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单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,第,4,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：,NO,2,CO,NO,CO,2,反应物,NO,2,分子和,CO,分子经过一次碰撞就转变成为产物,NO,分子和,CO,2,。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,第 4 节反应机理所谓基元反应是指反应物分子一步直接转化,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 NO,2,is proposed to be,Step 1NO,2,+NO,2,NO,3,+NO(an,elementary,reaction),Step 2NO,3,+CO NO,2,+CO,2,(an,elementary,reaction),Add these two equations led to the overall reaction,NO,2,+CO=NO+CO,2,(,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,.,Elementary reactions are steps,例如：,H,2,(g),I,2,(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：,I,2,I,I,（快）,H,2,2 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。,例如：H2(g)I2(g)2,分子数,Molecularity of Elementary Reactions,The total order of rate law in an elementary reaction is,molecularity,.,The rate law of 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);t,wo molecules collide and react is a,bimolecular,reaction(step);&three molecules collide and react is a,termolecular,reaction(step).,O,3,O,2,+O,rate,=,k,O,3,NO,2,+NO,2,NO,3,+NO,rate,=,k,NO,2,2,Br+Br+Ar Br,2,+Ar*,rate,=,k,Br,2,Ar,Caution,:Derive rate laws this way,only,for,elementary,reactions.,分子数Molecularity of Elementary,单分子反应,SO,2,Cl,2,的分解反应,SO,2,Cl,2,SO,2,Cl,2,双分子反应,NO,2,的分解反应,2 NO,2,2 NO,O,2,三分子反应,HI,的生成反应,H,2,2 I,2 HI,四分子或更多分子碰撞而发生的反应尚未发现。,单分子反应双分子反应三分子反应,Elementary Reactions are Molecular Events,N,2,O,5,NO,2,+NO,3,NO+O,2,+NO,2,NO,2,+NO,3,Elementary Reactions are Molec,A,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 reaction,2 NO,2,(g)+F,2,(g),2 NO,2,F(g),the rate law is,rate,=k NO,2,F,2,.,Can the elementary reaction be the same as the overall reaction?,If they were the same the rate law would have been,rate,=k NO,2,2,F,2,Therefore,they the overall reaction is not an elementary reaction.Its mechanism is proposed next.,A mechanism is a collection of,The,rate determining 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 NO,2,F,2,for the reaction,2 NO,2,(g)+F,2,(g),2 NO,2,F(g),and a two-step mechanism is proposed:i NO,2,(g)+F,2,(g),NO,2,F(g)+F(g)ii,NO,2,(g)+F(g),NO,2,F(g),Which is the rate determining step?,Answer,:The rate for step i is rate=,k,NO,2,F,2,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！,The rate determining step is t,二、如何由给出的反应机理推导出速率方程,例,1,、,The decomposition of H,2,O,2,in the presence of I,follow this mechanism,iH,2,O,2,+I,k,1,H,2,O+,IO,slow ii H,2,O,2,+IO,k,2,H,2,O+O,2,+,I,fast,What is the rate law?,Solution,The slow step determines the rate,and the rate law is:,rate=,k,1 H,2,O,2,I,Since both,H,2,O,2,and I,are measurable in the system,this is the rate law.,二、如何由给出的反应机理推导出速率方程例1、The deco,例,2,、,Derive the rate law for the reaction,H,2,+Br,2,=2 HBr,from the proposed mechanism:,i Br,2,2 Br,fast equilibrium(,k,1,k,-1,),iiH,2,+Br,k,2,HBr+H slow iii H+Br,k,3,HBr fast,Solution:,The fast equilibrium condition simply says that,k,1,Br,2,=,k,-1,Br,2,and,Br,=,(,k,1,/,k,-1,Br,2,),The slow step determines the rate law,rate=,k,2,H,2,Br Br,is an,intermediate,=,k,2,H,2,(,k,1,/,k,-1,Br,2,),=,k,H,2,Br,2,;,k=k,2,(,k,1,/,k,-1,),M,-,s,-1,total order 1.5,explain,快速平衡假设法！,例2、Derive the rate law for the,例,3,、,The decomposition of N,2,O,5,follows the mechanism:1N,2,O,5,NO,2,+NO,3,fast equilibrium,2,NO,2,+NO,3,k,2,NO+O,2,+NO,2,slow,3NO,3,+NO,k,3,NO,2,+NO,2,fast,Derive the rate law.,Solution,:,The slow step determines the rate,rate=,k,2,NO,2,NO,3,NO,2,&,NO,3,are,intermediate,From 1,we have ,NO,2,NO,3,=,K,K,equilibrium constant,N,2,O,5,K,differ from,k,Thus,rate=,K,k,2,N,2,O,5,例3、The decomposition of N2O5 f,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,R,prod,is the same as its rate of consumption,R,cons,.,R,prod,=,R,cons,Intermediate,time,R,prod,R,cons,Be able to apply the steady-state
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