindicatedreagentstotesttubeslabeled1B,2B,

3B,and4B(AllvolumesareinmL).

Reaction TestTube

0.20M(NH)2S2O8

0.20M(NH4)2SO4

1 1B 0.40 0.402 2B 0.40 0.403 3B 0.40 0.404 4B 0.40 0.40

5. Preparetostartthestopwatchtorecordthetimeofthereaction.Press“start”as

soonasthereagentshavebeencombinedasdescribedbelow.6. Combinecontentsoftesttube1Aand1B.Startthestopwatch.Thoroughlymix

thecontentsofthecombinedsolutionbytransferringitbackandforthbetweenthetwotesttubes.

7. Whenthedarkblue/purplecolorforms,pressthe“stop”buttononthe

stopwatchandtherecordthetimeinsecondsinthedatatable.8. Repeatsteps6and7witheachoftheotherpairsoftesttubes(combine2Awith

2B,3Awith3B,and4Awith4B).9. Emptythereactionmixturesintoalargebeakerlabeledas“waste”.Emptythe

contentsofthewasteintoawastedisposalcontainerprovidedbytheinstructor.10. Performasecondtrialofeachreactionensuringthatthetimesarewithin10%of

eachother.

12

PART2:DETERMINATIONOFORDEROFTHEREACTIONWITHRESPECTTOS2O82−1. Obtainfourtesttubesandlabelthemas5A,6A,7A,and8A.2. Obtainfourtesttubesandlabelthemas5B,6B,7B,and8B.3. Addthefollowingvolumesoftheindicatedreagentstotesttubeslabeled5A,6A,

7A,and8A(AllvolumesareinmL).

Reaction TestTube

0.2%Starch

0.012MNa2S2O3

0.20MKI

0.20MKNO3

5 5A 0.10 0.20 0.40 0.406 6A 0.10 0.20 0.40 0.407 7A 0.10 0.20 0.40 0.408 8A 0.10 0.20 0.40 0.40

4. Addthefollowingvolumesoftheindicatedreagentstotesttubeslabeled5B,6B,

7B,and8B(AllvolumesareinmL).

Reaction TestTube

0.20M(NH)2S2O8

0.20M(NH4)2SO4

5 5B 0.80 0.006 6B 0.40 0.407 7B 0.20 0.608 8B 0.10 0.70

5. Preparetostartthestopwatchtorecordthetimeofthereaction.Press“start”as

soonasthereagentshavebeencombinedasdescribedbelow.6. Combinecontentsoftesttube5Aand5B.Startthestopwatch.Thoroughlymix

thecontentsofthecombinedsolutionbytransferringitbackandforthbetweenthetwotesttubes.

7. Whenthedarkblue/purplecolorforms,pressthe“stop”buttononthe

stopwatchandtherecordthetimeinsecondsinthedatatable.8. Repeatsteps6and7witheachoftheotherpairsoftesttubes(combine6Awith

6B,7Awith7B,and8Awith8B).9. Emptythereactionmixturesintoalargebeakerlabeledas“waste”.Emptythe

contentsofthewasteintoawastedisposalcontainerprovidedbytheinstructor.10. Performasecondtrialofeachreactionensuringthatthetimesarewithin10%of

eachother.

13

PART3:DETERMINATIONOFACTIVATIONENERGYOFTHEREACTION1. Obtainfourtesttubesandlabelthemas9A,10A,11A,and12A.2. Obtainfourtesttubesandlabelthemas9B,10B,11B,and12B.3. Addthefollowingvolumesoftheindicatedreagentstotesttubeslabeled9A,

10A,11A,and12A(AllvolumesareinmL).Reaction TestTube 0.2%

Starch0.012MNa2S2O3

0.20MKI

0.20MKNO3

Approximatetemperature

9 9A 0.10 0.20 0.40 0.40 10°C10 10A 0.10 0.20 0.40 0.40 20°C11 11A 0.10 0.20 0.40 0.40 30°C12 12A 0.10 0.20 0.40 0.40 40°C4. Addthefollowingvolumesoftheindicatedreagentstotesttubeslabeled9B,

10B,11B,and12B(AllvolumesareinmL).

Reaction TestTube 0.20M(NH)2S2O8

0.20M(NH4)2SO4

Approximatetemperature

9 9B 0.20 0.60 10°C10 10B 0.20 0.60 20°C11 11B 0.20 0.60 30°C12 12B 0.20 0.60 40°C

5. Prepareanicebathtoatemperatureofaround10°C.

6. Preparetostartthestopwatchtorecordthetimeofthereaction.Press“start”as

soonasthereagentshavebeencombinedasdescribedbelow.7. Combinecontentsoftesttube9Aand9B.Startthestopwatch.Insertthereaction

mixtureintotheicebathandconstantlymixthecontents.8. Whenthedarkblue/purplecolorforms,pressthe“stop”buttononthe

stopwatchandtherecordthetimeinsecondsinthedatatable.Recordthetemperatureofthereactionmixture.

9. Combinetesttubes10Aand10Batroomtemperatureandrecordthetimefor

thereactionandthetemperatureofthereactionmixture.10. Prepareahotwaterbathtoatemperatureofaround30°C.

14

11. Combinecontentsoftesttube11Aand11B.Startthestopwatch.Insertthereactionmixtureintothehotwaterbathandconstantlymixthecontents.

12. Whenthedarkblue/purplecolorforms,pressthe“stop”buttononthe

stopwatchandtherecordthetimeinsecondsinthedatatable.Recordthetemperatureofthereactionmixture.

13. Prepareahotwaterbathtoatemperatureofaround40°C.14. Combinecontentsoftesttube12Aand12B.Startthestopwatch.Insertthe

reactionmixtureintothehotwaterbathandconstantlymixthecontents.15. Whenthedarkblue/purplecolorforms,pressthe“stop”buttononthe

stopwatchandtherecordthetimeinsecondsinthedatatable.Recordthetemperatureofthereactionmixture.

16. Emptythereactionmixturesintoalargebeakerlabeledas“waste”.Emptythe

contentsofthewasteintoawastedisposalcontainerprovidedbytheinstructor.17. Performasecondtrialofeachreactionensuringthatthetimesarewithin10%of

eachotherandthetemperaturesarethesame.

15

PART4:STUDYTHEEFFECTOFACATALYSTONTHERATEOFTHEREACTION1. Obtainfourtesttubesandlabelthemas13A,14A,15A,and16A.2. Obtainfourtesttubesandlabelthemas13B,14B,15B,and16B.3. Addthefollowingvolumesoftheindicatedreagentstotesttubeslabeled1A,2A,

3A,and4A(AllvolumesareinmL).

Reaction TestTube

0.2%Starch

0.012MNa2S2O3

0.20MKI

0.20MKNO3

13 13A 0.10 0.20 0.40 0.4014 14A 0.10 0.20 0.40 0.4015 15A 0.10 0.20 0.40 0.4016 16A 0.10 0.20 0.40 0.40

4. Addthefollowingvolumesoftheindicatedreagentstotesttubeslabeled1B,2B,

3B,and4B(AllvolumesareinmL).

Reaction TestTube

0.20M(NH)2S2O8

0.20M(NH4)2SO4

0.0020MCu(NO3)2

13 13B 0.80 0.00 1drop14 14B 0.40 0.40 1drop15 15B 0.20 0.60 1drop16 16B 0.10 0.70 1drop

5. Preparetostartthestopwatchtorecordthetimeofthereaction.Press“start”as

soonasthereagentshavebeencombinedasdescribedbelow.6. Combinecontentsoftesttube13Aand13B.Startthestopwatch.Thoroughlymix

thecontentsofthecombinedsolutionbytransferringitbackandforthbetweenthetwotesttubes.

7. Whenthedarkblue/purplecolorforms,pressthe“stop”buttononthe

stopwatchandtherecordthetimeinsecondsinthedatatable.8. Repeatsteps6and7witheachoftheotherpairsoftesttubes(combine14A

with14B,15Awith15B,and16Awith16B).9. Emptythereactionmixturesintoalargebeakerlabeledas“waste”.Emptythe

contentsofthewasteintoawastedisposalcontainerprovidedbytheinstructor.10. Performasecondtrialofeachreactionensuringthatthetimesarewithin10%of

eachother.

16

DataTablePART1:DETERMINATIONOFORDEROFTHEREACTIONWITHRESPECTTOI−

Reaction Timeinseconds(Trial1) Timeinseconds(Trial2)1

2

3

4

PART2:DETERMINATIONOFORDEROFTHEREACTIONWITHRESPECTTOS2O82−

Reaction Timeinseconds(Trial1) Timeinseconds(Trial2)5

6

7

8

17

PART3:DETERMINATIONOFACTIVATIONENERGYOFTHEREACTION

Reaction Timeinseconds(Trial1)

Temperature,°C(Trial1)

Timeinseconds(Trial2)

Temperature,°C(Trial2)

9

10

11

12

PART4:STUDYTHEEFFECTOFACATALYSTONTHERATEOFTHEREACTION

Reaction Timeinseconds(Trial1) Timeinseconds(Trial2)13

14

15

16

18

DataAnalysisPART1:DETERMINATIONOFORDEROFTHEREACTIONWITHRESPECTTOI−1. CalculatethemolarityoftheI−inthereactionmixtureforeachreaction.Example:Reaction1-M1=0.20M V1=0.80mL M2=? V2=1.9mL

€

M1V1 = M2V2

M2 =0.20M × 0.80mL

1.9mL= 0.084M

Reaction2-Reaction3-Reaction4-2. CalculatethemolarityoftheS2O82-inthereactionmixtureforeachreaction.Reactions1-43. Calculatetheaveragereactiontimeinsecondsforeachofthereactions.4. Calculatetheaveragerateofreactionforeachofthereactions.

ReactionRate=

€

Δ[I3−]

Δt=

€

0.00063MΔt

19

5. Entertheinformationfromsteps1through4inthefollowingtable.Reaction [I−],M [S2O82-],M Averagetime(s) ReactionRateMs-1

1

0.084

2

3

4

6. RateLaw:ReactionRate=k×[I−]x×[S2O82-]y.Substitutetheinformationfor

Reactions1-4fromabovetableintotheratelawexpressionandsolvefor“x”.Reaction1Reaction2Reaction3Reaction4

TheorderofthereactionwithrespecttoI−=

20

PART2:DETERMINATIONOFORDEROFTHEREACTIONWITHRESPECTTOS2O82−1. CalculatethemolarityoftheS2O82-inthereactionmixtureforeachreaction.Example:Reaction5-M1=0.20M V1=0.80mL M2=? V2=1.9mL

€

M1V1 = M2V2

M2 =0.20M × 0.80mL

1.9mL= 0.084M

Reaction6-Reaction7-Reaction8-2. CalculatethemolarityoftheI−inthereactionmixtureforeachreaction.Reactions5-83. Calculatetheaveragereactiontimeinsecondsforeachofthereactions.4. Calculatetheaveragerateofreactionforeachofthereactions.

ReactionRate=

€

Δ[I3−]

Δt=

€

0.00063MΔt

5. Entertheinformationfromsteps1through4inthefollowingtable.

21

Reaction [I−],M [S2O82-],M Averagetime(s) ReactionRateMs-1

5

0.084

6

7

8

6. RateLaw:ReactionRate=k×[I−]x×[S2O82-]y.Substitutetheinformationfor

Reactions5-8fromabovetableintotheratelawexpressionandsolvefor“y”.Reaction5Reaction6Reaction7Reaction8

TheorderofthereactionwithrespecttoS2O82-=

22

DETERMININGTHERATELAW1. RateLaw:ReactionRate=k×[I−]x×[S2O82-]y.SubstitutethevaluesofReaction

Rate,“x”,“y”,[I−],and[S2O82-]forreactions1-8andsolvefortherateconstant.

€

k =Reaction Rate[I−]x[S2O8

2−]y Equation11

Reaction [I−],M [S2O82-],M ReactionRateMs-1 Rateconstant“k”

1

0.084

2

3

4

5

0.084

6

7

8

2. Findtheaveragevalueoftherateconstantandgivethecorrectunit.3. GivetheRateLaw

TheRateLaw=

23

PART3:DETERMINATIONOFACTIVATIONENERGYOFTHEREACTION

€

k = Ae−EaRT

lnk = lnA − Ea

RT

lnk = −Ea

R⎛

⎝ ⎜

⎞

⎠ ⎟

1T

+ lnA

Acomparisonoftheaboveequationwiththeequationofastraightline,y=mx+b,indicatesthefollowing:

lnkisanalogoustoy

€

1Tisanalogoustox

€

−Ea

Risanalogoustom

lnAisanalogoustob

Thereforealinearregressionofaplotoflnkvs.

€

1Tshouldresultinastraightline

whoseslopewillbe

€

−Ea

Randwhosey-interceptwillbelnA.SinceRistheuniversal

gasconstant,thevalueofEa canbedeterminedfromtheslopeofthebestfitline.

1. Calculate“k”forreactions9-12usingEquation11.Usetheaveragevalueofthetimeandtemperatureofthetwotrialsforeachcalculation.

ReactionRate=

€

Δ[I3−]

Δt=

€

0.00063MΔt

€

k =Reaction Rate[I−]x[S2O8

2−]y

24

Reaction Averagetime

(seconds)ReactionRate,Ms-1

RateConstant,k

9

10

11

12

2. Completethefollowingtable(NOTE:Youcanuseaspreadsheetsuchas

MicrosoftExceltodothis,SeeBelow).3. Enterthedata(temperaturein°Candk)incolumnsAandB.UseRow1for

columnheadings.

4. EnterformulasinRow2foreachcalculation,asshowninthetableabove.

5. Ineachcolumn,pointthecursortothebottomrightcornerofacell(sayC2)anddragdown(tillRow5)theplussigntocopytheformulatotheothercells.RepeatthisforallthecolumnsDthroughE.

6. Todrawagraph,selectthexandydata,whichwouldbedatainfieldsD:2-5andE:2-5.

7. Click“Insert”andthen“Chart”.Choose“XY”scatterandselect“MarkedScatter”

8. Whenthegraphisdisplayed,clickonanydatapointonthechartandfromthetoolbar,select“Chart”andthen“InsertTrendline”.

A B C D E1 Temperature,T

°C k T,Kelvin 1/T ln(k)2 10 =A2+273.15 =1/C2 =LN(B2)3 20 4 30 5 40

25

9. Fromthepop-upbox,selectthe“Options”tabandchecktheboxes:1)Displayequationand2)DisplayR-squaredvalueandclickOK.

10. Fromtheequationofthestraightline,obtaintheslopeandsetthatequalto

€

−Ea

R.UsingthefactthatR=8.314J/molK,calculatethevalueofEa.

TheActivationEnergyEa=

€

kJmol

26

PART4:STUDYTHEEFFECTOFACATALYSTONTHERATEOFTHEREACTION1. Usingtheaveragetimeforthetwotrials,calculatetheReactionRateandthe

RateConstant“k”accordingtoequation11forthereactions13-16.

ReactionRate=

€

Δ[I3−]

Δt=

€

0.00063MΔt

€

k =Reaction Rate[I−]x[S2O8

2−]y

Reaction Averagetime(seconds)

ReactionRate,Ms-1

RateConstant,k

13

14

15

16

2. Calculatetheaveragerateconstantofthecatalyzedreactions(13-16).

Averagerateconstantofthecatalyzedreactions=

27

3. Comparetheaveragerateconstantfortheuncatalyzedreactions(1-8)withtheaveragerateconstantofthecatalyzedreactions(13-16)todeterminetherateenhancementduetothecatalyst.

Average“k”fortheuncatalyzedreactions(1-8)=k(uncatalyzed)=Average“k”forthecatalyzedreactions(13-16)=k(catalyzed)=

Rateenhancement=

€

k(catalyzed)k(uncatalyzed)

=