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Chapter 10 Chemical Kinetics II. Composite Mechanisms

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## Chapter 10 Chemical Kinetics II. Composite Mechanisms

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**Chapter 10 Chemical Kinetics II. Composite Mechanisms**~Chapter 9 dealt with the more important basic principles of chemical kinetics. To a considerable extent, but not entirely, it dealt with elementary reactions that occur in a single step; the reactant molecules form an activated complex, which passes directly into products. ~However, the majority of reactions with which the chemist deals are not elementary; instead, they involve two or more elementary steps and then are said to be composite, stepwise, or complex. ~This chapter is concerned how the rates of composite reactions are related to the characteristics of elementary steps.**There are various types of composite reactions. In some of**them, relatively stable molecules occur as intermediates. A simple example is the reaction between hydrogen and iodine monochloride. If this reaction occurred in a single elementary step, it would be third order- first order in hydrogen and second order in ICI-since a molecule of hydrogen and two molecules of ICI would come together to form an activated complex. This can be explained if there is initially a slow reaction between one molecule of H2 and one of ICl, In fact, the reaction is experimentally observed to be second order, being first order in hydrogen and first order in iodine monochloride: followed by a rapid reaction between the HI formed in this step and an additional molecule of ICI, Addition of these two reactions gives the overall equation. The HI produced in first reaction is removed as rapidly as it is formed. The rate of the second process has no effect on the overall rate, which is therefore that of the first step; the kinetic behavior is thus explained. In this scheme the first step is said to be the rate-determining or rate-controlling step.**Another reaction that involves a fairly stable intermediate,**and which has rate-determining step, is the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution, The experimental rates vary with concentration according to the expression The concentration of H2O+-OH, which is a protonated hydrogen peroxide molecule, is proportional to [H+][H2O2]. The rate of reaction 2 is proportional to [H2O+-OH] and therefore to [H2O2][H+] [Br-]. The kinetic equation is thus explained if reaction 2 is the slow and rate-controlling step. This result is explained if the reaction occurs in the following three stages: This explanation is supported by the fact that if solutions of HOBr and Br- are mixed, and the mixture is acidified, bromine is produced very rapidly; reaction 3 is therefore fast. On the basis of arguments of this kind, we draw conclusions about the elementary reactions that occur in composite mechanisms.**Several other examples will be given to illustrate the**methods used in arriving at a reaction mechanism. It cannot be emphasized too strongly that kinetic evidence-or indeed any other kind of evidence-can never prove a reaction mechanism, although the evidence may disprove a mechanism! The point is that one may suggest a mechanism that is consistent with all the kinetic evidence available at any time, but one can always find another mechanism, perhaps a more complicated one, that explains the experimental results equally well. It has often been found that a kinetic mechanism that has been accepted for many years is proved by later evidence to be quite wrong. Of course, it is true of all science that one can never prove any theory with absolute certainty: One can say only that the theory is consistent with all the known facts, but one has to admit that newly discovered facts may disprove the theory entirely. Nowhere is this more true than in the search for the mechanism of a chemical reaction.**Example 10.1**Suppose that a reaction of stoichiometry A+ BY + Z is zero order in the reactant A, and first order in B. Suggest a mechanism that is consistent with this result. Solution The obvious suggestion is that B forms some intermediate X by a slow step, and that X then reacts rapidly with A. BX (slow) X +AY+ Z (fast) To avoid complications for the time being we will suppose that the first reaction is highly exergonic (i.e., the equilibrium lies far to the right), so that we do not have to worry about the back reaction, which would complicate the situation. In the mechanism proposed for this reaction, the rate of the overall process is controlled by the rate of the first step; as soon as X is formed it reacts with A, and the amount of A does not affect the rate. Of course, there must be enough A present for this to happen; if there is no A there will be no reaction, and if there is very little A the rate will depend on the amount of A.**Problem 10.4**A reaction of stoichiometry A+B=Y+Z is found to be second order in A and zero order in B. Suggest a mechanism that is consistent with this behavior. Solution**Example 10.2**The kinetics of the reaction has been studied and the rate equation found to be Suggest a possible mechanism to explain this behavior. Solution What we notice at once about this rate equation is that the concentration of a product of the reaction appears in it, and to a power of -1. This can be explained if we assume that the product Fe(CN)64- is involved in a pre-equilibrium. The equation for the pre-equilibrium, which being fast is undisturbed by the sequent reaction, is The following is a possible scheme: This may occur rapidly, and the fact that the concentration of Fe(CN)63- occurs to the second power suggests that this pre-equilibrium is followed by a slow reaction of I2- with another molecule of Fe(CN)63- The rate of formation of the products is**Example 10.3**An investigation was made by M. J. Haugh and D. R. Dalton [J. Amer. Chem. Soc., 97, 5674(1975)] of the reaction of hydrogen chloride with propene at high pressures. They found that under some circumstances the reaction was first order in propene and third order in hydrogen chloride: Suggest a mechanism that is consistent with this result. Solution One thing we can be sure of is that the reaction does not involve three molecules of HCl interacting with one molecule of propene in one step. No case is known of four molecules coming together and reacting; the probability of even three molecules coming together is low. To explain the third-order kinetics in HCl we can suppose instead that two molecules first come together and form a complex (HCl)2, and that another molecule forms a complex with propene:**10.1 Evidence for a Composite Mechanism**~An obvious piece of evidence is when the kinetic equation does not correspond to the stoichiometric equation; several examples of this have just been noted. ~In other cases the kinetic equation is more complicated, sometimes involving reactant concentrations raised to nonintegral powers and with reactant concentrations in the denominator. For example, the gas-phase reaction between hydrogen and bromine would be first order in hydrogen and first order in bromine if it were an elementary reaction. In fact, the reaction follows an equation of the form where k and m are constants. This rate equation will be derived in terms of the rate constants for the individual steps.**~Another indication of a composite mechanism is the**detection, by chemical or other means, of reaction intermediates. When this can be done, a kinetic scheme must be developed that will account for the existence of these intermediates. ~Some times these intermediates are relatively stable substances; in other cases they are labile substances such as atoms and free radicals. ~Free radicals can sometimes be observed by spectroscopic and other methods, and evidence for their existence also be obtained by causing them to undergo specific reactions that less active substances cannot bring about. ~In particular, substances that react rapidly with intermediates bring about inhibition of the overall reaction. ~When the nature of the reaction intermediates has been determined, the step is to devise a reaction scheme that will involve these intermediates and account for the kinetic features of the reaction. ~If such a scheme fits the satisfactorily, it can be tentatively assumed that the mechanism may be the correct one. ~It should be emphasized, however, that additional kinetic work frequently leads to the overthrow of schemes that have previously been supposed to be established firmly.**Problem 10.1**Suppose that a reaction of stoichiometry A + 2B = Y + X is believed to occur by the mechanism X is an intermediate. Write the expression for the rate of formation of Y. Solution Problem 10.2 Suppose that a reaction A + 2B = 2Y + 2Z is believed to occur according to the mechanism Obtain an expression for the rate of formation of the product Y Solution**10.2 Type of Composite Reactions**Simultaneous Reactions ~ occurring in parallel B and C compete with one another for A Opposing Reactions ~ occurring in forward and reverse directions Consecutive Reactions ~ occurring in sequence Reactions are said to exhibit feedback if a substance formed in one step affects the rate of a previous step. For example, in the scheme the intermediate Y may catalyze reaction 1 (positive feedback) or inhibit reaction 1 (negative feedback). A final product as well as an intermediate may bring out feedback.**10.3 Rate Equations for Composite Mechanisms**In dealing with composite mechanisms it is necessary to consider the rates of elementary reactions as if they occurred in isolation in one direction. For example, for the composite mechanism there are four elementary reactions The rate for reaction 1 can be denoted by v1 and is the rate of AX if no other reaction were occurring; it is given by The total rate into X is the sum of the rates of all reactions that produce X The total rate out of X is the sum of the rates of all the reactions that remove X For a system at complete equilibrium the total rate into each species is equal to the total rate out of it.**Consecutive Reactions**The simplest consecutive mechanism is If the initial concentration of A is [A]0 and its concentration at any time t is [A], the rate equation for A is integration The net rate of formation of X is integration The equation for the variation of [Z] is most easily obtained by noting that**~Figure 10.1a shows the time variations in the**concentrations of A, X, and Z as given by these equations. ~We see that [A] falls exponentially, while [X] goes through a maximum. ~Since the rate of formation of Z is proportional to the concentration of X, the rate is initially zero and is a maximum when [X] reaches its maximum value. ~For an initial period of time it may be impossible to detect any of the product Z, and the reaction is said to have an induction period. ~Such induction periods are commonly observed for reactions occurring by composite mechanisms. Figure 10.1a**~Suppose that the rate constant k1 is very large and that k2**is very small. The reactant A is then rapidly converted into the intermediate X, which slowly forms Z. ~Figure 10.1b shows plots exponentials e-k1t and e-k2t and of their difference. ~Since k2 is small, the exponential e-k2t shows a very slow decay, while e-k1t shows a rapid fall. ~The difference (e-k2t-e-k1t) is shown by the dashed line in Figure 10.1b. ~The rate of change of the concentration of X is, equal to this difference multiplied by [A]0 (since k1>>k2), and [X] therefore rises rapidly to the value [A]0 and then slowly declines. ~The rise in [Z] then follows approximately the simple first-order law. Figure 10.1b k1>>k2**Steady-State Treatment**k2>>k1 Figure 10.1c At t = 0, [X] = 0, but after a very short time, relative to the duration of the reaction, the difference (e-k1t-e-k2t) has attained the value of unity, and the concentration of X is then [A]0k1/k2, which is much less than [A]0. After this short induction period the concentration of X remains practically constant, so that to a good approximation k2>>k1 This is the basis of the steady-state treatment, which is very commonly applied to reaction mechanisms.**~The concentration of the intermediate X is always much**smaller than the reactant concentration, the concentration of X rapidly reaches a value that remains practically constant during the course of the reaction. ~It is not possible to give a formal proof of this hypothesis, applicable to any reaction mechanism, because the rate equations for more complicated mechanisms are often impossible to solve. ~The rate of change of the concentration of an intermediate can, to a good approximation, be set equal to zero whenever the intermediate is formed slowly and disappears rapidly. ~In other words, whenever an intermediate X is such that it is always present in amounts much smaller than those of the reactants, the total rate into X is nearly the same as the total rate out of X.**Consider, for example, the mechanism**The differential rate equations that apply to this set of reactions are In order to treat this problem exactly it would be necessary to eliminate [X] and to solve the resulting differential equation to find [Z] as a function of t. Unfortunately, however, in spite of the simplicity of the kinetic scheme, it is not possible to obtain an explicit solution. Steady-State Treatment**Problem 10.3**Suppose that a reaction of stoichiometry A + B = Y + Z is believed to occur according to the mechanism. Apply the steady-state treatment and obtain an expression for the rate. To what expressions does the general rate equation reduce if a. The second reaction, is slow, the initial equilibrium being established very rapidly? b. The second reaction is very rapid compared with the first reaction in either direction? Solution a. If k2[B] is small compared with k-1, b. If k2[B] is very large compared with k-1,**Rate-Controlling (Rate-Determining) Steps**A rate-controlling step may be defined as a step which has a strong influence on the overall rate of the reaction. k2>>k-1 The initial step is the rate-controlling step or the rate-determining step. k-1>>k2 reaction 2 is often regarded as the rate-controlling step**Example 10.4**The reaction between iodide ions and the cobalt complex Co(CN)5OH22-, for which the stoichiometric equation is is believed to go by the mechanism Assume that the intermediate exists in a steady state, and derive the general rate equation. Write the rate equation for the special cases of low and high iodide concentrations, and decide which is the rate-controlling step in each case. Solution The steady-state equation for Co(CN)52- is Note that [H2O] is included in the value of k-1 because, as solvent, its value is essentially constant.**The general expression for the rate is**If the concentration of iodide ions is sufficiently low reaction 2, of rate constant k2, is commonly called the rate-controlling step. the iodide concentration is high enough reaction 1 is rate-controlling step.**Great care must be taken in identifying a rate-controlling**step, as it is easy to make mistakes in the case of more complicated systems. Unless the situation is straightforward, it is better to avoid the concept of the rate-determining step altogether, as it is never essential to identify such a step. The following points about rate-controlling steps should be noted in particular: 1.It is wrong to define a rate-controlling step, as is sometimes done, as the slowest step in the reaction scheme. Sometimes it is, but often it is not. If the reaction proceeds by a chain mechanism (Section 10.5), the chain- propagating steps are proceeding at the same rate, but one of them may be rate controlling. 2.It is impossible to decide on a rate-controlling step if one have no information about the relative values of the rate constants. Students sometimes think that they must decide on a rate-controlling step before they can work out a rate expression on the basis of the steady-state hypothesis. The schemes just considered show that this is not the case: One decides on a rate-controlling step after one has analyzed the kinetics. 3.In general, the rate-controlling step depends on concentrations of reactants. In the example just considered the concentration of iodide ions played a critical role.**Problem 10.7**Nitrogen pentoxide reacts with nitric oxide in the gas phase according to the stoichiometric equation The following mechanism has been proposed. Assume that the steady-state treatment can be applied to NO3, and derive an equation for the rate of consumption of N2O5. Solution The steady-state equation for NO3 is**The rate of consumption of NO is**The rate of consumption of N2O5 is The rate of formation of NO2 is**Problem 10.8**The reaction 2NO + O2 2NO2 is believed to occur by the mechanism Assume N2O2 to be in a steady state and derive the rate equation. Under what conditions does the rate equation reduce to second-order kinetics in NO and first- order kinetics in O2? Solution The steady-state equation for N2O2 is**The rate is**reaction 3 is the rate-controlling step reaction 1 is the rate-controlling step**Problem 10.10**The following mechanism has been proposed for the thermal decomposition of pure ozone in the gas phase Derive the rate equation. Solution The steady-state equation for O is The overall reaction is 2O33O2, and therefore**10.4 Rate Constants, Rate Coefficients, and Equilibrium**Constants For an elementary reaction it is easy to show that the equilibrium constant must be the ratio of the rate constants in forward and reverse directions. Thus, consider the process in which the reactions in forward and reverse directions, are elementary. If the system is at equilibrium Kc is the equilibrium constant. The ratio of rate constants in the forward and reverse directions is equal to the equilibrium constant Kc is the equilibrium constant for the overall reaction.**For any mechanism, involving any number of elementary steps,**the overall equilibrium constant is the product of the equilibrium constants for the individual steps and is the product of the rate constants for the reactions in the forward divided by the product of those for the reverse reactions: If a reaction occurs by a composite mechanism, and we measure a rate coefficient k1 for the overall reaction from left to right and also measure a rate coefficient k-1 for the overall reaction from right to left, at the same temperature, the ratio k1/k-1 is not necessarily the equilibrium constant for the overall reaction. The reason for this is that rate equations for composite reactions change with the experimental conditions, such as reactant concentrations, and the rate coefficients also change. The ratio of the rate coefficients k1 and k-1 that apply when the system is at equilibrium is equal to the equilibrium constant, but rate coefficients determined away from equilibrium are not necessarily the same as those at equilibrium, and their ratio is not necessarily equal to Kc.**Problem 10.14**Show that the mechanism leads to the result that the rate equation for the overall reaction is v = k[H2][I2]. Solution The overall rate is**Problem 10.11**A reaction occurs by the mechanism and the concentration of X is sufficiently small compared with the concentrations of A and B that the steady state treatment applies. Prove that the activation energy Ea at any temperature is given by that is, is the weighted mean of the values E1+E2-E-1 and E1, which apply, respectively, to the limiting cases of k1>>k2 and k2>>k-1. Solution The steady-state equation for X is The rate is**10.5 Free-Radical Reactions**The oxygen-hydrogen bond in the water molecule may be split homolytically; that is, one of the electrons goes with one fragment and the other with the other: In this process, two electrically neutral free radicals are produced; an atom is a special case of a radical. The hydrogen atom consists of a proton and an electron, and the hydroxyl radical consists of nine protons (one in the nucleus of the hydrogen atom and eight protons in the oxygen nucleus) and nine electrons (seven valence electrons plus two 1s electrons). The hydrogen atom and the hydroxyl radical are both one electron short of the noble gas structures and, therefore, are very reactive species. Radicals combine with one another with very low or zero activation energies, and their reactions with stable molecules occur with quite low activation energies.**In the ionization of water, on the other hand, a bond is**split heterolytically and the electron pair remains with the oxygen atom: The hydroxide ion is negatively charged; it has nine protons and ten electrons. It has the same electronic configuration as neon and therefore is chemically stable and unreactive. Whereas hydroxyl radicals cannot be stored, solutions containing hydroxide ions can remain intact for long periods of time.**Chain Reactions**Ions play little part in ordinary gas-phase reactions, owing to the difficulty which they are formed in the absence of an ionizing solvent. Atoms and free radicals are produced more easily in the gas phase and, because they enter readily into further reaction, they are important intermediates in reactions. This rate equation can be explained by the mechanism A reaction of this type is known as a chain reaction. One essential feature of a chain reaction is that there must be a closed sequence, or cycle, of reactions that certain active intermediates are consumed in one step and are regenerated in another; these active intermediates may be atoms, free radicals, or ions. It is also essential feature that the sequence is, on the average, repeated more than once.**~The first reaction, the production of bromine atoms from a**bromine molecule is known as the initiation reaction, since it starts the whole process. ~Reaction 2 and 3, the so-called chain~propagation steps, play a very important role in reactions of this type. Bromine atoms disappear in reaction 2 and reappear in reaction 3; hydrogen atoms disappear in reaction 3 and come back again in reaction 2. ~Because of this feature a small number of Br atoms, produced in reaction 1, can bring about a considerable amount of reaction, since after producing two molecules of hydrogen bromide, one in reaction 2 and one in reaction 3, a bromine atom is regenerated.**~Reaction 4 accounts for the fact that in the rate equation**HBr appears in the denominator; HBr reduces the rate by removing H atoms. If it were not for reaction -1, a single pair of bromine atoms could bring abut the reaction of all the H2 and Br2 present. ~Because of the termination reaction -1, however, only a limited amount of reaction is broughtabout each time a pair of bromine atoms is produced. ~Bromine atoms are continuously formed by reaction 1 and this keeps the reaction going.**The net rate of increase of the concentration of hydrogen**bromide is equal to The concentration of bromine atoms can be obtained by use of the steady-state method, which must now be applied to the two unstable intermediates H and Br. The steady-state equation for H is The steady-state equation for Br is**We can see that the reason the term in [HBr]/[Br2] appears**in the denominator of the rate equation is that HBr inhibits the reaction, by undergoing reaction 4, and that Br2 reduces the amount of inhibition, since, in reactions 3 and 4, Br2 and HBr are competing with one another for H atoms.**Organic Decompositions**A typical organic free-radical chain reaction is the decomposition of ethane under most conditions this is a simple first-order reaction It occurred in one stage by what is called a molecular reaction; that a small fraction of the molecules have sufficient energy for two C-H bonds to be ruptured, so that a hydrogen molecule is liberated. However, the evidence indicates that such a mechanism plays an umimportant role and that practically all the decomposition occurs by the following chain mechanism:**The initiation process involves the breaking of a C-C bond,**which is the weakest bond in the molecule. Reaction 2 is in a sense part of the initiation reaction; it converts CH3 into a radical C2H5, which can be involved in propagation; reaction 2 is not a propagation reaction since CH3 is not regenerated from C2H5. Reactions 3 and 4 are chain-propagating steps: C2H5 disappears in reaction 3 and appears in reaction 4, while H disappears in reaction 4 and appears in reaction 3. The main products of the reaction, C2H4 and H2, are formed in these propagation steps. The termination step, reaction 5, forms butane, C4H10, which can be detected as a minor product of the reaction. Methane, formed in reaction 2, has also been observed as a minor product of the reaction.**Example 10.5**Work out the expression for the overall rate of the ethane decomposition according to this mechanism, on the assumption that the steady-state hypothesis applies to the free radicals CH3, C2H5, and H. Solution The steady-state equations are The rate of formation of ethylene is most experiments have shown the reaction to be of the first order modified**Example 10.6**The mechanism originally proposed in 1934 by E O. Rice and K. F. Herzfeld for the ethane decomposition was (Note that this differs from the previous scheme only in the chain-ending step.) Derive the rate equation corresponding to this mechanism, assuming the reaction orders to correspond to the molecularities. Solution The steady-state equations are**rate constant k1 is very small**The rate of formation of ethylene is The Rice-Herzfeld mechanism thus gives first-order kinetics, in agreement with experiment. Nevertheless it turned out not to be the correct mechanism. One difficulty is that it was found that the ethyl radical concentration is much higher than the hydrogen atom concentration, so that the termination process C2H5+C2H5 must be more important than C2H5+H.**When ethanal (acetaldehyde) decomposes thermally the main**products are methane and carbon monoxide, and under usual conditions the order of reaction is 1.5. A variety of experimental evidence has shown that the reaction occurs to a large extent by the mechanism The steady-state equation for CH3 is The steady-state equation for CH3CO is The rate of change of the concentration of methane, which is approximately the rate of change of the concentration of acetaldehyde,