Chapter Five Rates of Chemical Reaction Weather can occur ? ---chemical thermodynamics How fast (reaction rate)? ---chemical kinetics Chemical Reaction

化学反应速度的快慢 ——主要决定于反应的内在机理。 化学反应速度的快慢 ——主要决定于反应的内在机理。 反应机理：化学反应所经历的途径或具体步骤， 又称为反应历程。 化学动力学的基本任务就是研究反应的机理。

反应机理可以告诉我们： 某一反应究竟是经过哪些步骤完成的， 了解各个步骤的特征和相互联系， 揭示化学反应速度的本质，使人们能够自觉控制反应速度。

问题 1. 为什么有些口服药物的服用方法是2次/天？ 3次/天? 2. 为什么静脉滴注某些青霉素类药物要加快滴注的速度？

二、酶作为生物催化剂的特性：

2. )

Chapter Five Rates of Chemical Reaction 5-1 Rates and Mechanisms of Chemical Reactions 5-2 Theories of Reaction Rate 5-3 Reaction Rates and Concentrations 5-4 Effect of Temperature on Reaction Rates 5-5 Effect of Catalyst on Reaction Rates

Reaction rate: changes in a concentration of a 5-1 Rates and Mechanisms of Chemical Reactions Reaction rate: changes in a concentration of a product or a reactant per unit time. The rate is defined to be a positive number. The unit of reaction rate is mol·L-1·S-1, mol·L-1·min-1 or mol·L-1·h-1 et al

Define reaction rate : 1. Average reaction rate 2N2O5 = 4NO2 + O2 average reaction rate is obtained by dividing the change in concentration of a reactant or product by the time interval over which the change occurs

SAMPLE EXERCISE : H2O2 = H2O + 1/2 O2 0 〃: c1(H2O2) = 15.88×10-3 mol/L 5〃 : c2(H2O2) = 12.8×10-3 mol/L So:

Reaction Rates and Stoichiometry H2O2 = H2O + 1/2 O2 用不同物质浓度变化所表示的反应速率之间存在着一定的关系是：它们之间的速率比正好等于反应式中各物质分子式的系数之比，

To generalize, for the reaction aA + bB cC + dD This equation can be used to establish the relationship between rate of change of one reactant or product to another reactant or product. You have to be able to do this on the test, too!

Define reaction rate : 2. Instantaneous reaction rate 2N2O5 = 4NO2 + O2 The instantaneous rate of change at a point is the same as the slope of the tangent line. That is, it's the slope of a curve. c c V = lim —————— = -———— - Δc Δt0 Δt dc dt t v t

5-1.2 The Mechanisms of Chemical Reactions A reaction mechanism is a description of the path that a reaction takes. elementary reaction？ overall reaction？ types of elementary reactions

5-1.2 The Mechanisms of Chemical Reactions elementary reaction: is one that the reactants can convert directly into the products in a single step when they act each other. (one-step reaction) 基元反应：反应物间相互作用直接转化为生成物的反应。 （简称元反应）。 例如：

overall reaction: Many reactions are actually made up of several elementary steps, which are combined to yield the overall reaction. NO2 + CO → NO + CO2 NO2 + NO2 → NO3 + NO (slow) NO3 + CO → NO2 + CO2 (fast) This means that the rate of the overall reaction is dominated（控制） by the rate of the first reaction, this is the rate-determining step.

H2(g) + I2(g)  2 HI(g) I2(g) → 2I (fast) H2+ 2I → 2HI (slow) ——rate-determining step.

Types of Elementary Reactions molecularity --for Elementary Reactions: unimolecular reaction: an elementary reaction in which the rearrangement of a single molecule produces one or more molecules of product. I2(g) ─→ 2I(g) bimolecular reaction: the collision and combination of two reactants to give an activated complex in an elementary reaction. NO(g) + O3(g) ─→ NO2(g) + O2(g)

_ termolecular reaction: an elementary reaction involving the simultaneous collision of any combination of three molecules, ions, or atoms. 2 NO + H2 ─→ N2 + H2O2

5-2 Theories of Reaction Rate One is the collision theory : The collision theory is based on the kinetic theory and assumes a collision between reactants before a reaction can take place. 1918年 Lewis以气体分子运动论为基础提出 Another is the transition state theory: The transition state theory suggests that as reactant molecules approach each other closely they are momentarily in a less stable state than either the reactants or the products.

5-2.1 Collision Theory and Activation Energy ● Contents of Collision Theory: ⑴ reacting molecules must come so close that they collide. ⑵ " effective" collisions : not every collision between molecules creates products, only few collisions between reactant molecules will react.

● Contents of Collision Theory: According to this theory, product formation can only take place when there are "effective" collisions between reactant molecules involved in the rate determining step of the process.

Br(g) + HI(g) ─→ HBr(g) + I(g) Straight on collision, hydrogen facing incoming bromine. Reaction occurs. Straight on collision, hydrogen facing away from incoming bromine. Reaction does not occur. Straight on collision, bromine at 90 degrees. Reaction does not occur.

What constitutes an effective collision? 发生有效碰撞的两个基本前提: ● enough energy

● proper orientation 对 HCl 和 NH3 的气相反应

Activation molecule: is the molecule have enough energy and can produce effective collision the average molecules must absorb some energy to become activation molecules

活化分子一般只占极少数，它具有的最低能量为Ec 。通常把活化分子具有的平均能量与反应物分子的平均能量之差称为反应的活化能，用符号 Ea 表示。

在一定温度下，反应的活化能越大，活化分子的分子分数越小，活化分子越少，有效碰撞次数就越少，因此化学反应速率越慢； 反应的活化能越小，活化分子的分子分数越大，活化分子越多，有效碰撞次数就越多，化学反应速率越快。

Figure: As the activation energy of a reaction decreases, the number of molecules with at least this much energy increases, as shown by the yellow shaded areas.

▲ 每一反应的活化能数值各异, 可以通过实验和计算得 到 。活化能越小, 反应速率越快。Ea是动力学参数。 ▲ 一些反应的Ea 2SO2(g) + O2(g) = 2SO3(g), Ea=251 kJ·mol-1 N2(g) + 3H2(g) = 2NH3(g), Ea=175.5 kJ·mol-1 HCl + NaOH → NaCl+H2O, Ea≈20 kJ·mol-1 ▲ 离子反应和沉淀反应的Ea都很小 一般认为Ea小于63 kJ·mol-1的为快速反应 小于40 kJ·mol-1和大于400 kJ·mol-1的都很难测定出

★ 反应物的能量必 须爬过一个能垒 才能转化为产物 ★ 即使是放热反应 (△rH为负值)， 外界仍必须提供 最低限度的能量， 这个能量就是 反应的活化能

5-2.2 The Transition State Theory Transition state theory (TST) is also called activated complex theory. reactants pass through high-energy transition states before forming products, they are associated in an unstable entity called an activated complex, then change into products. 要点： 化学键重排、 活化络合物形成

Example : NO2(g) + CO(g) ─→ NO(g) + CO2(g) ΔH = Ea,f - Ea,r =358 kJ·mol-1 -132 kJ·mol-1 ΔH = –226kJ·mol-1.

酶作用的机制——中间产物学说 理论受限 活化络合物的结构无法在实验中加以确定 计算过于复杂 酶与底物形成酶－底物中间复合物， 中间复合物再分解成产物和酶。 E + S = E-S  P + E 许多实验事实证明了E－S复合物的存在。

Factors That Affect Reaction Rates Concentration of Reactants Temperature Catalysts ___Speed by changing mechanism

5-3 Reaction Rates and Concentrations Chemical reactions are faster when the concentrations of the reactants are increased . Because more molecules will exist in a given volume. More collisions will occur and the rate ofa reaction will increase.

Concentration and Rate Each reaction has its own equation that gives its rate as a function of reactant concentrations. this is called its Rate Law A rate law shows the relationship between the reaction rate and the concentrations of reactants. 说明在一定温度下，反应速度与反应物浓度之间的定量关系 ——质量作用定律

5-3.1 The Rate Law The rate of a reaction is proportional to the product of the concentrations of the reactants raised to some power. aA + bB → cC + dD v∝[A]m[B]n v = k[A]m[B]n 反应速率方程或称质量作用定律

v = k[A]m[B]n k is a rate constant that has a specific value for each reaction.The value of k is determined experimentally. “Constant” is relative here - k changes with T， 　　 the unit(量纲) depend on m + n ① when [A]=[B]=1mol·L-1, v =k ② the greater the k , the faster the rate

质量作用定律（law of mass action） 对于基元反应，反应速率与反应物浓度的幂乘积成正比。幂指数就是基元反应方程中各反应物的系数，这就是质量作用定律。

质量作用定律仅适用于基元反应。 对于复杂反应， 它适用于每一步的基元反应，它的反应速度取决于定速步骤。 2N2O5（g）→4NO2（g）＋ O2 (g) N2O5 → NO3＋NO2 （慢） NO2＋NO3 → NO2＋O2＋NO （快） NO＋NO3 → 2NO2 （快） 实验证明 速率方程应为

v = k[A]m[B]n [A], [B] are the concentration of A and B; aA + bB → cC + dD [A], [B] are the concentration of A and B; m and n are themselves constants for a given reaction, it must be determined experimentally in general, m and n are not equal to the stoichiometric coefficients a and b

5-3.2 Order of A Reaction The order of a reaction with respect to one of the reactants is equal to the power to which the concentration of that reactant is raised in the rate equation. The sum of the powers to which all reactant concentrations appearing in the rate law are raised is called the overall reaction order.

m+n is overall order of the reaction For rate equation v = k[A]m[B]n m is the order of the reaction with respect to A, n is the order of the reaction with respect to B. m+n is overall order of the reaction the exponents m and n are not necessarily related to the stoichiometric coefficients in the balanced equation, that is, in general it is not true that for a A + b B → c C + d D, m ≠ a and n ≠ b

The rate law for the thermal decomposition（热分解） of acetaldehyde (CH3CHO) CH3CHO(g) →CH4(g) + CO(g) has been determined experimentally to be v =k[CH3CHO]3/2 and not rate = k[CH3CHO] ×

Example 5-1: Given the following data, what is the rate expression for the reaction between hydroxide ion and chlorine dioxide? 2ClO2(aq) + 2OH-(aq) →ClO3-(aq) + ClO2-(aq) + H2O [ClO2] (mol.L-1) [OH-] (mol.L-1) Rate (mol.L-1 s-1) 0.010 0.030 6.00×10-4 v1 0.010 0.075 1.50×10-3 v2 0.055 0.030 1.82×10-2 v3

Solution: v = k[ClO2]m[OH-]n m=? By inspection, m = 2. The reaction is 2nd order in ClO2

v = k[ClO2]m[OH-]n n=? The overall rate expression is therefore By inspection, n = 1 The overall rate expression is therefore v = k[ClO2]2[OH-]

想一想： 反应速度与速率方程？ 反应分子数和反应级数？

First-order reactions A first-order reaction is a reaction whose rate depends on the reactant concentration raised to the first power. A →B the rate equation is Differential form:

Thus Integrate the left side from c = c0 to c and the right from t = 0 to t. 速度方程的积分表达式

Can be rearranged to give: - { lnc – lnc0} = k1( t - 0 ) c0 is the initial concentration of c (t =0). c is the concentration of c at some time t.

or The characteristics of first-order reactions： A plot of lgc versus t (time) gives a straight line with a slope of -k1/2.303. 2. The rate constant, k, has units of [time]-1.

3. half-life (t1/2) is the time it takes for the concentration of a reactant A to fall to one half of its original value. By definition, when t = t1/2, , so

—— 一级反应的半衰期与速率常数k1成反比，与反应物的起始浓度无关。 恒温下，一级反应的半衰期是一个常数。 半衰期越长，k1就越小，反应速率就越慢。 可以用半衰期衡量反应速度， t1/2越大，反应越慢。

常见药物的半衰期： 　 药物名称 t1/2（h） 青霉素 0.5 链霉素 2～3 安定 20～40 放射性同位素14C半衰期为5580年

一级反应的特点是： （1）1gc对时间t作图可得一条直线，直线的斜率为 －k1/2.303。因此可由斜率求得反应速率常数k1。 而与浓度无关。 （3）恒温下，一级反应的半衰期是与速率常数k1成反比，与反应物的起始浓度无关的一个常数。

Example 5-2 (a) What is the rate constant k for the first-order decomposition of N2O5(g) at 25℃ if the half-life of N2O5(g) at that temperature is 4.03×104 seconds? Solution: (a) -

(b) Under these conditions, what percent of the N2O5 molecules will not have reacted after one day? Solution: (b) Putting in the value for k and substituting t = 8.64×104 seconds (one day has 86,400 seconds) gives

Hence Therefore, 22.6% of the N2O5 molecules will not have reacted after one day at 25℃.

Example 5-3 SO2Cl2（磺酰氯 ）decomposes to sulfur dioxide and chlorine gas Example 5-3 SO2Cl2（磺酰氯 ）decomposes to sulfur dioxide and chlorine gas. The reaction is first order. If it takes 13.7 hours for a 0.250mol/L solution of SO2Cl2 to decompose into a 0.117mol/L solution, what is the rate constant for the reaction and what is the half-life of SO2Cl2 decomposition? Solution:

K = 0.0554 h-1 一级反应很多，许多热分解反应、分子重排反应、放射性元素的蜕变等都是一级反应，许多化合物的水解反应在低浓度的水溶液中进行时也表现为一级反应，许多药物在生物体内的吸收、分布、代谢和排泄过程，也常近似地看作是一级反应。

[例5-2] 已知四环素在人体内的代谢服从一级反应规律。设给人体注射0.5g四环素，然后在不同时间测定血液中四环素的含量，得如下数据： 服药后时间t/h 4 6 8 10 12 14 16 血中四环素含量c/mg·L-1 4.6 3.9 3.2 2.8 2.5 2.0 1.6 试求： ①四环素代谢的半衰期； ②若血液中四环素的最低有效量相当于3. 7mg·L-1， 则需几小时后注射第二次？

解： ① 先求速率常数k1，一级反应以log c对t 作图，得直线， 见图5-6。由前后两点或由直线回归得: ∴

②由图5-6知，t＝0时，lgc0＝0.81，最低有效量c＝3.7mg·L-1 lgc＝0.57, 将（5-8）重排得： 计算表明：要使血液中四环素含量不低 于3.7mg·L-1，应于第一次 注射后6.3h 之前注射第二次，临床上一般控制 在6h后注射第二次， 即4次/天。

另外，衡量药物分解的速率时，常用分解10％所需的时间，称为十分之一衰期，用t0.9表示，恒温下t0.9也是与浓度无关的常数。 了解药物的半衰期，对于合理用药有着重要意义。 常见药物的半衰期，如 磺胺甲恶唑（新诺明），半衰期12小时左右； 头孢氨苄，半衰期为1～2小时； 地西泮为20～40h， 保泰松为48～120h。

Second - order reactions 二级反应是一类常见的反应，溶液中的许多有机反应像加成、取代及消去反应等都是二级反应。 v = k2[A]2 , v = k2 [A][B] [A]=[B]

The integrated rate law for a 2nd order reaction can be easily shown to be

The characteristics of second- order reactions： 1. A graph of 1/c against time is a straight line , the slope of which gives the rate constant for the reaction； 2. The rate constant, k, has units of [c]-1[t]-1； 3. The half-life of 2th-order reactions t1/2=1 / kc0. (c = 2/c0代入上式)

Note that the half-life of a second-order reaction is not independent of the initial concentration, as in the case of a first-order reaction. This is one way to distinguish a first-order reaction from a second-order reaction.

二级反应有以下特点是： （1）1/c对时间t作图可得一条直线， 直线的斜率即为反应速率常数k2。 （3）二级反应的 。由此可见：二级反应 的半衰期与反应物的初始浓度成反比。反应物初始 浓度越大，半衰期越短。

Example 5-4 Butadiene(C4H6丁二烯) dimerizes(聚合) to form C8H12（二聚物） Example 5-4 Butadiene(C4H6丁二烯) dimerizes(聚合) to form C8H12（二聚物）. This reaction is 2nd order in butadiene. If the rate constant for the reaction is 0.84 Lmol-1min-1, how long will it take for a 0.500 mol/L sample of butadiene to dimerize until the butadiene concentration is 0.200 mol /L? （5-11） Solution: t = 3.6 (min)

［例5-3］乙酸乙酯在25℃时的皂化反应为二级反应： CH３COOC２H５十NaOH→CH３COONa十C２H５OH 乙酸乙酯和氢氧化钠的起始浓度均为0.0100mol·L－１，反应20min后，氢氧化钠的浓度消耗了 0.00566mol·L－１， 求： ①反应速率常数； ②反应的半衰期。 解： ① ②

Zero - order reactions c = - k0 t + c0 v = k0 [A]0 integrated： 固体界面上的分解反应 —— is one where the rate does not depend on the concentration of the species. v = k0 [A]0 integrated： c = - k0 t + c0

c = - k0 t + c0 The characteristics of zero-order reactions： 1. A graph of c against t is a straight line with a slope of -k0. 2. The rate constant, k, has units of [c][ t ]-1； 3. The half-life of a zero-order reaction is

零级反应的特点： （1）c对t作图得一直线，斜率为－k0 （2）k0的单位为［浓度］［时间］－１ 所以k0与时间单位和浓度单位有关 （3）零级反应的半衰期： 所以零级反应的半衰期与最初浓度成正比。 反应物的初始浓度越大，半衰期越长。  近年来发展的一些缓解长效药，其释药速率在相当长的时间范围内比较恒定，即属零级反应。如国际上应用较广的一种皮下植入剂，内含女性避孕药左旋18－甲基炔诺酮，每天约释药30μg，可一直维持5年左右。

Example 5-5 The decomposition of HI into hydrogen and iodine on a gold surface is 0th order in HI. The rate constant for the reaction is 0.050mol·L-1·s-1. If you begin with a 0.500mol/L concentration of HI, what is the concentration of HI after 5 seconds? c0= 0.500mol·L-1 k =0.050mol·L-1· s-1 t =5s Solution: [HI] = 0.500－0.050×5 = 0.250(mol·L-1)

5-4 Effect of Temperature on Reaction Rates Chemical reactions are faster when the temperature is increased. Why? Figure: At a higher temperature,T2, more molecules have an energy greater than E a , as shown by the yellow shaded area.

When we increase the temperature, kinetic energies and speeds will increase and so the average energy of a collision will also increase. As a result, the energy of any particular collision will be more likely to exceed the activation energy. Thus, chemical reactions are faster at higher temperatures than at lower temperatures.

Rule of thumb： The dependence of reaction rate on temperature has led to a common rule of thumb: the rate of a chemical reaction will double for each 10℃ increase in the temperature.

5-4.1 Rule of Thumb经验法则 (Van’t Hoff Law) The rate of a chemical reaction will double for each 10℃ increase in the temperature. T : v T = k T [A]a [B]b (T + 10) : v (T+10) = k T+10 [A]a [B]b a A + b B = c C + d D v (T+10) k (T+10) γ= ---------- = ----------- = 2～4 v T k T

5-4.2 The Arrhenius Equation In 1889 Svante Arrhenius showed that the dependence of the constant of a reaction on temperature can be expressed by the following equation, now known as the Arrhenius equation 通过实验制作速率常数k随温度T升高而变大的曲线图，并分析曲线，可以得出速率常数随温度变化函数的经验方程:

The Arrhenius Equation: Ea —is the activation energy of the reaction (in kJ/mol) R — is the gas constant (8.314 JK-1mol-1) T — is the absolute temperature e — is the base of the natural logarithm scale A — represents the collision frequency, and is called the frequency factor (指前因子). 阿伦尼乌斯方程式的重要假设：A和Ea是不随温度改变的特征参数。 大多数反应在一定温度区间内的这种近似是完全允许的，这种近似方 法被称为线性化。

阿伦尼乌斯方程式把速率常数k、活化能Ea和温度T三者联系起来，由此关系式可以说明： （1）对某一反应：Ea基本不变，温度升高， 增大，k值也增大，反应速率加快，说明了温度对反应速率的影响。 （2）对不同反应：当温度一定时，活化能Ea值越小， 则 越大，k值愈大，反应速率愈快。 反之，Ea愈大，反应速率愈慢。

Taking the natural logarithm of both sides, the equation becomes or in terms of common logarithms :

Example 5-6 The rate constants for the decomposition of acetaldehyde CH3CHO(g) ─→ CH4(g) + CO(g) were measured at five different temperatures. The data are show below. Plot lgk versus 1/T and determine the activation energy (in KJ/mol) for the reaction.   ────────────────────────────── T(K) 700 730 760 790 810 k [1/(mol/L)1/2s] 0.011 0.035 0.105 0.343 0.789 ──────────────────────────────

Answer: We need to plot lg k versus 1/T . From the given data we obtain 1/T(1/K) 1.43×10-3 1.37×10-3 1.32×10-3 1.27×10-3 1.23×10-3 log k -1.96 -1.46 -0.979 -0.465 -0.103 The slope of the line is calculated from two pairs of coordinates: lg k 1/T

Thus, a plot of lg k versus 1/T gives a straight line Ea = 2.303×(8.314J/K·mol)× (9.38×103K) = 1.80×105 J/mol = 1.80×102 KJ/mol Thus, a plot of lg k versus 1/T gives a straight line whose slope is equal to -E a / 2.303R and whose intercept（截距） with the ordinate(纵坐标) is log A.

5-4.3 Application of Arrhenius Equation According to this equation, we can calculate E a and k T1 → k1 T2 → k2 (5-14) - (5-13) ：

Example 5-7 The rate constant of a first-order reaction is 3.46×10-2 /s at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 KJ/mol? Answer:

5-5 Effect of Catalyst on Reaction Rates Catalyst: a substance that increase the rate of a chemical reaction , but is itself neither consumed nor produced in the reaction MnO2 Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. Catalysts change the mechanism by which the process occurs.

Reaction O3 + O = O2

Catalysis can be classified into three types Catalysis can be classified into homogeneous(均相) catalysis, heterogeneous (非均相) catalysis enzyme catalysis.

Homogeneous catalysis In homogeneous catalysis, the catalyst is present in the same phase as the reactants, as when a gas-phase catalyst speeds up a gas-phase reaction, or a species dissolved in solution speeds up a reaction in solution

The rate is very Slow, but it can be catalyzed by Ag+ or Mn2+:

Heterogeneous catalysis In heterogeneous catalysis, the catalyst is present as a distinct phase. The most important case is the catalytic action of certain solid surfaces on gas-phase and solution-phase reactions.

catalysis CH3–CH3 rapid (k larger) e.g., H2 + CH2=CH2 Pd e.g., H2 + CH2=CH2 no reaction (k very small) 25ºC C2H4 + H2 C2H6 PE  RC  Ea (uncatalyzed) Ea (catalyzed)

Enzyme Catalysis Enzymes are biological catalysts. An average living cell may contain some 3000 different enzymes 酶催化作用的本质：降低反应活化能 E + S = E-S  P + E activated complex

E+S P+ E ES 能量水平 反应过程 G E1 E2 酶（E）与底物（S）结合生成不稳定的中间物（ES），再分解成产物（P）并释放出酶，使反应沿一个低活化能的途径进行，降低反应所需活化能，所以能加快反应速度。 E…S

酶与底物结合形成中间络合物的方式（理论） (1) 锁钥假说(lock and key hypothesis)： 认为整个酶分子的天然构象是具有刚性结构的，酶表面具有特定的形状。酶与底物的结合如同一把钥匙对一把锁一样.

(2) 诱导契合假说（induced–fit hypothesis): 该学说认为酶表面并没有一种与底物互补的固定形状，而只是由于底物的诱导才形成了互补形。 酶与底物接近→结构相互诱导→变形和相互适应 →结合。

Zn 羧肽酶活性中心示意图 羧肽酶的诱导契合模式 Tyr 248 Arg 145 Glu 270 底物

与酶具有高效催化作用的因素 邻近效应与定向排列 亲电催化 多元催化 亲核催化 酸、碱催化 离子催化 表面效应

1. 邻近效应(proximity effect)与 定向排列(orientation arrange )： 邻近效应： 酶将诸底物结合到其活性中心，使它们的反应基团相互靠近，降低进入过渡态所需的活化能；局部底物浓度增加。 定向排列：酶使诸底物反应基团形成正确的几何定向关系，产生有效碰撞。

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2. 多元催化(multielement catalysis) 多个基元催化形式的协同作用。一般包括： 酸-碱催化，共价催化（亲核催化, 亲电子催化）等。 酶分子中广义的酸碱基团：氨基、羧基、巯基、酚羟基、咪唑基。 同种基团在不同的微环境下解离度亦不同。 这种多功能基团兼有酸、碱双重催化效能。

底物中典型的亲电中心：磷酰基、酰基、糖基 某些辅酶：焦磷酸硫胺素、 磷酸吡哆醛等也参与共价催化作用。 底物中典型的亲电中心：磷酰基、酰基、糖基 这类反应有 羰基的加成，酮基和烯醇的互变异构， 肽和酯的水解及磷酸和焦磷酸参与的反应

例:胰凝乳蛋白酶催化肽键水解的多种类型催化机制 “ 催化三元区 ”——活性中心的结构特性 Asp102 ——起稳定His57构象作用 His57 ——质子受体，促进Ser195羟基氧对底物肽酰基 亲核进攻作用 Ser195 ——羟基氧亲核进攻作用 稳定作用

(2) 亲核共价催化作用： ① 四面体结构中间体的形成： Ser195羟基氧对底物肽酰基亲核进攻时, 形成一个共价四面体中间体

b: 水亲核进攻羰基碳，另一个过渡态四面体产生； ② His-57广义酸催化过程 酰基一酶中间体生成 ： a: His-57 给出质子，底物接受质子，伴随着 C-N肽 键的断裂后 , R1NH2（产物1）从酶中离去； b: 水亲核进攻羰基碳，另一个过渡态四面体产生； 产物1 a b

③His-57作为广义碱-酸催化作用 His57 作为碱 接受H2O给予的质子， 随后它作为酸将质子转移给Ser195氧， 以此方式帮助酰基-酶中间体的解离。 酰基一酶中间体虽然 不稳定, 但 它的存在已经被X射线衍 射实验所证实。 产物2

3. 表面效应(surface effect)： 酶活性中心的疏水环境排除了水分子对酶和底物功能基团的干扰、吸引或排斥，防止酶和底物之间形成水化膜，有利于酶和底物的密切接触。 ——疏水的为微环境极大利于酶的催化作用 + -

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