1、化 学 反 应 工 程,Chapter 2 Kinetics of Homogeneous Reactions,Simple Reactor Types,Ideal reactors have three ideal flow or contacting patterns shown in Fig.2.1. We very often try to make real reactors approach these ideals as closely as possible. Because: (1) Ideal reactors are easy to treat (it is simple
2、 to find their performance equations) ; (2) One of them often is the best pattern possible (it will give the most of whatever it is we want).,化 学 反 应 工 程,化 学 反 应 工 程,The Rate Equation,Suppose a single-phase reaction. The most useful measure of reaction rate for reactant A is then,(1),In addition, th
3、e rates of reaction of all materials are related by,化 学 反 应 工 程,Experience shows that the rate of reaction is influenced by the composition and the energy of the material. By energy we mean the temperature (random任意的 kinetic energy of the molecules), the light intensity within the system (this may a
4、ffect the bond energy between atoms), the magnetic field intensity, etc. Ordinarily we only need to consider the temperature, so let us focus on this factor. Thus, we can write,(2),化 学 反 应 工 程,2.1 CONCENTRATION-DEPENDENT TERM OF A RATE EQUATION,Before we can find the form of the concentration term i
5、n a rate expression, we must distinguish between different types of reactions. This distinction(区别) is based on the form and number of kinetic equations used to describe the progress of reaction. Also, since we are concerned with the concentration-dependent term of the rate equation, we hold the tem
6、perature of the system constant.,化 学 反 应 工 程,Single and Multiple Reactions,When a single stoichiometric equation and single rate equation are chosen to represent the progress of the reaction, we have a single reaction.When more than one stoichiometric equation is chosen to represent the observed cha
7、nges, then more than one kinetic expression is needed to follow the changing composition of all the reaction components, and we have multiple reactions.,化 学 反 应 工 程,series reactions,parallel reactions, which are of two types,R,A,S,and,Multiple reactions may be classified as:,competitive,side by side
8、,化 学 反 应 工 程,A more complicated example is,Here, reaction proceeds in parallel with respect to B, but in series with respect to A, R, and S.,Multiple reactions,化 学 反 应 工 程,Elementary and Nonelementary Reactions,Consider a single reaction with stoichiometric equation,If we postulate(假定) that the rate
9、-controlling mechanism involves the collision or interaction of a single molecule of A with a single molecule of B, then the number of collisions of molecules A with B is proportional to the rate of reaction. But at a given temperature the number of collisions is proportional to the concentration of
10、 reactants in the mixture; hence, the rate of disappearance of A is given by,化 学 反 应 工 程,Such reactions in which the rate equation corresponds to a stoichiometric equation are called elementary reactions. When there is no direct correspondence between stoichiometry and rate, then we have nonelementa
11、ry reactions. The classical example of a nonelementary reaction is that between hydrogen and bromine,(3),which has a rate expression,化 学 反 应 工 程,Nonelementary reactions are explained by assuming that what we observe as a single reaction is in reality the overall effect of a sequence of elementary re
12、actions. The reason for observing only a single reaction rather than two or more elementary reactions is that the amount of intermediates formed is negligibly(可以忽略的)small and, therefore, escapes detection.,化 学 反 应 工 程,Molecularity and Order of Reaction,The molecularity(分子数) of an elementary reaction
13、 is the number of molecules involved in the reaction, and this has been found to have the values of one, two, or occasionally three.,Note that the molecularity of a reaction must be an integer(整数) because it refers to the mechanism of the reaction, and can only apply to an elementary reaction.,化 学 反
14、 应 工 程,Molecularity and Order of Reaction,Often we find that the rate of progress of a reaction, involving, say, materials A, B, . . . , D, can be approximated(近似) by an expression of the following type:,(4),Where a, b, . . . , d are not necessarily related to the stoichiometric coefficients.,化 学 反
15、应 工 程,We call the powers(幂数) to which the concentrations are raised the order of the reaction. Thus, the reaction is,ath order with respect to A,bth order with respect to B,nth order overall,Since the order refers to the empirically(经验地) found rate expression, it can have a fractional value and need
16、 not be an integer.,For rate expressions not of the form of Eq. 4, such as Eq. 3, it makes no sense to use the term reaction order.,化 学 反 应 工 程,Rate Constant k,When the rate expression for a homogeneous chemical reaction is written in the form of Eq. 4, the dimensions of the rate constant k for the
17、nth-order reaction are,(time)-1(concentration)1-n,which for a first-order reaction becomes simply,(time)-1,(6),(5),(4),化 学 反 应 工 程,Representation of an Elementary Reaction,In expressing a rate we may use any measure equivalent to concentration (for example, partial pressure), in which case,For brevi
18、ty(简短), elementary reactions are often represented by an equation showing both the molecularity and the rate constant. For example,Whatever measure we use leaves the order unchanged; however, it will affect the rate constant k.,(7),-,化 学 反 应 工 程,It represents a bimolecular irreversible reaction with
19、 second-order rate constant k1, implying that the rate equation of the reaction is,It would not be proper to write Eq. 7 as,for this would imply that the rate expression is,化 学 反 应 工 程,On the other hand, we should note that writing the elementary reaction with the rate constant, as shown by Eq.7, ma
20、y not be sufficient(充分的) to avoid ambiguity(含糊,不明确). At times it may be necessary to specify(指定,详细说明) the component in the reaction to which the rate constant is referred. For example, consider the reaction,(8),化 学 反 应 工 程,If the rate is measured in terms of B, the rate equation is,If it refers to D
21、, the rate equation is,Or if it refers to the product T, then,But from the stoichiometry ,(9),hence,化 学 反 应 工 程,In Eq.8, which of these three k values are we referring to? We cannot tell. Hence, to avoid ambiguity when the stoichometry involves different numbers of molecules of the various component
22、s, we must specify the component being considered.,To sum up, the condensed(被浓缩的,精简的) form of expressing the rate can be ambiguous. To eliminate any possible confusion, write the stoichiometric equation followed by the complete rate expression, and give the units of the rate constant.,Question:,Conc
23、lusion:,化 学 反 应 工 程,Representation of a Nonelementary Reaction,A nonelementary reaction is one whose stoichiometry does not match its kinetics. For example,This nonmatch shows that we must try to develop a multistep reaction model to explain the kinetics.,化 学 反 应 工 程,Kinetics Model for Nonelementary
24、 Reactions,To explain the kinetics of nonelementary reactions we assume that a sequence of elementary reactions is actually occurring but that we cannot measure or observe the intermediates formed because they are only present in very minute quantities. Thus, we observe only the initial reactants an
25、d final products, or what appears to be a single reaction. For example, if the kinetics of the reaction,化 学 反 应 工 程,indicates that the reaction is nonelementary, we may postulate a series of elementary steps to explain the kinetics, such as,where the asterisks(星号) refer to the unobserved intermediat
26、es. To test our postulation scheme, we must see whether its predicted kinetic expression corresponds to experiment.,The types of intermediates we may postulate are suggested by the chemistry of the materials. These may be grouped as follows.,化 学 反 应 工 程,Free Radicals. Free atoms or large fragments o
27、f stable molecules that contain one or more unpaired electrons are called free radicals. The unpaired electron is designated by a dot in the chemical symbol for the substance. Some free radicals are relatively stable, such as triphenylmethyl,C,化 学 反 应 工 程,but as a rule they are unstable and highly r
28、eactive, such as,Ions and Polar Substances. Electrically charged atoms, molecules, or fragments of molecules, such as,are called ion. These may act as active intermediates in reactions.,化 学 反 应 工 程,Molecules. Consider the consecutive reactions,Ordinarily these are treated as multiple reactions. Howe
29、ver, if the intermediate R is highly reactive its mean lifetime will be very small and its concentration in the reacting mixture can become too small to measure. In such a situation R may not be observed and can be considered to be a reactive intermediate.,Transition Complexes. The numerous collisio
30、ns between reactant molecules result in a wide distribution of energies among the individual molecules. This can result in strained bonds, unstable forms of molecules, or unstable association of molecules which can then either decompose to give products, or by further collisions return to molecules
31、in the normal state. Such unstable forms are called transition complexes.,化 学 反 应 工 程,Postulated reaction schemes involving these four kinds of intermediates can be of two types.,Nonchain Reactions. In the nonchain reaction the intermediate is formed in the first reaction and then disappears as it r
32、eacts further to give the product. Thus,Chain Reactions. In chain reactions the intermediate is formed in a first reaction, called the chain initiation step. It then combines with reactant to form product and more intermediate in the chain propagation step. Occasionally the intermediate is destroyed
33、 in the chain termination step.,化 学 反 应 工 程,(intermediate)* Initiation,(Intermediate) * + Reactant (Intermediate) * + Product Propagation (Intermediate) * Product Termination,The essential feature of the chain reaction is the propagation step. In this step the intermediate is not consumed but acts s
34、imply as a catalyst for the conversion of material. Thus, each molecule of intermediate can catalyze a long chain of reactions, even thousands, before being finally destroyed.,Thus,化 学 反 应 工 程,Free radicals, chain reaction mechanism. The reaction,with experimental rate,can be explained by the follow
35、ing scheme which introduces and involves the intermediates ,The following are examples of mechanisms of various kinds.,化 学 反 应 工 程,2. Molecular intermediates, nonchain mechanism. The general class of enzyme-catalyzed fermentation reactions,with experimental rate,is viewed to proceed with intermediat
36、e (Aenzyme)* as follows:,A + enzyme (enzyme)*,( enzyme)* R + enzyme,化 学 反 应 工 程,3. Transition complex, nonchain mechanism.The spontaneous decomposition of azomethane,exhibits under various conditions first-order, second-order, or intermediate kinetics. This type of behavior can be explained by postu
37、lating the existence of an energized and unstable form for the reactant, A*. Thus,A + A A* + A Formation of energized molecule A* + A A + A Return to stable form by collision A* R + S Spontaneous decomposition into products,Lindemann (1922) first suggested this type of intermediate.,化 学 反 应 工 程,Test
38、ing Kinetic Models,Two problems make the search for the correct mechanism of reaction difficult. First, the reaction may proceed by proceed by more than one mechanism, say free radical and ionic, with relative rates that change with conditions. Second, more than one mechanism can be consistent with
39、kinetic data. Resolving these problems is difficult and requires an extensive knowledge of the chemistry of the substances involved. Leaving these aside, let us see how to test the correspondence between experiment and a proposed mechanism that involves a sequence of elementary reactions.,In these e
40、lementary reactions we hypothesize the existence of either of two types of intermediates.,化 学 反 应 工 程,Type 1. An unseen and unmeasured intermediate X usually present at such small concentration that its rate of change in the mixture can be taken to be zero. Thus, we assume,This is called the steady-
41、state approximation. Mechanism types 1 and 2, above, adopt this type of intermediate, and Example 2.1 shows how to use it.,Type 2. Where a homogeneous catalyst of initial concentration C0 is present in two forms, either as free catalyst C or combined in an appreciable extent to form intermediate X,
42、an accounting for the catalyst gives,C0 = C + X,化 学 反 应 工 程,We then also assume that either,or that the intermediate is in equilibrium with its reactants; thus,where,=0,化 学 反 应 工 程,EXAMPLE 2.1 SEARCH FOR THE REACTION MECHANISM,The irreversible reaction,(11),A + B = AB,(10),has been studied kinetical
43、ly, and the rate of formation of product has been found to be well correlated by the following rate equation:,What reaction mechanism is suggested by this rate expression if the chemistry of the reaction suggests that the intermediate consists of an association of reactant molecules and that a chain
44、 reaction does not occur?,化 学 反 应 工 程,SOLUTION,If this were an elementary reaction, the rate would be given by,(12),Since Eqs. 11 and 12 are not of the same type, the reaction evidently is nonelementary. Consequently, let us try various mechanisms and see which gives a rate expression similar in for
45、m to the experimentally found expression. We start with simple two-step models, and if these are unsuccessful we will try more complicated three-, four-, or five-step models.,化 学 反 应 工 程,Model 1. Hypothesize a two-step reversible scheme involving the formation of an intermediate substance A, not act
46、ually seen and hence thought to be present only in small amounts. Thus,which really involves four elementary reactions,(13),化 学 反 应 工 程,Let the k values refer to the components disappearing; thus, k1 refers to A, k2 refers to A2*, etc.,(18),Now write the expression for the rate of formation of AB. S
47、ince this component is involved in Eqs.16 and 17, its overall rate of change is the sum of the individual rates. Thus,化 学 反 应 工 程,Because the concentration of intermediate is so small and not measurable, the above rate expression cannot be tested in its present form. So, replace A2* by concentration
48、s that can be measured, such as A, B, or AB. This is done in the following manner. From the four elementary reactions that all involve we find,(19),Because the concentration of is always extremely small we may assume that its rate of change is zero or,(20),化 学 反 应 工 程,This is the steady-state approx
49、imation. Combining Eqs.19 and 20 we then find,(21),which, when replaced in Eqs.18, simplifying and canceling two terms (two terms will always cancel if you are doing it right), gives the rate of formation of AB in terms of measurable quantities. Thus,(22),化 学 反 应 工 程,In searching for a model consistent with observed kinetics we may, if we wish, restrict a more general model by arbitrarily selecting the magnitude of the various rate constants. Since Eq.22 does not match Eq.11, let us see if any of its simplified forms will. Thus, if k2 is very small, this expression reduces to,
