Chapter 15 Principles Of Reactivity: Chemical Kinetics

John C. KotzPaul M. TreichelJohn Townsendhttp://academic.cengage.com/kotzChapter 15Principles of Reactivity: Chemical KineticsJohn C. Kotz State University of New York, College at Oneonta

Chemical Kinetics We can use thermodynamics to tell ifa reaction is product- or reactantfavored. But this gives us no info on HOW FASTreaction goes from reactants to products. KINETICS— the study of REACTIONRATES and their relation to the way thereaction proceeds, i.e., its MECHANISM. The reaction mechanism is our goal! 2009 Brooks/Cole - Cengage2

Reaction RatesSection 15.1 Reaction rate change inconcentration of a reactant orproduct with time. Three “types” of rates–initial rate–average rate–instantaneous rate 2009 Brooks/Cole - Cengage3

Determining a Reaction RateSee Active Figure 15.2 2009 Brooks/Cole - Cengage4

Determining a Reaction RateDye ConcPLAY MOVIESee Chemistry Now , Chapter 15Time 2009 Brooks/Cole - CengageBlue dye is oxidizedwith bleach.Its concentrationdecreases with time.The rate — thechange in dye concwith time — can bedetermined from theplot.5

6Factors Affecting Rates Concentrations and physical state of reactantsand products Temperature Catalysts 2009 Brooks/Cole - Cengage

Concentrations & RatesSection 15.3Mg(s) 2 HCl(aq) f MgCl2(aq) H2(g)0.3 M HCl6 M HClPLAY MOVIE 2009 Brooks/Cole - Cengage7

8Concentrations and RatesTo postulate a reactionmechanism, we study reaction rate and its concentrationdependence 2009 Brooks/Cole - Cengage

Concentrations and RatesTake reaction where Cl- in cisplatin[Pt(NH3)2Cl3] is replaced by H2ORate of change of conc of Pt compdAm't of cisplatin reacting (mol/L) elapsed time (t) 2009 Brooks/Cole - Cengage9

Concentrations & RatesRate of change of conc of Pt compdAm't of cisplatin reacting (mol/L) elapsed time (t)Rate of reaction is proportional to [Pt(NH3)2Cl2]We express this as a RATE LAWRate of reaction k [Pt(NH3)2Cl2]where k rate constantk is independent of conc. but increases with T 2009 Brooks/Cole - Cengage10

Concentrations, Rates, &Rate LawsIn general, fora A b B f x X with a catalyst CRate k [A]m[B]n[C]pThe exponents m, n, and p are the reaction order can be 0, 1, 2 or fractions must be determined by experiment! 2009 Brooks/Cole - Cengage11

Interpreting Rate LawsRate k [A]m[B]n[C]p If m 1, rxn. is 1st order in ARate k [A]1If [A] doubles, then rate goes up by factor of If m 2, rxn. is 2nd order in A.Rate k [A]2Doubling [A] increases rate by If m 0, rxn. is zero order.Rate k [A]0If [A] doubles, rate 2009 Brooks/Cole - Cengage12

Deriving Rate Laws13Derive rate law and k forCH3CHO(g) f CH4(g) CO(g)from experimental data for rate of disappearanceof 00.18240.400.318 2009 Brooks/Cole - CengageDisappear of CH3CHO(mol/L sec)

Deriving Rate LawsRate of rxn k [CH3CHO]2Here the rate goes up by when initial conc.doubles. Therefore, we say this reaction isorder.Now determine the value of k. Use expt. #3 data—0.182 mol/L s k (0.30 mol/L)2k 2.0 (L / mol s)Using k you can calc. rate at other values of [CH3CHO]at same T. 2009 Brooks/Cole - Cengage14

Concentration/TimeRelationsWhat is concentration of reactant as function oftime?Consider FIRST ORDER REACTIONSThe rate law is![A]Rate k [A]!time 2009 Brooks/Cole - Cengage15

Concentration/TimeRelationsIntegrating - ( [A] / time) k [A], we getln is naturallogarithm[A]ln - kt[A]0[A] at time 0[A] / [A]0 fraction remaining after time thas elapsed.Called the integrated first-order rate law. 2009 Brooks/Cole - Cengage16

Concentration/Time Relations17Sucrose decomposes to simpler sugarsRate of disappearance of sucrose k [sucrose]If k 0.21 hr-1and [sucrose] 0.010 MHow long to drop 90%(to 0.0010 M)?Glucose 2009 Brooks/Cole - Cengage

Concentration/Time RelationsRate of disappear of sucrose k [sucrose], k 0.21 hr-1. Ifinitial [sucrose] 0.010 M, how long to drop 90% or to0.0010 M?Use the first order integrated rate law! 0.0010 -1ln # (0.21h)t&" 0.010 %ln (0.100) - 2.3 - (0.21 hr-1)(time)time 11 hours 2009 Brooks/Cole - Cengage18

19Integrated 1st order rate law Use to find eithertime to reach aparticularconcentration or tofind a concentrationafter a certain time, t Zero order and 2ndorder have differentequations 2009 Brooks/Cole - Cengage[A]ln - kt[A]0

Using the Integrated Rate LawThe integrated rate law suggests a way to tellthe order based on experiment.2 N2O5(g) f 4 NO2(g) O2(g)Time (min)01.02.05.0[N2O5]0 (M)1.000.7050.4970.173ln [N2O5]00-0.35-0.70-1.75Rate k [N2O5] 2009 Brooks/Cole - Cengage20

Using the Integrated Rate Law212 N2O5(g) f 4 NO2(g) O2(g) Rate k [N2O5][N 2 O5 ] v s . t i m e1l n [N 2 O5 ] v s . t i m e00-20ti m e5Data of conc. vs.time plot do not fitstraight line. 2009 Brooks/Cole - Cengage0ti m e5Plot of ln [N2O5] vs.time is a straightline!

Using the Integrated Rate LawPlot of ln [N2O5] vs. time is astraight line!Eqn. for straight line:y mx bl n [N 2 O5 ] v s . t i m e0ln [N2O5] - kt ln [N2O5]o-20ti m e5conc attime trate const slopeconc attime 0All 1st order reactions have straight line plotfor ln [A] vs. time.(2nd order gives straight line for plot of 1/[A]vs. time) 2009 Brooks/Cole - Cengage22

23Properties of Reactions 2009 Brooks/Cole - Cengage

24Factors Affecting Rates Concentrations and physical state of reactantsand products Temperature Catalysts 2009 Brooks/Cole - Cengage

25Factors Affecting Rates Physical state of reactantsPLAY MOVIE 2009 Brooks/Cole - Cengage

26Factors Affecting RatesCatalysts: catalyzed decomp of H2O22 H 2O 2 f 2 H 2O O 2PLAY MOVIE 2009 Brooks/Cole - Cengage

27Catalysts1. CO2(g) e CO2 (aq)See Page 7022. CO2 (aq) H2O(liq) e H2CO3(aq)3. H2CO3(aq) e H (aq) HCO3–(aq) Adding trace of NaOH uses up H . Equilibrium shifts to producemore H2CO3. Enzyme in blood (above) speeds up reactions 1 and 2 2009 Brooks/Cole - Cengage

28Factors Affecting Rates TemperatureBleach at 54 CBleach at 22 CPLAY MOVIE 2009 Brooks/Cole - Cengage

Iodine Clock Reaction1. Iodide is oxidized to iodineH2O2 2 I- 2 H f 2 H2O I22.I2 reduced to I- with vitamin CI2 C6H8O6 f C6H6O6 2 H 2 I-When all vitamin C is depleted, the I2interacts with starch to give a bluecomplex. 2009 Brooks/Cole - Cengage29

Iodine Clock Reaction 2009 Brooks/Cole - Cengage30

31Half-LifeHALF-LIFE isthe time ittakes for 1/2 asample todisappear.For 1st orderreactions, theconcept ofHALF-LIFE isespeciallyuseful.See Active Figure 15.9 2009 Brooks/Cole - Cengage

32Half-Life Reaction is 1st orderdecomposition ofH2O2. 2009 Brooks/Cole - Cengage

33Half-Life Reaction after 1half-life. 1/2 of the reactanthas beenconsumed and 1/2remains. 2009 Brooks/Cole - Cengage

34Half-Life After 2 half-lives1/4 of the reactantremains. 2009 Brooks/Cole - Cengage

35Half-Life A 3 half-lives 1/8of the reactantremains. 2009 Brooks/Cole - Cengage

36Half-Life After 4 half-lives1/16 of thereactant remains. 2009 Brooks/Cole - Cengage

Half-LifeSugar is fermented in a 1st order process (using anenzyme as a catalyst).sugar enzyme f productsRate of disappear of sugar k[sugar]k 3.3 x 10-4 sec-1What is the half-life of this reaction? 2009 Brooks/Cole - Cengage37

Half-LifeRate k[sugar] and k 3.3 x 10-4 sec-1. What is the halflife of this reaction?Solution[A] / [A]0 fraction remainingwhen t t1/2 then fraction remaining Therefore, ln (1/2) - k · t1/2- 0.693 - k · t1/2t1/2 0.693 / kSo, for sugar,t1/2 0.693 / k 2100 sec 35 2009 Brooks/Cole - Cengagemin38

Half-LifeRate k[sugar] and k 3.3 x 10-4 sec-1. Half-lifeis 35 min. Start with 5.00 g sugar. How much isleft after 2 hr and 20 min (140 min)?Solution2 hr and 20 min 4 half-livesHalf-life Time Elapsed Mass Left1st35 min2.50 g2nd701.25 g3rd1050.625 g4th1400.313 g 2009 Brooks/Cole - Cengage39

Half-LifeRadioactive decay is a first order process.Tritium f electron helium3H0 e3He-1t1/2 12.3 yearsIf you have 1.50 mg of tritium, how much is leftafter 49.2 years? 2009 Brooks/Cole - Cengage40

Half-LifeStart with 1.50 mg of tritium, how much is left after 49.2years? t1/2 12.3 yearsSolutionln [A] / [A]0 -kt[A] ?[A]0 1.50 mgt 49.2 yNeed k, so we calc k from:k 0.693 / t1/2Obtain k 0.0564 y-1Now ln [A] / [A]0 -kt - (0.0564 y-1)(49.2 y) - 2.77Take antilog: [A] / [A]0 e-2.77 0.06270.0627 fraction remaining 2009 Brooks/Cole - Cengage41

Half-LifeStart with 1.50 mg of tritium, how much is left after 49.2years? t1/2 12.3 yearsSolution[A] / [A]0 0.06270.0627 is the fraction remaining!Because [A]0 1.50 mg, [A] 0.094 mgBut notice that 49.2 y 4.00 half-lives1.50 mgf 0.750 mg after 1 half-lifef 0.375 mg after 2f 0.188 mg after 3f 0.094 mg after 4 2009 Brooks/Cole - Cengage42

Half-Lives of Radioactive ElementsRate of decay of radioactive isotopes given interms of 1/2-life.238U f 234Th He4.5 x 109 y14C f 14N beta5730 y131I f 131Xe beta8.05 dElement 106 - seaborgium - 263Sg0.9 s 2009 Brooks/Cole - Cengage43

MECHANISMSA Microscopic View of ReactionsMechanism: how reactants are converted toproducts at the molecular level.RATE LAW fMECHANISMexperiment f theoryPLAY MOVIE 2009 Brooks/Cole - Cengage44

Reaction MechanismsThe sequence of events at the molecularlevel that control the speed andoutcome of a reaction.Br from biomass burning destroysstratospheric ozone.(See R.J. Cicerone, Science,Science, volume 263, page 1243, 1994.)Step 1:Br O3 f BrO O2Step 2:Cl O3 f ClO O2Step 3:BrO ClO light f Br Cl O2NET: 2 O3 f 3 O2 2009 Brooks/Cole - Cengage45

Activation EnergyMolecules need a minimum amount of energy to react.Visualized as an energy barrier - activation energy, Ea.PLAY MOVIEReaction coordinatediagram 2009 Brooks/Cole - Cengage46

MECHANISMS& Activation EnergyConversion of cis to trans-2-butene requirestwisting around the C C bond.Rate k [cis-2-butene] 2009 Brooks/Cole - Cengage47

MECHANISMSCisTransition state48TransActivation energy barrier 2009 Brooks/Cole - Cengage

MECHANISMS49Energy involved in conversion of cis to trans buteneenergyActivatedComplex 262 kJcis-266 kJ4 kJ/moltrans 2009 Brooks/Cole - Cengage

Mechanisms Reaction passes thru aTRANSITION STATEwhere there is anactivated complexthat has sufficientenergy to become aproduct.ACTIVATION ENERGY, Ea energy req’d to form activated complex.Here Ea 262 kJ/mol 2009 Brooks/Cole - Cengage50

MECHANISMSAlso note that trans-butene is MORESTABLE than cis-butene by about 4 kJ/mol.Therefore, cis f trans is EXOTHERMICThis is the connection between thermodynamics and kinetics. 2009 Brooks/Cole - Cengage51

52Collision Theory Molecules must collide in order toreact Must collide with sufficient energy tobreak bonds (Ea) Must collide in the proper orientation 2009 Brooks/Cole - Cengage

Collision TheoryReactions require(a) activation energy and(b) correct geometry.O3(g) NO(g) f O2(g) NO2(g)1. Activation energy2. Activation energyand geometryPLAY MOVIEPLAY MOVIE 2009 Brooks/Cole - Cengage53

Effect of Temperature Reactions generallyoccur slower atlower T.In ice at 0 oCRoom temperaturePLAY MOVIEIodine clock reaction. SeeChemistry Now, Ch. 15H2O2 2 I- 2 H f 2 H 2 O I2PLAY MOVIE 2009 Brooks/Cole - Cengage54

Activation Energy and TemperatureReactions are faster at higher T because a largerfraction of reactant molecules have enough energy toconvert to product molecules.In general,differences inactivationenergy causereactions to varyfrom fast to slow. 2009 Brooks/Cole - Cengage55

Mechanisms1.Why is cis-butene e trans-butene reactionobserved to be 1st order?As [cis] doubles, number of moleculeswith enough E also doubles.2.Why is the cis e trans reaction faster athigher temperature?Fraction of molecules with sufficientactivation energy increases with T. 2009 Brooks/Cole - Cengage56

More About Activation EnergyArrhenius equation —Rateconstantk AeFrequency factorTemp (K)-E a / RTActivation 8.31 x 10-3 kJ/K molenergyFrequency factor related to frequency of collisionswith correct geometry.Plot ln k vs. 1/T fEa 1ln k - ( )( ) ln A straight line.R Tslope -Ea/R 2009 Brooks/Cole - Cengage57

58Arrhenius equation Find k at different temp, then:Plot ln k vs 1/T—slope -Ea/RTo det’n slope, use “rise over run” (y/x)R 8.314 X 10-3 kJ/mol-K or8.314 J/mol-K. See overheady-intercept ln A 2009 Brooks/Cole - Cengage