级物理化学-4、参考已双语reaction kinetics

Chapter5CatalyticReaction5.1Catalystsand5.2Acid-base5.3Enzymatic5.45.55.6HeterogeneousCatalysis:SurfaceMaiReturnCatalystsandConceptsconcerningHomogeneousandheterogeneousHowacatalystMaiReturnWhatareCatalystand许多化学反应，当在反应体系中加入少量反应物种以外的其它组分()这些添加物在反应终了时，并不进入产物，而且其数量及化学性质不因反应而改变。这种添加物被称为催。A+ MaiReturnHomogeneousandHeterogeneous多相催化反MaiReturnHowaCatalyst1.1.ThecatalystcannotchangetheΔrG$,equilibriumconstant(K)andmconversionoryieldcannotbeTheroleofthecatalystistoreducethetimetoreachanequilibrium。H(g)+

Δ

$＝-237kJmol- MaiReturnHowaCatalyst2Theessentialfunctionofthecatalystistodecreasetheactivationenergybychangingthereactionmechanism性介质旁观反应而是反应的积极Non-catalytic CatalyticA+B→ A+Cat.→M+B→P+ 2HO→HO+ H2O2+I－→H2O+ IO－+H2O2→H2O+O2+IMaiReturnCatalystDecreasesActivationCatalyticCatalyticAK∪ AKB[AKB]rkkkkfbEaEslowrkkkkfbMaiReturnknon-example（T=3001MaiReturnRateExpressionforaCatalyticConsiderAr=r0+If[cat.]isnotsorcat=kcat[cat][A]r0=k0[A]r=(k0+kcat[cat])[A]=kapp=k0+MaiReturnSpecialacid-baseGeneralacid-baseMaiReturnAcidandDefinitionofAcidandBrønstedacidandbase:proton(H+)donorandacceptorLewisacidandbase:electronacceptoranddonorSpecialacidandbaseinaqueousH3O+andMaiReturnConsiderasubstrate(reactant)capableofbeingconvertedbybothnon-catalyticandacid-basecatalyzedmechanismSS+OH－→P+r=kOH－[OH－]rk0[S]kH[H][S]kOH[OHS+H+S+H+→P+r=kH+[H+]S→r=k0k0kH[H]kOH[OHMaiReturn Kappvs kapp=k0+kH+[H3O+]+kOH- ([HO+][OH-]=Klogkapp=logk0,independentofreactionslogkapp=logkH+-logkapp~pH,straightlines，slope=-log =log -- + MaiReturnMaiReturnBrønstedAcid-catalyzedBrønsted kapp=k0+kH+[H3O+]+ CH3COCH3＋I2→r=CH3COCH3→CH3C(OH)=CH2CH3C(OH)=CH2＋I2→CH3CI(OH)-CH2I CH3CI(OH)-CH2I→CH3COCH2I＋HI kapp=k0+kH+[H+]+MaiReturn弱酸催化的速率常数与酸离解常数不同弱酸催 碘化反应的有关常催化（酸的离解常数(Lmol-1min-)(Lmol-盐5.1×10-二氯乙5.1×10-α，β一二酸6.7×10-一氯乙1.5×10-乙1.8×10-

Mai

Return

BrønstedBrønsted关系（Brønsted法则数常与其离解平衡常数Ka有如下关系ka=kb=MaiReturnEnzyme-catalyzedFeaturesforenzyme-catalyzedKinetics:Michaelis－MentenMaiReturnFeaturesforEnzyme-catalyzed一种酶只能催化一种或一类特定的反MaiReturnFeaturesforEnzyme-catalyzedMaiReturnMichaelis-MentenThesimplestenzymaticreactionisthatwithonlyonesubstrateandoneproduct.IthasbeenfoundthattheinitialratechangeswiththeconcentrationofthesubstratewiththefollowingcurveMaiReturnMichaelis-MentenAsimplemechanismwasproposebyMichaelisandMententoaccountthis E+ E+k-严格地说第二步也应该是可故此机理只运用于反应初始阶r k0 dt 0MaiReturnMichaelis-MentenSSapproximationforESUsingrelationship:[E]0=[E]+[ES]

d[ES]k[E][S] [ES]k[ES]d [ES]k1[Ek2k10=[E]+[ES][ES] k1[E]0(k1k2)k1[S]rk[ES] k2k1[E]0 ( k)k Michaelis-MentenMaiReturnImportantParametersinMichaelis-MentenAssumAssume:k1kMichaelisconstantMk1rm=k2[E]0MaximumrateMichaelis-Mentenequationcanbesimplifiedr rm[S]KM[S]MaiReturnMichaelis-Menten rm[S] [S]MAsr=umAs 1storderwithrespecttoAs[S]= equalstothesubstrateconcentrationM MaiReturnHowtoGetrmandrmandKMareimportantparametersforaspecificenzyme-catalyzedreactionTheycanbesimplyevaluatedfromtheLineweaver-Burkequation1r1r1KM1[S11以r

作图，从斜率及截距可以求出MaiReturnMaiReturn其中[A]0、[B]0分别为AB初始浓度，δMaiReturn若以[B]对反应时间t作图可得一S型曲线，显示有诱导期。反应开始B长时间以后，A很小，反应速率有极大值tmt=tmS的动力学曲线上恰极点。由反应动力学曲线切线交于t轴ti，可称此时间 自动催化反应的速率曲线(a)与动力为诱导 曲线MaiReturn2O－+2

O+H 今在E的初始浓度为4.00×10-7moldm-3pH9.1c0/moldm-r0/moldm-3s-MaiReturnO－+

O+ O－+ O2-+ 其中E－为中间物，可视为自由基，已知k22k1,试求k1与k2(1)r0kc0根据实验数据得：n= k=500.72s-MaiReturn(2)根据机 O－+ O+ O2－+E－ O22-+对E－d[E－]k[E][O－－k[E－][O－ [E]0=[E]+k1[E]= k1[E]=k2([E]0[E] k+ r=d[O]/dt=k

—][E]

[E]

—]=k[Ok k1+

k=2k1+

[E]0= k1=1.878×109dm3mol-1s-k=2k=3.755×109dm3mol-1s- MaiReturn总结本题虽为酶催化的题目,但并未涉及Michaelis-Menten机理,[E]0=[E]+[E－]这在Michaelis-Menten机理推导中也是一关键，MaiReturnFivetypicalstepsinheterogeneousActivesitesandMaiReturnFiveTypicalStepsinHeterogeneous发生在固体表面上的多相催化反应可分为以下5吸附作用在多相催化中起重要MaiReturnActiveSitesandSurface固体表面上吸附反应物的部位称为活性部SNumberofadsorptionsitesCoverage()Numberofadsorpitonsites表面反应速率主要决定于表面上反应物种的覆盖度覆盖度与反应物在气~pAdsorptionisotherms吸附MaiReturnLangmuirAdsorptionThesimplestadsorptionisothermforAdsorptioncannotproceedbeyondmonolayerAllsitesareequivalentandthesurfaceisTheabilityofamoleculetoadsorbatagivensiteisindependentoftheoccupationofneighboringsitesMaiReturnLangmuirAdsorption A kap1 rd=Atequilibrium，ra= 1 ka K=kkk 1 1LangmuiradsorptionMaiReturnLangmuirAdsorption=V 1=1 1+ VmKMaiReturnLangmuirIsothermforDissociativeForadsorptionwith AkA2 S kdkap1 K1/2p1/ r=k 1K1/2p1/ Atlow =MaiReturnLangmuirIsothermforCompetitive若有i种物质同时被同一活性位吸附，则对pi的第i组分的气体，其覆盖度i Ki 1 K 如A、B二种气体 aA aB 1a a 1a a TheincreaseinpAwilldecreaseTheincreaseinpBwilldecrease这是因为A和BMaiReturnHeterogeneousCatalysis:SurfaceUnimolecularreactions:rdnA Bimolecularreactions:rdnAk Ar k MaiReturnAssumenoAA

B r k 2 KA1KAr KA21 MaiReturnrrKA21 (1)IfpAisverylow,r=k2KApA=1stkapp=Ea=IfpAishighorKAislarge(strongadsorption),r= zerothEa=表 吸已，相力响附表.MaiReturnifaD(maybeaproductoranon-reactant)isalsoadsorbedonthecatalystsurfacer KA21 if1+KApA<<KD r

KA 速率与D的分压成反比，即D抑制A的反MaiReturnRateofbimolecularSurfaceTherearetwomechanismsforbimolecularsurface A A(g)+B(g) S