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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