化學(xué)動(dòng)力學(xué)第二章習(xí)題和答案

...wd......wd......wd...化學(xué)反響動(dòng)力學(xué)第二章習(xí)題Thefirst-ordergasreactionSO2Cl2SO2+Cl2hask=2.2010-5s-1at593K,WhatpercentofasampleofSO2Cl2wouldbedecomposedbyheatingat593Kfor1hour?HowlongwillittakeforhalftheSO2Cl2todecompose?解：一級(jí)反響動(dòng)力學(xué)方程為：反響達(dá)1小時(shí)時(shí)：=0.924=92.4%已分解的百分?jǐn)?shù)為：100%-92.4%=7.6%當(dāng)時(shí)，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)設(shè)反響級(jí)數(shù)為n，則假設(shè)n=1，則t=9，t=18t=24，t=40t=54，t=72，t=105假設(shè)n=2，則t：918244054k：0.10400.12290.14870.15090.1701假設(shè)n=1.5t：91824k：0.01650.01890.0222假設(shè)n=3t：91824k：2.0672.603.46反響為一級(jí)。(2)k=0.0123min-1=2.05×10-4s-1(3)=56.3min=3378s3、復(fù)雜反響：的速率方程為，推導(dǎo)其動(dòng)力學(xué)方程。要求寫出詳細(xì)的推導(dǎo)過程。解：設(shè)時(shí)，，，時(shí)，，，代入得：令α=，β=，γ=則，移項(xiàng)積分：令，得動(dòng)力學(xué)方程：4、復(fù)雜反響由以下兩個(gè)基元反響組成：求反響進(jìn)展過程中，A1物種濃度與A3物種濃度間的關(guān)系。要求寫出詳細(xì)的推導(dǎo)過程。解：速率方程：〔1〕〔2〕，得：設(shè)時(shí)，，，移項(xiàng)積分：考慮物料平衡：，代入上式，得[A1]~[A3]關(guān)系式為：即：假設(shè)時(shí)，，，則：ConsiderthereactionmechanismWritechemicalrateequationsfor[A]and[X].Employingthesteady-stateapproximation,showthataneffectiverateequationfor[A]isd[A]/dt=-keff[A][B]Giveanexpressionforkeffintermsofk1,k-1,k2,and[B].解：ⅰ.ⅱ.對(duì)X進(jìn)展穩(wěn)態(tài)近似，則即：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〕機(jī)理為：據(jù)快速平衡：(b)〔1〕k1〔2〕k2〔3〕k3對(duì)[N2O2]進(jìn)展穩(wěn)態(tài)近似假設(shè)只有很少量的N2O2轉(zhuǎn)變?yōu)镹O2，而絕大局部轉(zhuǎn)變?yōu)镹O，即k2>>k3[O2],則：Ea(overall)=Ea1+Ea3-Ea2=79.5+84-205=-41.5kJ·mol-1(c)(1)檢測中間體N2O2或NO3〔2〕大大增加O2的濃度第一歷程為：第二歷程為：測定速率常數(shù)大小是否與O2濃度有關(guān)?！?〕作r~[O2](固定[NO],測不同[O2]下的反響速率)第一歷程為直線，第二歷程不為線性。7、復(fù)雜反響常用來描述酶催化反響和熱活化單分子反響，假設(shè)其總包反響的反響速率方程為：推導(dǎo)總包反響活化能與各元反響活化能的關(guān)系。解：(1)當(dāng)k-1>>k2時(shí)，反響機(jī)理中的第二步為決速步，總包反響速率常數(shù)為：總包反響的活化能：(2)當(dāng)k-1<<k2時(shí)，反響機(jī)理中的第一步為決速步，總包反響速率常數(shù)為：總包反響活化能：嚴(yán)格來說應(yīng)為：8、H2+Cl2→2HCl反響的速率方程為：擬定該反響的反響機(jī)理。要求寫出詳細(xì)推導(dǎo)過程。解：寫出反響體系可能存在的各基元反響：Ea〔kJ/mol〕Ea〔kJ/mol〕117.220.94360.025.10.08.4~15.10.0根據(jù)活化能大小，擬定可能的反響機(jī)理：117.2(1)k125.1(2)k28.4~15.1(3)k30.0