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化学反应工程 第 2 章 均相反应动力学基础. 均相反应的基本概念( 1 )  均相反应:参与反应的各物质处于同一相内  均相反应动力学:研究各种因素如温度、压 力、催化剂、反应物组成等对反应速率、反 应产物分布的影响,并确定表达这些影响因 素与反应速率之间定量关系的速率方程。 均相反应动力学是解决工业均相反应器选型、操作与设.

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Presentation on theme: "化学反应工程 第 2 章 均相反应动力学基础. 均相反应的基本概念( 1 )  均相反应:参与反应的各物质处于同一相内  均相反应动力学:研究各种因素如温度、压 力、催化剂、反应物组成等对反应速率、反 应产物分布的影响,并确定表达这些影响因 素与反应速率之间定量关系的速率方程。 均相反应动力学是解决工业均相反应器选型、操作与设."— Presentation transcript:

1 化学反应工程 第 2 章 均相反应动力学基础

2 均相反应的基本概念( 1 )  均相反应:参与反应的各物质处于同一相内  均相反应动力学:研究各种因素如温度、压 力、催化剂、反应物组成等对反应速率、反 应产物分布的影响,并确定表达这些影响因 素与反应速率之间定量关系的速率方程。 均相反应动力学是解决工业均相反应器选型、操作与设 计计算的理论基础。

3  反应速率表示方法  The Rate Expressions  均相反应速率定义:单位时间内,单位体 积反应混合物中某一组分 i 的反应量(或生 成量)。

4 The minus sign means disappearing  Suppose a single-phase reaction A P The most useful measure of rate for A is then The Rate Expressions

5 例如,有一简单反应 A P

6 Caution 1 & 2  (-r A ) 为一整体符号,恒为正值  Suppose a single-phase reaction aA+bB pP+sS The rates of all materials are related by

7 Caution 3  According to the rate equation

8 Caution 4  Or For the ideal gases where In a constant-volume system the measure of reaction rate of component i becomes

9 Caution 5  The rate of reaction in its various forms is defined as follows  Based on unit volume of reaction fluid

10  Based on unit mass of solid in fluid-solid systems Based on unit interfacial surface in two-fluid systems or based on unit surface of solid in gas-solid systems

11  Based on unit volume of solid in gas-solid systems Based on unit volume of reactor

12  In homogenous systems, the volume of fluid in the reactor is often identical to the volume of reactor. In such case V and V R are identical and are used interchangeable.  In heterogeneous systems all above definitions of reaction rate encountered, the definition used in any particular situation often being a matter of convenience.

13  These intensive definitions of reaction rate are related by W —— 固体(催化剂)的重量 S —— 相界面面积 V P —— 固相(催化剂)占体积; V —— 液相占体积; V R ­­—— 反应器的有效体积, V R = V P + V 。

14 The conversion 转化率  Suppose that N A0 is initial amount of A in reactor at time t=0, and that N A the amount present at time t. then the conversion of A in the constant volume system is given by

15  微分上式 for the constant volume system 式中 C A0 ——A 的初始浓度。

16 Kinetics of Homogeneous Reaction  均相反应:反应物系中,所有反应物及生成 物(包括催化剂在内)都处于同一相中。  影响反应速率的参数:浓度、温度、催化剂 等,因此,反应速率与上述这些参数成函数 关系。

17  Suppose a homogeneous irreversible reaction The rate of progress of the reaction can be approximated by an expression of the following type

18  n= α+β ( n 为总反应级数)  Where α,β are not necessarily related to the stoichiometric coefficients. We call the powers to which the concentrations are raised the order of the reaction. Thus, the reaction is αth order with respect to A βth order with respect to B nth order overall

19  反应的级数:动力学方程中浓度项的幂数, (如  ,  ),由实验确定的常数。  基元反应级数  ,  等于计量系数值,  =a ,  =b  非基元反应,应通过实验来确定  级数的大小反映了该物料浓度对反应速率影响 的程度。级数越高,表明该物料浓度的变化对 反应速率的影响越显著,如是负值,表明抑制 反应,使反应速率下降。

20 反应速率常数 k Rate Constant k  When the rate expression for a homogenous chemical reaction is written in the form in book, the dimensions of the rate constant k for n th -order reaction are [time] -1 [concentration] 1-n which for a first-order reaction becomes simply (time) -1

21  反应速率常数值的大小决定了反应速率的高低 和反应进行的难易程度  不同反应有不同的速率常数,  对于同一反应, k 随温度、浓度和催化剂的变化 而变化。其中,温度是影响反应速率的主要因 素之一。

22 Arrhenius’ Law  For many reactions, and particularly elementary reaction, the rate expression can be written as a product of a temperature-dependent term and a composition-dependent term, or (-r A )=f 1 (temperature) · f 2 (composition) =k·f 2 (composition)

23  For such reaction the temperature-dependent term, the reaction rate constant, has been found in practically all cases to be well represented by Arrhenius’ Law: k=k 0 e -E/RT Where: k 0 frequency or pre-exponential factor E Activation energy At two different temperatures, Arrhenius’ Law indicates that

24 Activation energy and temperature dependency  From Arrhenius’ Law a plot of ln k vs 1/T gives a straight line, with large slope for large E and small slope for small E  Reactions with high activation energies are very temperature- sensitive; reaction with low energies are relatively temperature- insensitive.  Any given reaction is much temperature-sensitive at a low temperature than at a high temperature.  From the Arrhenius’ Law, the value of frequency factor k 0 does not affect the temperature sensitive.

25 等温恒容过程 反应动力学方程的建立  实验得到不同反应时间的各物料浓度数据  或者测定压力、密度、折射率等换算成各物料 的浓度  数据处理,有积分法和微分法

26 单一反应 The Single Reactions 一级反应 ( 不可逆单分子一级反应) Irreversible Unimolecular-Type First-Order Reaction  Consider the reaction A  P at constant- volume and constant-temperature processes, the first-order rate equation is For this reaction. Separating and integrating we obtain x A =(C A0 -C A )/C A0

27  A plot of ln(1/1-X A ) or ln(C A0 /C A ) vs. t, as shown on the right, gives a straight line through the origin for this form of rate of equation.

28 二级反应 Irreversible Bimolecular-Type Second- Order Reaction  Consider the reaction A+A  P with corresponding rate equation Which on integration yields

29  Consider the reaction A+B  P. The definition second-order differential equation becomes Noting that the amounts of A and B that have reacted at any time t are equal and given by C A0 x A, we may write above equation in terms of x A as Letting M=C B0 /C A0 be the initial molar ratio of reaction, we obtain

30 Which on separation and formal integration becomes After breakdown into partial fractions, integration and rearrangement, the final result in a number of different form is

31  The figures show two equivalent ways of obtaining a linear plot between the concentration function and time for this second-order rate law.

32 n 级反应 Irreversible nth-order Reaction  Consider the reaction nA  P. When the mechanism of reaction is not known, we often attempt to fit the data with n th -order rate equation of the form Which on separation and integration yields

33  The order n cannot be found explicitly from the equation, so a trial-and error solution must be made. This is not too difficult, however. Just select a value for n and calculate k. The value of n which minimizes the variation in k is the desired value of n.

34 气相反应A  3P 为一级反应,速度常数k=0.5min -1 ,反应在恒容 间歇式反应器中进行,求1min后体系的总压,进料状况如下:  a) 纯A,0.1Mpa;  b) 纯A,1Mpa;  c) 10%的A和90%的I(惰性气体)混合物,1MPa  解:

35 对比( 1 )、( 2 ),得

36 当 t=0 ,  =  0 积分( 3 ), 得

37 等温恒容不可逆反应的速率方程积分式

38 可逆反应 Reversible Reaction  可逆反应是指正方向、逆方向同时以显著速度 进行的反应,也叫对峙反应。  First-order Reversible reaction Let us consider the opposed unimolecular-typed reaction K C = K = k 1 /k 2 equilibrium constant Starting with a concentration ratio M=C R0 /C A0 the rate equation is

39 Now at equilibrium dC A /dt =0. Hence we find the equilibrium constant of A at equilibrium conditions to be

40  Combining the above equations we obtain, in terms of the equilibrium conversion This may be looked on as pseudo first-order irreversible reaction which on integration gives

41  A plot of –ln(1-x A /x Ae ) vs. t, as shown in the Fig., gives a straight line

42 二级可逆反应 Second-order Reversible Reaction  For the bimolecular-type second-order reaction With the restriction that C A0 =C B0 and C R0 =C S0, the integrated rate equation for A and B is as follows A plot can then be used to test the adequacy of this kinetics

43 等温恒容可逆反应速率方程积分式 (产物起始浓度为 0 )

44 动力学的实验和数据处理 Kinetics Experiment and its Data Analysis  A rate equation characterizes the rate reaction, and its form may either be suggested by theoretical consideration or simply the result of an empirical curve-fitting procedure. In any case, the value of the constants of the equation can only be found by experiment; predictive methods are inadequate at present.

45  The determination of the rate equation is usually a two-step procedure; first the concentration dependency is found at fixed temperature and then the temperature dependence of the rate constants is found, yielding the complete rate equation.

46  Equipment by which empirical information is obtained can be divided into two types, the batch and flow reactors. The batch reactor (釜式反应器 或间歇式反应器) is simply a container to hold the contents while they react.

47  The experimental batch reactor is usually operated isothermally and at constant volume because it is easy to interpret the results of such runs. This reactor is a relatively simple device adaptable to small- scale laboratory set-ups (装 置), and it needs but little auxiliary (辅助设施) equipment or instrumentation. Thus, it is used whenever possible for obtaining homogenous kinetic data.

48  The flow reactor( 连续式反应器 ) is used primarily in the study of the kinetics of heterogeneous( 非均相 ) reactions. Planning of experiments and interpretation of data obtained in flow reactors are considered in later chapters.

49  There are two procedures for analyzing kinetic data, the integral and the differential methods( 积分法和微分法 ).

50  In the integral method of analysis we guess a particular form of rate equation and, after appropriate integration and mathematical manipulation, predict that the plot of a certain concentration function versus time should yield a straight line. The data are plotted, and if a reasonably good straight line is obtained, then the rate equation is said to satisfactorily fit the data 用积分法分析实验数据 Analyzing the experiment data by integral method

51 Half-time t 1/2 method 半衰期法  For a single nth-order reaction  Integrating for n  1 gives  Defining the half-time of reaction,t 1/2, as the time needed for the concentration of reactants to drop to one-half the original value, we obtain

52  This expression shows that a plot of log t 1/2 vs. log C A0 gives a straight line of slope 1-n, as shown in Fig 2.7.1

53  The half-time method requires a series of runs, each at a different initial concentration, and shows that the fractional conversion in a given time rises with increased concentration for orders greater than one, drops with increased concentration for orders less than one, and is independent of initial concentration for reactions of first order.

54 用微分法分析实验数据 Analyzing the experiment data by differential method  The differential method of analysis deals directly with the differential rate equation to be tested,evaluating all terms in the equation including the derivative dC i /dt, and testing the goodness of fit of the equation with experiment.

55 用微分法分析实验数据  微分法求动力学方程是直接利用某一类动力学方 程的微分式,以反应速度对浓度的函数作图,然 后与实测的数据相拟合的一种方法。一般也是设 法把图形线性化,把实验数据代人。若得出一直 线,便认为所假设的动力学方程是正确的。否则, 重新选定另一个动力学方程进行猜算,直到得出 一条直线为止。

56  The procedure is as follows.  1) Plot the C A vs. t data, and then by eye carefully draw a smooth curve to represent the data. This curve mostly likely will not pass through all the experimental points/  2) Determine the slope of this curve at suitably selected concentration values. These slopes -dC A /dt=(-r A ) are the rates of reaction at these compositions

57  3) Now search for a rate expression to represent this (-r A ) vs. C A data, either by  a) picking and testing a particular rate form, (-r A )=kf(C A )  b) testing an nth-order form (-r A )=kC A n by taking logarithms of the rate equation

58  在处理实验数据时,最小二乘法特别适用于如下 形式的方程式: k,α,β是待测定的,为此,可对上式取对数: 写成:

59  根据最小二乘法法则: 当a 0 、a 1 、a 2 取得最佳值时,有:

60  也就是用 分别对 a 0,a 1,a 2 求偏导数并令其为零,可以得到方程组 | : 这是一个三元一次方程组,可以用行列式求解

61

62 复合反应 Multiple Reaction  Single reaction requires only one rate expression to describe its kinetics behavior whereas multiple reactions require more than one rate expression.  Multiple reactions can be considered to be combinations of two primary types: parallel reactions and series reactions.

63 收率: Yield, fraction yield 得率: Operation yield

64 选择性: Selectivity

65 一级平行反应 First-order Parallel Reaction  Consider the decomposition of A by either one of two paths: With corresponding rate equations

66 Integrating the equation (1)at t=0,C A =C A0 , C P =C P0 , C S =C S0 Integrating the equation (2)

67 Integrating the equation (3)

68 一级连串反应 Irreversible First-order Reaction in series We consider consecutive unimolecular-type first-order reaction such as Whose rate equations for the three components are

69  Let us start with a concentration C A0 of A,no P and S present, and see how the concentrations of the components change with time. By integration of Eq.(1) we find the concentration of A to be To find the change concentration of P, substitute the concentration of A from Eq.(4) into the differential equation governing the rate of change R, Eq.(2); thus

70  Which is a first-order linear differential equation( 一阶线形微分方程) of the form By multiplying through with the integrating factor( 积分因子 )R= the solution is

71  Applying this general procedure to the integration of Eq.(5), we find that the integrating factor is. The constant of integration is found to be –k 1 C A0 /(k 2 -k 1 ) from the initial conditions C R0 =0 at t=0, and the final expression for the changing concentration of P (6)

72  Noting that there is no change in total number of moles, the stoichiometry relates the concentrations of reacting components by which with Eqs.(4) and (6) gives Thus, we have found how the concentrations of components of A,P,and S vary with time.

73  If k 2 is much larger than k 1, Eq.(7) reduces to In other words, the rate is determined by k 1 or the first step of the two-step reaction. If k 1 is much larger than k 2, then Which is a first-order reaction governed by k 2, the slower step in the two –step reaction. Thus, in general, for any number of reactions in series it is the slowest step that has the greatest influence on the overall reaction rate (7)

74

75 t opt

76  As may be expected, the value of k1 and k2 also govern the location and maximum concentration of P. This may be found by differentiating Eq.(6) and setting dC P /dt=0. The time at which the maximum concentration of P occurs is thus  The maximum concentration of P is found by combining Eqs.(6) and the above to give

77  If k 1 =k 2, the concentration of P and S should be re-found Page 16 figure 2.4.3

78 自催化反应 Autocatalytic Reaction  A reaction in which one of the products of reaction acts as catalyst is called an autocatalytic reaction. The simplest such reaction is A+P  P+P for which the rate equation is Because the total number of moles of A and P remain unchanged as A is consumed, we may write that at any time C 0 =C A +C P =C A0 +C P0 =constant

79  Thus, the rate equation becomes Rearranging and breaking into partial fraction, we obtain

80  For an autocatalytic reaction in batch reactor some product P must be present if the reaction is to proceed at all. Starting with a very small concentration of P, we see qualitatively that the rate will rise as P is formed. At other extreme, when A is just about used up the rate must drop to zero. This result is given in fig.below, which shows that the rate follows a parabola, which a maximum where the concentrations of A and P are equal.

81

82 反应前后分子数变化的气相反应 Isothermal Gas-phase Reaction with varying-molecular  变容反应系统:反应前后分子数发生变化,如果过 程恒压,则为变容反应系统 ;  恒容变压过程:分于数发生变化的气相反应,如果 反应器的容积恒定,其结果使反应系统的总压变化, 称之为恒容变压过程。

83 膨胀因子 δ 和膨胀率 ε A 膨胀因子 δ A 的意义是反应物 A 每消耗 lmol 时,引起整个物系总 物质的量的增加或减少值。 aA+bB  pP+sS  δ A 的大小只取决于化学计量式本身,与是否存在惰性气体无关;  δ A 表示反应过程中 mol 数的变化,与体系体积变化无关;  对于复杂反应,其数值大小随转化率而变化,没有明确意义;  δ A 数值可正可负,可以为分数。

84  A capillary tube reactor can be used for isothermal constant pressure operation, of reactions having a single stoichiometry. For such system the volume is linearly related to the conversion, or  V=V 0 (1+ ε A x A ) whereε A is the factional change in volume of the system between no conversion and completely conversion of reactant A.

85  As an example of the use of ε A and δ A, consider the isothermal gas-phase reaction A  4R, (a) by starting with pure reactant A, ε A =(4-1)/1=3 δ A =(4-1)/1=3 (b) But with 50% inerts present at the start, two volumes of reactant mixture yield, on completely conversion, five volumes of product mixture. In this case ε A =(5-2)/2=1.5 δ A =(4-1)/1=3

86 等温等压变容过程 Isothermal constant pressure varying-volume processes 例题:设有一级气相反应 A  2P , 分别在等容、等 压条件下进行到 x A =0.5 ,求二者的残余 A 组分浓度 及反应速度。 等容:体积不变,压力增加 50% C A =C A0 (1-x A )=C A0 /2 (-r A )=kC A =kC A0 /2 等压:压力不变, 体积增加 50% C A =C A0 /3 (-r A )=kC A =kC A0 /3

87  Noting that n A =n A0 (1-x A ).  We have which is the relationship between conversion and concentration for isothermal varying-volume systems satisfying the linearity assumption

88  The rate of nth-order reaction (disappearance of component A), is  For the first-order reaction n=1

89 等温等容变压过程 Isothermal Constant- Volume Varying-pressure processes  At Isothermal Constant-Volume, ε A =0,δ A may not be equal to 0

90 [例题2.6-1](p 19) 总压法测定气相反应的速度常数。 设在一间歇反应器内进行等温等容反应 已知速率方程:

91 由式( 2.6-16 )得 代入速率式 将上式积分,得

92 作业  1 、 3 、 6 、 8 、 10  交作业时间: 3 月 28 日


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