Catalysis and Catalytic

Reactions

2

3

Scope:• Catalyst & Catalysis

• Limited to gas phase reactionscatalyzed by solids

• Mechanism & rate laws

• Interpretation of data – estimation of rate law parameters

• Physical properties of catalysts –estimation

• Catalytic reactors

• Design of Fixed bed reactor

4

What is a catalyst ??

• Alters the rate of reaction

• Highly selective

• Does it participate in the reaction ??

• How does it change the rate ?? – Offers an alternate path with low E.

• Does it affect ∆HR, ∆GR, and Eq. constant ??

• Does it affect yield & selectivity ??

• Does it initiate a reaction ??

5

Every catalytic reaction is a sequence of elementarysteps, in which reactant molecules bind to the catalyst,where they react, after which the product detachesfrom the catalyst, liberating the latter for the nextcycle

6

Potential energy diagram of a heterogeneous catalyticreaction, with gaseous reactants and products and a solidcatalyst. Note that the uncatalyzed reaction has toovercome a substantial energy barrier, whereas thebarriers in the catalytic route are much lower.

7

Homogeneous:

Catalytic:

At 600 K the ratio of catalytic to homogeneous rate is1.44x1011

Example: Boudart compared the homogeneous versuscatalytic rates of ethylene hydrogenation.

2

43000exp1027 Hp

RTr

−=

2

13000exp102 27

HpRT

xr

−=

8

What is a catalyst ??

• Were in use for making wine, cheese etc.

• Small amounts of catalyst

• Efficiency depends on activity, properties &

life of the catalyst

• Examples:

• Ammonia synthesis – Promoted iron

• SO2 oxidation – Venadium Pentaoxide

• Cracking – Sylica, alumina

• Dehydrogenation – Platinum, Molybdenum

9

Classification:

• Homogenous catalysis

• Heterogeneous catalysis

Catalysts are generally used to:

• Speedup reactions

• Change the operating temperature level

• Influence the product distribution

10

Promoter: is an additive which has nocatalytic properties of its own butenhances the activity of a catalyst

Promoter results in:

• Increase of available surface area

• Stabilization against crystal growth andsintering

• Improvement of mechanical strength

Examples: Alumina, Asbestos

11

Carrier: principally serve as a framework onwhich catalyst is deposited - no catalyticproperties of its own

Carrier results in:

• Highly porous nature - increase ofavailable surface area

• Improve stability

• Improves the heat transfercharacteristics

Examples: Alumina, Asbestos, Carborundum, Iron oxide, Manganese, Activated carbon, Zinc oxide

12

Accelerator: are substances which can beadded to a reacting system to maintainthe activity of a catalyst by nullifying theeffects of poisons

Poisons: substances which reduce theactivity of a catalyst. They are notdeliberately added but are unavoidablydeposited during the reaction.

Examples: Sulfur, Lead, Metal ions such as Hg, Pd, Bi, Sn, Cu, Fe etc.

13

Inhibitor: substances added to the catalystduring its manufacture to reduce itsactivity.

Coking/Fouling: deposition of carbonaceousmaterial on the surface of the catalyst -Common to reactions involvinghydrocarbons

14

Activity: of a catalyst depends on thetexture and electronic structure. Activityof a catalyst can be explained by:

• Active centers on the surface of thecatalyst

• Geometry of surface

• Electronic structure

• Formation of surface intermediates

Efficiency of a catalyst depends on :Activity, Selectivity and Life

15

Active site: is a point on the catalystsurface that can form strong chemicalbonds with an adsorbed atom/molecule.These sites are unsaturated atoms in thesolid resulting from:

• Surface irregularities

• Dislocations

• Edges of crystals

• Cracks along grain boundaries

16

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Mechanism of Heterogeneous Catalysis:

1. Bulk Diffusion of reacting molecules to thesurface of the catalyst

2. Pore Diffusion of reacting molecules into theinterior pores of the catalyst

3. Adsorption of reactants (chemisorption) onthe surface of the catalyst

4. Reaction on the surface of the catalystbetween adsorbed molecules

5. Desorption of products

6. Pore Diffusion of product molecules to thesurface of the catalyst

7. Bulk Diffusion of product molecules

18

Mechanism of Heterogeneous Catalysis:

19

Mechanism of Heterogeneous Catalysis:

20

21

Pore and film resistances in a catalyst particle

22

Rate-Determining Step (rds)In a kinetics scheme involving more than one step, it may be that one change occurs much faster or much slower than the others (as determined by relative magnitudes of rate constants).

In such a case, the overall rate, may be determined almost entirely by the slowest step, called the rate-determining step (rds).

The rate of the rds is infinitesimal when compared to the rates of other steps.

Alternately the rates of other steps are infinite compared to the rate of rds.

23

Bulk Diffusion:

• Diffusion controlled reactions are usuallyfast

• Design of reactors – design of masstransfer equipment

• Increase in mass velocity increases the rate

• High L/D ratio reactors (narrow) arefavored

24

Pore Diffusion:

• Pore diffusion controlled reactions are few

• Design of reactors – most complicated

• Approaches bulk diffusion if the pore size islarge

• Approaches Knudsen diffusion if the poresize is small.

• No effect of temperature or mass velocity

• Low L/D ratio reactors (wide) may be usedwith consequent reduction in pressure drop

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Chemisorption:

• Chemisorption controlled reactions areusually fast

• Rate increases rapidly with increase in temp.

• Permits the use of wide reactors

Surface reaction:

• 70% of the reactions which are notcontrolled by diffusion falls under this case

• Rate increases rapidly with increase in temp.

• Permits the use of wide reactors

27

Desorption:

• Desorption of a product could also be ratecontrolling in a few cases

Complexities:

• Theoretically more than one step can berate controlling

• Too many possible mechanisms

• Experimental data is normally fitted to anysingle rate controlling step, which is thencalled the most plausible mechanism

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Item Physical Adsorption Chemisorption

Forces of attraction

Weak – VanderWaals forces Strong valency forces

Specificity Low High

Quantity Large Small

Heat Effects Exothermic, 1-15 kCal/mol Exothermic, 10-100 kCal/mol

Activation energy

Low High

Effect of Temp. Rapid at low temperatures & reach equilibrium quickly. Beyond TC of the gas, no ads.

Slow at low temp., Rate increases with temp.

Effect of Pressure

Increases with increase in pressure

Little effect

Surface Whole surface active Fraction of surface only

Layers Multi-layer adsorption Mono-layer adsorption

Physical Adsorption Vs. Chemisorption

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Chemisorption rates:

• Adsorption data is reported in the form of isotherms

• Chemisorption may be considered as a reaction between a reactant molecule and an active site resulting in an adsorbed molecule

A + σ ⇔ Aσ (or) A + S ⇔ AS

Turnover Frequency (N): defined as the number of molecules reacting per active site per second at the conditions of the experiment – a measure for the activity of the catalyst

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Langmuir Isotherm - Assumptions:

• Surface is uniformly active

• All sites are identical

• Amounts of adsorbed molecules will not interfere with further adsorption

• Uniform layer of adsorption

Site balance:

t

vv

sitestotal

sitesvacantofNositesvacantofFraction

σ

σθ ===

.

t

AA

sitestotal

sitesoccupiedofNoAbyoccupiedsitesofFraction

σ

σθ ===

.

1=+ Av θθ

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Alternately:

numbersAvogadro

massunitsitesactiveofNositesactiveofconcMolarCt

'

/.. ==

numbersAvogadro

massunitsitesvacantofNositesvacantofconcMolarCv

'

/.. ==

numbersAvogadro

massunitAbysitesofNoAbysitesofconcMolarCAS

'

/.. ==

tASv CCC =+

Though other isotherms account for non-uniform surfaces, they

have primarily been developed for single adsorbing components.

Thus, the extensions to interactions in multi-component systems is

not yet possible, as with the Langmuir isotherm. Langmuir isotherms

are only used for developing kinetic rate expressions. However, not

all adsorption data can be represented by a Langmuir isotherm.

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Chemisorption rates (molecular adsorption):

A + σ ⇔ Aσ

Forward rate = k1pAθv

Backward rate = k2θA

At equilibrium: k1pAθv = k2θA

(k1/k2)pAθv = θA

AA

AAA

pK

pK

+=1

θ

AA

tAAAS

pK

CpKC

+=1

1=+ Av θθ

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Chemisorption rates (Atomic adsorption):

A2 + 2σ ⇔ 2Aσ

Forward rate = kApAθv2

Backward rate = k-AθA2

At equilibrium: kApAθv2 = k-AθA2

(kA/k-A)pAθv2 = θA2

AA

AA

ApK

pK

+=1

θt

AA

AA

AS CpK

pKC

+=1

What would be θA if chemisorption does not reach equilibrium ??

35

Effect of increasing temperature

Volume of gas

adsorbed

How to check for Molecular Adsorption / Atomic adsorption ??

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Langmuir adsorption isotherm for associative adsorptionfor three values of the equilibrium constant, K

AA

AAA

pK

pK

+=1

θ

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Surface Reaction: reaction between the adsorbed molecules on the surface of the catalyst may proceed in a number of ways:

Single site mechanism: Aσ ⇔ Rσ

Dual site mechanism: Aσ + σ ⇔ Rσ + σ

Aσ + Bσ ⇔ Rσ + Sσ

Aσ + Bσ ⇔ Rσ + σ

Langmuir-Hinshelwood kinetics

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Aσ + B(g) ⇔ Rσ

Aσ ⇔ Rσ + S(g)

Aσ + B(g) ⇔ Rσ + S(g)

Aσ + Bσ ⇔ Rσ + σ + S(g)

Eley Riedel Mechanism

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Surface reaction rates:

(1) Aσ ⇔ Rσ

Forward rate = kSθA Backward rate = k-SθR

At equilibrium: kSθA = k-SθR ARSK θθ /=

(2) Aσ + σ ⇔ Rσ + Sσ VASRSK θθθθ /=

(3) Aσ + Bσ ⇔ Rσ + Sσ BASRSK θθθθ /=

BASRS ppK θθ /=(4) Aσ + B(g) ⇔ Rσ + S(g)

(5) Aσ ⇔ Rσ + S(g) ASRS pK θθ /=

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Desorption rates:

Rσ ⇔ R + σ

(Desorption of R is the Reversal of adsorption of R)

Forward rate = kDθR Backward rate = k-DpRθV

At equilibrium: kDθR = k-DpRθV

VRRDVRR pKKp θθθ == /

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Synthesizing a rate law – Algorithm (Langmuir-Hinshelwood Approah)

1. Assume a sequence of steps

2.Write rate laws for each step assuming all steps to be reversible

3.Assume a rate limiting step

4.Equate the rate of rds to the overall rate

5.The rates of other steps are equated to zero (equilibrium)

6.Using the rates of other steps eliminate all coverage dependent terms

7.If the derived rate law does not agree with expt., goto (3)

42

• Approach is similar to non-elementary reactions

• In the case of non-elementary reactions there is only one rate law for a given mechanism

• But in the case of solid catalyzed gas phase reactions, there could be many (equal to the number of steps) rate laws for a given mechanism

• In a given mechanism, even after assuming each of the steps as rds, and none of them satisfy the experimental data, start with a new mechanism and repeat

43

Example: C6H5CH(CH3)2 → C6H6 + C3H6

Cumene → Benzene + Propylene

Suggested Mechanism:

C + σ ⇔ Cσ (Adsorption of cumene)

Cσ ⇔ Bσ + P(g) (Surface reaction)

Bσ ⇔ B + σ (Desorption of Benzene)

Rate laws for each of the steps:

CAVCA kpkAdsorptionofrateNet θθ −−=

PBSCS pkkreactionSurfaceofrateNet θθ −−=

VBDBD pkkDesorptionofrateNet θθ −−=

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Case-I: Adsorption is Rate limiting step

CAVCAC kpkAdsorptionofrateNetr θθ −−==−

)/( ACVCAC Kpkr θθ −=−

0Re =−= − PBSCS pkkactionSurfaceofrateNet θθ

0=−= − VBDBD pkkDesorptionofrateNet θθ

SPBC Kp /θθ =

DVBB Kp /θθ =

V

DS

BPC

KK

ppθθ =

Site balance: θC + θB + θV = 1

D

B

DS

BPV

K

p

KK

pp++

=

1

1θ

45

)/( ACVCAC Kpkr θθ −=−

)()(eq

BPCVAV

DSA

BPVCAC

K

pppk

KKK

pppkr −=−=− θθθ

D

B

DS

BP

eq

BPCA

C

K

p

KK

pp

K

pppk

r

++

−

=−

1

)(

Final rate law for Case-1

Case-II: Surface reaction is rate limiting step

CAVCA kpkAdsorptionofrateNet θθ −−=

PBSCS pkkreactionSurfaceofrateNet θθ −−=

VBDBD pkkDesorptionofrateNet θθ −−=

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VCACCAVCA pKkpkAdsorptionofrateNet θθθθ =⇒=−= − 0

PBSCSC pkkreactionSurfaceofrateNetr θθ −−==−

DVBBVBDBD KppkkDesorptionofrateNet /0 θθθθ =⇒=−= −

)/( SPBCSC Kpkr θθ −=−

Site balance: θC + θB + θV = 1

D

BCA

V

K

ppK ++

=1

1θ

)/()/( SDAPBCVASSDVPBVCASC KKKpppKkKKpppKkr −=−=− θθθ

D

BCA

eq

BPCAS

C

K

ppK

K

pppKk

r

++

−

=−

1

)(

Final rate law for Case-2

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Case-III: Desorption is rate limiting step

PCSBPBSCS pKpkkreactionSurfaceofrateNet /0 θθθθ =⇒=−= −

)/( DVBBDVBDBDC KpkpkkDesorptionofrateNetr θθθθ −=−==− −

VCACCAVCA pKkpkAdsorptionofrateNet θθθθ =⇒=−= − 0

Site balance: θC + θB + θV = 1PCSACA

VppKKpK /1

1

++=θ

PVCASB ppKK /θθ =

)//( DVBPVCASDC KpppKKkr θθ −=−

)//( DSABPCVASDC KKKpppKKkr −=− θ

CSACPAP

eq

BPCSAD

PCSACA

eq

BPCSAD

CpKKppKp

K

pppKKk

ppKKpK

K

pppKKk

r++

−

=++

−

=−

)(

/1

)/(

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D

B

DS

BP

eq

BPCA

C

K

p

KK

pp

K

pppk

r

++

−

=−1

)(

D

BCA

eq

BPCAS

C

K

ppK

K

pppKk

r

++

−

=−1

)(

CSACPAP

eq

BPCSAD

CpKKppKp

K

pppKKk

r++

−

=−

)(

Case-1

Case-2

Case-3

)(

)(

termAdsorption

ForceDrivingtermKineticRate =

What would be the effect of an inert ??

49

Effect of increasing reactant concentration:

Increasing the reactant concentration increases both the driving force and adsorption inhibition terms.

CA

rate

Volcano shape results from a competition between kinetic

driving force and adsorption inhibition terms.

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Term Case-1 Case-2 Case-3

Kinetic kA kSKA kDKAKS

Driving Force

Adsorption

eq

BPC

K

ppp −

eq

BPC

K

ppp −

eq

BPC

K

ppp −

D

B

DS

BP

K

p

KK

pp++1

D

BCA

K

ppK ++1 CSACPAP pKKppKp ++

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VCA pK θVCA pK θV

DS

BP

KK

ppθ

PVCAS ppKK /θDVB Kp /θ DVB Kp /θ

Coverage Case-1 Case-2 Case-3

θC

θB

52

53

54

55

• For a given mechanism, the driving force is unique, irrespective of RDS

• The product of equilibrium constant of all steps in the mechanism yield the overall eq. constant

• In the kinetic term, the rate constant of RDSwill appear

• If adsorption of A is not RDS, then KApA will appear in the adsorption term

• If desorption of B is not rate limiting, then pB/KD will appear in the adsorption term

• If SR is RDS, then the adsorption term will be raised to the power equal to the number of sites involved in the SR step.

Remarks:

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Exercise: Al2O3

N-pentane ⇔ I-Pentane

Suggested Mechanism:

N + σ ⇔ Nσ

Nσ + σ ⇔ Iσ + σ

Iσ ⇔ I + σ

Rate laws for each of the steps:

NAVNA kpkAdsorptionofrateNet θθ −−=

vISvNS kkreactionSurfaceofrateNet θθθθ −−=

VIDID pkkDesorptionofrateNet θθ −−=

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Case-I: Adsorption is Rate limiting step

NAVNAN kpkAdsorptionofrateNetr θθ −−==−

)/( ANVNAN Kpkr θθ −=−

0Re =−= − vISvNS kkactionSurfaceofrateNet θθθθ

0=−= − VIDID pkkDesorptionofrateNet θθ

SIN K/θθ =

DVII Kp /θθ =

V

DS

IN

KK

pθθ =

Site balance: θC + θB + θV = 1

D

I

DS

IV

K

p

KK

p++

=

1

1θ

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)/( ANVNAN Kpkr θθ −=−

)()(eq

INVAV

DSA

IVNAN

K

ppk

KKK

ppkr −=−=− θθθ

D

I

DS

I

EqINA

N

K

p

KK

p

Kppkr

++

−=−

1

)/(

Rate law for Case-1

Case-II: Surface reaction is rate limiting step

NAVNA kpkAdsorptionofrateNet θθ −−=

vISvNS kkreactionSurfaceofrateNet θθθθ −−=

VIDID pkkDesorptionofrateNet θθ −−=

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VNANNAVNA pKkpkAdsorptionofrateNet θθθθ =⇒=−= − 0

vISvNSN kkreactionSurfaceofrateNetr θθθθ −−==−

DVIIVIDID KppkkDesorptionofrateNet /0 θθθθ =⇒=−= −

)/( SINvSN Kkr θθθ −=−

Site balance: θC + θB + θV = 1

DINA

VKppK /1

1

++=θ

)/()/( 2

SDAINvASSDVIVNAVSN KKKppKkKKppKkr −=−=− θθθθ

2)/1(

)/(

DINA

EqINAS

NKppK

KppKkr

++

−=− Rate law for Case-2

60

Case-III: Desorption is rate limiting step

NSIvISvNS KkkreactionSurfaceofrateNet θθθθθθ =⇒=−= − 0

)/( DVIIDVIDIDN KpkpkkDesorptionofrateNetr θθθθ −=−==− −

VNANNAVNA pKkpkAdsorptionofrateNet θθθθ =⇒=−= − 0

Site balance: θC + θB + θV = 1NSANA

VpKKpK ++

=1

1θ

VNASI pKK θθ =

)/( DVIVNASDN KppKKkr θθ −=−

)/( DSAINVASDN KKKppKKkr −=− θ

NSANA

EqINSAD

NpKKpK

KppKKkr

++

−=−1

)/(

Rate law for Case-3

61

• How to verify which one is rate limiting step ??

• For this initial rate data is normally used

D

BCA

eq

BPCAS

C

K

ppK

K

pppKk

r

++

−

=−1

)(

C

C

CA

CAS

bp

ap

pK

pKkr

+=

+=−

110

• In the absence of any products initially, the rate law simplifies to:

62

Case-1

Case-2

NSANA

eq

INSAD

NpKKpK

K

ppKKk

r++

−

=−1

)(

2)/1(

)(

DINA

eq

INAS

NKppK

K

ppKk

r++

−

=−

D

I

DS

I

eq

INA

N

K

p

KK

p

K

ppk

r

++

−

=−

1

)(

NA pkr =− 0

20)1( NA

NAS

pK

pKkr

+=−

Case-3NSANA

NSAD

pKKpK

pKKkr

++=−1

0

63

Simplified rate laws:

• Over limited pressure range, the Langmuir isotherm θ = Kp/(1+Kp) can be replaced by an approximation θ = kpn

• In such cases the rate law assumes the form:

r = k pAm pBn pCo

Such rate laws may be reasonably accurate

Example: CO + Cl2 ⇔ COCl2 (Over charcoal)

2)1(2222

22

2

COClCOClClCl

ClCOClCO

COClpKpK

ppKkKr

++=− L-H approach

2/1

22 ClCOCOCl pkpr =− Simplified equation