厦大课件物化reaction-kinetics

1、1Reaction Kinetics (3) Xuan ChengXiamen UniversityPhysical Chemistry2Determination of the Rate LawPhysical ChemistryThe rate law(17.48)1. Half-life method(17.49)(17.29)For n 1(17.49)Reaction Kinetics3半衰期法确定反应级数用半衰期法求除一级反应以外的其它反应的级数。以lnt1/2lnAo作图从直线斜率求n值。从多个实验数据用作图法求出的n值更加准确。根据 n 级反应的半衰期通式： 取两个不同起始浓度

2、Ao，Ao作实验，分别测定半衰期为t1/2和 ，因为同一反应，常数相同，所以：Physical ChemistryReaction Kinetics4Physical ChemistryDetermination of the Rate Law2. Powell-plot method(17.50)the fraction of A unreactedFor n 1(17.28)For n 1(17.13)For n = 1For n = 1(17.51)Reaction Kinetics5Physical ChemistryDetermination of the Rate LawFor n

3、 1For n = 1(17.51)For a given n, there is a fixed relation between and for every reaction of order n.Plot versus log10 for commonly occurring values of n to give a series of master curves. (Fig. 17.6) The Powell-plot method requires the initial investment of time needed to make the master plots.Tabl

4、e 17.1 gives the data needed to make the master plots.Reaction Kinetics6Physical ChemistryDetermination of the Rate Law3. Initial-rate methodThe rate law(17.48)The ratio of initial rates for runs 1 and 2Measure r0 for two different initial concentrations A0,1 and A0,2 while keeping B0, C0, fixed. ca

5、n be foundThe orders , can be found similarlyReaction Kinetics7Physical ChemistryDetermination of the Rate Law4. Isolation methodThe rate law(17.48)Make initial concentrations of reactant A much less than the concentrations of all other species B0 A0, C0 A0, The rate law es(17.52)Where j is essentia

6、lly constant.The reaction has the pseudo-order .The orders , can be found similarly.Reaction Kinetics8孤立法确定反应级数 孤立法类似于准级数法，它不能用来确定反应级数，而只能使问题简化，然后用前面三种方法来确定反应级数。1.使AB先确定值2.使BA再确定值Physical ChemistryReaction Kinetics9Physical ChemistryRate Laws and Equilibrium Constants for Elementary ReactionsShow th

7、at for a reaction that takes place in a sequence of steps, the overall equilibrium constant is a product of ratios of the rate constants for each step.It is sufficient to consider a reasonably general but simple two-step reaction sequence, such as(second-order in each direction, k1, k-1)(first-order

8、 forwarded, second-order reverse, k2, k-2)(overall)Reaction Kinetics10Physical ChemistryRate Laws and Equilibrium Constants for Elementary ReactionsAt equilibrium, all the reaction are individually at equilibrium, and setting the net rates each equal to zero givesThe equilibrium constant of the over

9、all reaction is thereforeelementary reaction(17.53)*Reaction Kinetics11Physical ChemistryReaction MechanismsThe Rate-Determining-Step ApproximationThe reaction mechanism is assumed to consist of one or more reversible reactions that stay close to equilibrium during most of the reaction, followed by

10、a relatively slow rate-determining step, which in turn is followed by one or more rapid reactions.unimolecularThe number of molecules that react in an elementary stepThe molecularity of the elementary reactionbimoleculartrimolecular (termolecular)Reaction Kinetics12Physical ChemistryReaction Mechani

11、smsSuppose now that Then whenever a B molecule is formed it decays rapidly into C.(17.41)reduces toThe formation of C depends on only the smaller of the two rate constantsis called the rate-determining step of the reaction. For the consecutive unimolecular reactionsReaction Kinetics13Physical Chemis

12、tryThe Steady-State ApproximationAssumes that during the major part of the reaction, the rates of change of concentrations of all reaction intermediates are negligibly smallReaction MechanismsReactants Products Intermediates Time Concentration (17.35)Reaction Kinetics14Physical ChemistryReaction Mec

13、hanismsC is formed by a first-order decay of A, with a rate constant k1, the rate constant of the slower, rate-determining step.(17.38)The same result as before, but obtained much more quickly. Reaction Kinetics15Physical Chemistry(the rate-determining step)Reaction MechanismsConsider the following

14、mechanism composed of unimolecular reactionsis slower than remains close to equilibriumis not in equilibriumD is rapidly formed from CReaction Kinetics16Physical ChemistryReaction MechanismsExample 17.4is observed to be A proposed mechanism is(17.55)The rate law for the Br-catalyzed aqueous reaction

15、rapid equilib.(17.56)slowfastReaction Kinetics17Physical ChemistryReaction MechanismsExample 17.4(17.55)Deduce the rate law for this mechanism and relate the observed rate constant k in (17.55) to the rate constants in the assumed mechanism (17.56)rapid equilib.(17.56)slowfastthe rate-determining st

16、ep(1)(2)(3)The formation of ONBr in (2)(17.57)Step (1) is near equilibrium. Equation (17.53) givesReaction Kinetics18Physical ChemistryReaction MechanismsExample 17.4elementary reaction(17.53)*(17.57)(17.55)Example 17.5Reaction Kinetics19Physical ChemistryReaction MechanismsMore examples in using th

17、e steady-state approximationon the basis of the following mechanism:Account for the rate law for the position of N2O5First identify the intermediatesNO and NO3Reaction Kinetics20Physical ChemistryReaction MechanismsReaction Kinetics21Physical ChemistryReaction MechanismsAccording to the steady-state

18、 approximation, set both rates equal to zeroReaction Kinetics22Physical ChemistryReaction MechanismsThe net rate of change of concentration of N2O5 isReaction Kinetics23Physical ChemistryReaction MechanismsbecauseIt follows that the reaction rate iswhereReaction Kinetics24Physical ChemistryReaction

19、MechanismsPre-equilibriaFrom a simple sequence of consecutive reactions we now turn to a slightly more complicated mechanism:Where C denote the intermediate.This scheme involves a pre-equilibrium, in which an intermediates is in equilibrium with the reactants. A pre-equilibrium arises when the rates

20、 of formation of the intermediate and its decay back into reactants are much faster than its rate of formation of products; thus, the condition is possible when kakb but not when kb ka. Because we assume that A, B, and C are in equilibrium. Reaction Kinetics25Physical ChemistryReaction MechanismsPre

21、-equilibriaWe can write:In writing these equations, we are presuming that the rate of reaction of C to form P is too slow to affect the maintenance of the pre-equilibrium (see the following example). The rate of formation of P may now be written: This rate law has the form of a second-order rate law

22、 with a composite rate constant:whereReaction Kinetics26Physical ChemistryReaction MechanismsPre-equilibriaExample: Analyzing a pre-equilibriumRepeat the pre-equilibrium calculation but without ignoring the fact that C is slowly leaking away as it forms P. The net rates of change of P and C arewhere

23、Reaction Kinetics27Physical ChemistryReaction MechanismsPre-equilibriawherewhereWhen the rate constant for the decay of C into products is much smaller than that for its decay into reactantsReaction Kinetics28HomeworkPhysical ChemistryPage 592Prob. 17.28Prob. 17.29Prob. 17.33Page 593Prob. 17.39Prob. 17