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1、Chapter 8化学反应速率 Chemical kinetics,授课对象:10级药学本科,掌握内容,有效碰撞、活化分子、活化能的概念及活化能与活化分子数对反应速率的影响。温度、浓度、催化剂对化学反应速率的影响及浓度影响的定量关系质量作用定律、化学反应的温度因子。反应级数与反应分子数概念。一级反应的化学反应速率方程式的计算。,了解内容,化学反应速率的基本概念。二级、零级反应的特征均相催化作用的特点。,难点,活化能与活化分子数对反应速率的影响根据给出的实验条件写出化学反应速率方程式及进行有关的计算。一级反应的有关计算。,Chemical kinetics 化学动力学,The area of c
2、hemistry that is concerned with the speeds,or rates,at which reactions occur is called chemical kinetics,第一节 化学反应速率的表示方法,Reaction rate 化学反应速率,Definition:The ratio of the amount of decreasing in reactant or the amount of increasing in product to a specific time interval.一定时间内反应物浓度的减少量或生成物浓度的增加量与时间间隔的
3、比值Unit:molL-1 s-1(min-1,h-1),(1)Average rate 平均速率,The change in the concentration of reactant or product during a time interval.,The change in the concentration of reactant or product at one specific instant time.在反应时间间隔无限小时，反应物浓度或生成物浓度的变化与时间间隔的比值,(2)Instantaneous rate 瞬时速率,说明,用反应物表示速率变化时应标上负号,以保证速率
4、为正值当反应速率稳定时，反应式中用不同物质表示的速率间比值等于反应式中各物质的系数比VA:VB:VC:VD=a:b:c:d 化学反应速率值必须注明是以反应体系中的哪种化学物质表示而得,化学反应速率的表述之二,反应进度：,化学反应速率,单位体积内反应进度随时间变化率,此时的v代表整个化学反应的速率，与反应体系中选择何种物质表示反应速率无关,例：N2+3H2 2NH3,aA+bB=cC+dD,第二节 反应机理,Reaction processes occur in a single event or step are called elementary steps.一步就能完成的化学反应,A,B,
5、C,reactant,product,react,Elementary reaction 元反应：,A overall reaction consists of a sequence of elementary reaction.由两个或两个以上的元反应组成的化学反应叫做非元反应,Multistep mechanisms(composite reaction)非元反应（复合反应）,Molecularity of reaction 反应分子数,1SO2Cl2=SO2+Cl2,The sum of the stoichiometric coefficient of the reactants in
6、 elementary reaction.元反应中反应物系数之和称反应分子数,单分子反应:If a single molecule is involved,the reaction is said to be unimolecular.,methyl isonitrile rearrange,1NO2+1CO=NO+CO2,2NO+1H2=N2O+H2O 2I+1H2=2HI,三分子反应:Elementary steps involving the simultaneous collision of three molecules are said to be termolecular rea
7、ction,双分子反应:Elementary steps involving the collision of two reactant molecules are said to be bimolecular reaction,（注意：反应分子数仅适用于元反应）,e.g.2N2O5(g)4NO2(g)+O2(g)(A)N2O5 NO3+NO2(slow，rate limiting step，速率控制步骤）(B)NO3 NO+O2(quick)(C)NO+NO3 2NO2(quick)(A)(B)(C)三步就是该的反应机理,Reaction mechanism反应机理,The process
8、by which a reaction occurs is called reaction mechanism.It involve one or more intermediates化学反应进行的实际步骤。即实现化学反应的各步元反应组合的微观过程,Rate limiting step or rate-determining,H2(g)+I2(g)=2HI(g)(A)I2(g)=2I(quick)(B)H2(g)+2I(g)=2HI(g)(slow)（Rate limiting step/rate-determining step）,The slowest step in mechanism.
9、在化学反应的实际步骤中，由许多元反应组成，其中最慢的那步元反应限制着整个复杂反应的速率大小，这一较慢的元反应称为速率控制步骤，简称速控步骤,第三节 化学反应速率理论简介,碰撞理论活化过渡状态理论,碰撞理论与活化能,理论碰撞认为：化学反应的发生是反应物之间发生有效碰撞的结果1、有效碰撞：能发生化学反应的碰撞2、弹性碰撞：不能发生化学反应的碰撞,(a)弹性碰撞(b)有效碰撞,反应物分子互相接近，发生碰撞，碰撞后发生化学反应,仅有分子间的能量交换，不发生化学反应,H2+I2 HI,At room temperature the reaction proceeds very slowly.Only a
10、bout 1 in every 1013 collisions produces a reaction effective collision.,Each molecule undergoes about 1010 collisions per second,For example,The prerequisites of effective collision有效碰撞条件,(1)The molecules of the reactant must have enough energy.to stretch,bend,and ultimately break bonds,leading to
11、chemical reactions.,(1)反应物分子要有足够高的能量以克服分子或离子外层价电子云间的排斥力而充分接近，产生电子重排，使旧的化学键破裂，而形成新的化学键，即形成新的分子,The prerequisites of effective collision有效碰撞条件,(2)The colliding molecules with high energy must have a favorable orientation.,(2)具有高能量的分子间碰撞要有合适方向 分子在碰撞时，只有相互反应的原子碰撞在一起才可能发生化学反应，才能使分子或离子的反应部位间旧键破裂，新键形成,CO+N
12、O2 CO2+NO,Example,The orientation factor,Collisions must occur not only with sufficient energy but also with suitable orientation to cause reaction.,Conditions of effective collision,Activated molecule and activation energy活化分子与活化能,Activated molecule A molecule which has enough kinetic energy to tak
13、e part in a reaction.具有较大的动能并能够发生化学反应的反应物分子Activation energy(Ea)The minimum energy of the activated molecule or the minimum energy required to initiate a chemical reaction.活化分子所具有的最低能量与反应物分子的平均能量之差,Ea=E-E平均气体分子能量分布曲线,E平均,E,Ea,活化分子,动能,The characteristic of activation energy,(1)Ea must be positiveEa d
14、epends on the nature of the reactant and the route of the reaction.Ea is different in different reaction.Ea is different in different route of a reaction.(2)Ea independ on temperature and concentration.(3)Ea，the fraction of activated molecule,the reaction rate.,Relationship among activation energy,a
15、ctivated molecules and reaction rates,Ea2,Ea1,Activated molecules,Kinetic energy,In a given temperatureEa，activated molecules，V Ea，activated molecules，V,In general,(1)Ea400KJmol-1 反应极慢，难察觉The reaction rate is too slow to be feel.(2)Ea40 KJmol-1 反应极快，V几乎测不出来The reaction rate is too quick to be measur
16、ed(3)Ea 40400KJmol-1 一般化学反应在此范围 The Ea of a ordinary chemical reaction is in this range(4)Ea 60250KJmol-1 大部分反应在此范围 The Ea of most of chemical reaction is in this range,化学反应速率有效碰撞与能量及合适的碰撞方位关系,Z:碰撞频率 P:方位因子 f:能量因子,活化过渡状态理论,活化分子要按一定的取向发生碰撞发生碰撞的分子在彼此靠近至剩下几个分子的距离时，便引起内部结构的变化，即反应物分子间的价电子云发生相互穿透，先被活化成一种高
17、能量的中间过渡状态：活化络合物活化络合物分解成为产物或反应物,Transition-state theory,.The colliding activated molecule must have a favorable orientation.The inner structure of the colliding activated molecules change when they are close to each other at several molecular distance,that is,they form an activated complex with high
18、energy.The activated complex decomposed,it will change into the product or reactant.,活化络合物,活化络合物是一种处于过渡状态高能物质，它的价键结构处于旧键部分断裂、新键部分地形成，反应分子的动能大部分暂时地转变为势能的状态。它即可分解为反应物，也可形成生成物。当旧键削弱，新键加强时，就可以形成生成物，使整个体系的势能降低，而完成反应,Activated complex,The activated complex is not a stable molecule,and so it exists for onl
19、y an instant before flying apart one way or the other.It flies apart because its potential energy is higher than that of either the reactant or the product molecules.,Give out energy,Absorbenergy,A|A,B|B,+,2A-B orA-A+B-B,分子变化,能量变化,Kinetic energy,Potential energy,反应进程,Reactant,Activated complex,Produ
20、ctor reactant,The activated complex,Kinetic energy,H,反应进程的势能图,A2+B2Reactants,2ABProducts,Eb,Eb,Complex actived,Potential energy,E=H0 Exothermic reaction放热反应,活化能与反应热：H=Ea-Ea,EaEa放热反应,EaEa吸热反应,碰撞理论：简单的把化学反应看成活化分子间碰撞到合适部位即可发生，未考虑分子内部结构的变化活化过渡状态理论：考虑反应时分子内部结构的变化,优缺点,The effect of the concentration on th
21、e reaction rate,The expression of reaction rate is connective with the concentration directly.化学反应速率的表达式就与浓度直接有关,第三节 浓度对化学反应速率的影响,质量作用定律与速率方程式,1.质量作用定律(law of mass action)温度一定时，元反应的反应速率与各反应物浓度以计量系数为指数的幂的乘积成正比：aA+bBdD+eE v=k ca(A)cb(B)例如 v=k c(NO2)c(CO),The rate law(the law of mass action),When the t
22、emperature is fixed,the rate of a elementary reaction is proportional to the product of the molarity of the reactants,each raised to the power of its stoichiometric coefficient in the equation of the elementary reaction.,质量作用定律意义,温度一定时，增加反应物浓度可增大反应速率碰撞理论解释：增加了单位体积内活化分子分数，从而增加了分子间的有效碰撞次数，因而使反应速率增加,反应
23、速率方程式：表示反应物浓度与反应速率之间定量关系的数学式称为反应速率方程式。对元反应，根据质量作用定律可直接写出速率方程式：如上例 v=k c(NO2)c(CO)一般地aA+bBdD+eE v=k ca(A)cb(B)反应分子数=a+b,对复合反应，速率方程式需由实验测出如反应实验证明:v=k c(N2O5)研究表明上述反应分3步进行一般地 aA+bBdD+eE v=k cm(A)cn(B),3.反应速率常数(rate constant)反应：aA+bBdD+eE v=k cm(A)cn(B)k称为速率常数。与反应物浓度无关，与反应物本质及温度有关。k的物理意义：k在数值上等于各反应物浓度均为
24、1molL-1时的反应速率，故又称为比速率。即k=V/ca(A)cb(B)=V/1a1b=V,Explanation,(1)When the temperature is fixed,increasing creactant increases the rate of the reaction.(2)Collision theory:increasing the fraction of activated molecule in one volume unit increases the effective collision between the molecules,that will r
25、esult in increase of the rate of the reaction.,相同条件下，k愈大，表示反应的速率愈大。反映了反应的本质决定着反应的速率大小。k的量纲则根据速率方程式中浓度项上幂次的不同而不同。正逆向的反应k值不一样（因正逆向Ea不同）,Rate constant:k,(1)k:is only dependent on the nature of the reaction and the temperature,is independent on creactant.(2)Comparing the rates of different reactions und
26、er the same condition,the greater the k,the greater the rate of the reaction,that is to say,the rate of the reaction depends on the nature of the reaction.,(4)For a reversible reaction,different direction of the reaction has different k.(because their Ea are different),(3)The number of k amount to t
27、he number of rate when the concentration of every reactant is 1 molL-1,k is called specific rate,Attention,(1)The rate law only can be applied to elementary reaction.只有元反应，才可利用质量作用定律直接写出速率方程式,(2)The rate equation of a non-elementary reaction depends on the results of experiment.对非元反应，速率方程式由实验测定,Exam
28、ple,(3)The concentration of pure solid or pure liquid can be considered as 1.纯固态或纯液态反应物的浓度可看作常数，不必写入速率方程式(4)If the solvent was involved in a reaction in a dilute solution,the concentration of the solvent also can be considered as 1.稀溶液中溶剂参与的化学反应，可认为溶剂的浓度无变化，所以在速率方程式不必标出溶剂项,Attention,二、浓度对化学反应速率的影响,反
29、应类型：单分子反应、双分子反应、三分子反应反应级数：反应速率方程式中各反应物浓度方次之和反应：aA+bB=cC+dD化学反应速率方程：v=kcx(A)cy(B)反应级数 n=(x+y)反应级数可以是整数、零、分数或小数,n can be a integer,zero or fraction even negative.,The order of a reaction,It is the sum of the exponents on the concentration terms in the rate equation.,aA+bB=cC+dD,the order of reaction:n
30、=(x+y),若为元反应，则化学反应速率方程：V=kc a(A)c b(B)x=a；y=b反应级数=a+b=反应分子数如：NO2+CO=CO2+NO(元反应)V=kc(CO)c(NO2)反应级数=1+1=2反应分子数=1+1=2,aA+bB=cC+dD,If it is a elementary reaction:,The order of the reaction=a+b=molecularity of the reaction,x=a；y=b,e.g：NO2+CO=CO2+NO(elementary reaction),The order of the reaction=1+1=2=mol
31、ecularity of the reaction,Reaction mechanism of CO+Cl2=COCl2(phosgene)：,(1)Cl22Cl（quick）,(2)Cl2+Cl Cl3（quick）,(3)Cl3+COCOCl2+Cl（slow，rate-limiting step）,The overall reaction order is 1+3/2=2.5,反应级数与反应分子数的比较,1、一级反应,反应速率与反应物浓度的一次方成正比的反应 定积分得：一级反应方程式,First-order reactions,A first-order reaction is one
32、whose rate depends on the concentration of a single reactant raised to the first power,Definite integral,Y=mx+b,Characteristic of first-order reaction一级反应特征,lgc,t,0,The curve of first-order reaction,(1),(2)k is independent on the concentration.units：time-1 e.g.s-1,min-1,h-1(3)The half-life(t1/2)of a
33、 reaction is the time required for the concentration of a reactant to drop to one half of its initial value,c1/2=1/2c0.,、k的数值与浓度无关 量纲：时间-1 s-1、min-1、h-1、半衰期：当反应物浓度由CO 时所需时间为半衰期，用t1/2表示一级反应:c0 ln=kt1/2 t1/2=ln2/k 1/2cot1/2=0.693/k,Second-order reactions 二级反应,is one whose rate depends on the reactant
34、concentration raised to the second power or on the concentrations of two different reactants each raised to the first power.反应速率与反应浓度的二次方成正比,Definite integral,Y=mx+b,二级反应方程式,1/c,t,0,Slope=k,Characteristic of second-order reaction二级反应特征,The curve of second-order reaction,(1),(2)k Units:Lmol-1s-1（min-
35、1、h-1）(3)Half life(t1/2)：,Zero-order reactions反应速率与反应浓度无关,The rate of reaction is independent on the concentration of the reactant.,Definite integral,c=-kt+co,Y=mx+b,零级反应方程式,c,t,0,Characteristic of zero-order reaction,(1),Slope=-k,The curve of zero-order reaction,(2)kUnits:mol L-1s-1（min-1、h-1）(3)Ha
36、lf life(t1/2):,简单反应级数的特征,Summary,例1,反应2N2O5NO2+O2服从速率方程式V=kCN2O5，k=1.6810-2S-1,，如在一个5.00L的容器中放入2.5mol N2O5，在该温度下反应进行1min，问N2O5的剩余量及O2的生成量各为多少？解：从速率方程式V=kCN2O5知，反应符合 一级反应，且C0=2.5/5=0.5molL-1 根据一级反应方程式：lgC0/C=-kt/2.303 得lg0.5/C=-1.6810-260/2.303,c=0.182 molL-1则知N2O5剩余量n(N2O5)=cV=0.1825=0.91mol参加反应N2O5
37、量n(N2O5)=2.5-0.91=1.59mol设O2的生成量为x，则由反应方程式得：2N2O5NO2+O2 2 11.59 x生成氧的量：x=1.59/2=0.795 mol,Example,k is 1.6810-2s-1.Add 2.5mol N2O5 in a container of 5.00 L,please calculate how much N2O5 remained and how much O2 created after 1 min.,The rate equation of the reaction:2N2O5NO2+O2 is listed below,accor
38、ding to,The reaction is a first-order reaction:,Answer:,Assume the amount of O2 is x，2N2O5NO2+O2 2 1 1.59 xx=1.59/2=0.795 mol,N2O5 remains,n(N2O5)=cV=0.1825=0.91mol,The consumption of N2O5,n(N2O5)=2.5-0.91=1.59mol,例题2,在肺部血液中存在Hb+O2HbO2反应，对Hb及O2均为一级反应，血中Hb正常浓度8.010-6 molL-1，O2 1.610-6 molL-1,在37时，k=2
39、.1106mol-1L-1S-1计算：（1）在肺脏血液中，HbO2的生成速率?（2）在某种疾病中，V(HbO2)=1.110-4 molL-1S-1,为保持cHb不变，需输氧，此时c（O2）应高达多少才行？,解：由于 Hb及O2均为一级反应，所以(1)V(HbO2)=k cHb c（O2）=2.110-68.010-61.610-6=2.710-5 molL-1S-1(2)c（O2）=V(HbO2)/k cHb=1.110-4/2.110-68.010-6=6.510-6 molL-1,Example,There is a reaction:,in the blood of lung,it i
40、s first-order in Hb(hemoglobin)and also first-order in O2,the normal concentration of Hb and O2 are 8.010-6 molL-1 and 1.610-6 molL-1 respectively.at 37，k=2.1106mol-1L-1s-1,(1)please calculate how quickly the HbO2 was created in the blood of lung(2)if suffering some kind of disease,V(HbO2)=1.110-4 m
41、olL-1s-1 in order to keep cHb unchanged,how much the c(O2)is in oxygen therapy.,(1)V(HbO2)=k cHb c(O2)=2.110-68.010-61.610-6=2.710-5 molL-1S-1(2)c(O2)=V(HbO2)/k cHb=1.110-4/2.110-68.010-6=6.510-6 molL-1,Answer:,Because it is first order in Hb and also in O2,Unit of k is s-1，first-order reactiont1/2=
42、0.693/k=0.693/1.2410-4=5589.9s,Example,It is known that the k of a reaction is 1.2410-4s-1,please calculate its half life.,Answer:,例题3,已知2，2-偶氮二异丁睛的热分解反应：k=1.2410-4s-1,计算：（1）t1/2（2）转化率达75%时所需时间解：（1）单位为s-1，则为一级反应t1/2=0.693/k=0.693/1.2410-4=5589.9s（2）转化率达75%时，反应物的浓度c=c0-c075%=0.25c0则由lgc=-kt/2.303+lgc
43、0lg0.25c0=-1.2410-4t/2.303+lgc0t=11179.8s,第四节 温度对化学反应速率的影响,温度对化学反应速率的影响：不管是吸热反应还是放热反应，只要温度升高，均可使化学反应速率加快 如反应 H2+O2H2O 时才开始反应，需天才完全化合，时需小时，时可瞬时完全。,Temperature dependence of reaction rate,e.g.;H2+O2H2O begin to react finish in 80 days，finish in 2 hours finish in a twinkling,No matter the reaction is e
44、ndothermic or exothermic,its rate increase with increasing temperature.,The Arrhenius Equation,k is the rate constantEa is the activation energy R is the gas constant 8.314J/mol-KT is the absolute temperature.,A:Called the frequency factor,T2 T1 时 k2 k1，反应速率,当T一定 时,lgA一定,则Ea愈大,k愈小，v愈慢,对不同的反应,温度对反应速率
45、影响程度不同。当T相同 时，Ea愈大，斜率Ea/T愈大，k的变化程度愈大,Example,The following table shows the rate constants for the rearrangement of certain reaction at various temperatures.(a)from these data calculate the activation energy for the reaction.(b)what is the value of the rate constant at 430.0K?,：化学反应温度因子（温度系数）同一反应在低温时
46、较大（温度对化学反应速率影响程度大）高温时较小（高温时温度对化学反应速率影响程度小）,同一反应，在温度变化范围不大时，变化很小，对不同反应的变化值也不大，一般为24倍。,例,某反应在t时的化学反应速率为2molL-1s-1，当温度升高至(t+40)时，此时的化学反应速率有可能为：224 244=32512 molL-1s-1,At t,the rate of a reaction is 2molL-1s-1，when the temperature increases to(t+40),how about its rate?Its rate will increase to 224 244=3
47、2 512 molL-1s-1,Example,The reasons of temperature dependence of reaction rate,Collision theory:(1)Increasing temperature increases the fraction of activated molecules and then increases the rate.(leading reason)(2)Increasing temperature increases the kinetic energy of the molecules and then increas
48、es the rate.(secondary reason),e.g.the decomposition reaction of N2O5Ea=103KJmol-1 at 25 temperature:2535:Fraction of activated molecules:3.8 timesRate:3.8 timesAverage kinetic energy of molecules:only 3.3%.,温度使化学反应速率增大的原因：,碰撞理论：温度升高，使分子获得更多能量而成为活化分子。即活化分子分数增加，从而使反应速率加快(主要影响因素)温度升高,分子的动能增加(次要影响因素)气体
49、分子能量分布曲线：,例,N2O5分解反应的活化能为103KJmol-1温度:2535:活化分子百分率:3.8倍反应速率V 3.8倍分子平均动能仅 3.3%.,温度升高活化分子增多曲线图,动能,T1,T2,T2T1,Ea,High temperature,Low temperature,Molecular kinetic energy,Fraction of molecules having a given kinetic energy,Minimum energy needed for reaction,Ea,复习,化学反应速率受何种因素影响？用何指标表示？,Summary,Lower tem
50、perature,Higher temperature,Kinetic energy,Minimum energy needed for reaction,Ea,In a given temperatureEa，activated molecules，V Ea，activated molecules，V,第 五节 催化剂对化学反应速率的影响,一、催化剂catalyzer与催化作用catalysis催化剂是能改变化学反应的化学反应速率、在反应前后该物质的质量和化学组成不变的物质因催化剂存在而改变反应速率的现象称为催化作用,如CH3CHOCH4+CO未加催化剂时活化能为190 KJmol-1，加入
