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Catalysis and Catalytic

Reactions

A. SARATH BABU

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Scope: Catalyst & Catalysis ??

Limited to gas phase reactionscatalyzed by solids

Mechanism & rate laws

Interpretation of data estimation ofrate law parameters

Physical properties of catalysts

estimation

Catalytic reactors

Design of Fixed bed reactor

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What is a catalyst ??

Alters the rate of reaction

Highly selective

Does it participate in the reaction ??

How does it change the rate ??

Offers analternate path with low E.

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

Does it affect yield & selectivity ?? Does it initiate a reaction ??

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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 next

cycle

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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 the

barriers in the catalytic route are much lower.

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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

43000exp10 27 Hp

RTr

2

13000exp102 27 Hp

RTxr

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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

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Classification:

Homogenous catalysis

Heterogeneous catalysis

Catalysts are generally used to:

Speedup reactions Change the operating temperature level

Influence the product distribution

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Promoter: is an additive which has nocatalytic properties of its own but

enhances the activity of a catalyst

Promoter results in:

Increase of available surface area

Stabilization against crystal growthand sintering

Improvement of mechanicalstrength

Examples: Alumina, Asbestos

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Carrier: principally serve as a framework onwhich catalyst is deposited - no catalytic

properties 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

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Accelerator: are substances which can beadded to a reacting system to maintain

the 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.

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

activity.

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

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Activity: of a catalyst depends on thetexture and electronic structure. Activity

of a catalyst can be explained by: Active centers on the surface of the

catalyst

Geometry of surface Electronic structure

Formation of surface intermediates

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

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Active site: is a point on the catalystsurface that can form strong chemical

bonds with an adsorbed atom/molecule.These sites are unsaturated atoms in thesolid resulting from:

Surface irregularities Dislocations

Edges of crystals Cracks along grain boundaries

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Mechanism of Heterogeneous Catalysis:

1. Bulk Diffusion of reacting molecules to the

surface of the catalyst2. Pore Diffusion of reacting molecules into the

interior 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

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Mechanism of Heterogeneous Catalysis:

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Mechanism of Heterogeneous Catalysis:

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Pore and film resistances in a catalyst particle

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Rate-Determining Step (rds)In a kinetics scheme involving more than one step, itmay be that one change occurs much faster or much

slower than the others (as determined by relativemagnitudes 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 tothe rates of other steps.

Alternately the rates of other steps are infinite

compared to the rate of rds.

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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

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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 are

usually fast Rate increases rapidly with increase in temp.

Permits the use of wide reactors

Surface reaction:

70% of the reactions which are not

controlled by diffusion falls under this case Rate increases rapidly with increase in temp.

Permits the use of wide reactors

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Desorption:

Desorption of a product could also be rate

controlling 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

Activationenergy

Low High

Effect of Temp. Rapid at low temperatures &reach equilibrium quickly.

Beyond TC of the gas, no ads.

Slow at low temp., Rate increaseswith temp.

Effect ofPressure

Increases with increase inpressure

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 moleculeA + A (or) A + S AS

Turnover Frequency (N): defined as the

number of molecules reacting per active siteper second at the conditions of theexperiment a measure for the activity ofthe catalyst

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Langmuir Isotherm - Assumptions:

Surface is uniformly active

All sites are identical

Amounts of adsorbed molecules will notinterfere with further adsorption

Uniform layer of adsorption

Site balance:

t

v

v sitestotal

sitesvacantofNo

sitesvacantofFraction

.

t

AA

sitestotal

sitesoccupiedofNoAbyoccupiedsitesofFraction

.

1Av

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Alternately:

numbersAvogadro

massunitsitesactiveofNositesactiveofconcMolarCt

'

/..

numbersAvogadro

massunitsitesvacantofNositesvacantofconcMolarCv

'

/..

numbersAvogadro

massunitAbysitesofNo

AbysitesofconcMolarCAS '

/..

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 + A

Forward rate = k1pAv

Backward rate = k2A

At equilibrium: k1pAv = k2A(k1/k2)pAv = A

AA

AAA

pK

pK

1

AA

tAAAS

pK

CpKC

1

1 Av

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Chemisorption rates (Atomic adsorption):

A2 + 2 2A

Forward rate = kApAv2

Backward rate = k-AA2

At equilibrium: kApAv2

= k-AA2

(kA/k-A)pAv

2= A

2

AA

AA

A

pK

pK

1

t

AA

AA

AS CpK

pK

C 1

What would be A if chemisorption does notreach equilibrium ??

Eff t f in sin t mp t

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Effect of increasing temperature

Volumeofgas

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

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Surface Reaction: reaction between the adsorbedmolecules on the surface of the catalyst mayproceed in a number of ways:

Single site mechanism: A R

Dual site mechanism: A + R +

A + B R + S

A + B R +

Langmuir-Hinshelwood kinetics

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A + B(g) R

A R + S(g)A + B(g) R + S(g)

A + B R + + S(g)

Eley Riedel Mechanism

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Surface reaction rates:

(1) A R

Forward rate = kSA Backward rate = k-SR

At equilibrium: kSA = k-SR ARSK /

(2) A + R+ S VASRSK /

(3) A + B R+ S BASRSK /

BASRS ppK /

(4) A

+ B(g)

R

+ S(g)(5) A R+ S(g) ASRS pK /

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Desorption rates:

R R +

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

Forward rate = kDR Backward rate = k-DpRV

At equilibrium: kDR = k-DpRVVRRDVRR pKKp /

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Synthesizing a rate law Algorithm(Langmuir-Hinshelwood Approah)

1. Assume a sequence of steps2.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 allcoverage dependent terms

7.If the derived rate law does not agree with

expt., goto (3)

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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 givenmechanism

In a given mechanism, even after assuming eachof the steps as rds, and none of them satisfy

the experimental data, start with a newmechanism and repeat

E l ( )

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Example: C6H5CH(CH3)2 C6H6 + C3H6

Cumene Benzene + Propylene

Suggested Mechanism:

C + C (Adsorption of cumene)

C B + P(g) (Surface reaction)B B + (Desorption of Benzene)

Rate laws for each of the steps:

CAVCA kpkAdsorptionofrateNet

PBSCS pkkreactionSurfaceofrateNet

VBDBD

pkkDesorptionofrateNet

C s I: Ads pti n is R t limitin st p

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Case-I: Adsorption is Rate limiting step

CAVCAC kpkAdsorptionofrateNetr

)/( ACVCAC Kpkr 0Re PBSCS pkkactionSurfaceofrateNet

0 VBDBD pkkDesorptionofrateNet

SPBC Kp /

DVBB Kp /

V

DS

BPC

KK

pp

Site balance: C + B + V = 1

D

B

DS

BPV

K

p

KK

pp

1

1

)/( Kk

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)/( ACVCAC Kpkr

)()(eq

BPCVAV

DSA

BPVCAC

K

pppk

KKK

pppkr

D

B

DS

BP

eq

BPCA

C

K

p

KK

pp

Kpppk

r

1

)(

Final rate law for Case-1

Case-II: Surface reaction is rate limiting step

CAVCA kpkAdsorptionofrateNet

PBSCS

pkkreactionSurfaceofrateNet

VBDBD pkkDesorptionofrateNet

pkkreactionSurfaceofrateNetr

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VCACCAVCA pKkpkAdsorptionofrateNet 0

PBSCSC pkkreactionSurfaceofrateNetr

DVBBVBDBD KppkkDesorptionofrateNet /0

)/( SPBCSC Kpkr

Site balance: C + B + V = 1

D

BCA

V

K

ppK

1

1

)/()/(SDAPBCVASSDVPBVCASC

KKKpppKkKKpppKkr

D

BCA

eq

BPCAS

C

K

ppK

K

pppKk

r

1

)(

Final rate law for Case-2

Case III: Desorption is rate limiting step

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Case-III: Desorption is rate limiting step

PCSBPBSCS pKpkkreactionSurfaceofrateNet /0

)/( DVBBDVBDBDC KpkpkkDesorptionofrateNetr

VCACCAVCA pKkpkAdsorptionofrateNet 0

Site balance: C + B + V = 1PCSACA

VppKKpK /1

1

PVCASB ppKK /

)//( DVBPVCASDC KpppKKkr

)//( DSABPCVASDC KKKpppKKkr

CSACPAP

eq

BPCSAD

PCSACA

eq

BPCSAD

CpKKppKp

K

pppKKk

ppKKpK

K

pppKKk

r

)(

/1

)/(

BPppk )(

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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 bethe effect of

an inert ??

Effect of increasing reactant concentration:

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Effect of increasing reactant concentration:

Increasing the reactant concentration increases

both the driving force and adsorption inhibitionterms.

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

DrivingForce

Adsorption

eq

BPC

K

ppp

eq

BPC

K

ppp

eq

BPC

K

ppp

D

B

DS

BP

K

p

KK

pp1

D

B

CA K

p

pK

1 CSACPAPpKKppKp

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VCApK VCApK VDS

BP

KK

pp

PVCAS ppKK /DVB Kp / DVB Kp /

Coverage Case-1 Case-2 Case-3

C

B

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Remarks:

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For a given mechanism, the driving force isunique, irrespective of RDS

The product of equilibrium constant of all stepsin the mechanism yield the overall eq. constant

In the kinetic term, the rate constant of RDS

will appear If adsorption of A is not RDS, then KApA will

appear in the adsorption term

If desorption of B is not rate limiting, thenpB/KDwill appear in the adsorption term

If SR is RDS, then the adsorption term will beraised to the power equal to the number of

sites involved in the SR step.

Exercise: Al O

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Exercise: Al2O3N-pentane I-Pentane

Suggested Mechanism:

N + N

N + I + I I +

Rate laws for each of the steps:

NAVNA kpkAdsorptionofrateNet

vISvNS kkreactionSurfaceofrateNet

VIDID pkkDesorptionofrateNet

Case-I: Adsorption is Rate limiting step

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Case I: Adsorption is Rate limiting step

NAVNAN kpkAdsorptionofrateNetr

)/( ANVNAN Kpkr

0Re vISvNS kkactionSurfaceofrateNet

0 VIDID pkkDesorptionofrateNet

SIN K/

DVII Kp /

V

DS

IN

KK

p

Site balance: C + B + V = 1

D

I

DS

IV

K

p

KK

p

1

1

)/( Kpkr

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)/( ANVNAN Kpkr

)()(eq

INVAV

DSA

IVNAN

K

ppk

KKK

ppkr

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

vISvNS kkreactionSurfaceofrateNet

VIDID pkkDesorptionofrateNet

vISvNSN kkreactionSurfaceofrateNetr

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VNANNAVNA pKkpkAdsorptionofrateNet

0

vISvNSN ff

DVIIVIDID KppkkDesorptionofrateNet /0

)/( SINvSN Kkr

Site balance: C + B + V = 1

DINA

VKppK /1

1

)/()/(2

SDAINvASSDVIVNAVSN KKKppKkKKppKkr

2)/1(

)/(

DINA

EqINAS

NKppK

KppKkr

Rate law for Case-2

Case-III: Desorption is rate limiting step

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Case III: Desorption is rate limiting step

NSIvISvNS KkkreactionSurfaceofrateNet 0

)/( DVIIDVIDIDN KpkpkkDesorptionofrateNetr

VNANNAVNA pKkpkAdsorptionofrateNet

0

Site balance: C + B + V = 1 NSANAV

pKKpK 1

1

VNASI pKK

)/( DVIVNASDN KppKKkr

)/( DSAINVASDN KKKppKKkr

NSANA

EqINSAD

NpKKpK

KppKKkr

1

)/(

Rate law for Case-3

How to verify which one is rate limiting step ??

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How to verify which one is rate limiting step ??

For this initial rate data is normally used

D

BCA

eq

BPCAS

C

K

ppK

KpppKk

r

1

)(

C

C

CA

CAS

bp

ap

pK

pKk

r

110

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

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Case-1

Case-2

NSANA

eq

INSAD

N pKKpK

K

ppKKk

r

1

)(

2)/1(

)(

DINA

eq

INAS

N KppK

K

ppKk

r

D

I

DS

I

eq

INA

N

K

p

KK

p

K

ppk

r

1

)(

NApkr 0

20)1(

NA

NAS

pK

pKkr

Case-3 NSANANSAD

pKKpK

pKKkr

10

Simplified rate laws:

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p

Over limited pressure range, the Langmuirisotherm = Kp/(1+Kp) can be replaced by anapproximation = kpn

In such cases the rate law assumes the form:

r = k pAm pB

n pCo

Such rate laws may be reasonably accurate

Example: CO + Cl2 COCl2 (Over charcoal)

2)1(

2222

222

COClCOClClCl

ClCOClCO

COClpKpK

ppKkK

r L-H approach

2/1

22 ClCOCOClpkpr Simplified equation

How to verify whether the rate law confirms

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How to verify whether the rate law confirmsto experimental data or not ??

D

BCA

eq

BPCAS

C

K

ppK

K

pp

pKkr

1

)(

AS

D

BCA

C

eq

BPC

Kk

KppK

r

Kppp

1

BC

C

eq

BP

C

cpbpar

K

pp

p

Use regression

How to decide whether the fit is reasonable ??

If the fit is reasonable, evaluate theconstants

Design of Fixed Bed Reactor:

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g

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Design Equation

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Design Equation

General Mass Balance Equation:

Rate of input = rate of output + accumulation+ rate of disappearance

FA = FA + dFA + 0 + (-rA) dW- dFA = (-rA) dW

FA0

dxA

= (-rA

) dW

dW

AA rdWdF /

G l d i i f FBR

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Ax

AAA rdxFW0

0 //

General design equation for a FBR:

dt

dN

WWeightunittimeunit

ddisappeareAofMolesr AA

1

))((

Definition of rate of reaction:

When the rate is expressed in terms of catalyst weight,

mass transfer effects between the catalyst and the bulkfluid & also within the catalyst are ignored. Such masstransfer aspects could be important in some cases.

Fixed Bed Reactor Integral form

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Fixed Bed Reactor Integral form

Ax

AAA rdxFW0

0 //

1 /-rA

xA

W / FA0

Fixed Bed Reactor differential form

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Fixed Bed Reactor differential form

AA rdWdF /

FA

W

-rA xA

W/FA0

-rA

AAA rdWdxF /0

How to find the rate data ??

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How to find the rate data ??

How can we calculate the weight of the

catalyst needed for obtaining the givenconversion ??

D

BAA

eq

BCAAS

A

K

ppK

K

pppKk

r

1

)(

Ax

AAA rdxFW0

0//

Express the partial pressures in terms of xA

AA

A

A

A

xx

pp

11

0 AA

AR

A

R

xxarM

pp

1)/(

0

)( AA xfr Use numerical / graphical integration

Physical properties of catalysts:

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Physical properties of catalysts Bulk density Surface area Pore volume Pore size distribution

For Silica-Alumina catalyst:Surface area = 200 500 m2/gmPore volume = 0.2 0.7 ml/gm

Measurement of Surface area:

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Measurement of Surface area:Measuring the surface area active for

chemisorption is difficult because of: highly selective nature fraction of surface physical adsorption + chemisorption presence of promoter, carrier etc.

Universally surface area of a catalyst is

measured using physical adsorptionprinciples. It is approximated that the morethe area the more would be the activity ofthe catalyst.

Experiment:

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Experiment: The amount of N2 adsorbed at equilibrium at the

normal boiling point temp (-195.8 0C) is measured

over a wide range of N2 partial pressures below1 atm.

Identify the amount required to cover theentire surface by a mono-layer

VSTP

pNitrogen

Linear regionMono Layer ads

p/p0 < 0.1 Mono layer

0.1 < p/p0 < 0.4 Multi layer0.4 < p/p0 < 1.0 Capillary condensation

1 Langmuir Isotherm:

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mmmN

N

N

v

p

Kvv

p

v

v

Kp

Kp

1

12

2

2

1. Langmuir Isotherm:

p/v

p

Slope = 1/vm

2 BET Isotherm:

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75

00

)1(1

)( pcv

pc

cvppv

p

mm

2. BET Isotherm:

p/[v(p0-p)]

p/p0

Slope = (c-1)/cvm

P0 = vapor pressure / Satn pressure

1/cvm

vm = 1/(slope + Intercept)

Convert v to no of molecules

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Convert vm to no. of molecules = area covered by one molecule

22400

0NvS m3/2

0

09.1

NM

For Nitrogen: = 0.808 g/cc at -195.8 0C = 16.2x10-16 cm2 = 16.2 (A0)2

mvS4

1035.4 vm is in CC at STP

Specific Surface area = S/W cm2/gm

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ANY CLARIFICATIONS ?

Colton, Charles Caleb

Examinations are formidable, even to the best prepared, for the

greatest fool may ask more than the wisest man can answer