2020高中化学化学反应机理和催化剂

1、第4节反应机理、基本概念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：N2+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CO20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。Elementaryreactionsarestepsofmoleculareventsshowinghowreactionsproceed.Thistypeofdescriptionisamechanism.ThemechanismforthereactionbetweenCOandNO2isproposedtobeStep1NO2+NO2tNO3+NO(

2、anelementaryreaction)Step2NO3+COtNO2+CO2(anelementaryreaction)AddthesetwoequationsledtotheoverallreactionNO2+CO=NO+CO2(overallreaction)Amechanismisaproposaltoexplaintheratelaw,andithastosatisfytheratelaw.Asatisfactoryexplanationisnotaproof.例如：H2(g)+l2(g)=2HI(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

3、元反应：I2=I+I(快)H2+21=2HI(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MolecularityofElementaryReactionsThetotalorderofratelawinanelementaryreactionismolecularity.Theratelawofelementaryreactionisderivedfromtheequation.Theorderisthenumberofreactingmoleculesbecausetheymustco

4、llidetoreact.Amoleculedecomposesbyitselfisaunimolecularreaction(step);twomoleculescollideandreactisabimolecularreaction(step);&threemoleculescollideandreactisatermolecularreaction(step).3tO2+0rate=kO3NO2+NO2tNO3+NOrate=kNOJ2B+B+AtBr2+Ar*rate=kBr2ArCaution:Deriveratelawsthiswayonlyforelementayreactio

5、ns.单分子反应SO2C12的分解反应SO2C12=SO2+Cl2双分子反应NO2的分解反应2NO2=2NO+O2三分子反应HI的生成反应H2+21=2HI分子或更多分子碰撞而发生的反应尚未发现。ElementaryReactionsareMolecularEventsN2O5二N02+N03NO+O2+NO2JNO2+NO3Amechanismisacollectionofelementarystepsdevisetoexplainthethereactioninviewoftheobservedratelaw.Youneedtheskilltoderivearatelawfromamech

6、anism,butproposingamechanismistaskafteryouhavelearnedmorechemistryForthereaction,2NO2(g)+F2(g)t2NO2F(g),theratelawis,rate=kNOF.Cantheelementaryreactionbethesameastheoverallreaction?Iftheywerethesametheratelawwouldhavebeenrate=kNOJ2FJ,Therefore,theytheoverallreactionisnotanelementaryreaction.Itsmecha

7、nismisproposednext.反应机理中的慢反应步骤决定总反应的速率!Theratedeterminingstepistheslowestelementarystepinamechanism,andtheratelawforthisstepistheratelawfortheoverallreaction.The(determined)ratelawis,rate=kNOF，forthereaction,2NO2(g)+F2(g)-2NO2F(g),andatwo-stepmechanismisproposed:NO2(g)+F2(g)-NO2F(g)+F(g)NO2(g)+F(g)t

8、NO2F(g)Whichistheratedeterminingstep?AnswerTherateforstepiisrate=kNO2F,whichistheratelaw,thissuggeststhatstepiistherate-determiningorthes-l-o-wstep.二、如何由给出的反应机理推导出速率方程例1、ThedecompositionofH2O2inthepresenceofkfollowthismechanism,H2O2+卜一时一H20+IO-slowH2O2+IO一k2-H2O+O2+|-fastWhatistheratelaw?SolutionThe

9、slowstepdeterminestherate,andtheratelawis:rate二H2O2I-SincebothH2O2andI一aremeasurableinthesystem,thisistheratelaw.快速平衡假设法！例2、Derivetheratelawforthereaction,H2+Br2=2HBr,fromtheproposedmechanism:Bq二2BrH2+Br一k2-HBr+HH+Brk3HBrfastequilibrium（k“k.JslowexplainfastSolution:Thefastequilibriumconditionsimplys

10、aysthatBr2=心BrandBr=（可心回）Theslowstepdeterminestheratelaw,rate=k2HJBr=k2H2=kHJBr2沧totalorder1.5Brisanintermediatek=k2lkM-s-1例3、ThedecompositionofN2O5followsthemechanism:N2O5二NO2+NO3fastequilibriumNO2+NO3您NO+O2+NO2slowNO3+NOfNO2+NO2fast/Derivetheratelaw./Solution:NO2&NO3areintermediateK,equilibriumcon

11、stantKdifferfromkTheslowstepdeterminestherate,rate=k2NO2NO3From1,wehaveNO2NO3/=K/N2O5Thus,rate=Kk2N2O5稳态近似法！Intermediate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。RRzprod八consR二R/Aprod一八consRVR八prod八consRateofproducingtheintermediate,Rprod,isthesameasitsrateofconsumption,Rcons.Beabletoapplythesteadystate

12、approximationtoderiveratelaws假设H?+I2t2HI的反应机理如下：Step(1)I22IStep(1)2Ik、I?Step(2)H2+21S2HIDerivetheratelaw.Derivation:rate=k2H2I2(causethisstepgivesproducts!)butIisanintermediate,thisisnotaratelawyet.SincekxI2(=rateofproducing=kI2+k2HJI2(=rateofconsumingSteadyStateThus,k、I9I2=+*2H?/.rate=kk2H2I2/kA+k2H2Fromthepreviousresult:&k2H2l2rate=仮1+k2H2Discussion.If心k2H2then旳+k2H2=k2