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Combustion Chemistry

Hai Wang Stanford University

2015 Princeton-CEFRC Summer School On Combustion Course Length: 3 hrs

June 22 – 26, 2015

Copyright ©2015 by Hai Wang This material is not to be sold, reproduced or distributed without prior written

permission of the owner, Hai Wang.

4-1

Lecture 4

4. Bimolecular Reaction Rate Coefficients In the last lecture, we learned qualitatively the reaction mechanisms of hydrocarbon combustion. To make this description quantitative, we will need to have a basic knowledge of reaction rate theories. While the rate coefficients of a large number of combustion reactions are experimentally measured, reaction rate theories are often necessary to interpret the experimental data. We shall focus our discussion here to bimolecular reactions of the type

A + B → C + D and leave the discussion for unimolecular reactions to a later time. 4.1 Hard Sphere Collision We discussed earlier that an elementary chemical reaction requires molecular collision. The rate coefficient of a bimolecular reaction is essentially the product of reaction probability of two reactants (A and B) upon collision at a given temperature γ(T) and the frequency of collision ZAB, k(T) [A][B] = γ(T) ZAB (4.1) Here we assume that molecules may be described by rigid spheres. Figure 4.1 shows several scenarios of molecular encounter. Starting from the head-on collision, an offset of the two axes of molecular motion leads to off-center collision. Suppose the diameter of A and B are σA and σB, respectively. The limiting off-center collision would have a spacing equal to (σA+σB)/2 between the two axes of molecular motions. It may be concluded from this simple analysis that two molecules with their axes of motions lie within a cylindrical volume of cross section equal to

Figure 4.1 Various scenarios of molecule collision.

A B

Head-on collision

A B

Head-on collision

Off-center collisionOff-center collision

Limiting off-center collision

σA σB

σAB=(σA+σB)/2

Stanford University ©Hai Wang Version 1.2

4-2

S = πrAB2 = π

σ A +σB2

!

"#

$

%&2

(4.2)

would collide with each other. Here ABr is the collision radius. Suppose an NB number of B molecules are at rest and confined to an arbitrary volume V. An A molecule travels through the volume with a mean velocity equal to vA . The cylindrical collision volume that A sweeps through per unit time is πσ AB

2 vA (see, Figure 4.2). The number density of B is NB/V. Therefore the number of collision per unit time is

ZB s−1( ) = πσ AB2 vA NBV . (4.3)

If we have NA A molecules in the same volume, the collision rate is equal to

ZAB cm−3s−1( ) = πσ AB2 vA NAV

NBV

. (4.4)

Figure 4.2 Schematic illustration of the cylindrical collision volume in a total volume V.

The above derivation is simple, but it has one problem. That is, it treats the molecules like “ghost” particles since collision does not lead to changes in the direction of motion of the A molecule. This problem is easily mended by realigning the cylindrical volume with the velocity vector of the A molecule every time it undergo scattering with a B molecule. Since the number densities of A and B are independent of the orientation of the cylindrical volume, the result is not affected by the “ghost” particle assumption. A much more important problem in the derivation comes from the assumption that the B molecules are at rest while A sweeps through the volume. Actually it is the mean, relative velocity between A and B ( vAB ) that determines the collision rate, i.e.,

NB, V

2ABps

Av


4-3

ZAB cm−3s−1( ) = πσ AB2 vAB NAV

NBV

. (4.5)

In other words, the rate coefficient of bimolecular reaction at the collision limit (i.e., γ(T) ≡ 1) may be given by

kcoll cm3mol−1s−1( ) = πσ AB2 vABNavg . (4.6)

The remaining derivation will have to be centered on the relative, mean velocity ABv . 4.2 Mean Molecular Velocity Recall from Lecture 2 that the distribution of translational energy is given by the Maxwell-Boltzmann distribution,

ρ Ei( ) =NiN

=e−Ei kBT

e−E j kBT

j∑

. (2.41)

We wish to transform the energy distribution in terms of momentum, i.e.,

E =p2

2m=px2 + p y

2 + pz2

2m (4.7)

NiN

= exp −px2 + p y

2 + pz2

2mkBT

"

#$$

%

&'' qtrans . (4.8)

Here m is the molecular mass. The ratio Ni/N is proportional to the probability density function of momentum distribution,* i.e.,

( )2 2 2

, , exp2

x y zx y z

trans B

p p pcf p p p

q mk T

⎛ ⎞+ += −⎜ ⎟⎜ ⎟⎝ ⎠

. (4.9)

where c is the normalization constant, which can be found by recognizing that the probability of a molecule having any momentum must be unity. Therefore,

f px , p y , pz( )dpx dp y dpz

0

∞

∫0

∞

∫0

∞

∫ = cqtrans exp −px2 + p y

2 + pz2

2mkBT

$

%&&

'

())dpx dp y dpz0

∞

∫0

∞

∫0

∞

∫= 1

.(4.10)

* The probability density function describes that the probability to find a molecule having momentum in the ranges of px to px+dpx, py to py+dpy, and pz to pz+dpz is f(px, py, pz) dpx dpy dpz.


4-4

The above integral gives

( )3 22

transqcmkTπ

= (4.11)

( ) ( )2 2 2

3 2

1, , exp

22x y z

x y zB

p p pf p p p

mk TmkTπ⎛ ⎞+ +

= −⎜ ⎟⎜ ⎟⎝ ⎠ . (4.12)

Recognizing that the velocity distribution function is proportional to the momentum distribution function, i.e.,

( ) ( ), , , , y zxx y z x y z x y z x y zx y z

dp dpdpf v v v dv dv dv f p p p dv dv dv

dv dv dvv⎛ ⎞⎛ ⎞⎛ ⎞

= ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠ , (4.13)

we obtain

( ) ( )2 2 23 2

, , exp2 2

x y z

x y z x y z x y zB

m v v vmf v v v dv dv dv dv dv dv

kT k Tv π

⎡ ⎤+ +⎛ ⎞ ⎢ ⎥= −⎜ ⎟⎝ ⎠ ⎢ ⎥⎣ ⎦ . (4.14)

We are, in fact, interested in the total velocity of a molecule rather than the component velocity,

2 2 2x y zv v v v= + + . (4.15)

In the spherical coordinate, we may write ( )( )sinx y zdv dv dv dv vd v dθ θ φ= . (4.16) Therefore the probability density function of the total velocity is

( ) ( )

( )

2 2 23 2 22

0 0

2 2 23 22

exp sin2 2

4 exp2 2

x y z

B

x y z

B

m v v vmf v dv dv v d d

kT k T

m v v vmv dv

kT k T

π π

θ φ θπ

ππ

⎡ ⎤+ +⎛ ⎞ ⎢ ⎥= −⎜ ⎟⎝ ⎠ ⎢ ⎥⎣ ⎦⎡ ⎤+ +⎛ ⎞ ⎢ ⎥= −⎜ ⎟⎝ ⎠ ⎢ ⎥⎣ ⎦

∫ ∫ . (4.17)

The mean velocity of a molecule is

( )0

8 Bk Tv vf v dvmπ

∞

= =∫ . (4.18)


4-5

The mean, relative velocity of two types of molecules is

2 28 81 1B B

AB A BA B

k T k Tv v v

m mπ πµ⎛ ⎞

= + = + =⎜ ⎟⎜ ⎟⎝ ⎠ , (4.19)

where µ is the reduced mass (see, p. 18 of Lecture 2 notes). It follows that the collision rate constant is

( )3 1 1 28cm mol s Bcoll AB avgk Tk Npspm-‐ -‐ = . (4.20)

Take the example of H• + O2 → products. We have σH i = 2.05 (Å); σO2 = 3.46

(Å); σH i-O2= 2.755 (Å);

µ =1×321+32

"

#$

%

&'

1

6.023×1023= 1.61×10−24 (g)

In cgs units, the collision rate coefficient is

kcoll cm3mol−1s−1( ) = 8×1.3806×10

−16

π ×1.61×10−24×π × 2.755×10−8( )

2×6.023×1023 T

At 300 K, we have kcoll = 3.7×10

14 (cm3mol–1s–1); and at 1000 K, collk = 6.7×1014

(cm3mol–1s–1). In other words, the collision rate coefficient is of the order of 1014 cm3mol–1s–1. The order of magnitude may be compared favorably with the pre-exponential factor of the experimental rate expression for reaction (3.41), H• + O2 → O• + OH• (see Figure 3.7),

k cm3 mol-s( ) = 2.65×1016T −0.671 exp − 71.3 kJ/mol( )RuT#

$%%

&

'((

.

which gives 2.65×1016T −0.671 = 5.8 and 2.6×1014 (cm3mol–1s–1) at 300 and 1000 K, respectively. Therefore, we may conclude that the pre-exponential factor is related to the collision frequency, and the exponential term is related to the reaction probability γ(T). Of course, the reaction probability is roughly determined by the fraction of molecules with combined energy greater than the activation energy Ea.


4-6

4.3 Collision of Non-Rigid Sphere All gas molecules attract each other at a long range of separations, and they repel against each other at short separations. These forces are commonly known as the van der Waals interactions and may be modeled with the Lennard-Jones (L-J) 12-6 potential function,

VvdW ,AB r( ) = 4εABσ ABr

!

"#

$

%&12

−σ ABr

!

"#

$

%&6(

)

**

+

,

--

, (4.21)

where εAB is the collision well depth,σ AB is the collision diameter, and r is the distance of intermolecular separation (see, Figure 4.3). Table 4.1 lists the L-J 12-6 potential parameters for the self collision of some typical species. For dissimilar molecules, the pair potential parameters may be estimated from the self-collision parameters

σ AB ≈σ A +σB

2 , (4.22)

εAB = εAεB , (4.23)

Of course, the force on each molecule is related to the potential energy,

F r( ) =dVvdW ,AB(r )

dr . (4.24)

Let the positions of two colliding molecules be r1 and r2. The center-of-mass of reference frame may be defined as

rc =mArA +mBrBmA +mB

. (4.25)

Figure 4.3 Schematic illustration of the Lennard-Jones 12-6 potential function.

V(r)

r

0

0

σAB

εAB


4-7

Table 4.1 Lennard-Jones 12-6 potential parameters for self collision

Species (K)Bke s (Å) Species (K)Bke s (Å) H 145 2.05 C2H2 209 4.1 H2 38 2.92 C2H4 280.8 3.971 H2O 572.4 2.605 C2H6 252.3 4.302 HO2 107.4 3.458 C3H8 266.8 4.982 O 80 2.75 C4H10 357 5.176 O2 107.4 3.458 benzene 464.8 5.29 OH 80 2.75 CO 98.1 3.65 C 71.4 3.298 CO2 244 3.763 CH 80 2.75 N2 97.53 3.621 CH2 144 3.8 He 10.2 2.576 CH3 144 3.8 Ar 136.5 3.33 CH4 141.4 3.746

This reference frame is not very convenient for our subsequent analysis since the equation of motion involves two degrees of freedom. A relative coordinate system may be defined by fixing the origin of the coordinate on particle B, thus resulting in a single degree of freedom. The trajectory of A may be schematically depicted in Figure 4.4. The relative position of A with respect to B is r = rA − rB . (4.26)

The acceleration of A relative to B may be obtained by differentiating Eq. (4.25) twice with respect to time,

r = rA −rB =FAmA

−FBmB

. (4.26)

Figure 4.4 The trajectory of a collision in the relative coordinate system.

b

b

rm θ

r θm χ

vr


4-8

By Newton’s law we have FA = −FB and

r = 1mA

+1mB

!

"##

$

%&&FA =

1µFA . (4.27)

Here we see that in this relative coordinate the motion of A is as if it follows a force field with its mass equal to the reduced mass. Because of the attractive force at the long range of separation, the effective collision cross section is larger than that of hard-sphere collision, at least for low temperatures. Let the closest distance of approach be b. This is called the impact parameter of the collision. The angle of reflection is defined by χ (see, Figure 4.4). It may be shown that the collision cross section* is

S = 2π 1− cosχ( )0

∞∫ bdb , (4.28)

where

χ = π − 2bdr

r 2 1−2VvdW ,AB r( )

µvr2

−b2

r 2rm

∞

∫ , (4.29)

vr is the relative velocity. Here we introduce the reduced collision integral as

Ω* =S

πσ AB2

. (4.30)

The corresponding collision rate constant is

kcoll cm3mol−1s−1( ) = 8kBTπµ πσ AB

2 Ω*Navg . (4.31)

Compared to the hard-sphere model, non-rigid collision introduces the reduced collision integral Ω* , a unit less quantity, that accounts for the effect of intermolecular forces on the collision cross section. Equations (4.28) and (4.29) cannot be solved analytically. The reduced collision integral is usually obtained numerically and tabulated as a function of reduced temperature * See, Mcquarrie, D. A. Statistical Mechanics, Harper & Row, New York, 1973, Chapter 16.


4-9

T * = kBT εAB . (4.32)

The Ω* may also be written in a parameterized form,

Ω* = 1.16145T *−0.14874 + 0.52487e−0.7732T*

+ 2.16178e−2.437887T*

. (4.33) A consideration of intermolecular forces causes only a small modification to the collision cross section. For H•-O2, equation (4.33) gives Ω* = 1.1 at 300 K and 0.85 at 1000 K. We see that at low temperature, the force causes the collision cross section to increase because of the long range attractive force. At high temperatures we see a net decrease in the collision cross section because the cross section equation (4.28) effectively eliminates grazing-like collision, where the momentum of the reflected molecule is hardly changed, from the total cross section. An important aspect of the collision theory is that it sets the upperlimit of rate coefficients in the range of 1013 to 1014 cm3mol–1s–1 for bimolecular reactions that does not involve the H• atom as one of the reactants, and 1014 to 1015 cm3mol–1s–1 for reactions that involve the H• atom. This fact is found to be very useful for extrapolating experiment rate coefficient data to low and high temperatures. 4.4 Potential Energy Surface We stated earlier that a chemical reaction fundamentally involves changes in the nature of chemical substances by rearranging certain chemical bonds in the reactants. It is this re-organization of chemical bonds that is responsible for the heat of reaction. The same observation may be made during the course of a chemical reaction. Consider one of the simplest reactions H2 + D• → HD + H• , (4.34) where D is deuterium, an isotope of hydrogen. The overall reaction does not release or absorb heat, yet for the reaction to proceed to the products, it must overcome an energy barrier as the old chemical bond (H–H) is being broken and a new bond (H–D) is being formed. There are three independent variables (degrees of freedom) that define the relative positions of the three atoms: the H–H distance rH-H, the H–D distance rH-D, and the H–H–D angle, γH–H–D. The landmark work of Eyring and Polanyi of 1931 showed that the minimum energy (or favorable) path follows a linear configuration as the two reactants approach each other, i.e., γH–H–D = π. The potential energy may be plotted as functions of rH-H and rH-D as shown in Figure 4.5. Here the third degree of freedom is fixed with γH–H–D = π. The lowest electronic energy path is along the bottom of the “valley” as depicted in Figure 4.5. This path is known as the


4-10

Figure 4.5 Potential energy surface (PES) of reaction (4.34). Upper-left panel shows the “trajectory” of the reaction along the valley of the PES. Upper-right panel shows the saddle area of the PES. Lower-left panel shows the contour plot of PES. The PES is computed with the Density Functional Theory (B3LYP/6-311G*).

reaction coordinate. On this coordinate, a [H…H…D]† complex forms as D• approaches H2 from infinite separation or as H• approaches HD, i.e., H2 + D• ↔ [H…H…D]† ↔ HD + H• , (4.34a) The highest point on this coordinate is a saddle point, which quantifies the minimum potential energy that the complex must overcome to proceed from reactants to products. This saddlepoint, denoted by the symbol X in the contour diagram of Figure 4.5, is sometime called the transition state, or critical geometry, depending on which textbook you read. The potential energy is more conveniently plotted by unfolding the reaction coordinate on a linear axis, as shown in Figure 4.6. The electronic energy is shown by the thick solid line. On the top of the saddle point, we added a U shaped potential energy curve to denote the motion along the axis perpendicular to the reaction coordinate. Here this motion is the H…H…D bending, i.e., the vibrational fluctuation of the angle γH–H–D. Recall that the vibrational energy is quantized and

50

100

150

200

0.70.8

0.91.0

1.11.2

1.31.4

0.7

0.8

0.9

1.0

1.1

1.2

1.3

1.4

Pote

ntia

l Ene

rgy

(kca

l/mol

)

rH-H (≈)

rH

-D (≈) 17

18

19

20

21

22

0.8

0.9

1.0

0.8

0.9

1.0

Pote

ntia

l Ene

rgy

(kca

l/mol

)

rH-D (≈)

r H-H

(≈

)

rH-H (≈)0.8 1.0 1.2 1.4 1.6

r H-D

(≈

)

0.8

1.0

1.2

1.4

1.6

6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 22.0 24.0 26.0 28.0 30.0 32.0 34.0 36.0 38.0 >40.0

X †††


4-11

Evib ,n = n +12

!

"#

$

%&hν n = 0,1,… . (2.27b)

Therefore the vibrational energy is nonzero for zero vibrational quantum number, i.e., Evib ,0 = hν 2 . This energy is known as the zero-point energy. It follows that the lowest energy path of reaction (4.34a) must consider this zero-point energy. The dashed line in Figure 4.6 is a sum of the electronic energy resulting from bond breaking/formation and the zero-point energy of H…H…D bending vibration (note that this zero-point energy is negligible for infinite separation).

Figure 4.6 Potential energy along the reaction coordinate for H2 + D• ↔ [H…H…D]† ↔ HD + H•. The solid line denotes electronic energy and the dashed line denotes the sum of electronic and zero-point energy for H…H…D bending vibration.

The reaction energy barrier E0 may now be defined as shown in Figure 4.6. This energy is sometimes called the critical energy of reaction. Of course, reaction (4.34) is a only special case since the reaction is thermally neutral (i.e., Δ o,r TH = 0). The reaction energy barrier is equal in the forward and back directions. For an arbitrary reaction of the form AB + C → [A…B…C]† → A + BC, (4.35) the potential energy may be plotted along the reaction coordinate as shown in Figure 4.7. Since A, B, and C are of arbitrary molecular size, the sum of zero-point vibrational energies, Evib ,0 = hν i 2∑ , must be included in our consideration for reactants, products, and the complex . These are denoted by the U shaped curves in Figure 4.7. Since the reaction can be endothermic (ΔHr ,T

> 0) or exothermic (ΔHr ,T < 0), the energy barriers of the forward

and back reactions are not equal. They are related by the enthalpy of reaction at 0 K, i.e.,

E0, f = E0,b +ΔHr ,0K . (4.36)

Recall that the Law of Mass Action states that the forward and back reactions are related through the equilibrium constant.

Reaction coordinate

Pot

enti

al e

nerg

y

H2+D• HD+H•

[H…H…D]‡

E0

Evib,0=hυ/2


4-12

Figure 4.7 Potential energy along the minimum energy path. Left panel: an endothermic reaction; right panel: an exothermic reaction.

Kc =k fkb

=A fAbexp −

Ea , f −Ea ,bRuT

"

#$$

%

&'' , (4.37)

For reaction 4.35, the stoichiometric coefficients Δv = 0, so

Kc = K p = exp −ΔGr ,TRuT

#

$%%

&

'((

= exp −ΔHr ,TRuT

+ΔSr ,T

Ru

#

$

%%

&

'

((

= expΔSr ,T

Ru

#

$

%%

&

'

((exp −

ΔHr ,TRuT

#

$%%

&

'((

. (4.38)

A comparison of equations (4.37) and (4.38) yields

A fAb

= expΔSr ,T

Ru

"

#

$$

%

&

'', (4.39)

Here we see that the pre-exponential factor of a reaction rate coefficient is related to standard entropy change of reaction. Since entropy is a function of temperature, the pre-exponential factor is generally a function of temperature. In addition, we find that

Reaction coordinate

Pot

enti

al e

nerg

y

AB+C

A+BC

[A…B…C]‡

E0,fΔHr,0K>0

E0,b

Reaction coordinate

Pot

enti

al e

nerg

y

AB+C

A+BC

[A…B…C]‡

E0,f

ΔH r,0K


4-13

Ea , f −Ea ,b = ΔHr ,T

= ΔHr ,0K + − hT − h0( )AB − hT − h0( )C + hT − h0( )A + hT − h0( )BC#$

%&

= ΔHr ,0K +hT − h0( )[ABC ]† − hT − h0( )AB − hT − h0( )C

#$'

%&(

− hT − h0( )[ABC ]† + hT − h0( )A + hT − h0( )BC#$'

%&(

)

*+

,+

-

.+

/+

= ΔHr ,0K +Δ hT − h0( ) f†−Δ hT − h0( )b

†

. (4.40)

Combining equations (4.36) and (4.40), we obtain

Ea , f = E0, f +Δ hT − h0( ) f

†

Ea ,b = E0,b +Δ hT − h0( )b†

, (4.41)

which again shows that the activation energy is a function of temperature. This discussion led us to conclude that the Arrhenius equation is rather empirical. The fact that it sometimes fits well to the experimental data is merely an indication that ΔSr ,T

is small; and the sensible enthalpy change Δ hT − h0( )

† is of the order of RuT, which is generally much smaller than the

energy barrier E0. 4.5 Relation Between PES and Molecular Vibration Consider the reaction (4.34a). We have a total of Na = 3 atoms in the system. If we consider that the activated complex is a linear molecule, we must have 3Na – 5 = 4 vibrational degrees of freedom. These are

Clearly the asymmetric stretch is the reaction coordinate (Figure 4.6), since it corresponds to the making of the new H–D bond and the breaking of the old H–H bond. The symmetric stretch and the 2 bending modes are normal vibrations. These motions are perpendicular to the reaction coordinate. Consider a harmonic oscillator, the Hookes law states that the force of a harmonic oscillator is

F = −dVdx

= −κx , (4.42)

where V is the potential energy, and r is the displacement from the equilibrium distance re (x = r – re). It follows that the potential energy is a parabolic function with respect to x.

symmetric stretch asymmetric stretch bending (2)

HH HH DD HH HH DDHH HH DD


4-14

V x( ) = 12κx 2 , (4.43)

Of course, we may easily solve the dynamic equation of an oscillator and show that the force constant is related to the vibrational frequency by the following equation.

ν cm−1( ) = 12πckµ

, (4.44)

where c is the speed of light and µ is the reduced mass. For an oscillator at the equilibrium geometry (x = r – re = 0), we have F = (dV/dr)r=re = 0 and V = 0. Since equation (4.42) shows that the force constant is exactly equal to the second derivative of the potential energy V, we have

ν cm−1( ) = 12πc1µd 2V

dx 2

"

#$$

%

&''x=r−re=0

1 2

(4.45)

Equation (4.45) shows that the frequency of normal vibration may be calculated if the potential function is known. The asymmetric stretch in the complex, however, evolves into the motion that is responsible for the reaction. The potential energy of this motion around an activated complex or critical geometry is roughly an inverted parabola. The top of the parabola is the saddle point that divides the reactants and products. When we apply equation (4.45) to an inverted parabola at the saddle point, we found that

ν cm−1( ) = 12πc1µd 2V

dx 2

"

#$$

%

&''x=r−r †=0

1 2

, (4.46)

(where the equilibrium distance re in equation (4.45) is replaced by the characteristic distance r† which corresponds to the top of the inverted parabola). Since the second derivative of V is negative at x = r − r † = 0 , the vibrational frequency is imaginary. In other words, the asymmetric stretch in H…H…D is abnormal vibration. A critical geometry is characterized as one that has ∂V/∂x = 0 along the coordinates of all but one vibrational motions and exactly one vibrational frequency is imaginary.. 4.6 Transition State Theory We discussed earlier that the transition state is the highest potential energy point on the reaction coordinate. The reactants (or products) at this point is called the activated complex,


4-15

X† (see, e.g., Figure 4.5). Transition State Theory starts by assuming that for an arbitrary reaction, AB + C → [X†] → A + BC, (4.35a) the transition state is in equilibrium with the reactants,

[X † ] f = K†[AB ][C ] , (4.46)

where [X † ] f represents the concentration of activated complexes moving forward to products, and †K is the equilibrium constant for the formation of X† from the reactants AB and C. Let the frequency of X† decomposition into A and BC be k†. The overall rate of forming products is † †[ ] fk X , and the rate coefficient of the overall reaction is

k = K †k† . (4.47) The above equation basically implies that in most cases, an activated complex would not undergo reaction. Rather, upon formation the complex is most likely to revert to reactants. We may analyze this, again, using the steady state approach. Consider AB + C → [X†] (k1f) [X†] → AB + C (k1b) [X†] → A + BC ( †k ) The steady state population of [X†] is

X †!"#$%&=

k1 f

k1b +k†AB!" $% C!" $% .

The assumption that [X†] is in equilibrium with the reactants implies that k† k1b so

X †!"#$%&=k1 fk1b

AB!" $% C!" $%= Kc† AB!" $% C!" $% .

Therefore, the validity of the Transition State Theory lies in the fact that k† k1b , an assumption which we shall examine at a later time. For now, we note that to a first approximation, k1b is of the order of the vibrational frequency of an asymmetric stretch, say, 3000 cm–1, which gives k1b ~ 3000 (cm–1) × 2.998×1010 (cm/s) ~ 1014 (1/s).


4-16

4.6.1 Relation between Equilibrium Constant and Partition Function In equation (4.47) we expressed the rate coefficient of an overall bimolecular reaction to be the product of two physical quantities. In this section, we shall treat the equilibrium constant in equation (4.47). Consider the Gibbs function, G ≡H −TS =U −TS + PV . (4.48) Recognizing that the internal energy is the mean energy E (see, e.g., equation 2.61) and putting equations (2.69) into Eq. (4.48), we obtain

G = E −kBTE

kBT+ ln Q

"

#$$

%

&''+ pV = −kBT ln Q + PV , (4.49)

where Q is the partition function. Using equation (2.55) and the Stirling approximation, we may re-write equation (4.49) in terms of molar Gibbs function as

g = −RuT lnqN

, (4.50)

where q is the molecular partition function, and N is the number of molecules. For 1 mole of an ideal gas in the standard state of P = 1 atm,

g = −RuT lnq

N , (4.51)

where N is basically 1 mole of molecules (6.023×1023), and q is the partition function of 1 mole of an ideal gas in its standard state of P= 1 atm*

q mol( ) = 2πmkBTh2

!

"#

$

%&

32 (1 mol)RuT

Pqvibqrot qelec( ) . (4.53)

For a hypothetical chemical reaction at equilibrium AB + C → X , (4.35b)

* Note that the pressure dependence of the molecular partition function is entirely given by the translational part of the partition function.


4-17

we have

K p = exp −ΔGr

RuT

#

$

%%

&

'

((= exp ln

qX

N − ln

qAB

N − ln

qC

N

#

$

%%

&

'

((

=qX N ( )

qAB N ( ) qC N ( )

. (4.54)

The above equation is remarkably simple. There is, however, one small problem. That is, the earlier definition of the electronic partition function is referenced to the minimal electronic energy of each molecule. In equation (4.54), we are yet to account for the difference between the minimal electronic energies (plus the zero-point energy) of the product and reactants. In other words, equation (4.54) is applicable to a reaction that has the same potential energy between the reactants and product. Consider a simple endothermic reaction in which A isomerizes to B. The potential energy difference of A and B may be depicted in Figure 4.8. As we discussed before, the reaction enthalpy at 0 K ΔHr ,0K

> 0 is basically the energy difference from the zero-point energy of B to the zero-point energy of A, i.e.,

ΔHr ,0K = E+

12i

∑ hν i ,B −12i

∑ hν i ,A$

%&

'

()

= EB −EA +12i

∑ hν i ,B −12i

∑ hν i ,A$

%&

'

()

(4.55)

Let us assume for the time being that A and B are identical in terms of rotation, vibration, and electronic degeneracy. Therefore the molecular partition functions given in Lecture 2 are also identical. We let the electronic energy EB EA. If A and B are in equilibrium, equation (4.54) states that the equilibrium constant is unity, but this cannot be true since the potential energy of B is higher than that of A, and we expect that [B] [A] at equilibrium. So what went wrong here? The answer lies largely in the assignment of the electronic partition function. Equation (2.54) states that neglecting the excited electronic states, the electronic partition function is simply the electronic degeneracy of the ground state

qelec V ,T( ) = g1 . (2.54) since the reference energy of ground-state is set to be zero. To properly consider the electronic energy difference, we need make the following modification to equation (2.54):

qelec V ,T( ) = g1 exp −E kBT( ) . (4.56)


4-18

Now if A and B do not have the same zero-point vibrational energies, we must account for this difference by factoring out the zero-point energy term from the vibrational partition function of equation (2.79), i.e.,

qvib = exp − hν ii=1α

∑ 2kBT( ) 1− exp −hν i kBT( )#$ %&−1i=1α∏ , (4.57) such that

qvibqelec = 1− exp −hν i kBT( )"# $%−1g1 exp − hν ii=1

α

∑ 2kBT( )i=1α∏ exp −E kBT( )

= 1− exp −hν i kBT( )"# $%−1g1 exp − E+

hν i2i=1

α

∑(

)*

+

,- kBT

"

#.

$

%/

i=1

α

∏

. (4.58)

It follows that

qvibqelec( )Bqvibqelec( )A

=1− exp −hν i ,B kBT( )"# $%

−1g1,Bi=1

α

∏1− exp −hν i ,A kBT( )"# $%

−1g1,Ai=1

α

∏e−ΔHr ,0K

kBT . (4.59)

We now define the vibrational partition function as

!qvib = 1− exp −hν i ,B kBT( )#$ %&−1

i=1

α

∏ , (4.59)

ε0ε1ε2ε3ε4ε5ε6

ε0ε1

ε2ε3

ε4ε5ε6

,0r KHDo

E=EB–EA

A

B

ε0ε1ε2ε3ε4ε5ε6

ε0ε1

ε2ε3

ε4ε5ε6

,0r KHDo

E=EB–EA

A

B

Figure 4.8 Potential energy differences of an isomerization reaction A ↔ B.


4-19

and let the original definition of the electronic partition function stand as is. The expression for the equilibrium constant is*

K p =!qX N ( )

!qAB N ( ) !qC N ( )

exp −ΔHr ,0K

RuT

$

%

&&

'

(

))

, (4.60)

where !q reflects the change we made to the vibrational partition function. Equation (4.60) may be written in terms of Kc (recall that P = 1 atm),

Kc = K pRuT

=!qX N RuT( )

!qAB N RuT( ) !qC N RuT( )

exp −ΔHr ,0K

RuT

$

%

&&

'

(

))

=!qX V ( )

!qAB V ( ) !qC V ( )

exp −ΔHr ,0K

RuT

$

%

&&

'

(

))

. (4.61)

We now define a molecular partition function per unit volume

z(mol/cm3 )= !q V =2πmkBT

h2"

#$

%

&'

32

!qvibqrot qelec( ) , (4.62)

and thus

Kc =zXz AzB

exp −ΔHr ,0K

RuT

#

$

%%

&

'

((

. (4.63)

* This expression may be generalized for an arbitrary reaction

aA + bB ↔ cC + dD as

K p =!qC N ( )

c!qD N ( )

d

!qA N ( )

a!qB N ( )

bexp −

ΔHr ,0K

RuT

$

%

&&

'

(

))

where ΔHr ,0K is the enthalpy of reaction at 0 K.


4-20

Since the formation of activated complex in reaction (4.35a) and reaction (4.35b) are fundamentally the same and recognizing that the enthalpy of reaction ΔHr ,0K

for AB + C → [X†] is basically the energy barrier E0, the equilibrium constant in equation (4.47) is

K † =zX †

z ABzCexp −

E0, fRuT

"

#$$

%

&''

, (4.64)

and

k = k†zX †

z ABzCexp −

E0, fRuT

"

#$$

%

&''

, (4.65)

4.6.2 Transition State Theory Formulation In section (4.5), we discussed that in the case of reaction (4.34a) or other similar reactions the asymmetric stretch is no longer a normal vibration. Let us assume that there exists a small length l centered around X† along the reaction coordinate. The frequency of X† dissociation is determined by the motion of AB and C over this small length, i.e.,

k† = v l , (4.66) where v is the mean velocity of the point mass along the reaction coordinate. Since we have assumed that X† is in thermal equilibrium with AB and C, X† must also obey Boltzmann equilibrium. Therefore the average velocity in one dimension may be written as*

v =kBT2πµ

!

"#

$

%&

1 2

, (4.67)

where µ is the reduced mass of AB and C. Assuming that µ ~ 10 amu, we have v ~ 20,000 (cm/s). Typically, the length factor is of the order of Angstroms, so k† ~1012 1 s( ) . Compared with k1b ~ 1014 (1/s), we see that the assumption k† k1b is roughly valid.

* Equation (4.14) is partitioned into each separate coordinate as, for example,

f vx( )dvx =m

2πkT

!

"#

$

%&

1 2

exp −mvx

2

2kBT

(

)*

+

,-dvx

It follows that the mean velocity in one dimension is

v = vx f vx( )dvx0

∞

∫ = m2πkT#

$%

&

'(

1 2

vx exp −mvx

2

2kBT

*

+,

-

./dvx

0

∞

∫ = kBT2πm#

$%

&

'(

1 2

.

The mean approaching velocity of reactants AB and C is therefore

v =kBT2π

!

"#

$

%&

1 21mAB

+1mC

!

"##

$

%&&

1 2

=kBT2πµ

!

"#

$

%&

1 2

.


4-21

We now wish to eliminate the unknown quantity in equation (4.66), namely, the length factor l. Consider the partition function zX † . We know that one of the vibration motion in the activated complex is no longer a normal vibration. Rather, this motion represents pseudo-translational motions of AB relative to C. The corresponding partition function is that of the motion of a particle in a one-dimensional box,

zt =2πµkBT( )

1 2

hl . (4.68)

It follows that we can write zX † as

zX † = ztzX† =

2πµkBT( )1 2

hlzX† , (4.69)

where zX

† is the partition function of the remaining 3Na–1 degrees of freedom (e.g., 3 translational, 3 rotational, and 3Na–7 vibrational for a non-linear molecule). Combining equations (4.65), (4.66), (4.67) and (4.69), we obtain

k =kBT

h

zX†

z ABzCexp −

E0, fRuT

"

#$$

%

&''

, (4.70)

To make the above equation even more general, we consider a transmission coefficient κ to quantify the probability that the complex will dissociate into product rather than reverting to reactants. Though theoretically κ can vary from 0 to 1, it frequently assumes the value of unity. The resulting Transition-State-Theory expression, or the Eyring equation, for bimolecular reaction rate coefficient is

k =κkBT

h

zX†

z ABzCexp −

E0, fRuT

"

#$$

%

&''

, (4.71)

The kBT h factor is frequency factor, and equal to 6.25×10

12 and 4.16×1013 (1/s) for T = 300 and 2000 K, respectively. The use of partition function in equation (4.71) is quite attractive, since it allows us to extend the theory to molecules of arbitrary complexity. In addition, since the partition functions of the same equation have the units of mol/cm3, the reaction rate coefficient have the units of cm3/mol-s.


4-22

4.6.3 Relation between Transition State and Collision Theory Consider the reaction of two featureless particles 1 and 2 without an appreciable energy barrier. Each reactant contains only three translational degrees of freedom. The activated complex has three translational degrees of freedom,

zX † ,T =2π m1+m2( )kBT!" #$

3 2

h3 , (4.72)

and 2 rotational degrees of freedom due to a rigid rotor,

zX † ,R =8π 2IkBT

h2

!

"##

$

%&& , (4.73)

where the moment of inertia is given by the reduced mass and the collision diameter σ12 = σ1+σ 2( ) 2 , i.e.,

I = µσ1+σ 22

!

"#

$

%&2

= µσ122 . (4.74)

Putting equations (4.72) through (4.74) into (4.71) and assuming that the transmission coefficient is unity, we obtain

k =kBT

h

2π m1+m2( )kBT!" #$3 2

h38π 2µσ12

2 kBT

h2

!

"%%

#

$&&

2πm1kBT!" #$3 2

h32πm2kBT!" #$

3 2

h3

=8kBTπµ

'

()

*

+,1 2

πσ122

. (4.75)

As expected, the above expression is the collision rate coefficient (cf, equation 4.20). Clearly the collision theory is a special case of the Transition State Theory. 4.6.4 Transition State Theory – An Example Consider the classical reaction ClO• + ClO• → Cl2 + O2. (4.76) We have the following molecular parameters from spectroscopic data:


4-23

ClO•: B = 2.04 cm–1, σ = 1, ν = 800 cm–1, g = 2, m = 8.47×10–23 g

[ (ClOi)2† ]: BX

† B y†Bz†( )1 3

= 0.676 cm–1, σ† = 2, ν † = 1500, 800, 700, 600, 200 cm–1, g † = 1.

We assume that the reaction does not have an energy barrier (E0,f = 0), since both reactants are free-radical species and are expected to be highly reactive. We expand equation (4.71) to obtain

k =κkBT

h

z(ClOi)2

††

zClOi( )2

=κkBT

h

2π 2m( )kBT!" #$3 2

h3

2πmkBT( )3

h6

%

&

'''

(

'''

)

*

'''

+

'''

π 1 2

σ †kBT

BX†

,

-..

/

011

1 2kBT

BY†

,

-..

/

011

1 2kBT

BZ†

,

-..

/

011

1 2

kBTσB

,

-.

/

012

%

&

'''

(

'''

)

*

'''

+

'''

×1− exp −hν i

† kBT( )!"4 #$5−1

i=15∏

1− exp −hν kBT( )!" #$−2

%

&''

(''

)

*''

+''

g †

g 2

=κkBTh

2

πmkBT( )3 2

σ 2

σ †g †

g 2

!

"

444

#

$

555

πB4

kBTBX† BY

† BZ†

!

"44

#

$55

1 2 1− exp −hν i† kBT( )!"4 #$5

−1

i=15∏

1− exp −hν kBT( )!" #$−2

%

&''

(''

)

*''

+''

. (4.77)

We shall now calculate the value of the above equation one term at a time:

kBTh2

πmkBT( )3 2

σ 2

σ †g †

g

=κ

T

1.3806×10−16 g-cm2s−2K−1( ) 6.6262×10−27 g -cm2s−1( )2

3.1415926×8.4710−23 g( )×1.3806×10−16 g-cm2s−2K−1( )#$% &'(3 2

12

212

= 2.15×10−13 cm3s−1K1 2( ) κT

πB4

kBTBX† BY

† BY†

!

"##

$

%&&

1 2

=3.1415926× 2.044

0.695039×0.6763

(

)**

+

,--

1 21

T= 15.92 K1 2( ) 1

T


4-24

1− e−hν kBT( )−2= 1− e−1.4388×800 T( )

−2=

1

1− e−1151 T#

$%

&

'(2

1− e−hν i† kBT"

#$

%&'−1

i=15∏ =

1

1− e−2158 T( ) 1− e−1151 T( ) 1− e−1007 T( ) 1− e−863 T( ) 1− e−287.8 T( )

Collecting all the terms, we obtain

k cm3s−1( ) = 3.43×10−12 κT1− e−1151 T( )

1− e−2158 T( ) 1− e−1007 T( ) 1− e−863 T( ) 1− e−287.8 T( )

It may be calculated that for κ = 1 , k = 2×10−14 (cm3s–1) = 1.2×1010 (cm3mol–1s–1) at T = 300 K. 4.6.5 Transition State Theory in Thermodynamic Formulation The transition state theory expression (4.71) may be re-written in terms of thermodynamic properties. Let the changes of Gibbs function, enthalpy, and entropy between the transition state and the reactants be ΔG† , ΔH † and ΔS† , respectively. These changes are known as the Gibbs free energy of activation, the enthalpy of activation, and entropy of activation, respectively. Since

ΔG† = ΔH † −TΔS† = −RuT lnK p†

= −RuT lnKc†

RuT

, (4.78)

we have the equilibrium constant of activation

Kc† =RuT exp −

ΔH † −TΔS†

RuT

#

$%%

&

'((

. (4.79)

In the above equation, the RuT is purely a term that account for the units of the equilibrium constant. What is implicit in the relation between Kc and Kp is that they are related by a standard concentration, c = 1 atm RuT . In addition, equation (4.71) may be written in terms of this equilibrium constant (cf, equations 4.71 and 4.63),


4-25

k =κkBT

hKc† =κ

kBT

hRuT( )exp −

ΔH † −TΔS†

RuT

#

$%%

&

'((

=κkBT

hRuT( )exp

ΔS†

Ru

)

*++

,

-..exp −

ΔH †

RuT

#

$%%

&

'((

. (4.81)

The derivation of the above equation seems to be a meaningless thermodynamic exercise, since the resulting equation would still require the knowledge of molecular parameters to obtain the values of enthalpy and entropy of activation. However, this exercise is useful for the following reasons. Consider the usual Arrhenius expression

k = Aexp −EaRuT

"

#$$

%

&'' . (3.37)

where Ea is an experimentally observed activation energy. Taking the derivative of lnk with respect to T yields

d lnkdT

=EaRuT

2. (4.82)

In addition, we take the same derivative from equation (4.81) to obtain

Ea =RuT2 ddT

lnT 2 + lnKc†( )

=RuT2 2T+ΔH †

RuT2

"

#

$$

%

&

''

= 2RuT +ΔH†

. (4.83)

It follows that the A-factor in equation (3.37) is

A =κkBT

hRuT( )e2 exp

ΔS†

Ru

"

#$$

%

&'' . (4.84)

The above equation states that one of the governing factors of the magnitude of A is the entropy of activation. In the case when ΔS†=0, the A factor reduces to

A =κkBT

hRuT( )e2 =κ

kBT

hce2 . (4.85)

which at 500 K is equal to 3.2×1018 cm3mol–1s–1 for κ = 1. This is the upperlimit, of course, since the association of two reactants inevitably causes the entropy of the system to


4-26

decrease, i.e., ΔS†< 0. The closer the reactants are to cross the saddle point, the tighter the transition state is and the more negative ΔS† is. Conversely, if the two reactants need not to approach each other closely at the transition state, ΔS† could be close to zero. This type of transition states is known as the loose transition state. The rate coefficient associated with the loose transition state usually has a large A factor. 4.7 Tunneling In classical mechanics, a particle having energy below the rectangular potential energy barrier V0 would be “bounced” back right at a potential energy “wall” depicted in Figure 4.9. The wave-particle duality of quantum mechanics tells us that there is a finite probability for a molecule to tunnel through the potential energy barrier. For a one-particle, one-dimensional, time independent Schrödinger equation with the potential energy V(x) depicted in Figure 4.9,

−2m

d 2

dx 2ψ x( )+V x( )ψ x( ) = Eψ x( ) , (4.86)

the solution is*

ψ x( ) =α exp −2m V −E( )"# $%

1 2

x

&

'(

)(

*

+(

,(, (4.87)

where α is a constant. The transmission coefficient for a particle tunneling through the potential barrier of thickness δ is

t =e−2 dx 2m V −E( )"# $%

1 20δ∫

1+14e−2 dx 2m V −E( )"# $%

1 20δ∫

'()

*)

+,)

-)

2. (4.88)

It may be shown that when V(x) → ∞ in the potential barrier region, t → 0 (i.e., a wave packet cannot tunnel through an infinite potential barrier). In addition, the thicker the potential barrier, the smaller the transmission efficiency. The transmission efficiency is non-zero so long as the thickness δ and the potential barrier is finite.

Figure 4.9 Tunneling of a particle through a rectangular potential energy barrier.

* Razavy, Moshen (2003). Quantum Theory of Tunneling, World Scientific.

V(x)

x

V0( )xy

Tunneling below the energy barrier

0

δV(x)

x

V0( )xy


0

V(x)

x

V0( )xy


V(x)

x

V0( )xy


0

δ


4-27

Tunneling is known to play a role in reaction rate theory. The energy barrier of a reaction may be viewed as an imperfect rectangular potential barrier between the reactants and products. The reaction energy barrier is of finite width, typically of the order of 1 Angstrom. Thus there will be a non-zero probability for the reaction to undergo tunneling, even if classically there is insufficient kinetic energy for the reactants to react. In general tunneling is unimportant at high temperatures, since there usually are a large number of energized molecules to overcome the classical energy barrier. As the temperature falls, however, the fraction of energized molecule decreases. Since the tunneling efficiency (equation 4.88) is independent of temperature, tunneling contribution to the total rate coefficient will become increasingly important as the temperature is lowered. In this case, the rate coefficient would be flattened out at the low temperature end of an Arrhenius plot (see, Figure 4.10). Equation (4.88) also shows that the transmission efficiency increases exponentially as the particle mass is decreased. Therefore lower-mass molecules (e.g., the H• atom) tunnel more effectively than larger-mass molecules. It is for this reaction that the H-abstraction reaction of the type R-H + H i → R i + H2 , (4.89) requires the consideration of tunneling.

Figure 4.10 Comparison of reaction rate coefficient with/without the tunneling effect.

4.8 Final Remarks In this lecture, we discuss the fundamental reaction rate theory pertaining to bimolecular reactions. The collision theory sets the upper bound for the reaction rate coefficient, whereas the Transition State Theory gives a more detailed account of the reaction process that considers the influence of potential energy barrier and the nature of the activated complex.

1/T

log(k)

without tunneling

with tunneling

1/T

log(k)

without tunneling

with tunneling


4-28

The Transition State Theory (i.e., equation 4.71) is the foundation of modern reaction rate theory. It predicts that the rate coefficient of a bimolecular reaction is dependent on temperature only, if there is only one energy barrier separating the reactants and products. This type of reaction is usually known as the metathesis reaction, and includes among others, the H abstraction reactions and some of the reactions that exchange atoms or molecular groups between two reactants. The theory is useful for predicting reaction rate coefficient if the energy barrier and the molecular parameters can be accurately determined. Indeed quantum chemistry methods have become increasingly viable to provide the answers to reaction kinetics. At the present time, however, some experimental data are still needed since all quantum chemistry methods have their inherent uncertainties. To this end, the Transition State Theory is extremely useful for extrapolation of rate data from a limited temperature range of, say, a few hundred Kelvin, to the entire range of temperature of interest to combustion analysis, which usually span from 300 to 3000 K. In the next lecture, we will discuss yet another class of reactions, namely, unimolecular reactions. It will be shown that the theory of unimolecular reaction is equally applicable to a class of bimolecular reactions in which the reactants and products are separated by at least two potential barriers.

combustion chemistry - princeton university · 4.1 hard sphere collision we discussed earlier that...

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