1、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 N
2、O2 is proposed to beStep 1 NO2+NO2 NO3+NO(an elementary reaction)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 explanat
3、ion 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 o
4、f elementary reaction is derived from the equation.The order 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
5、 termolecular reaction(step).O3 O2+Orate=k O3NO2+NO2 NO3+NOrate=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 Molec
6、ular EventsN2O5 NO2+NO3 NO+O2+NO2 NO2+NO3A mechanism is a 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 reacti
7、on,2 NO2(g)+F2(g)2 NO2F(g),the rate law is,rate=k NO2 F2.Can 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 determi
8、ning step is the slowest elementary step in a mechanism,and 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
9、is the rate determining step?Answer:The rate for step i is 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,iH2O
10、2+I k1 H2O+IO slow ii H2O2+IO k2 H2O+O2+I fastWhat is the 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
11、Br2 2 Brfast equilibrium(k1,k-1)iiH2+Br k2 HBr+H slow iii 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 1.5expl
12、ain快速平衡假设法！例3、The decomposition of N2O5 follows the mechanism: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 dif
13、fer 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
14、IStep(1)2 I k-1 I2Step(2)H2 +2 I k2 2 HIDerive the rate law.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
15、stateFrom the previous result:k1 k2 H2 I2rate=k-1+k2 H2 Discussion:(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
