无机化学课件13ch14chemical kinetics

1、Chemistry: Atoms FirstJulia Burdge & Jason OverbyChapter 14Chemical Kinetics1414.1 Reaction Rates14.2 Collision Theory of Chemical Reactions14.3 Measuring Reaction Progress and Expressing Reaction RateAverage Reaction RateInstantaneous RateStoichiometry and Reaction Rate14.4 Dependence of Reacti

2、on Rate on Reactant ConcentrationThe Rate LawExperimental Determination of the Rate Law14.5 Dependence of Reactant Concentration on TimeFirst-Order ReactionsSecond-Order Reactions14.6 Dependence of Reaction Rate on TemperatureThe Arrhenius Equation1414.7 Reaction MechanismsElementary ReactionsRate-D

3、etermining StepExperimental Support for Reaction Mechanisms14.6CatalysisHeterogeneous CatalysisHomogeneous CatalysisEnzymes: Biological Catalysts14.1Chemical kinetics is the study of how fast reactions take place.Some happen almost instantaneously, while others can take millions of years.Increasing

4、the rate of a reaction is important to many industrial processes.Most reactions happen faster at higher temperature.Chemical reactions generally occur as a result of collisions between reacting molecules.According to collision theory of chemical kinetics, the reaction rate is directly proportional t

5、o the number of molecular collisions per second:14.2number of collisionsrate sCollisions that result in a chemical reaction are called effective collisions.The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction.Molecules must also be oriented in a way th

6、at favors reaction.Cl + NOCl Cl2 + NOAn effective collision results in reaction.Correct orientation to facilitate reactionCollisions that result in a chemical reaction are called effective collisions.The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction

7、.Molecules must also be oriented in a way that favors reaction.Cl + NOCl Cl2 + NOAn ineffective collision results in no reaction.Incorrect orientation does not favor reactionWhen molecules collide in an effective collision, they form an activated complex (also called the transition state).Measuring

8、Reaction Progress and Expressing Reaction RateChemical kinetics is the study of how fast reactions take place.14.3A BA decreasesAArate = rate = tBBrate = rate = tMeasuring Reaction Progress and Expressing Reaction RateA BBr2(aq) + HCOOH(aq) 2Br(aq) + 2H+(aq) + CO2(g)2 final2 initial2 final2 initialf

9、inalinitialfinalinitialBr Br Br Br BBaverage rate = average rate = ttt Measuring Reaction Progress and Expressing Reaction RateMeasuring Reaction Progress and Expressing Reaction RateBr2(aq) + HCOOH(aq) 2Br(aq) + 2H+(aq) + CO2(g)Measuring Reaction Progress and Expressing Reaction RateBr2(aq) + HCOOH

10、(aq) 2Br(aq) + 2H+(aq) + CO2(g)Measuring Reaction Progress and Expressing Reaction RateBr2(aq) + HCOOH(aq) 2Br(aq) + 2H+(aq) + CO2(g)-5-50.0101 0.01200.0101 0.0120BBaverage rate = 3.80average rate = 3.80101050.00.050.00.0/MMsts -5-50.008460.01200.008460.0120BBaverage rate = 3.54average rate = 3.5410

11、10100.00.0100.00.0/MMsts First 50 seconds:First 100 seconds:Measuring Reaction Progress and Expressing Reaction RateBr2(aq) + HCOOH(aq) 2Br(aq) + 2H+(aq) + CO2(g)Measuring Reaction Progress and Expressing Reaction RateThe instantaneous rate is the rate for a specific instant in time.Measuring Reacti

12、on Progress and Expressing Reaction RateTime (s)Br2(M)Rate (M/s)0.00.01204.20 x 10550.00.01013.52 x 105250.00.005001.75 x 105Br2(aq) + HCOOH(aq) 2Br(aq) + 2H+(aq) + CO2(g)2 250 s50 s2 2250 s250 sBrBr2 2BrBr5 55 5rate at 50.0 s3.52 10rate at 50.0 s3.52 102 2rate at 250.0 s1.75 10rate at 250.0 s1.75 1

13、02 2rate Br rate Br 2 2rate Br rate Br kk is called the rate constant.Measuring Reaction Progress and Expressing Reaction RateTime (s)Br2(M)Rate (M/s)0.00.01204.20 x 10550.00.01013.52 x 105250.00.005001.75 x 105Br2(aq) + HCOOH(aq) 2Br(aq) + 2H+(aq) + CO2(g)2 2rate Br rate Br kat t = 50.0 s2 2raterat

14、e Br Br k 5 531313.52 10 s3.52 10 s 3.49 10 s 3.49 10 s0.0101 0.0101 /MkMMeasuring Reaction Progress and Expressing Reaction Rate2H2O2(aq) 2H2O(l) + O2(g)2 2O O2 2 O O 1 1r ra at te e = = PtRTtMeasuring Reaction Progress and Expressing Reaction RateaA + bB cC + dD1A1B1C1D1A1B1C1Drate = rate = atbtct

15、dt Measuring Reaction Progress and Expressing Reaction RateWrite the rate expressions for the following reaction:CO2(g) + 2H2O(g) CH4(g) + 2O2(g)Solution:Use the equation below to write the rate expressions.1A1B1C1D1A1B1C1Drate = rate = atbtctdt 22422242CO H OCH O CO H OCH O 1111rate = rate = 2222tt

16、tt Worked Example 14.1Strategy For reactions containing gaseous species, progress is generally monitored by measuring pressure. Pressures are converted to molar concentrations using the ideal gas equation, and rate expressions are written in terms of molar concentrations.Write the rate expressions f

17、or each of the following reactions:(a) I-(aq) + OCl-(aq) Cl-(aq) + OI-(aq)(b) 2O3(g) 3O2(g)(c) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)Solution (a) All the coefficients in this equation are 1. Therefore,rate = = = =I-tOCl-tCl-tOI-t Worked Example 14.1 (cont.)Solution (b) rate = =(c) rate = = = =O3tO2t1213N

18、H3t14O2t15NOt14H2Ot16Think About It Make sure that the change in concentration of each species is divided by the corresponding coefficient in the balanced equation. Also make sure that the rate expressions written in terms of reactant concentrations have a negative sign in order to make the resultin

19、g rate positive. Worked Example 14.2Strategy Determine the rate of reaction and, using the stoichiometry of the reaction, convert to rates of change for the specified individual species.Consider the reaction4NO2(g) + O2(g) 2N2O5(g)At a particular time during the reaction, nitrogen dioxide is being c

20、onsumed at the rate 0.00130 M/s. (a) At what rate is molecular oxygen being consumed? (b) At what rate is dinitrogen pentoxide being produced?Solutionrate = = =We are given = 0.00130 M/swhere the minus sign indicates that the concentration of NO2 is decreasing the time.NO2t14O2tN2O5t12NO2t Worked Ex

21、ample 14.2 (cont.)Solution The rate of reaction, therefore, israte = = (0.00130 M/s)= 3.2510-4 M/s(a) 3.2510-4 M/s = = 3.2510-4 M/sMolecular oxygen is being consumed at a rate of 3.2510-4 M/s.(b) 3.2510-4 M/s = 2(3.2510-4 M/s) = = 6.5010-4 M/sDinitrogen pentoxide is being produced at a rate of 6.501

22、0-4 M/s.NO2t1414O2tO2tN2O5t12N2O5tN2O5tThink About It Remember that the negative sign in a rate expression indicates that a species is being consumed rather than produced. Rates are always expressed as positive quantities.Dependence of Reaction Rate on Reactant ConcentrationThe rate law is an equati

23、on that relates the rate of reaction to the concentrations of reactants.aA + bB cC + dDk: the rate constantx: order with respect to Ay: order with respect to Bx + y: represents the overall reaction order.14.4rate = kAxBy determined experimentallyF2(g) + 2ClO2(g) 2FClO2(g)The initial rate is the rate

24、 at the beginning of the reaction.Dependence of Reaction Rate on Reactant ConcentrationExperimentF2(M)ClO2(M)Initial Rate (M/s)10.100.0101.2 x 10320.100.0404.8 x 10330.200.0102.4 x 103Initial Rate Data for the Reaction between F2 and ClO2F2(g) + 2ClO2(g) 2FClO2(g)Dependence of Reaction Rate on React

25、ant ConcentrationExperimentF2(M)ClO2(M)Initial Rate (M/s)10.100.0101.2 x 10320.100.0404.8 x 10330.200.0102.4 x 103Initial Rate Data for the Reaction between F2 and ClO2F2 doublesClO2 constantRate doubles2321F0.20 2F0.10 MM3331rate2.4 10 / s2rate1.2 10 / sMMThe reaction is first order in F2; x = 1rat

26、e = kF2ClO2y rate = kF2xClO2y F2(g) + 2ClO2(g) 2FClO2(g)Dependence of Reaction Rate on Reactant ConcentrationExperimentF2(M)ClO2(M)Initial Rate (M/s)10.100.0101.2 x 10320.100.0404.8 x 10330.200.0102.4 x 103Initial Rate Data for the Reaction between F2 and ClO2F2 constantClO2 x 4Rate x 42221ClO0.040

27、4ClO0.010 MM3231rate4.8 10 / s4rate1.2 10 / sMMThe reaction is first order in ClO2; y = 1rate = kF2ClO2y rate = kF2ClO2 aA + bB cC + dDDependence of Reaction Rate on Reactant ConcentrationExperimentA(M)B(M)Initial Rate (M/s)10.100.0152.1 x 10420.200.0154.2 x 10430.100.0308.4 x 104Initial Rate Data f

28、or the Reaction between A and Brate = kAxBy Dependence of Reaction Rate on Reactant ConcentrationExperimentA(M)B(M)Initial Rate (M/s)10.100.0152.1 x 10420.200.0154.2 x 10430.100.0308.4 x 104A x 2B constantRate x 2aA + bB cC + dDInitial Rate Data for the Reaction between A and B 12A0.10 0.5A0.20 MM41

29、42rate2.1 10 / s0.5rate4.2 10 / sMMThe reaction is first order in A; x = 1rate = kAxBy rate = kABy Dependence of Reaction Rate on Reactant ConcentrationExperimentA(M)B(M)Initial Rate (M/s)10.100.0152.1 x 10420.200.0154.2 x 10430.100.0308.4 x 104A constantB x 2Rate x 4aA + bB cC + dDInitial Rate Data

30、 for the Reaction between A and B 31B0.030 2B0.015 MM4341rate8.4 10 / s4rate2.1 10 / sMMThe reaction is second order in B; y = 2rate = kABy rate = kAB2 Dependence of Reaction Rate on Reactant ConcentrationThree important things to remember about the rate law: The exponents in a rate law must be dete

31、rmined from a table of experimental data. Comparing changes in individual reactant concentrations with changes in rate shows how the rate depends on each reactant concentration. Reaction order is always defined in terms of reactant concentrations, never product concentrations.The reaction of peroxyd

32、isulfate ion (S2O82) with iodide ion (I) is:S2O82(aq) + 3I(aq) 2SO42(aq) + I3(aq) Determine the rate law and calculate the rate constant, including its units.Dependence of Reaction Rate on Reactant ConcentrationExperimentS2O82(M)I(M)Initial Rate (M/s)10.0800.0342.2 x 10420.0800.0171.1 x 10430.160.01

33、72.2 x 104Initial Rate Data for the Reaction between S2O82 and ISolution:Step 1:In experiments 1 and 2, S2O82 is constant. The I is doubled, and rate doubles. Dependence of Reaction Rate on Reactant ConcentrationExperimentS2O82(M)I(M)Initial Rate (M/s)10.0800.0342.2 x 10420.0800.0171.1 x 10430.160.0

34、172.2 x 104Initial Rate Data for the Reaction between S2O82 and I 12I0.034 20.017 IMM4142rate2.2 10 / s2rate1.1 10 / sMMThe reaction is first order in ISolution:Step 2:In experiments 2 and 3, S2O82 is doubled, I is constant, and rate doubles. Dependence of Reaction Rate on Reactant ConcentrationExpe

35、rimentS2O82(M)I(M)Initial Rate (M/s)10.0800.0342.2 x 10420.0800.0171.1 x 10430.160.0172.2 x 104Initial Rate Data for the Reaction between S2O82 and I22832282S O0.16 20.080 S OMM4342rate2.2 10 / s2rate1.1 10 / sMMThe reaction is first order in S2O82Solution:Step 2:In experiments 2 and 3, S2O82 is dou

36、bled, I is constant, and rate doubles. Dependence of Reaction Rate on Reactant ConcentrationExperimentS2O82(M)I(M)Initial Rate (M/s)10.0800.0342.2 x 10420.0800.0171.1 x 10430.160.0172.2 x 104Initial Rate Data for the Reaction between S2O82 and IThe rate law is: rate = k S2O82 I Solution:Step 3:Use t

37、he data from any experiment to calculate k.Dependence of Reaction Rate on Reactant ConcentrationExperimentS2O82(M)I(M)Initial Rate (M/s)10.0800.0342.2 x 10420.0800.0171.1 x 10430.160.0172.2 x 104Initial Rate Data for the Reaction between S2O82 and I411322833rate2.2 10 / s0.081 s(0.16 )(0.017 )S OIMk

38、MMM Worked Example 14.3Strategy Compare two experiments at a time to determine how the rate depends on the concentration of each reactant.The rate law is rate = kNOxH2y.The gas-phase reaction of nitric oxide with hydrogen at 1280C is2NO(g) + 2H2(g) N2(g) +2H2O(g)From the following data collected at

39、1280C, determine (a) the rate law, (b) the rate constant, including units, and (c) the rate of the reaction when NO = 4.810-3 M and H2 = 6.210-3 M.ExperimentNO (M)H2 (M)Initial rate (M/s)15.010-32.010-31.310-521.010-22.010-35.010-531.010-24.010-31.010-4 Worked Example 14.3 (cont.)Solution The rate o

40、f reaction, therefore, is= 4 = Canceling identical terms in the numerator and denominator givesTherefore, x = 2. The reaction is second order in NO.Dividing the rate from experiment 3 by the rate from experiment 2, we get= = 2 = Canceling identical terms in the numerator and denominator givesTherefo

41、re, y = 1. The reaction is first order in H2. The overall rate law israte = kNO2H25.010-5 M/s1.310-5 M/srate2rate1k(1.010-2 M)x(2.010-3 M)yk(5.010-3 M)x(2.010-3 M)y(1.010-2 M)x(5.010-3 M)x= 2x = 41.010-4 M/s5.010-5 M/srate3rate2k(1.010-2 M)x(4.010-3 M)yk(1.010-2 M)x(2.010-3 M)y(4.010-3 M)y(2.010-3 M

42、)y= 2y = 2 Worked Example 14.3 (cont.)Solution We can use data from any of the experiments to calculate the value and units of k. Using the data from experiment 1 givesk = =(c) Using the rate constant determined in part (b) and the concentrations of NO and H2 given in the problem statement, we can d

43、etermine the reaction rate as follows:rate = (2.6102 M-2s-1)(4.810-3 M)2(6.210-3 M)= 3.710-5 Ms-11.310-5 M/s(5.010-3 M)2(2.010-3 M)rateNO2H2= 2.6102 M-2s-1Think About It The exponent for the concentration of H2 in the rate law is 1, whereas the coefficient for H2 in the balanced equation is 2. It is

44、 a common error to try to write a rate law using the stoichiometric coefficients as the exponents. Remember that, in general, the exponents in the rate law are not related to the coefficients in the balanced equation. Rate laws must be determined by examining a table of experimental data.Dependence

45、of Reactant Concentration on TimeThe rate law can be used to determine the rate of a reaction using the rate constant and the reactant concentrations:A rate law can also be used to determine the concentration of a reactant at a specific time during a reaction.14.5rate = kAxByraterate constantrate la

46、wDependence of Reactant Concentration on TimeA first-order reaction is a reaction whose rate depends on the concentration of one of the reactants raised to the first power.C2H6 2 CH3rate = kC2H62N2O5(g) 2NO2(g) + O2(g)rate = kN2O5Dependence of Reactant Concentration on TimeIn a first-order reaction

47、of the typeA productsThe rate can be expressed as the rate of change in reactant concentration,as well as in the form of the rate law:rate = kASetting the two expressions equal to each other yields: Arate = t AAktDependence of Reactant Concentration on TimeUsing calculus, it is possible to show that

48、:ln is the natural logarithmA0 and At refer to the concentration of A at times 0 and tThe equation above is sometimes called the integrated rate law for a first order reaction. 0AlnAtkt Dependence of Reactant Concentration on TimeThe rate constant for the reaction 2A B is 7.5 x 103 s1 at 110C. The r

49、eaction is first order in A. How long (in seconds) will it take for A to decrease from 1.25 M to 0.71 M?Solution:Step 1:Use the equation below to calculate time in seconds.t = 75 seconds 3100.71ln7.510s1.25tt 0AlnAtkt Worked Example 14.4Strategy Use ln (At/A0) = kt to find H2O2t where t = 3 h, and t

50、hen solve for t to determine how much time must pass for H2O2t to equal 0.10 M. H2O20 = 0.75 M; time t for part (a) is (3 h)(60 min/h)(60 s/min) = 10,800 s.The position of hydrogen peroxide is first order in H2O2.2H2O2(aq) 2H2O(l) + O2(g)The rate constant for this reaction at 20C is 1.810-5 s-1. If

51、the start concentration of H2O2 is 0.75 M, determine (a) the concentration of H2O2 remaining after 3 h and (b) how long it will take for the H2O2 concentration to drop to 0.10 M.Solution (a)ln(b)ln H2O2tH2O20= ktH2O2t0.75 M= (1.810-5 s-1)(10,800 s) = 0.1944 Worked Example 14.4 (cont.)Solution Take t

52、he inverse natural logarithm of both sides of the equation to getH2O2t = (0.823)(0.75 M) = 0.62 MThe concentration of H2O2 after 3 h is 0.62 M.(b) lnThe time required for the peroxide concentration to drop to 0.10 M is 1.1105 s or about 31 h.H2O2t0.75 M= e0.1944 = 0.8230.10 M0.75 M= 2.015 = (1.810-5

53、 s-1)t2.0151.810-5 s-1= t = 1.12105 sThink About It Dont forget the minus sign. If you calculate a concentration at time t that is greater than the concentration at time 0 (or if you get a negative time required for the concentration to drop to a specified level), check your solution for this common

54、 error.Dependence of Reactant Concentration on TimeRearrangement of the first-order integrated rate law gives:Rearrangement in this way has the formof the linear equation y = mx + b. 0AlnAtkt lnAt = kt + lnA0lnAt = kt + lnA0Slope = k Intercept = lnA0Dependence of Reactant Concentration on TimeThe ra

55、te of position of azomethane is studied by monitoring the partial pressure of the reactant as a function of time.CH3N=NCH3(g) N2(g) + C2H6(g)The data obtained at 300C are listed in the following table:Time (s)Pazomethane (mmHg)0284100220150193200170250150300132Dependence of Reactant Concentration on

56、 TimePlotting the data gives a straight line, indicating the reaction is first order.lnAt = kt + lnA0Slope = 2.55 x 103 s1 Intercept = 5.654.804.905.005.105.205.305.405.505.605.700100200300400ln PTime (s)Dependence of Reactant Concentration on TimeEthyl iodide (C2H5I) poses at a certain temperature

57、in the gas phase as follows:C2H5I(g) C2H4(g) + HI(g)Determine the rate of the reaction, after verifying that the reaction is first order.Time (s)C2H5I (M)00.36150.30300.35480.19750.13Dependence of Reactant Concentration on TimeSolution:Plot lnC2H5I vs time. If a straight line results, the reaction i

58、s first order. The slope is equal to k.Slope = 1.3 x 102 s1; k = 1.3 x 102 s1Time (s)C2H5I (M)lnC2H5I00.36-1.02150.30-1.20300.35-1.39480.19-1.66750.13-2.04-2.50-2.00-1.50-1.00-0.500.00020406080ln C2H5ITime (s) Worked Example 14.5Strategy We can use ln (At/A0) = kt only for first-order reactions, so

59、we must first determine if the position of azomethane is first order. We do this by plotting ln P against time. If the reaction is first order, we can use ln (At/A0) = kt and the data at any two of the times in the table to determine the rate constant.The rate of position of azomethane is studied by

60、 monitoring the partial pressure of the reactant as a function of time:CH3N=NCH3(g) N2(g) + C2H6(g)The data obtained at 20C are listed in the following table:Time (s)Pazomethane (mmHg)0284100220150193200170250150300132 Worked Example 14.5 (cont.)Solution The table expressed in ln P isPlotting these dat