I :i   A (.1111)1 I Oil SIUDI Nľ,
When you rvail lliť hook you should pay nlli'iilioii lo u lew key thlngl,
• Look for equations in boxes, They are the mosl Importanl equations In the chapter, and you should prohahly incmori/.c Ihcm.
• Be sure to look at the solved examples. I put the solved examples al the end of the chapter so that you ean find them easily when you are doing the homework Be sure to look at them.
• Read the chapter twice. I usually recommend thai students (I) read ihe chapter through once, skipping the derivations, (2) go through the solved examples, and then (3) read the chapter a second time to concentrate on the details.
. Download the computer programs. I have posted computer programs lor many of the examples on my Website, http://www.uiuc.edu/~r-masel/. Be sure to download them and try them out.
. Remember to read the words as well as the equations. I know that students often use a book to help figure out how to solve their homework. Still, in my experience a qualitative understanding of kinetics is more important than knowing how to put numbers into the equations. Be sure to read the words as well as the equations.
• Do not trust everything in the textbook. I know that the statement "Do not trust everything in the textbook" sounds strange to a student, but remember that kinetics is an evolving field. If I look back to 10 years ago, things that I was sure then were right are now known to have serious limitations. I expect that 10 years from now, some of the ideas that I discuss here will be known to not work as well as I thought when 1 wrote this book. Evolution of knowledge is part of science; and in 10 years something in this book is bound to be wrong.
I have tried to make this book as useful of a teaching tool as possible given the length limitations. I hope that you enjoy it.
In Chapters 3-14, I will list a series of bibliographic works for further information. Reviews of the history of kinetics include:
V. A. Kritsman, G. E. Zaikov, and M. M. Emanuel, Chemical Kinetics and Chain Reactions:
Historical Aspects, Nova Science Publishers, 1995. K. J. Laidler, The World of Physical Chemistry, Oxford University Press, New York, 1996. M. C. King, Experiments with time, progress and problems in the development of chemical kinetics.
Ambix 28, 70 (1981). M. C. King, Ambix 29, 49 (1982).
Specific references mentioned in this chapter are given at the end of this book.
2
REVIEW OF SOME ELEMENTARY CONCEPTS
fill CI8
1In nh|i i live ol this chapter is to provide some background material that will be used lit till the rcsl of the book. Specifically, we review the definition of the reaction rate,
.....hiometric coefficient, the rate equation, the order of reaction, the preexponential,
II i I hi activation energy. We will then discuss empirical rate laws, and the elicit ol li lltpi inline and catalysts on rates. Discussions of these topics are necessarily brief, since hi iIn ii-.nli-i-s of this book will already be familiar with these topics. However, the i i i live ill ilns chapter is to bring the student's understanding of these topics to the 116X1 i   i I Wi will provide precise definitions of terms that were defined loosely in freshman in mi n \ We will point out the historical basis for ideas, and describe how the ideas i Ml i    ippllcd in nonstandard problems. We will also give real examples. There is necessarily
.......cvlcw material in the chapter. However, we directed the chapter toward someone
.....In nl\ has seen the material and who wants to understand it better, rather than to
■■in.....' learning the material for the first time.
I DIFINITIONS
i    i hi nil, it is useful to define some key terms. In this section we will define
. I he stoichiometric coefficient
. I he rate of reaction
• I loinogeiieous reactions
. Heterogeneous reactions
......i ■ useful to start with some history. Chemical reactions were first studied in detail
■ i ili. end of ihe eighteenth century. The initial work was done before people knew that
B
there well' .limns in iniilii uli".....il llir nliii 11 i.i I I hi' 11- weir i I it' III li 111 i < > 1111 x il II It I s wns
sill I ill dispute. 11 WHS tie.II lh.il i hen in .il ii\n lions tin I OCCUI Ilimnri. ill. ii.ilinr nl I lull'.k linns was siill uncleai
One of the difficulties in the early wink was thai when one did an experiment, everything seemed to change, for example, Lavoisier | 1 /K')| did a series ol experiments where he oxidized tin to tin-calx (SnO,). The volume of the lin changed during the reaction and the mass changed during reaction. All of the properties of the tin seemed to change during the reaction. Lavoisier initially had trouble finding anything that was constant. However, in a series of famous experiments, Lavoisier oxidized the tin in a sealed flask and weighed the flask as the reaction proceeded. The experiment required the most sensitive balance that had ever been constructed, and the work was possible only because the French Academy of Science decided to finance the work. Lavoisier found that the weight of the sealed flask did not change during reaction. Instead, the weight change occurred only when air was able to rush into the vessel. This work showed, for the first time, that mass was conserved during chemical reactions.
Lavoisier also found that a fixed amount of air reacted with a fixed amount of tin. This work lead to the law of fixed proportions, which states that chemicals react in fixed proportions during chemical reactions.
Finally, Lavoisier found that only part of the air was reactive. This finding lead Lavoisier to propose that air was made up of different elements: a reactive element and an unreactive element.
Lavoisier called the reactive element in the air the "atmospheric principle". Later, Joseph Priestley visited France at Lavoisier's invitation. Lavoisier and Priestley discussed Priestley's discovery [1790] of 'dephlosgenated air', which was liberated when mercury oxide was heated. Lavoisier quickly recognized that the dephlosgenated air was the reactive element in the air. Lavoisier renamed the reactive element oxygen. This work lead to the idea that air, and all mixtures and chemical compounds are made up of fundamental components called elements. Lavoisier also showed that reactions happen when an element moves from one chemical compound to another chemical compound.
Lavoisier then did a series of experiments where he reacted oxygen with a series of metals, and found that in most cases, a fixed amount of oxygen reacted with a fixed amount of metal to yield a fixed amount of oxide. Soon thereafter, Lavoisier proposed that molecules react in fixed proportions. That is, if one has a given reaction between A and B, one can define coefficients o<i and a2 such that during a given reaction, a, molecules of A always react with a2 molecules of B to yield a? molecules of C and a4 molecules of D:
a, A + a2B ==> a3C +a4D (2.1)
In Lavoisier's case, he said that 1 kilogram (kg) of air would always react with 0.78 kg of tin to yield 0.99 kg of tin-calx and leave 0.79 kg of inerts. Lavoisier decided to call the study of these fundamental proportions stoichiometry, from the Greek word stoichio-, which means fundamental. Lavoisier also proposed that one could write an equation for each chemical reaction by balancing the production of all of the essential elements in the reaction. In other words, during the reaction between oxygen and tin, the number of kilograms of oxygen used up during the reaction would be equal to the number of kilograms of oxygen in the product calx.
Lavoisier also defined the stoichiometric coefficient |3„ for molecule n participating in a given reaction, where Lavoisier defined the stoichiometric coefficient as the number of molecules of n produced when the given reaction goes once. Lavoisier's definition
is dilli'ii'iit lioni lin- oik' you nti' ir.'d in in lii.luii.il> < lniiii.liv The stokhlotnelrli
ciH'iiuu-nt is negative foi thi rawuuiti and poiltive foi the producti Foi example, In the
ii .ii lion
2CO + 02 => 2COj (2.2)
pYo = —2, p\), = - I, pVo, = 2. The advantage of this definition is that it is much imsici
to use ni a computet program and a spreadsheet than a definition where all stolci.......etrit
. hi in. nuts are positive, l ,ater in this book, we will find thai this definition also allow I you to simplify the equations for a complex reaction pathway. The disadvantage ol the di linition is thai you have to watch out tor the negative numbers.
(>ne thing to be careful about is that (he value of the stoichiometric coefficient changes .ii nulling to how one writes the equation. For example, reaction (2.2) can be rewritten
CO + i02 ==> C02 (2.3)
in reaction (2.3), Pco = —1. Po2 = ~~ J> an^ fico2 =
Dallon I 1805] extended Lavoisier's idea with his atomic theory. Dallon proposed thai each of Lavoisier's elements were composed of tiny particles called atoms. Dallon showed thai the number of each type of atom was conserved during chemical reactions Bai h particle had a fixed weight. Although Dalton could not weigh each atom, he could define .i ulalive scale by arbitrarily defining the relative weight of hydrogen to be one atomil mass unit (1 amu), and then calculating the weights of all of the elements accordingly. Dalton's calculations were not particularly accurate, but they did allow Lavosiei to calculate stoichiometric coefficients for many reactions.
Inspired by Lavoisier and Dalton's findings, many gentlemen-scientists set out to explore the nature of molecules. A particularly famous person at that time was Louis lliiiKird, [1818] who discovered hydrogen peroxide. Thénard thought thai hydrogen peroxide was a wondrous molecule. It was composed of hydrogen and oxygen the same components as ordinary water. Yet hydrogen peroxide behaved much differently than ordinary water. It could change the colors of dyes and would convert alcohol to vinegar. Thénard, being a wine connoisseur, was particularly interested in how hydioj'in peroxide could convert alcohol into vinegar. He studied the reaction and measured the rate Thénard found that when he diluted the alcohol, the rate went down. Additional hydrogen peroxide made the reaction go faster. Thénard then showed that by eliminating peroxide! he could make wine last longer. Thénard never quantified his findings, but his work showed for the first time that rates of chemical reactions varied with the concentration) of the reactants.
Many people expanded on Thénarďs findings. In 1869, Guldberg and Waage proposed that rates of reaction were proportional to the "active masses" of the reactants, when- m most cases the active mass roughly corresponded to the concentration of the spw Many other rate laws were proposed between 1860 and 1880. Still, it was not until Van I lloff published his famous book, Etudes de Dynamique Chemie in 1884 that rates were put into a quantitative framework.
Van't Hoff was considered to be the greatest chemist of his day. Van't Hoff started the field physical chemistry and was the first person to really push the idea that quantitative reasoning was useful to a chemist. Van't Hoff wrote several influential books, including Theoretische Chemie, which set out the field of physical chemistry for the first time, and Etudes de Dynamique Chemie, which established chemical kinetics as a separate subdiscipline.
1 Ml 'HI" «1  I I'm I  l nw"
lii lttulr\ ill- D\n<tmi<iu? ( lumir, Vim * I Hull i|iiiiiililli'il Ills ulcus iihoul rules ol chemical rciiclions, Van'l Moll (IK/K) noled is that il our wmils Id compare rules ol different reactions, one has to do so on ,i consistent basis Ai thai time, people had used .ill different kinds of reactors to examine rates ol reactions. Some people had winked in liny reactors, while other people hud worked on big reactors. Generally, big reactors produced products faster than did small reactors, independent of what reaction was occurring in the reactor. Van't Hoff asserted that if one wanted to understand reactions, one would have to remove the effects of the size of the reactor from the analysis. For example, Van't Hoff noted that if a reaction is occurring uniformly over the volume of a vessel, then if one doubles the size of the vessel, one will produce twice as many moles of products in a given time. Consequently, Van't Hoff noted that it was not very useful to compare the total number of moles of product produced in one reactor to the number of moles of a different product produced in a different-sized reactor. Rather, it was important to consider how the rate per unit volume varied from one reaction to the next. Van't Hoff defined rA, the rate that a given molecule A reacts as the rate of production of A in moles per volume per time. With this definition, the reaction rate, rA, for a species A is negative if A is a reactant and positive if A is a product. Van't Hoff showed that if a reaction is run in a closed vessel with perfect mixing, one can determine rA by measuring CA, the concentration of A, as a function of time, t, and then substituting into the following equation:
dCA dt
(2.4)
Van't Hoff also noted that equation (2.4) is useful only when a reaction is occurring uniformly over the volume of a reactor. Van't Hoff considered several reactions that were instead occurring on the surface of his reactor. In that case, it is useful to define RA, a rate per unit surface (i.e., mol/(cm2 second)).
In the work that follows, we will use a small r to indicate a rate per unit volume and a capital R to indicate a rate per unit area. We will also define heterogeneous reactions as reactions that take place on the walls of the reactor or on other surfaces within the reactor. We will also define homogeneous reactions as reactions that occur in the gas phase, in liquids, or throughout the bulk at a solid.
Again, these definitions are slightly different from the definitions you learned in freshman chemistry. In freshman chemistry, one usually talked about homogeneous reactions occurring in the same phase as the reactants and heterogeneous reactions as occurring in the interface between two phases. Note, however, that the freshman chemistry definition is not accurate. For example, consider the hydrogenation of ethylene:
H2 + C2H4
CiHfi
(2.5)
One can run ethylene hydrogenation by bubbling ethylene and hydrogen through a liquid containing rhodium atoms or by bubbling the same mixture through a reactor containing a solid platinum catalyst. In the former case, the ethylene and hydrogen absorb (i.e., dissolve) into the liquid and react while they are in solution. In the latter case, the ethylene and hydrogen adsorb onto (i.e., bind to) the surface of the catalyst and react on the surface. Therefore, both processes are very similar. Still, we call the former case a homogeneous reaction because the reaction is occurring in a liquid. We call the latter case a heterogeneous reaction because the reaction is occurring on a surface. Nevertheless, in both cases the reaction is occurring in the same phase as the absorbed or adsorbed
Mllll IllMll
lilliln '.' I    ' .1 if mi nil v ■>( III" lo'V ili'lloilioiľ.
Iliľ ai in hi iti 11I |iiuiliiii 1 >i 1 u In, ill when ilľ miction )'">'■ nine, I tit* ■toll 11 it>■ 111-1114 ini'llu it-ill is positive fill .1 I>■ 1 ><liit t .mil ni'i'iilivi' Im 11 11 .11 I.ml
tin- ih'I 111I1- 11I production nl 11 species A. i,\ is posiliir !m .1 piotlm 1 .m.l llt'i'.lIlM   I.H .1 Iľ.li I.ml
I .... 1 i,\ lor any species A partiiipaliiif.' 111 reaction I
A reailmn III.it hap|K-ns throughout the reacting mixture, A reaction lli.il happens near the boundary ol a reacting phase
IAI
K.m ol reaction 1.11
||iiiiiii|'.eneuus reaction Heterogeneous reaction
n .u 1.nils, so one cannot distinguish between homogeneous and heterogeneous reaction! in looking io sec if the reaction occurs in the same phase as do the reactants,
I'm I he material later in this book, we will have to distinguish between 1 \. the rate 0Í production ol species A; and. n, the rate of reaction I. Van't Hoff defined i|. the rate "i reaction l. as
ri = Tj-rA l.'.t.i Pa
vi Inn- 11 is the rale of reaction 1, rA is the rate of formation of A during reaction I. and | is ilu- stoichiometric coefficient of A. Note in general that rA ^ ľ). Again this definition II -. 11 j ■ 1111 n different from ihe one you learned in freshman chemistry. In Example 2.A at Ihe end ol iliis chapter, we show how to use measurements on a simple reactor to calculate 1 \ and i|. The reader may want to look at this solved example before proceeding.
fable 2.1 summarizes the key definitions so far in this chapter. The reader si.....1.1
memorize these definitions before proceeding.
2.2  VARIATION IN RATE WITH CONDITIONS
Nexl we want to discuss how rates of reaction vary with conditions. Van't Hoff showed thai generally rates vary with
• Concentrations of all of the reactants, products, and other species in the system
t Temperature
. The presence of solvents
. The presence of catalysts
In the next several sections, we will discuss the effects or concentration and temperatUN at length. The role of solvents and catalysts will be discussed in Chapters 12-14.
2.3   EMPIRICAL RATE LAWS
We will first consider the role of concentration on the rate of reaction. Our objective ll to define the rate equation and the order of a reaction.
Studies of the role of concentration on rates of chemical reacttons started tn the riuddk par, of the nineteenth century. Van't Hoff summarized the effects tn hts 1896 book
I Ml'IIIII Al HAM I AWM 11
Viin'I Hull reported ilain mi tunny dillcicut mulleins, and showed how In Correlate tin*
meuuremenU to the concentration! ol the various ipw lei Vun'i Hull proposed thai the rate was a function oi only the temperature ol the reactor, the concentration ol all ol
the species, the concentration of solvents 01 catalysis, ami the total picssiuc Further, Van'l Holt asserted that if one lixed the temperature and solvent, one could humiliate an equation for the rate as a function ol concentration, An equation thai describes the relationship between the rate of reaction and the concentration of all ol the species in the reactor is called a rate equation.
Van't Hoff published the findings described above in a series of books that were classics for the day. The interest generated by Van't Hoff s books and the fact that he won the first Nobel Prize in chemistry stimulated many other scientists to examine the effect of concentration on rates of reactions.
The most famous person to follow in Van't Hoff's footsteps was Friedrich Wilhelm Ostwald. Ostwald had a very strong character. On one hand, he argued vehemently that while atoms were useful theoretical constructions, atoms do not really exist. On the other hand, Ostwald was the first person to propose that ions formed when species dissolve in solution. Ostwald created electrochemistry and did much of the early work on thermodynamics of solutions.
Ostwald was fascinated with rates of reaction in solution. A particularly interesting example was the hydration of (+)sucrose (C12H22O11) to glucose (C6Hi206), and fructose, (C5H905CH2OH):
C12H22O1 I   + H2O '
C6H|20„ + CsHoCCHjOH
(2.7)
Reaction (2.7) is a common natural process called sugar inversion. Honeybees use sugar inversion to convert sugar from plants into honey. Humans invert sucrose as part of digestion.
Experimentally, sucrose is stable in water for months. Still, if one adds a small amount of either an acid such as HC1, or an enzyme called invertase, the sucrose is quickly converted into glucose and fructose.
Wilhelmy [1850] and Berthelot (1862) examined the reaction in some detail and found that it obeyed a simple rate law as follows:
t-(,
-rs = k3CsCA
(2.8)
where rG is the rate of formation of glucose, rs is the rate of production of sucrose, Cs is the sucrose concentration, Ca is the acid concentration, and k3 is a constant.
Ostwald reexamined reaction (2.7) and found that he could catalyze the reaction with a wide range of acids. Equation (2.8) did not explain his data very well. Still, he could fit the data with the following rate form:
rG — kfiCsC^
(2.9)
where CH+ is the concentration of hydrogen ions in solution.
Ostwald also generalized his results by noting that if one takes an arbitrary reaction of the form A => B, one might be able to fit the data to a form like equation (2.8). Thus rA, the rate of production of A, is given by
-k,CA
(2.10)
i     i   iln i 0111 rill I al Ii in ol the species A that is heme leaded away, iA is the late In. Hon ol A. mid k| is a constant, Note that lA is negative, i i,in it,I did extensive measurements ol rales ol many reactions in solution lie also i     I ill ol ihr i,ne data that had appealed prior to IK'M and found that the rale
i,, ,1.....     (ilven li\ equation 1.' Ill) sometimes works greal. However, more often than not
..... i poot representation ol the rate ol reaction. Ostwald also noted thai a
,1   muddied equation could represent a wide ranee ol data
rA= k„«'A>"
CID
1.....1   a lilliii'' constant.
1 11.   ' 1 illustrates some rale data that one might lit to equation (2.11). In the d.u.i Ihc 1.lie doubles when die concentration doubles. Consequently, 11 - I. One 1 11        1 hi ,1.H.1 in Table 2.2 back into equation (2.1 1) to calculate k„. k„ works out In In II "1/111111
Dulwilld 1 ailed 11 die order of die reaction, and k„ the rate constant. The relationship Mih, mlč . 11111 ilu- concentrations of the reactanls is called the rate equation  I In
..............I .1 i.ilr equation as the rate as a function of concentration is quite importB.nl
11.  ,  idei  hou Id memorize this definition before proceeding.
i'i ild also proposed that heterogeneous reactions could also be lilted by a similai i ,1. i<a 111
RA = kn(CA)n i' I'i
1, 11 1    r      sin lace concentration of A in mol/cm2. 1 ,n inline reference, we note that reactions that obey equation (2.11) with n = I an-11 1 in .1 order reactions, reactions that obey equation (2.11) with n = 2 arc called •n mid order reactions, while reactions that obey equation (2.11) with n = 3 are called 0....1 ,.11I11 reactions, The terms rate constant, rate equation, first-order, and vein//,/ ,1, quite important. The reader should memorize the definitions before proceeding ni exorcise, the reader might want to show that the rate constant for a homogeneou
, i,,.....gencous first-order reaction has the dimension (time)-1 (e.g., hour 1). the rati
.....mi tm a homogeneous second-order reaction has the dimensions volume/(molecul(
.......leVtime [e.g., Iiter/(molhour)], while the rate constant for a heterogeneous second
,,l.....1, hun has dimensions of area/(molecule or mole)/time [e.g., cm:/(mol hom >|
......I.ulv. ihc rale constant for a homogeneous third-order reaction has the dimensions
it.......       7i molecule or mole)2/time [e.g., liters2/(mol2-hour)]. While the rate constant foi
, linul order heterogeneous catalyst has dimensions of area2/time/(moles or molecule) in '/(mol2 minute)].
I In idea (hat there was a rate equation was a sizable advance. At the time this woik ,  going on, people did not know whether chemical processes would be amenable to quantitative analysis. The fact that the rate could be written as a function (i.e., equation)
I ulilo 2.2 Sample rate data to illustrate equation (2.11)
i \ nnil/liler
Rate, (molliter/hour)
CA, (mol/liter)
Rate, (mollilei I/limn
o
0.3
0.13 0.25
0.5 l.O
I II VII Y» 1 II     II ilVU   I I I Ml I I I 'M I I   I   I II II   I I ' I ' .
11 II I Kl'l IIIMI NI Al MI IIA III IN
ill llli' colli 'I'tlllillltms ill .ill ill till' spi'dl'S Ill I hi' ll'llllol llll'lllll llllll 111.HIS illrllih.ll processes would he aiiicuahlc to qiianlilalnc analysis Musi students do DO) iiali/c lh.il the
observation thai the rate can be written aa a rate equation la important In I'ai t, however, this observation forms the basis ol all studies ol kinetics.
In 1896, it was not obvious thai the rale was a function ol the concentration where we use the word function in a mathematical sense. Recall from freshman calculus, thai certain criteria must be satisfied before it is mathematically correct to say that one variable is a function of another. However, Van't Hoff (1884, 1896) examined all of the reactions thai had been studied prior to 1896, and found that in all cases, he could express the rate as a function of the concentration. Therefore, Van't Hoff asserted that one could write a rate equation for any reaction.
More recent data show that Van't Hoff is usually correct: one can usually fit rate data to a rate equation. However, there are some exceptions where the rate is not a mathematical function of just the temperature, pressure, and composition (see Section 2.8). Still, those are rare exceptions. In most cases, the rate is a function of the temperature, pressure, and composition.
Rate equations, of the form in equation (2.11), apply only to cases where there is only one reactant. When there are two reactants, A and B, one often finds rate equations such as:
r=kn,(CA)n(CB)m (2.13)
In the case of reaction (2.13), one defines the overall order of the reaction to be n + m. If n = 1 and m = 2, the reaction will be third-order. One might also say that the reaction is first-order in A and second-order in B.
Table 2.3 illustrates some data that one might fit to equation (2.13). In the data in the table, the rate goes up a factor of 2 when CA doubles, so n = 1. However, the rate goes up by a factor of approximately 22 = 4 when CB doubles. Therefore, m = 2. If one plugs back into equation (2.13), one finds km equals 0.5 liter/(molminute).
Table 2.4 summarizes the definitions in this section; one should memorize these definitions before proceeding.
Table 2.3  Sample data to illustrate equation (2.12)
Ca,	CB,	Rate,	CA,	CB,	Rate,
mol/liter	mol/liter	(molliter)/minute mol/liter		mol/liter	(molliter)/minute
I	0.25	0.031	0.25	1	0.13
1	0.5	0.13	0.5	1	0.25
1	1	0.5	1	1	0.5
1	2	2.0	2	1	1.0
Table 2.4	The key definitions from Section 2.3				
Rate equation Order
First-order reaction Second-order reaction Overall order of reaction
The rate as a function of the concentration of the reactants The exponent n is the expression
A reaction whose rate is preparation to the reactant concentration to the first power [e.g., n = 1 in equation (2.11)]
A reaction whose rate is proportional to the reactant concentration to the second order
The sum of the orders for all of the reactants
•   I  I Nlll.llKlll
In ih. in \i section, we will be discussing the rule equations loi a variety ol retCtiOfll it. I., I, 111 du no, we will need to define .nine notation We will define small k w it 11 I Mi. cilpi ,r. linn)' the rate constant foi a reaction; so k|, k., kt, ami k , will be
ih. .....iinslanls for reactions I, 2. 3, and    I, respectively. Similarly, large K wilh u
......i.....ii subscript will be an equilibrium constant for a reaction; so K|, K., and K,
nill In ilic equilibrium constants for reactions I, 2, and 3. Ijn, where the k has a tail, .ii hi Bolt/.mann's constant where Boltzmann's constant equals the gas law constant (R) ii Hi. J in Avogadro's number. We will use two different notations for a concentration
I   i" ' ic,n A: Cai and |A| (concentration Of A), where:
|A| = CA
(2.14)
\\. will use the |A| notation mainly where we have a molecule reacting. In the latOI IHI......... equation (2.9) will be rewritten
rG = k<i[sucrose][H4
(2. IS)
\iiiu we note that we define rA as the rate per unit volume of a homogeneous reaeimn ' lull K \ is the rale per unit surface area of a heterogeneous reaction. The reader should in. mori/c this notation before proceeding.
.' I   THE EXPERIMENTAL SITUATION
i low thai we have the notation out of the way, it is interesting to go back and compare . .piiiion (2.13) to the data. Our key point will be that equation (2.13) works for a wide v ii lety of reactions provided the data are being taken over a limited range of concentration In i.ui, it is useful to recall that Ostwald first proposed equation (2.13) in 1890. At Milium- ihere were reliable data for three gas-phase reactions and six reactions in solution I ihle 2.5 shows the rate equations for these reactions as cited by Van't Hoff (1896). Noiu ( ili.ii all of these rate equations are of the general form of equation (2.13). Since IK'io, thousands of reactions have been examined. Most follow simple first- or second ofdoi rule laws over moderate concentration ranges. There are some exceptions. Generally, the rule equation is different or there are concentrations of 0.001 and 10 mol/liter. Howevei equation (2.13) works for 95% of the reactions that have been examined so far. The main exceptions are catalytic reactions. They will be discussed later in this chapter.
Equation (2.13) has also proved to be useful in a wide range of systems outside nl chemistry. For example, Figure 2.1 shows data for the reproduction rate of E COh la common bacterium) in a sugar solution. Notice that the reproduction rate of E I'Oli follows
rcoi = kCoi [E. coli][sugar| (2.16)
where rCoi is the rate of production of E. coli, [E. coli] is the E. coli concentration, [sugai | is ihe sugar concentration, and kCot is a constant. The plot is linear over most of the range, indicating that the reproduction of E. coli is first-order in the sugar concentration.
Similarly, Figure 2.1 shows how the reproduction rate of Paramecium varies with the population of the Paramecium. Notice that at Paramecium concentrations below
M        III VII W (II 'K IMI I I I Ml Ml
11 li i «i'i iiimi ni/\i   .111 ir\ i ii in
l.lhli' .'       Sonic ill Ihi- i.ill
Reaction
.'(|ii.ilioir. Ih.il won. .Ir.. iiunnxl hutnili IIMIIť'
i'.....i. i.......i
.i'll. -
2AsH3 = 2PH3 + 402 -
Ci2H220n +H20
P4 + 6H2
> As2 + 3H2 => P205 + 3H20 H+
C6H120(,+c5h905CH2OH H
CH3COOR + H20 =
CH3COOH + ROH = ClCH2COOH + H20
CH3COOH + ROH CH3COOR + H20
2FeCl3
KC103 + 6FeO
■ HOCH2COOH + HCI SnCl2 > 2FeCI2 + SnCI4
KC1 + 3Fe203
(2.T 11 (2.1 I) (2.T.5) (2.T.7)
(2.T.9)
(2.T.11) (2.T.I3)
(2.T.15) (2.T.17)
1 ni,
I'Asll,
i'I'I 1,
rs =
k3[PHj]
ki|.V.II,|
k.ll'll.lio.l1 kfitsucrose]! 111 I
rAc = +k7|CH3COORl|H' I
rAc = -kslCHjCOOHJIROHHH'
rc2H,cio, = -k9|C2H3C102]
: -kl0[Fe3+]2[Sn2+] ^nlFe^JIClOj-]
(2.T.2) (2 1 4) (2.T.6)
(2.T.8)
(2.T.10)
(2.T.I2) (2.T.I4)
(2.T.16) (2.T.18)
"In these equations, rpH^, rAH3, r$, rAc. rc2H3Cio2. and fpit are tne rates °f formation of phosphine, arsine, sucrose, acetic acid, chloroacetic acid, and Fe3+, respectively; [PH3], [AsH3|, [02], [sucrose], [H+], [CH3COOR], [CH3COOH], [ROH], [Sn2+J, [C103_], and [C2H3C102] are the concentrations of phosphine, arsine, oxygen, sucrose, hydrogen ion, acetate, acetic acid, alcohol, Sn2+, C103~, and chloroacetic acid, respectively; and k3, k4, ks, k^, kj, kg, k9, kio, and kn are constants.
Uj
0 10 20 30 40 50 60 70 80 Paramecium concentration, #/cm3
0        50      100     150 200 Sugar concentration, mg/liter
Figure 2.1 The reproduction rate of Paramecium as a function of the Paramecium concentration and the rate of Escherichia coli growth in sugar solutions as a function of the sugar concentration. [Paramecium data of Meyers (1927) 1; E. coli data from Monod (1942).]
20/cm3 the reproduction rate of the Paramecium increases linearly with the Paramecium population:
rpar — kparfpar]
(2.17)
where rpar is the rate of Paramecium reproduction, [par] is the concentration of Paramecium, and kpar is a constant. Consequently, the reproduction of Paramecium is a first-order process. Equation (2.13) fits data in a wide variety of systems, which is why it has proved to be so useful.
Of course, data taken over a wider range of concentrations show that reactions do not follow equation (2.13). For example, the growth of E. coli follows so-called Monod (after Jacques L. Monod, biochemist) kinetics:
1,,,i,
1.1 K \l   . i'h |f.ii- 11 I
11 Ms >|Nugur|)
(2 IHl
«Inn- k 1 nud K . ure constunts,
according in equation (2.18), the rate goes up .nul then saturates as shown In 1 inu. ' ' 11 one plots the log of the rate versus concentration, one gets ;i charai leri itli
httpcd curve The one special Feature of reaction (2.16) is thai as the reaction proccedN
....... bacteria are born, so the rate increases. In the usual case, the rate decreases .is
ii.....i. nun proceeds because the reactants arc used up. One calls reactions where thi
1 1I1 nu reases as products build up aiiloculalvtic reactions. In autocatalytic reactions Hi. product catalyses (i.e., speeds up) the reaction. Many biological processes .in ......I .ilalylic.
Kale data lor most biological growth processes can lit lo equation (2.13) when
.......'titrations arc small. Such a result shows the power of equation (2.13). Still, equation
i'lt) seldom fits a) high concentrations. Instead, one needs a more complex rale equation lil 1 ih.ii in equation (2.18).
In the previous paragraph, we noted that the E. coli growth curve does 110I follow ■ .pillion (2,13) at high concentrations. Similar effects are seen in most chemical reactions I leme 2, i shows a plot of the rate of CH3NC isomerization as a function of pressure 1«' keep ihc plol in perspective, if we take the log of equation (2.11), wc obtain
ln(-rA) = ln(Kn) + nln(C,)
I ! I'M
1 liercfore, the plot of the log of the rate versus the log of the pressure should be lineal W iili .1 slope of 1 for a first-order reaction and linear with a slope of 2 for a second order .> 1. lion. Notice that the slope is a 1.0 at high pressure, implying a first-order reaction. However, lite reaction changes to second-order at low pressure. Experimentally, changes in the order of a reaction with pressure are quite common. In fact, it is unusual lo sec ,1 1 hange in the kinetics of a gas-phase reaction when one works over a wide range ol I -i. v.mes.
Ii is also common to find that the rate data for a given reaction cannot be fit lo a rale equation of the form in equation (2.13). For example, the rate of the reaction
H2 + Br2
2HBr
(2.20)
1     1.5 2 Concentration Figure 2.2 A plot of the rate for Monod kinetics, for ki
0
0.0001
0.001     0.01 0.1 Concentration 1,2 and K2 = 10,20,50,100,200.
1..
II Ml i MA I < II II   MM'   I 17
II   i I
1EtO
1E-2
cd
co 1E-4
DC
1E-6
1E-8
I ' *"""t '......1 .......1 ' ""'"1	
Slope=2	
r    Slope=1 Slfrk	
	1
J	
' ......... ........i .................i	.................1 ......"
1E-2 1E-1 1E+0 1E+1 1E+2 1E+3 1E+4 1E+5 Pressure.torr
Figure 2.3 The rate of CH3NC isomerization to CH3CN as a function of the CH3NC pressure. [Data of Schneider and Rabinovitz (1962).]
is thought to obey a complex rate equation, as follows:
k,[H2][Br2]
rHBr —
1 +K2
[HBr]
TbivT
(2.21)
where rHBr is the rate of formation of HBr, [H2] is the H2 concentration, [Br2] is the Br2 concentration, [HBr] is the HBr concentration, and, k| and K2 are constants. One cannot fit HBr rate data to equation (2.13).
Note also that one cannot easily define a reaction order when one has a complex rate equation such as that in equation (2.21). In my experience, equation (2.13) works great most of the time. However, it can fail, especially when the kinetic data are taken over a wide range of concentration.
2.4.1   The Relationship between Stoichiometry and Rate
Next we want to consider whether there is any relationship between the rate equation for a reaction and the stoichiometry of the reaction. The experimental data shown in Table 2.3 say no. Notice that in reaction (2.T.5), (in Table 2.5), two oxygens react with each phosphine. Yet reaction (2.T.5) is only half-order in oxygen. Similarly, the rate equation for reaction (2.T.7) is first-order in the H+ concentration, even though no H+ is produced or consumed in the reaction. Reaction (2.T.1) seems to be an exception to the rule. If one rewrites reaction (2.T.1) as
ph3
jp4 + |h2
(2.22)
One might say that the kinetics of the reaction bear some relationship to the stoichiometry. However, the data in Figure 2.3 show that the first-order behavior does not persist over a wide range of conditions. Experimentally, there is seldom any relationship between the
i,Hi i'/ ii rem linn mul the \lnlt humnu \ oj the rem linn (iMiseqiieully, sloii lllnineli y dors
mil i > in in I in vt* .in iiiipiiii.ini 11111111 * i it i* on tin- 11 n in ol the mil' equation
w. will dr.i ii/. thin point lurthci In Chuptoi I However, the Ice) thing to rcmcitihei
i.......w In thai rate data i an often be Hi to .i rate equation bul thai the rate equatli......ij
mil In m l.iinl lo (he sloiehionielry of the reaction in a simple- way.
i •   Siimm.uy ol the Effect of Composition on Rates
i......iin.iiy, then, so far in this chapter we have considered how rates ol reaction vtrj
llll concentration. We found that in most cases the rale of reaction varies in a simple ill composition. We often observe first- or second-order data. There are some i |in.mis. especially when we lake data over a wide range of conditions. However, Wt ......ii.illy find a rate equation that works even over a wide temperature range.
f.   TEMPERATURE EFFECTS
m litis point, we will be changing topics. So far, we have been talking about the
ill |.....Icncc of rale on composition. Now, we want to change topics and discuss how
li ii lion rates depend on temperature. Our objective will be to present Arrhenius' law, I ml Ihc I larcourt-Essen equation.
Sin.lies of the influence of temperature effects in reactions date back lo the nie.li.'\.il ill hi mists. The alchemists mixed substances together and saw what happened. One ol ih. iluiigs that was discovered was that many reactions turn on suddenly over a ni.ul. si
.....pcrature range. For example, the alchemists discovered that oil will suddenly ignite
vhen the oil is heated to 320°C. Similarly, a hydrogen-oxygen mixture is stable .ii KM) ( . Inii ignites suddenly at about 440°C. In a series of important experiments, Mey.i in.I Raum (1895) showed that if they held a mixture of hydrogen and oxygen at 300 <' I'oi 65 days, l -5% of the hydrogen and oxygen is converted into water. Therefore, it M II i ..in Imled that reactions do not really turn on suddenly. Instead, rates of reaction change with temperature.
People explored the effects of temperature on rates of reaction in the latter pari Ol Ihc nineteenth century. The studies were usually limited to a modest temperature range i 'ii ( I. The variations in rate were so small that no one was able to accurately test modell ini the temperature dependence of rates. Still, considerable data were generated.
In 1893, Kooij and Van't Hoff examined the decomposition of phosphine between Ho and 5I0°C. This was the first time that the kinetics of a single reaction was measured ovei li wide temperature range. Figure 2.4 shows a plot of the rate of phosphine decompoMii.ni ii reaction (2.T.1) (in Table 2.2) as measured by Kooij and Van't Hoff. Notice thai tin i.lie increases rapidly with temperature, doubling once every 35 K.
Prior to Kooij and Van't Hoff's work, there were two competing models in the literature .hi the effects of temperature on rates: Perrin's model and Arrhenius' model. Perrin (l')l()) had proposed that reactions were activated because the reactants had to accumulate enough energy to break bonds before reaction could occur. Perrin noted that the energy transfet could come from radiation or convection. If one calculates the rate of energy trangfei in a cold molecule, the rate of energy transfer from the walls of the vessel can be approximated by
ET = FJpT" (2.23)
II MM IIAIIIIII I I I I I  I', 10
550      600      650      700      750 800 Temperature
Figure 2.4 The rate of PH3 decomposition as a function of temperature. [Data of Kooij (1893).]
where Ex is the rate of energy transfer, T is the absolute temperature, Er is a constant, and n is a constant that varies from 1 (when forced convection dominates) to 4 (when radiation dominates). Consequently, Perrin suggested that if the rate of energy transfer from the walls of a vessel controls the rate of reaction, the rate constant for a given reaction k| can he written as
k, = k T"
(2.24)
where k[ is a constant. Equation (2.24) was also proposed by Harcourt and Essen on empirical grounds (the model worked).
An alternative model was proposed by Arrhenius. Arrhcnius suggested molecules needed to go over a barrier before reaction could occur. Wigner [1932] and Polanyi [1931] later represented the barrier as a hill between reactants and products as seen in Figure 2.5. Only hot molecules (i.e., molecules with a total energy greater than Ea) can react, and there are more hot molecules at higher temperatures. Consequently, rates increase with temperature.
Arrhenius quantified his ideas by assuming that the hot molecules were in equilibrium with the reactants. At equilibrium, FEa, the fraction of the molecules that arc hotter than
o c
LU
Reactants
Products
Reaction coordinate
Figure 2.5  Wigner and Poianyi's representation of Arrhenius' model of activation barriers to reactions.
I     i't hIvimi by
p.    e '*/k„i
(.' .'M
In it i ( is called ihc <i< tivatlon tntrgy, kn is Boltzmann'i constant, and. I is the ibtolutl
.....1» I ill lire N"ir thai Hull/in.inn's constant is jusl Ihc ideal gas law conslanl l R l di\ nlrd
i<t  tvogadro's number Also note the notation k|(. We use ihc tail on Ihc > to Indlottl
iiini kit is not a rate constant.
I he rule constant becomes
k, = k','e
„/kMr
(.' '61
lion I',' In i ailed the inrr\i><niciilitil. liquation (2.26) is called Arrhenius' law, ll is a key
i.....on for the remainder of this book.
I • 111111.>11 (.V.'K) is correct for Ihc case when the activation energy is measured In i Hoi ulorlcs per/molecule. If you measure activation barriers in kilocalories per mole, then ih. Inlliiwing equation should be used:
k" exp
Ea
RT
(2.27)
hi re again R is the gas law constant.
\ .hi i I loll" (1896) compared equations (2.24) and (2.26) to his phosphine data and httintl thai ai low pressure, equation (2.24) worked best. However, when he examined a kvlilei daia set, equations (2.24) and (2.26) worked equally well much of the time, Still, in i I loll found it hard to believe that radiation played an important role in chemical i. hi nuns Later, Langmuir [1920] showed that the emissivity of the walls did not affeel . lunik al reaction rates, so energy transfer from the walls of the vessel does nol affei I
ii urns As a result, Perrin's equation was discarded in the literature. Arrhenius' equation mi Ihc other hand is seen in most textbooks.
Still, Dunbar and McMahon (1998) have shown that radiation from the walls can have ,in Important influence on the rate of unimolecular reactions. Later in this book, we will iiinl thai energy transfer barriers to reactions are just as important as activation barriers I in energy transfer barriers are not barriers to energy transfer to the walls. Rather, the) iri barriers to energy transfer within a molecule. Reactions with energy transfer barriers .In nol follow Arrhenius' law [equation (2.26)]. Instead, they follow
k, = kum(Tf e
-eA/kbt
(2.28)
where k| is the rate constant, EA is the activation barrier, kB is Boltzmann's constant, I In temperature, and m and k^ are constants. According to Arrhenius' law, a plot of ihc log nl the rate versus 1/T should be linear. However, Figure 2.6 shows some data lor ,i typical reaction, and one observes significant curvature. Equation (2.28) fits the data in I igure 2.6 much better than does Arrhenius' law. Nevertheless, Arrhenius' law equation (2 !6) is cited much more often than equation (2.24) or (2.28). Therefore, the reado should memorize equation (2.26) before proceeding.
In my experience, Arrhenius' law works great when one is working over perhaps ,i 50-100 K temperature range. It only fails when one measures data over a much widei temperature range.
«U Ml VII W Ml m IMI I I ľ Ml N I AMY I ' H II III
II Ml M MAI IIIII I I I I I I
100
10
0 1
0.5
I r»r n| it tt Mil n it. K
	KMX) M00/00 <><	II 1.....	inn inn	
■	1  1   1 1	—r-	-1- 1-	
:			^<r^^*62qiorr_____•-- ---- 1 00 tOTr____g--	
I		—	---""^20 torr	
:	1 1	i	i           i i	
1.5
2.5
1000/T, K
3 5
Figure 2.6 The rate of the reaction CH + N2 -+ HCN + N as a function of the temperature. [Data of Becker, etal. (1996).]
2.5.1   Implications of Arrhenius' Law
In freshman chemistry, you learned that Arrhenius' law can be used to calculate activation barriers. The idea is to measure the rate of reaction as a function of temperature, and then make a plot of the log of the rate versus one over the temperature. The slope of the plot is proportional to the activation barrier. Figure 2.7 shows an example of that. Arrhenius' law equation (2.26) fits the data okay, although not as well as does equation (2.28).
In my view, if the only thing you could do with Arrhenius' law were to fit data, Arrhenius' law would not be as important. After all, equation (2.28) fits data much better. However, the real strength of Arrhenius' law is that it allows one to make predictions without doing very many experiments. In this section, we will show how Arrhenius' law can be used to make useful predictions from very little data.
According to Arrhenius' law, the rate constant for a reaction, k], will be given by
O.-E./kuT
k^e
(2.29)
where k" is the preexponential, Ea is the activation energy, kB is Boltzmann's constant, and T is temperature.
1000
100
ra 10
0.1
0.5
Temperature, K 1000 800700 600   500 400
300
I I I   I    I      I         I ^^^^^J,				
			^----        620tori-— ♦ ^^^^^ ^-—^^100 torr	♦
				
\	a__	♦ ^--'— ■	---9 20torr	:
!	♦ I	■ I 1	i i	
1.5
2 2.5 1000/T, K~1
3.5
Figure 2.7  A fit of the data in Figure 2.6 to Arrhenius' law.
In < I ■ 111 > 11 - ■ /, we will Inn! 11 tut fin an elementary reaction, k, is usually nhoul III1 " 7*01 "lid loi lust onlei reKtiOM, Ml'" ' A '/(molecule second) hu second oidei . . iiniis. and lO""' A /(molecule"' second) hu thud oidei reactions The reailei should on iiiiiil/e 11 icsc typical values ol the preexponenlials before proceeding,
Now, lei's consider a lirsl order reaction A   > It In Chapter 3, we will luid that if we id A nun ,i heakci and let il react, then 50% of the A will be reacted alter a lime i| •. I'll, ii by
t,/j = ^ (2.30)
i..... ' S shows a plot of \\ti versus liA calculated from equation (2.30) lot k"
'.'lid Notice thai at 300 K, a reaction with an activation barrier of 14 kcal/mol I i     ii'iini one millisecond (I ms) to go to 50% completion. A reaction with an activation
i....... ol IK kcal/mol takes about 1 second to go to 50% completion. A reaction wilh
.....malum harrier of 20.4 kcal/mol takes about 1 minute to go to 50% completion
'  ii lion with an activation barrier of 23 kcal/mol lakes about an hour to go to 509 ■ ,111| lb i it hi A reaction wilh an activation barrier of 24.8 kcal/mol will take a day to go 'i■. completion, A reaction with an activation barrier of 28 kcal/mol will take aboul , n in go to 50% completion. A reaction with an activation barrier of 31 kcal/mol ■ ii lake a century to go to 50% completion. A reaction with an activation harrici ol i' ul/mol will take about a million years to go to 50% completion, rii   n illy, this is very important because it means that if you run a reaction at room ii mpcrature and the rate is easily measurable (i.e., the reaction takes between 1 second mil I hour), the activation barrier for the reaction will usually be 20 ± 6 kcal/mol In Professor Masel's lab, students often do experiments where they slowly heal I i, in and watch for reaction. Generally, if one observes a reaction taking a minute ,ii .....i lemperature Tmimlle, then the activation energy for the reaction will be approximately
Ea = (1/15 kcal/mol-°K)T„
(2.31)
18
1E+14
1E+10
1E+6
1E+2
1E-2
1E-6
1,000,000 yi
1 min 1 sec
0 10 20 30 40
Ea, kcal/mol
Figure 2.8  A plot of r1/2 versus. Ea at 100, 200, 300, 400, and 500 K.
Ill VII W I II  i II 1MI   I I I Ml IN I nil T I ,1 "INI .1 I' ITI
II Ml'I IIA 11IMI    iih i 23
Nott iliai  I.....„„,  in <'t|iiiiiii>n 12 Ml Is U'iii|H'iuliiit' iii ili'i'in ■ Kelvin  Siniiliiily, II
the reaction taken h second al norm letnperatun   I .......,. then the activation barrlei 1«
approximately
Ea = kc-:il-mol)K|'l
(2.32)
We use the expression "the reaction takes a second" very loosely in equations (2.31) and (2.32). The equations were derived using the time where 50% of the product is used up. However, if you would instead consider the time it takes for 10% to be used up. or the time it takes for 90% to be used up, you could derive almost the same expression, except that the constant would change by perhaps 10%. Equations (2.31) and (2.32) are very important because they allow one to get a rough estimate of the activation energy from very little data. If you see a reaction at room temperature, you know that the reaction probably has a barrier of 20 ± 6 kcal/mol, independent of the details of the reaction. As an example, most of the chemical reactions occurring in the metabolic processes in living cells have activation barriers of 15-25 kcal/mol. So do the reactions involved with digesting your food or bleaching your clothes in a clothes washer. One can use equations (2.31) and (2.32) to get a reasonable value of the activation barrier for reaction for a wide variety of reactions without knowing much about the reaction. That makes equations (2.31) and (2.32) very powerful (although sometimes inaccurate.)
Still, equations (2.31) and (2.32) are approximations that assume that there is only one reaction occurring. If you have two competing reactions, or other complicated effects, then equations (2.31) and (2.32) will not work.
In my experience, equations (2.31) and (2.32) work in about 90% of the cases. In most of the other cases, there is something complicated going on, either there are competing reactions, or some other complexities. If you measure an activation barrier and it is way off from that predicted from equations (2.31) and (2.32), it is safe to assume that some complex process is happening during the reaction. You need to be very careful with your data in such a case.
Another application of equations (2.31) and (2.32) is to solve the equations to see at what temperature a given reaction will have a measurable rate. For example, one can solve equation (2.31) for TITlinute to show
15 K-mol
t -      — _f
1 minute — kcal
(2.33)
where Tminute is the temperature you would have to go to get a half-life of about a minute. According to equation (2.33), if a reaction has an activation barrier of 30 kcal/mol, you need to run the reaction at 450 K to get a reasonable reaction rate. Similarly, a reaction with an activation energy of 50 kcal/mol needs to be run at 750 K to get a reasonable reaction rate. Consequently, equation (2.33) allows one to plan experiments for a wide variety of systems.
Equation (2.33) can also be used to estimate ignition temperatures. If you have a gas burner in your kitchen, the gas usually stays in the (lame for about a tenth of a second. If the ignition process has an activation barrier of 50 kcal/mol, then, according to equation (2.33), you need to heat the system to about 800 K before ignition will occur.
K«|UMl.....I ■' 11 ) (•' 11) itiv very uselul in e.Hliinaling ncliviilion ban iris ami planning
i  nun ills The reiulei should incnioii/e these equations beloie piocrrdini!.
'1 11'111 ■1. io It.lie with Ii-mpci.ituie
1 H In ii.eliil io quantity Hie exlenl lo which rales go up will) increasing temperature
.......11 ' ' "    lion occurring al a rale ial temperature T,. According lo equation I.V.'O),
In! dm nib oidei reaction
r, = kfexp ^    '." j (CA)" (2.34)
.......lei i hanging the temperature to T... Ai thai temperature, the rate is given bj
r2 = k2,exp(-~^j (CA)" (2 13)
.......iniiioii (2.34) by equation (2.35) and rearranging yields
E, / 1 I
r2 = ri exp
kB
(2.36)
............ 1      allows one to calculate how the rate of a given reaction changes with
■ li111>- temperature.
H.   ' 'i shows the fractional change in the rate of an nth-order reaction when the
'  "i" 1......        is changed from 25 to 35°C as a function of the activation energy of the
' ...... 1 he increase varies from a factor of 1.1 to a factor of 10 for reasonable values
III I In in nvalion barrier. To put the plot in perspective, according to equation (2.32) S I  ii Hon thai lakes about a second at room temperature should have an activation energ) i inoul 'ii kcal/mol. Plugging into equation (2.36) shows that the rate of a reaction wiih <1 i ■ al/mol barrier should go up about 2| times for every 10°C.
20 30 Ea, kcal/mol
i iiiiini 2.9 The fractional change in the rate of an nth-order reaction when the temperature is changed from
Mto36*C.
t   ■  ■ ItIMMI
í uhlo ľ.ll    Ihn viill.itliiii In nit.' ul n not.......I i.....      Ilunn wilh n III K . IiiiimI" Iii
temperature
Reaction
CH3COOCH2CH1 + H20 CH3CH2CI + NaOH = CH3CH2CH2CI + NaOH = HPO3 + H20 > H3PO4
> (IM OOII i CH, CH, oil
H2C=CH2 + NaCI + H2() => CH3CH=CH2 + NaCI
I.....iŕl
w, 1111
23.5-43.6 24.5-43.6 0-61
K.11, i hange unii ii m K
lľlll|H-|||||llľ ( 'llllllgĽ
2.03 2.K7 2.68
1.0
Source: Data from Van't Hoff (1884).
Table 2.6 shows how the rates of several room-temperature reactions actually vary with temperature. Notice that the rate of room-temperature reactions increase by a factor of 2 or 3 for every 10 K increase in temperature, just as you learned in freshman chemistry. This happens because all room-temperature reactions have activation barriers of 20 ± 6 kcal/mol, and when the activation barrier is 20 ± 6 kcal/mol, the rate doubles or triples every 10°C.
You probably did not learn this in freshman chemistry, but the fact that rates double every 10°C is a very universal phenomenon. Plant growth follows Arrhenius' law. Growth of yeast or bacteria follows Arrhenius' law. Many other processes in nature also follow Arrhenius' law.
For example, Clausen (1890) examined the respiration rate of plants. He found that the respiration rate of wheat went up 2.47 times every 10°, lilac went up 2.48 times; and lupine (a blue flower) went up 2.46 times (see Table 2.7). Bailey and Ollis (1977) report the rate of E. coli (bacteria) growth as a function of temperature. Figure 2.10 shows their results. Notice that again, the rate of E. coli growth goes up about 2.5 times every 10 K until the yeast begins to die at 40°C.
A different example is the chirping of crickets. Male crickets chirp to attract mates, and the chirping rate changes as their temperature changes. Figure 2.11 shows how the rate of cricket chirping varies with temperature. Again the rate doubles every 10°C.
There are similar data on the growth of insects, lizards, and other cold-blooded animals. Even the speed at which ants walk doubles every 10°C. Another example includes the fact that bleach cleans clothes twice as fast when one raises the water temperature by 10°C. These examples show the idea that the rate of room temperature reactions goes up by about a factor of 2-3 for every 10°C increase in temperature; this rate change applies to a wide range of systems, and is not limited to things that happen in a chemical reactor.
Another application of Arrhenius' law is to learn how to control the rate of a chemical reaction. If you have a reaction that is too slow, the easiest way to speed it up is to increase the reaction temperature. Heating up a room temperature reaction usually doubles
Table 2.7  The variation in the respiration rate of plants with a 10° change in temperature
Wheat
Lilac
Lupine
2.47
2.48 2.46
II Ml 'I IIA 11IIII Illicit! 26
I
I
3.1        3.2        3.3        3.4        3.5 3.6 1000/T, K_1
I ii......."HI   I lii> n id - nl I. coli growth as a function of temperature. [Adapted from Bailey and (litis (l'l//| |
25"C 20'C
I OX
15C
I» 1
1 ......-'-11   The rate that crickets chirp as a function of temperature. Data for field crickets (Grylly:
1 Imiw.us). From Heinrich (1993).
......pies the rate of reaction. When you are running a reactor in industry, you often adjuil
I In i. inperature to control the rate. Equation (2.33) will allow you to estimate the required I. mperature.
.' !>3   Exceptions to Arrhenius' Law
I hi re is one key subtlety in Arrhenius' law, in that it applies only to elementary reactions, III M overall reactions. The terms elementary reactions and overall reactions will be defined in < hapler 3.) If you consider a complex reaction, Arrhenius' law will work for each ol tin . lementary reactions. However, the overall reaction might not show Arrhenius behavioi In this section, we want to work out a simple example, to illustrate the fact that overall i .iii-, might not show Arrhenius behavior even though elementary rates do. (consider a reversible reaction A ^ B. In Chapter 3 we will find
Source: Data from Clausen (1890).
rB = k2[A] - k2[B]
(2.37)
where Isi is ilit- rnio iimisi.iiii liii ilic loiward roucllou and k is die rulo conslanl loi iho revoisc i cat I um llnlh k| .nul k. lollow Aiiliciiius' law
k" exp
k2 = k" cxp
r' k„T
k„T
where k" and k2 respectively are the preexponentials for the forward and reverse reactions and Eg and E2d are the activation barriers for each reaction. In Chapter 3 we will show that for a reversible reaction
El = El + AHr
where AHr is the heat at reaction.
Figure 2.12 shows a plot of the rate calculated from equation (2.37) as a function of temperature. Notice that the overall rate of the endothermic reaction shows significant deviations from Arrhenius' law even though k| and k2 follow Arrhenius' law. This example illustrates that Arrhenius' law applies only to elementary reactions.
2.5.4  Important Exceptions to Arrhenius' Law
The example in the previous section was not a typical example, but in fact all reactions that follow complex rate laws do not also follow Arrhenius' law. Instead, a plot of the log of the overall rate versus one over temperature shows nonlinear behavior.
For example, earlier in this chapter we noted that the growth of bacteria follows Monod kinetics:
k,K2[£. coli]
I"ecoli —
(1 +K2[£. coli])
(2.38)
Figure 2.12 The rate of the reaction A^B with AH = +3 kcal/mol, and [A] = 1 mol/liter [B] = 0.01 mol/liter.
15 kcal/mol,
II i'I
ii i.l
ii 11
1Ei0
•1MIK     '100 K II!>(l K
		iii	
			
	[8]		
			
	[sl	0.1	
1,1.			
2.5
3.5
10007T, K
NyiH* V 1.1   Iho temperature dependence of a reaction that follows Monod's law rs = kiK2[S]/(1 I K^[S|| in' '/second -20 (kcal/mol)/kBT and K2 = 6 x 10 6 (10 kcal/mol)/kBT.
Ill   i..........2.38), k| and k2 follow Arrhenius' law. However, E. coli shows nonlineai
hvllHvlor.
i I im 2 I < shows a plot of the rate versus temperature calculated from Monod's law lilt oini ivpical values of the parameters. Notice that the rate shows no linear bcliavioi
.ii......perature. Our rules of thumb from Section 2.14 do not work in this example,
In my experience, most simple reactions follow Arrhenius' law. However, the overall i niiisi industrially important reactions cannot be accurately fit with Arrhenius' law. mi we note that each elementary reaction will follow Arrhenius' law. Ii is jusi the ■ ||i   ii overall reaction that shows derivations from Arrhenius' law.
CATALYSTS
In,In ,ii lally, the deviations from Arrhenius' law are quite important. Recall thai moil
"i i.....il reactions are run with catalysts. Well, most catalytic reactions do nol follow
\ 11 Ii, urns' law. These exceptions are so important that I wanted to highlight them early in tin lMink. In the remainder of this chapter, I want to discuss catalytic reactions, and show how the kinetics at catalytic reactions are different from the kinetics of simple reaction! in lolution.
In start, it is useful to review some background material on catalysis. Ostwald defined i ii.ilyst as "a substance one adds to a chemical reaction to speed up the reaction withoui Ihi ' atalyst undergoing a chemical change itself." Later in this book, we will find thai this ill llnition is not quite accurate. Catalysts do undergo chemical changes during the course ill reaction. It is just that the changes are reversible, so that the catalyst is not consumed i the reaction proceeds. Examples of catalysts include the acids in your stomach thai hi use to break down food, and the enzymes that people put in detergents to make iii. ilciergent work better. Most chemical processes use catalysts at some stage in the production process.
I here are two kinds of catalysts: homogeneous catalysts and heterogeneous catalysts.
I liiuiogeneous catalysts are things that one dissolves in a liquid or a gas to promote a
ni vil vv i M   li mni ill mi ninny i i ir
CAt/VI i ' ■ I' ■
n .i' I.....        HoltlOgl.........il il\  i   ill' III.I'  III hi', .liul I...... in      li .............. I.il i.hi'
anil alkyls. In sonic cases, solvent! tan ail as homogeneous catalysis
A heterogeneous catalyst is a solid thai one adds i" .1 re* in.....ilxture 1 he solid does
not dissolve in die mixture, hut still is able to promote a desired reaction 01 a scries of desired reactions. Typical heterogeneous catalysis include transition metals (e.g . platinum on alumina) or solid acids such as silica/alumina.
Heterogeneous and homogeneous catalysts are fundamentally different. A homogeneous catalyst dissolves into the gas phase or solution and acts uniformly throughout the liquid or gas mixture. On the other hand, heterogeneous catalysts do not dissolve. Instead, the reaction occurs on the surface of the catalyst.
Table 2.8 shows some typical homogeneous catalysts. Generally, homogeneous catalysts consist of transition metal atoms, peroxoradicals, and acids or bases. Industrially, one uses homogeneous catalysts in most polymerization and carbonylation reactions, and in some partial oxidations.
Homogeneous catalysts are also used in the home. For example, most detergents contain enzymes to allow the oxidation and dcpolymerization of stains (i.e., dissolved proteins). Cleansers contain metal atoms to speed up the action of bleach. Most biological reactions are also controlled by homogeneous catalysts, such as enzymes or RNA.
Heterogeneous catalysts are more important industrially. Figure 2.14 shows some typical heterogeneous catalysts. Heterogeneous catalysts are generally powders or pellets that one can add to a reacting mixture to speed up the reaction. Heterogeneous catalysts are used extensively in chemical processing because heterogeneous catalysts are easier to separate from the products of a reaction mixture than a homogeneous catalyst. Table 2.9 shows some typical heterogeneous catalysts. Over 90% of all bulk chemicals and petroleum products are made via catalytic processes.
Heterogeneous catalysts are also used in people's homes. For example, people used to sell catalytic igniters that one could use to start a gas range or fireplace. Now the most common heterogeneous catalyst in your possession is the catalytic converter in your car. The catalytic converter reduces the pollution and toxic emissions produced by your car. People also make catalytic converters to destroy pollutants in homes or in woodstove chimneys. Heterogeneous catalysts are not as common as homogeneous catalysts in the home but they are equally important.
Catalysts arc so important because they make tremendous differences in reaction rate. For example, a hydrogen/oxygen mixture may be stable for years at 25°C. However, if one adds a platinum wire to the mixture, the mixture explodes. Table 2.10 shows some
Table 2.8  Some Common Homogeneous Catalysts
Reaction
Catalyst
Ethylene -»• polyethylene (polymerization) Ethylene —> polyethylene, styrene —> polystyrene C2H4 + \(h -*■ CH3(C=0)H (Wacker process) Olefins + CO + H2 —>■ aldehydes (hydroformylation) CH3OH + CO -s- CH3COOH SO2 -> SO3 (lead chamber process) CH3COOH + CH3OH -»■ CH3COOCH3 + H20 (sucrose -*■ glucose + fructose)
TiCl4/Al(C2H5)3 (Ziegler-Natta catalyst)
Peroxides
PdEt3
Co(CO)6
RhCb
NO/N02
Acids or bases; invertase
- -ptw""
mmmĚSĚmKĚImmm
1.....rt ^.14  Photographs of some heterogenous catalysts. (From Wijngaarden and Westerterp, 1998).
Initio 2.9  Common Heterogeneous Catalysts
l iilulyst
Reaction
11........11 mi alumina, nickel on	Hydrogeneration/dehydrogenation
'.......la	
ri.........i/lln on acidic alumina	Reforming
guild .11 ills (zeolites)	Hydrocarbon isomerization, cracking
Silver	C2H4 + j02 -> ethylene oxide
lll-(t,l,(M03)y	CH2=CHCH3 + §02 + NH3 -> CH2CH-CHO 1 1 M >
v |Oj	so2 + |o2^so3
II ii......11 gauze	2NH3 + 402 -> N205 + 3H20
Dthei examples. Notice that in the examples shown, the catalyst lowers the activation i' irrier for the reaction by 19-30 kcal/mol. It is not unusual for a catalyst to lower the h irrier for a reaction by 19-30 kcal/mol, and Masel (1996) gives some special examples here the catalyst lowers the activation barrier for a reaction by more than 50 kcal/mol. \ I 9 30-kcal/mol change in the barrier for a reaction changes the rate tremendously. The . samples in Table 2.10 show rate enhancements of 108—1013. Similar changes in rate are quite common. Masel (1996) reports some special cases where the rate enhancements are .r. large as 1021.
from a practical standpoint, one of the key roles of a catalyst is to lower the temperature where a reaction takes place. According to Table 2.10, the reaction
(C2H5)20 :
2C2H4 + H20
(2.39)
Ill VII WIM   Ml IIVII    I I I I II I I I'M I ,   i   i H II .| I'l'i
i Ai/m , n. mih m ■ :n
IiiIiIii 7 10 Mm ■ 11 11 m |i in niln ul •linuli lypii ni 11 ii n liuiľ, 'Huni wlnii li i nt.ily.l c. iiililml In lliľ n.n In m 11 u x 11 n i
Reaction	Catalyst	I'„ 1 Incatalyzed, kcal/mol	U. Catalyzed, kcal/mol	Rate "i lull.111. illli'lll ('alculated a) 500 K
H2 + I2 -> 2HI	Pt	44	14	1013
2N20 -> 2N2 + 02	An	58		1(1"
(C2H5)20^ 2C2H4 + H20	h	53	34	I0N
Source: Table adapted from data in Bond (1987).
has an activation barrier of 53 kcal/mol in the absence of a catalyst and 38 kcal/mol in the presence of a catalyst. According to equation (2.33), you need to run a reaction with an activation barrier of 53 kcal/mol at about 800 K. While you can run a reaction with an activation barrier of 38 kcal/mol at 600 K, that temperature difference is huge if you need to pay the heating costs. Besides, there is an explosion hazard at the higher temperature. In industry, one hardly ever runs a reaction without a catalyst, which is why catalysis is so important. In 1995, almost 700 billion dollars of products were made with catalysts in the United States.
2.7   CATALYTIC KINETICS
One of the key features of catalysts is that they change the form of the rate equation. Over the years, there have been many attempts to determine how rates of reactions on heterogeneous catalysts vary with the partial pressure of the reactants. One of the things that has been discovered is that rates of catalytic reactions do not bear any simple relationship to the stoichiometry. In the gas phase, rates of reactions are often proportional to the reactant concentrations to some simple powers. However, catalytic reactions follow much more complex rate equations. It is common for the rate of a catalytic reaction to be constant or even go down as the concentration of one of the reactants increases. This is quite different from gas-phase reactions, where rates generally increase with increasing reactant pressure.
For example, Figure 2.15 shows data on the rate of CO oxidation on a catalyst called Rh(l 11). Notice that the rate increases linearly with the CO concentration up to a CO partial pressure of 10~7 torr. However, then the rate decreases again. One cannot fit data with a rate equation like that in equation (2.11). However, one can fit them with a more complex rate form:
reo
k|PcoPo2
(1 +K2Pco + K3Po2)2
(2.40)
where rCo is the rate of reaction; PCo is the CO pressure; P0 is the oxygen pressure; and, ki, K2, and K3 are constants.
In the discussion that follows, we will provide examples of this effect for two separate cases: unimolecular surface reactions, where a single reactant rearranges or decomposes to yield products; and bimolecular surface reactions, where two or more reactants combine and rearrange to form products.
in'
, in'
P
10'
'   1 1 1 1 11 n in }r	
	
:425	
- 415 K	
10 «
10 1
CO pressure, torr
10"6 10"
10-7
02 pressure, torr
Mum* J Itt Ilm inllnence of the CO pressure on the rate of CO oxidation on Rh(111). [Data of Schwnil/, ■I Hl 11 Ulm) I
1E+20
1E+19
■§ 1E+18
I o E
of 1E+17 rr
TIM 1     1-1  I I I I
1670
1E+16 0.01
0.1 1 10
Ammonia pressure, torr
100
i      «• .' Ill   I he rate of the reaction NH3 => ±N2 + §H2 over a platinum wire catalyst. [Data of Lofllni niul
Hi.I ii nl! II (tt)/6a,b).]
I inure 2.16 shows some data for the rate of a simple unimolecular surface reaction: iin .I. i omposition of ammonia over a polycrystalline platinum wire catalyst.
NH3 —► \N2 + |H2
(2.41)
in i  thai at moderate temperatures, the rate of reaction increases with increasing
............a pressure and then levels off. At higher temperatures the rate continues to increase
nvei the pressure range shown. However, other data show that the rate will eventually i ..II at sufficiently high pressures.
i i)iiu' 2.17 shows how the reaction rate varies with temperature at a series of fixed tires. Notice that the rate increases with increasing temperature and then begins to level ..II
Mi »Ii f» .i rivil   I I I ivil II I ni 11 li UNI .1 rill
(Al Al Y IM  MNI II
I
II i IM
E
o,
A o
3
O QJ O
E
ai 5 cr
1E+17
1E+16
' I	i       ] i b /b	
: a/		
		/ N
		/ l
	ř i/i) lZ-—i--"""l	/ Arrhenius'
c A		' Law i     i     i i
800
1000
1200 1400 Temperature, K
1600
1800
Figure 2.17  The variation in the rate of the reaction in Figure 2.16 with temperature: (a) Pnh3 =0.3,
Ph2 =0.15; (b)Pnh3 =0.3, Ph2 Loftier and Schmidt (1976a,b).]
= 0.44; (c) Pnh3 = 0.05, PH2 = 0.15; (d) PNH3 = 0.05, PHz = 0.45. [Data of
The fact that the data do not follow Arrhenius' law shows that there is some complication during the reaction. Later in this book, we will find that the ammonia sticks onto the platinum, and then there are two competing pathways: (1) reaction and (2) desorption back into the gas-phase. The presence of two competing pathways causes the curvature in Figure 2.17.
The results in Figures 2.16 and 2.17 are typical of those for unimolecular reactions on heterogeneous catalysts. Generally, one observes rates that increase with increasing temperature and pressure. However, the rate seldom increases monotonically with pressure. Rather, the rate increase slows with increasing pressure, and eventually plateaus at high pressures. The rate also increases with temperature. However, the slope of the curve decreases with increasing temperature.
Note that gas-phase reactions show behavior very different from that seen in Figures 2.16 and 2.17. In the gas-phase, the rate of a simple unimolecular reaction increases continuously with increasing pressure and never levels off. The rate also increases continuously with increasing temperature. According to Arrhenius' law
Rate = r0 exp
f'-.i
(2.42)
According to equation (2.42), the slope of the rate-temperature curve should increase with increasing temperature. However, Figure 2.17 shows that the slope actually decreases. Therefore, it seems that the kinetics of unimolecular catalytic reactions are quite different from the kinetics of unimolecular reactions in the gas-phase.
Even larger differences are seen for bimolecular reactions. For example, Figure 2.18 shows how the rate of the reaction
CO+ \o2
CO,
(2.43)
varies with temperature over a Rh( 111) catalyst. Notice that at fixed reactant pressure, the rate reaches a maximum with increasing temperature and then declines. Figure 2.15 shows the pressure dependence of the reaction rate. Notice that the rate first increases with
II i i:i
II 11:
II 111
1   I   1   1   1   1   '   I   '   1 ' P,;„     •' 1      1 l"ll	
	
	
II	1               .               1               .               1               .               1              .              1              . ■
•100 600 800
Temperature, K
400
600
Temperature, K
(«)()
ftfUf«l.ia   the  rate of the  reaction  C0+j02=>C02  on  Rh(111):  (a) PCo = 2.5 x 10 " torr
ľ.. H......n. «») I'm,    1 • 10 ' torr, P„, ' 2.5 x 10 " torr: (c) P,-„    8 ■ 10 ' Ion. I',,.    2!» -
m ....... (■!} i'm   2»10 'torr,  P0, = 4 x 107 torr; (e) Pco = 2 x 10 7 torr,  P0, = 2.5 x 10 " torr
|ft ř>Mi - 2,6 x 10 " torr, P0j = 2.5 x 10 8 torr. Data of Schwartz et al. (1986).
......II < O pressure and then declines. At lower temperatures, the rale continuously
wiih increasing CO pressure, while at higher temperatures the rate increases nli mi leasing CO pressure.
Mi. behavior in Figures 2.15 and 2.18 is typical of that for a bimolecular reaction H i In lerogeneous catalyst. The reaction rate generally shows complex behavior with ' iii|n i.iiiue and concentration. The behavior is usually quite different than thai ol
i.......li i ular gas-phase reactions.
11 |iu ally, one fits data for the rate of a surface reaction with a complex rate equation i 'i'll   'II shows a selection of the rate equations that have been fit to data on supported il i   Is I lie table is extracted from a longer compilation of Mezaki and lnouc (19911 Ihul die rale equations are often quite complex and the rate can go up or down w uli mi n rani' leaelant pressure.
I in example, (he rate of the reaction CO + 2H2 => CH3OH increases with increasing ■ 1 1 pi. . uie and then reaches a plateau. However, the rate continuously increases with I
......11''tis increase in H2 pressure. In contrast, the rate of the reaction S02 + j02 => S<),
i i id iun as the partial pressure of either reactant increases while the rate of the read.....
INI 11 I 6NO => 5N2 + 6H20 reaches a maximum with increasing reactant pressure and On ii 'li i lines. Generally, the kinetics of reactions on surfaces are quite different from die
i in.....■ "I reactions in the gas-phase or in solution, so much different rate equations are
in in i.illy used.
Ruble 2.1 I shows several other examples. Notice that the rate equations for surface . i. linns are generally more complex than the rate equations for typical gas-phase li ii nuns Generally, the rate equation has a term for each reactant in both the numerate! hi.I the denominator. The term in the numerator can cause the rate to decrease with ... i. using reactant concentration. By comparison, the rate of gas-phase reactions always ......a .es with increasing concentration.
I Ins unusual behavior arises because catalytic reactions are generally surface reactions. 11" reactants adsorb on the surface of the catalyst. Then there is a reaction between the '  ii iiuis lo form products.
CAIAI YIIC KINI lil '. 36
lithiu :• I I A -.olo, lion ul •.0111« ul Um into .«h 1.1Il>>mm lul -.111110 « miminu 1 nliilylli. 10.u linii'. i-xll.n Ind ll 1)111 lili.....llplliltlnn lil Mi/.iki .nul In.....1 (111(11)
Reaction
so2 + 302 -»- so3
N2 + §H2 -> NH3 CO + 2H2 -*• CH3OH C2H4 + H2 ^ C2Hň C2H4 + 202 -> C2H4O
1 latalysl
V2O5
Fe/Al203
CuO/ZuO/Al203
Ni/Al203
Ag/Al203
CO + H20 ^ H2 + C02 Fe203 /Cr203
4NH3 + 6NO -> 5N6 + 6H20 Pt
Knír Equation
k| K1 K.|I'si),l'(,, k'Ksl'sn, I I Kil',',,' I K4l'so, I Kjl'so,
kiPN,
P2
k-
1'nii,
"ph7
k|PcoPH2-l-2PCH3oH l+k3PH2+k4Pco + k5PcoPH'2
k]K2K3PH2Pc2h4
(1+K2Ph2 + K3PC2H4)2
kiK2Po2Pc2h4
1 + K2Po2Pc2h4
k, KatQPcoPH.o - k2K5K6PH2Pco2 (1 + K3PC0 + K4Ph2o + K5PCo2 + K6Ph2)2
k|K2K3PNOPNH3
(1 +K2PNO+K3Pj£3)2
Figure 2.19 Langmuir's model for the adsorption of gas on a solid catalyst. The light gray area represents the surface of the catalyst. The black dots are the sites on the catalyst are sites that are available to adsorb gas. The white ovals are the adsorbed A molecules. The dark ovals are adsorbed B molecules.
Figure 2.19 is a snapshot of what the surface during the reaction A + B => C. The light gray area represents the surface of the catalyst. The black dots are the sites on the catalyst are sites that are available to adsorb gas. The white ovals are the adsorbed A molecules. The dark ovals are adsorbed B molecules. During the reaction, an adsorbed A molecule reacts with an adsorbed B molecule to produce products. The rate of the reaction is proportional to the surface concentration of A and B.
An important detail is that the surface has a finite capacity to adsorb A and B. When you put more A on the surface, there is not as much room to hold B. Consequently, the B concentration will decrease.
Now consider a simple reaction A + B => C that follows
Rc = 1 Á2/(molecule-second) [A(ad)][Bud)]
(2.44)
In this equation Rc is the rate of formation of C, and [A(ad)] and [B(ad)] are the surface concentrations of A and B, in molecules/per square angstrom.
'   'i-      I '    lint 1 liniimm In II10 mlo n( pmili.......1 »,l <; ,r. |A......      | v.n n     .i-.-.i.....no
1      il i I"......I      1 '•'........lo/A'.ll,       I A    (1111 ill m 1 lie •.<>< < 111< I) |A,„,|| ||llllll(||
.....iri ule/A	III,,, (' |A(.	mnlci ulr/A ilt 111 ni i-, 1 from l)l + |B„ld)l= 1	K( . (molecule A i/mooikI (calculated from Equation I 1 11 >
111		(i n	0.09
11 •		0 g	0.16
0 J		0.5	0.25
o s		0	0.16
0,9		0 1	0.09
I .hi.   ! I 1 .hows how the rate of formation of C changes as |A(ad)] varies, assu......g
I i IH, „i,|     I molecule/A . Notice that initially the rale goes up as the A
........"| is increased. Hut then the rate reaches a maximum and decreases again with
.....ases in A. The decrease in rate causes the unusual behavior in Figure .'. IS
w, will derive a qualitative equation for this effect in Section 12.17.1. The thing to
.....ml" ' hn now is that rates of catalytic reactions show complex behavior because
..... ■ 1 the concentration of one reactant goes up, the surface concentrations of the
III III i leiulanls go down.
lilt othci thing that is special about a heterogeneously catalyzed reaction is thai the
.......ale scales as (he surface area of the catalyst, not the volume of the reactor. The
often measured as R, a rate in (molcm2)/min. That is different than the rate of a
......... a fluid, where r is the rate in (mol-cm3)/min or (moUitcr)/hour.
I     rimentally, the rate of a heterogeneously catalyzed reaction is proportional to the
........irea ol the catalyst. A catalyst with a surface area of 1000 nr will produce twice
.......h product as a catalyst with a surface area of 500 m2.
IVople often call reactions on heterogeneous catalysts surface reactions since the
■  ii........i rurs on the surface of the catalyst.
People have developed a number of special materials to squeeze as much surface I possible in as little volume as possible. Figure 2.20 shows some of the materials 1 Ii '" rally, il is possible to squeeze the surface area of 5000 m2 (a football field) into KKI i in' or less of material!
1 .....^-20  The amount of Linde molecular sieve, activated carbon, and y-alumina needed to get a surface
fii lout 5000 m2 (i.e., about the area of a football field). [From Masel (1996).]
Ml H I in i ',ll /viiy
ni', "in 11 ni n n i i iMľi i »im
1/
ľoi II typical catalytic reaction, míľ might havo a lalalysl Willi an ellei live surface .in .i Ol III1 ii ľ/lilii ni i ,il.11 \ -.1 .u ni ,i n.n i n n i i ale ni Ml    linul iiii/.i......I
In the literature, people rarely report a reai in m rate m mi ml m i/m-i ond Instead, they use a unit called a turnover number, TN. <>i turnovei frequenc) The turnovei number is defined as the rale ihai molecules are converted per atom in the surface ol the catalyst per second. One can calculate the turnover number from
(2.45)
where RA is the rate that molecules are converted in (molecule-cm2 )/second and Ns is the number of surface atoms/cm2. One usually sees turnover numbers expressed in units of reciprocal seconds (second-').
Figure 2.21 shows some typical values of the turnover numbers for some typical catalytic reactions. Note that typically, the turnover numbers are on the order of 1/second (i.e., 1 second1). Rates of 1/sec are also seen in semiconductor growth.
Students usually have trouble relating to a turnover number of 1/second. Just to give you an idea, a liter of a typical catalyst might have an effective surface area of 104m2. If you run a reaction over that catalyst, then, at a turnover rate of 1/second, you would produce about 10 mol/minute of product. If you had 2 liters of catalyst, you would produce 20 mol/minute of product. In general, we can write the reaction rate as 1 (molminute)/liter of catalyst.
102
10°
0) .£1
E
io~
10
~i-1-r
Hydrogenation
Dehydrogenation
7
Olefin isomerization
Silicon deposition
GaAs deposition
f
Alkane
hydrogenolysis cyclization
J_I_I_I_I_L
J_I_I_I_I_L
200      400      600      800     1000     1200 1400 Reaction temperature, K Figure 2.21  Turnover numbers for some typical reactions. [From Masel (1996).]
Nnw think Inn k lo your Organic 11 nil ii st r v days and recall i hi' icaclioiis you lau in
....... i ii' ii might hi v ■ i.ii rn to minutes to o I mol ol produi i In d !(K) i m
'   i    V i.....Iin Hun ol (I I mol in HI iiiiniiles in i .'00 cm vessel cmlespuuds to a leaclion
IMll- iii
" ......1 mol
0.01(1/   ;- I  ' I'll
I) I mol i m i........cs 1(0 ' litci i
lilci mm
.......ii nun ■ generally go 1000  10,000 limes faster than the reactions you run in
In mi ii\ I.ill Further, they will speed up reactions thai are too slow in n......I
i   .1 11 \ lah As a result, catalysis are very useful.
ill I" ill si us .in)' catalysts in detail in Chapter 12 in 111 is hook. For now. the ke\
.........ii« i is ih.ii catalysts tremendously change rates and also change the rati
i.........        lui n reaction
Mill riPLE STEADY STATES, OSCILLATIONS, AND OTHER IHIMI'I I XI TIES
...... "iln i subtler with catalytic reactions: sometimes, one cannot write a rate
n lui ii i atalylic reaction. All of the examples so far in this chapter had relatively ' in   i qualions. The rate could be written as a function of easily measured Ii    in h as the concentrations of the rcactanls and products and perhaps a catalj I
............i- )■... Ill1 |). There were no other variables in the rate equation. In most
.......in write the rale of reaction as a function of a scries of easily measurable
.....i.....        I here arc some exceptions. In this section, we will discuss the exceptions
Mi......(two kinds of systems that do not follow simple rate equations: those that shi IV
.....idled multiple steady states, and those that show something called osclllatloni
■ ii 'Ii ' ii'/. multiple steady states first, i..... 1.22  hows some data for the rate of CO oxidation:
CO+ {o2
CO,
(2.47)
.  ' rhodium catalyst. In the experiment, Schwartz et al. (1986) fed a fixed amounl
......nl <>i over a rhodium catalyst in a well-stirred reactor. Then the rale of ('<>■
......« is measured as a function of temperature. The conversion of the system was
1        " I hi i oncentrations of CO and O2 were constant within the reactor. Schwartz el al. ill ii u hen they heated the surface from a low temperature, the reaction rale followed 1 curve in the figure. In contrast, when they cooled from a high temperature, the ' ill lollowcd Ihc higher curve in the figure. Notice that at temperatures between 500 ami 1   Ihc 1.He is about an order of magnitude lower when the system is being heated up In 11 the system is being cooled down. Yet the concentration of all of the species hi.I iln lemperalure is constant.
Now   ilunk about how to write a rate equation for the data in Figure 2.22. One
......ii.iiiiie expressing the rate as a function of the reactant concentration, the product
titration, and temperature. However, that will not work for the data in Figure 2.22. I ill one observes two different rates at what are nominally the same conditions. As ■ 11 nil niie cannot write a simple rate equation for the data in Figure 2.22.
1 Ira) and Scott (1990) review several examples showing two different rates al whal
............lally the same conditions. The feature that all of these systems have in common
. ii. a u I1111■ the compositions are constant, some other internal variable is different when
M i| vi M I XAMI'I I 'i
IM
300
900
500 700 Temperature, K
Figure 2.22   Rate data for CO oxidation on Rh(100) catalyst. [Data of Schwartz et al. (1986).]
H« il Illinium nl liiiic I Ins milium is spci ml Ix-t iiusi- il one lunils n hxcd iiiiioiiiiI nl lllli) ) m,I in,lien ,u nl min 11 it- ii'.h loi, 11 it- leacloi does uol CORIC tO sleadv stale Instead.
ii......iipiisiiiiui in the retctoi oaclllatei with time, The rite nevei reichei iteady itate,
I In u.i illations in ľnú- seen wild the HZ reaclion an' more complex versions ol the
..... physics ili.il is causing the CO oxidation read ion to exhibit multiple Iteady Itattl
i ||i i in di |iciids on lho oxidation slate ol the cerium calalysl. I he cerium can he I I oi ' lad and the reactivity is different for ihc i t ami i 4 siaies In addition, there axe al i . i ílu., reaction pathways that occur simultaneously.
i Im i .111 write a rale equation lor reaction (2.48). However, Ihc rale is a lunclion ol (he
' . 1  and malate concentration. One cannot express the rate as a simple fund.....
..I till concentrations of ihc rcaclants and products. The result is a very complex iale
i......... Sec < i i a y and Scotl (I WO) for details.
I hi I . \ poml in remember for I he discussion in this book is thai in a complex cast thl
.....i n ill lion can depend on variables other than the concentration of the reaclanls ami
.....I the temperature. The additional variables make any kinetic analysis difficult
1 .m, i. ills. Van'l llolf's assumption that the rate is a function of the concentration ill   lot '<'>')', ol the reactions which have been studied so far. However, there are I
i. m i xi eptlons.
the rate is high than when the rate is low. The fact that you have an extra variable complicates the rate equation. For example, the rhodium catalyst in Figure 2.22 shows two different crystal structures. One crystal structure is more reactive than the other. The crystal structure then becomes another variable that one needs to consider in the rate equation. A surface with 30% of the first crystal structure and 70% of the second will have a reactivity quite different from that of a sample, which has 70% of the first crystal structure and 30% of the second crystal structure.
One can define a new internal variable, the percentage of the catalyst in each crystal structure. The different crystal structure is an internal variable which alters the rate. The presence of the extra internal variable makes the analysis of rate data difficult.
Some other examples have other complications. For example, Figure 2.23 shows data for the rate of the Belousov-Zhabotinskii (BZ) reaction:
HOOCCH2COOH + HBrO
products
(2.48)
0.35
1,00
Time, seconds
Figure 2.23 The concentration versus time measured during the Belousov-Zhabotinskii reaction. (Unpublished data of G. E. Poisson, D. A. Tuchman, and R. I. Masel.)
•il   ' .1IMMARY
III iimmary, then, in this chapter, we reviewed some of the elementary concepts In i in. in •. dial most of our readers had seen before. We defined the rate of reaction, ihc i.iic
• i.....on .iihI the rate constant. We saw what rate equations are like, apd we defined flrsl
....I ,i i ond-order reaction. We described how temperature affects rates and how lo use
il......loi mation. We briefly discussed catalysis. Most of the material was qualitative. We
ii quantify the material in the next several chapters.
1,10   SOLVED EXAMPLES
i   iniple 2.A   Illustration of Some of the Concepts from Section 2.1   Ethane emit
..... I ii mi cars are one of the major sources of air pollution in the United States. Mosi cai I
i...l.i\ are equipped with a catalytic converter, which, among other things, oxidizes ethane In i ii  and water. Consider the oxidation of ethane in the 10 liter catalytic convertei hown in Figure 2.A.I. The overall reaction is
702 + 2C2H6
4C02 + 6H20
(2.A.I)
V= 10 liters
0.16 mol/hour ethane 0.80 mol/hour oxygen 80.3 mol/hour water — dX.3 mol/hour carbon dioxide ()54 mol/hour nitrogen
I Inure 2.A.1   The flow rates into and out of a catalytic converter for the complete combustion of ethnnn
0.0 mol/hour ethane
0.24 mol/hour oxygen
80.8 mol/hour water
68.6 mol/hour carbon dioxide
654 mol/hour nitrogen
(a) Use a muss Imluncc In show Ihul tin- avenge lille ol |noilin lion ol in) ipw loi A in ili   icailanl 1 \ is
(mol/hour of A oni ol reactor)   (mol/houi ol A Into the reai tion) volume of ihe reactor
(b) Calculate rA for all of the species in the reactor.
(c) Calculate all of the stoichiometric coefficients (i.e., p\ values).
(d) Calculate the rate of the reaction from the equation
1
12.a = ^-rA Pa
Do a separate calculation for each species.
(e) How do the calculations compare?
Solution
(a) A mass balance on species A in the reactor is
Moles of A out per hour — moles of A in per hour = moles of A produced per hour
The definition of rA the rate of production of species A is
moles of A produced per hour volume of reactor
Combining (2.A.2) and (2.A.3) yields
moles of A out per hour — moles of A in per hour
rA = -
(2.A.2)
(2.A.3)
(2.A.4)
volume of reactor
(b) Equation (2.A.4) is the key equation for this problem plugging in the numbers 0 mol/hr — 16 mol/hr
ethane —
ro2 = rH2o = i"co2 =
Tn2
10 liters
0.24 mol/hour - 0.80 mol/hour
10 liters 80.8 mol/hour - 80.3 mol/hour
10 liters 68.6 mol/hour — 68.3 mol/hour
10 liters 654 mol/hour — 654 mol/hour
10 liters
(c) po2 = -7, pc2h6 = -2, f3co2 = +4, Ph,o = +6.
0.016 mol/(literhour)
= -0.056 mol/(literhour) = 0.05 mol/(literhour) = 0.03 mol/(liter-hour) = 0 mol/(litcrhour)
'II 11 VI II I XAMI'I 1
11
lilt Ni \i 1 llotllulc 11 a 1 lite tale ol lem lion .' A I  An online to equation I.' <>) I
i,,
(2.A.3)
I   1) 056 inol/thiei hum i|     0.008 mol/llilci 11..111 1
' ' \ 1
I
p\>,
I I
11 .11, I  0.016 mol/(literhour)]    0.008 mol/(liter-hour)
1V..11,, (—2)
I I
iio. =   10.03 mol/(liter hour)] = 0.0075 mol/(lilei limn)
Pcoj 4
; '   in,,, = ' |0.()5 mol/(literhour)| = 0.0083 mol/lliter lion.)
PH20 6
1, 1 Notice that all of the values of r2/u are the same lo within the measurement emu
.....1»I»- '.It   Comparing First- and Second-Order Reactions   Consider a first ordei
.......mil 1 second-order reaction: How much would the rate change if you diluted the
.....mils In a factor of 3?
Holulhn   lor a first-order reaction
r — kACA
let two different concentrations Ca and CA. The rates are
ra — kACA rA = kACA
Dividing (2.U.2) by (2.B.3) yields
r1 C 'a _ '-a
Ta Ca
(2.11.1 )
(2.B.2) (2.B.3)
(2.B.4)
1 In rclbre, if we dilute the reactants by a factor of 3, the rate will go down by a factoi
nl I:
,1      /A 1
rA
11 ni .lead we have a second-order reaction, then
rA = k2(CA)2 rA = k2(CA)2
rA C[
ra    \ cA
(2.B.5) (2.B.6)
(2.B.7)
I liricloie. il WC tliluli
ol 9:
ii .ii I.nils liy it I in lot ol I. Ihr mil- will (Jii ilnwn hy 11 lncloi
The difference between these results gives one a quick way to determine lite order of a reaction.
Example 2.C   Using Equation (2.32) to Estimate Ignition Temperatures   How hot
would you need to heat a methane-oxygen mixture before the mixture ignites? Assume that the activation barrier for the reaction is half the bond energy of the weakest bond in the methane molecule (bond energy = 104 kcal/mol).
Solution Ignition is a complex problem to solve exactly because one needs to consider the increases in temperature during ignition, how much heat is lost to the walls, and other effects. However, one can approximate the autoignition temperature from equation (2.34) assuming an ignition time of about 1 second.
T —
1 sec —
0.06 (kcal/(mol-K)
If we assume Ea = 52 kcal/mol (i.e., half the C-H bond energy), we calculatate
^(52 kcal/mol)^
0.06
kcal mol-K 1
866 K
By comparison, the experimental value is 810 K.
Example 2.D Fitting Data to Equation 2.28 In Section 2.6, we noted that if one measures rate data over a wide temperature range, it is often necessary to fit the data to
k, = k?TNexp
E, kBT
(2.D.1)
rather than Arrhenius' law. The objective here is to do an example of the fitting procedure. Yang et al. (1995) examined the rate of the reaction
NCO + NO:
C02 + N2
(2.D.2)
As a function of temperature. Their data are given in Table 2.D.I. In the table, the apparent rate constant is defined as rate/([NCO][NO|) and is measured in units of 10"cmV(moleculesecond).
Fit the data in Table 2.D.1 to equation (2.D.1).
Solution The easiest way to solve this problem is to use the linear regression tool in Microsoft Excel or Lotus 1-2-3. I used Excel.
M H VI 111 KAMI'I I 'i 43 Ii I    II»' i,ili> ol nun lion :•<::'......Iiiiii linn öl li'iii|k'i.iliiii.
1 l!l|ll'llllllll\	Ai>|mit-iit Rite	Icinpi i ......	A|i|iiin'iil Rate	li'iii|iiiiiiiiii',	....... H lit
K	(lomuuil	K	( IMIKlilllt	K	( 'mislaid
JUt	1 x	432	192	632	1.72
111	1 )6	538	2.28	836	1 's
I ii iii i .ii i j'ei I equation (2.1). I) as follows:
111 < k, )    111 (k V) I N In('l')
Fi, ki.T
t.'l» 11
.......'i> I) is a linear equation, with two variables, ln(T) and I/T. Therefore a lineal
..... p.n k.ig.e can be used to III the unknown parameters ln(k"). N. ami E,/kg I
'   hied i" solve I lie equation in Microsoft Excel.
t 11 up m\ spreadsheet as shown in fable 2.D.2. The temperature is listed in column i ii. i iinsI.ml K is listed in column B. I want to do a linear in of ln(k) versus 11n I i Ml  i . > 11111111 (' gives ln(k), column D gives ln(T), and column E gives l/T.
111.......1 iln- regression analysis tool (under analysis tools) in Excel. I called c2 through
.....I il' iluougli c7 X. I then used the analysis tool lo solve for the coefficients, I he
......il gives lols of output. The key part of the oulput is
Coefficients
Standard Error
Intercept 14.362 X variable 1 -2.0136 X variable 2 -469.29
Hu ii line the best lit of the line is
ln(k) = (14.4 ± 1.0)- (2±0.15)lnT +
1.07 0.1486 70.0
(-469 ± 70)
I In ii line
k = exp(14.4)T~2 exp
-469
I" .' D.2 The formulas used to solve Example 2.D.1
A	B	C	D	E
	k	In k	ln(T)	1 n
	3.8	=LN($B2)	=LN(A2)	= 1 /$A2
	3.36	=LN($B3)	=LN(A3)	= 1 /$A3
	2.92	=LN($B4)	=LN(A4)	= 1 /SA4
	2.28	=LN($B5)	=LN(A5)	= 1 /$A5
	1 . 72	=LN($B6)	=LN(A6)	- 1/SA6
	1 .28	=LN($B7)	=LN(A7)	= 1 /$A7
'.i M VI III XAMI'I I II 46
I lull- is nnr nllu'i ilcliill ol utile  Dilliiiiil . 1111 . ii 1.1 nil piogiiims give different usiills
I'm example <.iii.iii.iihi> gives
('ocflicicni
I Mill
Intercept 13.634 0.0467
X variable 2 -1.92033 0.3562
X variable 2        -408.09 166.479
This difference arises because of a bug in Excel.
Example 2.E Determining Reaction Order Under some conditions the platinum-catalyzed reaction CO + ^02 ^ C02 follows the rate equation
fco2 =
kl[C0][02]l/2 (1 + K2[CO])2
(2.E.1)
(a) What is the order of the reaction?
(b) If K2[CO] » 1, what is the order of the reaction?
(c) If K2[CO]     1, what is the order of the reaction?
Solution
(a) No general order.
(b) If K2[CO] S> 1, the rate equation becomes
ico2
ki[CO][Q2]'/2 = k,[CO][Q2]'/2 = k, [Q2]l/2 (; + K2[CO])2      (K2)2[CO]2      (K2)2 [CO]
Therefore, half order in 02; negative one order in CO; negative half-order overall, (c) If K2[CO] « 1, then
_ k,[CO][02]^ _ 1/2
rc02-(i+iyeo])2-k,[C01[021
Thus, first-order in [CO]; half-order in 02; |-order overall. Examples of fitting catalytic rate constants will be given in Chapter 3.
Example 2.F Numerical Integration of a Rate Equation Assume that you are running a reaction A => B that follows
,'-20 kcal/mol \ 10''/second) exp (-—-j [A]
(2.F.1)
where rA is the rate of reaction, T is temperature, kB is Boltzmann's constant, and [A] is the A concentration. The temperature varies during the course of the reaction according to;
i     Hill k I iii k sin
( ' ) \ iii seconds /
(2.F.2)
hIh'H1 i is nine. How long will u lake to reduce the A concentration from I mol/litei to ii i iniil/iiiii'
\ntintini    \i i Hiding lo equation (2.4)
d[A] .Ii
= rA
•   "iiiiiilii|t equations (2.F.1)-(2.F.3) yields
.11A |
(I x It)' '/second) exp
(20 kcal/mol)
kB(300 K + 10 Ksin(l/I() seconds))
Mi .niiiii('iii).' and integrating
fc* dlAI ,, f
tlx I (("/second) exp 11  IA | Jo
20 kcal/mol
i ' I i.
|A| (2.1 h
.Ii
kB(300 K+ 10 Ksin(l/I() seconds))
(2.F.3)
Ii ii i is ihe initial concentration of A and CA is the concentration a( time t. Performing ........|i.il mi the left-hand side yields
In [   A ]     (I x 10''/second) I exp
(20 kcal/mol).
,li
kB(300 K+ 10 Ksin(t/I() seconds))
(2.F.6)
■ iii hand side can be integrated with the trapezoidal rule. In the trapezoidal rule, divide lime into small segments At, then approximate the integral as a sum. For llliple, it you want to integrate a function F(t), you say
F(t)dt = Y, Fdi)At - -(F(to) + F(t„))At
(2.I-.7)
i|iicnily, according to the trapezoidal rule (20 kcal/mol)
exp
kB(300 K+ 10 Ksin(t/10 seconds)) (20 kcal/mol)
dl
^exp
- 0.5 (ex| + exp
kB(300 K+ 10 Ksin(ti/10 seconds)) (20 kcal/mol)
At
kB(300 K+ 10 Ksin(to/10 seconds)) (20 kcal/mol)
At
kB(300 K+ 10 Ksin(t„/10 seconds)).
At
(2.F.8)
ii vit vv ' ',    » -rvn   I I I tv1i n mi IT I .1 (Ml ,! I'll
where i, Is thi huh......ne Im n mi........ i1........... »nd 1« 1« Ihe llnal lime
(lombining (2.F 6) and i ' F 8) yield!
In
y^(l x 10l3/second)exp
-0.5 x 10" (exp + exp
20 kcal/mol k,t(300 K I 10 KsinO./IO seconds))
(20 kcal/mol)
Al
kB(300 K+ 10 Ksin(t«/I0 seconds)) (20 kcal/mol)
Al
Al
(2.F.9)
kB(300 K+ 10 Ksin(tn/10 seconds)). It is useful to simplify this expression by defining the temperature at time ti by
T; = 300 K + 10 Ksin(ti/10 seconds) (2.F.10) Combining equations (2.F.9) and (2.F.10) yields
In
= ]jP(l x 10l3/second)exp
20 kcal/mol \
20 kcal/mol \ kBTi )
At-0.5
x 10'
exp -
kBT0
20 kcal/mol \
(2.F.11)
Rearranging
f   „                             /   20 kcal/mol \ CA =CAexp^-^(l x 10l3/second)exp(--—-J
+0.5 x 10'
AI
20 kcal/mol \ /   20 kcal/mol \
^-JAt + eXPl--k-f^j
At
(2.F.12)
Therefore, I can integrate equation (2.F.12) numerically to calculate the concentration versus time. I used a Microsoft Excel worksheet to solve the problem. Table 2.F.1 shows the formulas in my worksheet; Table 2.F.2 shows the numerical values. I named cell bl to be dt so I could set delta time. I also named cell Dl CaO. I made column A time and column B, temperature. I defined column C as the term I needed to sum
Term = (1 x 10l3/second)exp
(20 kcal/mol)
kBT,
A!
(2.F.13)
Next I want to compute the integral from 0 to t„ where n varies. I defined column E to be the approximation to the integral
Integral
_                          /  20 kcal/mol \ ■ ]T(1 x 10l3/second)exp (--—-J At
+0.5 x 10'
(20 kcal/mol) \             /   20 kcal/mol \ exp | ---—--        ) At + exp I--—-       I At
kBT(,
kBTn )
(2.F.14)
■i
o — o o
co ^-
												ij				CM				■1	
•1		ii)		ill		1		CP				»—				.		,		,	
iii		iii		iii		iii		hi		iii		111		111		in		111		111	
ii		1)		11		11		ii		,1		ii		ii		ii		ii		ii	
				X				<				X		-		-		-		-	
111		UJ		111		in		,11		Ill		iii		Ill		in		hi		in	
1 >		1 1		I 1		O		' 1		d		1 ,		, 1		, >				i i	
<?		10		■11				,11		hi		,11		,11		t				,h	
i 1		1 1		1 i		, >		, >		, >		o		, >		c )				i i	
ii		ii		"		ii		11		n		ii		ii		ll		ii		ii	
												o		,		,m		,.,		-1	
																					
•1		in		CD		i		,11		,ii		n		1 1		Q		c 1		( l	
1 )		i 3		1 1		i i		a		i i		1		1		1		1		1	
1		1		1		1		i		1											
												o		1		im		,o		•1	
-i		in		<D		i		CO		05		T		.							
()		o		o		o		o		U		o		co							
1		+		1		1		1		1		+		1		+		+		l	
m		CO		CO		CO		co		(O		CO		<o		CO		n		C9	
a,		in				in		,/)		co		in		CO		1/1					
< >		o		o		C 3		c i		c )		C 1		C 1		c )		£			
in		m		m		m		in		m		in		in				in		"in	
< i		o		o		o		o		o		o		o		o		o		c 1	
ii		ii		ii		ii		n		n		ii		ii		ii		ii		ii	
												o				CM		CO		■1	
■l		in		CD				CO		O)		T—				,		T—			
o		o		O		o		o		o		O		o		o		o		,' 1	
co		CO		CO		CO		CO		CO		CO		CO		CO		C)		,•>	
in		to		00		CO		00		00		(ft		CO		vy		,/i			
o		o		O		o		CD		C J		O		c >		(i		< i		Q	
a		13		15		:z-				rz		33				ZL-		a		3-	
ZD		ZD		ZD		ZD		ZD		ZD		ZD		=1		ZD		ZD		D	
co		CO		CO		if)		GO		CO		CO		<n		CO		in		in	
"		ii		ii		ii		ii		ii		11		ii		ii		ii		ii	
		#								#		d				i '		CO		•t	
		10		CD				CO		cn											
CD		CO		CO		CO		CO		CO		CO		CO		CO		m		111	
																					
CO		CO		CO		CO		CO		CO		CO		CO		CO		co		lO	
cn		CO		CD		CD		cn		a>		cn		o>		it1		05		,1,	
																					
o		o		O		c		o		o		o		o		o		o		CD	
o		o		O		o		o		o		o		o		o		O		i ,	
o		o		d		o		d		d		d		d		o		o		o	
o	4-1	o	+-I	o	■H	o		o	•w	o	4-,	o	-w	o		o	4-,	o	1 '	c3	1 '
CM	■D	CM	13	CM	"O	CM	13	CM	■o	CM	"a	CM	TJ	CM	TD	CM	n	CM	ri	IN	11
						1		1		1		1		1		1		1		1	
CL	CO	CL	CO	11.	CO	CL	CO	CL	CO	CL	CO	CL	CO	CL	CO	CL	CO	qT	CO	cl	,■1
=EX	OJ	X uj	01	X uj		X LU	CO	=EX	CD	X uj	CD	X uj	CD	=EX	CD	=EX	CO	X in	,l>	x 111	
		in		CD		t>-		CO		CT>											
<		<		<		<		<		<		<		<		<		<		•I	
																21					
CO		CO		co		CO		CO		CO		CO		CO		(SI		Ul		1/1	
																					
o		o		o		o		o		o		o		o		O		CD			
+		+		+		+		+		+		+	_	+	_	+	_	+	.	+	
o	-—■	o	—	o	-—.	o	.—.	o	,—	o	-—,	o	O	o	O	o	o	O		, ,	
o	o	o	o	o	o	o	o	o	o	o	o	o	,—	o	T—	o		O		o	
CO	1—	CO		CO	1—	CO	T—	CO	T—	CO	T—	CO		CO		CO	-------	CO		CO	
ii	~~	ii		ii		ii		ii	.......	ii		ii	o	ii		ii	: .	ii	CO	ii	•1
																				+^	
■H				-H		4-,		-H		+J		■H		T3				D		ci	
■o						T3J		o		"O		T3		+		+		+		+	
+		+		+		+		+		1-		+		o				CM		CO	
CO				in		CD		iv		CO		cn		-.—							
<		<		<		<		<		<		<		<		<		3		•'l	
ii		II		ii		ii		ii		ii		ii		ii		ii		ii		ii	
				cd		t~		00		cjj		o				cm		co			
																					
Ill
VII W I Ii M IMI I I I Ml I II /vil , l i i| II I IT
i,ii>ii-;'i     Mu- iiiiiiii'iii .ii v.iIim". in niť
iI'iIkkiI In i ,ili ill.....      i A
	A	B	c	D	I	I
1	di	2	Ca0=	I		
2	time	temp	term	sum	integral	Ca
'~3~~	0	300	0.047678	0.047678	0	1
4	2	301.9867	0.0595	0.107179	-0.05359	0.947821
i 5	4	303.8942	0.073401	0.180579	-0.12004	0.886885
6	6	305.6464	0.088809	0.269388	-0.20114	0.817794
7	8	307.1736	0.104667	0.374055	-0.29788	0.742389
8	10	308.4147	0.119477	0.493532	-0.40995	0.663681
9	12	309.3204	0.131501	0.625033	-0.53544	0.58541
10	14	309.8545	0.139115	0.764148	-0.67075	0.511324
11	16	309.9957	0.141197	0.905345	-0.81091	0.444454
12	18	309.7385	0.137427	1.042772	-0.95022	0.386656
13	20	309.093	0.128379	1 .171 152	-1.08312	0.338537
14	22	308.085	0.115361	1.286513	-1.20499	0.299694
	spreadsheet continues					
38	70	306.5699	0.098104	2.295655	-2.22276	0.108309
39	72	307.9367	0.113554	2.409208	-2.32859	0.097433
I summed all of the terms from 0 to t and subtracted that sum from the second term in equation (2.F.14). I defined column F to be the concentration at time t. I used equation (2.F.12) to calculate it.
Therefore it takes between 70 and 72 seconds to get to 90% conversion.
Additional information about Arrhenius' law is given in
K. J. Laidler, The development of Arrhenius's equation. J. Chein. Educ. 61, 494 (1984).
B. Heinrich, Hot Blooded Insects, Harvard University Press, Cambridge, MA, (1993).
J. R. Heulett, Deviations from the Arrhenius equation, Q. Rev., Chem. Soc. 18, 277 (1964).
2.11 PROBLEMS
2.1   Define the following terms in three sentences or less:
(a) Rate equation
(b) Order
(c) Rate constant
(d) Preexponential
(e) Activation barrier
(f) Arrhenius' law
(g) Catalysis
(h) Oscillating reaction
(i) Multiple steady state (j) Catalyst
(k) Stoichiometric coefficient (1) First order (m) Second order (n) Half order (o) Heterogeneous reaction (p) Homogeneous reaction
I'I Ii Uli I Ml
IM
\\ lull did \ini Irani lhal was new In vt......llns i lupin ' Weic you avvaie ol nil nl llie
.I.........mi. ' Was llit- liisliiiual minim.ilnm new'' Wcie you aware ol Ihc linulalioiis
• a iln Aiihiiiius equation and ill lepicscnlui)' dala hv I rale law? Were you aware lhal Aiilienius' law applies in so many practical situations? Did you know lhal you
■ "ulil i a mi.it. .in activation barriet without doing man) experiments? Whal elsi did you learn that was new'.'
' • I nul 'I' examples ol kinetic processes in your home, such as cooking ditleieiii I nul-, nl meals, washing clothes, you digesting different types of food, or plains I'liiwiii).' on your window sill.
' I   Identify Ihc mills lor
lul i \. the rate per unit volume
iin R ,, llie rule per unit surface area
0 l a lir.I ouler gas-phase reaction (ill a lusl order surface reaction
ii i  \ second order gas-phase reaction
ill a second-order surface reaction
\ in i oulei reaction has a rate constant of 0.15/minute.
mil all ulale the rate of reaction at a reactant concentration of 0.57 inol/liier.
1 l>i Kepeal lor a second-order reaction with a rate constant of 0.73 Liter/(mol-minute) ' A   1 ali ulale the activation barrier for a reaction whose rate exactly doubles from
mi lilt) to I 10 K
H.i 200 to 210 K
i. i 100 to 310 K
till 500 to 510 K
(•) 750 to 760 K
(f) 1000 to 1010 K
\i i ording to Table 2.3, the oxidation of phosphine (PH3) obeys
nu, = k5[PH3][02]
1/2
ll"     I i
• H
(ill What is the overall order of the reaction?
il>) What is the order with respect to phosphine?
(c) What is the order with respect to oxygen?
i«l) Mow much will the rate increase if you double the oxygen pressure keeping
Ihe phosphine pressure constant?
(e) How much will the rate increase if you double the phosphine pressure?
(I*) How much will the rate increase if you double the total pressure constant?
Hint: First define xPH3 = Pph3/PTotai- Show P02 = PTotai - Pph3 = Pibttll I XpH3). Then substitute into equation (P2.7.1).
later in this book, we will provide some methods to estimate the preexponenlials and activation barriers for reactions. Assume that we solve a problem and make a mistake, and calculate a 25 kcal/mol barrier for a reaction that actually has | barrier of 30 kcal/mol. The preexponential was calculated accurately.
ňu nil I Mil
(n) II von need In i .ill ill,ilľ llu" lull' constant ill Mill K. how huge .hi cnni in the rale will you make?
(I)) How much would you have in change the temperature to i ompensate foi the error (i.e., to increase the rate to the rate you calculate)?
2.9   Milk goes sour bceause baeleria in the milk converts the lactose in the milk into lactic acid.
(a) The reaction goes 40 times faster at 25°C than at 4°C. Estimate Ihe activation barrier.
(b) How does your estimate compare to that from Figure 2.9? Assume a half life of 12 hours.
(c) How do you account for the differences?
2.10 Your taste buds work by a complex pathway. First the receptors in your tongue bind to some of the components in food. That sets off a complex chain reaction leading to a change in polarity of the cell, and eventually the sensation of taste. The object of this problem is to use the material in this chapter to guess how changes in the temperature of your tongue will affect the perception of taste.
(a) How much would you expect the sensitivity of your tongue to change when you change the temperature of your taste buds by 10°C?
(b) Guess or measure how much the temperature of your tongue changes when you drink a glass of soda pop (carbonated beverage).
(c) Guess or measure how much the temperature of your tongue changes when you eat a dish of ice cream.
(d) Ice cream manufacturers say that they need about twice as much sugar to sweeten ice cream as to sweeten soda pop to the same perceived level of sweetness. Are your expectations consistent with these findings?
(e) Try an experiment to verify that this works. Take a sample of ice cream and store it in a sealed container in your refrigerator overnight. Let the ice cream warm up to I0°C. Now do a taste test. Taste the frozen ice cream. Taste the melted ice cream. Which tastes sweeter?
2.11 When you cook rice, you hydrolyze the cellulose in the rice to starch via the following chemical reaction:
Cellulose + H2Q
starch
Brown rice takes twice as long to cook as white rice because there is twice as much cellulose to hydrolyze. What does that tell you about the kinetics of the hydrolysis reaction?
(a) If you had a first-order reaction, how would the cooking time vary with the cellulose concentration?
(b) What does the fact that the cooking time doubles as the cellulose concentration doubles tell you about the kinetics of the reaction?
(c) How would the cooking time vary if you were working at an altitude of 10,000 feet (ft)? (Hint: The boiling point of water at 10,000 feet is 92°C.)
i 11   II.....n nun ll'l
(ll I > lollow s
11    O.OK/second I'll
ml Calculate im,. I>., and rM. al 500 (', and a I'll, partial pressure ol 5 Ion.
|/V<>/<•  /(if) Ion     I aim. (atmosphere ol pressure). | i,i How in.im milligrams ol phosphorous do you deposit in loo seconds''
M<   In 'in lion .' I. we reported that Lavoisier found that I kg of air always reacted uli n /K kg ol tin to yield 0.99 kg of tin calx (SnOl ami leave 0.7') kg ol Inert!
i.ii Hid I iivoisiet measure things correctly? Does I kg of air react with 0.79 kg
,,i mi in s ield tt 99 kg ol tin calx? ih) How many kilograms of the following substances will react with I kg ol
hi  aluminum forming Al2()i, iron forming Fc20, Ni forming NIOj, silicon
ll Hllllllg SiOi?
ill What are Ihe stoichiometric coefficients in the reaction to form tin i.il\' V.'.iimc a basis of one mole of tin.
ii   I,, Si, him ' I, we noted that Lavoisier showed that mass is conserved dining
■ i......, .il reactions by sealing a flask filled with tin and air and measuring the
hi i hange as the oxygen in the air reacted with the tin.
nil II you have a scaled I liter flask filled with air and 10 grams of fin. how much ■ ni i can you produce if you use up all of the oxygen in the air?
llil By much would the weight of the scaled flask change?
ii I lly much would the weight change if you opened the flask?
nil Now think about doing this in the year 1796. How would you measure Ihe weigh) change?
.....isidci the reaction 4PH3 -> P4 + 6H2. (P2.15)
ini What are the stoichiometric coefficients of each of the species in the reat tlon 11". 15)?
tin II we load 2 mol of PH3 into a reactor, and convert 50% of the I'll, into
pioducts, how much P4 and H2 would be produced? n i assume that we are using reaction (P2.15) to deposit phosphorous onto I
silicon wafer. Assume that we deposit 10 4 mol/(hourcm2) of P4 onourwafet
( alculate
11) The rate of reaction (P2.15)
(2) The rale of hydrogen production
(3) The rate of phosphine (PH3) production
2.15a ,
till II we write the reaction PH3->■ ^P4 + |H2, how will the stoichioinctiit
coefficients change?
• ii.    \k [he following reactions homogeneous or heterogeneous?
In) Cooking a steak (i.e., converting the collagen to glycogen and oxidizing Ihe myoglobin)
i In Bleaching your clothes (i.e., reactions of bleach with dirt)
UV
MIVIIWiil MIMI I I I MI NI AMY i I INI I ľľi
III I In ill lii| i i11 i ir,I mi y0U i ,11 lir , ir. ii Imp OXyjlCM Willi ii i hi I
(d) Combustion in youi car's engine (i.e . burning gasoline)
(e) Burning a candle
(f) Digesting the sugar in a glass of beer (i.e., turning sucrose lo glucose in your stomach)
2.17   In Example 2.A we considered the oxidation of ethane in a catalytic converter, However, another important reaction is the oxidation of ethanol:
CH3CH2OH + 302
2C02 + 3H20
Assume that you measure the oxidation of ethanol in a catalytic converter and the data in Table P2.17 are obtained.
(a) Calculate the rate of formation of each species.
(b) Calculate the overall rate of reaction.
(c) Notice that the nitrogen and oxygen do not quite balance. This indicates that another reaction is occurring. What is the stoichiometry of other reaction? (Hint: How much excess oxygen and nitrogen are in the exit? What other species could decompose or react to form the excess oxygen and nitrogen?)
2.18 In Problem 2.D we fit the data in Table 2.D.1 to equation (2.D.1).
(a) Set up the problem yourself and see if you can reproduce the findings. The spreadsheet is available from Professor Masel's website. http://www.uiuc.edu/~rl-masel or Wiley's website.
(b) What is the activation energy (with error bars)?
(c) By much would the activation barrier change if the point at 293 K were misrecorded so k(273) = 3.08?
(d) What does the result in (c) tell you about the importance of experimental error in the measurement of activation barriers?
2.19 In Example 2.F we used a numerical procedure to integrate a rate expression.
(a) Set up the problem yourself and see if you can reproduce the findings. The spreadsheet is available from Professor Masel's website.
(b) Assume that the temperature instead varies as T = 300 K + 5 K sin(t/ 10 seconds). How will your answer change?
Table P2.17 The inlet and outlet flow rates (in mol/ hour) from an experimental catalytic converter with a volume of 10 liters
Species	Inlet Flow Rate	Exit Flow Rate
Ethanol	0.21	o.ot
Oxygen	0.80	0.30
Water	80.5	81.1
Carbon dioxide	60.9	61.3
Nitrogen	637	637.1
I'lli Hli I MM
• III
I i I
hi   \ iMiinr   lhal   11 it*   I i*i ■ 11 tt-i ill i ii i-   instead   v.uii's  as   I      10(1 K I I K'.inll/ i.....nul.i How will youi nnswei change?
nil v ume lhal ii" tempen.....t Instead varies ns i    KM) K i o i K»in(l
Mi .ii mills). How will youi nnswei change?
hi w Ital .in lín si results tell you uboul the Importam e ol temperature control In kiin in i measurements?
1 11 ii" ilnlii in i Mim   1 / and 2.10 to Arrhenius' law How well do the data ňl In uli   ' .   (Hint Draw i lun- mi the graph. Do nol use a spreadsheet.)
lili I*' 'I .lnu*,'. V  Van Spacinloiik and R, I. Masel's unpublished d.il.i foi lln
■ 1  ' ni'i.....Kture of ethylene and hydrogen over a platinum catalyst, Di.....|
i-.....i ni. Van Spaendonk and Mascl adsorbed ethylene and hydrogen onto
■ ill, i .nul then healed the catalyst and looked for reaction Bach ol thi in I i j • 111«• I'.' .'I corresponds lo a different reaction pathway; For Instanci
'i'   ii.....     peak .ii no K corresponds to a reaction where two adsorbed species
li i. i in tin in ethane.
IN) I .....i Ihc data shown, how many chemical reactions do you see? [Hint  Bai h
III mil il read ion will produce one peak.)
lit) i   i  lín data in figure P2.2I to estimate the activation barrier foi each ol
i......i al reactions. Assume that the peak temperature corresponds to the
Ii inpi rulure when- the half-life is one second.
......I)|fi'live ol (his problem is to use a spreadsheet (e.g., Lotus or Excel) lo
........ tin implications of Arrhenius'law.
tunc thai you are running a first-order reaction in a batch reactor, and iii ii Ihc reaction has a half-life of 1 minute at some temperature I. Calculate
x1° i	\ Hydrogen \Methane
■	Ethylene
- j	Ethane
I    i    I    i I.....	
100 200 300 400 500 600 700 800 9001000 Temperature, K Figure p2.21   Van Spaendonk and Masel's data.
M        IIIVIIWul MiMI M I Ml NIAMY Cl INCI I'lfl
I'lH Hll I Mr. BS
Ihc iictiviiiinii energy ol ihc reaction as a function ol I, tissuiuing thai Ihc reaction lollows Arrheiiius' law with a |>ioex|>onenlial of III1'/second. For the purposes of this problem, considei l to range between KM) and I (MM) K.
(b) Vary the preexponenlial from 10" to I015/second. To what extent do your estimates of the activation barrier change?
(c) Change your estimate of the half-life from a minute to an hour. By how much do your estimates of the activation barrier change?
(d) Now go back to your estimate of the activation barrier in part (a). At each temperature, calculate how much you would have to increase the temperature in order to double the rate. {Hint: At 500 K you would need to increase the temperature to perhaps 514 to double the rate.)
(e) Use your results to see if the idea that rates double every 10 K works for this example.
2.23   Yang, Lee and Wang, International J. Of Chemical Kinetics, 27, (1995), examined the rate of the reaction
1111
NCO + N02
CO + 2NO
(P2.23)
2.24
as a function of temperature. Their data are given in Table P2.23. In the table, the rate constant is defined as rate/([NCO][N02]) and is measured in units of 10~" cm3/(moleculesecond).
(a) Are the units of the rate constant correct?
(b) Fit the data to Arrhenius' law. How well do they fit?
(c) Is the fit improved with equation (2.28)? Use the methods in Example 2.D.
(d) How do you interpret the sign of the activation barrier?
(e) On the basis of your fits and the error bars in Excel, how much confidence do you have in the value of the activation barrier?
Eberhard and Howard, International J. Of Chemical Kinetics, 28, (1996) 731 examined the rate of the reaction
CH3CH200 + NO ==> CH3CH20 + N02
(P2.24.1)
as a function of temperature. Their data are given in Table P2.24. In the table, the rate constant is defined as rate/([CH3CH200][NO]) and is measured in units of 10~12 cm3/(molecule-second).
Table P2.23  The rate of reaction (P2.23) as a function of temperature
Temperature,	Apparent Rate	Temperature,	Apparent Rate	Temperature,	Apparent Rate
K	Constant	K	Constant	K	Constant
294	2.68	384	2.08	579	1.14
320	2.43	431	2.02	665	1.12
351	2.57	493	1.58	774	1.05
i .,1,1.. IV .■•!    IIk. nit,, ol i„n< lion (!■:' ;••!) .r. .i Inn, lion ol Imnpmiitiim
i iii|»....... Apparent Riilc
i Constant
•in,
.MIK "I ■ i I
16.9 I/') i i o II I
11-1111 >,........
K
252 '<,l
298 298
Apparent Rule
i   , Ml i",
11.9 11 i.
9.06 9.67
i bmpei Bturc K
325 355 III.' 403
Appaiinl Kale i i hi .l.inl
9,2 8.02 676 6.58
i n
ini Aic the units of the rate constant correct? tli) I n the data to Arrhenius' law. How well do (hey lit? o i In the lil improved with equation (2.28)? Use the methods in Example 2 l> nil I low do you interpret (he sign of the activation barrier? lei (In the basis of your fits, how much confidence do you have in the value "I the activation barrier?
V. noied earlier in this chapter, crickets chirp to attract males; the cricket metabolic rate varies with the cricket's body temperature, so the chirping rate
il.ii varies with the cricket's temperature. [Walker, Evolution 16, (1962) 407 I .....lined the rate that a group of crickets chirped and obtained the following date
temperature, Chirp rate, Temperature, Chirp rate, Temperature, Chirp rate, I chirps/minute       °F       chirps/minute       °F chiips/iiunui.
•ii
•iS
60 65
64
73
\22 I 36
77 81
136 148
In) Estimate the activation energy for the process leading to chirping. (Ii) A cricket often chirps from dusk to dawn. Assume that during the course ol the night, the temperature varied as follows:
Temperature,
Temperature,
Temperature,
lime	°F	Time	F	Time	"F
9 p.m.	80	12 p.m.	73	3 a.m.	67
10 p.m.	77	1 a.m.	68	4 a.m.	67
1 1 p.m.	75	2 a.m.	67	5 a.m.	67
How many chirps would a cricket make from 9 p.m. to 5 a.m.? (Hint: Set up a problem as an integral that needs to be solved, then use the methods in Example 2.F to do the integration.) (c) Next time you are walking at night, listen for the crickets. How fasl are tin \ chirping? Does the chirp rate agree with the data above? Can you use the rati at which crickets chirp to estimate the air temperature? How accurate is youi measurement?
A.M. Smith, Environ, Entomol. 21, (1992), 314 examined the speed that pea weevil (Brochus pisorum) eggs matured as a function of temperature. The following data were obtained:
....
,„ .,,    . .,   ,. rivii mi mi rjimir i < ini i it.
I'lli illl I M'l
	1 lays for		1 i.i\■. Im		1 >.l\ N Ii n
Pemperature,	Weevils Id	temperature,	Weevils to	I'cnipcraturi	SVee\ lis in
°C	Hatch	°C	II.Hi Ii	"C	Hatch
10.7	38	18.1	15.6	23.7	7.3
14.4	19.5	18.1	9.6	24.7	4.5
16.2	15.6	21.4	9.5	28.6	7.1
(a) Fit these data to Arrhenius' law and estimate the activation energy lor the process. For the purposes of this problem, assume that the time to hatch is the time where the weevils are 90% developed (i.e., XA = 0.9).
(b) Assume that a weevil lays her eggs on your garden on a Monday. How long will it take before the weevil larvae hatch and start to eat your plants? Temperature data follow:
	Temperature,		Temperature,		Temperature,
Day	High/Low, °C	Day	High/Low, °C	Day	High/Low, °C
Mon.	18/24	Fri.	18/21	Tues.	17/27
Tues.	14/21	Sat.	18/25	Wed.	19/21
Wed.	17/23	Sun.	16/21	Thürs.	18/28
Thürs.	22/28	Mon.	19/28	Fri.	17/25
2.27
For the purposes of discussion, assume that the daily temperature varies linearly from a high at 2 p.m. to a low at 5 a.m. Use the Methods in Example 2.F to do the integration. A more detailed discussion of the temperature effect can be found in P.G. Allsopp, Agricultural Forecasting and Meteorology, 41, (1987).
Harlow Shapley was a famous astronomer who discovered the Milky Way galaxy. As a sideline, on cloudy days, he measured the walking speed of ants outside the Mount Wilson Observatory. Shapley, Proc. National Academy of Science, 6, (1920, 1924) 436 reported the following data:
	Running		Running		Running
Temperature,	Speed,	Temperature,	Speed,	Temperature,	Speed,
°C	cm/second	°C	cm/second	°C	cm/second
9.0	0.44	21	2.23	33	4.32
10.3	0.64	22	2.16	33.5	4.77
12.5	0.77	23.5	2.29	34	4.35
14.5	1.10	24.5	2.65	35	5.08
17	1.28	25.5	2.94	36	5.57
18.6	1.48	26	2.56	37	5.67
19	1.62	30	3.05	38	6.06
20	1.90	31	3.06	38.5	6.60
(a) Fit these data to Arrhenius' law and estimate the activation energy for the process. Assume that the running speed corresponds to the rate of a metabolic process in the ants.
i ••»
I Ml
tin h\ .m experiment in sec ii Shapley is cornel I'm lomc .mis mi .1 sklllei mul
vv.il« h how quickly (hey move as you heal die skillel Do (he anls lollow inn 1 \|>i'i lations''
HI Assume that you and your boyfriend or girlfriend go Im a picnic in (he park mi a warm i.'K "(') day. You lay out the picnic basket on a blanket and si.nl lo e.ii I Inbeknownsi lo you. I here is an anthill 10 meters away. How long will U
talu before the anti discover your picnic basket and begin to raid your basket? nil Helnrich'i book, the Hot Blooded Insects, Harvard University Press, 11993) hai Othei examples of changes in insecl behavior with temperature. Find anoihci . Kample to suggest a problem like tins one. Be sme to solve the problem to
make sine thai il works.
i <plaln when you can represenl rate data with a rate expression (i.e., determine Which criterion needs lo be satisfied for the rate to be a function ol onl)
m.......centrations ol all the species in the reactor, pressure ami temperature)
111 mi   flunk back lo the phase rule from your thermodynamics class. When an
11 mpen.....<■. pressure, and composition insufficient to uniquely define all ol the
properties of the system such as density?)
II,.  growth ol Illinois follows kinetic expressions similar lo those discussed m
.....haptei I ook at the following papers and describe the findings Hasení levi i
, i il  \nnals of Oncology, 7, (1996); Chaplain el al. Wounds 8 (1996) l.\ < lioebe.
h, i. uluuhhiI Journal oj Radiation Oncology. 34 (1996) 395; Wassciiiiaiin el ,il Mulhematk •// Biosciences, 136 (1996) 111; Costa et al. Mathematical Blost iem $, i • . , ľ»•»•.( I'll. Skip any papers not available in your university's library.
Sponge el al. Mathematical Bioscience, 138 (1996) I, described a kinetic model
i. ,i ili, spread of an HIV (human immunodeficiency virus) infection. I ook up thi Ii.i|n'i .iml describe the findings.
Wh,ii did you learn new doing this homework set?