无机课-化学反应速率-2014-考研试题文档资料系列.ppt

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1、化学反应速率Chemical Kinetics2第二章How reactions proceed?What determines their rates?How to control those rates?有的化学反应，从热力学的角度考虑，进行的趋势很大，但因其反应速率太小，事实上几乎不可能发生。有的化学反应，自由能降低不大，但其正逆两个方向的反应速率都很大。Rate may be expressed in three main ways:1.Average reaction rate:a measure of the change in concentration with time2.I

2、nstantaneous rate:rate of change of concentration at any particular instant during thereaction3.Initial rate:instantaneous rate at t=0-that is,when the reactants are first mixed第第1节节 反应速率的定义反应速率的定义一、平均反应速率(Average reaction rate)定义：单位时间内反应物浓度的减少或生成物浓度的增加表示。单位：moldm-3s-1，moldm-3min-1或或 moldm-3h-1。the

3、surroundings 例1、乙酸乙酯的皂化反应 CH3COOC2H5 OH CH3COO CH3CH2OHReaction rate:changes in a concentration of a product or a reactant per unit time.concentrationReaction rate=t t tchangeReaction rate:changes in a concentration of a product or a reactant per unit time.concentrationReaction rate=tchangeDefine re

4、action rate and explainAverage reaction rateInstantaneous reaction rate(2 tangents shown)Initial reaction rateConsider the hypothetical reactionaA+bB cC+dD对于一般的化学反应 aA bB gG hH原则上，用任何一种反应物或生成物的浓度变化均可表示化学反应速率，但我们经常采用其浓度变化易于测量的那种物质来进行研究。二、瞬时反应速率(Instantaneous reaction rate)瞬时速率：某一时刻的化学反应速率称为瞬时速率。AB的斜率

5、表示时间间隔 t tBtA 内反应的平均速率。例3、利用表中反应物的浓度对时间作图。dDetermination of the rate of deterioration of penicillin during storage at two different times.Note that the rate(the slope of the tangent to the curve)at 5 weeks is greater than the rate at 10 weeks,when less penicillin is present.四、如何测得反应速率(Determine Reac

6、tion Rates)To measure reaction rate,we measure the concentration of either a reactant or product at several time intervals.The concentrations are measured using spectroscopic method or pressure(for a gas).For example,the total pressure increases for the reaction:2 N2O5(g)4 NO2(g)+O2(g)Because 5 mole

7、s of gas products are produced from 2 moles of gas reactants.For the reactionCaCO3(s)CaO(s)+CO2(g)The increase in gas pressure is entirely due to CO2 formed.barometer三、初始反应速率(Initial rate)Initial rate:instantaneous rate at t=0-that is,when the reactants are first mixed。The orange curves show how the

8、 concentration of N2O5 changes with time for five different initial concentrations.The initial rate of consumption of N2O5 can be determined by drawing a tangent(black line)to each curve at the start of the reaction.This graph was obtained by plotting the five initial rates against the initial conce

9、ntration of N2O5.The initial rate is directly proportional to the initial concentration.This graph also illustrates how we can determine the value of the rate constant k by calculating the slope of the straight line from two points.Initial rate of consumption of N2O5=k N2O5initial第第2节节 反应速率与反应物浓度的关系

10、反应速率与反应物浓度的关系Initial rate of consumption of N2O5=k N2O5initialThe constant k is called the rate constant for the reaction，速率常数。，速率常数。At any stage of the reaction,and provided products do not participate in the reaction,Rate of consumption of N2O5=k N2O5我们把用来表达反应速率与反应物浓度关系的方程式就叫做反应速率定律或质量作用定律（Rate La

11、w）。不同的化学反应，其速率表达式不同。(a)When the rates of disappearance of NO2 are plotted against its concentration,a straight line is not obtained.(b)However,a straight line is obtained when the rates are plotted against the square of the concentration,indicating that the rate is directly proportional to the squar

12、e of theconcentration.Differential Rate LawsDependence of reaction rate on the concentrations of reactants is called the rate law,which is unique for each reaction.For a general reaction,a A+b B+c C productsthe rate law has the general formorder wrt A,B,and C,determined experimentally reaction rate

13、=k AX BY CZ For example,the rate law israte=k Br-BrO3-H+for 5 Br-+BrO3-+6 H+3Br2+3 H2OThe reaction is 1st order wrt all three reactants,total order 3.Use differentials to express rates基元反应：指反应物分子一步直接生成产物的反应。质量作用定律：基元反应的速率与反应物浓度以其化学计量数为幂指数的连乘积成正比。对于基元反应 a A b B g G h H质量作用定律的数学表达式：r k c(A)m c(B)nk 称为

14、速率常数m，n 称反应物A，B的反应级数k，m和n 均可由实验测得Estimate the orders and rate constant k from the results observed for the reaction?What is the rate when H2O2=I-=H+=1.0 M?H2O2+3 I-+2 H+I3-+2 H2OExprmtH2O2I-H+Initial rate M s-110.0100.0100.00501.15e-6 20.0200.0100.00502.30e-6 30.0100.0200.00502.30e-640.0100.0100.010

15、01.15e-6Learn the strategy to determine the rate law from this example.Figure out the answer without writing down anything.例子：Estimate the orders from the results observed for the reaction H2O2+3 I-+2 H+I3-+2 H2OExprmtH2O2I-H+Initial rate M s-110.0100.0100.00501.15e-6 20.0200.0100.00502.30e-6 1 for

16、H2O2 30.0100.0200.00502.30e-6 1 for I-40.0100.0100.01001.15e-6 0 for H+1.15e-6=k H2O2x I-y H+z 1.15e-6 k(0.010)x(0.010)y(0.0050)z exprmt 111-=-=-2.30e-6 k(0.020)x(0.010)y(0.0050)z exprmt 22 2 x=1Other orders are determined in a similar way as shown before.Now,lets find k and the rateThurs,rate=1.15e

17、-6=k(0.010)(0.010)from exprmt 1 k=1.15e-6 M s-1/(0.010)(0.010)M3=0.0115 M-1 s-1And the rate law is therefore,d H2O2 k rate=0.0115 H2O2 I-a differential rate lawd ttotal order 2The rate when H2O2=I-=H+=1.0 M:The rate is the same as the rate constant k,when concentrations of reactants are all unity(ex

18、actly 1),doesnt matter what the orders are.当反应物的浓度都为1.0M时，其反应速率的值和反应级数无关！The reaction rate dH2O2/dt=0.0115 H2O2 I,forH2O2+3 I-+2 H+I3-+2 H2OWhat is dI/dt when H2O2=I=0.5?Solution:Please note the stoichiometry of equation and how the rate changes.dI/dt=3 dH2O2/dt=3*0.0115 H2O2 I=0.0345*0.5*0.5=0.0086

19、 M s-1In order to get a unique rate constant k,we evaluate k for the reaction a A+b B product this wayrate=-1/a dA/dt=-1/b dB/dt=k Ax By第第3节节 反应物浓度与时间的关系反应物浓度与时间的关系一、一、0 0级反应级反应(a)The concentration of the reactant in a zero-order reaction falls at a constant rate until the reactant is exhausted.(b)T

20、he rate of a zero-order reaction is independent of the concentration of the reactant and remains constant until all the reactant has been consumed,when the rate falls abruptly to zero.速率常数的单位与反应级数有关：一级反应 s1 二级反应 dm3mol1s1 n级反应 dm3(n1)mol1(n1)s1化学反应速率常数k是在给定温度下，各反应物浓度皆为 1 moldm-3时的反应速率，因此也称比速率常数。速率常数

21、是温度的函数。One reactant A decomposes in 1st or 2nd order rate law.Differential rate lawIntegrated rate law dA/dt=kA=Ao k t dA =k AA=Ao e k t or ln A=ln Ao k t d t dA 1 1 A conc at t =k A2 =k t d t A Ao Ao conc at t=0二、二、1 1级反应级反应AtlnAtA=Ao e k tln A=ln Ao k tt dA =k AA=Ao e k t or ln A=ln Ao k t d tHalf

22、 life&k of First Order Decomposition半衰期和速率常数The time required for half of A to decompose is called half life t1/2.SinceA=Ao e k t or ln A=ln Ao k tWhen t=t1/2,A=AoThus ln Ao=ln Ao k t1/2 ln 2=k t1/2k t1/2=ln 2=0.693 relationship between k and t1/2Radioactive decay usually follow 1st order kinetics,a

23、nd half life of an isotope is used to indicate its stability.Evaluate t from k or k from t dA 1 1 A conc at t =k A2 =k t d t A Ao Ao conc at t=0三、三、2 2级反应级反应Dimerization of butadiene is second order:2 C4H6(g)=C8H12(g).The rate constant k at some temperature is 0.100/min.The initial concentration of

24、butadiene B is 2.0 M.Calculate the time required for B=1.0 and 0.5 MCalculate concentration of butadiene when t=1,5,10,and 30.例子：例子：A 2nd Order ExampleDimerization of butadiene is second order:2 C4H6(g)=C8H12(g).The rate constant k at some temperature is 0.100/min.The initial concentration of butadi

25、ene B is 2.0 M.Calculate the time t required for B=1.0 and 0.5 MCalculate concentration of butadiene when t=1,5,10,and 30.1 1 =k tBBo1 1 BBo t=kBoB=Bo k t+1t=1510153035B=1.67 1.0 0.670.500.290.25Work out the formulas and then evaluate values 第第 4 节反应机理节反应机理所谓基元反应是指反应物分子一步直接转化为产物的反应。如：NO2 CO NO CO2 反

26、应物NO2 分子和CO分子经过一次碰撞就转变成为产物NO分子和CO2。基元反应是动力学研究中的最简单的反应，反应过程中没有任何中间产物。一、基本概念Elementary reactions are steps of molecular events showing how reactions proceed.This type of description is a mechanism.The mechanism for the reaction between CO and NO2 is proposed to beStep 1 NO2+NO2 NO3+NO(an elementary re

27、action)Step 2 NO3+CO NO2+CO2(an elementary reaction)Add these two equations led to the overall reactionNO2+CO=NO+CO2(overall reaction)A mechanism is a proposal to explain the rate law,and it has to satisfy the rate law.A satisfactory explanation is not a proof.例如：H2(g)I2(g)2 HI(g)实验上或理论上都证明，它并不是一步完成

28、的基元反应，它的反应历程可能是如下两步基元反应：I2 I I （快）H2 2 I 2 HI （慢）化学反应的速率由反应速率慢的基元反应决定。基元反应或复杂反应的基元步骤中发生反应所需要的微粒（分子、原子、离子）的数目一般称为反应的分子数。分子数Molecularity of Elementary ReactionsThe total order of rate law in an elementary reaction is molecularity.The rate law of elementary reaction is derived from the equation.The ord

29、er is the number of reacting molecules because they must collide to react.A molecule decomposes by itself is a unimolecular reaction(step);two molecules collide and react is a bimolecular reaction(step);&three molecules collide and react is a termolecular reaction(step).O3 O2+Orate=k O3NO2+NO2 NO3+N

30、Orate=k NO22Br+Br+Ar Br2+Ar*rate=k Br2ArCaution:Derive rate laws this way only for elementary reactions.单分子反应SO2Cl2 的分解反应 SO2Cl2 SO2 Cl2双分子反应NO2 的分解反应 2 NO2 2 NO O2三分子反应 HI 的生成反应 H2 2 I 2 HI 四分子或更多分子碰撞而发生的反应尚未发现。Elementary Reactions are Molecular EventsN2O5 NO2+NO3 NO+O2+NO2 NO2+NO3A mechanism is a

31、collection of elementary steps devise to explain the the reaction in view of the observed rate law.You need the skill to derive a rate law from a mechanism,but proposing a mechanism is task after you have learned more chemistry For the reaction,2 NO2(g)+F2(g)2 NO2F(g),the rate law is,rate=k NO2 F2.C

32、an the elementary reaction be the same as the overall reaction?If they were the same the rate law would have been rate=k NO22 F2,Therefore,they the overall reaction is not an elementary reaction.Its mechanism is proposed next.The rate determining step is the slowest elementary step in a mechanism,an

33、d the rate law for this step is the rate law for the overall reaction.The(determined)rate law is,rate=k NO2 F2,for the reaction,2 NO2(g)+F2(g)2 NO2F(g),and a two-step mechanism is proposed:i NO2(g)+F2(g)NO2F(g)+F(g)ii NO2(g)+F(g)NO2F(g)Which is the rate determining step?Answer:The rate for step i is

34、 rate=k NO2 F2,which is the rate law,this suggests that step i is the rate-determining or the s-l-o-w step.反应机理中的慢反应步骤决定总反应的速率！反应机理中的慢反应步骤决定总反应的速率！二、如何由给出的反应机理推导出速率方程例1、The decomposition of H2O2 in the presence of I follow this mechanism,iH2O2+I k1 H2O+IO slow ii H2O2+IO k2 H2O+O2+I fastWhat is the

35、rate law?SolutionThe slow step determines the rate,and the rate law is:rate=k1 H2O2 I Since both H2O2 and I are measurable in the system,this is the rate law.例例2、Derive the rate law for the reaction,H2+Br2=2 HBr,from the proposed mechanism:i Br2 2 Brfast equilibrium(k1,k-1)iiH2+Br k2 HBr+H slow iii

36、H+Br k3 HBr fastSolution:The fast equilibrium condition simply says thatk1 Br2=k-1 Br2andBr=(k1/k-1 Br2)The slow step determines the rate law,rate=k2 H2 Br Br is an intermediate =k2 H2(k1/k-1 Br2)=k H2 Br2;k=k2(k1/k-1)M-s-1total order 1.5explain快速平衡假设法！例3、The decomposition of N2O5 follows the mechan

37、ism:1N2O5 NO2+NO3fast equilibrium 2NO2+NO3 k2 NO+O2+NO2slow3NO3+NO k3 NO2+NO2fastDerive the rate law.Solution:The slow step determines the rate,rate=k2 NO2 NO3 NO2&NO3 are intermediateFrom 1,we have NO2 NO3 =KK,equilibrium constant N2O5 K differ from kThus,rate=K k2 N2O5稳态近似法！以假设中间产物的浓度恒定不变为基础！即、中间产

38、物的生成速率与其消耗速率相等。Rate of producing the intermediate,Rprod,is the same as its rate of consumption,Rcons.Rprod=RconsIntermediatetimeRprod RconsBe able to apply the steady-state approximation to derive rate laws假设 H2+I2 2 HI的反应机理如下：Step(1)I2 k1 2 IStep(1)2 I k-1 I2Step(2)H2 +2 I k2 2 HIDerive the rate la

39、w.Derivation:rate=k2 H2 I 2(cause this step gives products！)but I is an intermediate,this is not a rate law yet.Since k1 I2(=rate of producing I)=k-1 I2+k2 H2 I2(=rate of consuming I)Thus,k1 I2 I2=k-1+k2 H2 rate=k1 k2 H2 I2 /k-1+k2 H2 Steady stateFrom the previous result:k1 k2 H2 I2rate=k-1+k2 H2 Di

40、scussion:(i)If k-1 k2 H2 then k-1+k2 H2=k2 H2,then rate=k1 k2 H2 I2 /k2 H2 =k1 I2(pseudo 1st order wrt I2)using large concentration of H2 or step 2 is fast(will meet this condition).(ii)If step(2)is slow,then k2 k1,and if H2 is not large,we have k-1+k2 H2=k-1 and rate=k1 k2 H2 I2 /k1=k2 H2 I2第第5 5节反

41、应速率理论简介节反应速率理论简介 20世纪，反应速率理论的研究取得了进展；1918年路易斯（Lewis）在气体分子运动论的基础上提出的化学反应速率的碰撞理论；30年代艾林（Eyring）等在量子力学和统计力学的基础上提出的化学反应速率的过渡状态理论。一、碰撞理论碰撞理论认为：反应物分子间的相互碰撞是反应进行的先决条件。反应物分子能量高；碰撞频率越大；碰撞方向有利；有效碰撞次数多，反应速率越大。即即 Z*Z f P f 为能量因子为能量因子；P为取向因子；为取向因子；P 取值在取值在1109之间。之间。Ea 称活化能，一般的化学反应称活化能，一般的化学反应 Ea 为每摩尔几十到几百为每摩尔几十到

42、几百 千焦千焦。Not all collisions leads to a reaction For effective collisions proper orientation ofthe molecules must be possible二、过渡状态理论过渡状态理论认为：当两个具有足够能量的反应物分子相互接近时，分子中的化学键要发生重排，即反应物分子先形成活化配合物的中间过渡状态，活化配合物能量很高，不稳定，它将分解部分形成反应产物。该理论认为，活化配合物的浓度；活化配合物分解成产物的概率；活化配合物分解成产物的速率均将影响化学反应的速率。例如例如 反应反应 NO2 CO NO CO2

43、 正反应活化能正反应活化能Ea 活化配合物的势能活化配合物的势能 反应物平均势能反应物平均势能逆反应活化能逆反应活化能Ea 活化配合物的势能活化配合物的势能 产物平均势能产物平均势能反应的热效应反应的热效应 rHm Ea Ea aaO结论：结论：若正反应是放热反应，其逆反应必定吸热。不论是若正反应是放热反应，其逆反应必定吸热。不论是放热还是吸热反应，反应物必须先爬过一个能垒反应才能放热还是吸热反应，反应物必须先爬过一个能垒反应才能进行。进行。如果正反应是经过一步即可完成的反应，则其逆反如果正反应是经过一步即可完成的反应，则其逆反应也可经过一步完成，而且正逆两个反应经过同一个活化应也可经过一步完

44、成，而且正逆两个反应经过同一个活化配合物中间体。这就是微观可逆性原理。配合物中间体。这就是微观可逆性原理。化学反应的热效应化学反应的热效应 rHm Ea Ea 当当 Ea Ea 时，时，rHm 0 反应吸热；反应吸热；当当 Ea Ea 时，时，rHm 0 反应放热。反应放热。三、温度对化学反应速率的影响过渡状态理论认为：在反应过程中反应物必须爬过一个能过渡状态理论认为：在反应过程中反应物必须爬过一个能垒才能进行。升高温度，反应物分子的平均能量提高，减垒才能进行。升高温度，反应物分子的平均能量提高，减小了活化能的值，反应速率加快。小了活化能的值，反应速率加快。碰撞理论认为：温度高时分子运动速率增

45、大，活化分子的碰撞理论认为：温度高时分子运动速率增大，活化分子的百分数增加，有效碰撞的百分数增加，反应速率增大。百分数增加，有效碰撞的百分数增加，反应速率增大。1899年年Arrhenius总结了大量实验事实，归纳出反应总结了大量实验事实，归纳出反应速率常数和温度的定量关系速率常数和温度的定量关系式中式中k为反应速率常数，为反应速率常数，Ea为活化能，为活化能，A为指前因子。为指前因子。通常温度升高，化学反应速率常数增加。通常温度升高，化学反应速率常数增加。七、催化剂与催化反应简介 H2(g)1/2 O2(g)H2O(l)rGm 237.1 kJmol1 H2(g)O2(g)H2O2(l)rG

46、m120.4 kJmol13/2H2(g)1/2 N2(g)NH3(g)rGm16.4 kJmol1在热力学上看，均为常温常压下可以自发进行的反应。但是由于反应速率过慢，在通常的条件下不能得到人们所希望的反应 释放的能量和反应、的产物。催化剂的作用机理改变反应历程催化剂的作用机理改变反应历程aaaacccO多相催化在化工生产和科学实验中最为常见。催化剂通常是固体，不论反应物为气体还是液体，反应都是在催化剂表面进行的。反应的主要步骤是：反应物分子扩散到固体催化剂表面并被吸附；被吸附在催化剂表面的反应物分子发生反应，生成产物；产物分子从固体催化剂表面脱附并扩散离开催化剂。化工生产中的合成氨反应，就

47、是由被吸附在铁催化剂表面上的N2 和 H2完成的，生成的NH3 分子解吸后扩散离开铁催化剂表面。The reaction between ethene,CH2=CH2,and hydrogen on a catalytic metal surface.In this sequence of images,we see the ethene molecule approaching the metal surface to which hydrogen molecules have already adsorbed:when they adsorb,they dissociate and sti

48、ck to the surface as hydrogen atoms.Next,after the ethene molecule also sticks to the surface,it meets a hydrogen atom and forms a bond.At this stage,the CH2CH3 radical is attached to the surface byone of its carbon atoms until it meets another hydrogen atom;then ethane is formed and escapes from th

49、e surface.酶催化反应广泛地存在于生物体内，蛋白质、脂肪和碳水化合物的合成及分解均属此类反应。酶是生物体内反应的催化剂，酶本身就是一种蛋白质，粒子的直径约为 10 100 nm。酶催化反应介于均相催化和多项催化反应之间，既可以看成是底物（酶催化反应中称反应物为底物）与酶形成中间产物，又可以看成是酶的表面上吸附了底物，然后底物在酶表面上完成反应。In the lock-and-key model of enzyme action,the correct substrate is recognized by its ability to fit into the active site l

50、ike a key into a lock.In a refinement of this model,the enzyme changes its shape slightly as the key enters.(a)An enzyme poison(represented by the blue sphere)can act by attaching so strongly to the active site that it blocks the site,thereby taking the enzyme out of action.(b)Alternatively,the pois