化学动力学第二章习题和答案

...wd......wd......wd...化学反响动力学第二章习题Thefirst-ordergasreactionSO2Cl2SO2+Cl2hask=2.2010-5s-1at593K,WhatpercentofasampleofSO2Cl2wouldbedecomposedbyheatingat593Kfor1hour?HowlongwillittakeforhalftheSO2Cl2todecompose?解：一级反响动力学方程为：反响达1小时时：=0.924=92.4%已分解的百分数为：100%-92.4%=7.6%当时，s=31500s=8.75hourT-butylbromideisconvertedintot-butylalcoholinasolventcontaining90percentacetoneand10percentwater.Thereactionisgivenby(CH3)3CBr+H2O(CH3)3COH+HBrThefollowingtablegivesthedatafortheconcentrationoft-utylbromideversustime:T(min)091824405472105(CH3)CBr(mol/L)0.10560.09610.08560.07670.06450.05360.04320.0270Whatistheorderofthereaction?Whatistherateconstantofthereaction?Whatisthehalf-lifeofthereaction?解：(1)设反响级数为n，则假设n=1，则t=9，t=18t=24，t=40t=54，t=72，t=105假设n=2，则t：918244054k：0.10400.12290.14870.15090.1701假设n=1.5t：91824k：0.01650.01890.0222假设n=3t：91824k：2.0672.603.46反响为一级。(2)k=0.0123min-1=2.05×10-4s-1(3)=56.3min=3378s3、复杂反响：的速率方程为，推导其动力学方程。要求写出详细的推导过程。解：设时，，，时，，，代入得：令α=，β=，γ=则，移项积分：令，得动力学方程：4、复杂反响由以下两个基元反响组成：求反响进展过程中，A1物种浓度与A3物种浓度间的关系。要求写出详细的推导过程。解：速率方程：〔1〕〔2〕，得：设时，，，移项积分：考虑物料平衡：，代入上式，得[A1]~[A3]关系式为：即：假设时，，，则：ConsiderthereactionmechanismWritechemicalrateequationsfor[A]and[X].Employingthesteady-stateapproximation,showthataneffectiverateequationfor[A]isd[A]/dt=-keff[A][B]Giveanexpressionforkeffintermsofk1,k-1,k2,and[B].解：ⅰ.ⅱ.对X进展稳态近似，则即：即：ⅲ.(a)Thereaction2NO+O22NO2isthirdorder.AssumingthatasmallamountofNO3existsinrapidreversibleequilibriumwithNOandO2andthattherate-determiningstepistheslowbimolecularreactionNO3+NO2NO2,derivetherateequationforthisthemechanism.(b)Anotherpossiblemechanismforthereaction2NO+O22NO2is(1)NO+NON2O2k1(2)N2O22NOk2(3)N2O2+O22NO2k3Applythesteadystateapproximationto[N2O2]toobtaintheratelawford[NO2]/dt.IfonlyaverysmallfractionoftheN2O2formedin(1)gosetoformproductsinreaction(3),whilemostoftheN2O2revertstoNOinreaction(2),andiftheactivationenergiesareE1=79.5kJ/mol,E2=205kJ/mol,andE3=84kJ/mol,whatistheoverallactivationenergy?(c)Howwouldyoudistinguishexperimentallybetweenthemechanismsuggestedinpart(a)and(b)?解：〔a〕机理为：据快速平衡：(b)〔1〕k1〔2〕k2〔3〕k3对[N2O2]进展稳态近似假设只有很少量的N2O2转变为NO2，而绝大局部转变为NO，即k2>>k3[O2],则：Ea(overall)=Ea1+Ea3-Ea2=79.5+84-205=-41.5kJ·mol-1(c)(1)检测中间体N2O2或NO3〔2〕大大增加O2的浓度第一历程为：第二历程为：测定速率常数大小是否与O2浓度有关。〔3〕作r~[O2](固定[NO],测不同[O2]下的反响速率)第一历程为直线，第二历程不为线性。7、复杂反响常用来描述酶催化反响和热活化单分子反响，假设其总包反响的反响速率方程为：推导总包反响活化能与各元反响活化能的关系。解：(1)当k-1>>k2时，反响机理中的第二步为决速步，总包反响速率常数为：总包反响的活化能：(2)当k-1<<k2时，反响机理中的第一步为决速步，总包反响速率常数为：总包反响活化能：严格来说应为：8、H2+Cl2→2HCl反响的速率方程为：拟定该反响的反响机理。要求写出详细推导过程。解：写出反响体系可能存在的各基元反响：Ea〔kJ/mol〕Ea〔kJ/mol〕117.220.94360.025.10.08.4~15.10.0根据活化能大小，拟定可能的反响机理：117.2(1)k125.1(2)k28.4~15.1(3)k30.0