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A. Collision Theory

Overview – Collision TheoryIn Chapter 3 we presented a number of rate laws that depended on

both concentration and temperature. For the elementary reactionA + B C + D

the elementary rate law is

€

−rA = kCACB = Ae−E RT CACB

We want to provide at least a qualitative understanding as to why the rate law takes this form. We will first develop the collision rate, using collision theory for hard spheres of cross section Sr,

€

πσAB2 When all collisions occur

with the same relative velocity, UR the number of collisions between A and B molecules,

€

˜ Z AB, is

€

˜ Z AB = SrUR˜ C A ˜ C B {collisions/s/molecule]

Next we will consider a distribution of relative velocities and only consider those collisions that have an energy of EA or greater in order to react to show

€

−rA = Ae−E A RT CACB

where

€

A = πσ AB2 8k BT

πμ AB

⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

NAvo

with σAB = collision radius, kB = Boltzmann’s constant, AB = reduced mass, T = temperature and NAvo = Avogadro’s number. To obtain an estimate of EA, we use the Polyani Equation

€

E A = E Ao + γPΔHRx

Where HRx is the heat of reaction and

€

E Ao and P are the Polyani

Parameters. With these equations for A and EA we can make a first approximation to the rate law parameters without going to the lab.

1CollisionTheoryDownload.doc

I. FUNDAMENTALS OF COLLISION THEORYThe objectives of the development that follows are to give the reader insight as to why the rate laws depend on the concentration of the reacting species (i.e., –rA = kCACB) and why the temperature dependence is the form of the Arrhenius law, k=Ae–/RT. To achieve this goal we consider the reaction of two molecules in the gas phase


A + B


⏐ → ⏐


C + D

We will model these molecules as rigid spheres.

A. Rigid SpheresSpecies A and B are modeled as rigid spheres of radius σA and σB, respectively

2σA

AB

2σB

Figure PRS.3A-1 Schematic of molecules A and B

We shall define our coordinate system such that molecule B is stationary wrt molecule A so that molecule A moves towards molecule B with a relative velocity UR. Molecule A moves through space to sweep out a collision volume with a collision cross section πσAB

2 illustrated by the cylinder shown below.

UR

σAB

σAB

Figure PRS.3A-2 Schematic of collision cross section

The collision radius is σAB .


If the center of a “B” molecule comes within a distance of σAB of the center of the “A” molecule they will collide.The collision cross section of rigid spheres is Sr =πσAB

2 . As a first approximation we shall consider Sr constant. This constraint will be relaxed when we consider a distribution of relative velocities. The relative velocity between two gas molecules A and B is UR.†

€

UR = 8k BTπμ AB

⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

(1)

wherekB = Boltzmann’s constant = 1.381 10–23 J/K/molecule

= 1.381 x 10–23 kg m2/s2/K/moleculemA = mass of a molecule of species A (gm)mB = mass of a molecule of species B (gm)

† This equation is given in most physical chemistry books, e.g., see Moore, W. J. Physical Chemistry, 2nd Ed. P.187, Prentice Hall, Englewood Cliffs, NJ


AB = reduced mass =


mAmBmA + B


(g), [Let AB]MA = Molecular weight of A (Daltons)NAvo = Avogadros’ number 6.022 molecules/molR = Ideal gas constant 8.314 J/mol•K = 8.314 kg • m2/s2/mol/K

We note that R = NAvo kB and MA = NAvo • mA, therefore we can write the ratio (kB/AB) as

€

k B

μ AB= R

MAMB

MA + MB

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

(2)


An order of magnitude of the relative velocity at 300 K is


UR ≅3000 k hr12CollisionTheoryDownload.doc

, i.e., ten times the speed of Indianapolis 500 Formula 1 car. The fo lowing c

Table PRS.3A-1 Molecular Diameters†

Molecule

Average Velocity,

(meters/second)

Molecular Diameter (Å)

H2 1687 2.74CO 453 3.12Xe 209 4.85He 1200 2.2N2 450 3.5O2 420 3.1

H2O 560 3.7C2H6 437 5.3C6H6 270 3.5CH4 593 4.1NH3 518 4.4H2S 412 4.7CO2 361 4.6N2O 361 4.7NO 437 3.7

Consider a molecule A moving in space. In a time t the volume V swept out by a molecule of A is

† O’Hanlon, J. F., A User’s Guide to Vacuum Technology, Wiley, New York (1980)13

CollisionTheoryDownload.doc

V = U Rt( )l6 7 4 8 4

πσAB2


V

A

Figure PRS.3A-3 Volume swept out by molecule A in time t

The bends in the volume represent that even though molecule A may change directions upon collision the volume sweep out is the same. The number of collisions that will take place will be equal to the


number of B molecules, V


C B18CollisionTheoryDownload.doc

, that are in the volume swept out by the “A” molecule, i.e.,

˜ C BV =No. of B oleculeσ in V[ ]

where C B is in moleculesdm3 [ ] rather than [moles/dm3]In a time t the number of collisions of this one A molecules with many B molecules is

€

UR˜ C Bπσ AB

2 Δt . The number of collisions of this one A molecule with all the “B” molecules per unit time is


Z 1A• B =πσAB2 ˜ C BU R


(3)

However, we have many A molecules present at a concentration

€

˜ C A, (molecule/dm3). Adding up the collisions of all the A molecules per unit volume, C A , then the number of collisions

€

˜ Z AB of all the A molecules with all B molecules per time per unit volume is

€

˜ Z AB = πσ AB2

Sr6 7 8 UR

˜ C A ˜ C B = SrUR˜ C A ˜ C B

(4)

Where Sr is the collision cross section (Å)2. Substituting for Sr and

UR

€

˜ Z AB = πσ AB2 8k BT

πμ ⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

˜ C A ˜ C B [molecules/time/volume] (5)

If we assume all collisions result in reactions then

€

− r A = ˜ Z AB = πσ AB2 8k BT

πμ ⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

˜ C A ˜ C B [molecules/time/volume] (6)

Multiplying and dividing by Avogadrósnumber, NAvo, we can put our equation for the rate of reaction in terms of the number of moles/time/vol.

€

− r ANAvo

⎛ ⎝ ⎜

⎞ ⎠ ⎟

−rA

1 2 3 NAvo = πσ AB

2 8k BTπμ

⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

˜ C A

NAvoC A

1 2 3

˜ C BNAvo

C B

1 2 3 NAvo

2

(7)

€

−rA = σ AB2 8πk BT

μ

⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

NAvo

A1 2 4 4 4 3 4 4 4

CACB [moles/time/volume] (8)

where A is the frequency factor

€

A = σ AB2 8πk BT

μ AB

⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

NAvo (9)

€

−rA = ACACB (10)


Example Calculate the frequency factor A for the reaction

€

H + O2 → OH + O

at 273K. Additional Information

Using the values in Table PRS.3A-1Collision Radii

Hydrogen H σH=2.74 Å/4 = 0.68Å = 0.68 x 10–10m

Oxygen O2

€

σO 2= 3.1

1 Å 1.55Å = 1.5 x 10–10m

€

R = 8.31J mol K = 8.314 kg m 2 s2 K mol

Solution

€

A = SrURNAvo (E-1)

€

A = πσ AB2 URNAvo = πσ AB

2 8k BTπμ

⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

NAvo (9)

The relative velocity is

€

UR = 8k BTπμ

⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

(1)

Calculate the ratio (kB/AB) (Let AB)

€

k B

μ= R

MAMB

MA + MB

= 8.314 kg ⋅m 2 s2 K mol1g mol( ) 32g mol( )1g mol + 32g mol

⎡

⎣ ⎢

⎤

⎦ ⎥

= 8.314kg• m 2 s2 K mol

0.97 g mol × 1kg1000g

€

k B

μ= 8571m 2 s2 K

Calculate the relative velocity

€

UR =8( ) 273K( ) 8571( ) m 2 s2 K

3.14

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

1 2

= 2441m s = 2.44 ×1013 Å s (E-2)

€

Sr = πσ AB2 = π σ A + σ B[ ]

2 = π 0.68 ×10−10 m +1.55 ×10−10 m[ ]2

=15.6 ×10−20 m 2 molecule

Calculate the frequency factor A

€

A = 15.6 ×10−20 m 2

molecule2441m s[ ] 6.02 ×1023 molecule mol[ ] (E-3)


€

A = 2.29 ×108 m 3

mol⋅s= 2.29 ×1011 dm3

mol⋅s(E-4)

€

A = 2.29 ×106 m 3

mol⋅s× 1mol

6.02 ×1023 molecule1010 Å

m

⎛

⎝ ⎜

⎞

⎠ ⎟3

€

A = 3.81×1014 Å( )3

molecule s (E-5)

The value reported in Masel† from Wesley is

€

A =1.5 ×1014 Å( )3

molecule s

Close, but no cigar as Groucho Marx would say.

For many simple reaction molecules the calculated frequency factor, Acalc is in good agreement with experiment. For other reactions, Acalc, can be an order of magnitude too high or too low. In general, collision theory tends to over predict the frequency factor A

€

108 dm 3

mol• s< Acalc <1011 dm3

mol• s

Terms of cubic angstroms per molecule per second the frequency factor is

€

1012 Å3 molecule s < Acalc <1015 Å3 molecule s

There are a couple of things that are troublesome about the rate of reaction given by Equation (10), i.e.

€

−rA = ACACB (10)

First and most obvious is the temperature dependence. A is proportional to the square root of temperature and so therefore is –rA, i.e.

€

−rA ~ A ~ T

However we know that the temperature dependence of the rate of chemical reaction on temperature is given by the Arrhenius equation

€

−rA = Ae−E RT CACB (11)or

€

k = Ae−E RT (12)This shortcoming of collision theory along with the assumption that all collisions result in reaction will be discussed next.

† Masel, R. I. Chemical Kinetics and Catalysis, p367, Wiley, New York (2001).24


II. SHORTCOMINGS OF COLLISION THEORY

A. The collision theory outlined above does not account for orientation of the collision, front to back and along the line-of-centers. That is, molecules need to collide in the correct orientation for reaction to occur. Figure PRS.3A-4 shows molecules colliding whose centers are off set by a distance b.

b

B

b = impact parameter

A

Figure PRS.3A-4 Grazing collisions

B. Collision theory does not explain activation barriers. Activation barriers occur because bonds need to be stretched or distorted in order to react and these processes require energy. Molecules must overcome electron-electron repulsion in order to come close together†

C. The collision theory does explain the observed temperature dependence given by Arrhenius Equation

€

k = Ae−E RT

D. Collision theory assumes all A molecules have the same relative velocity, the average one.

€

UR = 8k BTπμ AB

⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

(1)

However there is a distribution of velocities f(U,T). One distribution most used is the Maxwell-Boltzmann distribution.

III. MODIFICATIONS OF COLLISION THEORYWe are now going to account for the fact that we have (1) a distribution of relative velocities UR and (2) that not all collisions result in a reaction, only those collisions with an energy EA or greater. The goal is to arrive at

€

k = Ae−E A RT

† Masel, R. I., lit cit25


A. Distribution of VelocitiesWe will use the Maxwell-Boltzmann Distribution of Molecular Velocities (A6p.26). For a species of mass m, the Maxwell distribution of velocities (relative velocities) is

€

f U,T( )dU = 4π m2πk BT

⎛ ⎝ ⎜

⎞ ⎠ ⎟3 2

e−mU2 2kBT U2dU (13)

A plot of the distribution function, f(U,T), is shown as a function of U in Figure PRS.3A-5.

€

f U,T( )

fT1T2T2 > T1U

Figure PRS.3A-5 Maxwell-Boltzmann distribution of velocities

Replacing m by the reduced mass of two molecules A and B.

€

f U,T( )dU = 4π μ2πk BT

⎛ ⎝ ⎜

⎞ ⎠ ⎟3 2

e−μU 2 2kBT U2dU

The term on the left hand side of Equation (13), [f(U,T)dU] is the fraction of A molecules with relative velocities between U and (U + dU). Recall from Equation (4) that the number of A–B collisions for a reaction cross section Sr is

€

˜ Z AB = Sr U( )U˜ k U( )

1 2 3 ˜ C A ˜ C B (14)

except now the collision cross section is a function of the relative velocity.

Note we have written the collision cross section Sr as a function of velocity U, i.e., Sr(U). Why does the velocity enter into reaction cross section, Sr? Because not all collisions are head on, and those that are not, will not react if the energy, (i.e., U2/2), is not sufficiently high. Consequently, this functionality, Sr = Sr(U), is reasonable because if two molecules collide with a very very low relative velocity it is unlikely that such a small transfer of kinetic energy is likely to activate the internal vibrations of the molecule to cause the breaking of bonds. On the other hand for collisions with large relative velocities most collision will result in reaction.

We now let k(U) be the specific reaction rate for a collision and reaction of A-B velocities with a velocity U.

€

˜ k U( ) = Sr U( )U m 3 molecule s[ ] (15)


Equation (15) will give the specific reaction rate and hence the reaction rate for only those collisions with velocity U. We need sum up the collisions of all velocities. We will use the Maxwell Boltzmann distribution for f(U,T) and integrate over all relative velocities.

€

˜ k T( ) = k0

∞∫ U( )f(U,T)dU = f U,T( )0

∞∫ Sr U( ) UdU (16)

Maxwell distribution function of velocities for the A/B pair of reduced mass AB is†

€

f U,T( ) = 4π μ2πk BT

⎛ ⎝ ⎜

⎞ ⎠ ⎟3 2

U2 e− μU 2

2k BT (17)

Combining Equations (16) and (17)

€

˜ k T( ) = Sr0

∞∫ U 4π μ2πk BT

⎛ ⎝ ⎜

⎞ ⎠ ⎟3 2

U2 e− μU 2

2k BT dU (18)

We let Sr=Sr(U) for brevity. We will now express the distribution function in terms of the translational energy eT.

We are now going to express the equation for

€

˜ k (T). Equation (18) in terms of kinetic energy rather than velocity. Relating the differential translational kinetic energy, e, to the velocity U.

† Moore, lit.cit p185, Atkins, P. W. Physical Chemistry, 5th ed. p.36 (1994).27


et =

U 2

2


noting and multiplying and dividing by 2 and we obtain

€

dε t = μ UdU

and hence the reaction rate

€

˜ k T( ) = 4π μ2πk BT

⎛ ⎝ ⎜

⎞ ⎠ ⎟3 2

Sr0

∞∫ 2μ

μU2

2e

−μU 2

21

kBT

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 1

μ ⎛ ⎝ ⎜

⎞ ⎠ ⎟ μUdU

dμU 2

2dε t

1 2 3

1 2 3

Simplifying

€

=4π μ2πk BT

⎛ ⎝ ⎜

⎞ ⎠ ⎟3 2

2μ 2 Sr0

∞∫ ε t e− ε t

k BT dε t

€

˜ k T( ) = 8

πμ k BT( )3

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

1 2

Sr0

∞∫ ε t e− ε t

kBT dε t m 3 s molecule[ ] (19)

€

˜ k T( ) = 8

πμ k BT( )3

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

1 2

Sr ε t( )0

∞∫ ε t e−ε t kBT dε t

Multiplying and dividing by kBT and noting

€

e t k BT( ) = E RT( ), we obtain

€

˜ k T( ) = 8k BTπμ

⎡ ⎣ ⎢

⎤ ⎦ ⎥1 2

Sr E( )0

∞∫ Ee−E RT

RTdERT ⎛ ⎝ ⎜

⎞ ⎠ ⎟ (20)

Again recall the tilde, i.e.,

€

˜ k (T) denotes that the specific reaction rate is per molecule (dm3/molecule/s). The only thing left to do is to specify the reaction cross section, Sr(E) as a function of kinetic energy E for the A/B pair of molecules.

B. Collisions that Result in ReactionWe now modify the hard sphere collision cross section to account for the fact that not all collisions result in reaction. Now we define Sr to be the reaction cross section defined as


Sr =PrπσAB2


Where Pr is the probability of reaction. In the first model we say the probability is either 0 or 1. In the second model Pr varies from 0 to 1 continuously. We will now insert each of these modules into Equation (20).

B.1 Model 1In this model we say only those hard collisions that have kinetic energy EA or greater will react. Let E et. That is, below this energy, EA, the molecules do not have sufficient energy to react so the reaction cross section is zero, Sr=0. Above this kinetic energy, all the molecules that collide react and the reaction cross section is

€

Sr = πσ AB2

€

Pr = 0∴ Sr E,T[ ] = 0 for E < E A

Pr =1 Sr E,T[ ] = πσ AB2 for E ≥ E A

€

(21)

(22)

Figure PRS.3A-6 Reaction cross section for Model 1

Integrating Equation (20) by parts for the conditions given by Equations (21) and (22) we obtain

€

˜ k = 8 k BTπμ

⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

1+ E A

RT ⎡ ⎣ ⎢

⎤ ⎦ ⎥ e−E A RT πσ AB

2 (23)

€

=UR πσ AB2 1+ E A

RT ⎡ ⎣ ⎢

⎤ ⎦ ⎥ e−E A RT Derive

Derive

€

˜ k T( ) = 8

πμ k BT( )3

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

1 2

Sro

∞∫ εT( )εT e−εT kBT dεT

€

Sr = 0 ε < ε*

Sr = πσ AB2 ε ≥ ε*


€

˜ k T( ) = 8

πμ k BT( )3

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

1 2

πσ AB2 ε tε*

0∫ e−εT kBT dεT

€

=πσ AB2 8

πμ k BT( )3

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

1 2

k BT( )2 ε T

k BT

E*

∞∫ e−εT k BT dε T

k BT

€

=πσ AB2 8k BT

πμ

⎡ ⎣ ⎢

⎤ ⎦ ⎥1 2

ERT

EA

∞∫ e−E RT dERT

€

X = εT

k BT , dX = dεT

k BT , X = E

RT , εT

k BT= E

RT

€

udv∫ = uv− vdu∫

Derive cont’d

€

˜ k T( ) = πσ AB2 8k BT

μπ ⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

Xu{E A RT

∞∫ e−XdXdv

1 2 3 = −Xe−X ⎤

⎦ ⎥ ⎥E A RT

∞

− e−XdXE A RT

∞∫

€

=πσ AB2 8k BT

μπ

⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

E A

RTe−EA RT + e−EA RT ⎡

⎣ ⎢ ⎤ ⎦ ⎥

€

=πσ AB2 8k BT

μπ

⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

E A

RT+ 1

⎡ ⎣ ⎢

⎤ ⎦ ⎥e−EA RT

€

E A

RT> 1

€

k T( ) = NAvo˜ k T( ) = πσ AB

2 8k BTμπ

⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

NAvoE A

RT

€

=A E A

RT€

Over predicts thefrequency factor

€

Sr = Pr πσ AB2

Generally

€

E A

RT>>1, so

€

˜ k = UR πσ AB2 E A

RT′ A

1 2 4 4 3 4 4 e−E A RT

Converting

€

˜ k to a per mole basis rather than a per molecular basis we have


€

′ A = E A

RT ⎛ ⎝ ⎜

⎞ ⎠ ⎟σ AB

2 π8k BTμ AB

⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

A1 2 4 4 3 4 4

NAvo = E A

RTA

€

˜ k = ′ A e−E A RT = E A

RTAe−E A RT

The good news and the bad news. This model gives the correct temperature dependence but predicted frequency factor A′ is even greater than A given by Equation (9) (which itself is often too large) by a factor (EA/RT). So we have solved one problem, the correct temperature dependence but created another problem, too large a frequency factor. Let’s try Model 2.

B.2 Model 2 Line of CentersIn this model we again assume that the colliding molecules must have an energy EA or greater to react. However, we now assume that only the kinetic energy directed along the line of centers E<<, is important. So below EA the reaction cross section is zero, Sr=0. The kinetic energy of approach of A towards B with a velocity UR is E = AB

€

UR2

( ). However, this model assumes that only the kinetic energy directly along the line of centers contributes to the reaction.

Here, as E increases above EA the number of collisions that result in reaction increases. The probability for a reaction to occur is†

€

Pr = E − E A

E ⎡ ⎣ ⎢

⎤ ⎦ ⎥ for E > E A (24)

and

€

Sr E,T( ) = 0 for E ≤ E A

Sr = πσ AB2 E − E A( )

E for E > E A

Derive

€

(25)

(26)

Derive

A UR

ULC

B

b

† Steinfeld, J. I., J. S. Francisco, and W. L. Hayes, Chemical Kinetics and Dynamics, (p250) Prentice Hall, Englewood Cliffs NJ (1989).

Masel lit.cit. p483.35


The impact parameter, b, is the off-set distance of the centers as they approach one another. The velocity component along the lines of centers, ULC, can be obtained by resolving the approach velocity into components.

At the point of collision the center of B is within the distance σAB

UR

b

σAB

θ

The energy along the line of centers can be developed by a simple geometry argumentsDerive cont’d

€

sin θ = bσ AB

(1)

The component of velocity along the line of centers

ULC = UR cos θ (2)

The kinetic energy along the line of centers is

€

E LC =ULC

2

2μ AB=

UR2

2μ ABcos2 θ = E cos2 θ (3)

€

E LC = E 1− sin 2 θ[ ] = E 1− b2

σ AB2

⎡

⎣ ⎢

⎤

⎦ ⎥ (4)

The minimum energy along the line of centers necessary for a reaction to take place, EA, corresponds to a critical value of the impact parameter, bcrit. In fact,this is a way of defining the impact parameter and corresponding reaction cross section

€

Sr = πbcrit2 (5)

Substituting for EA and bcrit in Equation (4).

€

E A = E 1− bcrit2

σ AB2

⎛

⎝ ⎜

⎞

⎠ ⎟ (6)

Solving for

€

bcrit2

€

bcr2 = σ AB

2 1− E A

E ⎛ ⎝ ⎜

⎞ ⎠ ⎟ (7)


The reaction cross section for energies of approach E > EA, is

€

Sr = πbcrit2 = πσ AB

2 1− E A

E ⎛ ⎝ ⎜

⎞ ⎠ ⎟ (8)

The complete reaction cross section for all energies E is

€

Sr = 0 E ≤ E A

Sr = πσ AB2 1− E A

E ⎛ ⎝ ⎜

⎞ ⎠ ⎟ E > E A

€

(9)

(10)

A plot of the reaction cross section as a function of the kinetic energy of approach,

€

E = μ ABUR

2

2is shown in Figure PRS.3A-7.

Model 2

Model 1Sr

EA E

€ πσAB2

Figure PRS.3A-7 Reaction cross section for Models 1 and 2

Derive

€

E ≤ E A Pr = 0 Sr = 0

E > E A Pr =1− ε*ε

Sr = πσ AB2 1− ε*

ε ⎛ ⎝ ⎜

⎞ ⎠ ⎟

Recalling Equation (19) and substituting for Sr

€

˜ k T( ) = 8

πμ kBT( )3

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

2

πσ AB2 1− ε*

ε ⎛ ⎝ ⎜

⎞ ⎠ ⎟ε*

∞∫ ε e−ε kT dε

= πσ AB2 8

πμ k BT( )3

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

1 2

kT( )3 ε

kTε*

∞∫ e−ε RT dε − ε* e−ε kTε*

∞∫ dεkT

€

ek BT

= ERT

€

˜ k T( ) = πσ AB2 8k BT

πμ ⎡ ⎣ ⎢

⎤ ⎦ ⎥1 2

E A

RT+1− E A

RT ⎡ ⎣ ⎢

⎤ ⎦ ⎥ e−E RT


€

˜ k T( ) = πσ AB2 8k BT

πμ ⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

e−E A RT

Multiplying by Avogodro’s number

€

k T( ) = ˜ k T( )NAvo

€

−rA = πσ AB2 URNAvoe−E A RT CACB (28)

Which is similar to the equation for hard sphere collisions except for the term

€

e−E A RT

€

k T( ) = πσ AB2 8πk BT

μ ⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

NAvo e−E A RT (29)

This equation gives the correct Arrhenius dependence and the correct order of magnitude for A.

€

−rA = Ae−E RT CACB (11)

Effect of Temperature on Fraction of Molecules Having Sufficient Energy to React

We now will manipulate and plot the distribution function to obtain a qualitative understanding of how temperature increases the number of reacting molecules. Figure PRS.3A-8 shows a plot the distribution function given by Equation (17) after it has been converted to an energy distribution.

We can write the Maxwell-Boltzmann distribution of velocities

€


⎛ ⎝ ⎜

⎞ ⎠ ⎟3 2

U2e−μU 2

k BT dU

in terms of energy by letting

€

e=U2

2 to obtain

€


⎛ ⎝ ⎜

⎞ ⎠ ⎟3 2

U2e− μU 2

2k BT dU

€

f ε t ,T( )dε = 2

π1 2 k BT( )3 2 ε1 2 e

− εk BT dε t

where f(e,T) de is the fraction of molecules with kinetic energies between e and (e+de). We could further multiply and divide by kBT

€

f ε,T( )dE = 2π1 2

εk BT

⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

e−E RT dεk BT

Recalling

€

ek BT

= ERT


Bingo!

€

f E,T( )dE = 2π1 2

ERT ⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

e−E RT dERT

€

f E,T( ) = 2π1 2

ERT ⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2 e−E RT

RT

Fraction of molecules with energy ΕA or reater

Ε Ε

f(Ε, Τ)

By letting

€

X = ERT

, we could have put the distribution in dimensionless form

€

f X,T( )dX = 2π1 2 X1 2e−XdX

f(X,T) dX is the fraction of molecules that have energy ratios between

€

ERT

and

€

E + dERT

€

f E,T( )dE = 2π1 2

ERT ⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

e−E RT dERT

Fraction of collisions that have EA or above

€

= 4π(RT)3

⎡ ⎣ ⎢

⎤ ⎦ ⎥E A

∞∫1 2

E1 2 e−E RT dE

This integral is shown by the shaded area on Figure PRS.3A-8

Figure PRS.3A-8 Boltzmann distribution of energies


As we just saw only those collisions that have an energy EA or greater result in reaction. We see form Figure PRS.3A-10 that the higher the temperature the greater number of collision result in reaction.

However, this Equations (9) and (11) cannot be used to calculate A for a number reactions because of steric factors and the molecular orientation upon collision need to be considered. For example, consider a collision in which the oxygen atom, O, hits the middle carbon in the reaction to form the free radical on the middle carbon atom CH3

€

˙ C HCH3

€

CH2|

O → CH3 → CH3˙ C H CH3 + • OH

|CH2

Otherwise if it hits anywhere else, say the end carbon,

€

CH3 ˙ C H CH3 will not be formed†

€

O → CH3CH2CH3 → CH3˙ C HCH3

| → • CH3CH2CH2 + • OH

Consequently, collision theory predicts a rate 2 orders of magnitude too high for the formation of CH3

€

˙ C HCH3.

IV. OTHER DEFINITIONS OF ACTIVATION ENERGYWe will only state other definition is passing, except for the energy barrier concept which will be discussed in transition state theory.

† Masel, Lit cit p.3640


A. Tolman’s Theorem Ea = E* E

* ≡e*( )


ea =

Averae Εneryof MoleculeσUnderoinReaction

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥−

Averae Εneryof ColloidinMoleculeσ

⎡

⎣ ⎢

⎤

⎦ ⎥+ 12kΤ


The average transitional energy of a reactant molecule is


32

kT45


.

B. Fowler and Guggenheim


ea =

Averae Εneryof MoleculeσUnderoinReaction

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥−

Averae Εneryof ReactantMoleculeσ

⎡

⎣ ⎢ ⎤

⎦ ⎥48


C. Energy BarrierThe energy barrier concept will be discussed in transition state theory, Ch3 Profession Reference Shelf B.

€

A+ BC → AB+ C

E

A, BC

EA

AB, C

A – B – C#

ProductsFigure PRS.3A-9 Reaction coordinate diagram

For simple reactions the energy, EA, can be estimated from computational chemistry programs such as Cerius2 or Spartan, as the heat of reaction between reactants and the transition state

€

E A = HABC# − HA

o − HBCo

V. ESTIMATION OF ACTIVATION ENERGY FROM THE POLYANI EQUATIONA. Polyani EquationThe Polyani equation correlates activation energy with heat of reaction. This correlation

€

E A = γPΔHRx + c (33)

works well for families of reactions. For the reactions

€

R + H ′ R → RH + ′ R where R = OH, H, CH3 the relationship is shown below in Figure

PRS.3A-10.


Figure PRS.3A-10 Experimental correlation of EA and HRx from Masel (Lit cit). Courtesy of Wiley.

For this family of reactions

€

E A =12 kcal mol + 0.5 ΔHR (34)

For example, when the exothermic heat of reaction is

€

HRx = −10 kcal mol

The corresponding activation energy is

€

E a = 7cal mol

To develop the Polyani equation we consider the elementary exchange reaction†

€

A+ BC → AB+ CWe consider the superposition of two attraction/repulsion potentials VBC, VAB similar to the Lennard-Jones 6-12 potential. For the molecules BC, the Lennard-Jones potential is

€

VBC = 4εLJr0

rBC

⎛ ⎝ ⎜

⎞ ⎠ ⎟12

− r0

rBC

⎛ ⎝ ⎜

⎞ ⎠ ⎟6 ⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ (35)

rBC = Distance between molecules (atoms) B and C. In addition to the Lennard-Jones 6-12 model, another model often

used is the Morse potential which has a similar shape

VBC = e−2β rBC−r0( )−2e−β rBC−r0( )

[ ] (36)

When the molecules are far apart the potential V (i.e., Energy) is zero. As they move closer together they become attracted to one another and the potential energy reaches a minimum. As they are

† after R. I. Masel Lit cit.50


brought closer together the BC molecules begin to repel each other and the potential increases. Recall that the attractive force between the B and C molecules is

€

FBC = − dVBC

dx(37)

The attractive forces between the B–C molecules are shown in Figure PRS.3A-11.

A potential similar to atoms B and C can be drawn for the atoms A and B. The F shown on the figures represents the attractive force between the molecules as they move in the distances shown by the arrows. That is the attractive force increase as we move towards the well (ro) from both directions, rAB>ro and rAB<ro.

VBC

eLJ

rBCrAB

r0

kJolecule

BCÅ

FBC FBC

rBCrAB

r0

VAB

kJmolecule

ÇAB

FAB

(a) (b)Figure PRS.3A-11 Potentials (Morse or Lennard-Jones)

One can also view the reaction coordinate as a variation of the BC distance for a fixed AC distance,

€

| A ← B → C |l6 7 4 4 4 8 4 4 4

| A BC |l6 7 4 4 4 8 4 4 4

| AB C |l6 7 4 4 4 8 4 4 4

rAB rBC

For a fixed AC distance as B moves away from C the distance of separation of B from C, rBC, increases as B moves closer to A. As rBC

increases rAB decreases and the AB energy first decreases then increases as the AB molecules become close. See point in Figure PRS.3A-11. Likewise as B moves away from A and towards C similar energy relationships are found. E.g., as B moves away from A towards C, the BC energy first decreases due to attraction then increases due to repulsion of the BC molecules as they come close together at point in Figure PRS.3A-11. The overlapping Morse potentials are shown in Figure PRS.3A-11. We now superimpose the potentials for AB and BC to form the following figure


Reference

Energy

rBC rAB

E1R

E2P

HRx=Ε2P – Ε1P

r2r1

BC S2

S1AB

Εa

rBC = rAB

r** *

Reaction Coordinate

Figure PRS.3A-12 Overlap of potentials (Morse or Lennard-Jones)Let S1 be the slope of the BC line between r1 and rBC=rAB. Starting at E1R at r1, the energy E1 at a separation distance of rBC from r1 can be calculated from the product of the slope S1 and the distance from E1R. The energy, E1, of the BC molecule at any position rBC relative to r1 is

€

E1 = E1R +S1 rBC − r1( ) (38)

€

e.g., E1 = −50kJ +10kJnm

× 5 − 2[ ]nm = −20kJ ⎡ ⎣ ⎢

⎤ ⎦ ⎥

Let S2 be the slope of the AB line between r2 and rBC=rAB. Similarly for AB, starting on the product side at E2P, the energy E2 at any position rAB relative to r2 is

€

E 2 = E 2P +S2 rAB − r2( ) (39)

€

e.g., E 2 = −80kJ + − 20kJnm

⎛ ⎝ ⎜

⎞ ⎠ ⎟× 7 −10( )nm = −20kJ

⎡ ⎣ ⎢

⎤ ⎦ ⎥

At the height of the barrier

€

E1∗ = E 2

∗ at rBC∗ = rAB

∗ (40)

Substituting for

€

E1∗ and

€

E 2∗

€

E1R +S1 rBC∗ − r1( ) = E 2P +S2 rAB

∗ − r2( ) (41)

Rearranging

€

S1 rBC∗ − r1( ) = E 2P − E1R( )

ΔH Rx

1 2 4 3 4 +S2 rAB

∗ − r2( ) (42)


€

rBC∗ = rAB

∗ = r∗

€

S1 r∗ − r1( ) = ΔHRx +S2 r∗ − r2( )

Rearranging

€

S1 −S2( )r∗ = ΔHRx +S1r1 −S2r2 (43)

Solving for r*

€

r∗ = ΔHRx

S1 −S2( )+ S1r1 −S2r2

S1 −S2(44)

Recalling Equation (43)

€

Ea = E1∗− E1R = S1 r∗− r1( ) (45)

Substituting Equation (44) for r* in Equation (45)

€

E a = S1ΔHRx

S1 −S2+ S1r1 −S2r2

S1 −S2

⎡ ⎣ ⎢

⎤ ⎦ ⎥−S1r1 (46)

Rearranging

€

E a = S1

S1 −S2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ΔHRx +S1

S1r1 −S2r2

S1 −S2− r1

⎡ ⎣ ⎢

⎤ ⎦ ⎥ (47)

€

= S1

S1 −S2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ΔHRx +S1

S1r1 −S2r2 −S1r1 +S2r1

S1 −S2

⎡ ⎣ ⎢

⎤ ⎦ ⎥ (48)

Finally, collecting terms we have

€

E a = S1

S1 −S2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

γP

1 2 4 3 4 ΔHRx + S1S2

S1 −S2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ r1 − r2[ ]

E a∗

1 2 4 4 3 4 4 (49)

Note: S1 is positive, S2 is negative, and r2 > r1; therefore Ea∗ is positive

as is P.

Ea =Εa∗ +πH Rx

Solving for r* and substituting back into the Equation Equation (8) gives the Polyani correlation relating activation energy and heat of reaction. Values of

€

E ao and P for different reactions can be

found in Table 5.4 on page 254 of Masel. One of the more common correlations for exothermic and endothermic reactions are given on page 73 of Laidler.(1) Exothermic Reactions

€

Ea = 48.1+ 0.25 ΔHRx( ) kJ mol (50)If

€

HRx = −100kJ molthen


€

E a = 48.1+ 0.25( ) −100 kJ mol( )

E a = 23.1kJ mol

(2) Endothermic Reactions

€

Ea = 48.1− 0.75 ΔHRx( ) kJ mol (51)

Szabo proposed an extension of Polyani equation

€

E a = ∑D j Breaking( ) − α ∑D j Forming( ) (52)

€

D i = Dissociation Energies

B. Marcus Extension of the Polyani EquationIn reasoning similar to developing the Polyani Equation Marcus shows (see Masel pp.584-586)

€

E A = 1+ ΔHRx

4E A0

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

E A0

C. Blowers-Masel RelationThe Polyani Equation will predict negative activation energies for highly exothermic reactions Blowers and Masel developed a relationship that compares quite well with experiment throughout the entire range of heat of reaction for the family of reactions

€

R + H ′ R → RH + ′ R as shown in Figure PRS.3A-13

Figure PRS.3A-13 Comparison of models with data. From Masel.

Courtesy of Wiley

We see the greatest agreement between theory and experiment is found with the Blowers-Masel model.


VI. CLOSUREWe have now developed a quantitative and qualitative understanding of the concentration and temperature dependence of the rate law for reactions such as

A + B C + Dwith

–rA = k CACB

We have also developed first estimates for the frequency factor, A from collision theory

€

A = SrURNAvo = πσ AB2 8k BT

πμ AB

⎛ ⎝ ⎜

⎞ ⎠ ⎟1 2

NAvo

and the activation energy, EA from the Polyani Equation

€

E A = E Ao + γPΔHRx

these calculated values for A and EA can be substituted in the rate law to determine rates of reaction.

€

−rA = kCACB = Ae−E A RT


References for Collision Theory, Transition State Theory and Molecular DynamicsP. Atkins, Physical Chemistry, 6th ed. (New York: Freeman, 1998)P. Atkins, Physical Chemistry, 5th ed. (New York: Freeman, 1994).G. D. Billing and K. V. Mikkelsen, Introduction to Molecular Dynamics and

Chemical Kinetics (New York: Wiley, 1996).P.W. Atkins, The Elements of Physical Chemistry, 2nd ed. (Oxford: Oxford

Press, 1996).K. J. Laidler, Chemical Kinetics, 3rd ed. (New York: Harper Collins, 1987).G. Odian, Principles of Polymerization, 3rd ed. (New York: Wiley 1991).R. I. Masel, Chemical Kinetics and Catalysis, Wiley Interscience, New York,

2001.

As a short hand notation we will use the following references nomenclatureA6p701 Means Atkins, P. W. Physical Chemistry, 6th ed. (1998) page 701.L3p208 Means Laidler, K. J., Chemical Kinetics, 3rd,ed. (1987) page 208.This nomenclature means that if you want background on the principle, topic, postulate, or equation being discussed, go to the specified page of the text referenced.


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