Ch.12:ChemicalKinetics

Recall

8

ReactionRates•  ReactionRateàThechangeinconcentrationofareactantorproductperunitoftime

Factors That Affect Rxn Rate 1. Nature of the reactants

•  Adding an inert gas has no effect 2. Concentration of reactants 3. Temperature 4. Surface Area of Reactants 5. Pressure 6. Catalysis

ReactionRate

LudwigWilhelmy(1812-1864)•  Wilhelmyiscreditedasthefirstpersontomeasuretherateofachemicalrxncarefully

•  Measuredhowfastsucrosehydrolyzedintoglucoseandfructose

•  Showedhowtherxndependedontheamountofsugarpresent–thegreatertheamount,thefastertheinitialrate

Recall:ReactionRate•  Asareactionproceedsovertime,theconcentrationofthereactantdecreases,andtheconcentrationofaproductincreases.

ReactionRates•  Therateofachemicalrxnisameasureofhowfastarxnoccurs

•  TheratecanbedeterminedfromthedisappearanceofreactantsORtheappearanceofproducts

•  However,thevalueshouldalwaysbepositive(byconvention),soyouwillneedtoaddanegativesigntoyourcalculationsforreactants

Rate = [A]t 2−[A]t1t2 − t1

=Δ[A]Δt

ReactionRatesH2(g)+I2(g)à2HI(g)

Rate = −Δ[H2 ]Δt

ReactionRatesH2(g)+I2(g)à2HI(g)

Rate = −Δ[I2 ]Δt

ReactionRatesH2(g)+I2(g)à2HI(g)

Rate = + 12Δ[HI ]Δt

ReactionRate•  Therxnratetendstodecreasewithtime.Asthereactantstransformtoproducts,theirconcentrationsdecrease,andtherxnslowsdown.

MeasuringReactionRate

MeasuringReactionRates•  Recall:spectroscopyisthemostcommonmethodtomeasurereactionrates.Why?

MeasuringReactionRatesH2(g)+I2(g)à2HI(g)

•  I2isviolet,whileH2andHIarecolorless.•  AsI2reactswithH2,thevioletcolorfades.•  Thiscanbemonitoredwithaspectrometer.•  Theintensityofthelightabsorptionwilldecreaseastherxnproceeds,providingadirectmeasureof[I2]asafunctionoftime.

TheRateLaw

TheRateLawForAàproducts

Rate=k[A]n•  k=rateconstant(constantofproportionality)•  n=reactionorder•  Mostcommonrxnorders:

–  n=0àZeroorderrxnàrateisindependentof[A]–  n=1à1storderrxnàrateisdirectlyproportionalto[A]–  n=2à2ndorderrxnàrateisproportionalto[A]2

RateLaws•  Forazeroorderreaction…•  ChangingtheconcentrationofreactantAhasnoeffectontheoverallrate

RateLaws•  Forafirstorderreaction…•  [A]hasadirecteffectontherate.Forexample,doubling[A]willdoubler.

RateLaws•  Forasecondorderreaction…•  Changing[A]hasanexponentialeffectontherate.

RateLaws•  Iflookathowconcentrationchangeswithtime,wecanlookattheslopetodeterminewhattherategraphswouldlooklike.

RateLaws•  Ratevs.concentration•  Whatkindoffunctionsarethese?

RateLaws•  Theorderofareactioncanbedeterminedonlybeexperiment

•  Fromdata,wecanusethemethodofinitialratestodeterminereactionorder

•  Forthismethod,areactionisrunseveraltimeswithdifferentinitialreactantconcentrationstodeterminetheeffectofconcentrationontherate.

RateLaws•  Whathappenswhen[A]doubles?•  Theratealsodoubles•  Thereforen=1andrate=k[A]

[A](M) InitialRate(M/s)0.10 0.015

0.20 0.030

0.40 0.060

RateLaws•  Wecandeterminethevalueofkbysubstitutingvaluesfor[A]andtherateusingthedatatable

[A](M) InitialRate(M/s)0.10 0.015

0.20 0.030

0.40 0.060

[A](M) InitialRate(M/s)0.10 0.015

0.20 0.030

0.40 0.060

rate = k[A]

k = rate[A]

=0.015M / s0.10M

k = 0.15s−1

RateLaws•  Determinetheratelawforthedatabelow:

[A](M) InitialRate(M/s)0.10 0.015

0.20 0.015

0.40 0.015

RateLaws•  Rate=k[A]0=k•  Rate=0.015M/s

[A](M) InitialRate(M/s)0.10 0.015

0.20 0.015

0.40 0.015

RateLaws•  Determinetheratelawforthedatabelow:

[A](M) InitialRate(M/s)0.10 0.015

0.20 0.060

0.40 0.240

RateLaws•  Rate=k[A]2•  Note:Youcouldalsodeterminetheordermathematically

[A](M) InitialRate(M/s)0.10 0.015

0.20 0.060

0.40 0.240

RateLaws•  Todeterminetheordermathematically,setuptheratiooftwooftheratesandpluginvalues

[A](M)

InitialRate(M/s)

0.10 0.0150.20 0.0600.40 0.240

rate2rate1

=k[A]n

k[A]n

0.240M / s0.60M / s

=k(0.40)n

k(0.20)n

4.0 = 2n

log4.0 = log(2n )n = 2

RateLaws•  Thismethodisusefultoknowbecausetheorderdoesnothavetobeaninteger.

[A](M)

InitialRate(M/s)

0.10 0.0150.20 0.0600.40 0.240

rate2rate1

=k[A]n

k[A]n

0.240M / s0.60M / s

=k(0.40)n

k(0.20)n

4.0 = 2n

log4.0 = log(2n )n = 2

TryThis:•  ThereactionAàBhasbeenexperimentallydeterminedtobesecondorder.Theinitialrateis0.0100M/sat[A]=0.100M.Whatistheinitialrateat[A]=0.500M?

RateLaws•  Let’sextendourratelawtomultiplereactants:•  Givenareactionintheformof:

aA+bBàcC+dD•  Theratelawwouldbe:

Rate=k[A]m[B]n•  Wherekistherateconstant,[]=molarity,andmandnarethereactionorders.

•  Theoverallorderofthereactionisthesumoftheexponents(m+n)

RateLaws•  Ex/

C3H6O+Br2àC3H5OBr+HBr

•  Theratelawwouldhavetheform:Rate=k[C3H6O]m[Br2]n

•  Andthenwewouldusedatatodeterminetheorderofthereaction(lookforwardtothisinthenotsodistantfuture).

•  Ratelawsarealwayswrittenintermsofthereactants.

TryThis:•  Considerthereactionbetweennitrogendioxideandcarbonmonoxide:NO2(g)+CO(g)àNO(g)+CO2(g)

•  Fromthedata,determine:a.  Theratelawforthereactionb.  Therateconstant(k)forthereaction

[NO2](M) [CO](M) InitialRate(M/s)

0.10 0.10 0.0021

0.20 0.10 0.0082

0.20 0.20 0.0083

0.40 0.10 0.033

TryThis:•  Rate=k[NO2]m[CO]n•  IfIdouble[NO2]butkeep[CO]constant,whathappenstotherate?

•  Som=2•  (remember–youcanalsofindthismathematically)

[NO2](M) [CO](M) InitialRate(M/s)

0.10 0.10 0.0021

0.20 0.10 0.0082

0.20 0.20 0.0083

0.40 0.10 0.033

TryThis:•  Rate=k[NO2]2[CO]n•  IfIdouble[CO]butkeep[NO2]constant,whathappenstotherate?

•  Son=0

[NO2](M) [CO](M) InitialRate(M/s)

0.10 0.10 0.0021

0.20 0.10 0.0082

0.20 0.20 0.0083

0.40 0.10 0.033

TryThis:•  Rate=k[NO2]2[CO]0•  Rate=k[NO2]2•  Nowfindkthroughsubstitution•  k=0.21M-1s-1

[NO2](M) [CO](M) InitialRate(M/s)

0.10 0.10 0.0021

0.20 0.10 0.0082

0.20 0.20 0.0083

0.40 0.10 0.033

PracticeQuestion•  ForthereactionA+B→products,thefollowingdatawereobtained:

– Whatistheexperimentalratelaw?a.  Rate=k[A] d.Rate=k[A]2[B]b.  Rate=k[B] e.Rate=k[A][B]2c.  Rate=k[A][B]

Initialrate 0.030 0.060 0.060

[A]0 0.10 0.20 0.20

[B]0 0.20 0.20 0.30

TheIntegratedRateLaw

IntegratedRateLaw•  Ourpreviousratelawsrelatetheratetoconcentration

•  Whatifwewanttoshowtherelationshipbetweenconcentrationandtime?

•  Weusetheintegratedratelawwhenwewanttoknowhowlongareactionneedstoproceedtoreachaspecificconcentrationofareagent

IntegratedRateLaw•  Recall

– Forafirstorderreaction,theratelawcanbeexpressedasrate=k[A]1orsimplyrate=k[A]

•  Fortheintegratedratelaw(whichincorporatestimeasourvariable),afirstorderreactionbecomes

ln[A]t=-kt+ln[A]0•  [A]0=theinitialconcentrationofA•  [A]t=theconcentrationofAattimet•  k=rateconstant•  Keepinmindthattheonlyvariablesherearetand[A]t–everythingelseissimplyanumber

IntegratedRateLaw•  Thesearesimplyrelatedtoourfriendlylinearfunction:

IntegratedRateLaw•  Lookingattheseequations,tellyourneighborwhatyouwouldexpectthegraphstolooklike

IntegratedRateLaw•  Withsomevariation,generallywe’relookingat:•  (Notetheslopesare,respectively,-k,-k,andk)

ZeroOrder•  Wecangraphdatatodeterminetheorderofareaction

FirstOrder•  Wecangraphdatatodeterminetheorderofareaction

SecondOrder•  Wecangraphdatatodeterminetheorderofareaction

SampleQuestion•  ThefollowingdatawasobtainedforthedecompositionofSO2Cl2.Showthattherxnis1storder,thendeterminetherateconstantfortherxn.

Time(s) [SO2Cl2](M) Time(s) [SO2Cl2](M)

0 0.100 800 0.0793

100 0.0971 900 0.0770

200 0.0944 1000 0.0748

300 0.0917 1100 0.0727

400 0.0890 1200 0.0706

500 0.0865 1300 0.0686

600 0.0840 1400 0.0666

700 0.0816 1500 0.0647

Half-Life

Half-Life•  Recall:

– HalfLife(t1/2)=timerequiredfortheconc’ofareactanttofalltohalfofitsinitialvalue

•  Forafirstorderreaction,half-lifeisindependentoftheinitialconcentration,andwecanuse

t1/2=0.693/k

Half-Life

Half-Life•  Note:•  Youcanstilldeterminehalf-lifeforzero-orderandsecond-orderreactions;however,theyaredependentontheinitialconcentrationsandthereforenotconstant

SampleQuestion•  Forafirst-orderreactionaA→Products,thefirsthalf-lifeis20minutes– Whatisthesecondhalf-life?a. 10minutesb. 20minutesc.  40minutes

SampleQuestion•  Forasecond-orderreactionaA→Products,thefirsthalf-lifeis20minutes– Whatisthesecondhalf-life?a.  10minutesb.  20minutesc.  40minutes