Collision Theory

H. Scott Fogler 1/24/06

Chapter 3

Professional Reference Shelf A. Collision Theory Overview – Collision Theory In Chapter 3, we presented a number of rate laws that depended on both concentration and temperature. For the elementary reaction A+B→C+D the elementary rate law is

"rA = kC A C B = Ae "E RT C A C B We want to provide at least a qualitative understanding of why the rate law takes this form. We will first develop the collision rate, using collision theory for hard spheres of 2 cross section Sr, "#! AB . When all collisions occur with the same relative velocity, UR, the ˜ AB , is number of collisions between A and B molecules, Z

˜ AC ˜B Z˜ AB = Sr U R C

[collisions/s/molecule]

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

!

"rA = Ae "E A

RT

C AC B

where

!

A=

'1 2

$

8k T "# 2AB & B ) % "µ AB (

N Avo

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

E A = E oA + " P #H Rx Where ΔHRx is the heat of reaction and E oA and γP are the Polyani Parameters. With these equations for A and EA we can make a first approximation to the rate law ! to the lab. parameters without going

!

1 CD/CollisionTheory/ProfRef.doc

HOT BUTTONS I. Fundamentals of Collision Theory II. Shortcomings of Collision Theory III. Modifications of Collision Theory A. Distribution of Velocities B. Collisions That Result in Reaction 1. Model 1 Pr = 0 or 1 2. Model 2 Pr = 0 or Pr = (E–EA)/E IV. Other Definitions of Activation Energy * A. Tolman’s Theorem Ea = E B. Fowler and Guggenheim C. Energy Barrier V. Estimation of Activation Energy from the Polyani Equation A. Polyani Equation B. Marcus Extension of the Polyani Equation C. Blowers-Masel Relation VI. Closure References for Collision Theory, Transition State Theory, and Molecular Dynamics P. 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 (New York: Wiley Interscience, 2001). As a shorthand notation, we will use the following references nomenclature: A6p701 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 referenced text.

I.

FUNDAMENTALS OF COLLISION THEORY The objective of this development is to give the reader insight into why the rate laws depend on the concentration of the reacting species (i.e., –rA = kCACB) and –/RT why the temperature dependence is in the form of the Arrhenius law, k=Ae . To achieve this goal, we consider the reaction of two molecules in the gas phase !" C + D A+B !

We will model these molecules as rigid spheres.


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

A B 2!A

2!B

Figure R3.A-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, !" 2AB , illustrated by the cylinder shown in Figure R3.A-2.

!AB

!AB UR Figure R3.A-2 Schematic of collision cross-section.

The collision radius is ! AB .

! AB = ! A + ! B If the center of a “B” molecule comes within a distance of ! AB of the center of the “A” molecule they will collide. The collision cross section of rigid spheres is Sr = !" 2AB . 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 gas molecules A and B is UR.†

# 8k B T &1 2 UR = % ( $ "µ AB ' where kB = Boltzmann’s constant = 1.381 × 10 = 1.381 kg m2/s2/K/molecule

–23

(R3.A-1) J/K/molecule

!

†

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


mA = mass of a molecule of species A (gm) mB = mass of a molecule of species B (gm) m Am B µAB = reduced mass = (g), [Let µ ≡ µAB] mA + mB MA = Molecular weight of A (Daltons) NAvo = Avogadro’s number 6.022 molecules/mol 2 2 R = Ideal gas constant 8.314 J/mol•K = 8.314 kg • m /s /mol/K We note that R = NAvo kB and MA = NAvo • mA, therefore we can write the ratio (kB/µAB) as

kB µ AB

" $ R =$ $ M AM B $ # MA + MB

% ' ' ' ' &

(R3.A-2)

An order of magnitude of the relative velocity at 300 K is U R ! 3000 km hr , i.e., ten times the speed of an Indianapolis 500 Formula 1 car. The collision diameter and velocities at 0°C are given in Table R3.A-1.

!

Table R3.A-1 Molecular Diameters† Molecule

Average Velocity, (meters/second)

Molecular Diameter (Å)

H2 CO Xe He N2 O2 H2 O C2H6

1687 453 209 1200 450 420 560 437

2.74 3.12 4.85 2.2 3.5 3.1 3.7 5.3

C6H6 CH4 NH3 H2S CO2 N2 O NO

270 593 518 412 361 361 437

3.5 4.1 4.4 4.7 4.6 4.7 3.7

Consider a molecule A moving in space. In a time Δt, the volume ΔV swept out by a molecule of A is

†

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


!l4 6 47 8 !V = (U R !t )"#2AB

!V A

Figure R3.A-3 Volume swept out by molecule A in time Δt.

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, ˜ B , that are in the volume swept out by the A molecule: ΔV C

[ C˜ B!V = No. of B molecules in !V] ˜ B is in molecules dm 3 rather than [moles/dm3] where C

[

]

In a time Δt, the number of collisions of this one A molecule with many B ˜ B "# 2AB$t . The number of collisions of this one A molecule molecules is U R C with all the B molecules per unit time is

!

˜ 1A• B = !"2AB C ˜ BU R Z

(R3.A-3)

! ˜ A, However, we have many A molecules present at a concentration, C 3 (molecule/dm ). Adding up the collisions of all the A molecules per unit ˜ AB of all the A molecules with all ˜ A , then the number of collisions Z volume, C B molecules per time per unit volume is Sr 67 8 2 ˜ C ˜ = Sr U R C ˜ AC ˜B Z˜ AB = "# AB U R C ! A B

!

(R3.A-4)

2

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

$ '1 2 ! ˜Z AB = "# 2AB & 8k B T ) C ˜ AC ˜ B [molecules/time/volume] % "µ (

(R3.A-5)

If we assume all collisions result in reactions, then 12 2 % 8k B T ( ˜ ˜ AC ˜ B [molecules/time/volume] (R3.A-6) "r˜A = Z AB = #$ AB ' * C & #µ )

!

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

!

12 ˜A C ˜B # "˜rA & C 2 # 8k B T & N 2Avo % ( N Avo = )* AB % ( N Avo ' N Avo N Avo $ )µ ' $1 12 3 123 23 CA CB "rA


!

(R3.A-7)

"rA =

% 8#k B T ( #$ 2AB ' *

12

(R3.A-8)

N Avo C A C B [moles/time/volume] µ# ) & 1444 424444 3 A

where A is the frequency factor

'1 2

$

8#k B T A = " 2AB & ) % µ AB (

!

(R3.A-9)

N Avo

(R3.A-10)

"rA = AC A C B ! Example

Calculate the frequency factor A for the reaction !

H + O 2 " OH + O at 273K. Additional information: Using the values in!Table R3.A-1 Collision Radii Hydrogen H

σH=2.74 Å/4 = 0.68Å = 0.68 x 10–10m

Oxygen O2

"O 2 =

3.1 Å 1.55Å = 1.5 x 10–10m 2

R = 8.31J mol K = 8.314 kg m 2 s 2 K mol !

Solution

(R3.A-E-1)

A = Sr U R N Avo

!

A=

"# 2ABU R N Avo

=

$ 8k T ' "# 2AB & B ) % "µ (

The relative velocity!is

!

12

N Avo

# 8k B T &1 2 UR = % ( $ "µ '

Calculate the ratio kB/µAB (Let µ ≡ µAB)

kB R 8.314 kg " m 2 s 2 K mol 8.314kg • m 2 s 2 K mol = =! = # (1g mol)(32 g mol)& M AM B 1kg µ 0.97 g mol ) % ( 1000g MA + MB $ 1g mol + 32 g mol '

!CD/CollisionTheory/ProfRef.doc

6

(R3.A-9)

(R3.A-1)

kB = 8571m 2 s 2 K µ Calculate the relative velocity

" (8)(273K )(8571) m 2 s 2 K %1 2 ! ' = 2441m s = 2.44 (10 13 Å s UR = $ 3.14 $# '& 2

[

Sr = "# 2AB = "[# A + # B ] = " 0.68 $10 %10 m +1.55 $10 %10 m

!

(R3.A-E-2)

]

2

=15.6 $10 %20 m 2 molecule Calculate the frequency factor A

! A=

15.6 "10 #20 m 2 [2441m s] 6.02 "10 23 molecule mol molecule

[

]

(R3.A-E-3)

3 m3 11 dm A = 2.29 "10 = 2.29 "10 mol# s mol # s

(R3.A-E-4)

8

!

$10 10 Å ' 3 m3 1mol A = 2.29 "10 " & ) mol# s 6.02 "10 23 molecule % m ( 6

!

A = 3.81"10 14 (Å)

3

molecule s

(R3.A-E-5)

† The value ! reported in Masel from Wesley is

!

A =1.5 "10 14 (Å)

3

molecule s

Close, but no cigar, as Groucho Marx would say.

! reaction molecules, the calculated frequency factor Acalc, is in For many simple 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 overpredict the frequency factor A

10 8

dm 3 dm 3 < A calc 1 RT

!

k (T) = N Avo k˜ (T) =

$ 8k T ' "# 2AB & B )

N Avo

% µ" (

!

=A

!

12

EA RT

EA RT

Over predicts the frequency factor

Sr = Pr "# 2AB

E Generally, A >>1, so RT !

!

! 2

U "# E k˜ = R AB A e %E A 14 4RT 244 3

RT

A$

!

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

!

12 # & # & E *8k T EA 2 B ! A" = % A () % A ( N Avo = AB $ RT ' µ AB ' RT $ 144424443 A

k = A"e #E A

RT

=

EA Ae #E A RT

RT

We have ! good news and bad news. This model gives the correct temperature dependence but predicted frequency factor A′ is even greater than A given by Equation (R3.A-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. 13 CD/CollisionTheory/ProfRef.doc

B.2 Model 2 In 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 E A

E

(R3.A-26)

(Click Back 2) UR

A

!

ULC

b

B

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 !AB

†

Courtesy of J. I. Steinfeld, J. S. Francisco, and W. L. Hayes, Chemical Kinetics and Dynamics, (Englewood Cliffs NJ: Prentice Hall, 1989, p.250); Mp483.


(Click Back 2 cont’d) The energy along the line of centers can be developed by a simple geometry argument

sin " =

b

(1)

# AB

The component of velocity along the line of centers ULC = UR cos θ

!

(2)

The kinetic energy along the line of centers is

U 2LC U 2R E LC = = cos 2 " = E cos 2 " µ AB µ AB % b2 ( E LC = E 1" sin 2 # = E'1" 2 * & $ AB )

[

]

(3) (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 ! (5) Sr = "b 2crit Substituting for EA and bcrit in Equation (4)

!

$ b2 ' E A = E &1" crit ) 2 % # AB (

(6)

$ E ' b 2cr = " 2AB &1# A ) % E (

(7)

Solving for b 2crit

!

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

% E ( Sr = "b 2crit = "# 2AB '1$ A * & E )

!

(8)

The complete reaction cross section for all energies E is

!

Sr = 0

E " EA

(9)

& E ) Sr = #$ 2AB (1% A + ' E *

E > EA

(10)

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

!

U 2R E = µ AB 2 15

CD/CollisionTheory/ProfRef.doc

!

is shown in Figure R3.A-7.

"# 2AB Model 2 Sr

!

Model 1

EA

E

Figure R3.A-7 Reaction cross section for Models 1 and 2.

Recalling Equation (20)

# &1 2 ˜k (T) = % 8k B T ( $ "µ '

ESr (E )e )E RT dE

*

+0

(RT)

(R3.A-20)

2

Substituting for Sr in Model 2

# &1 2 ˜k (T) = % 8k B T ( $ "µ '

!

") 2AB (E * E A )e *E RT dE

+

,E

(RT)

A

(R3.A-27)

2

Integrating gives

k˜ (T) =

!

$ 8k T ' "# 2AB & B ) % "µ (

12

e *E A

Derive

RT

(Click Back 3)

(Click Back 3)

!

!

E " EA

Pr = 0

Sr = 0

E > EA

Pr =1#

# &2 8 ( ") 2AB k˜ (T) = % 3 %$ "µ (k B T) ('

$ '1 2 8 ) (kT) 3 = "# 2AB & &% "µ (k B T) 3 )( !

! CD/CollisionTheory/ProfRef.doc

' E * Sr = %& 2AB )1# A , ( E +

$* $

+

*

,** kT

16

, +* / *+ kT d+ .1* 1 + e A+ 0

2

3E

e -* RT d* - **

+

,** e -* kT

d* kT

(Click Back 3 cont’d)

" k BT =

=

E RT

'1 2 $ E A E ' + 1* A ) e *E RT & RT ( % "µ ( % RT $

8k T "# 2AB & B )

!

$ 8k B T '1 2 *E 2 ˜ = "k ( T) = "# AB & ) e A % "µ (

!

RT

Multiplying both sides by NAvo

k˜ ( T) = Ae "E A

!

RT

Multiplying by Avogodro’s number

!

k (T) = k˜ (T)N Avo

"rA = #$ 2AB U R N Avoe "E A

RT

C AC B

(R3.A-28)

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

e "E A Bingo!

!

!

RT

!

k (T)

$ 8#k B T ' = " 2AB & ) %

µ

(

12

N Avo e *E A

RT

(R3.A-29)

This equation gives the correct Arrhenius dependence and the correct order of magnitude for A. (R3.A-11) "rA = Ae "E RT C A C B ! Effect of Temperature on Fraction of Molecules Having Sufficient Energy to React Now we will manipulate and plot the distribution function to obtain a ! qualitative understanding of how temperature increases the number of reacting molecules. Figure R3.A-8 shows a plot of the distribution function given by Equation (R3.A-17) after it has been converted to an energy distribution. We can write the Maxwell-Boltzmann distribution of velocities µU 2

# µ &3 2 2 ) k T f (U,T)dU = 4"% ( U e B dU $ 2"k B T '

µU 2 in terms of energy by letting " = to obtain 2 ! ! 17 CD/CollisionTheory/ProfRef.doc

Derive (Click back) µU 2

# µ & 3 2 2 ) 2k T f (U,T)dU = 4"% ( U e B dU 2"k T $ B ' f (" t ,T)d" =

!

2 #

12

(k B T)

32

"1 2 e

$

" k BT

d" t

where f(ε,T) dε is the fraction of molecules with kinetic energies between ε and (ε+dε). We could further multiply and divide by kBT. 12 2 $ " ' *E RT d" f (",T)dE = 1 2 & ) e k BT # % k BT(

! Recalling

" k BT

=

E RT 12 2 # E & dE f (E, T)dE = 1 2 % ( e )E RT RT " $ RT '

!

!

12 2 # E & e )E RT f (E, T) = 1 2 % ( RT " $ RT '

! f(!, #)

!

Fraction of molecules with energy ! " or greater

!

!

E , we could have put the distribution in dimensionless form RT 2 f (X,T)dX = 1 2 X 1 2 e #X dX " E E + dE f(X,T) dX is the fraction of molecules that have energy ratios between and ! RT RT By letting X =

!

12 2 # E & dE f (E, T)dE = 1 2 % ( e )E RT " $ RT ' !RT

Fraction of collisions that have EA or above

!

# 4 &1 2 1 2 +E RT = *E % e dE ( E A $ "(RT) 3 ' )


!

!

Derive

This integral is shown by the shaded area on Figure R3.A-8.

Figure R3.A-8 Boltzmann distribution of energies.

As we just saw, only those collisions that have an energy EA or greater result in reaction. We see from Figure R3.A-10 that the higher the temperature the greater number of collision result in reaction. However, this Equations (R3.A-9) and (R3.A-11) cannot be used to calculate A for a number pf reactions because of steric factors and because 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 ˙ HCH3 reaction to form the free radical on the middle carbon atom CH3 C

CH 2 | O " CH 3 " CH 3 C˙ H CH 3 + • ! OH | CH 2

˙ H CH 3 will not Otherwise, if it hits anywhere else (say the end carbon) CH 3 C † be formed !

O " CH 3 CH 2 CH 3 " CH 3 C˙ HCH 3 | ! CH + • OH " • CH 3 CH 2 2

Consequently, collision theory predicts a rate 2 orders of magnitude too high ˙ HCH3. for the formation of CH3 C

! IV. OTHER DEFINITIONS OF ACTIVATION ENERGY

! other definitions in passing, except for the energy barrier We will only state concept, which will be discussed in transition state theory. †

M1p.36.


(

*

*

)

* E !e A. Tolman’s Theorem Ea = E "Average Energy % "Average Energy % of Molecules 1 ! a = $ Undergoing ' ( $ of Colloiding ' + kT $ ' # Molecules & 2 Reaction # & 3 The average transitional energy of a reactant molecule is kT . 2

B. Fowler and Guggenheim

"Average Energy % "Average Energy % of Molecules ! a = $ Undergoing ' ( $ of Reactant ' $# '& # Molecules & Reaction C. Energy Barrier The energy barrier concept is discussed in transition state theory, Ch3, Profession Reference Shelf B.

A + BC " AB + C #

A–B–C E

! EA AB, C

A, BC

Products Figure R3.A-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 = H #ABC " H oA " H oBC V.

ESTIMATION OF ACTIVATION ENERGY FROM THE POLYANI EQUATION A. Polyani Equation! The Polyani equation correlates activation energy with heat of reaction. This correlation (R3.A-33) E A = " P #H Rx + c works well for families of reactions. For the reactions

R + HR" # RH + R" ! where R = OH, H, CH3, the relationship is shown in Figure R3.A-10. !

CD/CollisionTheory/ProfRef.doc

20

Figure R3.A-10 Experimental correlation of EA and ΔHRx. Courtesy of R. I. Masel, Chemical Kinetics and Catalysis (New York: Wiley Interscience, 2001).

For this family of reactions (R3.A-34)

E A =12 kcal mol + 0.5 "H R For example, when the exothermic heat of reaction is

"H Rx = #10 kcal mol !

The corresponding activation energy is

E a = 7 cal mol !

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

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

VBC

12 6*# & # & r r = 4" LJ ,% 0 ( ) % 0 ( / ,+$ rBC ' $ rBC ' /.

(R3.A-35)

where 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 = D e

!2" (rBC ! r0 )

! 2e

!"( rBC !r0 )

]

(R3.A-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 brought closer together, the BC molecules

†

After R. I. Masel (Loc cit).


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

FBC = "

dVBC dx

(R3.A-37)

The attractive forces between the B–C molecules are shown in Figure R3.A-11. A potential similar to atoms B and C can be drawn for the atoms A and B. The F shown in figures a and b represents the attractive force between the ! in the distances shown by the arrows. That is the attractive molecules as they move force increase as we move toward the well (ro) from both directions, rAB>ro and rAB