ChapterChemicalKinetics(化学动力学—反应速率)

1、1Chapter 3 Chemical Kinetics(化学动力学化学动力学反应速率反应速率) 3.1 The Concept of Chemical Reaction Rate3.2 The Theories of Reaction Rate(反应速率理论)3.3 The Dependence Factor of Reaction Rate(影响反应速率的因素) 1. Concentration (浓度) 2. Temperature (温度) 3. Catalysis(催化剂)23.1 The Concept of Chemical Reaction RateChemical kinet

2、ics(化学动力学化学动力学) is the study of how rapidly chemical reactions occur. The reaction rate is defined in terms of the rate of appearance of a product(产物) or the rate of disappearance of a reactant(反应物). We can define the average rate of reaction as the change in moles of B (or A) per unit time. For the

3、 time interval(间隔) between 0 and 20 minutes, the average rate is?for the reaction A B 3We can also express this average rate in terms of the disappearance of A. Because the number of moles of A decreases, it is necessary to introduce a minus sign(减号) into the formula in order for rate to be expresse

4、d as a positive quantity.The rate of appearance of B is equal to the rate of disappearance of A because they have a one-to-one(1比1) stoichiometric(化学计量) relationship. V(average) = B/ t = (0.46-0)mol/(20-0)min = 0.023 mol/min V(average) = A/ t = (0.54-1.00)mol/(20-0)min = 0.023 mol/min 4Most often, r

5、eaction rates are expressed in units of concentration, usua-lly M/s (Ms1), rather than moles.(M =molL-1) Consider the reactionfor the reaction aA + bB cC + dD Although we can calculate the average rate, it is usually desirable to know the instant instantaneous rate (瞬间速度)of a reaction at a particula

6、r time. V(average) = C4H9Cl/ t = C4H9OH/ t 5EXAMPLEEXAMPLEAt 340K2N2O5(g)4NO2 (g) +O2 (g) t/min 0 1 2 3 4 5C(N2O5)/molL-1 1.00 0.70 0.50 0.35 0.25 0.17From t = 0 to t = 2minaverage V(N2O5) = -(0.50 1.00) molL-1 /(2-0)min = 0.25molL-1 min -1 average V(reaction) = V(N2O5)/ 2t = V(NO2)/ 4t = V(O2)/ t =

7、 0.25molL-1 min -16Molecules must collide to react, the more collisions per second, the faster the reaction occurs. The higher the absolute temperature, the more collisions there will be per second . Higher temperature contributes to more forceful collisions because molecules are moving faster. More

8、 collisions will occur at higher reactant concentrations.3.2 The Theories of Reaction Rate(反应速度理论反应速度理论)碰撞模型碰撞模型-Lewis,1918)7EaEm*E0Em Distribution of molecular energy of gas (气体分子能量分布图气体分子能量分布图) Maxwell分布曲线分布曲线Em 气体分子平均能量气体分子平均能量E0 活化分子最低能量活化分子最低能量Em* 活化分子平均能量活化分子平均能量活化能活化能 Ea = Em* - EmE E0 活化分子活化

9、分子 General the Ea are 40 400 kJ mol-1. In constant temperature, the higher the Ea, the slower the rate of reaction.8The colliding molecules must have a total kinetic energy(动能) equal to or greater than the activation energy, Ea. The activation energy is the mini-mum(最小) energy required to initiate(开

10、始) a chemical reaction. It can be thought of as the hill in below. Regardless of whether the elevation(平面) of the ground on the other side is lower than the original position, there must be enough energy imparted(给予) to the golf ball to get it over the hill. Activation energy(活化能）活化能）- Ea9In order f

11、or a collision to result in a reaction, it must be an effective collision(有效碰撞). The molecules must be moving fast enoughand oriented properly so that a reaction can occur.- (速度与方向) Effective collision(有效碰撞有效碰撞)10Figure in above illustrates the case for a chemical reaction. The difference between th

12、e energy of the original molecule and the highest point in the reaction pathway is called the activation energy. At the point of greatest energy, the species present is called the transition state(过渡状态过渡状态). It is also referred to as an activated complex(活化络合物活化络合物). Only molecules with sufficient k

13、inetic energy can get over the activation energy “hill.”methyl isonitrile甲基异腈过渡状态模型过渡状态模型 Eyring,1930s)113.3 The Dependence Factor of Reaction Rate(影响反应速率的因素影响反应速率的因素)Reaction rates depend on:The concentrations of the reactants(反应物) : Most chemical reactions proceed faster if the concentration of on

14、e or more of the reactants is increased. The temperature at which the reaction occurs: The rates of chemical reactions increase as the temperature increases. The presence of a catalyst(催化剂): The rates of many reactions can be increased by adding a substance known as a catalyst. The surface area of s

15、olid or liquid reactants: Reactions that involve solids often proceed faster as the surface area of the solid is increased.12The Dependence of Rate on Concentration1. Elementary reaction(基元反应基元反应)Most of reactions are not elementary reaction, they are complex reactions by two or more elementary reac

16、tion step from the reactant to products. The rate equation can only get by experimental results. The reaction in which reactants change to products by one step only.V = kAaBbaA + bB gG + hH For Elementary reactionIt also designated as Mass Action Law(质量作用定律), whichOnly suitable for Elementary reacti

17、on,In the complex reactions, the slowest step designated as Rate-Determining Step(速控步骤). Rate equation(速率方程). k is rate constant(速率常数) 132. Rate constant and reaction order(速率常数与反应级数速率常数与反应级数)The data in the table indicate that the rate of this reaction does depend on the concentrations of both reac

18、tants. In fact, the rate appears to be proportional (比例)to each of the concentrations. 14The constant k is the rate constant (速率常数). Using the data from the experiment number 2 In general, reactions have the rate lawRate = kreactant 1mreactant 2n15The exponents(指数) m and n must be determined experi-

19、mentally. The exponent to which a reactants concentration is raised in the rate law is referred to as(称为) the reaction order (反应级数) with respect to that reactant. In the ammonium ion and nitrite(亚硝酸盐) ion example, the exponents in the rate law were both 1. We would say that the reaction is first ord

20、er in ammonium ion and first order in nitrite ion. The overall reaction order (总反应级数) is the sum of the exponents in the rate law. We would then say that the ammonium and nitrite reaction is second order overall(二级反应).The rate law for this reaction is rate = kAmBn. We will use the experimental data

21、NOT equation to determine the values of m, n, and k.16when the concentration of A remains constant and the concentration of B is doubled, the rate doubles. This tells us that the reaction is first order in B.When the concentration of B remains constant and the concentration of A is doubled, the rate

22、 increases by a factor of four.overall reaction order (总反应级数总反应级数) ism + n = 2 + 1 = 3 (三级反应三级反应)17Units(单位) of the rate constant depend on the overall order of the reaction. In the ammonium and nitrite example the units of k are M 1s1. These are the units necessary to cancel properly and give the r

23、ate in units of Ms1.Overall Order of ReactionUnits of k0Ms 11s12M 1s13M 2s1QUESTION：How to use the units of k to conclude the overall order of a reaction?18EXAMPLEWe knowed by experimentH2PO2-(aq) + OH- (aq) HPO3-(aq) + H2 (g) c (H2PO2-)/moll-1 ; c (OH- )/moll-1; v /moll-1s-1(1) 0.10 0.10 5.3010-9(2

24、) 0.50 0.10 2.6710-8(3) 0.50 0.40 4.2510-7AnswerSuppose: V = k cx (H2PO2-) cy (OH- )From (1): 5.3010-9moll-1s-1 = k (0.10 moll-1)x(0.10 moll-1)yFrom (2): 2.6710-8moll-1s-1 = k (0.50 moll-1)x(0.10 moll-1)yFrom (3): 4.2510-7moll-1s-1 = k (0.50 moll-1)x(0.40 moll-1)y(2) (1), we can get: x = 1; (3)(2),

25、we can get: y = 2So the Order of Reaction is 3; V = k c (H2PO2-) c2(OH- )You can get the k in any equation abovek =5.3010-6( l2mol-2s-1)19The relationship between reaction rate and temperature is expressed quantitatively(数量上) by the Arrhenius equation(阿累尼乌斯方程)k - rate constantA - frequency factor(频率

26、因子), a sort of constant associated with a particular reactionEa - activation energy (活化能)R - gas constantT - absolute temperature Arrhenius equation(阿累尼乌斯方程阿累尼乌斯方程-1889) Vant Hoff rule(范荷夫规则范荷夫规则)k = Ae-Ea/RTrkkVVtttt1010r temperature modulus(系数) of reaction rate, usually r = 24 20We can write this

27、equation for two different temperatures. Subtracting(减法) one from the other giveswhich simplifies to give us a useful equationIn general, as temperature increases, the rate of a chemical reaction increases too. Taking the natural log(自然对数 ) of both sides gives an equation with the form of a straight

28、 line(直线方程). lnk1= -Ea/RT1+lnA; lnk2= -Ea/RT2+lnAlnk2 - lnk1 = (-Ea/RT2) - (-Ea/RT1)ln(k2 /k1) = Ea/R (1/T1 - 1/T2)log(k2 /k1) = Ea/2.303R (1/T1 - 1/T2)21EXAMPLEQ(question) What is the activation energy Ea ? log(k2 /k1) = Ea/2.303R (1/T1 - 1/T2)We knowed by experimentCO(g) + NO2(g)NO(g) + CO2(g) T /

29、 K k /moll-1s-1 600 0.0280 650 0.220A(answer)log(0.220/0.0280) = Ea/2.3038.314 (1/600 - 1/650)Ea = 134 kJ mol-1ATTENTIONR = 8.314 J K-1 mol-122催化作用催化作用)A catalyst(催化剂) is a substance that is added to a chemical reaction to increase the rate of that reaction. The decomposition of hydrogen peroxide (过

30、氧化氢的分解) occurs very slowly.A catalyst speeds up a reaction by lowering the activation energy. It does this by providing a different mechanism (机制) by which the reaction can occur. In the case of the decomposition of peroxide, instead of taking place in one very slow step, the reaction takes place (i

31、n the presence of the NaBr catalyst) in two steps, both of which are relatively fast. 23Energy of reaction pathwayreaction pathwayEEa1Ea2rHrHEa2Ea1Ereaction pathwayrH = Ea1 Ea2rH 0exothermic reaction(放热反应放热反应) rH 0endothermic reaction(吸热反应吸热反应) 24The energy profiles for both the catalyzed and uncatalyzed reactions of decomposition of hydrogen peroxide are shown in Figure below. 2H2O+O2+2Br-+2H+25Catalyst (催化剂)催化剂只改变动力学速度（具体途径），并未改变热力学的状态，即始、终态，(rH and rG)，因此催化剂不能用来改变热力学判断不能进行的反应（rG0）26A homogeneous catalyst(均相催化剂均相催化剂) is one that exists in th