7
INTRODUCTION TO REACTION RATE THEORY
nil cis
111 i ii hi this book, we have been discussing what is called classical kinetics: rale laws fill i in inical reactions and how those rate laws arise from the mechanism of the reaction. In iIh next several chapters, we will be changing topics and starting to discuss how h ii iMhis happen on a molecular level. Our objective will be to formulate an expression i    ilir rate of reaction in terms of the rates of all of the elementary processes thai OCCUI
■ ii.....y reaction and the forces on all of the molecules. This is a large topic. There are two
i ni   predicting the preexponentials and predicting the activation barriers. The objecl 1)1 i hapter 7 is to provide a brief overview of the prediction of preexponentials wilh olllsion theory, transition state theory, and the RRKM model so that the reader can gel ......i induction to the material. Chapters 8-11 will fill in the details.
/ I INTRODUCTION
In lliis chapter, we will be discussing what is called reaction rate theory. The objective • i reaction rate theory is to relate the rate constants for reactions to the properties of the fi ii i.mis, products, and transient intermediates.
The earliest work on reaction rate theory came from Arrhenius (1889). Arrhenius was interested in why activation barriers arose in chemical reactions. Arrhenius considered a Dimple reaction:
A->B (7.1)
.mil proposed that if one looked at a chemical system containing A and B, there were Iwo kinds of A molecules in the system: reactive A molecules (i.e., A molecules thai had ilie right properties to react), and unreactive A molecules (i.e., A molecules that did not have the right properties to react).
359
Aiilieniiis noted III.il Ihcie ľ. .i ilľ.iiiliiilinii nl tuolei nim m-Iih ItlCN US nullified | ľinmv 7.1. Must 41I t lit- molecules niv loo cold In relet, Howevoi. there lire .i lew mnlcciileá in the tail ni the Boltzmann distribution thai can react
AiThenius then derived an equation Im tlie nite nl reaction The derivation assumed, that equilibrium was maintained between the reaelive and unreaclive A molecules W» reproduce the arguments below.
Recall that at equilibrium, the concentration of molecules of A that have the iifdll properties to react, CA, is related to the overall concentration of the unreaclivi \ molecules by
Ca = CAe-<AGt/feT>
(7.1)
ta = -K„CA = -K0CAe^(AG+/bT)
k, = K0e~<AGt/!íBT)
(7.41
Note:
(7..1)
AGf = AHr -TAS*
on^^TLZZyTŽľ 10 COnVert an UnreaCtÍVe A molecule a reac J (7.4) yields ^ °f ^ ^ ****** equation (7.5) into ec„,.......
(7.0)
2 3 4
Velocity, cm/second x 1000 Figure 7.1   The Boltzmann distribution of molecular velocities.
i.........(7,6) Ii equivalent to
k,    k„e "' "
i -m n mi i mi,       n, t
i / /1
ko = Koe(ASt>/kB
(7.8)
In , ,|n.iiinii (7.6), k| is the rate constant for a reaction (7.1), k0 is the preexponenlial loi ii inhi, Iv, is the activation energy for the reaction, kB is Boltzmann's constant, and I ii temperature.
I i|tiution (7.7) was Arrhenius' key result. At the time this work was done, there were mimical rules to predict how rates vary with temperature. Arrhenius was the lust in in derive a theoretical expression. When the expression was found to lit data, .....\ expression, which was renamed Arrhenius' law, was universally adopted in
km. Ill s
Vnluiiius's law was one of the early successes of physical chemistry. It applied to
.....In- behavior of chemical reactions and the growth and reproduction of plants and
lllipli annuals as described in Chapter 2. Consequently, Arrhenius' law was an important
•i.n .nice.
I .'   COLLISION THEORY
arrhenius never was able to provide a model for K0 in equation (7.8). Fortunately, Irani/ i I'MS) and Lewis (1918) independently proposed the collision theory of reactions Tin i,l, it in the original version of collision theory was that molecules need to collide il die
.....leeules are going to react. Figure 7.2 shows a few trajectories that we have calculated
....... a program from Chapter 8. In the top case, A comes in to hit BC, then A Mies
IWUy again. According to Trautz and Lewis, this would be an unreactive collision. In (he
•iid ease, A comes in and hits B-C, and C flies away. That is a reactive collision 11.mi/ and Lewis proposed that one could calculate the reaction rate by looking at how in an collisions occurred and multiplying by a reaction probability that is a function of
.....gy. As a result, one can calculate the reaction rate from the number of times that the
lini molecules collide.
I rautz and Lewis then used the kinetic theory of gases to calculate a collision rate. If "in assumes that all hot molecules react when they collide, then one can show that the nile of the reaction
A + BC->AB + C (7.9)
ľA-BC
jABC rreaction
(7.10)
where Zabc is the rate of collisions between A and BC in units of molecules/(A3 second) and Preaction is the probability of reaction once the molecules collide.
11h i iMi in him in, :m»:i
Figure 7.2  A collision between an A molecule and BC molecules.
Using Arrhcnius' model in Section 7.1, we write
P   ,    - e-AOt/feT
r reaction — *~
Combining equations (7.10) and (7.11) shows
_ 7 p-AOVtfnT
1*A—>BC — ^ABC
(7.11)
(7.12)
Next, let's derive an equation for ZABc. First consider the reaction between the molecules A and BC. To simplify the analysis, we will treat the molecules as two billiard balls. Figure 7.3 shows some of the possible collisions between the billiard balls. In tin-figure, A is moving toward BC. A starts at the upper left, and the motion of A is indicated as a series of circles. In the top case, a hot A molecule approaches a hot BC but it misses. In the second case, the hot A just misses B-C. In the third case, the hot A collides with the hot BC. Trautz and Lewis assumed that in the first two cases, no reaction occurs, bin in the third case, reaction happens with unity probability, provided the molecules have a minimum energy Ej.
Notice that any A molecule that gets within a distance bcon of the BC molecule will collide with the BC molecule. Therefore the collision diameter becomes a distance. Any collisions with a distance less than bc„n react.
Next, we will derive an expression for the rate of molecular collisions. For the purposes of derivation, consider the motion of a given BC molecule toward an A molecule as shown in Figure 7.4. During the collision BC moves toward A as shown in this figure. We wil define vA_>BC as the velocity of A toward BC.
Figure 7.3  Some typical billiard ball collisions.
Now let us consider a small increment of time tc. If we treat A as a fixed point, then .lining the collision BC will move a distance LABc given by
LaBC = VA->BCtc
(7.13)
Next, we want to note that if there is an A molecule within the cylinder in Figure 7,4, Ihcre will be a collision between A and BC sometime during the time slice tc.
If there are two molecules within the cylinder, there will be two collisions. In general, Ihc number of collisions will be
/Number of collisions\ /volumeofN
of the given = I gr  ) CA
V     BC molecule     /      v 3
(7.14)
where CA is the concentration of the A molecules in the reacting mixture, measured in iuolecules/cm3.
1.1 H I P.U IN till i illi
lit',
Figure 7.4  The collision of two molecules A + BC.
The volume of the cylinder is
Volume of
cylinder J ~ ^"^abc For the purposes of derivation, it's useful to define a new variable by
°a->bc = ^(bcoii)2
where bcon is the collision diameter combining equations (7.13)—(7.16). Then
(7.15)
(7.16)
(Number of collisions\ of the given BC I molecule /
- va_bcCao^_,.bc
(7.17)
Now consider a volume v of gas. The number of BC molecules in the gas is given by
Number of BC molecules = vCBC (7.18)
where CBC is the concentration of BC in the reacting mixture, measured in molecules/cm1. By definition
total number of   \       /   total number of \ collisions at A with        ( collisions of A with ) all BCs       /      V        allBCs /
W\BC
(volume) (time)
(v)tc
(7.19)
I In       iii llllllllll'l  iii l (llllsllllls is
....... hi......v i >ii       \ / ,, \
,, \      / llllllllici ill collisions al \
\ molecules with all ,  ,     , ,
}    \ a given lu molecule I in molecules    /     v '
•   ■ i ■ i • i,i,,i;■ equations (7.17)-(7.20) yields
Zabc = CaCbc(ja^BVA^BC
numbei ol H(' molecules in the ] volume
(7.21)
i |ii limn (7.21) gives us the collision rate if we know the velocity of A toward BC, In I n ii system, there will be a distribution of velocities, so we have to average equation i i over the velocity distribution. The result is
Zabc — Va^bcCaCbc"a^bc
(7.22)
In ii \ ,\ ,bc is the average velocity of A toward BC. Equation (7.22) gives Ihe loial rale ll ■ ollisions between hot A molecules and hot BC molecules.
Substituting equations (7.5) and (7.19) into equation into equation (7.12) yields
ra—>bc = (vA-*Bca^BceASt/fe)(e-aht/ibT)CACBc
- here vA .bc is the average velocity of A moving toward BC.
instituting equation (7.16) into (7.23) and setting AH* = EA yields
(7.23)
ta^bc = 7rdJonVABce~W(eASt/fe)CAC
lit'
(7.24)
In "' i'a »bc is the rate of reaction between A and BC, Vabc is the average velocit] mI A toward BC, CA and Cbc are the concentrations of A and BC, and dn,n is the iiillision diameter of A-BC as indicated in Figure 7.4. EA is the average amount of
ii' rgy needed to get the reaction to happen in kJ/molecule, kB is Boltzmann's constant 11 IK x 10~23 J/(molecule-K)], T is the absolute temperature, and AS* is the entropy of 11 it- molecules that react in kJ/(molecule-K).
li|iiation (7.23) is just a second-order rate law, with an activation barrier of AH* and i preexponential, k0, which is given by
ko = vA^Bc^BceASt/fe (7-25)
Neither Trautz nor Lewis had a way to calculate AS*. Consequently, they set AS* to zero hi i-i|iiation (7.25) to obtain
ko = Va^bcctL
(7.26)
I .iiiation (7.26) is the key result for simple collision theory.
Trautz and Lewis also asserted that the molecular velocity in equation (7.26) should be calculated, ignoring the fact that we are considering hot molecules. We will derive an
......M' HI Mil i HIY 367
i'<|iiiilioii lui vA .,„ is Section K. 16.4. The u-sull is
where
v\ -ik
I
_ / 8W1 Wmab(
1/2
+
I
Rabc     rnA     mB + mc
(7.27)
(7.28)
and mA, mB, and mc are respectively the masses of A, B, and C in atomic mass unit] (1 amu = 1.66 x I IT24 grams).
I find it convenient to rewrite equation (7.27) as
vABC = 2.52 x 10'
A
I 2
second \ 300 K For example, for the reaction F + H2 -* HF + H at 400 K
mF = 19 amu mH2 = 2 amu 1
+
1 amu
Habc
(7.29)
!
21
M-abc     19 amu    2 amu     38 amu 38
M-abc = 2| amu
vABC = (2.52 x 1013 A/second) (4(^C
\300 K
1/2
1/2
amu
38
amu
= 2.2 x 1013 A/second
There is one other detail. It is not clear whether you should use van der Waal's radii, covalent radii, or some other radii in the calculation. Generally, covalent radii give better predictions, although Lewis' original papers suggest van der Waal's radii instead. Trautz and Lewis picked numbers for dcon, but did not have a theory to predict it, so this was a limitation of their model.
7.2.1   Predictions of Collision Theory
Equations (7.23) and (7.26) were big advances. Equation (7.23) correctly predicted that an elementary reaction between two molecules would show second-order kinetics. That was quite important to people deriving rate equations. Equation (7.26) gave a numerical value for the prcexponential for the first time. Again, that was quite important. Further, the numerical value was close to the experimental values.
It is interesting to put some numerical values into equation (7.26) so that we can compare to experiments. Table 7.1 shows the molecular velocities of a number of molecules computed at 273 K. Notice that most of the molecular velocities in the table are between 2 x I0l2 and l x 1013 A/second. If we assume an average velocity of
1......./ 1	Mud-. iii.ii vol... iii.". .111.1	.■Ih-.ioii
(lliinmtmi	i.ii .1 iiiiiiiIhii .ii imili'i Hi.1, .ii -'/:i k	
	Moleculti Velocity,	1 ..II......
Molecule	A/second	I liameti i \
He	1.2 x 10''	2.2
N2	4.5 x 1012	3.5
Oj	4.2 x l()12	1 I
IM)	5.6 x l()12	3.7
(Ml,,	4.37 x I012	(.5
C„H„	2.7 x 1012	5.3
I    Hi1' A/sccond and an average collision diameter of 3 A, substitute into equal ion ' Idi, and substitute the resultant value of a*A into equation (7.26), we predict a |.n. s|n>iK'iitial of about (4 x 10'2 A/second) x [7r(3 A)2] = 1.1 x 1014 A'/seeond. Il is useful to compare this result to the compilation of preexponentials in 'fable / .' i   that most of the preexponentials in Table 7.2 are between 1012 and 2 x \ '/(molecule-second). Equation (7.26) predicts 1 x 1014. Therefore, il seems thai
.......mi (/.26) predicts preexponentials that are usually within an order of magnitude ol
iii. experimental value.
\. i ending to equation (7.26), preexponentials for reaction are determined by two 11. inis: the size of the molecules, as given by the cross section; and the average velocih ..I the molecules, as given in Table 7.1. According to equation (7.26), if everything else . i qual, faster-moving molecules will react more quickly than will slower-moving ones i In . trends generally agree with experiments. Equation (7.26) also predicts that bigger
.....lecules (i.e., those with larger cross sections) will react more quickly than will smallii
molecules, Again, that agrees with experiment. Collision theory also gives preexponentiall ih ii .ue generally within the right order of magnitude. That is quite a significant ivsuli mi i.Kiing that collision theory was first proposed in 1918. [able 7.3 shows a few additional values of the preexponentials calculated from equation (7.30). Details are given in the solved examples. Generally, collision theory lends to overestimate the preexponentials for reaction, although the errors are nlu-u i than a factor of 10. To put this in perspective, in 1918, one could not measute preexponentials to within a factor of 10, so this was as good an approximation as the data llsell.
table 7.2 A selection of the preexponentials reported by Westley (1980)
Reaction
Preexponential, ^/(molecule-second)
Reaction
Preexponential. A3/(moleculeseconcl)
II t-C2H6 II i CH — II | ("H4 -I) + H2 -» () l- OH -() + CH4 ■ () + CH3 ■ i) + HCO
-> C2H5 + H2
H2 + C ■» H2 + CH3
OH + H ■ 02 + H ■> CH3 + OH -> H + CH30 ->■ H + c02
1.6 x 1014 1.1 x 1012 1 x 1014 1.8 x 1013 2.3 x 1013 2.1 x 1013 5 x 1013 5 x 1012
O + C2H6 -»• OH + C2H5 O + C3H8 -»■ (CH3)2CH + OH 02 + H -> OH + O OH + OH -* H20 + O OH + CH4 -> H20 + CH3 OH + H2CO -> H20 + HCO OH + CH3 -» H + CH3O OH + CH3 -> H20 + CH2
13
2.5 x 10
1.4 x 1010
1.5 x 1014 1 x 1013 5 x 10'3 5 x 1013 1 x 1013 1 x 1013
. I II M>|  M l I II '.......I l       M     I Ml I If IT
mr mami ;i 11 in wiuNin .....anyimihui
II .0
I.lhli- /  I    I 'i < ■ x i >■ >■>•-■ ill. il-. i .ill iil.il, ,1 Iiihii <>i|ll<lll<m (/  III) Iim .1 illiillliiil III in.ii lloiiri
compared to oxporimont.il datii
Calculated Preexponential Assuming bcon = van der Waals Radius,
('.iK ni,iir.I Preexponential Assuming bn,n = Covalenl Radius.
Reaction
Experimental PreexponcntUdj
Á3/(molecule-second)    Á3/(moleculesecond)   ÁVduolcculesei.....I)
H + C2H6 -> C2H5H2 H + CH -> H2 + C O + C2H6 -* OH + C2H5 OH + OH -> H20 + 0 H + 02 ->• OH + O
6.2 x 1014 4 x 1014 1.9 x 1014 1.25 x 1014 4.0 x 1014
2.0 x 10
14
2.0 x 1014 7.6 x 1013 5.8 x 1013 2 x 10
14
1.6 x 1014
1.1 x 1()12
2.5 x II)11
1 x 10"
1.5 x 1014
I't 111111V i iniicd 1l1.1i um' 11111 n| i| n < >\......111\ .111 • 111111 lui thcNC oil'octN by i inisiilii nic the
1 mi 111 i . 111.11 ii >ii ( / 25)  Ret .ill Iiihii < IlliptCI 6 lli.il ttlC inliii|i\ \n .1 imi.i'.mic nl liuw 1 iiiili^iiinluiiis are available 111 1111- system, specifically:
(7.11)
Configurations    exp (
11    > luiik ai reaction (7.32), there are many configurations of llie readmits ih.ii Jo nol .....'action. We can define AS1 by
AS
configurations which leaJ to reaction average number of configurations of the reactants
(7.34)
7.2.2  Failure of Collision Theory
Unfortunately, there are a few examples where the precxponentials differ substantially from those preJicted by collision theory. The reaction
CH3CH2CH3 + O:
-> CH3 C HCH3 + .OH
(7..Mil
has a preexponential of 1.4 x 1010 A3/(moleculesecond). That is over two orders nl magnitude lower than predicted from collision theory [equation (7.26)]. In contrast, the reaction
202 --> 20. + 02 (7.31)|
has a preexponential of 5.8 x 1015 AV(moleculesecond). That is about two orders of magnitude larger than predicted from collision theory. These are special cases. In mosl cases, collision theory gets about the right answer. Still, there are a number of examples where Trautz and Lewis' version of collision theory cannot explain the data.
The Trautz-Lewis model failed because it treated the collision between the reactants as a billiard ball collision. Every molecule of A that collided with BC was assumed to react with a fixed probability. In reality, the reaction probability varies with how th collision occurs. For example, consider the following two reactions:
CH3CH2CH3 + O: CH3CH2CH3 + O:
-> CH3CHCH3 + .OH -+ .CH2CHCH3 + .OH
During reaction (7.32a), the oxygen reacts with the hydrogen on the propane's middle carbon. If the incoming oxygen hits the hydrogen on the middle carbon, then the reaction can occur. However, if the oxygen hits anywhere else, a »CH2CH2CH3 species will form. The Trautz-Lewis model ignores the fact that you need special collision geometry to allow the reaction to happen. As a result, their model tends to overestimate rates of reaction such as reaction (7.32a).
This example shows that, in some cases, one needs molecules to collide in the correct way for the desired reaction to occur. The Trautz-Lewis model ignores the need for a special collision geometry to get a desired reaction, so it does not always give a good prediction of the rate.
(7.32a) I (7.32b) I
•. culled the entropy of activation. Generally, exp(ASVkB) >s 'css ln;m •lni1 " III 1» much less then 1 in a case such as reaction (7.32), where only special conligiiralinus nl to reaction. There are also a few cases where exp(ASt/^B) is greater ihan I, bei aui I
ii.....in- more accessible configurations in the transition state than in the reactanls
I'olanyi also defined loose transition states and tight transition states. Loose
......lion states have AS   near zero, so any collision can lead to reaction, Tighl
......lion slates have negative values of ASf. Tight transition states arise when a special
nnllguration of the reactants is needed in order for a reaction to occur.
i In fortunately, in 1931, Polanyi was not able to get an expression of AS1, so he could mil i lined collision theory to account for this effect.
Vnother weakness of the model is that it does not explain activation barriers. Neithei \iibi iiiiis, nor Trautz, nor Lewis was able to explain why the reactants needed to be 1ml
......der for reaction to occur. Trautz and Lewis just assumed — without explaining ilns
i lumption — that reactions had barriers.
/ I  THE MARCELLIN-WIGNER-POLANYI MODEL
M.ucellin (1914-1920), Wigner (1932), and Polanyi (1931, 1932) provided the first model ih ii explained why reactions have barriers. Their idea was that the reactants need to crosi .....'iiiilain in a potential energy surface in order for the reaction to occur. See Figure 7..5.
Reactants
Products
Reaction coordinate Figure 7.5 Polanyi's picture of excited molecules.
IMI MAMI I II IN WNlNIII I'l II ANVI Ml IIII I
1/ 1
A""'8 ",M'""iWigner.MdPolinyl.theactlvi..........trríei I. an.....atedwiththj
energy to gel ovci the barrier.
Figure 7.6 shows what the barrier really looks like for the reaction
H + CH,OH
H2 + CH2OH
(7.11)
There is a two-dimensional barrier in Figure 7.6 rather than the one-dimensional barrier 111 Figure 7.5. The two-dimensional barrier is called a potential energy surface. The potential energy surface looks like a gorge through the mountains with two broad valleys and 11 saddle point in between. Point X in the figure corresponds to the minimum in the potential at the reactants, Y is the minimum at the products, and there is a rise corresponding in the saddle point. Figure 7.7 shows the saddle point in more detail. Notice that there is a hill. The hill is the barrier to the reaction.
People usually plot the potential energy surface as a contour plot. The contour plot is shown on the right of Figure 7.6. The lines are contours of constant energy, and are
Energy
Transition state
1.5
(a)
2 2.5 C-H Distance (angstroms) (b)
Figure 7.6 A potential energy surface for the reaction H + CHiOH -»(+> + PH r>H ftr™ th= Blowers and Masei. The iines in the f,ure are contours of+«j£ Z^^^^SSSSA
Transition state Figure 7.7  A blowup of the saddle point in Figure 7.6.
1 1 ii.il/iiuii apurl Poltil X In ílu llgi......... .ponds to the ml..........1 In the potcntlul
11 1 mi. ^ la the minimum Hi Ihi produi ts, and there In .1 rise corresponding 10 '|    iddlc point.
.....pii". / <i and s.A show more Information about potential energj surfaces Please
il these two examples before proceeding with this chapter, Marcel I in, Wigner, and Polanyi noted that molecules need to get over the rise in the , h .in m energy surface in order for reaction to occur. Therefore, they proposed tint I ......... barriers are associated with the rise in the potential energy surfaces. In dull.......e, this is called rise the transition state, or the col (saddle), in the potential
m I.m- Generally, it is still believed that activation barriers are associated with .1. hills in the potential energy surface.
In iin literature, people sometimes say that the barrier height is equal to the a< tivatlofl In fact this is only an approximation. For example, Table 7.4 compares th( II inou barriers for a number of reactions to the energy of the saddle point on the 1  |i Mii.il energy surface. One sees qualitative agreement, that reactions with largei saddli
........i-rgies also have higher activation barriers. However, the activation barriei II not
11 ll) equal to the saddle point energy.
1 1 I Tunneling
illy, when a hydrogen atom is being transferred during a reaction, the barriei is a llllli less than the saddle point because of a quantum-mechanical process called tunneling in 1 'hupter 8, we will show that quantum-mechanically an atom that does not have quite ■ Hi nigh energy to go over a barrier still has some probability of going through that barriei III Mi.".r eases, the activation barrier is lower than the barrier in the potential energy surface In 1 imsc some atoms can get through the barrier.
Tunneling arises because of the uncertainty principle. Classically one might say thai the . 11 i.uiis are just approaching the barrier. However, quantum-mechanically there is sunn
.....ertainty in the position of the molecule, and so when you think that the reactants are
|ll 1 approaching the barrier, the wavefunction for the molecules will have a component .hi the other side of the barrier as shown in Figure 7.8. This uncertainty in the atomic
Table 7.4 A comparison of the saddle point energy from accurate ab initio calculations and the activation barrier measured experimentally for a number of reactions
	Saddle Point	
	Energy,	
Reaction	kcal/mol	kcal/mol
H + H2 -> H2 + H	9.60	8.0
D + H2 -> DH + H	9.60	7.78
H + CH4 —► H2 + CH3	13.5	11.9
H + CH3CH3 -> H2 + C2H5	11.8	9.7
F + H2 -> HF + H	5.8	1.7
H + CH3CH2CH3 -> H2 + C3H7	10.4	8.2
H + ch3oh ->■ CH2OH + H	9.8	9.0
CH3 + cf3i -> ch3i + CF3	4.95	7.5
11 IAN? It III IM 'HAM IMI i IIIY .173
I.ill ..I
wavefunction extends through barrier
( lllNllll (ll
wavnliini li<hi Iii'Idic' b.imri
Barrier
Figure 7.8 A diagram showing the extent of the wavefunction for a molecule. In (a) the molecule is by itself. In (b) the molecule is near a barrier. Notice that the wavefunction has a finite size (i.e., there is some uncertainty in the position of the molecule). As a result, when a molecule approaches a barrier, there is a component of the molecule on the other side of the barrier.
positions allows a molecule to transport through a barrier. We call the transport through the barrier tunneling.
Tunneling is most important for reactions in which a hydrogen atom is being exchanged. The effect is that activation energies for hydrogen transfer reactions are generally less than the energy of the barrier on the potential energy surface.
7.3.2  Dynamic Effect
Most ligand transfer reactions have activation energies that are slightly higher than the saddle point energies. Physically, when the reaction A + BC -> AB + C occurs, the BC bond needs to accumulate enough energy to break. Generally, it is easier to accumulate enough energy if there is a little excess energy in the system. Consequently, the average energy of the reacting molecule is usually slightly higher than the saddle point energy.
We will discuss tunneling and dynamic effects in detail in Chapter 8. However, the thing to remember for now is that activation energies are usuaily slightly different from the energy of the saddle point in the potential energy surface.
7.4  TRANSITION STATE THEORY
Marcellin was killed in World War I, and neither Wigner nor Polanyi were able to derive a simple expression for the rate. However, Eyring (1937) took an equation that appeared in Tolman (1927), and modified it using Wigner's ideas to derive what is called transition state theory. We will reproduce the key arguments in Chapter 9. The idea builds on Arrhenius' model described in Section 7.1. First you use statistical mechanisms to calculate an expression for Kequ, the equilibrium concentration of hot molecules. Then you plug back into collision theory to get the rate constant. Consider the reaction
A + BC
AB + C
(7.36)
\......lui)' in equation (6 '), K,,|u, il» cqiiililiMiiiii constant loi ihc production Ol Um IkX
. - li'ilvc molecules, is
Maim c iif/|Bi
K,„„
(7.37)
'U'lrn
..... li1 is the average energy needed to traverse the barrier, qA,„  is the partition
...... per unit volume lor the hot molecules, qA and qu< are the partition function*
......ii volume for A and BC, kB is Bolt/.niann's constant, and T is the icmperaluii- II
. 111 >.111111equations into Arrhenius' model, assuming that the reaction probability ll i|ti il in the probability that a molecule is hot when the collision occurs, we obtain
IaMbc
, i/    QaBC „-e+/|j„t
(7.38)
III 11 kA .bc >s the rate constant for reaction (7.36), and K0 is the rate constant foi the i lion of hot molecules. Equation (7.38) is exact, but we will need an expression fol i n We can get it from collision theory.
I list, let us define a new partition coefficient q+, by
QaBC
1a—bc
(7.39)
III . quation (7.39) q'A^BC is the partition function for the translation of A toward BC and 11   Is the partition function for all of the other modes of the reacting A-B-C complex. ('ombining equation (7.38) and (7.39) yields
kA—*bc = (K<>qA-.Bc)
qAqBc
(7,10)
I'.yring (1935, 1937, 1944) proposed that one could replace q+ in equation (7.5) with m Ihc partition function of the transition state, and E+ the average energy of the molecules will react with E*, the energy of the transition state. See Eyring (1944, 1980). The result is
kA—bc — (K()qA_>Bc)
_sl_
QaQbc
.e-e*/*bt
In Chapter 9 we will use collision theory to show
Koq^bc = (^) Substituting equation (7.42) into equation (7.41) yields
(7.41)
(7.42)
(7.43)
where kB is Boltzmann's constant, T is the temperature, hP is Planck's constant, EJ- is the energy of the saddle point in the potential energy surface, qA and qBC are the partition
linn lions loi I In- ii.ii i.ml', .mil i|, is Ihc ptJtltlon I u ih I ii hi lul .ill ill 1111- Hindi's ol liansilioltl
sink- except Ihc motion over Ihc hiirrii-r.
Equation (7.43) is the key result foi transition state theoi \ < >in- should also memoril equation (7.43) before proceeding through the rest of this chapter.
7.4.1   Key Predictions of Transition State Theory
Transition state theory was a major advance. It provided a way to predict the magnitude of the preexponential. It also gave lots of parameters (i.e., the q values) that could he used to fit data.
It is useful to use transition state theory to get an order of magnitude for the ratal constant. In Example 7.B, we show that for most first-, second-, and third-order reactions, qx/qAa.BC ls between O.l and 10 in units of molecules, angstroms, and seconds, while kBT/hp ~ lO'Vseconds. Consequently, transition state theory predicts
Preexponentials for elementary reactions are usually between 1012 and 10'4 in units of A, molecules, and seconds independent of the order of the reaction or other details.
Other key predictions of transition state theory include
• Transition state theory identifies the barrier to reaction with the saddle point in the potential energy surface. Saddle point energies can be calculated exactly as described in Chapter 11, which then allows activation barriers to be calculated.
. Transition state theory allows you to calculate bcon in equation (7.16) rather than guess it. In Chapter 9, we will show that for a simple collision bcon is just the distance between the reactants at the transition state.
• Transition state theory suggests that Pactions can be less than e^^1. For example, in the reaction
D + CH —r-> C + HD
the incoming deuterium has to hit the hydrogen atom for reaction to occur. Transition state theory accounts for that by saying that the transition state must be a linear CHD molecule.
• Transition state theory also predicts that small changes in the shape of the potential energy surface produce large changes in q^-, which, in turn, changes the rate.
We work out solutions in Examples 7.B-7.E. One should read those examples before proceeding with this chapter.
Transition state theory was a major advance. It gave an equation that could be used to make practical calculations before the days of computers. It gave an explanation of barriers to reactions that could be explained to college freshmen. The model contained lots of parameters (i.e., the q terms) that could be fit to data. When these ideas were first proposed, no one knew what potential energy surfaces for reactions were like. Transition state theory allowed people to fit their data to an empirical model, and so transition state theory explained a lot of data that had not been explained before.
Transition state theory has dominated our thinking about reactions since the early 1950s, so it was quite an important advance.
iiiANtilin m maii iiiionv 375
' i I  Relationship between Tranaltlon State Theory and Collision Theory
I hit' ol the details is llial liansilion slale llieoiy is closely related to collision llieoiv n. <i «i will show thai transition stale theory and collision theory are Identical lot línii.....i ol structureless particles. I.el's assume thai Ihc liansilion stale lor Ihc reaction
Id (he hard-sphere collision in figure 7.4. Next, we will obtain expressions lor qA, q,,, . mul q',
\ only has Iranslalional motion. According to Table 6.6
, (27rmAkBT)3/2
i / I h
.........."""""
................
KÍ ,,.....e rotation of the BC molecule. Therefore
(7.45)
(2jr(mB + mc)^bt)3/2 n
n     =---ť--Qv.bc
Ibc h3 nP
..... mB and mc are the masses of B and C, respectively, and qv.Bc is the par,,,.....
......"on f°r the BC vlbrat,on- There is also a partition function foi the
q* = (q*)3(qí)2qv,BC
(7.46)
»' nave ,    (27r(mA + mB + mc)WT)3/2
=-hT
(7.47)
^■cording to the results in Chapter 6
. 87t2IkBT
v\ nli
I = UABc(bcoll)2
Substituting all of the q values into equation (7.43), we obtain
(7.48)
(7.49)
(27r(mA + mB + mc)kBT)3/2
kA_Bc —
hpT a^&rFi^
----u6
hp
/qVABC\ ^Wc^b«^ (7.50) * \qv.ABC J hP
CrOllIng (Mil like linns yields
kA^BC = (kBT)l/2
Hl\   I   Ml||   I 111,
niA(nin I in, )
3/2
Maw x (h,
.....'(;)
1/2
c 'i'1"'1 (7.51)
Note that according to equation (7.28)
Mabc —
1
I
+
I
(mA)(mB +mc) mA + mB + mc
mA    mB + mc Substituting equation (7.52) into (7.51) yields
kA^E
(kBT)
1/2.
1
Ma^c V*
i 2
^(bcoll)2exp-4/faT
Further rearrangements yields
kA-.ec = f-^1) 7r(bco„)2e-4/feT V ^Mabc /
According to equation (7.27)
(7.52)
(7.53)
(7.54)
i  i    I lllllllllionr. ol luillMtlon M.iln Ihttniy
i Itliillunutcly, tramltlon lUU iheuiy llio makes errors, so one has lo be curel'ul when .....ľ n  I in example, ľ'igurc 73) gives .i plot ol the experimental values ol 11 u null .....      divided by kBT/hp lot a numbei oi reactions versus the value ol the lami
li . .iK ui.iicd from transition state theory. Notice thai there is very little correlation .n iiansiiiou slate theory and the experiment. Physically, Figure 7.') tests win Mu i .i.  .|, i|xi||„ term in equation (7.43) is correctly modifying the rale as the shape "I il» i  i nihil energy surface changes. The figure shows that transition state theory is noi qmie
.....r the right answer.
h m.umu slate theory docs not work because transition state theory assumes thai thl H u hints remain at a pseudoequilibrium as the reactants traverse the potential OMfgy . and that tunneling can be ignored. A typical A + BC collision might lasi only m 1  n mills. There is no time for all of the normal modes to reach thermal equilibrium i> nil. there are some dynamical corrections to the rate equation. When the re* tion
A + BC
-> AB + C
(7.57)
in   one has to get energy into the B-C bond to break the bond. One also has to
..... i.i momentum into C to carry C away. It happens that the energy and momentum
|l in i.i rales play a significant role in reactions. Those influences are ignored in transition ..i,He theory.
(3*
1/2
Va^bc
V nMabc,
Substituting (7.55) into (7.54) yields
ka^bc = vA^Bc7r(bc„ii)2e"et/teT
(7.55)
(7.56)
Notice that equation (7.56) is identical to equation (7.26) (collision theory). Therefore, transition state reduces to collision theory when the transition state geometry is a hard-sphere.
Transition state theory differs from collision theory in two key ways:
• One can actually make a calculation. With transition state theory the problem with collision theory is that one never knows bcon in equation (7.16) and thus one cannot make any detailed calculations. Transition state theory replaces bco]] with dcon. The distance between the reactants is calculated using transition state geometry. In Chapter 11 we will show that one can use ab initio methods to calculate the transition state geometry. Therefore dcon can be calculated exactly.
• Transition state theory allows you to consider reactions like reaction (7.32a) where only special configurations lead to the desired products. In such a case the partition function for the transition state will be reduced since fewer configurations lead to reaction. See equation (7.34).
c o E i~
ti x
üj
0.05
0.01 0.02 0.05   0.1    0.2   0.5     1      2      5 10 Transition state theory
I Igure 7.9 A comparison of the kh/rjBT measured experimentally for the reactions in Table 7.4 to tho ii.insition-state approximation of the same quantities.
There is ii simple wiiy >•> llimk uhoul litis process ('onsiilei u hypolliclh nl icaclion
su( 11 as thai shown In Figure / 2 I •......g reai lion, the molci ulcs collide «iili one anothel
as shown in figure 7.2. A comes in ami then collides with IK'. Then (' flics awa) Marcellin showed that one can view the process as a trajectory on a potential energ] surface, where RAB, the distance from A to B, and R,M-, the distance from I) to (', chann with lime. Figure 7.10 shows a typical trajectory. Notice thai the trajectory needs to y<> around a sharp turn on the potential energy surface in order for reaction to happen I lie sharp turn corresponds to converting momentum from the A-B motion to B-C motion Consequently, one needs to have enough momentum transfer and sidewise motion to gel a reaction to happen.
Sidewise motion comes from vibrational energy of the BC molecule. In Chapter 8, we will find that one has to carefully balance the translational energy from the incoming A molecule with the vibrational energy in the B-C molecule in order for the reaction In happen. If there is too little vibrational energy, the reactants never make it around the turn in the potential energy surface. If there is too little translational energy, the reactants never make it up the hill. All of these effects can be explored by computing the trajectories of molecules over potential energy surfaces, as described in Chapter 8.
At present, in general, there are no simple ways to predict a rate of reaction when energy transfer is important. People discuss the effect by rewriting equation (7.43) as
kA-*BC =
k(T)
Qt QaIbc
(7.58)
where k(T) is called the transmission factor. Physically, k(T) accounts for the fact thai some collisions with enough total energy do not make it over the transition state. k(T) also accounts for the fact that some molecules can go through the barrier rather than over the barrier due to a tunneling process as discussed in Section 7.3.1. It is possible to do complex calculations to look at each state of the system and use an approximation to
vi
Mlliuai. the probability thai the mole, ulce u-m I I Ik calculations are complicated, b......
i.iiii. | pic one can calculate Ml)
.......uiuv. k is often Healed as a "I.uIk.c" factor. Il is temperature depended and
... In ,,.,„•. than I when tunneling In important. Generally, il is haul to predict t ......Is  which limits the ulility of equation (7.58). Accurate Methods ... calculate II
...... discussed in Chapter l>.
I it   MICE-RAMSPERGER-KASSEL-MARCUS (RRKM) THEORY
I *c wanl to change topics and start to discuss unimolecular isomerization real I.....
......„,l,, ular isomerizations are reactions in which a single molecule rearranges to form
, ihli products. They can be written in general as
B
i   miplcs of unimolecular reactions include
CH2 / \ H2C—CH2
CD2CO + hpv
=> CH2=CH—CH3
-+ CD2CO* -► CD^+CO*
(7.5'))
(7.60)
(7.61)
Where hPv is a photon and the asterisks (♦) represent excited species. Unimoleculai u nous are different from ligand exchange reactions. Uecall that an elementary reaction
B
i / (..'i
Reactants
BC
Figure 7.10  A trajectory on a potential energy surface.
, impossible. Instead, we need a collision partner X to get the reaction to occur. In i Iwpter 5, we wrote the reaction with a collision partner as
A + X
B + X
(7.63)
I uuation (7 63) is an approximation. It says that the collision partner collides with the n uctant to immediately produce the product. Lindemann, (1922) however, found thai the process does not occur immediately. Instead, it was better to think about the reaction as ... .lining in following two steps:
A + X A* + X
2
a
-» B
(7.64)
First the collision partner collides with the reactant A, to form A*, an excited A molecule then the excited A decomposes to yield products.
Next it is useful to calculate a rate equation for reaction (7.64). The derivation lollows that in Chapter 4. First one writes equations for the rate of formation of all the species
IIHI   MAMMt't Kill M KA'ľ.l I   MAI M .11!, II IIIKM) 11 II ( HIY 381
Hun oilľ lipplies Ihr Mc.stale .i| )| 11 < > x i ■ ■ í.it ■< >ii In A*    I he icstill is Hie lollowinu
approximation to the rale equation
k,k3[A][X]
k3 + k;|X|
I / (»1|
Equation (7.65) is Lindemann's approximation to the rale of a unimolecular reaction.
In order to understand equation (7.65), we will consider the special case that the onl) collision partners are other reactants so that [X] = [A]. Equation (7.65) becomes
rB =
k,k3[A]2 k3 + k2[A]
(7.66)
According to equation (7.66), the reaction should be second-order at low pressure and first-order at high pressure (i.e., when k2[A] » k3). Figure 7.11 shows a log-log plot of the rate data for the reaction
CH, NC
CH3CN
(7.67)
and indeed the log-log plot is second-order at low pressure (i.e., slope = 2) and first-order at high pressures (slope = 1).
If one compares to the Lindemann model in detail, one finds that it fits very well at high and low pressures, but that there are some small deviations at intermediate pressures. Also, the value of k3 that you need to fit the data is way loo large. In Figure 7.9, we used k3 = 10l7/second, while according to transition state, the k3 should be 10'4/second, at most.
The reason why k3 is so large is that the excited complex lasts for a long time in comparison to usual molecular collisions. The complex might last 1(T7 seconds, compared to 1(T13 seconds for a transition state. With a bimolecular collision, the hot molecules
m.IIi.I.....it. and it iliev do mul timing Ihc collision. Ilie readmits My apart   In .i
..........I, . ill.n leitelion, though, mutants iiillliol lly apart Instead, the leading atoms bump
• .....Ii olhei multiple limes In ellcci, you gel multiple collisions. That increases
Ilie mil
Knsscl i l92Ku,b,e), Rice, and Ramsperger (1927, 192«) worked out a classical ei|iinlion ........itc of reaction lor cases where the molecules stay hot lor a long time. Transition
iI hi ii \ assumes that reaction occurs only when there is enough energy in the
III lianslalion lo carry the reactants over the transition slate. However, Rice, i mi porger, and Kasscl assumed that the activation energy could he distributed <>wi
,l.........,il modes and one would still get reaction. They then asserted thai the fttl
ililtl In enhanced by a factor of
Number of ways to put an energy Ea in s modes
Number of ways to put an energy of Ea in one mode
|h then counted the modes, assuming that the spacing at the vibrational level* Wil ......I, smaller than kBT to estimate the enhancement. After some algebra, they obtained
h„
_J_ /_Ea_
(s- 1)! \kBT
_3_e-Ea/!řBT
QaQbc
(7.68)
In re s is the number of modes that can store energy. According lo equation (2 1 11
' '   is on the order of 20. Methylisocyanide (CH3NC) has 12 vibrational modes, and il
, I
■ i arbitrarily assume that six of them contribute to reaction, then according to equation i I 6K), the rate will be enhanced by a factor of
I
(6-1)!
(20)6-' = 26,667
(7.69)
1E+2
1E+0
2 1E-2 B
ô
I 1E-4
CD (0
cr
1E-6 r
1E-8
!-1—t i i i m P	-1-1   1   1 T T 1T j-1-1   1  1 1 HI Second order			-1-1—t—t-TTI 1	1     1   1  1 1 III
I	Lindemanr	i model —			
■   First ore !					
1			Dala		
					■
0.01
0.1
10 100 Pressure, torr
1000 10000
Figure 7.11 The rate of the reaction CH3NC — CH3CN as a function of pressure. (Data of Schneider and Rabinowicz (1962).
Physically, if one can store up energy in all of the modes, the rate will be faster than ii .ill of Ihc energy needs to be deposited in the breaking bond during a single collision
\ a result, the rate constant for a unimolecular reaction tends to be bigger than the rite .. instant for a bimolecular reaction.
II one tests equation (7.68) in detail, one finds that it explains why the k, in the I indemann model is so large, but it does not explain the errors in the Lindemann model
n medium pressures. The reasons for the errors are
. The rate constant varies according to how much energy is put in the molecule. II you use a laser to put considerable excess energy in the reactants, the reactants read more quickly than if there is only a small amount of excess energy.
. There are quantum effects, the assumption that the level spacing is less than kBT is not accurate for vibrations.
Marcus derived a modification of the Rice-Ramsperger-Kassel theory for that case According to the model, the rate of reaction for an molecule excited with an energy E* is given by
rA^B =
_q|i_
hpqA
(7.70)
- i it ri» . I 1« Mi I \fy I I   Mil "III
•II H VI I I I
win n hi Is Planck •■ constant. *i\ Is the partition Function pci nun volume ol thi rcuctunl A. and q^ is the partition luiiction loi the molecules llial have m\ iiniulalctl enough energy to react. Noie thai i|M is different from q^. q' includes only stutcs ih.u have energy in the mode that translates ovet the saddle point in the potential encrg) surface. These are states where energy is localized in the bond that breaks during 11 it-reaction, qj^, on the other hand, includes all of the slates ol the system, which, added up together, have enough energy to cross the barrier independent of whether the encrg] is localized in any specific bond. The key assumption in equation (7.70) is that il ilu-excited complex has enough energy to react, the excited complex will eventually decaj into products.
Equation (7.70) is one version of the RRKM model. Figure 7.12 shows how well the model fits in the best case. It happens that in this case the RRKM model predii i-that the rate should show a staircase pattern with increasing energy for reasons that anj described in Chapter 9. The data seem to show the same staircase pattern. We will give other examples in Chapter 9 and will find that the theory does not always do as well as shown in Figure 7.12. Still, with unimolccular reactions, we can predict very accural! rate constants easily. One can also make accurate predictions with bimolecular reactions, but in those cases the calculations are much more elaborate.
There is one other important detail: the rate constants for unimolecular reactions change according to what collision partner is used. Table 7.5 gives examples of this. Generally, it is not uncommon for the rate constant for a unimolccular reaction to vary by more than an order of magnitude according to which collision partner is used. Unfortunately, then is not enough room here to describe all of the effects.
| .1.1.   I '.    Ilui |>iooK|iononllnl» |lii c iii"/|mi,p|   •......loi ii notion of lllllllioloi llllll
100
200
300
400
Energy above transition state, cm-'
Figure 7.12 A comparison of the rate of the photolysis of ketene (CH2CO) as a function of energy to the predictions of the RRKM model.
••m.tlona,	tin you chmigc	II10 colllnlnil 11111 turn			
I'..........		ko When x Ai^oii	kiiWIu-n x	W.ilri	ko When x - Nj
Nil) I X	• NO I 0 I X	1.7 x I014	6.7 x	10"	1.57 x 10"
.....X	• OH I II I X	2.1 x 10"	3.5 x	10"	5.1 x I0W
Id'  i X	• O. | 11 I X	1.5 x 10"	3.2 x	It)1"	2 x 10"
II. l x .	It I II 1 X	6.4 x 1017	2.6 x	10"	
Mi I X »	20 1 x	1.9 x 10"			1.0 x 10"
I iiilii ill Wcsllcy <I')X<».
1.   ' .1IMMARY AND PLAN FOR THE NEXT TWO CHAPTERS
1.........lary, in this chapter we reviewed three key methods that can be used lo estimate
111 ui.-niials for reaction: collision theory, transition state theory, and the RRKM
id I All three of these models are based on Arrhenius' idea that a reaction happens I.. 11 Iwo hot molecules collide with the right configuration to react. Collision llicoix
.......is that reaction happens whenever two hot molecules collide. The assumption is
.....Ii thai it does not matter how the two molecules come together. For example, In
Mi. reaction A + BC -> AB + C, collision theory assumes that the rate will be the same windier A first collides with B or A first collides with C. Transition slate theory lakes ilu-M|i|insiie view and says that a reaction occurs only if the collision reaches the transition 1 in transition state theory also assumes that equilibrium is manufactured throughout Hi. . ullision process. The RRKM model goes back to the collision theory assumptions, lllll says that if you have a unimolecular reaction, a reaction will occur whenever the Initial excitation of the reactant A puts enough energy into the molecule independent "I ■ In 11 the initial energy is deposited. Each of these models are approximations. Ilowevei ih. \ do work in a wide variety of systems.
In the next two chapters, we will be expanding these ideas in detail. In particular, III 1 h.ipier 8, we will consider collision theory and try to understand what happens when niiiiiis or molecules collide. We will see how one calculates the reaction probability and w ill provide software so that students can use the results to calculate rates. We will then .1 . .mi findings to derive transition state theory, the RRKM model, and related results 111 Chapter 9. I divided up the discussion in this way because, when I teach the material
......idergraduates, I find that I cannot get to catalysis unless I skip Chapter 8 and 9
However, I believe that collision theory is essential to our understanding of reactions 1 herefore, I urge the readers to read the chapters and use the findings to learn about how .....Icailes react.
7.7  SOLVED EXAMPLES
1 sample 7.A A Collision Theory Calculation Use collision theory to calculate the pieexponential for the reaction
H + CH3CH3-► H2 + CH2CH3
(7.A.I 1
.11 500 K.
Solution   Accoidmg in collision itu-oi>
k<> = H'ollVAIIC
First let us calculate vABC. According to equation (7.26)
vABC = 2.52 x 10" Ä/second (-J—)1" (^)
\300 Kj    \ uABC )
1/2
\\ III]
Uabc = H—j-r
(7.A.2);
(/.AH
(7.A.4)
MA + MBC
For reaction (7.A. 1)
M-ABC
I
I
I
= 0.968 amu
(7.A..
1 amu    30 amu Plugging in the numbers shows that at 500 K:
'500      1/2 /   ' —«   ^ '/2
vABC = 2.52 x 10'
300 K
0.968 amu
: 3.31 x 1013 Ä/second (7.A.6)
There is some question about what values of dcoM to use in the calculation. Hydrogen has a van der Waals diameter of 1.5 A, while ethane has a van der Waals diameter of 3.5 A. One approximation to dco,i is
1.5 A+ 3.5 A dcoii =---= 2.5 A
Substituting (7.A.5) and (7.A.6) into equation (7.A.2) yields (2.5 A)2
A3
k„ = nx~'"      (3.31 x 1013 A/second) = 6.49 x 1014 -
molecule molecule-second
(7.A.7)
(7.A.8)
Example 7.B   Approximate Evaluation of
q
«JaObc
Earlier in this chapter, we said
that typically - Is within an order of magnitude or two of unity. The object of this
4a9ibc problem is to evaluate
qAqBc
using the approximations given in Chapter 6 for a case
where A, B, and C are atoms, not molecular ligands.
Solution Consider a second-order reaction A + BC -+ AB + C. Recall from Chapter 6 that if a molecule contains na atoms, the molecule will have 3na total vibrational, translational, and rotational modes. All molecules can translate in three directions, so every molecule has three translational modes. A linear molecule can rotate in two directions, so
li linn iwo rotational modes Such i molti ula will have lna   1-2« 3nt — 5 mIh.iihhi.iI
\ n.>iiiiiir,n mole< iMe i .in mi.ii.....iMi' directions, so It will have three rots......al
i    Such a molecule will have ln„    I    t    .hi,    6 vibrational modes.
II \ .mil.mis ;i single alom, llien 11 u- alum can only translate, not vibrate. Therefore
9a = <Mi >a
(7.1*. I )
li. M .|, is ;i iranslational partition function and the parentheses designate thai we arc
......I, line molecule A. The BC molecule can translate and rotate in (wo direction!, li DM
■ Intnl modes. There are three translational modes and two rotational modes. Therefore, ii......nisi be one vibrational mode. Consequently
Mbc = (qfav^BC
i / H
..... .1, is a iranslational partition function per unit length, qv is a vibialional pal........
...... n.hi. and q, is a rotational partition function and the parens are used to designate I
IK molecule.
i in A BC transition state has three atoms, so it must have 3na = 9 total modes, II I nr;ii ABC as linear, then the molecule will have three translational modes and iwo ml ii,,ii.il modes. That leaves 9-3 — 2 = 4 vibrational modes. Therefore
q* = (q,3qtq?)TST
(7.H.3)
hi. q* is the partition function for the transition state and the parentheses arc used to in.In ate that this is for the transition state. The transition state approximation is to licit dI the vibrational modes as a translational mode:
qA = (q?qvq?qA-Bc>rsT
ill I I
where qA^BC is the partition function for the motion of A toward BC. In transition sin. theory, we want q*, not q*. By definition
q* = —-— = (q,3qvqr)TST
qA^BC
('oinbining equations (7.B.1)-(7.B.5) yields
q* (qv<i?q?>rsT
qAqBc (q13)A(q13qvqr2)Bc
Now we will make the approximation that
(q,)TST    (q,)A ~ (q,)BC = q, (qr)TST » (qr)BC (qv>TST « (qv)BC = qv
Substituting (7.B J)-(7.B.9) into equation (7.B.6) yields
q* q?
qAqß
(7.H.5)
(7.B.6)
(7.B.7) (7J3.8) (7.B.9)
(7.B.I0)
•Ii II VI II I XAMI'I I M
Ai ( mi,line in Inlilc (i V i|v      I,,],     I A Siilisliliilni)' min i'i|iiiilinii ( / II 10) yields
( /.H.II)
One can do the same analysis for a nonlinear transition slate. The result is
= q^lx50      g5 A,
qAiBc    q? (3/Ä)
(7.H.IJ
Equations (7.B.11) and (7.B.12) are estimates of the partition functions from transition1 state theory.
Notice that in this chapter we saw that qV^A^ABc 's between 0.1 and 10 but we got smaller values. This is an artifact that is created because we have ignored the facj that the transition state is bigger, than the reactants, so its rotational partition function is bigger, too. Also, vibrational partition functions in transition states tend to be biggd than in the reactants. Those two factors produce another factor of 5-10 in the partition function. Multiplying the results in (7.B.11) and (7.B.12) by 5 shows that q*/qAqB should be between 0.1 and 10 in angstrom units (A).
Example 7.C A True Transition State Theory Calculation In Example 7.B, we used the approximation in equations (7.B.7)—(7.B.9) to calculate the partition functions. Here we want to do the calculation more carefully. The reaction
F + H2
HF + H
C7.C.11
is one of the most heavily studied reactions in the literature. It is important in chemical lasers. Stark and Werner [1996] calculated a potential energy surface using exact methods, while Stechleretal [1985) fit a potential energy surface to the reaction, and adjusted the potential energy surface so that transition state theory fits the experimental results.
Solution In the solution here, we will first compute the rate using transition state theory with a potential energy surface that has been adjusted so that transition state theory works. Then, we will use the exact results and see how things change.
Table 7.C.1 shows the parameters used in the calculations. Generally, the adjusted potential energy surface was calculated by extending the H-F bond in the transition state by 30% and reducing the barrier by 3 kcal/mol. We also assumed a curvature of 310 cm-1 for the barrier, that is we assumed the barrier looks like an inverted parabola with a vibrational frequency of 310 cm"'. All of the other parameters do not matter to the calculation.
According to transition state theory:
kF-»H2 =
kjT hp
<jf-h2(
qH2
-E*/kBT
(7.C.2)
It is useful to divide up the partition functions in equation (7.C.2) into the contributions from the translation, vibration, rotation, and electronic modes:
III. IIIKIIIV I it" 1 oirtlmil '"I
,    ,.........(,„•, u'kmI I" i nl' Hi"'" ""' Iii'"-'"1"" •
....,im......•«..<«.....•.....•»....................,,,w",m,:onm■"m>
I " Ki'.ui.inls TuiiiMlinn SUM ___——
BxttCl
Used for Transition State t !alculations
illii
UK it «wiiii im. I.....I
C|ll, Itwiil
I lilt mini' burlier
1.34 A 0.801 A 3750 cm 1
7 7
kcal/mol 21 amu 5.48 amu-A 4
1.602 A 0.756 A 4007 cm 1 397.9 cm 1 397.9 cm 1
310 cm 1
1.7 kcal/mol
21 amu
7.09 amu-A
4
19 amu 4
II
0.7417 \ 1195 ' I "'
2 amu
0.275 amu K I
I ii
h„
qH2qF
q;
q^qF
qHlqn
-Er/ki.1 (7.C.31
:::     »« ^ »*. ,n a*-6» **** ** *........
,.........        (7.C.3). According to Table 6.7
/ 2jrmkBT
1/2
(7.C 11
.,, „ „ m„s,a„„„,l P.f ion-fl^-^S „,, ma ot the molecule, Ü, « B° ™° s c™ „„rinc M ,ransi„c in ,l„v.-
nvri1: str-: -......- -.....-.......
...... directions. Consequently
	^27rmtkBT^ V     h2 )		3/2	
^2jrmFkBTyV2		^2jrmH,kBT^ v     h2 )		
(7.C.5)
,,,e ,„,, mH„ and mt are the masses of fiuonne, H2, and the transition state, reforming the algebra; we obtain
VqFqH2/,
mFmH;
V2^kBT
(7.C.6)
" " ".........I» 'I* I % ' iii m   I I' 'PJ m/m i   m ii i im y
•Ii ii vi 111 xami'i i •'.
111   1   ulnte iiic i.i.i term In equation (7.C.6) Rearranging thi laut term showi
•V2        , „.„     „ .1/2   , . ■> v j/j
ii,;
,27rk,t'l'
Substituting the numbers yields h2   \3/2    /300 KVV2
^ till
too K
T-
h
*7rknl UM) K)
(7.C1
27rkBT
V   T )
( /in10 Ä\2
(6.626 x 10~34 (kg-meter2/second)2'
meter) \\.66x 10~27 2jr(1.381 x 10"23 kgmeter2)/(second2K)(300 K)
uniu
Doing the arithmetic yields
/   h2   \3/2    ,300 K\3/2        j3 _„
Combining equations (7.C.6) and (7.C.9) yields
kg)
(7.C.H)
(7.C.9J
qrqH
M,
3/2
/300 K
3/2
(1.024 Ä3-amu3/2)
,(Mf)Mh2
Setting T = 300 K, Mt = 21 amu, MP = 19 amu, MH2 = 2 amu yields
^   x /     21 amu     n3/2
(7.C.HH
VqFqH-
(19 amu)2 amu
1.024 Ä3-amu3/2 = 0.42 Ä3 (7.C.U)
Next, let's calculate the ratio of the rotational partition functions. The fluorine atom doc, not rotate, so
According to table (6.7)
iH2qF2 / rot
87rkBTI
q
qr =
hp
(7.C.I2)
(7.C.13)
where kB is Boltzmann's constant, T is the temperature, hp is Planck's constant, and I is the moment of inertia of the molecule. Combining (7.C.12) and (7.C.13) yields
_S1
qH2/,
8jrkBTI*/h3
,87rkBTIH2/hpy IH2 Substituting in the adjusted value of I* and IH2 from Table 7.C.1 yields
I*     7.091 amu-A2
_q_ qH2
!h2     0.275 amu-ä:
= 25.8
(7.C.14)
(7.C.15)
I let's calculate th< vibrational pi.......in function* According to Table 6.3
(7.C.I6)
I in.i gel an expression for the exponential term in equation (7.C.16), M is easj hIiow
h„(l cm-')
ly
)J\   T yVlcm-'/
k„T     V(kB)(300 K)J\   T   ; VI cm lug in values of hp and k„ from the end of the book yields h,.v     (2.85 x 1()-3 kcal/(molcm_1))(l errr1) /300 K
i ,. I        (1.980 x 10--3 kcal/(mol-K))(300 K)   V T
) ( 1 cm 1 )
(7.C.17)
(7.C.IK)
vc actually used hPC/Na and kB/Na in equation (7.C.16), and not hp where N, ll idro's number and ! is the speed of light, to get the units right. Doing the arithmetil .....juation (7.C.18) yields
hpv ,n-3,f300K
= (4.784 x 10"3)
V T
(l cm"1)
(7.C.19)
lablc 7.C.2 shows numerical values for various values of v. The vibrational partition linn lion ratio equals
qH2/
q^gLAnd = (Dd ■ 19)(1.19)
(qH-H)h2 1
1.42
(7.C.20)
I let's calculate the ratio of the partition functions for the electronic stale. Let's nil lider only the ground electronic state:
= 1
^qF/elec      (ge>H2(ge)F 1x4
I in.illy, let's calculate kBT/hP:
k„T     (1.381 x 10~23 (kgmeter2)/(second2molecule-K)(300 K) ^ T 6.626 x 10~34 (kgmeter2)/second
= 6.25 x 10l2/moleculesecond ^3qq k)
Table 7.C.2 The vibrational partition function
(7.C 'I I
»b' hp
300
i)
(7.C.22)
Mode	v, cm 1	hpv/kBT	qv
	4395.2	21.0	1.0
(qHH>H2	4007	19.2	1.0
Mhcml	379.9	1.82	1.19
'I VI It I XAMI'I i
I'll
Putting thin .mi togethoi .111<>w our in , m, ulute .1 preexp.........11i
')( "'
■ / Vqii.q
'i    \   /  <r   \   1 q1
Ariins V-lll.-Ml /„„ V <-lll..*|l / sih Vqil,l|l /,.,,.,
(7.C.23I
Plugging in the numbers yields
k<) = 2(6.25 x 1012/(moleculesecond))(0.42 A3)(25.8)(l.42)( I)
= 1.92 x 1014 A3/(molecule-second) (7.C.24V
If one uses the actual transition state geometry, the only thing that changes significantlj is the rotational term. One obtains
qn2/,
5.48 (amu-A2)
= 19.9
M2jm    0.275 (amu-A2) Then k0 becomes
k„ = 2(6.25 x 10,2/molecule-second)(0.42 A3)(19.9)(1.4)(l) = 1.46 x 1014 A3/(moIecule-second)
a.cm
(7.C.26)
One can also calculate the preexponential via old collision theory. In collision theory, one considers the translations and rotations, but not the vibrations:
ko = 1
qH2qF
qH2qF
(7.C.27)
In equation (7.C.26), the rotational partition function should be calculated at the collision diameter, and not the transition state geometry. If we assume a collision diameter of 2.3 A (i.e., the sum of the van der Waals radii), we obtain
, / (2 amu)(19 amu)\ IT = (rF-H2r(uFH2) = (2.3 A)2 (--~-1 ) = 9.57 A2-an
21 amu
Substituting into equation (7.C.25) using the results given above yields
9.57 A2amu
(7.C.28)
k0 = 2(6.25 x 10l2/(mol-second))(0.42 A3) = 1.8 x 1014 A3/(mol-second)
0.275 A2-amu
(7.C.29)
Table 7.C.3 summaries these results. In this particular example, collision theory did pretty well. Transition state theory does not improve the predictions if you use the real transition state geometry. You can fit the real data if you "adjust" the transition state geometry. Overall, transition state theory is almost the same as collision theory.
Of course, one should expect that from the calculations. Transition state theory makes two corrections to old collision theory:
I11I1I11 / C :i   A 1 oiii|iiiiI'iiiii ol tlm |)l<iiix|iiiiieiill.il
I .1I1 ol.ilotl t > v Ifiiimll.....        lili- Ihooiy .iixl collision
II II I II V III 11 II- UK 1 MU.....11 itnl v.iloo 1111 A '/(11 ml M-i 01 i<l)|
kn transition sioii- theory with adjusted	1.92 x	l()14
transition state geometry		
ko transition state theory with exact	1.4d x	10"
transition stale geometry		
kn collision theory	1.8 x	10"
kn experiment	2.3 x	10"
I   I i.nisiiion state theory uses the transition state diameter rather than the colli......
diameter in the calculation. '   liansilion state theory multiplies by two extra terms: (a) the ratio of the vibrational
partition function, to the electronic partition function for the transition stale and
1 hi 1 lie reactants.
Illii 1 ol these corrections is very big, and these corrections tend to cancel one anothei
I In result in this example was that the prediction was less reliable than collision theory
i Ins is a typical result. Transition state theory makes small adjustments to collision
.......y, but ignores other key terms such as the coupling between vibrations and
..... i.iiions. The net result is that for many examples, transition state theory is noi more
II urate than collision theory in estimating preexponentials for reactions.
1 simple 7.D   Activation Barriers for Reactions According to Transition Mule
I henry   Use transition state theory to calculate the activation barrier for reaction / (' i
Solution   According to transition state theory
F — F* 4- F* — FHz — FF ii T  '  M)      Ml M)
( / II I I
where E* is the energy of the transition state relative to the reactants and Ej, Ii',' , Ej in the zero-point energies of the various species. Recall that E0 = hPv/2, where hp is ri.niek's constant. Therefore
Ea = Ej + ^hp(vhh + v^nd + v^end - \>lnnaiaK — vhh)
II we use the "adjusted" potential energy surface, we obtain
Ea = 1.7 kcal/mol + ^(2.859 x 10~3 kcal/(mol-cirr1))
x (4007 cm-1 +397.9 cm"1 +397.9 cm"1 - 310 cm"1 -4392 cm"1) (7.D.3) = 1.7 kcal/mol + 0.14 kcal/mol = 1.84 kcal/mol
()ne can also calculate the activation barrier based on the exact potential energy surface. Stark and Werner [1996] et al. did not report vibrational frequencies for the bending mode, so I will assume that they are the same as in the approximate potential energy surface. In
,H2
(7.1).
392       IN UM )| Uli tu IN H i III Al Iii )N MAU Ulli my
m n vi n i xami'i i •. :to:i
Table 7.D.1   A comparison ot the activation barrier calculated from transition state theory to the experimental value (in kcal/mol)
Ea from the exact potential energy surface 5.38 Ea from the "adjusted" potential energy surface 1.84
Ea from experiment 1.58
that case
Ea = 5.6 kcal/mol + ±(2.859 x KT3 kcal/(mol-cm-1))
x (3750 cm"1 + 397.9 cm"1 + 397.9 cm"1 - 310 cm" = 5.6 kcal/mol - 0.22 kcal/mol = 5.38 kcal/mol
4392 cm"1) (7.D.4)
Table 7.D.1 compares these calculations to the experimental results. Again, transition stale theory gets you the right order of magnitude. However, it is not exact. In this particulai case, tunneling lowers the barrier and tunneling is ignored in transition state theory. One can get exact results by adjusting the potential energy surface. An alternative is to do the calculation correctly including tunneling as described in Chapter 9.
Example 7.E Temperature dependence of the preexponential In Example 7.C, we calculated the preexponential term at 300 K to calculate k0. How does the preexponential change with temperature?
Solution The principal temperature dependence comes from the kBT/hP term and the ratio of translational partition functions in equation (7.C.24). If one substitutes equation (7.C.I0) into equation (7.C.24), one finds
kB300 K\ f 300 K\V2 ( q*
QH7qHF/ Irans VqiblHI1/ rot 1 n        300 K   N1 n     ' 300 K
qH2qHF/3(vihK VqH2qHF/3eleCK
(7.E.1)
where
qH2qHF/ w»
is the translational partition function at 300 K. Comparing equa-
31«) k
tions (7.D.1) and (7.C.22) shows
ko » kf K
300 K
1/2
(7.E.2)
In the homework set, we ask the reader to look at a case where the number of rotational modes also changes. In that case, one predicts some extra temperature dependence due to the temperature dependence of the rotational partition function.
Example 7.F Transition State Analogs In the examples so far in this chapter, we considered exact transition state for a reaction. However, another approach is to find a
> ii molecule that is an million ol Ihc imnsilion stale The idea is to use the billowing ' <i'n union:
k„    /A1.e,ASl/*" (7.F.1)
In n i,, is the preexponential, kB is Boltzmann's constant, I is the temperature, hp is
i.....t .....Maui, and AS1 is the entropy of formation of the transition state.
i.....In n estimates AS* by looking at a molecular analogy of the transition stale Foi
imnli  i onsider the reaction
CF3Br + CH3 -► CF3 + BrCH,
i In 11.01'.itii>ii stale for the reaction looks like
CF.vBr-.-CH, i mi an analogous molecule and use it to estimate AS+ .
(7.F 'i
(7.F 1)
Holullon Consider the stable molecule CF3-S-CH3. It has a structure like thai ol i I | Hi CH3. The bonds in CF3-S-CH3 are shorter than in the transition stale, and iii frustrated rotations have higher barriers, but CF3-S-CH3 is an analog of the transition llHlc lor reaction (7.F.2). Now consider the reaction
CF3S + CH3 ==> CF3SCH3 i in reaction is similar to the reaction
CF3Br + CH3 -► (CF3—Br—CH3)tst
(7.1 .4)
(7.F.5)
where (CF3—Br—CH3) is the transition state for reaction (7.F.2), and it happens that I • in look up (ASA) the entropy change for reaction (7.F.4). According to the NISI tables. iSa is -41.2 cal/(mol-K). Years ago people often used the approximation
i iniibining equations (7.F.1) and (7.F.6) yields
ko > ZABe+as^b
(7.F.6)
(7.F.7)
We put a "greater than or equal to" sign in equation (7.F.4) because ASA is a lower bound lo \S for the reaction. Generally, the analog has a smaller rotational partition function Mi.in the transition state, so the ASA for the analog is always less than AS*. For example. AS1 for reaction (7.F.2) is -14 cal/(mol-K), while the reaction
CH3S + CH3 ===> CF3SCH3
(7.1 .8)
has a AS of -41.2 cal.(mol-K). By comparison, collision theory gives an upper bound to the preexponential. The result is that one can get a range for the preexponential for the reaction.
•i( II VI II I XAMI'I I H
i n us use Mu- transit ion state analog to estimate k,, foi reai lion (7.B.2) at 300 K Wa| will use the approximation
k„ = ZABc ASaA» (7.1',91
First, we will estimate ZAli following the methods in Example 7.A:
Zab = Jr(b„,ii)"vABc
Habc =
( II It
1 I
+
I I
+
13.8 amu <7.MI|
mcF,Br    nicH,      '79 amu     15 amu At 500 K
., j /500 K \ "'
vabc = 2.52 x 1013 A/second -
' V300 K/
Assume a collision diameter of 2 Á:
1 amu
13.8 amu
8.8 x 1012 Á/second
(7.F.1Í
ZA1, = tt(2 Ä)2(8.8 x 1012 A/second) = 1.1 x LO1'
(7.F.1 I
mol/second Substituting into equation (7.F.9)
k„ = (1.1 x 1014 A3/(molecule-second)) exp (   412 cal/^molK^ (7.F.14)
V 1.98 cal/(mol-K) /
Doing the arithmetic yields
ko = 1 x 10s A3/(molecule-second) (7.F.15)
Benson (1986) also has one other good rule of thumb. If you have a transition state ol the form CF3-Br-CH3, then the CF3 and CH3 bending and rocking models will have vibrational frequencies of about 70% of the vibrational frequency of the rocking modes in a stable molecule. That typically adds a factor of 3 to the rate. Therefore the value based on an analog is k„ = 3 x 105 A /(molecule-second). By composition, the experimental value is 1012±0-5 A3/molecule.
In my experience, the transition state analogs grossly underestimate preexponentials. However, they provide useful lower bounds, so if you measure a lower preexponential than you calculate for the analog, you know that something is wrong.
We will see later in this book that even though transition state analogs are not that useful in predicting AS\ the analogs are very useful in designing catalysts or enzymes. Catalysts and enzymes enhance rates by lowering the energy of the transition state for a reaction. One way to design a catalyst is to find a transition state analog (i.e., one with the correct ASA) and then design a catalyst that strongly binds to the analog. If the catalyst binds strongly to the analog, the catalyst will usually also bind strongly to the transition state. The binding lowers AG*. Transition state analogs are very useful in designing catalysts even though they are not as useful for calculating rates.
There also is the converse, that if you want to stop a catalytic reaction, then a good way to do that is to add a transition state analog to block the transition state for the formation of an undesired product. For example, there is a class of AIDS drugs, called
hi Inhibitors, which woik by blinking the liaiisilion stale lot one ol the ke\ steps in
I   i,„ tioiiol the aids vims People design the drugs by rinding a transition slate anulog lii    iep in the reprodui tlon i yi le ol the vims, and then testing the transition state «tlHln|t lis a drug. The transition state analogs are very potent drugs. They really work!
.....,,1. 7.<;   Understanding Contour Hots   Consider a potential v delined by:
v(t-1. r2, r0. a, w, vp, wa, hr) = w*(Exp(-2'a'(r1 - rOn -2*Exp(-a*(r1 - rO))) + (w + hr)*(Exp(-2*a*(r2-rO)) -2*Exp(-a*(r2-rO))) +vp*Exp(-a*(r1 + r2-2*r0))+ w + wa*Exp(-4*a*a*((r1 - rO)A2 + (r2 -3*rO)A2)) + wa'Exp(-4*a*a*(«r1 -3*rO)A2) + ((r2-rO)A2)» If(v > 20 + abs(hr)) Then v = 20 + abs(hr)
(7.0.1)
hi
t ,i, i ontour plots for the following function of rl and r2 with parmeters rO, a, w. > |i «.i given by
in w 109 kcal/mol, wa = 0, hr = 0, vp = 258 kcal/mol, rO = 0.74 A, a = 1.2 A lb) w = 109 kcal/mol, wa = 0, hr = 0, vp = 218 kcal/mol, rO = 0.74 A, a = 1.2 A i, i w = 109 kcal/mol, wa = 0, hr = 0, vp = 158 kcal/mol, rO = 0.74 A, a = 1.2 A oh w = 109 kcal/mol, wa= -10, hr = 0, vp = 258 kcal/mol, rO = 0.74 A, a
1.2 A'1
(c) w= 109 kcal/mol, wa = 20, hr = 0, vp = 178 kcal/mol, rO = 0.74 A, a = I..' \ Hi w = 109 kcal/mol, wa = 0, hr = -20, vp = 233 kcal/mol, rO = 0.74 A. a 1.2 A"1
Solution   I used microsoft excel to do the calculations. First I defined a microsoft excel
......Ink- to calculate the v. My module is given in Table 7.G.I. The spreadsheet can be
i......,1 in the instructors materials.
I hen I set up a microsoft Excel spreadsheet to do the calculations. Table 7.G.2 shows Ihi formulas in part of the spreadsheet. I named cell Bl w, cell B2 vp, cell D2, and cell
i utile 7.G.1   The module used to calculate the function in equation (7.G.1)
Public Function v(M,r2,r0,a,w,vp,wa,hr) As Variant v   w*(Exp(-2*a*(r1 - rO))-2*Exp(-a*(M - rO)))
v + (w + hr)*(Exp(-2*a»(r2 - rO))-2'Exp(-a*(r2 - rO))) v   v + vp*Exp(-a*(r1 + r2-2*r0)) v   v + w
v   v + wa*Exp(-4*a*a'((r1 - r0)"2 + (r2 -3*r0)"2)) v   v + wa*Exp(-4*a*a*(((M - 3*r0)"2) + ((r2 - r0)"2))) 11 (v > 20 + Abs(hr)) Then v = 20 + Abs(hr) I nd If
I nd Function
CO
8
Table 7.G.2 The formulas in part of the spreadsheet3 used to solve Example 7.G.2. A copy of this spreadsheet is available at http:/www. wiley.com/chemicalkinetics
01
03
=A4+sp
=A5+sp
=A6+sp
:A7+sp
=A8+sp
=A9+sp
11 =A10+sp
12 =A11+sp
13 =A12+sp
14 =A13+sp
15 =A14+sp
109.4
258
0.5
=v(B$3,$A4,rO, a,w,vp,wa,h r)
r0=
:B3+sp
v(B$3,$A5,r0, a,w,vp,wa,hr)
v(B$3,$A6,rO, a.w,vp.wa,hr)
v(B$3,$A7,r0, a,w,vp,wa,hr)
=v(B$3,$A8,r0, a,w,vp,wa,hr)
v(B$3,$A9,r0, a,w,vp,wa.hr)
=v(B$3,$A10,rO, a,w,vp,wa.hr)
v(B$3,$A11,rO, a,w,vp.wa.hr)
v(B$3,$A12,rO, a,w,vp.wa.hr)
v(B$3,$A13,rO, a,w,vp.wa.hr)
v(B$3,$A14,rO, a,w,vp.wa.hr)
=v(B$3,$A15,rO, a.w,vp.wa,hr)
v(C$3 a.w,vp
,$A4,rO, .wa.hr)
v(C$3 a.w,vp
,$A5,rO, .wa.hr)
v(C$3 a.w,vp
,$A6,rO, .wa.hr)
=v(C$3 a.w,vp
,$A7,rO, .wa.hr)
=v(C$3 a,w,vp
,$A8,rO, .wa.hr)
v(C$3,$A9,rO, a,w,vp.wa.hr)
=v(C$3,$A10,rO, a,w,vp.wa.hr)
v(C$3,$A11,rO, a,w,vp.wa.hr)
=v(C$3,$A12,rO, a,w,vp.wa.hr)
v(C$3,$A13,rO, a,w,vp.wa.hr)
v (C$3,$A14,rO, a.w,vp.wa,hr)
=v(C$3,$A15,rO, a,w,vp.wa.hr)
aThe actual spreadsheet extends to cell AF40.
1 .2
0.74
=C3+sp
=v(D$3, a.w.vp,
$A4,rO, wa.hr)
v(D$3, a.w,vp,
$A5,rO, wa.hr)
v(D$3, a,w,vp,
$A6,rO, wa.hr)
=v(D$3, a.w,vp,
$A7,rO, wa.hr)
v(D$3, a.w.vp,
$A8,rO, wa.hr)
=v(D$3,$A9,rO, a.w,vp,wa,hr)
v(D$3,$A10,rO, a,w,vp.wa.hr)
v(D$3,$A11,rO, a,w,vp.wa.hr)
v(D$3,$A12,rO, a,w,vp.wa.hr)
v(D$3,$A13,rO, a.w,vp.wa,h r)
=v(D$3,$A14,rO, a,w,vp.wa.hr)
v(D$3,$A15,rO, a.w,vp.wa,hr)
sp=
=D3 + sp
=v(E$3,$A4,rO, a,w,vp.wa.hr)
=v(E$3,$A5,rO, a,w,vp.wa.hr)
=v(E$3,$A6,rO, a,w,vp.wa.hr)
=v(E$3,$A7,rO, a,w,vp.wa.hr)
=v(E$3,$A8,rO, a,w,vp.wa.hr)
v(E$3,$A9,rO, a,w,vp,wa,hr)
v(E$3,$A10,rO, a,w,vp.wa.hr)
=v(E$3 a,w,vp
,$A11,rO, .wa.hr)
v(E$3 a.w,vp
,$A12,rO, ,wa,hr)
v(E$3 a.w.vp
,$A13,rO, .wa.hr)
v(E$3 a.w,vp
,$A14,rO, .wa.hr)
v(E$3 a.w.vp
,$A15,rO, .wa.hr)
0.05
=E3+sp
=v(F$3,$A4,rO, a,w,vp.wa.hr)
=v(F$3,$A5,rO, a,w,vp.wa.hr)
=v(F$3,$A6,rO, a,w,vp.wa.hr)
=v(F$3,$A7,rO, a,w,vp.wa.hr)
v(F$3,$A8,rO, a,w,vp.wa.hr)
=v(F$3,$A9,rO, a,w,vp.wa.hr)
v(F$3,$A10,rO, a,w,vp.wa.hr)
v(F$3,$A11,rO, a,w,vp.wa.hr)
=v(F$3,$A12,rO, a,w,vp.wa.hr)
v(F$3,$A13,rO, a.w,vp.wa,h r)
=v(F$3,$A14,rO, a,w,vp.wa.hr)
hr=
0
=F3+sp
v(G$3,$A4,rO, a,w,vp,wa,hr)
v(G$3,$A5,rO, a.w,vp.wa,hr)
v(G$3,$A6,rO, a,w,vp.wa.hr)
=v(G$3,$A7,rO, a,w,vp.wa.hr)
v(G$3,$A8,rO, a,w,vp.wa.hr)
v(G$3, a.w.vp,
$A9,rO. wa.hr)
=v(G$3, w.vp,
$A10,rO, wa,hr)
v(G$3, a.w.vp,
$A11,rO, wa.hr)
v(G$3, .w.vp,
$A12,rO, wa.hr)
=v(G$3: a.w.vp.
$A13,rO. wa.hr)
v(GS3, a,w,vp,
$A14,rO, wa,hr)
v(F$3,$A15,rO, =v(GS3,SA15,rO, a,w,vp.wa.hr)     a,w,vp.wa.hr)
2 ^ > P
>° CD
i m - 3
ft r*
Q a
2 S
o £
3. -0- =r
_. <
o t3
s II
I
—
0
—
0
§ o
o 'S
Ü o
" c.
s> c
o 5
3 I
g o 5
3. Ef
Crt CD
0 —
1 6
2 n °- n = 13
o o
-
s
(
n
6 ~o
o
o'
—
S
—
—
a
r. □
rn a
r.
s
3 _
S- 3
-
7   x _
n ^ o
a> £ B.
<° £r o
•a jo o
o a
o 3 r~
3 <" 3"
B. b, »
E. 3 -o
es. °" o
Si 3- o
— - "3
5" a g
ss o o
32 § Ef »   S Q
a- c <
a3 ?
i —
*< ft
a
C   ... Cl
s. 5. «,
(TO
° 2-» 5 S
1 Ef ä
g     « M.
g. S. 3
a- & §
% « 5
£ 1- £
H H 3" ? f? 3
5- 2"
8. 5
2 O
•   P ?
pa
03
a
& ^ a. & *■ —
-s >• o
2. _~
3   • £
° o
3 ft
£ g
a &
n
S, ° 11 =. —
— -
=   > =
5. > r
1 8 f
= —
3 g oq  ö ~  _ Z
I I 1 s- as*
§ _ " ob a — b
O. g. ►< jj h =
3 -
00    ^_ £
Hl Mt . I.......I   i Ml I ii iy
?l( II VI 11 I KAMI'I I ',
|<)M
wcococowcowcocococo
Figure 7.G.2 A side view of the potential in Figure 7.G.1 (elevation = 0, rotation = 90, perspective = 30, chart depth = 100).
|| In very important iluii ilie iciuli'i I>»■ able id innpn/r the saddle point! mi ihcsc in plots Therefore the readct ihould eximlne these contoui plots carefully before I........hmt.
1 i I M   I solved part B using the same spreadsheet that I useil for part A. I changed the i \ |i hi,I then used the same plots as with pail A.
I..... <il -hows a plol of the potential. In this case, there is a deep valley, bill no
hill i" iwcen the reactants and products.
i.....        '.0.5 shows a top view of the potential. Notice that there are the hemisphcrii .ii
bill none of the parabolic contours, and none of the areas that look like a bod) ,ih urns and legs.
Vain Ihis problem was solved using the spreadsheet in Table 7.(i. I and the Ni.....
, i ....., , ommands. Figure 7.G.6 shows a diagram of the potential. In this case linn r. ,i
ill,   us before, but now the potential falls, showing a well. A well corresponds to the .......hi ol a stable molecule or molecular complex.
I Igurc 7.(1.7 shows a top view of the potential. In this case there is a bull ey< hi ol ellipses corresponding. The bull's-eye pattern corresponds (o a stable complex ........Iccule.
' \ii Ci   Again I used the same spreadsheet as before to calculate the potential, In this , , I set sp = 0.07 to get a better plot, and inputted the parameters accordingly. I ipiic 7.G.8 shows a top view of the potential for this case. Notice thai there an i ■■■ ' lliptical areas corresponding to complexes or intermediates and a saddle point in
In iwcen.
20.0 f8.0 f6.0 14.0
■■■■■ "
■■■■■   \ •-■ *t.
■■■■■ "9,% *|k
■■■■■■ . "fi,.
■■■■■■ ^
■■■!■■
■■■■■■■ i
5!!!!!!. !!!!!!!!-!!!!!!!!!_
Hssaasss:. .
5SSSSS5SS5SS.
■■■■■■■■!■■■■■■ ■■■■■■■■■■■■■■■■
SSSS55SSSSS55SSSSS«.
■■■■■■■■■■■■■■■■■■■■■■■«
12.0 --10.0
8.0 4
6.0
4.0 4-
2.0 +
0.0
••Saw
S31
■■■■■■■■■sb
-shsssss
Mt..   . ""■■»>.
"wÄjaiii,*. m^i0äBi—
m
.«■»%..____ ■
...1
:sssssssss:ssss:hsss:»s:....
CO CO o> CM i-     T-     T- CM
|- S28 S25 S22 S19 S16 S13 S10 S7 S4
ttffsi
io
cm
00 ■■-
cvi co
co
Figure 7.G.3 A top view of the potential in Figure 7.G.1 (elevation = 90, rotation = 0 perspective chart depth = 100).
,30,
I liiure 7.G.4 A diagram of the potential when w = 109 kcal/mol, wa = 0, vp = 218 kcal/mol, rO = 0.74 A, ,i    1.2 A 1 (elevation = 15, rotation = 60 perspective = 30, chart depth = 100).
•il II VI II I XAMI'I I ',
IUI
Figure 7.G.5 A top view of the potential in Figure 7.G.6 (elevation = 90, rotation = 0, perspective ;iu, chart depth = 100).
Muni« f.Q.7  A top view of the potential in Figure 7.G.6 (elevation = 90, rotation = 0, perspective 30,
ll II' ilnplh 100).
Figure 7.G.6 A diagram of the potential when w = 109 kcal/mol, wa = 0, vp = 158 kcal/mol, rO = 0.74 A, a = 1.2 A~1 (elevation = 15, rotation = 60, perspective = 30, chart depth = 100).
o   co   <o   o>   eg in
t-    t-    i-    i-    CM CM
I Igure 7.G.8 A top view of the potential for w = 109 kcal/mol, wa = -10, hr = 0, vp = 158 kcal/mol, hi   0.74 A, a = 1.2 A-1 (elevation = 90, rotation = 0, perspective = 30, chart depth = 100).
Many reactions show potential energy surfaces like those in Figure 7.G.8, and so one-should study the features carefully.
Part E I solved part E using the same spreadsheet as for part D with sp = 0.1(). Figure 7.G.9 shows the plot. In this case, there are two saddle points and an intermediate or complex in between.
Pert F I solved part F using the same spreadsheet as for part D with sp = 0.10. I'igure 7.G.10 shows the plot. There are parabolic contours showing that the potential i using, but there is no transition state between reactants and products.
Readers may want to reproduce these results themselves before proceeding to the next i li.ipter.
''''''' 'I'll,   i i1 'ri .....i    '   m.....■    i     i Ml I M iY
I'ln Hli I M'
III I
;■(> n 18.0 16.0 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0
v~ .ii:
or     \\ i
.v   ".......• . \.
if ItlSBK^ % K \ *k 1
is Xv-
at \mb£b
f-mm* ,
I I I I I I
• .11
:,;>n
S25
S22
S19
S16
S13
S10
S7
S4
S1
Figure 7.G.9 A top view of the potential for w = 109 kcal/mol, wa = 20, hr = 0, vp = 178 kcal/mol, rO = 0.74 A, a = 1.2 A~1 (elevation = 90, rotation = 0 perspective = 30, chart depth = 100).
Figure 7.G.10 A top view of the potential for w = 109 kcal/mol, wa = 0, hr = -20, vp = 233 kcal/mnl, rO = 0.74 A, a = 1.2 A"1 (elevation = 90, rotation = 0, perspective = 30, chart depth = 100, sp = 0.1).
Example 7.H Counting normal modes Consider the reaction H + CH3CH3 CH2CH3. How many modes should you include in q*?
H2
Solution Consider the transition state H ■ • • H • ■ • CH2CH3. The transition state li.r 9 atoms so it must have 3 x 9 = 27 normal modes. The transition is nonlinear, so has three translational modes, three rotational modes, so it must have 27 — 3 — 3 = 21 vibrational modes.
H
......d 11li" iiiodes is tbc II II (' imyiuincllie slicUli   ľhnl is tlie IflOdC thtl c.unes llie
m  hydrogon ulonj) the reai Ilon i oordlnute i hal mode ľ. alread) Im luded m il»' I k u I iiľi iľim in transition stale Iheory so wc sliould not include u in q'. Tlmi leiives i    'n \ iIii.iikiii.iI imidcs. Therefore, the transition state musí have 1 translational iluee rolational modes and .'o vibrational modes,
V J  SUGGESTIONS FOR FURTHER READING
i li inentary discussions of transition state theory and collision theory can be
I.....id III
i   i I mil, i. ( hemical Kinetics, 3rd ed.. Harper & Row, New York, 1987. .........ill suggestions for further reading are given in Chapters 8 and 9.
I I'MOBLEMS
i   I Icline the following terms:
(a) Arrhenius' theory	(k)	Potential energy surface
ili) ( ollision theory	(1)	Reduced mass
le) Transition state theory	(m)	Collision partner
ul) URKM model	(n)	Partition function
(e) 1 .indemann model	(0)	Moment of inertia
(f) Transmission coefficient	(P)	Transition state
(g) Wigner and Polanyi's model	(q)	Col
ilii Tunneling	(r)	Loose transition state
(i) lintropy of activation	(s)	Tight transition state
(j) Saddle point energy	(1)	Transition state analog
'   lii Section 7.4.1 we noted that transition state theory predicts that a bimolec ular collision has a preexponential in the order of 1013 A3/(moleculesecond i ('alculate a typical value of the preexponential in units of (a) cm3/(mol-second). (h) liter/(molsecond), (c) liter/(molhour), (d) dm3/(molsecond).
' * Compare and contrast: collision theory, transition state theory, and the RRKM model. What are the key assumptions in each model? What are the key predictions of each model? Where do the models fail? How is transition state theory different from Wigner and Polanyi's model?
' I Describe in your own words why the preexponentials for unimolecular reactions are so large. Where does the large term come into the RRKM model? How large is the term?
7.5 What lead Arrhenius to propose that the rate varied exponentially with temperature? That is, what in Figure 7.1 and equation (7.2) leads to an exponential temperature dependence?
7.6 One of the ways that enzymes work is by binding to the transition state of a reaction. Assume that you have an enzyme that forms a 10-kcal/mol bond to the
••«>•< ill 11 li pi n H . Ill IIN ii p 111 A< iii in MA 11  11II i HIV
1'in ilii I M\ 40n
readmits (i.e., A(i K) kcal/inol), n I.' kciil/niol bond In I Ik- products, and u 2lfl kcal/mol bond to the transition state. H s how much will the rati- ol i caelum i mi cum
al 310 K in 11 ie presence ol llie cn/yme?
7.7 Table 7.4 compares the saddle poinl energy to the activation harrier lor a iiiiinhof
of reactions.
(a) Plot the measured versus calculated activation barrier. Do you sec any pattern?
(b) Why are the barriers lower than the saddle energy for hydrogen transfer reactions?
(c) How do the other barriers compare?
7.8 Figure 7.6a shows a potential energy surface for the reaction
H + CH3OH-► H2 + CH2OH
Make a copy of the figure. Start with the picture on the left of the figure anil identify
(a) The minimum energy of the reactants when the reactants are far apart. U Inn When the reactants are far apart, the H-H distance is large.)
(b) The minimum energy of the products.
(c) The vibrational well of the reactants.
(d) The vibrational well of the products.
(e) The transition state.
(f) What does the C-H potential look like when the H-H distance is large?
(g) What does the H-H potential look like when the C-H distance is large?
(h) What does the potential look like near the transition state?
(i) Repeat all of the questions on the contour plot on Figure 7.6b. Can you see the C-H and H-H potentials? Can you see the shape of the transition state?
7.9 In Section 7.4, we discussed the Lindemann model. We assumed that a unimolec ular reaction A     B followed
X + A
A* + X
A*
and stated that
k|[A][X]
(P7.9.I)
k3
[X] +
(a) Use the steady-state approximation to derive equation (P7.9.1).
(b) Now let [X] = [A]. Convert equation (P7.9.1) to a linear form. Two linear forms are possible: one based on the Lineweaver-Burke transformation (see Example 3.A), and another based on the Eadie-Hofstee transformation. Derive an equation for both of them.
(c) Table P3.17 page 149 shows some data for the reaction HNC => HCN. How well do each of the lines fit?
Ill) Why do the hues look so pool when Figure / It) shows dint equation I'/ '» I tils the data quile well'' Mini look hm k to the discussion ol errors 111 Example VA
fill   11 so iiansition slate theoiy lo calculate the temperature dependence ol the rule 1 oiistaul for
In) llie reaction A I IK'   > AB ) C, where A, B, and C are atoms. Assume a
hue.n transition stale, ílu  \ 1 |t('   . AH I (', where A. B, and C are atoms. Assume a iioiilineai
Iiansition slate.
n 1  \H 1 CI)   • AliC I I), where A, B. ('. and I) are atoms. Assume .1 line.u transition slate.
oil \H 1 CD —► ABC + D, where A, B, C, and D are atoms. Assume a nonlineai transition state.
Hl I sinuate the value of the preexponential at 500 K in all cases. 'II   I uimple 7.CÍ showed values of the potential given by equation (7.(1.1).
in) Sei up the spreadsheet yourself and verify the findings. The original spreadsheet
is posted at http://www.wiley.com/chemicalkinelics. iln l'i nil out the numerical values of the spreadsheet for w = 100 kcal/mol, wa 0.
hr = 0, vp = 230 kcal/mol, rO = 0.74 Á, a = 1.2 A"1. (C) Consider the reaction A + BC —► AB + C, with n = the AB distance ami
1. = the BC distance. Find the place corresponding to the reactants and the
products. (Hint: Examine Example 8.A — this is similar except n and rj are
reversed.)
id) Trace a path from the reactants to products as indicated in Example 8.A.
lei Identify the saddle point on your plot.
ď) Show that your point is really a saddle point.
lg> What is the energy of the transition state?
1 1 '   I sample 7.G showed values of the potential given by equation (7.G.1).
la) Set up the spreadsheet yourself and verify the findings. The original spreadsheet
is posted at http://www.wiley.com/chemicalkinetics. lb) Make contour plots (i.e., top views) of the potential for the following
parameters:
(1) w = 100 kcal/mol, wa = 0, hr = 0, vp = 218 kcal/mol, rO = 0.74 A, a
1.2 A"1
(2) w = 100 kcal/mol, wa : a = 1.2 A"1
(3) w = 100 kcal/mol, wa = 1.2 A"1
(4) w = 100 kcal/mol, wa = 1.2 A"1
(5) w = 100 kcal/mol, wa = 1.2 A-'
(6) w = 100 kcal/mol, wa = 1.2 A-'
Use sp = 0.1 in all cases
: 0, hr = -20, vp = 218 kcal/mol, rO = 0.74 A, 0, hr = 20, vp = 218 kcal/mol, rO = 0.74 Ä, a -: 5, hr = 0, vp = 218 kcal/mol, rO = 0.74 A, a ■ : 0, hr = 0, vp = 188 kcal/mol, rO = 0.74 A, a ■ -5, hr = 0, vp = 218 kcal/mol, rO = 0.74 A. a
(ť) In ciuli i use 1. 11 <t' I tin- iciiclanls, prodlil i., li.insilion slnlcs. in complexes (HI
your plots lollowmg Problem / I Id) Id) Which oi the reactions arc uctivalcd In ihe forward direciioii'.' \llmi tin
reaction will be activated it < I > there is a saddle point Willi an energy abovlj
the reactants or (2) if the reaction is endothennic so that the sysieni inn i
uphill to products.] (e) Which of the reactions are activated in the reverse direction'.'
7.13   The reaction
CH2 / \ CHt—CH2
CH3CH=CH2
has an activation barrier of 66 kcal/mol.
(a) Use the Kassel-Rice-Ramsperger model to estimate the preexponential fo| i reaction at 1000 K. (Assume that you are working in the high-pressure limil Ml that k = k3.)
(b) Estimate the rate constant at 1000 K.
(c) Estimate the temperature dependence of the preexponential for the reaction
7.14 D. R. Hershbach, et al. (1956), reported preexponentials for a number of reaction Most were near 10'3 A/(moleculesecond), but some were not. Table PI. 14 gives . few examples where the preexponential differs substantially from 1013. Go thn>ii|ih each of the examples and tell us what you know about the transition state.
(a) Is it a loose (i.e., hard-sphere-like) or tight (i.e., have a very specific geomeln i
(b) Diagram a likely transition slate from the data given.
(c) Estimate the preexponential for your transition state. Can you find a transii.....
state with a small enough preexponential?
7.15 The preexponentials for surface reactions can be very different from 1013. Consiclr the reaction
CO + S
COad
(P7.15.I)
Table P7.14 A selection of reactions that show preexponentials well below 1013
Reaction
Ea, kcal/mol
log(ko). A3/(molecule-second l
N02 + 02 N03 +0 ► N02F + F
-> co + co2
NO + 03 -NO + 03 -N02 + F2 N02 + CO 2N02        > 2NO + 02
NO + N02Cl-► NOCI + N20
2NOC1 = F2 + CIO: 2C10 =
==> 2NO + Cl2
-► FC102 + F
=> Cl2 + o2
2.6 7
10.4
31.5 26.5
6.4 20.3
8.5 0
11.1 12
11.4 12.3
11.5
11.1
12.2 9.7
10.0
I'Mi Hli I M'
mi
n.
tlil Assiiiue lhal tlie liaiisilinii slnle is n siulncľ nimples lhal does uol liniislale. vlbrate, Ol roUte. ľstinmlľ ihc preexponeiitial loi thc reactiou. Mini: lle caiefiil
wltl......ts SVhen you repluce a translallonal with a vibrational partitlon funetlon
von icplacc qi wiih qr, llowevci c|, Inol q,) appears in cquation (7.U.3).
(h) Now assume thal the preexponential can translate in two dimensions and rotate around O ne axis. How woukl thc precexponenlial change?
i In objective ol ilns problém is to calculate the rate oi formation ol N<) from the ■......youi bedroom. Assume that the m,im reactiou is
N2 +02
2NO
i '
" nli an activation barrier of 85 kcal/mol.
in) What is the collision rale of N2 with 02 in air at 300 K and I aim'.'
(It) What fraction of the collisions have at least 85 kcal/mol?
(C) Assume that there are 2000 liters of air in your bedroom. Use collision theoi \
to estimate the NO production rate in molecules/per hour, (ill NO is poisonous. Will you ever produce enough NO to poison you ' Be sure
to consider the equilibrium concentration of NO.
Use transition state theory and collision theory and the data in Table P7.I7 to i innate the rate of the reaction
H + HBr
-> H2 + Br
(P7.I7.I)
ill (00, 400, 500, 600, 700, 800, 900, and 1000 K. In) Estimate the preexponentials at each temperature
(ID Make an Arrhenius plot of the rate constant as a function of temperature to estimate a preexponential. How well does your preexponential lit agree with the results in (a)?
(c) How well does your activation barrier compare to E*?
Table P7.17 Data for the transition state of reaction (P7.17.1)
	Linear Transition State	H	Reactants HBr
THH	1.5 A	—	1.414 A
rHBr	1.42 A	—	—
VH-Br	2340 cm"1	—	2650 cm'1
vHHBr bend	460 cm-1	—	—
vHHBr bend	460 cm"'	—	—
Curvature	320 cm1	—	—
I	10.4 amu-A2	—	1.99 amu-A2
I	1.21 kcal/mol	—	—
ge	4	1	4
»HUI IN II II II N II   I II IN 11 P Ml IV   I II IIN I I'M 1   11 II ' 'MT
l'.....him lll'l
(ill How do 11 Ii-  picexponcillillls Compute  I" ll»' expel lllieillill  value ul   I  ' -
10'' A'/inii.ii-. ule second)?
More Advanced Problems
7.18   The objective of this problem is to use collision theory and transition stale theory to model adsorption. Consider a simple chemical reaction:
CO + S
-*■ coa
(P7.IK.I)
where a CO molecule reacts with a surface site, S. Masel (1996) shows that oni can model the rate via collision theory:
Rad = (0.25vCoKCcoCsPs,ick
(P7.18 21
7.19
where Rali is the rate of adsorption in molecules/cm2, vCo is the average molecular velocity of the CO, C„, is the gas-phase CO concentration in molecules/cm2, C„ III the concentration of surface sites, as is the surface area taken up by one surface site, and Pstick is the probability that the CO sticks when it hits the surface. Notice that we use 0.25vCo in equation (P7.18.2) rather than vco. If all the CO would impinge perpendicularly to the surface, then vCo would have been the correct term, However, in reality, the COs impinge from all directions, and it works out that the correct velocity is 0.25vCo-
(a) Calculate the rate of adsorption of CO in molecules/(cm2 second) at 300 K and 1 atm pressure. Assume that Pslick is 0.5, and that Cs = 10" sites/cm2.
(b) How long will it take for all of the surface sites to be filled with gas?
(c) In reality, Pslick decreases as the surface coverage increases because gas cannol stick to sites that are filled with gas. Calculate how long it will take to fill 99fl of the surface with gas assuming that PstjCk = 0.5(1 — 9co)» where 8C0 is infraction of the surface covered by CO.
(d) How low of a pressure do you need to reach to slow down the adsorption process, so that the adsorption process takes 100 seconds?
(e) The calculations so far in this problem assumed that the sticking probability was 0.5. Use transition state theory to get a "better" value. Assume that there is no barrier to adsorption (i.e., Ea = 0). Also assume that at the transition state, the CO is stuck to the surface, so it does not translate or rotate. Instead the transition state vibrates, with vibrational frequencies of 45, 49, 130, 133, and 2085 cm1.
In this chapter we have been examining chemical reactions, but the equations in the chapter are also useful for nuclear reactions. Consider the simple nuclear reaction:
'Pb + n
BPb
where n is a neutron.
(a) Derive an equation for the rate of reaction as a function of the concentration of neutrons and lead atoms, the average velocity of the neutron and the lead atoms,
.nut Ibe ( losssei lion loi ueiiliou capture Noli- llie i insssei lion n,     »Mb,,,u )''.
where b.....is the collision dlimetei
llil Assume that you have a nuelcai leacloi producing I cV neutrons. Calculate the i.iic constant ol the reaction. Assume a crosssection ol o 17 barn/lead atom 11 bin n   it) '" A) [Hint: First calculate the velocity of the neutrons from
I ,,     ', in,,! \ „)'. where |{„ is the energy of the neutron, in,, is the mass ol ilu
neutron, and v„ is the velocity of the neutron], ii) Show thai the reaction is first-order in the neutron concentration, itl) Now consider shooting neutrons through a Hat plate. Show that if you slaii the
neutrons moving through the plate at time, t = 0, the concentration of neutrons
will obey
dC„
—- = rrxv„CnCpb (P7.19.1) dt
where C„ is the neutron concentration, CPh is the concentration of lead atoms, v„ is the neutron's velocity, and ax is the cross section, (e) Do a shell balance to show that the concentration of neutrons at any distant t x from the front of the plate obeys
dC„ _
—      _ <7xCnCph
dx
(P7.19
(f) Integrate equation (P7.19.2) to derive an equation for the fraction of the neutrons that are captured in passing through a lead plate as a function of the thickness of the plate. Assume that the plate is 22% 207Pb and that the other lead isotopes have small neutron cross sections.
((j) How thick of a lead plate do you need to capture 99.999% of the neutrons''
•u   Berry and Marshall (1918) studied the reaction
CH3 + C1CF,
CHjCl + CF3
They obtained the results in Table P7.20 (see also Figure P7.20). The objective of this problem is to use transition state theory to obtain a rate constant loi the reaction.
(.1) How many translational, vibrational, and rotational modes are there in the reactants, products, and transition state?
'CF
7
H--C-— CI —— C-- F
/   t      t \
H rCH3-CI        rCF3-CI F
Figure P7.20  Transition state geometry for Problem 7.20.
I'llt >lll I M'. 411
i.,i,i,- i'/ ;>o  D..i.. tm i'i<......in /
rcH rcF
rcF3-ci rcH,-ci Angle H-C-Cl Angle Cl-C-F
E*
Vibrations
ciii (planar) 1.078 A
( l( I
1.335 A 1.749 A
ISI
1.083 1 131 2.014 2.031
Curvature
			104.1	
		110.4°	109.2	
			19.4 kcal/mo	
384	cm-'	337 cm"'	14	cm
1407	cm-'	332 cirT1	7')	cm"'
1477	cm"'	402 cm"'	79	cm
3061	cm-1	531 cm"1	242	cm-1
3240	cm~'	531 cm"'	242	cm"'
3240	cm-'	752 cm"'	274	cm-1
		1095 cm-1	483	cm"'
		1220 cm-'	483	cm 1
		1220 cm"1	647	cm-1
			755	cm-'
			755	cm-1
			951	cm"'
			1131	cm"'
			1190	cm"'
			1190	cm"'
			1379	cm"'
			1379	citt'
			2909	cm"1
			3052	cm"'
			3052	cm"'
			101	cm"1
(b) Next, you need to calculate the three moments of inertia for each of the specie! Recall from freshman physics that
[x= ^mifXj-X™)2
atoms
iY = ]r>i(Yi - Ycm)2
atoms
Iz =      mi(-Z' ~~ Zcm^
lill ( alt ill.lie 11 ii- posilnili ni lín 11 iil*-i ol mass liom
X,,„ Vm.X, Ycm = Yl m'Y'
I
Zen = ^ niiZj
ii-i Next, calculate the ihrce moments of inertia.
ill Plug into transition state theory to calculate the rate constant for the reaction al 800 K.
i ' iiit ol the key steps in the reproduction cycle of the HIV-I virus is foi the vlrui in produce an enzyme protease that catalyzes the hydrolysis of a protein in youi blood cells. The enzyme produces fragments that act as the building block I'm ilic production of more viruses. One class of AIDS drugs work by mimicking iln- transition state for the reaction and binding to the active site on the enzyme / rickson and Wtodawer, (1993) describe the findings.
nil Explain how the transition state analogs block the transition state.
ill) Look up the structure of Indinavir, Saquinavir, and Ritonavir. How arc they
similar or different? icl Look up how each of these drugs bind to the enzyme. («l) What do these findings tell you about the transition state for the reaction on
the protease?
where X;, Y;, and Z, are respectively the X, Y, and Z positions of atoms I; and Xcm, Ycra, and Zcm are the positions of the center of mass, (c) Pick one of the carbon atoms as the origin, then calculate the X, Y, and Z positions of all of the other atoms in the reactants, products, and transition state.