REACTIONS AS COLLISIONS  
i" i hupter 7, we started to study collision theory and described how one can use Mull.....j to estimate reaction rates. All of the analysis in Chapter 7 was for a hard-sphere
Hi lion. The objective of this chapter is to extend the ideas to more realistic collisions i" i'.iiiicular, we will show how one can use trajectory calculations to calculate rales.
.....line ilie forces involved, and provide qualitative information about reactive collisions
i ih material will draw heavily on simulations. A suitable simulation program is available i......      Intp://www.wiley.com/chemicalkinetics.
II I   HISTORICAL INTRODUCTION
l in- idea that reactions were associated with collisions of molecules dates back to the i Hi nl kinetics. The objective of this section will be to provide an historic overview of iin Ideas so that we have a basis for further discussion.
Recall that in 1884 Van't Hoff had proposed that reactions would obey what is now . iilled Arrhenius law:
where ki is the rate constant for a reaction, ko is the preexponential for the reaction, I , is the activation energy for the reaction, kB is Boltzmann's constant, and T is the i1 inperature. As noted in Chapter 2, this rate form was not initially accepted. However, i quation (8.1) eventually was adopted by most investigators.
In Chapter 7, we noted that Arrhenius wrote a series of papers in which he tried ......iderstand why reactions obey equation (8.1). Arrhenius (1889) considered an ideal
••Mf CIS
k,
= koe-<E<^BT>
(8.1)
413
Ill
111 ai in nr. /v. i i n i inionn
CHI I If Ml IN till I ill . ■ii'.
n .1. in.ii
Arrhenius proposed thai if one looked al a chemical system containing A and It there were two kinds of A molecules in the system: reactive A molecules (i.i \ molecules that had the right properties to react), and unreactive A molecules (l.i
A molecules that did not have the right properties to react). Arrhenius then ass.......I
that equilibrium was maintained between the reactive and unreactive A molecule . in obtain
k, = (K0e(AS*)/fe)e-|(AM*^BT| is 11
Equation (8.3) is equivalent to equation (8.1) with
k0 = K0e<ASt>/fe
(K-ll
8.2  COLLISION THEORY
In 1899, Arrhenius did not have a model to estimate either K0, AS* or AH*. Howevei | few years after Arrhenius' work appeared, Trautz (1916, 1918) and Lewis (1916, l(>IN| independently proposed what is now called the collision theory of reactions in an attempl to get a value for K0. We already discussed the findings in Chapter 7. The objective <>| this section is to review the key equations.
Recall that the objective of collision theory is to use a knowledge of molecular collisions to predict K0 from equation (8.4). Trautz and Lewis proposed a model to do just that. The model builds on Arrhenius' concept that only hot molecules can react. Tlir model assumes that the rate of reaction is equal to the rate of collisions of the molecule Consequently, if one can calculate the collision rate, one will then get an expression lm K0. In the remainder of this section we will present an expression for K0 following thl derivation of Trautz and Lewis.
Trautz and Lewis considered ligand exchange reactions of the form
BC
A15 + C
(8..M
where a radical A abstracts an atom B from a molecule BC to yield products. Trautl and Lewis noted that the reaction occurs when a hot A molecule collides with a Imi BC molecule. Under most circumstances, collisions should be rare events. Consequents Trautz and Lewis assumed that the rate of reaction was equal to the rate of collision of hot A molecules with hot BC molecules. Trautz and Lewis then used statistical mechanics to estimate a value for the collision rate, and thereby derived a value for Kq.
The derivation is given in Chapter 7. The result is
i"a->bc = Za^bcC
liACi I k„Tl
where Za^bc is the rate of A — BC collisions. Equation (7.22) gives the total rale ol collisions between hot A molecules and hot BC molecules.
Za^bc = va-^bcCaCbc<7a->bc
(X.6)
(8.7)
Ii ii i,  ii, is the avenge velocity ol A toward M(. ('A and ('|)( lire the concentration! mil BC, "a -iir is the collision cross seciion, AC J1 is I he free energy ol activation, liiili/iiiaiin's constant, and I is I In- temperature, Combining equations (8.6) and IN ' i yields
t'a .iic = (vA .,«•< .,k.casi/»m)(c AM'/WI |CACV (8.8)
ii  iu is the average velocity of A moving toward BC. I i|UUlion (8.8) is a second-order rale law, with an activation barrier of All1 and a i.....enlial, k«, given by
ko = vA^Bc^a^rreAS*/liB
(8.9)
i haul/ nor Lewis had a way to calculate AS*. Consequently, they sel AS' to /cm pi iimn (8.9) to obtain
ko = VA^BCOi-
(8.10)
11.mi/ and Lewis also asserted that the molecular velocity in equation (8.10) should 1 ili ulaled, ignoring that we are considering hot molecules. According to the results in 1 It ipli i 6, I he average velocity of A moving toward BC is given by
:= /8kBT\'/2 Wabc/
i hire
I
l
M-abc     mA + mB + m
(8.1 I )
(8.12)
mil in \, mB, and mc are the masses of A, B, and C, respectively. A derivation of equation i I i is given in Section 8.16.4.
Equation (8.10) is the key result for simple collision theory. One should memorize the . i|iuiion before proceeding.
" .' i   Predictions of Collision Theory
it is useful to review the predictions of collision theory. We have already discussed iln key predictions in Section 7.2. According to collision theory
. Preexponentials for reactions are all about 1013 or 1014 units of angstroms, molecules, and seconds.
. The preexponentials scale as the molecular diameter. Larger molecules are more reactive than smaller ones.
. It does not matter how the molecules collide, they will always find the right configuration to react.
I lie first two predictions have excellent agreement with experiment. Preexponentials are UHUally within a factor of 10 of those predicted by collision theory. Larger molecules are
ITU Ml mil' llll I 'VI ' i II I I Hi ill i
generally more reactive, The third prediction does no) woik as well, Molecules often >Ui mil 111ict 1110 light configuration to react. All theite ideas are discussed in Section /..' Any ! reader who has not read the discussion of collision theory in Section / .' should 'i" I before continuing with (his chapter.
There are a few examples where the preexponentials differ suhstantially from iIiono predicted by collision theory. The reaction
CH3CH2CH3 + O
CH3CHCH3 + OH
(H.I.I)
has a preexponential of 1.4 x 1010 A3/(molecule-second). That is over two orders "I magnitude lower than one would expect from [collision theory from equation (8.I0)| In contrast, the reaction
202 -> 20 + Oz (8. lli
has a preexponential of 5.8 x 1015 A3/(moleculesecond). That is about two order. o| magnitude larger than one would expect from equation (8.10). These are special casi■•. In most cases, collision theory predicts reasonable preexponential factors. Still, there arc i 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 read,mi I as a billiard ball collision. Every molecule of A that collided with BC was assumed M react with a fixed probability. In reality, the reaction probability varies with how Mn collision occurs. For example, consider a simple reaction:
(CH3)2CH2 + 0:
(CH3)2CH. + .OH
(8.15)
During reaction (8.15), the oxygen reacts with the hydrogen on the middle carbon OJ pentane. If the incoming oxygen hits the hydrogen on the middle carbon, the reaction can occur. However, if the oxygen hits anywhere else, a .CH2CH2CH3 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 in reactions similar to (8.15).
This example shows that in some cases, one needs molecules to collide in the correi 1 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.
8.3  AN IMPROVED COLLISION THEORY
Figure 8.1   A typical collision between A and BC.
I usi, it is interesting to consider what a collision is like. Figure 8.1 shows two different
in .....s between A and BC. In the top case, A flew toward BC, A stuck, and (' flew
In this case, reaction happened. In the bottom case, A flew toward BC. A collided • nil IK ' but A left again. No reaction occurred.
Now imagine doing a thought experiment where one could watch thousands <>l
.....1.1 iilcs and see when collisions happen. If one could watch the collisions, one could
il, 11I.iic a reaction rate. In practice, one cannot watch the collisions of real molecules.
1......te can simulate the collisions in a computer.
hi ihe next three sections, we will derive equations for the key collision process hirst, ill derive an expression for the rate of reaction as a function of the probability tlnii two molecules react when the molecules collide at a given angle and velocity. We Will then show how to calculate the reaction probability by watching the dynamics ol fin particles. We will also need to discuss intermolecular forces. Some of the material 1 heavy going. We need to introduce several concepts over a short time. However, the i< nil 1 should work through the materials in order to understand what is said later in Ihe . hapter, and in the chapters that follow.
In the last section, we noted that the main weakness of the Trautz-Lewis version ol collision theory is that it ignores the fact that one needs a special geometry in order Im reaction to occur. In the remainder of this chapter, we will discuss some of the work that has been done to improve collision theory. The general approach will be to find a wa\ to determine which collisions actually lead to reaction, and then find a way to use that information to predict a rate. Qualitatively, we will determine which of the collisions lead to reaction by doing a calculation to see how the reactants change during the collision, and using the calculation to determine whether reaction occurs.
In this section we will describe the qualitative effects and the key equations. The quantitative details will be discussed in Sections 8.4 and 8.5.
B.3.1   The Reaction Probability
I ii i, it is useful to use statistical mechanics to derive an expressioa for the rate as ,1 I unction of the reaction probability. Recall that in statistical mechanics, one uses a i.iiistical ensemble to calculate an average value of a function. Well, a rate is a function, mi if one averages the rate over a statistical ensemble of all possible collisions, one can 1 .ileulate an average rate of reaction. In this section, we will derive an equation for all iln key quantities. The result looks formidable. However, in reality, the result is easier ill.in it initially appears.
I el's dílnu' ľ.....
uc molecule n 11 u-
varies with
„„, us the probability that u given A molecule lends with a fiivej two molecules collide. One can show lhal the reaction piohabihly
• va->bc, the velocity that the A molecule approaches the IK' molecule
• EBc, the internal state (i.e., vibrational-rotational energy) of the IK' molecule bcfofl collision occurs
• The impact parameter bA->Bc> which is a measure of how closely A collide! with BC
• The angle of approach, where § is a measure of the angle of the collision
• The initial position RBc and velocity vBc of B relative to C when collision occurs
Figure 8.2 page 420 shows how bA—bc and <t> are defined. The figure assumes thai
when atom A collides with the BC molecule, the A atom follows the trajectory show.....
Figure 8.2. The impact parameter is defined as the between the point that atom A would intersect the plane going through the center of the BC bond and the center of mass <>i BC if the atom A would fly in a straight line and not be attracted or repelled by BC. I In angle of approach is defined as the angle where the A impinges relative to the angle ol the B-C bond.
Next we want to briefly discuss why the reaction probability varies with vA—bc, EbcJ bA-cBC, <K Rbc, and vBc- The velocity term is obvious. The faster the reactants are moving, the quicker they go over the activation barrier. The EBC term is less obvious. Physically, though, if you put a lot of energy into the BC bond, the BC will stretch, and a bond thai is stretching is easy to break. Consequently, the reaction probability generally rises as f|„ increases. The impact parameter is a third term. Physically, the reactants need to hit eat fc other to react. Consequently, when bA—BC is large, the reaction probably is negligible While the reaction probability is significant when bA—Bc is small. The term is hanlei to understand. Consider a reaction such as
D + CH
C + HD
(8.1(0
Notice that the reaction probability will be higher if the incoming deuterium hits the hydrogen than when the deuterium hits the carbon. The 4> dependence accounts for thai.
If you have stationary BC molecules, the first four are all the variables you need However, if the reactants have internal motion, you need to consider that as well. Pol example, the CH could be rotating during reaction (8.16). The rotary motion could bring the hydrogen around to cause the hydrogen to collide with the deuterium. That will enhance the rate.
Generally, there are many variables that affect a reaction probability. One should memorize the list on the previous page before proceeding.
8.3.2  An Equation for the Rate as a Function of the Reaction Probability
In statistical mechanics, one calculates everything as an ensemble average. For example, let's assume that we have an expression for the reaction rate as a function of vA—bc, Ebc, <t>, bA-Bc R-bc, vBC:
\......„„, „, ,|„- .esul.s „, So lion Ml. Hu- oveinll i ate, i A .„, . is - ..I.......«1 l-V w»Wm
i      i, i, i<„    v,.,   In ('hai)lei <>. we louml that lite .......nun |X I I) ovei vA .,„■. I'.nc. <ľ. bA .m ■       • v»' 1,1
l in i.....i\piession is
r a >bc
E
i'a -1«
(8.18)
„ is the probability that a given stale n is occupied. i .ii ílu- work thai follows, it is convenient to replace the probability in equation (8 18) i    i distribution function, D(vA .no FBc,<t>, bA_.B> Rbc, Vbc)-
i hope thai you remember the velocity distribution function D(vA) from physical
i.....      ii \   The velocity distribution function is the probability that the velocity is
vA and vA +8(vA). In the same way, D(vA —► BC. EBc.<P. ba_«b. Rbc, vB( I is ;<■ is between vA      and vA^Bc +8(vA_Bc
l>. (w, in v ,\ anu vA ,i, probability that va_bc „„i | „,  | 8(Ebc). and so on.
You use D(vA-Bc.EBc.<l>,bA-B. Rbc.vbc) m
„„ple. recall from physical chemistry that if you wanted to calculate the v< lo< It) ,1 some function F(vA), you would calculate
). Fur is between In, the same way that you use D(vA) So
i HgC o
(8.1'))
t lnu
F = J F(vA)D(va)dvA
p is the average value of F. In the same way the average rate becomes
Ia-.bc(va^bc. Ebc. <P, ba^bc. R, vBC) x D(va^bc, Ebc, <K bA^Bc, Rbc, vBc)dvA^Bc dEBC d(|) dba-bc dRBC dvBr(8.2())
I .|ii.nion (8.20) looks pretty formidable, but it is merely saying that one needs III ueiage the rate over six variables. One should memorize this equation before i • i • n ceding.
u i :i
Derivation of an Expression for rA_Bc(vA-bc, Ebc,*, bA^Bc r-. vBc)
, wc want to derive an expression for rA-*Bc(VA^BC, Ebc, *, ba^bc, R. vw:)■ I" •„, Hon 7.2 we found that for a hard-sphere collision, the reaction rate is given by
rA—>bc
collision rate\ / probability of reaction \ \^   during a collision J
(8.21)
„ here the collision rate is the same as given by equation (8.7). Combining equations (8.7,
(8.21) shows
i'a—bc
= CaCbcVa->bcCaCbc<7a-»bcPh
(8.22)
fA—bc(va->bc. Ebc, 4>, bA->B, Rbc, Vbc)
(8.17)
| „„ ,non (8 22) applies only if you have a hard-sphere collision. If you do not have hara-sphe« collSon, then the reaction probability is going to vary with the distance u '"en the reactions prior to collision bA-BC, the angle d>. and all the rest of the vanahles
illHCUSsed in Section 8.3.2. ,..nl Equation (8.22) is not correct for this situation. However, one can denve a correct , J.Z by considering the differential slice of area shown in Figure 8.2, calculating a
.» »tvii i ill ml m . i n un i-mi ilimill 1 1 II 'i
•I.-1
■ ll III 11
ll i   ill ii ll'.l'llll III dl'llllc lilt.' I in.'.  .11 I........ii hs
Figure8.2 A typical trajectory for the collision of an A atom with a BC molecule as calculated by Mi" methods in Section 8.3.2.
"lit <Va .|U . I'-ik') = J ......1.....<Va -ik . I * hi . K\ .1« Kill . V,„ )
x D|(Rbci vhi )bA -iii ll»a .»( ll<l>llK|l( llvm IN '<•!
Mlti'M l)|(K|t( ,vm) is the distribution function lor Km- and V|1( . Physically, lite cross
. .......-. proportional to the rale constant for a set of molecules whose energy and
»I'd•• 11v arc lixed.
i ijuulions (K.25) and (8.26) are the statistical mechanics expression for the reaction I .111.1111>11 (8.25) looks very complicated, hut actually it is not. The equation simpl) I hill you need lo average (he reaction rate over (he velocity distribution Ol A thl imil i ncrg) of BC, the impact parameter, and the angle of approach. Readers should
....... themselves that Equation (8.25) computes a reaction rale as an average real i.....
ilnlily before proceeding.
differential rate, and then integrating the differential rate drA^Bc is given by
drA^Bc — CACBcVA->BcPn
, d(area)
(8.2.1)
where d(area) is the differential area, (bA^Bcdba^bc d<|>).
In reality, the A molecules have a distribution of velocities. Consequently, one has to
average over the distribution of velocities. One also has to average over the distribi.....%
of internal energies of the B-C molecule since the reaction probability changes if the B-C molecules starts out hot. Combining equations (8.21)—(8.23) and integrating over all Hie variables yields
ra—>bc
cAcB
I! (V/wbc, EBc, bA^BCi <t>, Rbc, vbc)va^bc
x D(va^bc, EBC, Rbc, vbc) dvA^ec dEBC(ba^bc dbA^Bc)d<t> dRBC dvBC (8.2 11
In equation (8.24) D(vA_BC, EBc, Rbc. vbc) is the probability that a given pan of molecules will have a velocity, vA^Bc; an internal energy, EAB; an internal coordinate, rbc; and an internal velocity, vBC: bA_^Bc is the impact parameter: Preaction(vA-»Bc, EBC, bA_„BCi Rbc, vBc) is the probability that the molecules will read when collision occurs, and (bA^BcdbA^Bc d<)>) is the differential area. Note that the integral in equation (8.24) goes over only positive values of the velocity, since a colli sion occurs if A moves toward BC, but no collision occurs when A moves awaj from BC.
Again, we have chosen a sign convention so that if vA_+Bc is a positive when atom A is moving toward BC, while if vA_>Bc is negative, atom A is moving away from BC. We consider only positive velocities in equation (8.24), because if the velocity is negative, the A atom never collides with the BC molecule.
One also needs an expression for k2 the rate constant for reaction (8.5). The result is
*-2 — ^reaction (va^bc. EBC, ba_>bc, 4\ Rbc, vbc)va_>bc
x D(va^bc, EBC, Rbc, ^bc)dvA_»Bc dEBc(bA_>Bc dbA_+Bc del?)dRsc, dvBC (8.25)
| i   MOLECULAR DYNAMICS AS A WAY TO COMPUTE REACTION PROBABILITIES
inn (8.25) is the key result from the last section. What it says is thai one CM inputc a rate by picking values of the initial velocity, energy, impact parameter, and nl approach, RBc, vBC, and then compute the reaction probability. One repeats i different angle or energy, and computes reaction probability again. The result is a ili   ol plots like those in Figure 8.1. One then plugs into equation (8.25) to compute ■ i iir  An example calculation is given in Example 8.C. This is an easy calculation lilt it computer. We ask the reader to do the calculation in the homework set. Problems H ' I 8..M).
I In- objective of this section is to describe how one computes the reaction probabilit) I in general approach is to compute thousands of trajectories as A approaches BC, One linn averages to calculate a rate.
Molecular dynamics (MD) is used to compute all of the trajectories. In Ml) 01* in A and BC molecule, gives the atoms their initial velocities and positions, and lines a computer to calculate how all of the atoms move in the force field created h\ all III the olher atoms in the system. For example, during the reaction
A + BC
AB + C
(8.27)
win start A moving toward BC, as shown in Figure 8.1, and then compute what happens as 1111 reaction proceeds. Generally, the calculations are done by solving Newton's equations • ■I motion for all the atoms in the system.
That is not as difficult as it sounds. Consider a particle, P, that is moving in a force Held. Think back to your days in freshman physics. In freshman physics, you probabl) '.nlved a problem where you computed the trajectory of a cannon ball. The approach was in integrate Newton's equations of motion. If a cannon ball experiences a net force. I i .
1 In i he literature people sometimes integrate over the angular momentum vector. This is equivalent to the derival.....
line
I
Ihm Ihr position and ,n i rlci .il n Ml iil Ihr i Million lull .ir nie niveil l'V
F,
(I Rp dl'
(K.JH)i
where rp is the position of the cannon ball, aV is the acceleration, m> is the cannon bull mass, and t is time. Note that the force, position, and acceleration in equation (8 'Hi are vectors.2 to a reasonable approximation atoms and molecules usually follow equation (8.28). Consequently, one can use equation (8.28) to determine whal happens when uloffl A collides with molecule B-C.
Equation (8.28) is a simple second-order differential equaiion. If one knows the initial conditions for the variables in equation (8.28) (i.e., the initial position and initial velocitj i one can use the numerical algorithms to the supplementary materials in Chaptei 6 10 numerically integrate equation (8.28) to calculate the position of particle P as a function of time. A computer program called ReactMD is available from Dr. Masel's Website and from http://www.wiley.com/chemicalkinetics. The reader should download that program before proceeding.
We call the position of a particle P as a function of time, the trajectory of the partil l> P. Trajectories are very important to the work in this chapter.
Now the next question is how to use equation (8.28) to calculate what happens durifll a reaction. Consider a molecule A reacting with a second molecule BC. Let's assuml that we know the following: vA_^Bc, the velocity with which the A molecule approach* the B-C molecule; EBc, the vibrational-rotational energy of the BC molecule beforl collision occurs; the initial velocity and position of BC; the "impact parameter," bA ,bi . and the angle of approach, 4>. Notice that if we know the forces in equation (8.28), wc can calculate the trajectory of the A molecule as it approaches BC. We can also integrate equation (8.28) to obtain the change in the position of B and C. A typical result is given in Figure 8.1. Generally, either the reaction will occur or it will not. If A reacts with BC, the A atom will stay attached to B and the C atom will fly off. In contrast, if no reaction occurs, A will fly away by itself. One can use MD to calculate whether a reaction occur, at any set of initial conditions. Hence, one can get all of the data one needs to numerically integrate equation (8.25).
According to equation (8.25), the reaction rate is equal to the reaction probability (0 or 1 for these collisions) averaged over all of the initial conditions. One can compute Unnecessary integral by changing the value of the initial positions, and velocities and do the calculation again. If one repeats that process for a wide range of values of all of the parameters, one can plug into equation (8.25) to calculate the reaction rate.
8.4.1   Molecular Dynamics Simulation of Reactive Collisions
The hard part in the computation is the calculation of Preaction(vA->Bc, EBc)- In this section we will describe how one calculates Paction(va^bc, Ebc)-
Consider the interaction between three atoms A, B, and C. If A, B, and C are confined to a line, then one can replace ra, rb, and Rc—the position vectors for A, B, and C — by scalar quantities, RA, RB, and Rc. The classical equations of motion for A, B, and C are
mA
-Ra
dt-
(8.29)
'■ We use R to designate a position to distinguish R from a reaction rate.
in,,
' M
,ir
d'lK, dl '
Ft
(8 I0)
(8.31)
hi ii i< \. K|i, and K, are the positions of atoms A. IJ and (': mA. nin. and in, are thl .......I A. It and ('; I is lime; and l-A, !•'„, and IV are the net forces on atoms A. B,
• Hill ( '
II one knows the forces, one can numerically integrate equations (X._'()| (8.31) to "I.in- I*..-...i„,„(vA .iic, Ebc) as a function of the initial velocity and internal energy Ol Ihr molecules.
1,4.9   How Do We Know the Forces?
i......lei to gel any further on this topic, we will need to introduce considerable new
..... i ml  First, we need to know all of the forces between the molecules. Then, tVl
ml to < alculate how each of the molecules move. Sections 8.4.2-8.5 discuss the fort < ■
...... 8.6 discusses how molecules move. Students generally find the materials in thin
■ lion pretty difficult. However, I advise that you stay with it. It will gel easier.
\ . noted above, in order to do any calculations, one needs to know the forces on ill ■ ■I Ihc particles. For the purposes of discussion, it is useful to define a quantity called the •••<• ntial energy surface for the system, VTotai(RA, Rb, Rc). which is the potential energy III Ihc system as a function of RA, Rb, and Rc, the position vectors of atoms A, li and i Recall from freshman physics that if one knows the potential energy of a system of i mil Ks, one can calculate the forces on each particle from a very complicated looking ininlion:
FA =	-vAvtota,
FB =	-vBvullal
iv =	-VcVtouü
(8.32)
where FA, FB, Fc are the forces on atoms A, B, and C and
9     9 3
V. =
9xA 9yA 9zA 9     9 9
9xB ' 9yB ' 9zB 9     9 9
9xc' 9yc' 9zc
(8.33)
where XA, YA, and ZA are the x, y, and z coordinates of A; XB, YB, and ZB are the x, y, and z coordinates of B; and Xc, Yc, and Zc are the x, y, and z of C. In actual practii e, one never has to worry about equations (8.32) and (8.33). The computer does all the work automatically.
However, the key thing to notice is that if one knows the potential energy surface, VM.,i, for a reaction, one can calculate the trajectories of all of the atoms, which in turn ran be used to calculate a reaction probability and hence a reaction rate.
Next, we need lo (Ir.i us', what Inn i ". luiwcen mm >li i nli". .iir like Al llns |iuinl. I ihh'tl
to discuss ;i little notation. When people lulk uboui "iniernioleculai lours   \Uc) u\n<\) plot the forces between molecules. Instead, they plol Ihc intermolei tilat potentials as you may recall from above, the potential is related lo the force by equation (8 I ' > I ii. force is the derivative of the potential, or the potential is the integral oi the fori t
In the literature, it is common to see plots of the potential energy of the system i | function of the intermolecular distance. For example, Figure 8.3 shows a potential cm n diagram for the interaction of two neon atoms with one another and the interaction belu 11II two fluorine atoms with one another. Notice that in both cases, the potential starts out ||
zero at long intermolecular distances, reaches a minimum at some intermediate disti.....
then rises again at short distances. In freshman physics you learned that the force is iclali il to the potential by
3V
FN. = --- (8 Ml
OlR-NeNe
At long distances, 3V/3RNeNe is positive. According to equation (8.34), the force l| negative. So the force pushes R-NeNe in the negative direction. R-NeNe shrinks, which num that the neons are pulled together. In contrast, at short distances, the 3V/3rNcNc is negativi according to equation (8.34), so the force is positive. The forces push RncNc in the positivt direction. The neons move apart.
Another way to look at this result is to say that the neon-neon potential is attractive at long distances and repulsive at short distances. Physically, what is happening is ili.il the neons attract at long distances because of dispersion (induced dipole-induced dipole i
forces. In contrast, they repel at short distances because of a repulsion between the elect.....
clouds and between the atomic cores.
There is a minimum in the potential of intermediate interatomic distances where all the forces balance. The minimum in the potential corresponds to the interatomic distantr in neon dimers in the gas phase.
The neon-neon potential looks similar to the potential for the interactions of any two nonpolar molecules. For example, the potential for interaction between two methanes m ethane and a methane looks similar to that in Figure 8.3. There are some slight differences
100
50
0
1 0
lu
-50
-100
-150
	ii         i         i i
	-V—► \Ne2
	
	\/f2
i         i         i i	
0
10
o
0 I
-5 2>
£= UJ
-10
15
12 3 4
Distance, angstroms Figure 8.3  A neon-neon potential and a fluorine-fluorine potential.
>ih pi>l.n in.ilr. nli". ie y . |\vo wiiU'iM Mm e the hilenliiliihi ol the ilipnle allci Is the new i.....        Still, ime linils that the poteiiliul looks qualitatively like the neon neon potential
in I ||UN 8.3.
I if nir 8.3 also shows a fluorine fluorine potential. Hie fluorine fluorine potential nl   |usi like the neon neon potential The fluorines attract ai long distances, so the
.....Ittl goes down, but then there is repulsion at short distance as the atomic cores In
iln II........e overlap. I he well is deeper with two fluorines than with two neons because
III  H........es can form a fluorine-fluorine bond. However, the fluorine fluorine potential
. i\ similar lo I ho neon-neon potential.
1,4.3   Intermolecular Forces
i......lei lo proceed, we will need to understand why intermolecular potentials look tin w a\
III    do Must ol you have already learned about intermolecular potentials in youi physical
i......-.ii v course. Here we will review some of the key ideas to help you understand
I It I.....olccular potentials.
I i.in key forces come into play when atoms or molecules interact with one anothei
. I lispersion forces
. Forces due lo electron exchange and bonding
- I'auIi repulsions
. Nuclear repulsions
(i '1.3.1   Dispersion Forces   Dispersion forces are weak, long-range physical interai
...... Dispersion forces occur whenever there are two molecules in moderate proximity,
..... iiler the two neon atoms shown in Figure 8.4. For the moment, we will treat the
.......classically as a nucleus surrounded by electrons. Figure 8.4 shows the instantaneous
i" i 11 n hi of the electrons in our idealized neon. If you start with a single isolated neon, then ill. re will be a distribution of electrons around each neon. The nucleus and the electrons li .i in an instantaneous dipole. The dipole is always fluctuating, and on average neon does nul have any net dipole. But there is still an instantaneous dipole.
Now consider bringing the two neons together. Each neon will have an instantaneous ilqinle. The dipole on one of the neons can attract the dipole on the other neon, pro\ ided Ihc two dipoles align. The dipole-dipole interaction lowers the energy of the system. Thai produces the lowering of the energy seen in Figure 8.3.
An important detail is that you get the lowering of the energy only when the dipoles are aliened. The electrons in the neons are each moving; in order to keep the dipoles aligned.
Key © Electron ® Nucleus
i Instantaneous í dipole
Figure 8.4  The interaction between two neon atoms.
when im election in one ol the neons moves, llie electrons on the otliet neons need to
move, loo   II llie rice tTOM do nol mow together, llie ihpolcs will go onl ol lllijl.........
People call tins simultaneous motion of the electrons on both neons correlated motion <"
correlation. Correlation allows the dipolcs to slay aligned, so it produces die I..we........I
energy shown in Figure 8.3.
8.4.3.2 Pauli Repulsions The neon-neon potential is not attractive at all distances, however. Instead, the potentials are attractive at long distances, but repulsive at short distances as shown in Figure 8.3. It is useful to consider why the repulsions arise in terms of the interactions between the arbitals on each neon. One can understand why the repulsions arise with the aid of Figure 8.5. Figure 8.5 is a plot of the changes in ih< highest occupied molecular orbitals (HOMOs) of the pair of neons. Positive orbitals aie lightly colored in the figure; negative orbitals are darkly colored.
The HOMOs on each neon start out as a p orbital on each neon. Recall that neon lu the structure ls22s22p6, so the p orbitals are the highest occupied orbitals of the sysiem The top part of Figure 8.5 shows what the p orbitals actually look like. In freshman chemistry you might have been told that the p orbitals look like a figure eight. Howevet when you actually calculate the shapes, you find that the p orbitals look more like two half-spheres with a space between.
Now consider moving the two neons together. The neons can come together with llie two positive parts together or apart. Quantum-mechanically, orbitals of the same sign have a bonding attraction while orbitals of the different sign have an antibonding repulsion. The picture in the middle of Figure 8.5 shows the antibonding orbitals, while the picture on the bottom shows the bonding orbitals.
If you had just the bonding and antibonding orbitals, there would be little net repulsion However, there is an additional effect — when you push the two neons together, die orbitals distort. I like to think of it as pushing two balloons together. The orbitals flatten at the point of intersection because the electrons in the orbitals repel as the result of .i quantum effect associated with something called "electron exchange" (see Chapter 1 I ). Ii
Separated
Ne-Ne collision
^^^^^
Anti-bonding
Bonding
Figure 8.5  The changes in the highest occupied molecular orbital of Ne2 as the neon atoms move together.
toil* energy to disloil the inlnlals As | lesull, when vou push two neon iiloms loo > I OK
......hi neon* repel Peopli call this Interaction a Paull repulilon becausi Pauli rlrW
ihowed how it would arise miilhcmiiliciilly
i hud. about pushing two balloons logethet, You gel a repulsion as the balloon flatteni
.....e thing happens when vou push two orbitals togethet The orbitals distort, so the)
.....I
iiu net result is a long-range attraction and a short range repulsion; therefore die
........imn potential looks like die potential in Figure 8.3. llie potential slarls out at /en.
I.....he neon atoms are far away, readies a minimum, and uses again. Hie general
h ,.„ m Figure 8.3 is typical lor the interaction of any molecule with any other; in othei urdN "lie always sees an attraction potential at long distances anil a repulsive potential
ni hoit distances.
Next, it is useful to ask how the interactions would change if we replace llie two
......., wiih two fluorines. Well, neon and fluorine are somewhat similar. Fluorine is next
111 in mi in the periodic table, and both neon and fluorine arc polarizable. However, there 1» , iundamental difference between the interaction of two neons and two fluorines. A ni on ,i, ,ni has one more electron than a fluorine. The two fluorines can form a fluorine fluorine ipn.i bond, and there is no antibonding interaction.
Figure 8.6 shows an orbital diagram of the other orbitals in the fluorines. Fot this CBSi With fluorine, the antibonding orbital is empty, so there is no repulsion.
i )l course, one can put only two electrons in a bond. If you try to put three electrons iiiio a bond, you will get a Pauli repulsion. Consequently, there are Pauli repulsions m Ii .ii lions such as
F + e" +F-> F2 (8 15)
F + F~
-> F2
(8. «.i
where e   is an electron.
Now if one pushes the fluorines together, there is another effect, because the nonbonding orbitals in the fluorine begin to overlap. Again, that produces a Paulirepulsion, no die energy rises, and there is no exchange (bonding) interaction to lower the barrier, Detailed calculations show that the fluorine-fluorine potential looks very similar to the neon neon potential in Figure 8.3. The only key difference is that the fluorine-fluorine potential is sharper and deeper than the neon-neon potential (see Figure 8.3).
Separated fluorines
CS3
Figure 8.6  The changes in the highest occupied molecular orbital when two fluorines come together.
All iiioiii atom, ilium molecule, mid molecule molecule poienliiils look (|tinllliilivcly like those in Figure 8 1 rhe potential is attractive al long range, and repulsive ui shorl range, and there is a minimum in between
8.4.4  Reactive Potentials
The discussion in Section 8.4.3 was for the interaction of two atoms, A and B. Next, we want to consider a generalized reaction:
A + BC
-> AB + C
(8.37)
Recall that reaction (8.37) is called an "exchange" reaction.
When an exchange reaction occurs, there are some extra forces to consider. Foi example, consider the reaction
Cl + F2
-> C1F + F
(8.38)
Here, the chlorine-fluorine and fluorine-fluorine potentials are both completely attractive. As a result, one might imagine that when a chlorine approaches an F2, the chlorine would be attracted to both of the fluorines to form a stable C1F2 complex. In fact, however, und« most conditions, the chlorine atom reacts with the F2 to form C1F and a fluorine atom.
8.5   POTENTIAL ENERGY SURFACES FOR REACTIONS
In this section, we will describe why you form C1F, and not a stable C1F2 molecule, and we will describe the active potential. Let's consider reaction (8.38) in more detail.
Figure 8.7 shows an orbital diagram for the interaction of a chlorine atom with a F2. The chlorine starts out with a p orbital, while the F2 starts out with an intact sigma bond. The sigma bond is formed from two p orbitals, so there are nonbonding lobes on either side of the two fluorines corresponding to the nonbonding portions of the p orbitals.
Separated reactants
Nonbonding lobe
During reaction
Nonbonding Fluorine-Fluorine lobe bond
Fluorine-Fluorine bond
Figure 8.7  The changes in the 3Aig orbital when a chlorine approaches a F2.
ľnil NIIAI I Nl Mi IV '.I MM Al I •• I i Ml III Al III IN'i
4iM)
When llic chlorine approaches llie lluoime, llieie is a I'auli lepulsiun belween I In* Mil.n il in llie chlorine and the nonbonding lobe in the lUionne Hie nonbonding lube i.. .1 up against the f F bond, so the F F bond disioiis. The F F bond pushes up
.....die nonbonding orbital m the othei fluorine. Thai pushes the second fluorine awtj
.......ii\ the forces become so strong thai die fluorine fluorine bond breaks,
Most exchange reactions (i.e., reactions of the form A i uc  • AH i O show similai in   • (no orbital pushes into another, KvenUially bonds break leading to produi ti
>   i   Description of Potential Energy Surfaces for Reactions
■ m represent (his information with a potential energy contour. Note thai il is a
.....Illdimcnsional potential energy contour since the potential varies wilh the positions ol
hli<i me- and both of the fluorines.
i we will describe what the potential energy diagram is like. Foi llie purpusi ol n   n m. we will consider the reaction
C1+I8F- 19F
+ C1I8F+19I
18 19)
I i siinie that the reaction occurs when the chlorine approaches the fluorine along the 1   ink- s bond in Figure 8.8.
I here are three atoms in reaction (8.39). If one puts the origin on one of the atoms, iii< ii die oilier two atoms can each move in three directions. Therefore, in principle, one mild need to consider a 2 x 3 = six-dimensional potential energy surface to fully model Mi. reaction. However, in fact, the key dimensions are the ones that involve breaking
i, n  llie distance from the incoming chlorine atom and the 18F; and Rpp, the length ol
ii. i   F bond. For future reference, it is important to note that rcif and RFr. are related
i. R.p8 and Rp9, the positions of each of the atoms, by
Rcif = |Rci -< I
Rpp = |Rp8 — Rp9|
(8.40)
i igure 8.9 shows several views of an appropriate potential energy surface for reaction i K V')). The top left figure is a three-dimensional (3D) plot of the potential energy surface i In plot is a plot of the energy of the system as a function of Rff and Rcif- The leli nie ol' the figure corresponds to the potential at large Rqf. At this side of the plot, llie . 11 loiine atom is far away from the F2, so the potential looks just like the F-F potential in I Igure 8.3. There is a well and a barrier. Similarly, the right side of the figure corresponds in llie large F-F distance. This corresponds to a stable C1F with a potential that is similar in the F-F potential. The back corner of the figure corresponds to all the atoms being
CI
	
RCI-HF	
Figure 8.8  The geometry considered in the potential energy surface shown in Figure 8.9.
•I 10
Ml Ai. IH iN'i A!i i ,i H I I'.I' nr.
I noii.iy
Figure 8.9  A potential energy surface for reaction (8.39).
close together. Notice that this region looks like a little hill in a miniature golf course. There is a saddle point on the top of the hill. In the literature, people call the top of the hill the col or the transition state.
Unfortunately, when you look in the literature, you find that you do not see potential plots like the one in the top left part of Figure 8.9. Instead, people plot contour plots like
I'l 'II IIIIAI I Nl HI IY
I At.I 11 I i HI III AI tic dr.
1 11
If 1« i ' in the In >il < ci 11 11)'hi nl PlguN K'l I lir 11111111111 pint represents the same ml on nation hi Ihc II) plot. To gel tioni llu- II > plot tu tin- contour plot, you first need to rotate the
11 ' hj 180 . and look si i .11111 down 11»' plot as shown In the top righl < > 1 Figure 8.9 1 in 11 you add contours as shown in the hcciicini leii Then you eliminate the shading i" llu I'li'i on the bottom right of Figure 8.9
iin plot on the bottom right provides the same information as the plot on the uppet in 1 in plot is just harder to see. The lines in the figure arc contours of constant energ) 1 It  1 ..in iiii.il is /cm in Ihc upper right corner. The potential goes down, moving alone the
.....I. ol Ihc figure, reaching a minimum at point A. The potential then goes up again
mm me lo the lower right corner of the figure. Initially, the potential is almost constant,
.....muč left from the lower right corner of the figure. Once one gels halfway across thi
...... the potential starts to rise again. The potential eventually reaches a maximum at
iin I.mci left corner of the figure. Similarly, the potential starts out at zero at the uppei . oiner of the figure. The potential goes down moving left from there, reaching 1
.............11 at point B. Then the potential rises again. Physically, the top left portion ol
llu llgure, Rci-f is large, while rff is small. Therefore, the top left portion of the figure
"   ponds to the initial state of the system where the chlorine atom is just beginning to ii'i'iii.ieh the fluorine molecule. When chlorine is far away from the fluorine, there will i" little interaction between the chlorine and the fluorine. As a result, the potential energ) ......c will look just like that in a fluorine molecule. The potential will be attractive 11
I i" I F distances and repulsive at short F-F distances. The potential will not depend in , 1 1 . In a similar way, in the bottom right portion of the figure, Rcif is small, while " 1 1 is large. Therefore, the bottom right portion of the figure corresponds to the final
1 iti ol (he reaction where the chlorine has reacted with the F2 to form a stable C1F
.....I.' ule and an isolated l9F. In that case, there will be little interaction between the ( II
.....let ule and the l9F. As a result, the potential energy surface will look just like that in
ill molecule. The potential will be attractive at long Cl-F distances and repulsive at In 111 Cl-F distances. The potential will not depend on rf_f.
Now, the lower left portion of the contour plot in Figure 8.9 is complex. The lowet I. li portion of the curve corresponds to the case where the chlorine and the fluorine are ■ lose together. Obviously, if the chlorine and the F2 get too close, the potential will be
ii 1H11 si vc. However, at intermediate distances, there will be a metastable Cl-F bond and
.....etaslable F-F bond. As a result, the potential energy of the system will be lower than
in the case when the chlorine, l8F, and l9F are all far apart. Still, Figure 8.7 shows that 1I1. 11' is a Pauli repulsion between the chlorine and the fluorine. The Pauli repulsion raises iin energy of the system. The Pauli repulsion goes away when the l9F moves away. The in 1 effect is that the potential goes up slightly when the chlorine is close enough to the I . lo react. The Pauli repulsion raises the energy, so it produces a barrier to the reaction.
Most potential energy surfaces for exchange reactions (i.e., reactions of the form A l BC -> AB +C) look just like the potential energy surface in Figure 8.9. There is usually a Pauli repulsion, so there is a small barrier to the reaction. The one exception is . 1.1 iron transfer reactions. For example, the reaction
Na + F2
NaF + F
(8.41)
.1. nially occurs via two steps. First an electron is transferred from the sodium to the F2:
Na + F2->Na+ + Fr (8.42)
432        Ml Al III INN AN I I II I INK INN
III Al III IN AN Ml HU IN I IN I'l III NIIAI I Nl III IV '.I Mil Ai I '. Ill
Then llio ions react:
Na1  I F
Naľ' I F
(Kill
These reactions arc very different from reaction (8.39) hecause the electron lianslct eliminates most of the Pauli repulsions. A detailed description ol these reactions will be given in Section 8.14. However, for now we will consider only reactions like reai lion (8.39), where there is a significant barrier to reaction.
For future reference, we will call the saddle point in the potential in Figure 8.9 I hi transition state for the reaction. Later in this chapter, we will note that an uiuleisiaiuliin of the properties of the transition states is critical to the understanding of rates of chémii || reactions.
Students often have reported some difficulty seeing the saddle point in the potcntiul energy surface, and so I thought that a numerical example would be helpful. Table 8.A.4 (in Section 8.17, page 470) shows some numerical values of a potential energy surface calculated in Example 8.A. In the table, we have defined function V(ri, r2) for the reaction A + BC —► AB + C. The potential is a function of r,, the B-C distance, and of r2, the A-B distance; the potential is also a function of a, r0. We then used a spreadsheet to calculale numerical values at V as a function of r, and r2. Table 8.A.2 shows the formulas in 11 it-spreadsheet. Column A gives the values of r2, row B gives values of ri, and the region from B4 to Q19 shows numerical values of the potential.
Now let's focus on the numerical values in Table 8.A.4. If we start out near the reactants, the potential is low. We started the table at a point where the energy is 3 kcal/mol. When the reactants come together, r2 is reduced. Notice that the energy goal up with decreasing r2.
Next, let us focus on the minimum-energy pathway from reactants to products in Table 8.1 (in Section 8.11). The shaded cells in Table 8.1 show the minimum-energy pathway going from the reactants to the products. Notice that as we move along the minimum-energy pathway from reactants to products, the energy goes up to 8.9 kcal/mol, and then back down to 3 kcal/mol.
The place where the potential is 8.9 kcal/mol is a saddle point in the potential energy surface. It is a local maximum if you approach along the shaded pathway in Table 8.A.4, but it is a local minimum if you go horizontally or vertically in the table. Thus the transition state is a saddle point in the potential energy surface.
We expand these ideas in Examples 7.G and 8.A. The reader should examine those examples before proceeding with this chapter.
8.6   REACTION AS MOTION ON POTENTIAL ENERGY SURFACES
i.......H reuctlon (8,44), radical A approaches molecule IK'. The A can icact in extract
i'l.....i BC or bounce without reacting i here are two key coordinated In the system
■ ,, iIn distance from A to B, and Rbc> the distance from B to C,
I 11-.I. it is useful to note that there are three ways lo visuali/e the reaction: \   ,i trajectory in space where all the atoms move
\ .i trajectory on the potential energy surface in Figure 8.9 where the bond lengths evolve
\ . .i plol ol the motion ol the atoms versus time
ii i   hard to explain what the motion of atoms is like without doing i llmu \  program called RcactMD is available from Dr.  Masel's Website .nnl liti|i //www.wiley.com/chemicalkinetics, which simulates several cases with pictures and 111» >iis of the trajectories, i Igure 8. ID shows what the real trajectory is like. In this case, we plotted the positions •i iln .iinins as a function of time. Generally, what happens is that an A atom comes up BC. The A atom stops because of the Pauli repulsion. If the A atom happens in
Time
Time
Now that we have the preliminaries out of the way, we can talk about what reactions between molecules are like. The general scheme will be to use molecular dynamics to integrate equation (8.28) to calculate the trajectory of all of the atoms as a reaction proceeds.
In this section we will describe what the trajectories are like and how they lead to reaction. We will only consider reactions with activation barriers in this section. We will consider barrierless reactions (e.g.. Na I F2 -> NaF + F) in Section 8.14. For discussion purposes, we will consider a general reaction
A + BC
AB + C
(8.44)
Time
I igure 8.10 A series of trajectories during the reaction A + BC ■nd various initial reactant configurations.
AB + C, with MA = Mc = 1, MB = 19,
4'M III Al  III IN'. AM,I l| I IIIMINM
1.11IAI II AIM I I Al I Mil ft I H till I I >l I I'll! INM I XI HANI II III Al III IN'I 430
conic up jusl when (' is moving nway. reaction happens II mil, A hounces oil  In Ihf lop in the Figure 8.10 case, A is far From n<' al the end ol pnx ess so no rem lion iu happened. In contrast, in the middle case, A and it are close togcthci and (' has llowf away. In this case reaction has occurred.
Do not worry about understanding the details of Figure 8.10 now. We will expli.....i
in further detail later in this chapter.
Marcellin (1912,1914a,b) showed that it is useful to plol the position lime data on it potential energy surface. Recall that when the A approaches the BC, Rah and Km' will change with time. One can plot that change as a trajectory (i.e., line) on (hi potential energy surface as shown in Figure 8.11. The solid dark line in Figure 8.11 starts ai 11■• reactants (i.e., the upper left) and ends at the products (i.e., the lower right). WV will call that line a reactive trajectory or a reactive collision, since if the system follows Mini trajectory, the system reaches the products so reaction occurs. Similarly, the heavy dashi d line starts out at the reactants (i.e., the upper left) and loops back to the upper left (ir . back to the reactants). We will call such a trajectory a nonreactive collision, since il iln system follows that line, the system never gets to products so no reaction occurs.
One can calculate a reaction rate from Figure 8.11 by starting with a distribution ol molecules, computing a series of trajectories like those in Figure 8.11, and adding up Ma-rate for the cases when the molecules react. Note that in the case shown in Figure 8,98 the potential goes up when the hydrogen comes close enough to me F2 to react because of the Pauli repulsion between the hydrogen and the F2. Therefore, it will take a certain amount of energy to get the hydrogen to react with the fluorine. This is a general result There is always a Pauli repulsion during any exchange reaction. Consequently, one can conclude that:
It usually costs energy to bring the reactants close enough that they can react. As a result, there is a barrier to most exchange reactions.
In I 'duplets III and I I. we will show Mini intililion.il bmtiers arise hcvuu.se one needs in n Ii ot break bonds during chemical reactions,
QUALITATIVE FEATURES OF THE COLLISIONS: EXCHANGE REACTIONS
in il.. in-si several sections, we will discuss what reactions arc like. We will considei three is pes of reaction: unimolecular reactions, bimolecular recombination reactions, Hi I < < hange reactions. The discussion in this section will consider exchange reactions
■ III.....'action types will be discussed starting at Section 8.10.
I u hange reactions are the most common elementary reactions. An exchange reaction I   ni\ reaction of the form
A + BC-> AB+C (8.45)
n ii radical or atom A, collides with a molecule BC to transfer a ligand B and produi I radical C. Examples of exchange reactions include
D. + CH3CH3 F. + CH3CI HO. + CH3CH2CH3
-> DH + .CH2CH
-» FCH3 + CI.
-> HOCH3 + .CH2CH3
is Ii.1
(8.47)
(8.48)
In all cases, one is breaking one bond and forming another. The formation ami ill ilruclion of bonds is one of the key features of ligand exchange reactions.
In the next several sections, we will describe what ligand exchange reactions are like
.....1 molecular level. For most of the discussion, we will consider a special example:
hi 1 (change reaction H + FH HF + H. We will assume that the reaction follows the mill nlial energy surface in Figure 8.12. The potential energy surface looks just like idc
Reactant
Rbc
Figure 8.11   A series of typical trajectories for motion over a potential energy contour.
Reactants
♦
u
0.5 Ä
Transition state 13.86 kcal/mol
24 kcal/mol
15 kcal/mol
£1
9 kcal/mol
Figure 8.12  An idealized potential energy surface for the reaction A + BC -* AB + C.
Ml Ai III iN'i A'i 1.1 II I I'.H IN',
IJIIAI IIAIIVI I I All III!
II  111| l I ll I I' ,11 IN'. I XI I l ■• M, II III AI 11< IN! i
4:t;
ones wc discussed curlier in the chuplei will) broad wells and a barrioi in lielween, have chosen parameters so thai the reaction lias a barriei ol I 1.86 kcal/mol, ami
equilibrium bond lengths are 1.04 A. A and (' each have a mass ol I amir It litis......
of 19 amu. For discussion purposes, I have sei ihe A I! and It (' bond energies to I 30 kcal/mol, not the 108 kcal/mol seen in HF. I have also artificially tripled Ihe uniili of the vibrational wells so that it will be easier to observe the vibrational motion,
Our general approach will be to start atoms moving, and calculate Ihe position, ,„, a function of time, to see whether reaction happens. A detail in the calculations is I one needs to specify the initial conditions to actually do the calculations. General)! specifics the initial rotational and vibrational states of the BC molecule, the initial vein of A toward BC, and the initial position of B and C as they vibrate in and out. In work that follows, we usually fix the total energy and vary the other variables. In tin calculations, the total energies is the sum of the rotational and vibrational energies ol in and the translational energy of A toward BC.
Figure 8.13 shows what typical trajectories are like. In this case, we pui a lolul 18 kcal/mol into the incident molecules, and calculated the position of the atoms as n function of time. In both cases we put the same amount of energy in the molecule buj choose different values for the initial B-C position, so we got different results. In the I up case (a), the atom A comes in and hits the BC molecule and reacts. Notice that A sltirl out far away from BC. However, as time evolves, A comes until it is in close proximllj to BC. Then there is some complex behavior. Then after some time, C moves away, whlll A and B stay together. This is a reactive collision because we end up with A bound |q B, and C far away.
The next case, (b), is a nonreactive trajectory. In this case, A + BC collides as befon However, A moves away at the end of the collision, so there is no reaction.
		(a)
c ____		
B		
		
Time
Time
Figure 8.13 Two typical trajectories for the reaction A + BC => AB + C, with MA = Mc = 1, Mc = 19, Etotai = 18 kcal/mol (Etrans = 14 kcal/mol, Eto, = 0 kcal/mol, Evib = 4 kcal/mol). Reaction occurs in the top trajectory but not in the bottom trajectory.
Mm. iMD I'.eneiales thousands ol liii|ci loiirs like litis   Ihe icadei should inn sonic
H i|i ■.....es before proceeding.
I |||ure 8 II shows a sei ics ol li a|ccloi ics calculated by pulling a total I 1.8 kcal/mol
i, .....i,i■ H-. .nici c .iii iii.linn' the traja tories ol the molecules, Nonce that in all cases.
......,„ . ins. To keep tins in perspective, we choose a case where Ihe incident eneij-s
, | I.H kcal/mol, while the barrier was 13.86 kcal/mol. In all cases, no reaction happens
..............derstand tins finding with the aid of Figure 8.15. Figure 8.15 is a replol ol
i 11 i in Figure 8.14 on a potential energy surface. The trajectories all start out moving ,i,l Ihe transition state. However, there is not enough energy to cross the barriei ,|in nllv, no reaction happens.
tin  i a general result. Classically, if the molecule does not have enough energj to Mi, hainci, no reaction occurs. Quantum-mcchanically, there is a small reaction , Hi due to a process call tunneling. See Sections 7.3.1, 9.3.1, and 9.5.
\ i higher energies, one does have some probability of reaction. Figure 8.1(> shows I ol trajectories calculated by setting the energy of the incident molecules, and then
I i,.....ig all of the other parameters to optimize the probability of reaction. At an energy
III nil kcal/mol, the A atom comes in and makes it almost all the way to the transition However, the A does not have enough energy to make it over the hill, so the A
o a.
	(b)
c	
B	
A /	
Time (x1)
Time (x1)
o 0.
(d)
*>SrVVV\/rVVVV\
Time (x1)
Time (x2)
Hume 8.14 A sampling of the trajectories taken by putting a total of 13.8 kcal/mol into the reactants,
.......,mg random initial positions for the atoms, and putting a random amount of rotational energy into the
in  I Kind. We chose the initial velocity of A toward BC so that the total energy was 13.8 kcal/mol, and
.......we integrated the equations of motion to see whether reaction occurs. In this example, the barrier is
i i nil kcal/mol, so no reaction occurs.
t.'lll III Al  IK ill'. A'. i.....Mi il ľ.
IJIIAI IIAtlVI I I Atlllll hl U llll      11 i i i1 'i i    i ■ ■ 11 ■ i i1 , i ii   . ii.ill'. 430
Figure 8.15  A replot of the data from Figure 8.14 on a potential energy surface.
Figure 8.16 A series of trajectories calculated by fixing the total energy of the reactants and then optimizing all of the other parameters. The barrier is 13.88 kcal/mol for this example.
.....\ lie. hhu k line) leads hum umlauts In piodiicls. One also uhscives icaclioii
Hl ľ M.I).
Niiliiľ dial when I lit.- energy is below die saddle point energy. 1.1.86 keal/uiol. no
.......i occurs. However, as soon as the energy goes above 13.86 kcal/mol, 11 it- reaction
Inn  i , msecjucntly, there is some probability of reaction whenevei the incident mole< tiled . iioiigh energy to traverse the barrier. Míľ. is mil obvious from figure 8.16, but, in fact, when the incoming molecules liavi I..m l\ enough energy to cross the barrier, the reaction probability is relatively small • hie • nm- excess energy to gel reaction to happen at a reasonable rale. I * I»s ii ally, the trajectory has to go through the pass of the lop of the potential cncfg) •m Im c figure 8.17 shows a picture of ihc pass. A molecule with energy E can go through III  ílu pari ol the pass where the energy is less than E. Figure 8.17 shows two diffcri ill \i ,i low energy, the molecule can go through only the bottom of the pass, so die low. However, as you raise the energy, the molecule can go through a wide portion I lln pass, Consequently, the rate is higher. I Igiirc H.I8 shows some actual data. The figure shows how the cross section loi the ■n II f Hi —► H2 + H varies with the energy of the incoming H atom, where, as 1 il previously, the cross section is a quantity proportional to the rate constant 101 Million Notice that the reaction probability is near zero until the molecules have enough in cross the barrier. Then the cross section increases rapidly. The cross section again at high energy because the reactants fly past each other before the reai lion
.....11111 Empirically, rates generally reach a maximum at energies 2-3 times the barnei
Did Ihcn decrease again.
Interestingly, there is a small probability of a reaction occurring when the energ) Il I., low the top of the barrier. This occurs because of a quantum effect called tunneling. >* Im Ii will be discussed in Chapter 9.
Ilieic are lots of other details that affect rales of reactions. When you do an Ml) il illation, you specify the initial positions and velocities of all of the atoms, and then numerically integrate equation (8.28) (Newton's equations of motion) to calculate how Mi.  lystem changes. It happens that when the energy is what is called "near threshold" just above the barrier), small changes in the initial positions and velocities oi the
.......s can affect whether reaction actually happens.
< ienerally, reactions are rare events until you are several kcal/mol above the ii>ľ ol 1i1. barrier as shown in Figure 8.18. One key parameter is the relative motion of all ol
Opening for higher energy
Opening for low energy
stops before it makes it over the barrier. In the particular case shown, the trajectory seems to double back on itself, so you see only one line in the figure. Actually, there are two lines, one on top of the other. The plots for E = 13.7 and 13.8 kcal/mol are similar to those for E = 13.6. There is no reaction. However, at E = 13.9, reaction occurs, so the
Figure 8.17  A blowup of the top of the barrier.
UMAMIAIM II AIIIMI Ulil Hli (IIIMľlloNtl I XI IMANIU lllAi IMIľ. 441
20     30     40     50 60 Collision energy, kcal/mol
70 80
Figure 8.18 The cross section for reaction H Levine and Bernstein (1987)].
+ H2 -> H2 + H. [Adapted from Tsukiyama et at. (1988) an|!
o
CL
O
o
Q.
__c__		
B		- ReDUlsion
		
Time		
C		"(c)
B -.		
-"-"a""		
Time		
C		
' "—'—		
B		
		
Time
Time
Time
Time
Figure 8.19 A series of cases calculated by fixing free energy at 18 kcal/mol, fixing the vibrational energy at 6 kcal/mol, and varying whether A hits when C is vibrating in toward B or out away from B.
the atoms at the moment of collision. Figure 8.19 shows a series of cases where we let B-C start vibrating, and then delayed the collision, so that A would hit B-C at different times along the B-C vibrational cycle. Sometimes A would hit when the B-C bond wtl stretched; at other times A would hit when the B-C bond is short. Notice that only aboul half of the trajectories lead to reaction even though the reactants have 18 kcal/mol, which is more than enough energy to cross the barrier. If you look carefully at the trajectories
ill m a ii c I hut the ci it u ul vín i,il ilť is the motion ol C when A CfOIMI the dashed line i \Mu.Ur-, up and i'ľ i. i tin reai lion proi ecds II A rem hcs the line In N IN when (' happens to be moving away, (' continues to move away from Ii ('.
...........i urs n <' is vibrating m inward u, no reaction occurs, ( use (hi is haul to
lnu ii 111.111 v (' starts to vibrate away just when A hits the dashed line. This example iii   thai you need coordinated motion of all of the atoms in order foi reaction to m n II <  is moving away from U anyway, C continues to move away. However, il ■ n  m toward B, the incoming A molecule does not have enough momentum to i\  consequently, no reaction occurs, i    pi. often ihink about this effect in terms of what happens on a potential encrgj hni  t igure 8.20 shows a replol of the data in Figure 8.19 on a potential energy surface Ihul ilie molecules need to go around a bend in the potential energy surface In
......u linn to happen. Well, the bend is like a bend in a racetrack, and the.....lei Lllc
in \m hroni/.e its turn in order to get around the track. If the molecule turns at the
....... Ilie molecule does not make it around the track even though the mold ulc
ion ili.m enough energy to cross the barrier. Think about what happens at a hum
......mi sc. If the driver turns too fast, the car does not make it around the t m\<
lluotigh it has plenty of power to keep going. In the same way, the molecules nerd ordinate their motion as the system reaches the transition state. In particular, the m ľil in be moving to the right as they approach the transition slate in figure 8 .'(I m in the right is equivalent to C moving away from B when A approaches BC, \ related effect is that the molecule needs to be moving in the right direction as n ii Ins the barrier. The molecules need to be moving up the hill as they approach the i m older for reaction to happen. Vibrations parallel to the hill arc less effective in ........ molecules over the barrier.
rBC —*■ rBC —" rBC —
Figure 8.20  A replot of the results in Figure 8.19 on a potential energy surface.
Ill Al ■ I II iN'iAM.olllMIl INN
(JIIAI IIAIIVI I I All llll m iII llll COIIINIONN I xi .hani II III ai.ih inn AA'.i
Trans E ■ 14.5 kcal/mol Vib E   0.01 kcal/mol
li.ni-. i     i.' k, .il ,.....
VlbE   2.5 kcul/iniil
Trans E = 7.25 kcal/mol Vib E = 7.25 kcal/mol
Trans E = 4.8 kcal/mol Vib E = 9.7 kcal/mol
Trans E 2.4 kcal/mol Vib E - 12.1 kcal/mol
Figure 8.21 A series of cases where the molecules have enough energy to get over the barrier Nnllnf that some of the molecules do not make it, because the partitioning of the energy between transition „.-i rotation is incorrect.
Figure 8.21 shows some examples where we varied the initial vibralion.il m translational energies on the molecules and tried to see if the molecules made il u the barrier. In these cases, there is more than enough energy to carry the molecule the barrier, but sometimes, the molecules are moving in the wrong direction as lit approach the barrier. There is no momentum to carry the reactants over the barrici reaction happens.
One might think that the way to optimize the reaction is to put all of the enery >
translation to give the molecules as much momentum as possible to go up the b......
However, the results in Figure 8.21 show that does not really work. The reason is hcciniit the trajectory needs to make a left turn on the potential energy before the trajecton p i to the transition state. Consequently, it is necessary to balance the forward motion ol lit atoms and the turning motion to optimize the reaction probability. Generally, some nl il energy needs to be in translation, but other energy needs to be in vibration to optiiiii/M the turning of the molecule. Again, there is a close analogy to a racetrack. If you i watch a race, you will notice that the racecars go wide before they go around a i in Well, the same thing happens with molecules. The trajectory needs to go wide on lit potential energy surface to make it around the curve. The only way that molecules, n wide is if they accumulate some vibrational energy before they collide. Consequents lltf
optimal trajectories always have some vibrational energy. If the molecules have too.....
vibrational energy, though, the molecules do not have enough momentum to gel ovei
barriers, so one has to carefully balance the vibrational and rotational energy to gel n m.....
to happen. There also is a timing issue. A race driver who turns too early will nevei in il it around the track. In the same way, when molecules vibrate too early, they nevei nml it over the barrier. Perfect timing occurs when C is vibrating out when A comes In
The trajectories in Figure 8.21 were plotted on a standard potential energy suil i There is some subtlety in looking at the motion on a potential energy surface in ilm
, i.....      . i.iii sumeliines seem In linn uround Willi no i cum ill. Tins is an uililiicl
«>,iii l.iinl wilh something subtle in llie in.illnni.Hus  Kccilll dial the atoms aie lenlly ., mi ,i nine 11iih111n.11e potential energy surface, where the nine coordinates hi llie ^  >   mil / coordinates ol atom A; the X. Y. anil /. coordinates ol atom Ii; ami die X. i / coordinates ol atom ('. When you collapse the nine dimensional space onto a
.......i i.ni.il plot, you can gel artifacts. In particular, atoms can seem to turn mound
(in no ii'iihoii.
Il happens Ilial if one uses some special coordinates, called iim.w wrixhlril ci>i>nlimit?\, .......Is vanish. The mass-weighted coordinates are defined by
Ri = Rah +
(iii^
mAmc_
iuH)(ni|, I in, )
Kill'
1111 {(111A   I   iii!,   I   iii, I
(niA I in,, Mm,, + trie)
i '
(1.49)
(H.W)
I i......       H     is a replol of the data in Figure 8.21 on mass-weighted coordinates Notice
ii,,. in, |iiiiiiiiial energy surface is slightly skewed. However, the trajectories now look HiiihiiiI In particular, the trajectories thai do not make il slop because Ihey are moving I || ,n, I in die lull, rather than up the hill.
i ii is Interesting to consider how the shape of the potential energy surface allecis Hi nl reaction. The effect is substantial. If you have a potential energy thai is banked
......hi, k, then the incoming molecule can ride around the curve. In sucli ■ I Ut
.......I energy is easily transferred from ihc incoming A molecule to the (' molecule
Him • mi. it you have an L-shaped potential energy, such as that in Figure 8.23, then you iii-iil vlliiutional energy to get around the curve. L-shaped potential energy surfaces aie ......ihi when the reactants A, B, and C are atoms. They are less common when
IninsE 14.5 kcal/mol Vib E ■ 0.01 kcal/mol
Trans E " 12. kcal/mol Vib E = 2.5 kcal/mol
Trans E 9 7 kcnl/ntnl Vib i    4 11 km......I
ans E = 7.25 kcal/mol Vib E - 7.25 kcal/mol
Trans E = 4.8 kcal/mol Vib E = 9.7 kcal/mol
Trans E 2 4 kciil/rnol Vib E - 12.1 kcal/mol
Figure 8.22 A replot of the data in Figure 8.21 in mass-weighted coordinates.
444       Ml ACIIONM A!l(
)l I Kill IN!I
UIIAI IIAIIVI I I Allllll (11 H lili COI I KlIONtl I XI.HANOI III AC I ION! I 44!)
-AB
Himi I.ml'.
0.5 A
24 kcal/mol
15 kcal/mol
'• BC
Figure 8.23  An L-shaped potential energy.
A and C are molecular ligands. Still, L-shaped potential energy surfaces do present somi particular difficulties during reaction.
8.7.1   Polanyi Rules
Polanyi rules pose some special issues. Polanyi showed that with an L-shaped potential energy surface, the position of the transition state also has a large effect on the encrg) transfer barrier to reaction. Polanyi noted that one can characterize reactions accordin| to whether they have an early, middle, or late transition state. Figure 8.24 shows 0 example of the potential energy surface with early, middle, and late transition stales. An early transition state is one that arises when the reactants have hardly changed. A lit transition state is a transition state where the species look similar to the products, .m,i middle transition states are somewhere in between. The reaction
CH3OH + O-y CH30 + OH (Si'
has an early transition state. The reaction
35C1 + CH336Cl-► 35ClCH3 + 36C1
(8.52)
AB
R-BC RBC RBC
Figure 8.24  Potential energy surfaces with early, middle, and late transition states.
i i.II, ii in Mi,,ii si,ilc The lenclion
I) I i II < M
DII I (II,
(8 33)
Ii,   , I.in transition state.
\ o-. 11 nun with an early transition state will reach the transition state before the
...... iignificantly deformed. The H C bond will not be stretched significant!)
,i. die A ii bond will be much larger than in a stable AH molecule Set 11 In contrast, a reaction with a late transition state will reach the saddle point III ,, 1 he end of the reaction. The B-C bond will be almost broken, while the A It bond ,11 l„  about the same length as in a stable AB molecule. A reaction with a middle
......lion slate is somewhere in between the two, where the bonds in both the reactants
1 products are stretched by similar amounts in the transition state.
In dus section we will discuss some of the special effects which occur with I -.li.ip, ,1
I ......ul energy surfaces. The work will build on findings of Polanyi.
Polanyi showed that Preaciion(vA^Bc, EBc) changes according to whether the reaction ,n early, middle, or late transition state. If the reaction has an early transition slanil.....iction probability is higher when the translational energy is large than when ih,
II 1  \ ibrational energy is large. In contrast, if the reaction has a late transition stale, the
.....hi probability is higher when the B-C vibrational energy is large than when die
II 111 lutional energy is large. Physically, when there is an early barrier to the reaction,
..... 1 ol the barrier is associated with getting A and BC in close proximity. In dial case,
llli litster A moves toward BC, the more momentum A will carry into the collision. As
........ nium allows it to get in close proximity to B-C. That increases the rate of reaction
in , ontrast, if the reaction has a late transition state, most of the barrier is associated with lili 'I nig the B-C bond. In that case, it is better to have energy in the B-C bond before Mi, 11 .ii lion, rather than having to transfer energy during the collision.
Polanyi showed that reactions generally retain energy during collision; thus, il you have an early transition state so that the A needs to be moving fast during the reaction, ih, 11 die collisional energy generally stays with A. In that case, the A-B molecules come ,,11 hot (i.e., with a high vibrational energy) while C comes off cold (i.e., moving slowly). In , 11111 rast, if you have a late transition state, so that A starts out cold while B-C is
iied, the A-B will end up cold while B and C will end up with a considerable amount ,,1 ii.uislational energy.
Polanyi showed that one can measure the vibrational and translational temperatures. In
......paring the distribution of vibrational energy levels and velocities to those predicted
h\ llolt/.mann's equation. For example, Figure 8.25 shows how the vibrational energy is p 11111 lonal in the DF produced during the reaction:
F + D2
DF + D
(8.54)
\,voiding to Boltzmann's distribution, the P„, the probability that a given state, h, is 111, upied should follow
p„ = e'En/feT (8.55)
where En is the energy of the nth state.
ligure 8.25 shows how P„ actually varies. Notice that the ground state (n = 0) has ,1 low probability of being occupied. The n = 3 vibrational level has a larger population 1I1.in does the ground state. This is quite different from the scenario expected from the llnltzmann distribution. The average energy of the HF is 21.3 kcal/mol. Therefore, it
Iin       III AI III (NM A'. < i il I I'.Ii ill
llll Nl INI INI All i AMI 44/
-1 ' I
Koll/m.inii distribution
Experiment
n 4
20
Energy, kcal/mol
30
n=5
40
Figure 8.25 The distribution of vibrational energy produced during the reaction F + H2 -> HF + H. [FteiuK of Polanyi and Woodall (1972)].
seems that the HF is much hotter than the hydrogen atoms, which is as expected foi ,m early transition state.
All of these trends can be explored with the ReactMD program. Any reader who has access to this program is advised to run the program and vary the parameters brim, proceeding with this chapter.
8.8  SUMMARY OF LINEAR COLLISIONS
At this point, it is useful to summarize what we have learned about linear A + UC collisions.
. First, we found that to a reasonable approximation one can treat the collision of two molecules as a collision between two classical particles following Newton's equations of motion.
. The reactants must have enough total energy to get over the transition state (or col) in the potential energy surface in order for reaction to happen.
. It is not good enough for the molecules to just have enough energy. Rather, tin-energy needs to be correctly distributed between vibration and transition.
. Coordinated motions of the atoms are needed. In particular, it helps to have C moving away from B when A collides with BC.
. We also find that we need to localize energy and momentum into the B-C bond for reaction to happen.
. The detailed shape of the potential energy surface has a large influence on the rate.
These effects mean that the system has very complex behavior. Note, however, that the latter four effects cause perhaps a factor of only 10 or 100 in rate. There are always some trajectories that make it over the barrier, even though the molecules have barely enough
io i niss ihr Ii.una  II I'f nl lite lfii|i'( lours make it. those n,i|o lours will have iin iinpoiiiinl elicit on Ihr i.ilr
UK NONLINEAR CASE
.  11 inn in the Sections 8.7 and 8.7.1 considered the dynamics ol the reaction III       Alt I (' where A, B, and C are confined to a line. With a linear collision, ill
i llu  \ uioms thai hit the BC molecule will react with a constant probability. However,
11 \   \. I!, and C arc not confined to a line. Instead, the reaction occurs ovet ■> dimensional space. With a three-dimensional system, the reaction probability varll Hit liov, i losely the molecules collide. Generally, reaction probabilities arc highet when iiIon collide head on than when the molecules miss. The object of the ncxl few iin ,ue lo discuss these effects. Ill i.nl il is useful to note that one can quantify this effect by seeing how the reaction i■ .1 .iin\ varies with the impact parameter bA-BC- Recall from Section 8.3.1 that the
.....i,, i parameter is a measure of how close the reactants come to one another. (lenerall)
ii lion probability is high if there is a head-on collision between the reactants \ mi collision corresponds to an impact parameter of zero. In contrast, the reaction liltih .i.iluv is near zero if the reactants miss. A large-impact parameter corresponds to the
. ......, missing. Generally, the reaction probability varies with the impact paramelei
qucntly, the impact parameter has a large effect on the reaction. I i nie X.26 shows how the reaction probability varies with the impact parameter at
i.......»diäte impact parameters. Generally, the reaction probability is highest with an
i parameter of 0. The reaction probability is relatively insensitive to the impact
.....,i, Iii when the impact parameter is small. However, there is a sharp cutoff at a
impact parameter. As a result, the reaction rate is negligible unless atoms hit with
ii small impact parameter.
i im- ran understand this behavior qualitatively with the aid of Figure 8.27. When the
.....i. , ules collide with an impact parameter of zero, all of the momentum of atom A is
.....led toward BC. In contrast, when the impact parameter is nonzero, one can divide
0 1 2
Impact parameter, A Figure 8.26 The variation in Pa^bc with changing impact parameter.
4411      ill ai in nr. a'.....I i.n nr.
IMI N( INI INI All I AM 44»
Figure 8.27  The typical trajectory for the collision of an A atom with a BC molecule.
A's momentum into two components as indicated in Figure 8.27. One component, vh. carries A toward BC, while the second component, Vj_, carries A away from BC \ increases as the impact parameter increases. As a result, the probability that atom A will move away from BC increases as the impact parameter increases.
In order to fully model this process for the reaction A + BC -»■ AB + C, one has to integrate the equations of motion [i.e., equations (8.29) to (8.32)] numerically. In generil that cannot be done analytically. However, one can get the structure of the solution In solving the equations of motion for a simpler reaction, A + B —>■ AB. A detailed derivation, of the key equations will be given in Section 8.16.1 at the end of this chapter. The km result is that when A and B collide with a nonzero impact parameter, the distance from A to B, the rAb. will act as though A and B are moving in an "effective potential" given In
V,
(bA^Bc)2EkE
+ VAB(RAB)
(8.56)
where V(rab) is the A-B potential, EkE is the kinetic energy of A moving toward U, bA-,Bc is the impact parameter, and rab is the distance from A to B. Equation (8.56) arises because whenever v± is nonzero, there is a component of angular momentum that carries the reactants apart. One needs extra kinetic energy to overcome that component of angular momentum. The first term on the right of equation (8.56) is the extra kinetic energy you need to overcome the component of the angular momentum that carries the reactants apart. Therefore, we will call this barrier the angular momentum barrier to reaction. The extra angular momentum barrier grows as the impact parameter increases. Consequently, the extra angular momentum barrier keeps the molecules from reacting when the impact parameter is large.
Figure 8.28 shows a plot of the effective potential as a function of Rab to see how that can work out. We choose a case where V(rab) is what is called a Lennard-Joncs potential (i.e., a potential like that in Figure 8.30). When b = 0, there is no barrier to prevent A from approaching B. Note, however, that as bA^ec increases, a barrier arises. This barrier prevents the reaction from occurring at large-impact parameters.
If one considers a more complex reaction, one finds that the arguments are more subtle. For example, with the reaction A + BC -* AB + C, one can transfer angular momentum from A to C, which slightly reduces the angular momentum barrier to reaction. Still,
n i'll  a plot of the effective potential as a function of rab for a modified Lennard-Jones Potential
*llli l>« .„,.    0,1.2,3,4,5,6.
.....I.ii momentum barriers arise in all gas-phase reactions:
I'lli .uigular momentum barrier prevents reactions from occurring when molecules ipproach with large-impact parameters. As a result, no reaction occurs unless the h hi i.mis get close to each other.
ii t» I   Changes in the Potential Energy Due to Nonlinear Collisions
I In i, is another effect that also changes the potential with nonlinear collisions. Recall thai 'ii  potential energy surface in Figure 8.9 was calculated for the case where the chlorine
Il.....     approached the fluorine directly along the fluorine-fluorine bond. In that case, the
pnli mial energy surface is simple. However, if the chlorine approaches along a different III |lc, ihe potential energy surface changes.
First note that the incoming chlorine needs to form a bond with the p-orbital. When the I hlinine approaches the F2 at some finite angle away from the fluorine-fluorine bond, Ihe .mi lap with the p orbital is reduced. That reduces the net attraction between the fluorine mil the chlorine. Simultaneously, the overlap with the nonbonding orbitals in the Fj .....cases. There is a Pauli repulsion between the nonbonding orbitals and the incoming
■ liloiine. The result is that the barriers increase with a nonlinear collision. There are
■ ipiations in the literature to quantify the increase, but when this book was being written, iin equations did not seem to have good correlation with experiment. Still, the qualitative ii tilure that the barriers to reaction increase with nonlinear collisions will be important to I hi analysis because the increase in barrier decreases the rate.
11.9.2  Influence on the Overall Rate
Now, it is useful to go back and see how the angular momentum barrier and the variation in ilu barriers to reaction will change the analysis in Section 8.2 (i.e., collision theory). Nole
■i Mr
in m i ii irgn am mil1 ill H ľ i
Mi illl Ci IMI'I I X III ACIK iN'. 401
thai the analysis m Section K.J assumed thai the reaction rate was equal to the ool rate times the probability thai molecules thai collided ui luall) rem leil l he probability assumed to be constant for all molecules thai hil each other. However, the resultN In previous section show that the probability is not constant. Thus, some modification pjj | results in Section 8.2 are needed.
One can account for the variations in the reactive probabilily by interiminjj ci|.....
(8.24). It is useful to define a reaction cross section, o^BC, by integrating ovei Ihc imps parameter, da-*bc, and <bABC, rABc, vabc:
0-a^Bc(Va^BC, EBc) = J J Preaction(vA^BC, EBC, bA_*BC. <t>, RbC, V|1(')
x D, (RBC, vBc)bA^BC, dbA^BC d<t> dRBC dvBC        (8.17) j
where D, (rbc, vBc) is the distribution function for the internal states of the BC moli cull bA->Bc is the impact parameter, 4>ABc is the angle of approach, and rab and vB(- arc ill internal positions and velocities of the BC molecule as defined in Section 8.3.1. For a spherically symmetric case:
<^bc(va^bc, EBC) = 27rJpA^BC(ba^bc, va^bc, EBC)bA_
bc
dbA_
(K.Mtl
where PA_>Bc is the reaction probability as a function of b averaged over rbc, vBC, an. I i The reaction probability is already in equation (8.58). Therefore, equation (8.25) i-the rate constant becomes
I^a—>bc =
va^bc^bc(va^bc,Ebc)D2(va^bc, EBc)dvA^BCdEBc       (8 1«)
where D2 is the distribution function for vA_*Bc and EBrj-
Equations (8.58) and (8.59) can be used to calculate the rate constant for read.....
if Preaction (bA->Bc, vA^Bc, EBC) is known. Hence, they are quite useful. An exampli calculation is given in Example 8.D.
in.|....i mi inlliieiue on the icaclioii rule. < ici.ei.illy. one lltuls thai when the syslem has
li.ui It .....u x h energy lo cross the aclivalion barrier, reaction rarely occurs, because Ihc
m i \ seldom gets localized m the n ( bond I he rea< tion rate Increases w Ith
iHiiin.....fi energy. Note, however, thai il A comes in too last, the A will leave before
Hu......ue Ibi reaction to occur. In that ease, the reaction raie will go down again
iln i Interesting effect is that when you have bigger molecules, the collisions last \   n result, the reaction probability increases. It is hard lo understand why thai n  u ithout detailed analysis. However, when you watch movies of collisions you 1111 il you have a reaction of little molecules:
H + F2
-> HF + F
(8(.(l)
il. lion might last only 10"12 seconds. However, if two moderatc-si/ed molecules 11, such as
C5H120 + HC5H|2-> C5Hl2OH + C5Hl2 (8.61)
ii i.mis might stick together for 10"'° seconds. There are more chances for reaction In M ihc reactants stick together. As a result, the preexponentials for reactions tend in go 11H reactants get bigger, in   u ally, collisions last longer with big molecules than with small molecules because i"\ partitioning in the molecules. Recall that, in general, molecules have to be i'1 pretty fast (i.e., molecules need significant translational energy) in order lo have ii energy to cross the barrier. When a collision occurs between two small molecules. |h  molecules continue to move quickly, and as a result, the collision lasts for only an
.....ni In contrast, with a big molecule, the translational energy can be converted into
,i.....onal energy. It is like bouncing a bowling ball on a mattress. The mattress lakes
n|. . ncrgy, so the ball does not fly away. In the same way, a small molecule transfers lo the bigger molecule. The molecules slow down, so the molecules stay in close |»ni muiy longer. The molecules can also get entangled with one another. As a result, the ii i.mis stay in closer proximity longer with the big molecules than with small molecules
8.10   MORE COMPLEX REACTIONS
All of the simulations we have presented were for cases where A, B, and C are singll atoms. However, one can generalize these effects to reactions where the individual spei ii A, B, and C are molecular ligands rather than atoms. Generally, everything we have said so far in this chapter also applies to the case where A, B, and C are molecular ligandl A reaction involves concerted destruction and formation of bonds. Both bond formation and bond destruction processes must occur simultaneously. One can view this process sj an energy transfer barrier to reaction. During the reaction A + BC -»■ AB + C, energj must accumulate in the B-C bond before any reaction occurs. No reaction occurs unless C leaves before A leaves, and A might stay around for only 10"14 seconds. As a result, il helps to have the atom B moving away from the rest of the molecule C when the attackinj group A moves in toward B.
Another issue is that in order for reaction to occur, the B-C bond that needs to break (i.e., the B-C bond) must somehow acquire enough energy to break before the attackinj' group A leaves. Therefore, the rate at which energy accumulates in the B-C bond has an
1,10.1   State-Selected Chemistry
I h. ni her thing you see with a more complex molecule is state-selected chemistry. The objective of state-selected chemistry is to use a laser to excite certain vibrational or ml itional modes in a molecule in a way that a desired reaction pathway is enhanced. For
unple, Crim [1996] examined the photodissociation of monodeuterated water (HOD).
II miii simply photolyze HOD, with a photon, v, you get two reactions:
HOD + v HOD + v
-> H + OD
-> HO + D
(8.62)
(8.63)
Kcaction (8.62) is very slightly favored over reaction (8.63) because of a quantum process i idled "tunneling." However, Crim (1996) found that if they excite the OD bond before photolysis, reaction (8.63) is favored. So far, the enhancements have been a factor of shout 200. Unfortunately, you get that only with very expensive lasers, and it works only with small molecules. However, these effects are important in special reactions.
IV
MI Al III IN'i AM I Ii I I'lH IN'
8.11   COMPLEXITIES: UNIMOLECULAR REACTIONS
So far, wc have been discussing exchange reactions, where a ligand is exchanged liuffl one species to the next. Those are the simplest reactions. However, there are olhei iiaelliMI types. In the next several sections, we will discuss these other reaction types.
In this section we will discuss unimolccular reactions. Unimolecular reactions air ilu< second most common type of reaction. Unimolccular reactions are reactions ol the hum
BC
products
where a stable molecule BC is somehow converted into products. Examples of unimoli ular reactions include
o2
CH3CH2CH3
CH2
/ \ = H2C—CH2
O + O CH3 + C2H5
CH2=CH—CH3
(S (iM (SMI
(S.(»7)
In Chapter 4, we noted that unimolecular reactions are never elementary reactions. < lltl can use molecular dynamics to see why that is so. If one starts with a stable mole, uli the molecule is by definition stable. It never dissociates.
Consider trying to dissociate F2. Figure 8.29 shows a plot of potential energy of I as a function of the F-F distance. The plot is a duplicate of the plot in Figure 8.3. Now consider trying to dissociate an F2 molecule with an initial state given by the line Ifl the figure. If you start with the F2 with a state on the line, the fluorines vibrate bai I and forth. However, the fluorines are trapped in the attractive well. In the state shown the fluorine-fluorine bond has a total energy of 40 kcal/mol (i.e., the fluorines arc held together by 40 kcal/mol. One needs to accumulate the 40 kcal/mol in order for the bond to break. If the F2 is isolated from its surroundings, energy is considered, so there is n.> way to accumulate the 40 kcal/mol. As a result, the fluorine cannot fall apart.
Simple unimolecular reactions are never elementary reactions.
0 12 3 4
Distance, angstroms Figure 8.29  A plot of the fluorine-fluorine potential from Figure 8.3.
,,1, IV,' ,1 i I .1 II /V I 111.......I I'
■r. 1
OIIIM", one does sometimes uhsei\e nveiull iciicliiius like
1 >
(S.dSl
1       ri. that is never an elementary reaction. Ill ii 1.1 reactions like (8.65), (8.66), and (8.67) occur via bimolecular collision
x + o2
O + O + X
(8.69)
\ is a collision partner, as discussed below. Tin ill ully, a hot X molecule collides with the 02, and transfers energy to the < >. Ihr
1.......ulates in the O-O bond. Eventually, after several collisions, the 11 < > bond
......I'll energy to break. Notice that one needs to accumulate energy in the () () bond
rem lion can happen, which is why a collision partner is needed. One ran ohm rvi I 11 lions of the form
AB + v
A + B
(S /0|
v is a photon, since the photon provides the energy (and momentum) needed to
......ale the molecules.
toother observation is that the rate constant varies according to the nature ol the
.....m partner. For example, Table 8.1 shows how the rate of the reaction ().. | X •
\ varies with the collision partner. The rate is zero in the absence of a collision
1 ....... Further, one finds that the rate constant varies by a factor of almost SO according
In Ihcr argon or oxygen is used as the collision partner.
Ihr result of this discussion is that unimolecular reactions are very complicated, ami
often needs detailed calculations to make useful predictions.
' 'in of the key things you observe with a unimolecular reaction is that the reaction 111 two stages. First, the collision partner collides with AB to form a hot AH
.....Ii 1 ulc. Then, the hot molecule decomposes to form product. The hot molecule often
|m 1 lot 10 seconds. By comparison, ligand exchange reactions last only 10 13 seconds ■ on icquently, the preexponentials for unimolecular reactions tend to be much highet than III |Mi exponentials for ligand exchange reactions.
\i this point, we will defer our discussion of unimolecular reactions to the next chaptei ii" dung to remember for now is that unimolecular reactions are never elementary III Ii nd, one needs a collision partner for reaction to occur.
Table 8.1 Arrhenius parameters for the reaction O2 + X —► 20 + X
Collision Partner	Preexponential, Ä"V(mol-second) at 4000 K	Activation Knergy. kcal/mol
None	0	7
Ar	7.5 x 10'4	118.8
o2	5.8 x I0l6	118.8
Source: Data from Wcslley (1980).
■ • i mi m iin 'mn
in;1  iiimoi i cm am mi comiiinaiion mi a< iion-.
There is one other common lype ol reaction called a bimolet iilai rcnmihiinilitm rem nun Bimolecular recombination reactions are any reaclion ol the form
B +C
IK'
(8 /I}
where two species bind together to form a stable molecule. Examples ol himolei aim reactions include
F + F-> F2
CH3 + CH2CH3 -> CH3CH2CH3
Na + F-> NaF
IK /.'1 (8 Ml (8.74)
One would think that reactions (8.72) and (8.74) would be quite rapid. After all, accordln| to Figure 8.3, there is no barrier to bringing two fluorines together. In the same way. tin 11 is no barrier to bringing a sodium up to fluorines. Actually, however, the rate of reac lion (8.72)-(8.74) and all other bimolecular recombination reactions is zero in the absence ol a collision partner.
8.13   ENERGY TRANSFER BARRIER TO REACTIONS
Next, it is useful to discuss why reactions (8.72)-(8.74) do not occur in the absence ol | collision partner. The difficulty is associated with an energy transfer barrier to reaction According to data in the CRC, reaction (8.74) is 181 kcal/mol exothermic. If a sodium would collide with a fluorine, one might think that one could form a sodium fhion.l, molecule. However, it would be a very hot sodium fluoride molecule since the heal ol reaction is 181 kcal/mol exothermic. In fact, heat balance shows that the sodium fluoridl would heat up to 34,000 K! Well, at 34,000 K, the sodium fluoride has a high probabihi\ of dissociating. The result is that the sodium and fluorine never stay together very long One can form stable sodium fluoride by cooling off the sodium fluoride before the sodium and the fluorine fly apart. However, one needs to transfer the heat in some way. Ihe only way that heat can be transferred is by convection (i.e., collisions) or radiation (i.e . photons). Consequently, if the reaction does not produce a photon, then one would need a collision partner for reaction to happen.
Reaction (8.74) is a special case, because the reaction is so exothermic. However, next, we want to show that any reaction of the form A + B —> AB, where two reactantl combine to form a single product, cannot occur in the absence of a collision partner 01 photon.
Consider the generalized reaction A + B —y AB. At the start of the reaction, A approaches B with some energy, EAB. For the purpose of discussions, assume that the A-B potential, V(rab), looks like the potential in Figure 8.3. As A approaches B, A is first accelerated since V(Kab) is attractive at long distances. However, A begins to slow down again as A and B become very close. One can calculate A's velocity from an energy balance:
Total \ _ / kinetic \ / potential \ energy J     \ energy J    \ energy J
(8.751
•*-All       ,MaiiVai, I V(ra„)
1 8 /'» i
VAl, is the velocity of A toward IK', MA|, is Ihe effective mass of Ihe AH pan. ami 1 Is the attractive potential between A and B. .....'lid to also calculate the net force on A from equation (8.29). Note thai in oidei
I 11 'in reaction A + B -> AB to occur, A must fall into B's attractive well. A's velocity
1     in.ill (i.e.. like a vibration). Further, the net force on A must be small, or else ,11 be pushed away from B. It works out, however, that when one actually calculates 101 11 y from equation (8.76) and the net force from equation (8.29), one finds thai in never gel to the situation where both the force and the velocity arc small ai the ..... time,
< onsider (he case where an A atom collides with a B atom. Figure 8.30 shows a plol III \    velocity and the net force on A during a collision. A starts out moving slowlj
II mi. it is accelerated by B's attractive well. The net result is that A's velocit) ll
1.....     lurge when A is sitting over the minimum in the A-B potential. As a result, A flies
1 , 1 ill, potential well. A will eventually stop when the A-B bond is highly compressed However, the net force on A is tremendous there, so A is pushed away. One can never
Bl lo the position where the net force on A is small, and where A is moving slowly ligh that A stays bound to B. As a result, A cannot react with B. I In analysis above was for a Morse potential. However, detailed simulations wilh the purl.mi KeactMD show that the reaction A + B -» AB never occurs. A always scatters ' 1 1   bounces away) from B independent of the form of A-B potential.
Physically, what is happening is that A is approaching B with some energy, Eab. A Iihn lo lose that energy in order to fall into B's attractive well. However, there is no place
	
1	1/ j^Z-^ Actual potential /                 Potential (a)
	
	Velocity (b)
	Force* 0.1 ^
	N——T"   i     i     i     i     i i
123456789 10
I Igure 8.30  (a) A typical potential for the reaction A + B -> AB. (b) The velocity of A toward B. (c) The net
t, nee on A.
im.......m
mi «.....i r ■
Iiii llic cneigv In )'u As ,i ii Mill. A sini|ilv llii". mil ,i|\iiu
One can generalize this analysis to conclude thai elementary reactions "i the form
A + B-► C
are impossible in the absence of collision partners or photons.
Reactions of the form A + B —*■ AB + v, where v is a photon, are possible bei in the photon can carry away energy and momentum. In practice, though, such reactions itn rare.
Still, experimentally, one often observes reactions of the form A + B => AB. Nuii however, that such reactions are never elementary reactions. Instead, one finds Mm the reaction actually follows a mechanism A + B + X —> AB + X, where X is again a collision partner. X can be another molecule in the system, or the walls (i.i surfaces) of the reaction vessel. The function of the collision partner is to removi energy and momentum from the reactive complex so that A can fall into B's at 11 in tive well. Experimentally, the rate of a reaction of the form A + B=>AB goes In zero at low pressures because there are no collision partners. However, one i m observe reactions of the form A + B => AB at moderate pressures. At moderate prei sures, collisions with other molecules in the system occur. Those collisions can) away energy and momentum from the AB complex. As a result, the reaction has | finite rate.
In the problem set, we ask the reader to show that the reaction looks elementary at high! pressures. However, the reaction is not really elementary. In particular, the rate constant! for the reaction vary with the environment since the collision partner plays an importunl role in the reaction.
For example, Table 8.2 shows the rate constant for the reaction NO + O + X N02 + X. Notice that the rate constant is about a factor of 60 lower when the reaction is run in argon than when the reaction is run in water vapor. Clearly, this is a sign i I ii ant effect. Therefore, it is incorrect to view reactions of the form A + B => AB as elemental") Rather, it is better to view the reactions as needing a collision partner.
The same arguments do not apply to ligand exchange reactions. Ligand exchangl reactions such as reactions (8.45)-(8.48) all have two species coming into the reaction.
Table 8.2	Arrhenius parameters for the	
reaction NO + O + X -» N02 + X		
	Preexponential,	
Collision	Activation	Energy,
Partner	Ä /(mol2-second)	kcal/mol
None	0	?
Ar	1.7 x 1014	1.88
o2	1.7 x 1015	1.88
co2	3.84 x 1015	1.88
H20	1.1 x 1016	1.88
Source: Data from Weslley (1980).
mid two species leaving it Ihe cncig\ In bieuk bunds can come limn the leiiclanls, lni. Mu pioilucls can cany oil ,in\ cneigy denied in Ihe piocess. As 11 result, the huge
imii'\ ii.inslci bauieis to reaction seen with recombination reaclions are not observed nl. llgHIld exchange icaclion. Instead, one gels collisions, and those collisions lead lo
IfHi Hull.
I lir. leads lo Ihe key point in this section:
II ii iiHiilary chemical reactions must have two or more reactants and two 01
i......      products. Reactions of Ihe form A + B —► AB, AB —> A t B. A   > C are not
■ I.....nl.iiy; collision partners or photons are needed before such reaclions can occiu
tin n iidci should memorize Ibis box before proceeding. | i i   ILECTRON TRANSFER REACTIONS
Hum   i   another class of reactions where angular momentum barriers play a key role:
...... dial do not have a barrier. We will discuss them in this section.
' . milkier the reaction
Na + F2 => NaF -I- F (8.77)
i .......(8.77) looks very similar to reaction (8.38). The only difference is that we have
placed the chlorine by sodium. At first glance one might think that the potential energ) Ini reaction (8.77) would look very similar to that for reaction (8.38). In fact, i i there is very little similarity. Reaction (8.77) actually occurs in two steps First, odium atom approaches the F2 and exchanges an electron.
Na + F2
Na+ + (F2r
(8.78)
I'll '' the sodium ion reacts with the F2 ion to yield products:
Na+ + (F2)~ -► NaF + F
(8.7<>)
i   i .....entally, the electron transfer occurs when the reactants are 50 A apart. Then the
n proceeds quite quickly. We will describe the features of the electron trnnslci in Chapter 11. However, a qualitative picture is given in Figure 8.31. i i inc 8.31 shows the energy of the neutral (Na + F2) and ionized state of a NaF.
......plex as a function of the distance between the sodium and the F2. When the molecules
.1. I a apart, the neutral species are stable. However, as the sodium approaches the F2, 111 I urged state has a lower energy. Quantum-mechanically, the electron always goes in llic lowest energy state and electrons move at speeds close to the speed of light
■ mi i quently, the electron jumps from the sodium to the F2 as soon as the charged stale '11   i lower energy than the neutral state. There are some details, based on something
■ Hi .1 an "avoided" crossing, which will be discussed in Chapter 10. However, the kc\ >i.1111■ in remember is that if the charged state has a lower energy than the neutral species, in. . Iiarged species will form. Experimentally, the charged species form when the reactants 11 A apart (i.e., before there are significant Pauli repulsions). i 'me the charged species form, the reaction happens very rapidly. Recall that Na' is
■ urn ion. The Na+ can move all the way up to the F~ without experiencing a significant
nwrri f Ml INI Al MAIIMIAI
IN..' I ,.|
[Na t F?]
Figure 8.31   The energy of the neutral and charged states of NaF2 as a function of rnaf the sodium II.........
distance.
Pauli repulsion. Further, the Na+ is drawn in because the positive charge on the soil......
is attracted to the negative charge on the . As a result, the potential energy surface lot the reaction does not show a barrier. Overall, reaction (8.79) is extremely rapid.
Still it is not instantaneous. Recall from Section 8.9 that every reaction has an angul u momentum barrier as shown in Figure 8.28. The barrier grows as bA—bc incren i This barrier prevents reaction (8.79) from occurring when bA—bc is large. Consequentl) reaction (8.79) has a finite rate even though there is no barrier for reaction.
In a more general sense, angular momentum barriers to reaction are very importlfll for any reaction with a negligible barrier. After all, the rate should be infinite withoul ihl angular momentum barrier to reaction. We will quantify these results in Chapter 9. Id-thing to remember for now is that reaction rates are finite for barriers reactions becau I of the angular momentum barrier to reaction.
8.15 SUMMARY
The three primary causes of reaction barriers are
• Hills in the potential energy surface . Energy transfer barriers
• Angular momentum barriers to reaction
The hills arise because the electron clouds need to be compressed in order for the read.mm to get close enough to react. If you do not have enough energy, no reaction can occui
The energy transfer barriers arise because one needs to put energy into a bond to jtel a bond to break. If a molecule has enough energy to react, the molecule still might no) react because the energy is in the wrong mode.
Similarly, one needs to remove energy to get a new bond to form. If there is im place for the energy to go, you could form only very hot molecules. Hot molecules an unstable. Angular momentum barriers are more subtle, because they are important onl) with large-impact parameters. Generally, if the impact parameter is large, the reactanll will fly apart before they can react.
Al Ihts |><11111 we need lo cud oill dim iimhIoii In this chapter, we have shown that their in dure pilltcipal causes barriers to leaclioii All three Ol thrsr pioeessrs air unpin i.nil
......n * >nr should stud) them In detail
We have also loiiinl dial our can e\ploir all ol the effects via trajectory calculations iiM'iloiy calculations are quite important lo reactions. I he reader should run some ii,M' i lours before proceeding further.
It HI   SUPPLEMENTAL MATERIAL
n ii) I   Derivation of the Angular Momentum Barrier to Reaction
\ iiiited previously, there is an angular momentum barrier to reaction. The purpose ol iin in \i several pages will bc to derive equation (8.56) for the barrier.
I In derivation starts by looking at the classical equations of motion of A and Ii durinj ii i ollision between A and B.
d2RA dt2
mB
dr
: FA = -VAV(ra, rb) = FB = -VbV(ra, rb)
(8.80)
(8.81)
I rout Newton's second law, the force that A exerts on B must be equal and opposite to I he force that B exerts on A:
FB = -FA (8.82)
It is useful to change coordinates to what is called center-of-mass coordinates. Let's ileline a new vector X via
marA-r-mbrb
-■- (8.83)
mA + mB
X
Lation rs0™0" °f 1^ CentCr °f maSS °f A R Takmg the Second d™"e of M|iiation (8.83), we obtain
d2X
I
d2ra d2rb ma—— + mB__
dtz      VmA + mBy V"" dt2   ' dt Substituting equations (8.80)-(8.82) into equation (8.84) yields d2X
dt2
mA + mB
(FA+FB) = 0
(8.84)
(8.85)
I hcrefore, the center of mass of the system moves with a constant velocity throughout Ihc collision. Now, let's develop an expression for rab, the distance from A to B. The definition of rab is
Rab = rb - ra (8.86) I.iking the second derivative of equation (8.86) yields
d2rab _ d2rb d2ra
dt2
dt2
dt2
(8.87)
Substituting (8.80) mul (8.81 ) into equation (K HI) yiflils
d2RAn
dť
but FA
-FR. Therefore
d2RAB dť
I'ii in,,
mA mB/
M-ab
where uAB is defined by
I 1 1 _ mA + mB u.AB     mB     mA mAmB
IS Mil
IK K'i)
(K 'l(l|
The quantity uAB is called the reduced mass of the system. Rearranging equation (K 89) yields
M-ab "
d2RAB
dť
but
FB = -VBV =
Rab = Rb - Ra d
FB =
dXB ' dYB'dZB
V
dXab dYAB dZA
V
(8.911 (8.92) I
(8.93] (8.94)
where XAB, YAB, and Zab are the x, y, and z coordinates of RAB.
It is useful to convert equation (8.91) to spherical coordinates; RAB, 0Ab, and 4>aBi where RAB, 0AB and <|>AB are related to XAB, YAB, and ZAB, respectively, the X, Y, and / components of RAB, by
XAB = rab cos0AB
YAB = RABsin0ABcos(j)ab
ZAB = R-ab sin0AB sin(|)ab
For the moment, we will consider only 4>Ab = °.
Let's define LAB as the angular momentum of the AB pair by
(8.95)
Lab = P-abva^bRab — Hab
d0AE
dt
(8.9(,l
where vA^B is the instantaneous velocity of A toward B. Next, let's demonstrate thai angular momentum is conserved during the collision. When (|>AB = 0, it follows that
0AB = arctan
(8.97)
d()Aii di
d / YAI,
I  ,  (Y'mA      ''I VXah
/ 1 W I "Yah Yai, ,»Xa„\ V I 1 YAI1; Uai.   dt      X2B   dt )
Multiplying die lop and bottom of equation (8.98) by XAB yields
y   dYAB ^ dXAB **» _    AB   dt dT     xab dYAB     YA„ dX„
(8.98)
dl
V 2     i Y2 aai)   '   1 ab
RAB dt
RAB dt
Now i onsider:
dLAB _ d /      2 90AB dt ^abrab^-
dl
lib .......ng equation (8.99) into equation (8.100) yields
dLAB _ d /      /     dYAB XAB dt      ďt \^     V       dt   _ ^dt""
(8,99)
(8 looi
(K.IOI )
ľ. iluiming the algebra, we obtain
II,m,         dXABdYAB              d2YAB dXab dYab
uab—------b HabXab—--|tAB -
dl dt     dt    ' r-""--™ dt2
III re the first and third terms cancel. Therefore
dt dt
uabY/i
dľ
dlab _,,   Y   d2YAB d2XAB dt   " dt2   ~ ^abYab^2-
I lin can use equations (8.92) and (8.94) to show
3V(RAB)       XAB 3V(RAB)
M-ab-
d2XAB
= costfAB
uAB-
dt2 3rab rab 9rab
|2Yab dnfl 3v<rab) Yab3V(rab) —t— — sin«AB—--=---
dt' 3Rab Rab 3rab
Substituting equations (8.104) and (8.105) into equation (8.103) yields dLAB (   Yab3V(rab)^ ( XAB3V(rab)
— — a* ~ 1
dt V   R-ab    3R-ab   J \   Rab 3Rab
Note that all of the terms on the right of equation (8.106) cancel. Therefore
dLAB
di
= 0
(8.102) (8 l<l<>
(8......
(8.105)
(8.106)
(8.107)
muri i mi niai mau imai
Ii. i
liquution (8.106) is n kcv irsiill because il shows Ihm iinguliii moiiirnlum is Ii conseiwil quantity, Next, we will .ho« thai the energj ol the system, i mi. is •' i onscrvcd quantity where the energy is given by
Eab = 5UabVa^b + v<rah)
(K KIH)
where vA^B is the instantaneous velocity of A toward B and V(rab) is the A B potential Note
'dYABN
Combining equations (8.108) and (8.109) yields
Now consider the quantity
:Mab
dEAB
dXA, dt
+
+
dY,
dt
is 109)
dt
+ V(RAB)
(8.1 1(1)
dE„
dt
= 2uA
dXAt dt
dt
d2XAB dt2
Differentiating (8.110), we obtain
+
dY»
dt
d YA| dt2
+
dv(RAB) dRAB
dRi
dt
Note
R-ab = \/xab + Y
2
KB
(8.111)
is.I I 'i
Therefore
dRA
dX,
+ YA
dY«
^       \/X2ab + Yab V"""  dt       — dt Substituting equation (8.112) into the first term in (8.113) and rearranging yields
dRAB     [Xab (dXAB\    Yar / dY^ Rab
dt
dt
+
Rab
dt
(8.113)
(8.114)
Substituting (8.114) into (8.111) yields
dEA
dt
= MAB
dXA
dt
dV(RAB) dRAB
Rearranging equation (8.115) yields
dEAB = /dXAB\ dt   ~ V dt )
d2XAB
dt2
Xab I dXAB
Rab
dt
dYAB^	1 /d2YAB
dt ,	t{ dt2
	(dYab\1
Rab	x  dt |
(8.115)
m-ab
d2XAI dt2
+
Mab
dt2
Xab dV(RAB) Rab dRAB
Yab dV(RAB)
+
Rab dRab
(8.116)
11 mi i, , online lo cquulioiis (8 1(1-11 und |K | OS)
d\\n      XA|i dV(KAn)
M Ml
Maii
i, rcforc
ll      '   Raii dKA||
dYAn ( YAiidV(RAii) dt       RAB dRAB
dEAi(
di
(8.117) (8.1 18)
(8.119)
I mi is a conserved quantity!
Next, we want to use conservation of energy to derive an equation for llie anpilui momentum barrier for reaction. Equation (8.108) can be rewritlen as
Eab = |mab(v2 + v^) + v(rAB)
(8 .120)
In ii  v|| and v_l arc the two components of the molecule's velocity as delined in 8.27. Next, it is useful to derive an expression for v± during the A-B collision, provides angular momentum to the A-B pair, so let's start by calculating the angulai
........eniiim of the system. Consider the situation long before a collision occurs. Atom
\ v. moving toward atom B with a velocity, v(). Combining equations (8.95) and (8.98)
I hows
dYab \    .,    / dXab
o 2   fde\ (<
LAB — MAB^aB I ^ I - XAB I
dt
dt
where l.AB is the initial value of LAB. Initially
dYAB
YAI( = b.
dt
= 0,
dX,
dt
= v0
(8.121)
(8.122)
Where v0 is the velocity of A toward B at the beginning of the reaction (i.e., when RAn i  large so that V(Rab) = 0J. Substituting equation (8.122) into equation (8.121) shows
Lab = -bMAßVo
(8.123)
Noie that angular momentum is a conserved quantity. Therefore, Lab is the value of the lllgular momentum at any point during the A-B collision must equal the initial value:
Lab — L.
bMABVo
(8.124)
I hi future reference, it is useful to note that when rab is large so that V(rab) is zero, •quation (8.120) becomes
Eab = 2"Mabv0
(8.125)
' lext, it is useful to derive an expression for vj. in terms of the angular momentum for die collision.
d0AB
v± = Rab-^ (8.126) dt
in iii-i i wi it i 'm pnn 11 Mini
i" '
Substituting equation ik 'Mil inlii ri|iiulnill (8 I .'01 yields
' ■ ■ ■     ' linvr Muled 111.11
I All
' ~ RaiiM
Substituting equations (8.124) and (8.127) into equation (8.120) shows
1 (dRAB ^ LAB        i  y,„ ,
E = -Hab I -   -    + x--y- + V(Rab)
2 V  dt  J 2uABRAB
Substituting equations (8.123) and (8.125) into equation (8.128) yields
„ 1 /dRAB E=2^Bl^T
EABb
* \1!
+ V(RAB)
It is useful to define an effective potential, Vetl, by
EAnb2
Veff(RAB)= +V(Rab) RAB
Substituting (8.130) into equation (8.129) and rearranging yields
dRAB
dt
±W-(E-Veff(RAB))
PAB
iH I '/l
<8.I.>H|
(8.1291
(8.130)
(8.131)
Equations 8.130 and 8.131 are the key results cited earlier in the text.
8.16.2  Mass-Weighted Coordinates and the Affine Transformation
In the main body of this chapter, we found that we could represent the behavior of the system as a trajectory on a potential energy surface. If the reactants are confined to a plane, the general equations of motion for the reaction A + BC => AB + C are
d2RA l
dt2 mA d2RB
dt2 ~~ mB
d2Rc _ j_
dt2 ~~ m
FA -Fb
where
FA
3V(RAB, Rbc)      , 9V(rab, rbc)
9Ra
+
;ir.
9V(RAB, RBc) 3V(RAB, RBC)      3V(RAB, RBC)
FB =--
9RB
3"<r
9k ,
Fc =
aV(RAB,RBC) 9V(RAB,RBC)
9RC
3rb
(8.132)
(8.133)
(8.134)
(8.135)
(8.136) |
(8.137
»* All       "ii '*A
k|m = k, k„
(8.1 18) (8 1)9)
II yon waul in lollow the trajectory, you need to see bow kab and kb<. evolve in lime. ......I.......)• equations (8.132) (8.134) yields
ll'lk \||	d:R„	d2RA
,ir	,11'	dt2
d2RBt	d2Rc	d2RB
ill'	dl'	dt2
= —FB mB
= —Fc mc
mA 1
mu
I'a
FB
' iiiniiii)' equations (8.I35)-(8.137) into equations (8.140) and (8.141) yields d2RAn
dt2
d'K„c dt2
V mA    mB /
(
9V(RAB, RBC)        1   9V(RAB, RB(:)
3d
ah
mB 3r
J_ + j_\ 3V(RAB,RBC)        1   3V(RAB, Rbc)
mB     mc/       9rbc mB 9rab
is 140) (8.141)
(8.14.') (8.143)
.....parison, if RAB and rbc followed Newton's equation of motion, you would expecl
iln ii mi to lollow
dRAB 1 9V(RAB, RBC)
- (8.144)
dt2
Pi 9ra
here Pi is a constant.
I Ihii fore, there is an extra term in equations (8.142) and (8.143) that one would Itnl expect from Newton's equation of motion. In actual practice, that is not much ol ■i problem, since people normally integrate equations (8.142) and (8.143) numerically 11 mi, in the literature, people sometimes change the coordinates so that the system MIows Newton's equation. So-called mass-weighted coordinates are used to eliminate in. . Mia terms in equations (8.142) and (8.143). The mass-weighted coordinates, R| and in- defined by
Ri = RAB +
mAmc
i n
R2 = RBC-
(mA + mB)(mB + mc)
mB(mA + mB +mc) x 1/2 (mA + mB)(mB + mc)
Rue
(8.145) (8.14b)
l quations (8.145) and (8.146) are called the affine transformation. After considerable llgcbra, one can show that R| and r2 obey
M Mi
dt2
d2R2 "dt2"
9V(rab, rbc)
3R,
3V(rab,rbc)
3r2
(8.147)
(8.148)
TOO III Al IK IN'. AM I II I nil INN
HI »1 VI III KAMI'I I II
1(1/
Willi
I
(>1
(K II'M
Mill
Figure 8.32 shows a diagram of the potential energy surface for the reaction F I II HF + H plotted in mass-weighted coordinates. One finds that the coordinate is uu lim d by an angle 9 where
mAmc
1 '/2
(mA + mB)(mB + mc)
(SI MU
The reaction F + H2 ->• HF + H has an unusually large tilt because the fluorine is Q much heavier than the hydrogens.
The advantage of mass-weighted coordinates is that if you think about the motion ol the system as rolling a marble over the potential energy surface, then the marble will follow the correct trajectory in a mass-weighted coordinate for an impact parameiei ol zero. One still has an effective potential when the impact parameter is nonzero.
8.16.3  MD/MC Calculations of Reaction Rate Constants
One can, in principle, combine Monte Carlo (MC) and molecular dynamics (Ml)I calculations to estimate rate constants for each type of reaction discussed in the last two sections. The general procedure is to start with an ensemble of reactant molecules, ami assume that the internal states of the reactant molecules are in equilibrium before reaction occurs. One then uses a Monte Carlo routine similar to that discussed in Chapter 6 tl I choose initial conditions for the molecules, while a molecular dynamics routine is used to calculate the reactive cross section. For example, if one were examining the read inn A + BC -> AB +C, one would use the Monte Carlo routine to pick initial values ol Va^bc and EBC, where vA^Bc and EBC are as defined at the beginning of Section 8.(1 The calculations are easier if one modifies the Monte Carlo routine to pick only those values of vA_>BC and EBC that meet the constraint that the total energy of the molecule be sufficient to traverse the activation barrier for reaction. One then picks several value, of the impact parameter b and integrates equations (8.29)-(8.31) to calculate Pa-^bc (b)
Figure 8.32 A potential energy surface for F + H2 -> HF + H in mass-weighted coordinates. After Bender, 1972.
Hue Ihi'ii plugs inlii equation (8"iKl In tillcilluU' a retCtlVC miss section I he pun ess ' " penled lor thousands ol values ol vA .111 and I•,„•. One then lakes an average using ■ .pialum (8.59) to calculate a rale Sin h 1 ah illations are easy in do provided one knows an
1 ■ ni.He potential energy surface im the reaction. < lenerally, the results of the calculation) quite well with experiment provided the potential energy surface is accurate and tunneling is negligible. As a result, such calculations are reasonably common, although Hill II  common as you might imagine since die computations are quite lime consuming
n 11. -1   The Velocity Distribution in Center at Mass Coordinates
i i we wanl to use the results in Section 8.16.1 to derive an expression for the vA .,„■. Mu average velocity of A toward BC.
\' cording to the analysis in Section 6.4.1 and Example 6.B, the average velocitj ol ......olecule is
JJJva^bc exp(-E/kBT)dvA^BC
va->bc —
exp(-E/kBT)dvA ^Bc
IS 1
\i miding to equation (8.107), during a collision of A with BC, the total energy is
gneii by
(8.152)
E = 5u.aB(vA^BC)2 + V(rabc)
We will consider only the case where rabc is so large that V(rabc) is negligible. Equation (8,152) then becomes
(8.153)
E = 5Uabc(va^bc)2 nbstituting equation (8.153) into equation (8.154) yields
vA^Bc =
Í
vA_bc exp(- ihabc(vA^Bc)2)dvA^BC
exp(-iu.ABC(vA^BC)2)dvA^BC
(8.154)
liquation (8.154) is identical to equation (6.106) except that the mass is u.AB. We do the iniegral in Example 6.B. The results are
vA->Bc
ttUAB
1/2
(8.155)
0.17  SOLVED EXAMPLES
Kxample 8.A Properties of Potential Energy Surfaces Assume that the potential energy surface for the reaction D + H2 -» HD + H is given by
V(r,, r2) = W(e~2a(r,-r<,)) - 2e-a(r,~r|)>
+ W(e~2a(r2~r()) - 2e"a(r2"ro,)+ + VP(e-a(r|+r2-2ro,)-|-W
(8.A.I)
in A( in iN'i A'n 11| I r.n in'
•HHI
where V < r,. distance, W
) is the potential energy sin like, i, is the II II distance, i   is the II l>
It)').I kcal/mol.
1.95 A
(I 14 A, and V,
258 kcal/mol
I  ,1,1., II A   I     AltollllllK •tpilltKllllllKll III I llll III,lie 111« • llllllotl.il
(a) Use a spreadsheet to calculate values lor Ihe potential energy surface lor i| ami i. varying between 0.5 and 2 A.
(b) Trace the minimum-energy pathway from reactants to products.
Solution I solved this in Excel. A piece of my spreadsheet is shown in Tables 8.A. I ami 8.A.2. The original spreadsheet is available at http://www.wilcy.eom/chemicalkineiics
First I defined my own function V using the macromodule. The macromodule allows you to define your own functions in Excel. You use it by first inserting a module workshi 11 in Excel, using the Insert-Macro-Module on the Excel menu. You then type in a program in a language called Visual BASIC.
Table 8.A. 1 shows my module. The first line defined the function V as public (available to the rest of the system) and of type variant (general). I could have also declared it as single (single-precision real number) or double (double-precision real numbci I | described in Chapter 2. I then defined the function and included an end statement.
I then used the function in my spreadsheet. Table 8.A.2 shows the spreadsheet. 1 listed ri values in cells A4, A5, A6,... and r2 values in cells B3, C3, D3,.... I nameil cell Bl = w, cell B2 = vp, cell Dl = a, and cell D2 = rt).
I then calculated a matrix of V(ri, r2) values, so cell B4 contains V(ri, r2) with rj = b$3 and r2 = $a4, cell B5 contains V(r1? r2) with T\ = B$3 and r2 = $A5, and cell C4 contain| V(r,, r2) with r, = C$3 and r2 = $A4.
Table 8.A.1   The module used to declare the function V in Excel
Public Function v(r1, r2, rO, a, w, vp) As Variant
v = w*(Exp(-2*a*(r1 - rO))-2*Exp(-a*(r1 - rO)))
v = v + w*(Exp(-2*a'(r2 - rO)) - 2*Exp(-a*(r2 - rO)))
v = v + vp*Exp(-a*(M + r2-2*r0))
v = v + w
End Function
Table 8.A.2 A part of the spreadsheet showing the formulas used to calculate the potential
A B		C	D
w=	109.4	a=	1 .954
vp=	258	rO=	0.74
I r2\N	0.5	0.6	0.7
0.5	=v(B$3,$A4,	=v(C$3,$A4,	=v(D$3,$A4,
	r0,a,w,vp)	rO,a,w,vp)	r0,a,w,vp)
To. 6	=v(B$3,$A5,	=v(C$3,$A5,	=v(D$3,$A5,
	rO,a,w,vp)	rO,a,w,vp)	rO,a,w,vp)
0.7	=v(B$3,$A6,	=v(C$3,$A6,	^v(D$3,$A6,
	r0,a,w,vp)	rO,a,w,vp)	rO,a,w,vp)
0.8	=v(B$3,$A7,	=v(C$3,$A7,	=v(D$3,$A7,
	r0,a,w,vp)	rO,a,w,vp)	rO,a,w,vp)
•\r1
(I. 7
n.a
109.4_
268_
0.5_
w*(EXP(-2* a'(B$3-rO))-2*EXP(-a*(B$3-i (III) iw'(EXP(-2*a*($A4-rO))-2'EXP(-a*($A4-rO)))+vp*EXP(-a'(B$3+$A4-2*r0))+w
w'(EXP(-2* a'(B$3-rO))-2'EXP(-a*(B$3-rO)))+w*(EXP(-2*a*($A5-r0))-2*EXP(-a*($A5-rO)))+vp*EXP(-a'(B$3+$A5-2*r0))+w
=w*(EXP(-2*
a*(B$3-rO))
-2"EXP(-a*(B$3-
rO)))+w*(EXP(-
2*a*($A6-rO))-
2*EXP(-a*($A6-
rO)))+vp*EXP(-
a*(B$3+$A6-
2*r0))+w
=w*(EXP(-2*
a*(B$3-rO))-
2*EXP(-a*(B$3-
rO)))+w*(EXP(-
2*a*($A7-r0))-
2tEXP(-a*($A7-
rO)))+vp*EXP(-
a(B$3+$A7-
2*r0))+w
a*_
r0=_
0.6_
=w*(EXP(-
2'a*(C$3-rO))-
2*EXP(-a*(C$3-
rO)))+w*(EXP(-
2*a*($A4-rO))-
2*EXP(-a'($A4-
rO)))+vp*EXP(-
a*(C$3+$A4-
2*r0))+w
I .9!>4 0. 74
=w*(EXP(-2*
a*(C$3-rO))
-2*EXP(-a*(C$3-
rO)))+w*(EXP(-
2*a,($A5-rO))-
2*EXP(-a*($A5-
rO)))+vp*EXP(-
a*(C$3+$A5-
2*r0))+w
=w*(EXP(-2*
a*(C$3-rO))-
2*EXP(-a*(C$3-
rO)))+w*(EXP(-
2*a*($A6-rO))-
2*EXP(-a*($A6-
rO)))+vp*EXP(-
a*(C$3+$A6-
2*r0))+w
=w*(EXP(-2*
a*(C$3-rO))-
2*EXP(-a*(C$3-
rO)))+w*(EXP(-
2*a*($A7-rO))-
2*EXP(-a*($A7-
rO)))+vp*EXP(-
a*(C$3+$A7-
2T0))+w
0.7
^w'(EXP(-2*a*(D$3-
rO))-2*EXP(-a*(D$3-
rO)))+w*(EXP(-
2*a*($A4-rO))-
2*EXP(-a'($A4-
rO)))+vp*EXP(-
a*(D$3+$A4-
2*r0))+w
=w*(EXP(-2*a*(D$3-
rO))-2*EXP(-a*(D$3-
rO)))+w*(EXP(-
2*a*($A5-rO))-
2*EXP(-a'($A5-
rO)))+vp*EXP(-
a*(D$3+$A5-
2*r0)) +w
=w*(EXP(-2*a*(D$3-
rO))-2*EXP(-a*(D$3-
rO)))+w*(EXP(-
2*a,($A6-rO))-
2*EXP(-a*($A6-
rO)))+vp*EXP(-
a*(D$3+$A6-
2*r0)) +w
=w*(EXP(-2*a*(D$3-
rO))-2*EXP(-a*(D$3-
rO)))+w*(EXP(-
2*a,($A7-rO))-
2*EXP(-a*($A7-
rO)))+vp*EXP(-
a*(D$3+$A7-
2*r0))
+w
0.8_
=W(EXP(-
2*a*(E$3-rO))-
2'EXP(-a*(E$3-
r0») iw1 (I Xl'i
2*a*($A4 iiin
2'EXP(-a'($A4
rOiD+vp'CXl'i
a'(E$3+$A4
2*r0))+w
=w'(EXP(-2* a*(E$3-rO))-2*EXP(-a'(E$3 rO)))+w*(EXP( 2*a*($A5- rO))
2'EXPi a i$AS rO)))+vp*EXP(-a*(E$3+$A5 2*r0))+w
=w*(EXP(-2'
a*(E$3-rO))-
2*EXP(-a*(E$3
rO)))+w*(EXP(-
2*a*($A6-rO))-
2*EXP(-a'($A6
rO)))+vp*EXP(-
a*(E$3+$A6-
2*r0))+w
=w*(EXP(-2'
a'(E$3-rO))-
2*EXP(-a'(E$3-
rO)))+w*(EXP(-
2*a*($A7-rO))-
2*EXP(-a*($A7-
rO)))+vp*EXP(-
a*(E$3+$A7-
2*r0))+w
The word macromodule does not work with all spreadsheets, so I have also included , iprcadsheet with all of the formulas inserted directly. It is Table 8.A.3.
The numerical values in the spreadsheet are presented in Table 8.A.4. The reactants i hi out with r2 large, so row 19 is close to the reactants. If we read across row 19, we Dud that the minimum comes in column E, so cell EI9 is close to the reactants.
Similarly, the products have n large, so column Q is close to the products. If we read down column Q, we find that the minimum comes in row 7, so cell Q7 is close to the products.
'iiii vi 111 xami'i i ', 4/1
Ni'xl, we need In I race n pnlh Inmi lencliinls In |.inducts We inn lake any path we nil  llowcvci, Ilk- lowest cneigy | >nt 11 encs up then dnwn. I lie tiansilinu stale lies between tells 11(1 and MM. Notice that the liaiisilmn stale r. • •ii nl in a Incal tiiiuiiiiiiiu. hecause iilnsl til the surrounding cells have a highci cncigy ii   • i mi. die eneii'v decreases alone the shaded lit) 111 I diagonal, so die transition stale i ft nil) i saddle point.
V\i nsk the reader to continue these results in the homework set.
I ......|>le H.ll   Numerical Integration of Newton's Equations of Motion    flu objl I
••i dir. example is to use (he algorithms from Example 4.1? (numerical iiilcgialinii
......' allv integrate Newton's equations of motion.
Inlegiale the c(|uations of motion for the hypothetical reaction A I B   > AH
Suliilinn   Consider two atoms, A and B, that satisfy
d2RA ^
mA
mB-
dt
d2RB dt
= -FA
(8.B.I)
(8 It 'i
In ic ka atul RB are the positions of A and B, mA and mB are the masses of A and It. and i  ,.i die force on A due to the presence of B. The negative sign in equation (8.B.2) anst i    hi ic nl Newton's second law (where any force is met by an equal and opposite force) K \. Kh, and FAB in equations (8.B.1) and (8.B.2) are vectors. However, it is useful i" ill i 11 them to scalar quantities. Let's define XA, YA, and ZA, and XB, YB, and /„ in ilu \. Y, and Z coordinates of A and B; vXA, vYA, vZA, vXB, vyb, and v/n to In- the ■  .uid /. components of the velocity of A and B; and Fx, FY, and F7 to be the X, Y, ami / components of FAB. Then equations (8.B.1) and (8.B.2) become
mA
mA
mA
dXA dt
dY, dt
dZA dt dvxa dt
dvYA dt
dvzA dt dXB dt
dYB dt
dZB dt
vxa
vya
vza
= FX
= vXB
= vYB
vzb
(8.1».\) (KM H 18 It M
(8.B.6) (8.B.7) (8.B.8) (8.B.1)) (8.B.10) (8.B.II)
.1 II VI 11 I
1/1
iii,,
mB
mB-
.Ii
<-l\N_|l dt
dvZB dt
Fx
= -FY = -Fz
ix it 12) (8.B I >)
is li iii
Therefore, one can calculate the motion of the two atoms by picking initial velocities and positions and numerically integrating equations (8.B.3)-(8.B.14).
We still need to get an expression for the sources. Let's define Rab as the distance from A to B:
Rab = v/(XA - XB)2 + (YA - YB)2 + (ZB - ZA)2 According to equation (8.32):
(8.6.15)
dVA
dXA
where VAB is the A-B potential.
Combining equations (8.B.14) and (8.B.15) yields
Fx =
dVAB       dRAB dVAB     (XB - XA) dVAB
Similarly
FY = F/ =
dXA dRAB (YB - YA) dVAB
dR,
Rab dRab (ZB - ZA) dVAB
dRA
(8.B.16)
(8.B.17) (8.B.18)
This gives us a system of equations that can be integrated numerically.
In order to do the numerical integration, one needs to convert the equations slightly. Let's define
W(l) = XA, W(2) = YA, W(3) = ZA, W(4) = XB, W(5) = YB W(6) = ZB, W(7) = vXA, W(8) = vY/. W(9) = VzA W( 10) = vXB, W( 11) = vXB, W( 12) = vZB
The equations of motion become dW(i)
ill
= W(i + 6)
dW(i) _ W(i-3)-W(i-6)dVAB dt Rabiha dRab
dW(i) W(i-9)-W(i-6)dVAB
for i = 1 -6 for i = 7-9 for i = 10-12
dt Rabma dRAB
In order to do a numerical integration, you need to also know the Jacobian:
dFdW(i,j):
d /dWi\
dW(j) V dt J
(8.B.19)
(8.B.20) (8.B.21) (8.B.22)
(8.B.23)
Ihr lollowui)' coinpulri piofin.....lor-........testation Shaded lines in ihr programs
.....i......it.....Iiurs
i'iioi.iiam nitegratemd
"ivr diffequ for md problem i N 11 (11 R kmaxx , nmax, nvars i'auami II R (kmaxx=200,nmax=50,nvars=12) tNlltilR kmax,kount,nbad,nok,I ui ai (Ixsav,eps,hstart,time1 ,time2,w(nvars) common /path/kmax,kount,dxsav,xp(kmaxx),yp(nmax,kmaxx) mi ai. ma,mb
rial r0, well, range,jab, rab.qab
i oniinon /potent/r0, well, range, jab, rAB, qab
common /masses/ma,mb i ■11 HNAL stiff,derivs,rkqs mi.i I.
mb=35. kmax=0
listart=1 .e-3 eps=1.e-6 i0=2. well=5. range=1
i
i lni(>2=0.
i
W(1)=-10
w(2)=0.
w(3)=0.
w(4)=0.
w(5)=0
w(6)=0.
w(7)=1.0
w(8)=0.
w(9)=0.
w(10)=0.
w(11)=0.
w(12)=0.
i:   next open data file and write out initial ionditions
open(unit=4,file= traject.txt ,
action= write ,status= replace )
write(4,100)vab,time2,  (w(i),i=1,12) write(6,100)vab,time2,w(1),w(7)
c next start solving the equation
(iii win ii! ((i i.....;•. 11 .  en.))
1 .and. (w(i).yt.( 10.1))
2 .and. (w(l).lt.(l0.1))) timel=time2 time2=time1+0.1
c
c odeint integrates the rate equation from
ctimel to time2
c
call odeint(w,NVARS,timel,time2, eps,hstart,0,nok,nbad,derivs,
100 format(1x,f5.2,12( ,2x,f7.3))
write(4,100)time2, (w(i),i=1,12) write(6,100)time2,w(1),w(7) 10 end do
write(*, (/1x,a,t30,i3) ) Successful steps: ,nok
write(*, (1x,a,t30,i3) )   Bad steps: ,nbad 999     write(*,*)    NORMAL COMPLETION STOP END
SUBROUTINE derivs(time,w,f)
c
c subroutine to calculate the C derivatives
REAL time,w(12),f(12) REAL ma.mb integer I
REAL rO, well, range,JAB, rab.QAB common /potent/rO, well,  range, JAB, rAB,QAB
COMMON /masses/ma,mb C Assume a morse potential, i.e. C      VAB= well*(exp(-2.*(rAB-rO)/range) C      1 -2*exp(-(rAB-rO)/range))
rAB=sqrt( (w(1) -w(4) )**2+ (w(2)-w(5))"2+ (w(3)-W(6))"2)
QAB= 2*well/range* (exp(-(rAB-rO)/range)
-exp(-2*(rAB-rO)/range))/rab
do 10 1=1 ,6
f(i)=w(i+6) 10 enddo do 20 i=7,9
f(i)= QAB*(w(i-3)-w(i-6))/ma
n miililo
du :i() I 10,12
1(1)    UAB-(w(i-9) -w( 1 (i) ) /ml) lil     i nut nun' i ntin n I NI I
iiiiiiiitii iní jacobn(time,w,dfdt,dfdw,n,nmax)
INI I (it R n.nmax.I
m Al   t i me ,w(*) ,dfdt ('), df dw(nmax,nmax) lit Al  i0, well, range,JAB, rab,QAB Bommon /potent/rO, well, range, JAB,
i All, (1AB
do 15 i=1, n • i r cit (i)=o.
do 11 j=1,n
(ll(lw(i,i )=0.0 I I niuldo In ciiildo
i Alt (   (w(1)-w(4))"2   + (w(2)-w(5) )"2 + (w(3)-w(6))«2) i Ali sqrt(rab) uAli   2. well/ranged exp (- (rAB-rO) / range) exp(-2*(rAB-r0)/range))/rAB .iAH   2. well/range/range1 ( exp(-(rAB-rO)/range) t2.*exp(-21(rAB-rO) / range) 'i AB/rAB-QAB/Rab ■u     do      i=1 ,6
dfdw(i,i+6)=1. .'ii enddo
do 30 i=7,9 do 29 j=1,3
dfdw(i,j)= JAB (w(i-3)-w(i 6))/ma •(w(j)-w(j+3))
dfdw(i,j+3)= JAB*(w(i-3)-w(i-6))/ma
*(w(j+3)-w(j)) "i enddo
dfdw(i,i-3)=dfdw(i,i-3)+QAB/ma
dfdw(i,i-6)=dfdw(i,i-6) -QAB/ma 10 enddo
do 40 i=10,12
do 39 )=1,3
dfdw(i,j)= JAB (w( i 9)-w(i-6))/ma *(w(j)-w(j+3))
dfdw(i,j+3)= JAB (w(i-9)-w(i-6))/ma T(w(j+3)-w(i)) i'i enddo
dfdw(i,i-9)=dfdw(i,i-9)+QAB/mb dfdw(i,i-6)=dfdw(i,i-6)-QAB/mb 40 enddo return END
ni «i in n-iii «ii i ,i ii i inn ir«ii
Example x.<!  Numeri< .ii Integration ol Equation (8.133)1 he advantage ol the compuM
program is thai il is easily generalized lo lime dimensions. However, a simplci icsull cat be obtained by integrating equations (8.1 14) and (8.1 15), The following program use the IMSL Runge-Kutta routine to do that, There is some subtletj in the program slnoi equation (8.135) has two roots, and you can get stuck when the velocity equals /cm However, the following equation takes care of those subtleties.
I
fll H VI 111 KAMI-II •;
Ml
PROGRAM radialinte C   Solve diffequ for MD problem USE numerical_libraries INTEGER NVARS PARAMETER (NVARS=1) INTEGER i,idO REAL timel,time2,w(NVARS) Real parm(50) REAL ma,mb,mu Logical flag
REAL rO, well, range,E,b common /potent/rO, well,range,E,flag,b COMMON /masses/ma,mb,mu EXTERNAL derivs ma=1 . mb=35.
mu=1./(1/ma+1/mb) parm(1)=1.e-3 eps=1.e-6 r0=2. well=5. range=1 ID0=1
flag=.false . b=0.
c
c The k's re the rate constants time2=0. w(1)=10
E=0.5*mu*(1 -**2) c next open data file and write out initial conditions open(unit=4,file='traject.txt',action='write' ,status='replace')
write(4,100)time2, w(1)
write(6,100)time2,w(1) c next start solving the equation
do while ((time2.lt. (20.))
1 .and. (w(1).gt.(-10.1))
2 .and.  (w(1).It.(10.1))) timel=time2 time2=time1+0.1
IVPRK integrates the rate equation
call ivprk (ido, nvars, timel,time2,eps,param,W)
derivs,
derivs (N,t,w,wprime)
inn      rnimnt(1x,f5.2,12(     , 2x,f7.3))
wi I I n(<1, 100) t lino^, w( 1 )
wi 11 c(i;, too)time2,w( i) in mid du
unit   wrlte(V)     NORMAL COMPLETION .Mil' I Nil
Millioni iiii-lnteger N Logical flag
ica 1 t ,w(') ,wprime(*), Veff , VAB, rAB REAL ma,mb,mu
Hl AI   rO, well, range,E,b,Eeff
common /potent/rO,well,range,E,flag,b
COMMON /masses/ma,mb,mu
rAB w(1)
VAB= well*(
exp( -2. '(rAB-rO)/range)
exp( - (rAB-rO) /range)) VrlI  VAB+E'b b/rAB/rAB Eeff= E-Veff if(Eeff.gt.0)then wprime(1)= -sqrt(2*mu*Eeff) if(flag)wprime(1) =-wprime(1) else
wprime(1)= 10. Endif
if(abs(wprime(1)).It.0.01) flag=.true.
in t urn 1'ini
I Knmple8.D   Calculating the Rate Constant Using Equation (8.60)   Figure 8.17 how s some data for the cross section for the reaction H + H2 -» H2 + H as a function nl liT, Ihe translational energy of H approaching H2. The energy is measured in center-nl mass coordinates as described in Section 8.9. Assume that the cross section follows:
0
For  ET < 0.35 eV
'a^bc - ,- ,       / 1 -e035 bv-et\       F„v (8.D.I)
12 A I —FTiö—7}—   x e  t/       for   et > 0.35 cV 5.08 eV /
u here ET is the translational energy. Calculate the rate constant for the reaction at 300 K. Solution
According to equation (8.60), if there is no EBc dependence, then
kA^BC = vA^Bco-^^BD(vA^Bc)dvA^E
(8.D.2)
•Mil VI III XAMI'I I •. 4/u
Ationline i" hmiIin in I xumplc 6 I I
(vA .n, )2exp
D(vA .bc)
(Va^bc)"
1
,2        /   M-abc . i\
1 exP ^--^-(va >bc) j
(-^-^;(va-bc)
2kBT
(8.D.3)
where uabc is the reduced mass of ABC, kB is Boltzmann's constant, and T is die temperature. Looking up the integral in the CRC yields
D(vA.BC)=-l(^): V* \ 2kBT /
-|^(va.bc)2
(8.D.4)
Combining equations (8.D.2) and (8.D.4) and substituting ET = ^HabcOa^bc)2 yields
kA
bc -
8kBT
*UabJo   \\KTJ  A-BC V^BT
Note
Therefore
8kgT ^Uab
— va->bc
(8.D.5)
(8.D.6)
kA-»Bc = vA_
For future reference, it is useful to define an average cross section, Ia-bc, by
Jo UBT,
Equation (8.D.7) then becomes
ka-+bc = Va^bcI^bc
Let's define a new variable, W, by
E, - 0.35 eV
W =
(8.D.8)
(8.D.9)
(8.D.10)
Substituting equation (8.D.5) into equation (8.D.4) yields
Ica-bc = vA^Bce-°-35 *W
WkBT + 0.35 eV
0.35 eV/|sBT
(8.D.11)
I ,,i Inline leleieiue. il is useliil lo uole
k„ I     ll (i ki ill mol     il li '(i eV/.....lei ule
1   _      1 I       _ 1.5
Uahc     2 iiinu/mnl     I amu/mol amu
(8.1) I !) (8.1). I U
\......line to equation (7.29)
1/2
vA -bc
I amu
i
,2.52 x I0'3 A/second) (^)
/300 K\l/2 /   I amu
(2.52 x 10" A/second) ——--— (8.11 I 11
' \300 KJ    VI amu/1.5/
3.08 x 1013 A/second
llh liluting equation (8.D.9) into (8.D.7) and adding the appropriate conversion I.mm.
Melds
kA .bc = (2.94 x 1013 A/second) e"035 eV/^ /W + 0.35 eV
0.15 eV/kBT V kBT
rr = 0 for ET < 0.35 eV. Therefore
°a-bc exp(-W)dW
(8.D.15)
kA^BC =    294 >< 10'3 A   e-0.35 eV/feT
~* molecule-second
x     (W + 0.35eV/kBT)ffA^BCe dw Jo
(8.D.K.)
I .is define Lwbc and F(W) by
/W + 0.35 eV
o   V kBT /W + 0.35 eV\ ,
(8.D.I7) (8.D.I8)
i .imbining equations (8.D.15) and (8.D.16) yields
Lwbc=     F(W) e'-w,dW
(8.D.19)
One can eonvciiiciilly mlegiale        i>>■■ (K I) I /) using Ihr I itgiictc int. )■'■«"*formula:
F(W)e' w,clW = ^B,h'(W,) (K.I >.>())
i
where the Bi and W; values are as given in the following spreadsheet for the calculation!
IIII VI III KAMI 'I I
	A	B	C	D	E	F '
02	kbT=	=D2'0.00198 /23.05	T= 2	300		
					1=	=SUM (F5:F10)
	W	Et	S	F(w)	Bi	term in sum
05	0.22285	=A5*kbT +0.35	= 12*(1-EXP((0.35-$B5)/5.08))* EXP(-$B5/2)	=$B5'$C5 /kbt	0.458964	=D5*E5
06	1.118893	=A6*kbT +0.35	= 12*(1-EXP((0.35-$B6)/5.08))* EXP(-$B6/2)	=$B6*$C6 /kbt	0.417	=D6*E6
07	2.99273	=A7*kbT +0.35	= 12*(1-EXP((0.35-$B7)/5.08))* EXP(-$B7/2)	=$B7'$C7 /kbt	0.113373	=D7*E7
08	5.77514	=A8*kbT +0.35	= 12*(1-EXP((0.35-$B8)/5.08))* EXP(-$B8/2)	=$B8'$C8 /kbt	0.0103991	=D8*E8
09	9.83747	=A9*kbT +0.35	= 12'(1-EXP((0.35-$B9)/5.08))* EXP(-$B9/2)	=$B9*$C9 /kbt	0.000261017	=D9*E9
10	15.98287	=A10*kbT +0.35	=12*(1-EXP((0.35-$B10)/5.08))* EXP(-$B10/2)	=$B10*$C10 /kbt	8.98547e-7	=D10*E10
Here are the results:
	A	B	C	D	E	'     F 11
	kbT=	0.02577	T=	300		
					1 =	0.747826
	W	Et	S	F(w)	Bi	term in sum
	0.22285	0.355743	0.011349	0.156665	0.458964	0.071904
	1 .118893	0.378834	0.056199	0.826153	0.417	0.34506
	2.99273	0.427123	0.146036	2.420454	0.113373	0.274414
	5.77514	0.498826	0.26998	5.225937	0.010399	0.054345
	9.83747	0.603512	0.43199	10.11683	2.61E-04	0.002641
	15.98287	0.76188	0.6385	18.87694	8.99E-07	1.7E-5
M II llllC
II / IS \
kA .,„- = 3.08 x        -2— x (0.748 A:)e °'M second
kA^BC = 2.3 x 10'
k3
molecule-second
. 0.11 cV/^,1
4(11
(SI) 1 I I (SI) "I
(SI) M|
•   thai Ihe activation barrier is about 0.35 eV (i.e., the minimum energy to I-. i reaction) even though the reaction probability is small below 0.5 eV. Ii is noi it'll) 0.35 eV, though. In the problem set, we ask the reader to calculate the rat* ..ii.i.mi .ii other temperatures. Both vA^BC and Ia->bc arc temperature-dependent "
.......ake an Arrhenius plot of the data, you find that the activation barmi i bill
i lose lo but not exactly 0.35 eV, even though the reaction probability is negligible ni I     0.35 eV.
I vuiiiple 8.E   Calculating the Cross Section   A program called ReactMl) is availabli
I......      Dr. Masel's website. Assume that you used the program to calculate Ihe reaction
probability as a function of impact parameter, and the data in Table 8.E.1 were obti.....id
i ii. ulale the cross section for the reaction.
Solution   The cross section is given by
o = 2tt\ P(b)bdb
(8 I l i
We can integrate using the trapezoid rule. I lere is a spreadsheet to do the calculations:
	A	B	C
03	b	P(b)	b P(b)
		0.84	=A4 B4
	0.2	0.83	=A5 B5
	0.4	0.85	=A6B6
07	0.6	0.78	=A7-B7
08	0.8	0.8	=A8'B8
(19	1	0.75	=A9-B9
10	1 .2	0.8	=A10*B10
1 1	1 .4	0.83	=A11*B11
12	1 .6	0.72	=A12-B12
13	1 .8	0.21	=A13"B13
14	2	0.1	=A14'B14
15	2.2	0	=A15'B15
16			
17		integral=	=0.5'(C4+C15)+SUM(C5 : C14)
18		a=	=2*PI()-C17
Ml Al .III IN'I A'l I III I r.li IN'.
Inlilitlll  I   ll.-.iili . li......ho i.iili.iiliilliui ill
reaction probability as a function of Impact parameters
P(b)
f>a .lie, A
I'd.)
0	0.84	IJ	0.80
0.2	0.83	11	0.8.1
0.4	0.85	1.6	0.72
0.6	0.78	1.8	0.21
0.8	0.80	2.0	0.10
1.0	0.75	2.2	0
Here are the results:
	A	B	C
03	b	P(b)	
04 05	0	0.84	0
	0.2	0.83	0.166
"W 07	0.4	0.85	0.34
	0.6	0.78	0.468
i 08	0.8	0.8	0.64
07	1	0.75	0.75
10	1.2	0.8	0.96
11	1 .4	0.83	1 .162
T2~	1 .6	0.72	1 .152
13	1 .8	0.21	0.378
14 hit 16 17	2	0.1	0.2
	2.2	0	0
			
		integral=	6.216
18		o=	39.05628
Example 8.F   Calculation of the Cross Section for a Hard-Sphere Collision.
Assume that you have two molecules colliding as shown in Figure 8.F.1, and assume that the reaction occurs whenever
Ip-ab(vi)2 < E* Derive an expression for the cross section.
(8.F.I)
Solution   From Figure 8.F.1, we obtain
v± = vab sin0
(8.F.2)
••"iv.....AMi'ir. 4li:i
Figure 8.F.1   A hard-sphere collision.
i nun geometry, we have
sin0 = Wl
ri lore
,|Iab(vl)- = -Hab(Vab)
III 11 I'i is ihe translational energy. Ilieiclore, reaction occurs whenever
i i i us define bc„n by
I -
1 -	( b V"			( b \2"
	z—	= ET	1 -	'-
	V Ocoll /			Vbcoii/
1 -	( b \2'
	-
	V bcoii /
> I
= E*
(8.F.3)
(8.11)
(8.F.5)
(8.F.6)
H\ definition
(ben,)2 = (bcoll)2 I Et
1 if b < bcrit 0   if  b > bcrit
w here Pa^bc is the reaction probability. From equation (8.59)
o£=jrf bPA^BC(b)db
Jo
I ombining equations (8.F.8) and (8.F.9) and integrating yields
(8.l;./)
(8.F.8)
(8.F.9)
where vab is the velocity of the molecule.
°x = ^(bcril)2
(8.F.10)
mi «i hi hi . /v.i i ii i mi in1.
I'lll Illl I M'.
Combining equations (8.F.7) and (8.F.9) yields
„| = jirdx,,,,)' ( '   (   '   )    lor ■ s 0 elsewhere
(HI' III
Equation (8.F.10) is called the hard-sphere cross section.
In the homework set, we ask the reader to show that if one plugs equation (H.I II) into equation (8.D.7), one obtains
kA^ec = Jr(bcon)2e e*^bt
(X.I 12)
8.18  SUGGESTIONS FOR FURTHER READING
A general discussion of the influence of dynamics on reactions can be found in:
A. J. Alexander and R. N. Zare, Anatomy of elementary chemical reactions, J. Chan I .In, ll 05 (1998).
R. D. Levine and R. B. Bernstein, Molecular Reaction Dynamics and Chemical Reactivity, (hliiril
University Press, Oxford, UK, 1987. R. Schinke, Photodissociation Dynamics, Cambridge University Press, Cambridge, UK, 1993 W. H. Miller, Quantum and semiclassical theory of chemical reaction rates, Faraday Discuss I in
1 (1998).
G. C. Schatz, Quantum effects in gas phase bimolecular reactions, In W. Hase, ed., Advant i I n,
Classical Trajectory Methods, JAI Press, Stamford, 1998, p. 205. D. M. Hirst, Potential energy surfaces and Reaction dynamics, Taylor & Francis, London (198
8.19 PROBLEMS
8.1   Define the following terms:
(a) Collision theory
(b) Molecular dynamics
(c) Newton's equations of motion
(d) Impact parameter
(e) Angle of approach
(f) Distribution function
(g) Trajectory
(h) Cross section
(i) Bimolecular exchange reaction (j) Recombination reaction
(k) Barrierless reaction (I) L-shaped potential surface (m) Polanyi rules (n) Collision partner (o) Reaction collision
(p) Unreactive collision
(q) Reduced mass
(r) Reaction probability
(s) Classical equations of motion
(t) Effective potential
(u) Potential energy surface
(v) Dispersion force
(w) Electron exchange and bondiaj
(x) Pauli repulsion
(y) Transition state
(z) Mass-weighed coordinates
(aa) Early transition state
(bb) Middle transition state
(cc) Late transition state
• J    I i|ii,ilinti i   .'"iI has a p.ml uilegial
ini Why do we need to Integrate ovei -ill ol these variables?
(In Wh.il dues llie distribution 11111> lion do in llie equation 7
H i   I uphlln in your own words why llie potential looks the way il does In Figure .". 1 win Is there a well? Why does the potential use again al short distances?
i   l (plain in your own words why figure 8.9 looks the way il does
in) Why is there a well in the I* E and CI  E potentials?
ilii Why is there a repulsion at short F F and Cl-F distances?
u I Why is there a barrier to reaction'.'
nil Identify the reactants, products, and transition slate on the figure.
(el How does the contour plot correspond to the two-dimensional original''
|      I lie objective of this problem is to review some of the results in this chapter,
la) Why did we say that the top trajectory in Figure 8.13 lead to reaction while
the bottom trajectory did not? d>) Which trajectories lead to reaction in Figure 8.15? (c) Why were there no reactions in Figure 8.15? Id) Which trajectories lead to reaction in Figure 8.16?
(e) Why did some of the trajectories in Figure 8.16 lead to reaction while others did not?
(f) Which trajectories lead to reaction in Figure 8.19?
(g) What is different about the trajectories that lead to reaction in Figure 8.19 and I hose that do not?
ih) Repeat (f) and (g) for Figure 8.20. (i) Repeat (f) and (g) for Figure 8.21.
H (.   Explain in your own words why Figure 8.18 looks the way it does.
(a) Why is there negligible reaction when ET < 9 kcal/mol? (Note: E 9.3 kcal/mol.)
(b) Why does the cross section rise from 10 to 40 kcal/mol?
(c) Why does the cross section drop again at high energies?
(d) If you work at a temperature of 500 K, what is the probability that a state at 9 kcal/mol, 12 kcal/mol, 25 kcal/mol, 50 kcal/mol, 200 kcal/mol will be occupied?
N.7   Explain in your own words why the trends summarized in Section 8.8 are observed.
H.H   Explain in your own words why the reaction probability decreases as the impact parameter increases.
K.'»   Explain in your own words why unimolecular reactions are impossible in the absence of a collision partner.
Mill   Explain in your own words
(a) How the angular momentum barrier to reaction arises? When is it important'.' How does it affect reaction?
-1111)
III Al ;ll( IN'I A!i I ;i (I I Kill IN'
I'lH Hli I M'
■iii/
I low tin- energj transfci harrlci to rem lion urisi 1 w id ii is ii important 1 11- iw docs ii affccl reactions?
8.11 Explain how ihc energy and angular momentum barriers lo reaction could iillctl the following reactions:
(a) CH3NC + X -> CH3CN + X (d) Na + CI -> Nat I
(b) F + H2     HF + H2 (e) CH3 + CH3CH3 -*■ CH,( 111, i II
(c) CI + F2 -» C1F + F (f) H' + C3HS ->■ H'C,H7 I 11
8.12 The reaction CH3CH3 =>• H2C=CH2 + H2 proceeds by the following approxinmk mechanism:
CH3CH3
CH3 + CH3CH3
2CH3
CH4 + CH2CH3
CH2CH3 —-* H2C=CH2 + H
H + CH3CH3 —H2 + CH2CH3
2CH2CH3
2H
CH3CH2CH2CH3
-> H2
2CH3
CH3CH3
On the basis of the results in this chapter:
(a) Which of the reactions need collision partners? If a reaction needs a coll......
partner, explain what the collision partner will do. If the reaction does nol.....1
a collision partner, explain why not.
(b) How do energy transfer barriers to reaction affect each of the reactions a ho
(c) How do momentum transfer barriers to reaction affect each of the read.....
above?
8.13   In Section 8.2.1 we noted that the reaction
CH3CH2CH2 + O-► CH3CHCH3 + OH
(P8.I I I i
has an unusually low preexponential [1.4 x 1010 A3/(moleculesecond)J, while iln reaction
CH3CH2CH3 + O-► CH2CH2CH3 + OH (P.8.13.2)
has a normal preexponential [7 x 1012 A3/(moleculesecond)].
(a) Make a diagram of the reactants. Based on the geometry alone, how mm h larger will the cross section for reaction (P8.13.2) be than for equalion (P8.13.1)?
(It) Now put m the van del Waals null! I low much will Ihc I'll, I'loups constiaiu reaction (P8.H 1)7 ilhnl   look al the I taction Ol the solid Bilglc thai r.
blocked).
to Are these effects large enough to explain the supposed fat toi ol stio different c m rate?
oil Do you have any other ideas what could cause the preexponential ot equation (P8.13.1)? (Hint: Qualitatively, how will rotational motion ol the ( I I ,< I I < I I , ailed the rate of both reactions?)
h I I   In Section 8.2 we noted that ihc reaction
-> 20 + 02
M. 15
202
has an unusually large preexponential |5.8 x 10ls A-'/(molecule-second)!.
la) Use the preexponential to estimate the average cross section for the reaction.
namely, cta^bc in equation (8.10). (I>) How can the cross section be that big? Would the attractive forces betwei n
(wo oxygen atoms be large enough to give this large of a cross section?
(c) What if a complex formed? Could that account for the results?
(d) What if the key reaction were:
o2 + o:
20 + 02
(P8.I4.1)
where 0| is an excited 02 molecule? How large would the 02 need to be to accounl for the preexponential? That is, how large of an oxygen-oxygen bond would be needed to account for the preexponential?
(e) Now consider the reaction
20
-> O,
(8.14.2)
Assume that the potential follows
Vab(R-ab) =119 kcal/mol
1.21 A - rab
e-=--1
2.58 A
In Figure 8.28, we found that when b is large there is a barrier to reaction. Calculate the barrier height as a function of b. [Hint: Use equation (8.5(0 to calculate Veff as a function of rab. Use the solver function on Excel lo calculate the barrier in the effective potential.]
(f) What value of b gives a barrier of E = 2 kcal/mol (i.e., 2kBT at 500 K)?
(g) Assume that all molecules react when b is such that the barrier is less than K and no reaction occurs. Estimate the cross section for reaction (P8.14.2.)
(h) How does the cross section in (g) compare to the value estimated in a)? How does the computer program in Example 8.B work?
(a) Show that the program is correctly plotting a trajectory.
(b) Where are the intermolecular forces calculated?
I'll! MU IM'. 4H0
U'l When is Ihť actual uilepallou ilunr '.'
(d) Why are Ihcic I.' v.m.ihli-s Willi onl\ lun alums?
(e) Whai wnulil you need to do to the program to enable it lo handle iiu- uaciioit
A + BC
AB + C?
(f) What extra equations would you need in (e)?
(g) What extra forces would you need? Be sure to consider the effects of orhitiil compression like that in Figure 8.7.
8.16 Describe in your own words the key features in the bottom picture in Figure H.7, Where are orbitals being compressed? How do the forces arise on the rightmost fluorine atom? How is the fluorine-fluorine bond distorted?
8.17 Refer to Table P8.17. How does the preexponential calculated in Example si compare to that estimated by:
(a) Collision theory (Chapter 7)
(b) Transition state theory (assume a linear transition state)
8.18 In Example 8.D, we derived an expression for the rate of the reaction H + H2 H2 + H. As part of the derivation, we calculated a messy integral. The integral i temperature-dependent.
(a) Calculate the preexponential and the rate of reaction at 200, 300, 500, 700 and 1000 K. The spreadsheet is available in the instructions materials.
(b) Make an Arrhenius plot of the reaction rate. How close is the activation ban hi to the 0.35 eV expected from equation (8.D.23)?
(c) How close is the preexponential inferred from your Arrhenius plot to the preexponentials you estimated in (a)?
(d) Use transition state theory to see how the preexponential should vary willi temperature.
(e) How well does the actual temperature dependence compare to that expected from transition state theory? (Hint: Make a log-log plot of the preexponential vs. T. How does your slope compare to what is expected?)
8.19 Assume that the cross section for a reaction follows
E > E*
Table P8.17  Data for Problem 8.17
r	H	H2	Linear TST
Mass	1 amu	2 amu	3 amu
Rh-h	—	0.7417 Á	0.87 Ä
Vibrations	—	4395 cm"1	2012 enr1
	—	—	965 enr1
	—	—	965 cm"1
E*	—	—	8.5 kcal/mol
lul Assume n„ - 1.5 A'and I       I I HO ki .il/niol Niimeiic.illy inlcpale equation IH I) /) In c.ili ul.lie the piccxptincut nil lale constant lot Ihe i cud it in to t alciilulc 11 it- i.lie constant ul Kill K lit) luii-pale Ihe expression analytically.
li ) Mow does your result compare lo collision theory (Section 8.2).' II III   We discussed chemical reactions in this chapter, hul the concepts are usclul lot .....li .ii reactions as well. Consider the reaction
(8 !0 i i
n + :,,7Pb
,'b
«licic n is a neutron.
,,, v.siiinc that you have a nuclear reactor that is producing neutrons with thl following energy distribution:
D(E) =
E \ 1/2 r*l*>
(8 .'() 'i
wilh Eo = 1.5 keV. Figure P8.20 shows data for the cross section lot the reaction. Estimate the rate constant, lit) How thick of a piece of lead will be needed to stop 99.99'/< of the neutrons ' Lead is 22% 2(,7Pb. Assume the other isotopes of lead do not react with neutrons.
ul Calculate the conversion of neutrons with an energy of 20 keV. Arc they all trapped?
Ul) In actual practice, the neutrons undergo something called "inelastic collisions", where they give up energy to the lattice. How will the inelastic collisions affect your answer in (c)?
1E-3
„■' ..........................i ........1
1E"14E^Tli^l,1E--3l1E--2  1E-1   1E+0 TeTi   1E+2  1E+3 1E+4
Energy, eV
. ,Mrti„n n Jr.207 Pb -» 208 Pb as a function of the energy of the
I iqure P8.20  The cross section for the reaction n +    kd -»
tiiiutron.
ď) Assume llml the crou lection foi tbc Inelastic > • >l I ■. u >i ľ. i. Mioo hiinti Independent Oi energy. I low uupoilant will tlie mel.islie collisions he m slow my down the .'(> keV neuiioiis so that the neutrons are more casil) Irappi d ' 8.21   Sala/.arl el al. (1997) measured the eross section lor the reaction:
Ca + HBr-► CaBr + H (I'K.'II)
and the following data in Table P8.21 were obtained.
(a) Use the procedure in Example 8.D to estimate the activation harnei mill preexponential for the reaction at 200, 300, 400, 500, 600, and 700 K.
(b) How does a plot of the cross section versus energy compare to thai in Figure 8.29?
(c) On the basis of the results, what is the barrier to the reaction?
(d) Make an Arrhenius plot of the rate constant versus 1/tempcrature to csliiuiilt the activation barrier for the reaction.
(e) Why didn't the wiggles in the plot of a versus E affect the activation hanlci or preexponential?
8.22   Knott, Proch and Kompa, j. Chem. Phys.. 108, 527 (1998) measured the cm section for the reaction
CO + H2-
-»> HCO- + H
For energies between 102 and 5 eV, the cross section followed
a =
18.2 Á2 (eV)"-:
(ET)
,0.37
(8.2.1,1)
(P8.22.2)
(a) Make a plot of the cross section versus energy.
(b) On the basis of these results, what is the barrier to the reaction?
(c) Use the procedure in Example 8.D to integrate the rate equation to calcuj the preexponential and activation energy at 100, 200, 300, 400, and 500 K
(d) Why is the preexponential so big?
Table P8.21   The rate of reaction (P8.21.1) estimated from the
Energy, kj/mol
Energy kJ/mol
ET < 19.5
20.5
20.7
21.0 21.3
21.6 21.9 22.2 22.5
22.8
23.1
results of Salazar (1997)
Energy kj/mol
0
0.39
0.81
1.00
0.78
0.51
0.42
0.48
0.59
0.61
0.56
23.4
23.7
24.0
24.3 24.6 24.9 25.2
25.5
25.8
26.1
26.4
0.50 0.47 0.48 0.50 0.50 0.48 0.47 0.46 0.46 0.46 0.45
26.7
27.0
27.3
27.6 27.9 28.2 28.5
28.8
29.1
29.4
29.7
o l>
n 11
0,11
0.43
0,1 I
0.43
0.42
0.42
0.41
0.41
0.41
(«•) Ate the leaelanls huge enough In an mint Im Ihe huge eioss seclion?
(I'l Aie Iheie strong enough nlliiu lions between the reiiclnnls lo pull Ihe hydrogen
in so dial Ihe cross seclion is huge'' (answer yes!) i i'l Make an Arihciuiis plot ol Ihe data
iii) Why docs the Arrhenius plot show a harrier when the reaction is haiiiciless''
n ' * I he program RcaclMI) is available from Dr. Masel's Website and in the instructions materials, Choose the reaction A + BC —► AB + C and ihe case "choose random trajectories".
(n) What does the potential energy surface look like? How does il compare to Figure 8.9?
Ih) Use the print command at the end of the trajectories. Make a plol of the
potential energy surface. Identify the reactants, products, and transition state (c) On your plot, label the trajectories that lead to reaction, id) Calculate the reaction probability for the trajectories on your plot.
(e) Now rerun the cases again, resi/Jng the plots so that you can see the animation and the motion on a potential energy surface side by side. Find a irajccloiy where C is moving away form B when A hits. What does that Irajccloiy look like on the potential energy surface? What does that trajectory look like on a plot of position versus time?
(f) Use your results in (e) to explain why we say that "in order for a reaction to happen, C must be moving away from B when A hits."
(g) Now compare the plots on a simple potential energy surface and on mass weighted coordinates. How are the trajectories the same or different?
H.24 This problem also uses ReactMD available from Dr. Masel's Website. Choose the reaction A + BC -> AB + C and the case "choose random trajectories".
(a) Run trajectories, record the energy of the trajectory, and determine whethei the trajectory lead to reaction.
(b) What is the highest energy you used where no reaction happened? What is the lowest energy? Where did you see a reaction?
(c) How do you rationalize your results given that the barrier is only 13.80 kcal/moľ
(d) Use your data to calculate the reaction probability versus energy. Divide up the energy scale into 0.5 kcal/mol increments (e.g., 13-13.5 kcal/mol, 13.5-14.0 kcal/mol) and calculate the reaction probability versus energy.
(e) An additional file, Does it react. TXT, is available from Dr. Masel's Website and in the instructions materials. Repeat your results in (d) using the data in this file.
(f) The calculations use a one-dimensional potential. Assume that you can convert to three-dimensions using o\ = 3 Ä2Preactjon(Ex).
(g) Make a plot of <rx versus ET. How does it compare to Figure 8.26?
(h) Fit your data to Equation 8.D. 1.
(i) Follow the procedure in Example 8.D to calculate the preexponential and activation barrier at 500 K. (Note: You may have to interpolate to estimate a cross section.)
ľiiiiiiiiM'. 40.'l
(J) Kľ|H'iil (g) Iim l     H III. /(Hl. ODO, mul I 1(H) K. Milke im Anliciiiiis plol ol your dala Mow do I lu' pi cc \ pouční i al anil the ailivalion baníci CSlillllllcd Itnin I lit-
slope ol the pioi compare lo youi result'
8.25 This example also uses ReactMD from Dr, Masel's Website. Pick the rem......
A + BC ->• AB + C.
(a) Choose the case "vary incident energy, optimize oilier parameters", and inn trajectories. What is the minimum energy that leads to reaction? How does dial compare to the barrier height? Plol out your results you will need llicin lalei in the problem.
(b) Now choose the case "vary incident energy, do not optimize." How do youi results differ?
(c) Look carefully at the trajectories to see how the reactive trajectories differ from the unreactive trajectories. First determine how much the trajectory vibrate! How much vibrational energy is there in the "optimal" trajectories? How mm h vibrational energy is there in the trajectories that do not react?
(d) Next, look at when the trajectory turns on the potential energy surface. When does the optimal trajectory turn? When do the nonreactive trajectories turn'.'
(e) Relate the turning motion to coordinated motion. Look carefully at the coordinated motion by cascading and resizing the windows so that you can see the animation at the same time that you see the trajectory on the potential energy surface.
8.26 This example also uses ReactMD from Dr. Masel's Website. Choose the reaction A + BC —> AB + C and the case "vary the timing of the collision."
(a) Run the case and see how the reactive trajectories differ from the unreaclive trajectories. Look particularly at when the trajectory turns on the potential energy surface. When do the reactive trajectories turn? When do the unreactive trajectories turn?
(b) Now vary the position of the transition state from an early transition state to a late transition state. How do your results change?
(c) Repeat (a) and (b) for a late transition state.
8.27 This example also uses ReactMD from Dr. Masel's Website. Choose the reaction A + BC -* AB + C and the case "vary the partitioning between translation and vibration."
(a) Run the case and look at the trajectories. What happens if you have no vibrational energy? What happens if you have too much vibrational energy and too little translational energy?
(b) Now change to an early transition state. How does the answer in (a) change?
(c) Now change to a late transition state. How does the answer in (a) change?
(d) How do your results compare to the Polanyi rules?
8.28 This problem uses ReactMD, available from Dr. Masel's Website, to calculate a rate constant for the reaction A + BC —> AB + C.
(a) First, in order to learn how to use the "fix the energy vary the initial conditions" option, the program calculates the reaction probability of an energy of 14.0 kcal/mol.
lb) Now assume thai you waul In i all ulale a lale constant billowing die pun eduie m Example XI) Sel up an expiession lor lA .»< following Ihe procedure III Bxample 8.1). Assume dial Ihe eioss section is zero lor F ■ I I 88 kcal/mol It) In order lo use l.agucic integration, you will need die cross seclion al a series Of energies, follow ihe procedure in Example 8.1) lo calculate those energies. (Hint: II you calculated dungs correctly, die first energy poinl should be III) kcal/mol.)
(<l) Use ReactMD lo calculate the reaction probability al each energy  You will
need al least 2(X)() trajectories of each energy lo gel an accurate answer, («•) An additional file, rcaclionprob.TXT, is available from Dr. Masel's Wcbsiie
Repeat your results in (d) using the data in this file. If) Assume     = (I Á)2 x Preaciion. Integrate equation (8.D.7) to calculate a rale constant assuming vABc = I x I013 A/sccond. n "'   Problems 8.24-8.28 assumed a collision where the reactants were confined 10 I line. Now consider a nonlinear collision.
la) Use the "vary the impact parameter" option in ReactMD to see how the reaclion
probability varies with impact parameter. What do you find? (Ii) Look carefully at the animation. How do the nonlinear collisions differ from
Ihe linear collisions? Do you see any "weird" trajectories, where the incoming
A molecule inserts directly into the B-C bond? (c) What do the "weird" trajectories look like when plotted on a potential energj
surface?
Id) What does a trajectory with a large-impact parameter look like on a potential energy surface?
(e) Can you see evidence that atoms are hitting the angular momentum ban ici to reaction? Describe what you see.
(f) Now try varying the angle of approach. How do the trajectories change'
H Ml The objective of this problem is to use ReactMD to calculate the collisional cross section. The procedure will closely follow Example 8.E. First you will use ReactMD to calculate the reaclion probability. Then you will integrate equation s ľ I lo calculate Ihe cross section.
(a) Use ReactMD to calculate the reaclion probability at several values of b for an energy of 10 kcal/mol.
(b) Integrate equation (8.D.4) to calculate the cross section.
(c) What would you need to do to calculate a rate constant from these results?
N.3I   The objective of this problem is to understand how trajectories change as you change properties. Consider the reaction A + BC —> AB + C confined to a plane (a) Assume that A starts with an impact parameter at 3 Á, and moves past BC. For the purposes of the problem we will assume
XA = 3 Á
YA = (1012 Á/second)t - 10 Á
xB = xĽ = o
Y„ l A i o I A mm IO"/second i Y, = I A   o.l A sin IOl5/second i
where t = time and XA, XB, X(, YA. Y|,, and Y< arc the X and Y position of A, B, and C.
(b) Show that these equations satisfy Newton's equations of motion, with I! < i harmonic oscillator. Assume that A does not interact with BC.
(c) Use a spreadsheet to plot XA, YA, XB, YB, and XC,Y, versus time.
(d) Next, use the spreadsheet to plot rab versus. rbc.
(e) Why does rab decrease then increase again given that there is no interaction between A and BC?
(f) Why are there all of the wiggles in the trajectories? More Advanced Problems
8.32 The objective of this problem is to use collision theory to make some prediction', about film growth. Assume that you are designing an evaporator (see Figure P8 5 ' to deposit copper. The evaporator works by heating copper in a crucible. The coppei sublimes, filling the crucible with copper vapor.
(a) Use collision theory to estimate how many copper molecules will escape from a 1-cm2 hole at the top of the crucible. Assume that the copper pressure inside the crucible is 10~9 Atm, and the temperature is 1500 K.
(b) Assume that the copper is being deposited on a 10-inch wafer and that all of the copper that leaves the crucible makes it onto the wafer. How long will it take to deposit a 10-|im thick layer of copper?
8.33 Example 8.A used the routine ODEINT to integrate the trajectories. However, ODEINT may not be available at your university. Convert the program so thai it can use the IMSL subroutine IVPRK. What do the trajectories look like using I PAK?
8.34 In the examples in this chapter, we used trajectory calculations to model the reactions of two simple molecules. The objective of this problem is to decide what would change if the molecules are big, such as proteins.
Moving copper atoms
Wafer
Figure P8.32  A rough diagram of the evaporator.
till How would the equations ol motion change? Would the same equations ol motion apply?
ili) llow would the forces be different? Would the forces still !»■ determined by path repulsions, oi would othei forces be important?
i< i How would the collisions be different? Would molecules just bounce, or would ihey slick together for a long time? Could you use the same computer codes to simulate the reactions?
I lie   Materials   Processing   Simulation   Center   at   Cal    lech (California Insiiiuie   of  Technology,   Pasadena)   has   posted   a   number   of Simula nous   of  the   changes   in   the   arbital   shapes   during   Diels-Aldcr reac nous   The  URL  (universal  (also  uniform)  resource  locator]   address is hi lp://www. wag.caltech.edu/gallery/gallery .quantum.html.
In) Download the simulations and look at them.
tli) llow do the Orbital distartions compare to those described in this chapter?
(c) What will the potential energy surface for a Diels-Alder reaction be like'.'
Id) What will you use as coordinates for a plot of the potential energy surface?
I Kl   < lei a copy of the following articles and write a three page description of the findings.
(n) Aoi/., F. J., Friedrich, B., Herrero, V. J., Rabanos, V. S., and Vcrdasco. .1. E, Effect of pendular orientation on the reactivity of H + DC1 — A quasiclassical trajectory study, Chem. Phys. Lett. 289, 132 (1998).
(li) Ansari, W. H., and Sathyamurthy, N. Classical mechanical investigation of collinear H-+H2-+H- dynamics, Chem. Phys. Lett. 289, 487 (1998).
(c) Bolton, K., Hase, W. L., Schlegel, H. B., and Song, K. A direct dynamics study of the F 4- C2H4 -*■ C2H3F + H product energy distributions, Chem. Phys. Lett. 288, 621 (1998).
(d) Bujnowski, A. M., and Pitt, W. G. Water structure around enkephalin near a PE surface — A molecular dynamics study, J. Colloid Interface Sci. 203, 47 (1998).
(e) Doubleday, C, Bolton, K., and Hase, W. L. Direct dynamics quasiclassical trajectory study of the thermal stereomutations of cyclopropane, J. Phys. Chem. 102, 3648 (1998).
(f) Kobayashi, H., Takayanagi, T., and Tsunashima, S. Studies of the N(D2) + H2 reaction on revised potential energy surfaces, Chem. Phys. Lett. 277, 20 (1997).
(g) Kumar, S., Kapoor, H., and Sathyamurthy, N. Dynamics of the Reaction He + H2+ —► HeH+ + H on the Aguado-Paniagua surface, Chem. Phys. Lett. 289, 361 (1998).
(h) Lightstone, F. C, Zheng, Y. J., and Bruice, T. C. Molecular dynamics siniu lations of ground and transition states for the S(N)2 displacement of CI- from 1,2-dichloroethane at the active site of Xanthobacter autotrophicus haloalkane dehalogenase, J. Am. Chem. Soc. 120, 5611 (1998).
(i) Simka, H., Willis, B. G., Lengyel, I., and Jensen, K. F. Computational chemistry predictions of reaction processes in organometallic vapor phase epitaxy, Prog. Crys. Growth Charact. Mater. 35, 117 (1997).
(J) Wiii.h. II U . IVslherbc, <i II. mid llnsc, W. I lui|ri loiy sliuhi's nl S(N)J nucleophillc substitution.^ Intramoleculai and i.....noleculm dynamics nl the
CI-.. . (II,Hi anil (KM, . . . lit '-Complexes. ./. Am. Clam. Sot   ll(>. ')M4
(1994).
(k) Pecina. O.. and Sehmiekler. W. On the dynamics of electrochemical ion transfer reactions, J. Electroanal. Chem. 450, 303 (IWK).
(1) Meijer, E. J., and Sprik, M. Ab initio molecular dynamics study of the react ion of water with formaldehyde in sulfuric acid solution, J. Am. Chan. Soc I 20 6345 (1998).
(m) Xie, J. Q., and Feng, J. Y. Molecular-dynamics simulation of silicon lilm growth from cluster beams, Nucl. Instrum. Methods Phvs. Res. Sec. B 142. 77 (1998).
(n) Valkealahti, S., and Manninen, M. Diffusion on aluminum-cluster surfaces anil
the cluster growth, Phys. Rev. B 57, 15533 (1998). (o) Bhatti, Q. A., and Matthai, C. C. Computer simulation of adatom dynamics on
single stepped SiC(OOl) surfaces, Thin Solid Films 318, 46 (1998). (p) Munger, E. P., Chirita, V., Sundgren, J. E., and Greene, J. E. Destabilization
and diffusion of two-dimensional close-packed Pt clusters on Pt(lll) during
film growth from the vapor phase, Thin Solid Films 318, 57 (1998). (q) Jager, H. U., and Weiler, M. Molecular dynamics studies of A-C-H film growth
by energetic hydrocarbon molecule impact, Diamond Relat. Mater. 7, 858
(1998).
(r) Yamahara, K., and Okazaki, K. Molecular dynamics simulation of the stun tural development in sol-gel process for silica systems, Fluid Phase Equilii 144, 449 (1998).
(s) Levine, S. W., Engstrom, J. R., and Clancy, P. A kinetic monte carlo study ol the growth of Si on Si(100) at varying angles of incident deposition, Surf. Sci 401, 112 (1998).
(t) Chatfield, D. C, Eurenius, K. P., and Brooks, B. R. HIV-1 protease cleavage mechanism — A theoretical investigation based on classical MD simulation and reaction path calculations using a hybrid QM/MM potential, THEOCHEM—./. Mol. Struct. 423, 79 (1998).
9
TRANSITION STATE THEORY, THE RRKM MODEL, AND RELATED
RESULTS
till CIS
in Chapter 8, we showed how one could calculate reaction rates exactly by keeping n n k of how molecules move. Often, however, it is too expensive to calculate i i 11, m rate exactly. Instead, one uses an approximation to estimate the rate of II m lion. In this chapter, we will discuss three approximations: transition state theory, ili, Kice-Ramsperger-Kassel-Marcus (RRKM) model, and phase space theory. We will i ipund on the findings in Chapter 7 and show how these approximations can be used to i innate preexponentials for reaction to reasonable accuracy.
(1 I   ARRHENIUS' MODEL AND TOLMAN'S EQUATION
I In subject that we new call reaction rate theory got its start when Arrhenius wrote a sei ies nl papers showing how an activation barrier arose in chemical reactions. We presented Vrrhenius' original derivation in Section 8.1. However, now it is useful to extend the Ht'guments to the generalized reaction:
A + BC
AB + C
(9.1 )
In the remainder of this section, we will provide a derivation of Tolman's equation for I he rate constant for reaction (9.1):
qAqBC
exp
E
(9.2)
where kA^Bc is the rate constant for reaction (9.1), kg is Boltzmann's constant, T is the absolute temperature, hP is Planck's constant, qA is the microcanonical partition function per unit volume of the reactant A, qBc is the microcanonical partition function per unit
497