Chapter7b:(Radical(Reactions(andEnzyme(Catalysis ...

Winter 2015 Chem 350: Statistical Mechanics and Chemical Kinetics

Chapter 7b: Radical Reactions and Enzyme Catalysis and Inhibition

95

Chapter 7b: Radical Reactions and Enzyme Catalysis and Inhibition ...................................................... 95 Radical reactions, Radical Polymerization Explosions ............................................................................... 95

A Famous example: The HBr Reaction ................................................................................................... 98 Radical Chain Polymerization .............................................................................................................. 100 Radical Chain Explosions ...................................................................................................................... 102 Lindemann Mechanism for Unimolecular reactions ............................................................................ 103

Catalysis, Enzyme Catalysis and Inhibition .............................................................................................. 105 Enzyme Inhibition ................................................................................................................................ 106 Heterogeneous Catalysis ..................................................................................................................... 108

Langmuir Adsorption Isotherms .......................................................................................................... 109 Langmuir Isotherm with dissociation .................................................................................................. 110


Radical reactions, Radical Polymerization Explosions

Radical chain reactions: Let me discuss radical chain reactions by looking at a (simple) example.

4 2 3CH Cl CH Cl HCl+ → +

The proposed reaction mechanism (a set of elementary reactions) is as follows:

a) Initialization: Cl2→

f12Cl i

b) Propagation: Cl i +CH4

f2

HCl +CH3i

CH3

i +Cl2f3

CH3Cl +Cl i

c) Termination: Cl i +Cl i →

b1

Cl2 (reverse of a))

But also CH3i +CH3

i → C2H6 (not discussed in R&E).

Other reactions involving radicals are possible eg. Cl i + HCl → Cl2 + H i (unlikely, much higher energy)

Cl i +CH3Cl → HCl +CH2Cl i These would definitely complicate the reaction mechanism.



96

You can see how clever the mechanism is. Cl i produces CH3i , and CH3

i produces Cl i radicals. One radical formed can hence make a large number of product molecules. This aspect governs all reactions involving radicals. The reaction terminates if 2 radicals collide and react. (This is why anti-‐oxidants that can remove radicals can be highly effective even in small doses.) Let me set up the reaction mechanism in the usual systematic fashion

1. Cl2!

f12Cl i ; R1 = f1[Cl2]− b1[Cl]2

Cl i +CH4!

f2

HCl +CH3i ; R2 = f2[Cl i ][CH4]− b2[HCl][CH3

i ]

CH3

i +Cl2!f3

CH3Cl +Cl i; R3 = f3[CH3][Cl2]− b3[CH3Cl][Cl i]

To discuss rates of reaction one applies the steady state approximation to the intermediates

CH3i , Cl i

1)

d Cl i⎡⎣ ⎤⎦dt

= 2R1 − R2 + R3 = 0

2)

d CH3i⎡⎣ ⎤⎦

dt= R2 − R3 = 0→ R2 = R3

Using the second equation in the first we see

d Cl i⎡⎣ ⎤⎦dt

= R1 = 0

2 f1 Cl2⎡⎣ ⎤⎦ − 2b1 Cl i⎡⎣ ⎤⎦2= 0

→

Cl i⎡⎣ ⎤⎦ =f1

b1

Cl2⎡⎣ ⎤⎦⎛

⎝⎜⎞

⎠⎟

1/2

d HCl⎡⎣ ⎤⎦dt

= R2

d CH3Cl⎡⎣ ⎤⎦dt

= R3 = R2 =d HCl⎡⎣ ⎤⎦

dt

This makes sense. If the propagation steps of the mechanism dominate the rates for the increase in cocentrations for concentrations HCl⎡⎣ ⎤⎦ , CH3Cl⎡⎣ ⎤⎦ are identical. This is really due to

the fact that we don’t allow CH3i to react with other radicals in termination reactions. This

simplifies matters, but is probably not quite correct. We could easily examine the effects in Matlab simulations. (You will see this statement more often in this section!) If we neglect the backwards rate in reaction 2, we find



97

= f2 CH4⎡⎣ ⎤⎦ Cl i⎡⎣ ⎤⎦ =



= f2 CH4⎡⎣ ⎤⎦ Cl i⎡⎣ ⎤⎦ = f2

f1

b1

CH4⎡⎣ ⎤⎦ Cl2⎡⎣ ⎤⎦1/2

= keff CH4⎡⎣ ⎤⎦ Cl2⎡⎣ ⎤⎦1/2

We can examine the rate law in matlab, and examine the concentration dependence and the value of the effective rate constant. We can also investigate what happens if we include backward rates. Let me discuss some more examples. The math gets complicated quickly. C2H6 C2H4 + H2 Mechanism:

o Initiation : 1. C2H6→

f12CH3

i

2. CH3

i +C2H6→f2

CH4 +C2H5i

o Propagation:

3. C2H5

i →f3

C2H4 + H i

4. H i +C2H6→

f4

C2H5i + H2

o Termination: a) Hi + H i → H2

b=5) Hi +C2H5

i f5⎯ →⎯ C2H6

c) Hi +C2H5

i → C2H4 + H2

d) C2H5i +C2H5

i → C2H4 +C2H6

It will be clear that to include all reactions is complicated. Easy to do by computer, hard by hand. If one only considers reaction b) for termination, one obtains reactions considered in R&E (9.15).

Apply steady state to the radical intermediates:

d CH3 i⎡⎣ ⎤⎦dt

= 0 = 2 f1 C2H6⎡⎣ ⎤⎦ − f2 CH3i⎡⎣ ⎤⎦ C2H6⎡⎣ ⎤⎦

→

CH3i⎡⎣ ⎤⎦ = 2

f1

f2

(constant!)

If we would also include reverse of reactions 1 and 2 this would no longer be true!

1)

d C2H5i⎡⎣ ⎤⎦

dt= 0 = R2 − R3 + R4 − R5



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2)

d H i⎡⎣ ⎤⎦dt

= 0 = R3 − R4 − R5

Adding 1) + 2)

−2R5 + R2 = 0

−2 f5 H i⎡⎣ ⎤⎦ C2H5i⎡⎣ ⎤⎦ + f2 CH3

i⎡⎣ ⎤⎦ C2H6⎡⎣ ⎤⎦ = 0

use

CH3i⎡⎣ ⎤⎦ = 2

f1

f2

H i⎡⎣ ⎤⎦ =

f1 C2H6⎡⎣ ⎤⎦f5 C2H5

i⎡⎣ ⎤⎦

Substitute in 2)

0 = f3 C2H5

i⎡⎣ ⎤⎦ −f4 f1

f5

C2H6⎡⎣ ⎤⎦2

C2H5i⎡⎣ ⎤⎦

− f1 C2H6⎡⎣ ⎤⎦

or 0 = f3 C2H5

i⎡⎣ ⎤⎦2− f1 C2H6⎡⎣ ⎤⎦

2C2H5

i⎡⎣ ⎤⎦ −f4 f1

f5

C2H6⎡⎣ ⎤⎦

This is a quadratic equation for

C2H5i⎡⎣ ⎤⎦ , which can be solved, using the positive root, and

simplified to

C2H5i⎡⎣ ⎤⎦ = C2H6⎡⎣ ⎤⎦

f1

2 f3

+f1

2 f3

⎛

⎝⎜⎞

⎠⎟

2

+f1 f4

f3 f5

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

[ ]2 6effk C H=

d C2H4⎡⎣ ⎤⎦dt

= f3 C2H5i⎡⎣ ⎤⎦ = f3keff C2H6⎡⎣ ⎤⎦

d C2H6⎡⎣ ⎤⎦dt

= ... ( ) [ ]22 6effk C H=

→ In the end the whole reaction is first order reaction for [ ]2 6C H , just as if we would react ethane as

given in the overall reaction. Kinetics is often quite a puzzle to sort out experimentally. Here one could easily go astray. Look back at how we eventually sorted out that the reaction is first order in [ ]2 6C H

!

Again it will be clear how quickly things become complicated. A Famous example: The HBr Reaction



99

H2 + Br2 2HBr The following reaction mechanism describes the reaction

1) Br2

b1

f12Br i

2) Br i + H2

b2

f2

HBr + H i (slowest)

3) H i + Br2→

f3

HBr + Br i (fast)

4) H i + Br i →

f4

HBr

5) H i + H i →

f5H2

Reaction 3) is very fast. H radicals are consumed as soon as they are produced. This means their concentration is very small [ ] 0H ≈ . For this reason we can neglect reactions 4) and 5)

irrespective of 4 5,f f (

Br i⎡⎣ ⎤⎦ is also small).

→Apply steady state to

H i⎡⎣ ⎤⎦

d H i⎡⎣ ⎤⎦dt

= r2 − r3 = 0

0 = f2 H2⎡⎣ ⎤⎦ Br i⎡⎣ ⎤⎦ − b2 HBr⎡⎣ ⎤⎦ H i⎡⎣ ⎤⎦ − f3 H i⎡⎣ ⎤⎦ Br2⎡⎣ ⎤⎦

H i⎡⎣ ⎤⎦ =f2 H2⎡⎣ ⎤⎦ Br i⎡⎣ ⎤⎦

f3 Br2⎡⎣ ⎤⎦ + b2 HBr⎡⎣ ⎤⎦

This relation can be checked in numerical simulation with 3f large

The reactions with Br i are slower. It is somewhat dubious if one can make further approximations, by simply assuming the steady state approximation.

If we do steady state: [ ]

1 2 30 2d Br

r r rdt

= = − +

Since (see above) 2 3r r= → 1 0r =

If we would use prequilibrium: Br2 2Br i (This is already at equilibrium as reactant! We would assume equilibrium is maintained during reaction. This seems to be

a reasonable approximation. ) → 1 0r = Both reasonings lead to same result → 1 0r =

f1 Br2⎡⎣ ⎤⎦ = b1 Br i⎡⎣ ⎤⎦

2



100

Br i⎡⎣ ⎤⎦ = f1 / b1 Br2⎡⎣ ⎤⎦

1/2

Also this relation could be checked in numerical simulation For the rate of reaction we find (use the formula above for

H i⎡⎣ ⎤⎦ ):

d HBr⎡⎣ ⎤⎦dt

= f2 H2⎡⎣ ⎤⎦ Br i⎡⎣ ⎤⎦ − b2 HBr⎡⎣ ⎤⎦ H i⎡⎣ ⎤⎦ + f3 H i⎡⎣ ⎤⎦ Br2⎡⎣ ⎤⎦

= f2 H2⎡⎣ ⎤⎦ Br i⎡⎣ ⎤⎦ 1+f3 Br2⎡⎣ ⎤⎦ − b2 HBr⎡⎣ ⎤⎦( )f3 Br2⎡⎣ ⎤⎦ + b2 HBr⎡⎣ ⎤⎦( )

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= f2 H2⎡⎣ ⎤⎦ Br i⎡⎣ ⎤⎦

2 f3 Br2⎡⎣ ⎤⎦f3 Br2⎡⎣ ⎤⎦ + b2 HBr⎡⎣ ⎤⎦


=2 f2 H2⎡⎣ ⎤⎦ Br i⎡⎣ ⎤⎦

1+ b2 HBr⎡⎣ ⎤⎦ f3 Br2⎡⎣ ⎤⎦

Finally we substitute

Br i⎡⎣ ⎤⎦ = f1 / b1 Br2⎡⎣ ⎤⎦1/2


=2 f2 f1 / b1 H2⎡⎣ ⎤⎦ Br2⎡⎣ ⎤⎦

1/2

1+ b2 / f3( ) HBr⎡⎣ ⎤⎦ Br2⎡⎣ ⎤⎦( ) You can observe the integrated rate law is really very complicated for such a simple reaction. We will examine the reaction in Matlab!. It will turn out that it is crucial to have Br2 , Br i to pre-‐equilibrate (as they would in practice). It is an interesting experiment to equilibrate reagents at temperature T1 and to run the reaction at temperature T2. The computer can simulate this case as effortlessly as ever, but the above tedious derivation may no longer hold! In particular Br2 , Br i

would initially be in equilibrium at T1, but then need to reach a new equilibrium at T2. At this stage the approximations might break down. We can verify using Matlab. Radical Chain Polymerization

Radical chain reactions are a major source for polymers, for example polymerization of polyethylene

Initiation: I2 → 2I i or CH3I → CH3

i + I i

Ii +C2H4 → ICH2CH2

i

or

CH3i +C2H6 → CH3CH2CH2

i → .... Propagation:



101

ICH2CH2 C2H4( )n

CH2CH2i +C2H4 → ICH2CH2 C2H4( )n+1

CH2CH2i

The polymer grows by one unit in each reaction. The head of the polymer stays a (similar) radical Termination:

a) Pni + Pm

i → Pn+m

b) PnC2H2i + PmC2H2

i → PnCH2CH3 + PmCH = CH2

c) C2H4 + PnCH2CH2i → C2H5

i +PnCH = CH2

→ start new chain from C2H5i

Cross chain formation is also possible. A radical might attack in the middle of the chain, or extract a hydrogen atom from the middle. Since radicals are very reactive, many different products can be formed. Of course the full kinetics is complicated. The simplest schematic to analyse a rate equation is

d M i⎡⎣ ⎤⎦dt

= 0 = 2ki I⎡⎣ ⎤⎦ − 2kt M i⎡⎣ ⎤⎦2= 0

M i⎡⎣ ⎤⎦ =

kt

ki

⎛

⎝⎜⎞

⎠⎟

1/2

I⎡⎣ ⎤⎦1/2

M i radical concentration of any kind

d M i⎡⎣ ⎤⎦dt

= −kp M⎡⎣ ⎤⎦ M i⎡⎣ ⎤⎦

= −kp

kt

k1

⎛

⎝⎜⎞

⎠⎟

1/2

I⎡⎣ ⎤⎦1/2

M⎡⎣ ⎤⎦

[ ]effk M= [ ]1/2~effk I

The [ ]I concentration then determines the average chain length n. The average chain

length can be defined as the ratio of the rate of monomer consumption divided by the rate of producing new radical (or reactive) centers:

n =

d M⎡⎣ ⎤⎦ dtd I i⎡⎣ ⎤⎦ dt

=−kp M⎡⎣ ⎤⎦ M i⎡⎣ ⎤⎦

2ki I⎡⎣ ⎤⎦

[ ] [ ]

[ ]

1/21/2

2

tp

i

i

kk I Mkk I

⎛ ⎞⎜ ⎟⎝ ⎠=

[ ][ ]1/2

~k MI



102

→ decreasing [ ]I , initiator of radicals, increases the chain length. Since each I

produces 2 radicals, it scales with [ ] 1/2I −

Radical Chain Explosions A famous explosion is 2 2 22H O H O+ →

The reaction is exothermic and hence the heat produced raises the temperature and increases the rate of the reaction. This is a plausible reasoning, but heat dissipates quickly, there is another mechanistic reason for the occurrence of the explosion: branching radials or radical multiplication. Reaction mechanism:

1) H2 +O2 → 2OH i

1b) Also O2 → 2Oi seems relevant (not quoted in R&E, O atoms are present always) 2) H2 +OH i → H i + H2O

3) Hi +O2 → i O i +OH i

4) iO i + H2 → OH i + H i 3) and 4) provide additional radicals!

5) Ri +Wall → R −Wall 6) R1

i + R2i → R1R2 5) and 6) act as termination

The key to a possible explosion is that a single radical might produce 2 radicals, including O atoms which are like a bi-‐radical. Then things can get out of hand! Let’s make a very rough schematic of kinetics:

A+ B→

k1

Ri +C

Ri →

kb

nRi + P1 (branching)

Ri →

kt

P2 Here n is an integer, the “branching efficiency”

d Ri⎡⎣ ⎤⎦dt

= k1 A⎡⎣ ⎤⎦ B⎡⎣ ⎤⎦ + nkb Ri⎡⎣ ⎤⎦ − kb Ri⎡⎣ ⎤⎦ − kt Ri⎡⎣ ⎤⎦

= Γ + keff Ri⎡⎣ ⎤⎦

Now effk can be greater or less than zero:



103

(Note I do not understand this model: I can only generate two radicals from an initialization, or take away two radicals in termination. Presumably this complicates matters such that one cannot solve the model anymore (I didn’t try). This seems a real issue with kinetics models: We postulate models we can solve, not necessarily models that are realistic. Anyway.)

If 0effk < radicals gradually disappear, but if 0effk > the number of radicals grows over time, it

is like a population growth and a “radical population explosion” can occur. Let’s obtain the integrated rate law:

d Ri⎡⎣ ⎤⎦dt

= Γ + keff Ri⎡⎣ ⎤⎦

d Ri⎡⎣ ⎤⎦Γ + keff Ri⎡⎣ ⎤⎦

= dt

d Ri⎡⎣ ⎤⎦Ri⎡⎣ ⎤⎦ + Γ keff

= keff dt

ln Ri⎡⎣ ⎤⎦ + Γ / keff( ) = keff t +C

Ri⎡⎣ ⎤⎦ = Cekeff t − Γ / keff At 0t = , Ri⎡⎣ ⎤⎦ = 0

(note: what about O atoms?)

→ Ri⎡⎣ ⎤⎦ =

Γkeff

ekeff t −1( )

The model only provides the long time behavior: growth or decay (that seems reasonable from my noted criticism). We can do quite realistic simulations in Matlab, and simply monitor the rate of radical production. Then we can analyse if explosive growth does occur. It becomes a mantra in these lecures: Numerical simulations are relatively straightforward. Making logical/consistent analytical models is much harder. They probably should be tested against numerical simulations.

Lindemann Mechanism for Unimolecular reactions



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Consider A P→ . Molecules A would need excess energy (from collisions) to react. One possibility is that the reaction is really bimolecular A A P A+ → +

d A⎡⎣ ⎤⎦dt

= b1 P⎡⎣ ⎤⎦ A⎡⎣ ⎤⎦ − f1 A⎡⎣ ⎤⎦2

This would yield second order kinetics which is typically not observed Lindemann suggested an alternative mechanism

A+ A

b1

f1A*+A

2

*f

A P→

*A : vibrationally excited (for example) due to collisions

Then [ ] [ ]2 *d P

f Adt

=

[ ] [ ] [ ][ ] [ ]2

1 1 2

** * 0

d Af A b A A f A

dt= − − = (steady state)

[ ] [ ][ ]2

1

2 1

*f A

Af b A

=+

[ ] [ ]

[ ]2

2 1

2 1

d P f f Adt f b A

=+

If [ ]1 2b A f>> → first order kinetics, if [ ]A is small (low pressure) → second order kinetics

An immediate generalization of the Lindemann mechanism is that *A is formed by collisions

with other species M (e.g. air molecules).

A+ M

b1

f1A*+M

A*→

f2

P

Then

d P⎡⎣ ⎤⎦dt

=f1 f2 M⎡⎣ ⎤⎦ A⎡⎣ ⎤⎦f2 + b1 M⎡⎣ ⎤⎦

= keff A⎡⎣ ⎤⎦

If [ ]1 2b M f<< (small [ ]M ) , keff = f1 M⎡⎣ ⎤⎦ . If [ ]1 2b M f>> , 1 2

1eff

f fkb

=



105

Catalysis, Enzyme Catalysis and Inhibition Let us discuss enzyme catalysis. The kinetics of other (homogenous) catalysts are similar

S + E

b1

f1ES

ES→

f2

P + E

Enzyme E binds to reactant/substrate S, yielding the complex ES. The complex ES reacts to product and regenerates enzyme/catalyst E

d P⎡⎣ ⎤⎦dt

= f2 ES⎡⎣ ⎤⎦

[ ] [ ][ ] [ ] [ ]1 1 2 0d ES

f E S b ES f ESdt

= − − = (steady state)

[ ] [ ][ ] [ ][ ]1

1 2 m

f E S E SES

b f K= =

+ 1 2

1m

b fKf+= (Michaelis Constant)

Further inside is gained by trying to express in terms of initial concentrations [ ] [ ] [ ] [ ]oS S P ES= + + [ ] [ ] [ ]oE E ES= +

[ ] [ ] [ ] [ ]oS S P ES= − − [ ] [ ] [ ]oE E ES= −

At initial stage of reaction [ ]P

is small (and we can neglect)

Km = [E][S]

[ES]

[ ] [ ][ ] [ ] [ ]( ) [ ] [ ]( )m o oK ES E S S ES E ES= = − −

[ ][ ] [ ] [ ] [ ]( ) [ ]2o o o oS E ES S E ES= − + +



106

If we also assume [ ]ES is small

→ [ ] [ ][ ][ ] [ ]o o

m o o

S EES

K E S=

+ +

[ ] [ ][ ]

[ ] [ ]2 o o

om o o

d P f S ER

dt K E S= =

+ +

[ ] [ ][ ][ ]2

1 m o o

o o o

K E SR f S E

+ +=

Typically [ ] [ ]o oE S<<

(very little amount of catalyst or enzyme)

[ ] [ ] [ ]2 2

1 1 1m

o o o o

KR f E S f E

= +

→ if one plots 1

oR vs

[ ]1

oS

So the maximum rate is [ ]2 of E : all of the enzyme is bound into [ ]ES and reacts to products. If

[ ]oS is decreased, the rate is reduced (not all of the enzyme is bound to S )

In enzyme kinetics 1

oR vs

[ ]1

oS is referred to as a Lineweaver-‐Burk plot and is used to extract

the Michaelis constant Km Enzyme Inhibition Elementary reactions:

1) E + S ES 1 1,f b 2) ES P + E 2f (fast)



107

3) EI E + I [ ][ ][ ]I

E IK

EI=

4) ESI ES + I [ ][ ][ ]II

ES IK

ESI=

5) ESI P + EI 3f (assume slow: inhibition) EI : Inhibitor I binds to the active site → competitive inhibition ESI : Inhibitor I binds to different site, but ESI does not react to P , or slowly ( 3 2f f<< )→ non competitive inhibition

Both processes can happen simultaneously (mixed inhibition). Let’s assume 3 0f ≈ Let me first state the result, and then derive. The reaction rate v satisfies

[ ]

[ ]2

/o

II I m o

f Ev

K Sα α≈

+ 1 2

1m

b fKb+= [ ]max 2 ov f E=

Or [ ]max max

1 1I mII

o

Kv v v S

αα= +

[ ]1II

IK

α = + [ ]1IIII

IK

α = + → plot 1v vs

[ ]1

oS → extract info on α I ,α II , KI , KII , Km

Derivation:

EI⎡⎣ ⎤⎦ = E⎡⎣ ⎤⎦I⎡⎣ ⎤⎦

KI

[ ] [ ] [ ]II

IESI E

K=

[ ] [ ] [ ] [ ] [ ]oE E EI ES ESI= + + +

[ ] [ ] [ ] [ ]1 1I II

I IE ES

K K⎛ ⎞ ⎛ ⎞

= + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

[ ] [ ] [ ]o I IIE E ESα α= +

Moreover: [ ][ ] [ ] [ ]1 1 2 0f E S b ES f ES− − =

[ ] [ ][ ] [ ][ ]1

2 1

oo

m

E SfES E Sf b K

≈ =+

[ ] [ ]( )[ ]o I II o mE S K Eα α= +

[ ] [ ][ ]( )o

I II o m

EE

S Kα α=

+



108

[ ] [ ][ ]( ) [ ]2

2o m

oI II o m

f S Kv f ES E

S Kα α= = ⋅

+

[ ]

[ ] [ ]2 maxo

II I m o II I m o

f E vK S K Sα α α α

= =+ +

[ ]max max

1 1I mII

o

Kv v v S

αα= +

We will explore enzyme catalysis using Matlab. Iα : Competitive → change of slope on Lineweaver-‐Burk

IIα : Non competitive → change of intercept We will see this works in a beautiful fashion, using a Matlab simulation without approximations. Heterogeneous Catalysis (a very brief glance at a complicated subject. Not discussed in class. Will not be on final exam) Many industrial processes are run using a solid surface as a catalyst, often a metal.

We first need to understand something about the adsorption of a gas on a surface. In reality, adsorption occurs preferentially near an edge or a kink on the surface, as there are more atoms to bind to

Let’s first estimate how often gas phase molecules collide with a surface atom. From kinetic theory of gases (not treated in these notes)

( )

( )( )( )1/2 1/21

/2 / /

o ow

Z P PPZmkT T k M gmolπ −

= =⎡ ⎤⎣ ⎦

24 2 12.63 10oZ m s− −= ⋅ Then for 1 atmP = , 300T K= , 129M gmol−= → 27 2 13 10wZ m s− −= ⋅ . This is the number of collisions per unit area per second

21m of surface 19 2~ 10 atoms m− → 27 19 83 10 /10 ~ 3 10⋅ ⋅ collisions per atom per second The main strategy to keep a surface clean would be to reduce the pressure (eg. To 810 atm− )



109

A large array of experimental techniques exists to investigate surfaces. (Visit Tong Leung’s lab to see the action!) Two types of adsorption are distinguished:

a) Physisorption: molecules are weakly attached to the surface (~ 20 /kJ mol ) eg. By Van der Waals or electrostatic forces.

b) Chemisorption: The molecules react with the surfaces, forming chemical bonds (~ 200 /kJ mol ). For molecules to adsorb HΔ is always negative, as SΔ (gas → surface) is always less than 0 it follows 0G H T SΔ = Δ − Δ <

To understand kinetics of catalysis on surfaces, we first need to understand adsorption itself. Fractional Coverage:

number of occupied adsorption sides

total number of adsorption sidesθ =

Langmuir Adsorption Isotherms Assumptions:

o Adsorp at most one monolayer o Surface is uniform o No interaction between adsorbates

All of these assumptions are somewhat dubious

A g( ) + M surface( )

kd

ka

AM surface( )

dθdt

= ka PAN 1−θ( )− kd Nθ

,a dk k : rates of ad/desorption, N : total number of adsorption sites

AP : partial pressure of A

At equilibrium 0ddtθ =

( )1a A dk P N k Nθ θ− =

( )a A d a ak P k k P θ= +

1

a A A

d a A A

k P KPk k P KP

θ = =+ +

/a dK k k= (equilibrium constant)

Or 1AKP θ

θ=

−

Describes fractional coverage as a function of partial pressure



110

Other expression for θ , ads

M

VV

θ = (volume of adsorbed gas/full monolayer coverage)

1

ads

M

V KPV KP

=+

( )1 ads MKP V KPV+ =

Vads = KPVM − KPVads

Divide by ads MKV V P

1KVM

⋅ 1P= 1

Vads

− 1VM

→plot 1P vs

1

adsV

K can also be measured as function of T . This gives access to adsHΔ

2

ln adsHKT RT

Δ∂ =∂

(thermo)

Langmuir Isotherm with dissociation If a molecule 2A dissociates when adsorbing on the surface, we have

A2 + M 2AM

dθdt

= ka PA2N 1−θ( )( )2

− kd Nθ( )2= 0 at equilibrium

(2 empty sides)

ka PA2

N 1−θ( )( )2= kd Nθ( )2

ka PA2

= θ1−θ

⎛⎝⎜

⎞⎠⎟

2

ka PA2( )1/2

= θ1−θ



111

( )( )

1/2

1/21a A

a A

k Pk P

θ =+

1/2P dependence rather than P

Chapter7b:(Radical(Reactions(andEnzyme(Catalysis ...

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