ChE 452 Lecture 20 Collision Theory

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So Far This Course Has Shown

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1) How to get rate equations from experiments

2) How to get rate equations from mechanisms

3) How to predict the mechanism Next part of the course: Theory of rate constants:

Theory Of Reaction Rates Has Two Parts

Theory of PreexponentialsCollision Theory, Transition State Theory, RRKM, Molecular Dynamics

Theory of Activation BarriersPolanyi Relationship, Marcus Equation, Blowers-Masel, Quantum Methods

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Models For Preexponentials

Collision theory (old collision theory) – simple model for preexponential - ~1013/sec, ~1013Å3/sec, ~1013A6/sec

Transition state theory – slightly better model for preexponential – bimolecular (small correction to collision theory).

RRKM – better model for preexponential – unimolecular-explains rate constraints at 1018/sec

Molecular Dynamics & Tunneling – accurate method, but time consuming

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Plan For Today

Describe Arrhenius’ Model (1889) Describe Trautz and Lewis model

(1918) Show limitations

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Arrhenius Model For AB

Cold unreactive molecules Hot reactive molecules

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0 1 2 3 4 5

Velocity, cm/sec x 1000

Num

ber

Of M

olec

ules

Hot Reactive Molecules

Cold Unreactive Molecules

Figure 7.1 The Boltzmann distribution of molecular velocities.

Divides molecules into two populations

Next Derive Equation For Rate

Equilibrium:

Rate equ

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B

ΔG†-

k T1 ok =k e

(7.4)

B

ΔG†-

k T† uA AC =C e

(7.2)

Derivation Continued

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††

B B

ΔHΔS-

k k T1 ok = k e e

(7.6)

†

B

(ΔS )

k

0 ok =k e

a

B

E-k T

1 0k =k e

(7.7)

(7.8)

††† STHG

(7.4)

(7.5)

B

ΔG†-k T

1 ok =k e

Result of Arrhenius’ Model

Rate constantvaries exponentially

with T-1

No expression for Ko

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(7.8)

†

B

(ΔS )

k

0 ok =k e

Collision Theory

Assume Ko

equals the collision rate

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(7.2)

B

ΔG†-k T† u

A AC =C e

Collision Theory

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A

A

B B

C

C

AA

B B

CC

Figure 7.2 A collision between an A molecule and BC molecules.

†

B

ΔG-

k TA BC ABCr =Z e

(7.12)

(7.10)

(7.11)

†

B

ΔG-

k TreactionP =e

A BC ABC reactionr =Z P

Next: Consider Billiard Ball Collisions

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BC

A

A

BC

A

BC

A

A

A

Collisions occur whenever molecules get close

Figure 7.3 Some typical billiard ball collisions

Next: Calculate How Many Collisions Occur

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A

A

BC

bcoll

XY

LABC

AA

A

A

AA

Consider the volume swept out by a BC molecule in time to

LABC = vABC tc(7.13)

Next: Calculate How Many Collisions Occur

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# collisions

of the given

BC molecule

Volume of

cylinderCA

(7.14)

where CA is the concentration of the A molecules in the reacting mixture, measured in molecules/cm3. The volume of the cylinder is

Volume of

cylinderb ) Lcoll

2ABC

(

(7.15)

Pages Of Algebra Yields Trautz & Lewis’ Approximation

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k = v 0 A BC A BCc

(7.26)

Equation (7.26) is the key result for simple collision theory.

Derivation

Additional Assumption

Calculate the molecular velocity ignoring that molecules are hot.

Where:

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v =8 T

A BCB

ABC

1 2/

(7.27)

1 1

m

1

m mABC A B C

(7.28)

and mA, mB and mC are the masses of A, B and C in

atomic mass units (1 AMU = 1.66 10-24g).

k

Simplified Equation

In lecture 14 we showed

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v 2.52 10Å

sec

T

300K

1AMUABC

131/2

ABC

1/2

(7.29)

Example 7.A A Collision Theory Calculation

Use collision theory to calculate the preexponential for the reaction:

H+CH3CH3 H2+CH2CH3

(7.A.1)

at 500K.

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Solution:

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According to collision theory:

ABCcoll2

0 vdk (7.A.2)

Step 1: Calculate VABC

According to equation (7.26):

with

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v 2.4 10 Å / secT

300K

1AMUABC

131/2

ABC

1/2

BCA

ABC

M

1

M

11

(7.A.3)

(7.A.4)

Step 1 Continued

For reaction (7.A.1)

(7.A.5)

Substituting the numbers shows that 500K:

(7.A.6)

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A-BC1

1

1AMU

1

30AMU

0.968AMU

1/213 13

ABC

500K 1AMUv 2.52 10 3.31 10 Å/sec

300K 0.968AMU

Step 2: Estimate dcoll

Trautz’s approximation

Were dA and dB are the Van der Waals radii of A and B

Therefore

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2

ddd BA

coll

Å53d Å,51d6H2C2H ..

Å522

3.5Å Å51dcoll .

. (7.A.7)

Solution Continued

Substituting (7.A.5) and (7.A.6) into equation (7.A.2) yields:

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(7.A.8)

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13 140

2.5Å Å Åk =π 3.31 10 =6.49 10

molecule second molecule second

Discussion Problem

Use collision theory to calculate the rate constant for the reaction

F + H2 H + HF

Assume a collision diameter of 2.3Å

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Solution: Step 1 Calculate

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AMU19

1

AMU2

1

m

1

m

11

FH2

= 1.81 AMU

Step 2: Calculate v

V xAMU T

K

V xK

K

V x

2 52 101

300

2 52 101

181

300

300

187 10

131 2 1 2

131 2 1 2

13

.sec

..

.

/ /

/ /

Å

Å

sec

Å

sec

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v

v

v

Solution

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2o collk =v π(d )

ko=(41012Å/sec) ((3Å)2) = 1.1 × 1014 Å3/sec

Key Predictions Of Collision Theory

Preexponentials always between 1013 and 1014/sec for small molecules

No special configurations effects Lighter species (i.e. H atoms tend to

react faster). Larger molecules have larger cross

sections than smaller molecules

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Preexponentials Usually The Same Order As Collision Theory?

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Table 7.2 a selection of the preexponentials reported by Wesley [1980]

Reaction Preexponential

Å3/molecule Sec

Reaction Preexponential

Å3/molecule SecH+C2H6

C2H5+H2 1.6 1014 O+C2H6 OH+C2H5 2.5 1013

H+CH H2+C 1.1 1012 O+C3H8 (CH3)2CH+OH 1.4 1010

H+CH4 H2+CH3 1 1014 O2+H OH+O 1.5 1014

O+H2 OH+H 1.8 1013 OH+OH H2O+O 1 1013

O+OH O2+H 2.3 1013 OH+CH4 H2O+CH3 5 1013

O+CH4 CH3+OH 2.1 1013 OH+H2CO H2O+HCO 5 1013

O+CH3 H+CH3O 5 1013 OH+CH3

H+CH3O 1 1013

O+HCO H+CO2 5 1012 OH+CH3 H2O+CH2 1 1013

Comparisons Between Collision Theory And Experiments

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Calculated Preexponential

assuming bcoll=van Der Waals radius

Calculated Preexponential assuming bcoll=covalent radius

Experimental

Å3/molec sec Å3/molec sec Preexponential

6.2 1014 2.0 1014 1.6 1014

4 1014 2.0 1014 1.1 1012

1.9 1014 7.6 1013 2.5 1013

1.25 1014 5.8 1013 1 1013

4.0 1014 2 1014 1.5 1014

Table 7.3 Preexponentials calculated from equation (7.30) for a number of reactions compared to experimental data.

Reaction

25262 HHCHCH

H CH H C2

O C H OH C H2 6 2 5

OH OH H O+O2

H O OH O2

Cases Where Collision Theory Fails

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CH CH CH O: CH C HCH + OH3 2 3 3 3

(7.30)

ko=1.41010 Å3/molecule-sec

2O 2O O2 2 (7.31)

ko=5.81015 Å3/molecule-sec

Why Does Collision Theory Fail For Reaction 7.30

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Reaction 7.30 requires a special collision geometry:

(7.33)

(7.34)

•

3 2 3 3 3

3 2 3 2 2 3

CH CH CH +O: CH C HCH +•OH (7.32a)

CH CH CH +O: CH CH CH +•OH (7.32b)

B

S

kConfigurations = e

†

B

ΔS

k configurations which lead to reactions e =

average number of configurations of the reactants

Summary

Collision theory: reaction occurs whenever reactants collide.

 Gives correct order of magnitude or

slightly high pre-exponential Some spectacular failures TST theory after exam

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Class Question

What did you learn new today?

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