ChE 553 Lecture 23 Catalysis By Surfaces

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Objective For Today

• Ask How Surfaces Can Catalyze Reactions

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Catalysis Definition

Ostwald defined a catalyst as a substance which changed the rate of reaction without itself being consumed in the process

Not being consumed catalyst does change

Freshman Chemistry View Of Catalysts

• “Catalysts lower barriers to chemical reactions”

• Catalysts often raise barriers to elementary steps in a reaction mechanism!

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Key Mechanism Of Catalyst Action

• Stabilizing intermediates by bonding to them

• The Increase in intermediate concentration leads to a substantial increase in rate.

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Catalytic Reaction Occurs Via A Catalytic Cycle:

reactants + catalyst complex

 

complex products + catalyst

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Example: Rhodium Catalyzed CH3OH+COCH3COOH

HI

H O2

CH OH3

CH I3

CH COOH3

CH COI3[Rh(CO) I ]2 2

-

RhI

H C3C I

I

O

RhI

CH3

CI

I

O

CO

CO

CO

Figure 12.1 A schematic of the catalytic cycle for Acetic acid production via the Monsanto process.

Methyl radicals are not stable in the gas phase or solution but they are stable when bound to rhodium. The higher concentration leads to a higher rate

Example To Illustrate How Increasing Barriers Can Increase Rates

consider AAadB with the formation of B rate determining

The rate of reaction, r, is given by

r= KadkBPA/(1+KadPA)

Where Kad is the equilibrium constant for adsorption, kB is the rate constant for the production of B and PA is the pressure of A

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Note the correct rate equation is

r= KadkBPA/(1+KadPA)

I am assuming that the last term is small

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Model That I will derive In Lecture 22

• Kad = K0* exp((ΔGad+X)/RT)

Where ΔGad is free energy of adsorption on some reference surface, and X is the change in free energy in moving to some other surface

kB=ko*exp(-(Ea+0.5 X)/RT

So as you increase X, you increase the activation energy for the formation of B.

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Now Combine The Equations

r = K0* exp((ΔGad)/RT)* ko*exp(-(Ea)/RT)* exp((0.5*X)/RT)

When X=0, r=r0 the rate on the reference surface, Combining

r = r0*exp((0.5*X)/RT)

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Now Consider How Increases In X Changes The Rate

From before

kB=ko*exp(-(Ea+0.5 X)/RT

r = r0*exp((0.5*X)/RT)

Notice that when we increase X, we increase the activation barrier for B formation, yet the rate of B formation goes up!

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Physical Interpretation

• The rate is given by

• r= kB θB

• When we increase X, we increase θB and decrease kB. As long as the increases in θB are larger than the decrease in kB the net rate will increase

• Most catalysts work by increasing θB. The rate increases even though the activation barrier for the rate determining step goes up. 13

Limit To The Analysis

• θB cannot be bigger than 1

• r= kB θB = KadkBPA/(1+KadPA)

• At low coverages increases in θB dominate so the net rate will increase

• Once θB gets close to 1, θB cannot increase any more. The rate decreases with increasing X.

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Plot of actual case from lect 24

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Heat Of Formation Of Intermediate, Kcal/mole

2R

ate

, Mol

ecul

es/

Cm

S

ec

-40 -20 0 20 40 60 801E+0

1E+2

1E+4

1E+6

1E+8

1E+10

1E+12

1E+14

1E+16

Net Result Is Volcano Behavior

Heat Of Formation Of Formate

Tem

pera

ture

For

50

% C

onve

rsio

n

Au

Ag

Pt

Pd

Ir

Rh

Ru

Cu

Co

Ni Fe

W

50 60 70 80 90 100 110 120

350

400

450

500

550

600

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HCOOH HCOO(ad) Had

H(ad) HCOOad CO2 H2

(12.75)

Sabatier’s Principle

The best catalysts are substances which bind the reactants strongly, but not too strongly.

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What Types Of Reactions Will Be Increased By Metal Surfaces?• Bulk metals have many free electrons

• Free electrons rapidly exchange with radicals – radical intermediates stabilized

• Free electrons neutralize carbocations– Carbocation reactions slowed down

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A Selection Of The Reactions Catalyzed By Supported Metals

Reaction Catalyst Reaction Catalyst Hydrocarbon

Hydrogenation, Dehydrogenation

Pt, Pd, Ni CO + H2 Hydrocarbons

(Fischer-Tropsch)

Fe, Rh

CO oxidation, total oxidation of

hydrocarbons

Pt, Pd, Cu, Ni, Fe, Rh, Ru

Steam reforming for

production of hydrogen

Ni plus additives

CO + 2H2 CH3OH

Cu/ZnO Reforming (Isomerization of

oil)

Pt/Re/Al2O3

Key intermediates, hydrogen atoms, oxygen atoms, methyls ethyls …

Insulating Oxides Stabilize Ionic Intermediates

• Insulating oxides have no free electrons– radical intermediates not stabilized

• Insulating oxides are ionic– Ionic intermediates, i.e. carbocations can

be stabilized

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Table 12.2-Some Reactions Commonly Catalyzed By Solid Acids And Bases

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Reaction Example Typical Application

Isomerization(Rearranging the

structure of a molecule)

CH2=CHCH2CH3

CH3CH=CHCH3

Octane EnhancementMonomer Production

Paraxylene Production

Alkylation(Making too little molecules into a

bigger one)

CH3CH=CHCH3 + CH3CH2CH2CH3

(CH3CH2)CH(CH3)(C4H9)

Pharmaceutical Production

Monomer ProductionFine Chemicals

Butane + olefin octane

Cracking(Taking a big molecule and making it into two

littler ones).

C12H24 C7H14 + C5H10 Crude Oil ConversionDigestion

Semiconductors Are In Between

• Semiconductors have fewer free electrons, and also be ionic

• Usually use semiconductors when you want to stabilize some radicals e.g. oxygen atoms or sulfur atoms but not other radicals, such as hydrogen atoms.

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A Selection Of The Reactions Catalyzed By Transition Metal Oxides, Nitrides, And Sulfides

(Semiconductors)

Reaction Catalyst Reaction Catalyst

2 SO2 + O2 2 SO3 V2O5 CO + H2O CO2+

H2 (Water Gas Shift)

FeO, CuO, ZnO

Hydrodesulfurization CoS, MoS, WS 2(CH3)3COH Þ(CH3)3COC(CH

3)3 + H2O

TiO2

CH3CH=CH2 + O2 (Bi2O3)x(MoO3)y 2 CH3CH=CH2 + 3

O2 + 2NH3

CH2=CHCHO + (Bismuth molybate)

2CH2=CHCN +

H2O Uranium Antimonate

6 H2O

(aminoxidation)

(FeO)x(Sb2O3)y

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A Selection Of The Reactions Catalyzed By Transition Metal Oxides, Nitrides, And Sulfides

Reaction Catalyst Reaction Catalyst

4 NH3 + 4 NO +O2

4N2 + 6 H2O

benzene+O2

maleic anhydride +

water(Selective catalytic reduction)

naphthylene+O2 phthalic anhydride +

waterCH3CH2(C6H5) +O2

CH2=CH(C6H5) + H2O

NiO, Fe2O3,

V2O5, TiO2

(styrene production) CuO, Co3, O4,

MnO2

Aromatiztione.g. Heptane Tolvene

H2 or H2O

Cr2O3/Al2O3 Hydrodenitrogenation

NiS,MoS

V2O5, TiO2 (V2O5)x(PO4)y

FeO Selective oxidation of hydrocarbons

Summary

• Most catalysts work by stabilizing intermediates

• Intermediate concentration goes up

• Activation energy for product formation goes up (not down)!– Rate constant for product formation goes down

• Net effect is an increase in rate

• Metals, semiconductors, insulators stabilize different intermediates – gives different chemistry 25