CHBE 551 Lecture 31 Mass Transfer & Kinetics In Catalysis

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Key Ideas For Today

Generic mechanics of catalytic reactions

Measure rate as a turnover number Rate equations complex Langmuir Hinshelwood kinetics

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Generic Mechanisms Of Catalytic Reactions

3

B

Langmuir-Hinshelwood Rideal-Eley Precursor

A

B A

A B

A

A B

AB

B AB

AAB

ABA

B

AB

ABA

B

AB

ABA

B

AB

Figure 5.20 Schematic of a) Langmuir-Hinshelwood, b) Rideal-Eley, c) precursor mechanism for the reaction A+BAB and ABA+B.

Turnover Number: Number Of Times Goes Around The Catalytic Cycle Per

Second

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HI

H O2

CH OH3

CH I3

CH COOH3

CH COI3[Rh(CO) I ]2 2

-

RhI

H C3C I

I

O

RhI

CH3

CO I

I

C

O

CO

CO

CO

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

Printing press analogy

•Reactants bind to sites on the catalyst surface•Transformation occurs •Reactants desorb

Typical Catalytic Cycle

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2

2

2

O O O O O O O O O O

+ H

H H

O O O O O O O O OH H

- H O

+ 1

/2 O

A 2

2

2

O

+1/2 O

H OH

- H

O

B

+ H

H OH

Figure 5.10 Catalytic cycles for the production of water a) via disproportion of OH groups, b) via the reaction OH(ad)+H)ad)H2O

Definition Of Turnover Number

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S

AN N

RT

(12.119)

Physically, turnover number is the rate that the catalyst prints product per unit sec.

Typical Turnover Numbers

200 400 600 800 1000 1200

Reaction Temperature, K

10-6

10-4

10-2

100

102

Tur

nove

r N

umb

er, s

ec-1 Hydrogenation

OlefinIsomerization

Dehydrogenation

AlkaneHydrogenolysis

Cyclization

SiliconDeposition

GaAsDeposition

1400

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Next Topic Why Catalytic Kinetics Different Than Gas Phase Kinetics

1012

1013

10-6

Ra

te, M

ole

cule

s/cm

-se

c2

1011

10-7

10-8

10-6

CO pressure, torr10

-710

-8

O pressure, torr2

390 K

410 K415 K

450 K450 K

440 K440 K

425 K

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Figure 2.15 The influence of the CO pressure on the rate of CO oxidation on Rh(111). Data of Schwartz, Schmidt, and Fisher.

Temperature Nonlinear

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Rat

e, M

olec

ules

/cm

-se

c2

Temperature, KTemperature, K

1E+13

1E+12

1E+11

600400 800600 800400

P =2.E-7 torrCO

A

B

C

2P =2.5E-8 torrO

DEF

Figure 2.18 The rate of the reaction CO + ½ O2 CO2 on Rh(111). Data of Schwartz, Schmidt and

Fisher[1986]. A) = 2.510-8 torr, = 2.510-8 torr, B) = 110-7 torr, = 2.510-8 torr, C) = 810-7 torr, = 2.510-8 torr, D) = 210-7 torr, = 410-7 torr, E) = 210-7 torr, = 2.510-8 torr, F) = 2.510-8 torr, = 2.510-8 torr,

Derivation Of Rate Law For AC

Mechanism

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S+BBad

7

8

Also have a species B

(12.122)

1 S + A Aad 2

3 AAd Cad 4

5 Cad C + S

6

(12.121)

Derivation:

Next uses the steady state approximation to derive an equation for the production rate of Cad (this must be equal to the production rate of C).

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Derivation Continued

rC k3[Aad ] k4[Cad ]

(12.123)

SS on [Aad] and [Cad] 0 rAad k1PA[S] k2[Aad ] k3[Aad ] k4[Cad ]

(12.124) 0 rCad k6PC[S] k5[Cad ] k4[Cad ] k3[Aad ]

(12.125)

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People usually ignore reactions 3 and 4 since their rates very low rates compared to the other reactions.

Dropping The k3 And k4 Terms In Equations 12.124 And 12.125 And

Rearranging Yields:

[Aad ]k1k2

PA [S]

(12.126)

[ ]Cad ]k6k5

PC[S

(12.127)

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Similarly for B

Badk8k

PB S7

(12.128)

Rearranging Equations (12.126), (12.127) And (12.128) Yields:

[Aad ]

PA[S]

k1k2

(12.129)

[Bad ]

PB[S]

k8k7

(12.130)

CadPc S

k6k5

(12.131)

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Derivation Continued

Equations (12.129) and (12.130) imply that there is an equilibrium in the reactions:

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1 A + S Aad 2 8 B + S Bad 7

6 C + S Cad 5

(12.132)

B

[Aad ]

PA[S]

k1k2

(12.129)

[Bad ]

PB[S]

k8k7

(12.130)

CadPc S

k6k5

(12.131)

Site Balance To Complete The Analysis

If we define S0 as the total number as sites n the catalyst, one can show:

Pages of Algebra

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(12.133)

S0 [S] [Aad ] [Bad Cad ]

[Aad ]KA PAS0

1 KA PA KBPB KCPC

(12.140)

[Cad ]KCPCS0

1 KA PA KBPB KCPC

(12.141)

Substituting Equations (12.140) And (12.141) Into Equation (12.123) Yields:

rk3KA PAS0 k4KCPCS0

1 KA PA KBPB KCPC

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(12.142)

In the catalysis literature, Equation (12.142) is called the Langmuir-Hinshelwood expression for the rate of the reaction AC, also called Michaele’s Menton Equation.

Qualitative Behavior For Bimolecular Reactions (A+Bproducts)

PA

0 10 20 30 40 500.0E+0

5.0E+13

1.0E+14

1.5E+14

2.0E+14

2.5E+14

Rat

e, M

olec

ules

/cm

/se

c 2

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Figure 12.32 A plot of the rate calculated from equation (12.161) with KBPB=10.

1012

1013

Rat

e, M

olec

ules

/cm

-se

c2

1011

10-6

CO pressure, torr10

-710

-8

390 K

410 K

450 K

440 K

Physical Interpretation Of Maximum Rate For A+BAB

Catalysts have finite number of sites.

Initially rates increase because surface concentration increases.

Eventually A takes up so many sites that no B can adsorb.

Further increases in A decrease rate.

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Qualitative Behavior For Unimolecular Reactions (AC)

P =25B

PA

0 10 20 30 40 500.0E+0

5.0E-9

1.0E-8

1.5E-8

2.0E-8

Rat

e, M

oles

/cm

/se

c 2

P =0B

20

0.01 0.1 1 10 1001E+16

1E+17

1E+18

1E+19

1E+20

770 K

1070 K

870 K

1270 K1670 K

Ammonia pressure, torr

Rat

e, M

olec

ules

/cm

-se

c2

Langmuir-Hinshelwood-Hougan-Watson Rate Laws: Trick To Simplify The

Algebra

Hougan and Watson’s Method: Identify rate determining step (RDS). Assume all steps before RDS in

equilibrium with reactants. All steps after RDS in equilibrium with

products. Plug into site balance to calculate rate

equation.

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Example:

The reaction A + B C obeys:

S A Aad (1)

S B Bad (2)

Aad Bad C 2S (3)

(12.157)

Derive an equation for the rate of formation of C as a function of the partial pressures of A and B. Assume that reaction (3) is rate determining.

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Solution

rC = k3[Aad][Bad] (1) Assume reaction 1 in equilibrium

1A

ad KPS

]A[ (2)

Similarly on reaction 2

2B

ad KPS

]B[ (3)

Combining 1,2 and 3 2

BA321C SPPkKKr (4)

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Solution Continued

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Need S to complete solution: get it from a site balance. So= S + [Aad] + [Bad] (5) Combining (2), (3) and (5) So= S + SK1PA + K2PB (6) Solving (6) for S

B2A1

o

PKPK1S

S

(7)

Combining equations (4) and (7)

2B2A1

2oBA321

CPKPK1

SPPkKKr

(8)

Background: Mass Transfer Critical to Catalyst Design

C HH

HCH

HH

DiffusionChannel

Cavity

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Figure 12.27 An interconnecting pore structure which is selective for the formation of paraxylene.

Introduction

In a supported catalyst reactants first diffuse into the catalyst, then they react The products diffuse out

Gives opportunity for catalyst design

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Reactant Concentration Drops Moving Into the Solid

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Reactant Concentration

Dis

tan

ce

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.2

0.4

0.6

0.8

1

Distance from the center of the pellet

Con

cent

ratio

n

Edge of Pellet

Center of Pellet

GasPhase

Gas Phase Conc

Avg conc in pellet

Conc in pellet

Thiele Derivation for Diffusion In Catalysis

Assume: Constant effective diffusivity First Order reaction per unit volume Irreversible reaction – Rate of diffusion of

products out of catalyst does not affect rate

28Reactant Concentration

Dis

tan

ce

Define “Effectiveness Factor”

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e

Actual rate for a catalyst pellet with mass transfer limitations

Rate in the absence of mass transfer limitations

Derivation: Mass Balance On Differential Slice

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Taking the limit as Δy → 0

Y

4 4

4

y DdC

dYy D

dC

dy

y y r

2 A

y y

2e

S

y

2A

e

d C

dy

2

y

dC

dy

r

D

2A A A

e

0

Spherical pellet

Long Derivation

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C Cy

y

Sinh 3 y/y

Sinh 3A A0 P P

P

PP A

e

y

3

k

D

e

P P P

1 1

3

1

3 tanh

Thiele Plot

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1.0

0.0

0.8

0.6

0.4

0.2

Mas

s T

rans

fer

Fac

tor

0 2 4 6 8 10Thele Parameter

Zero Order

First Order

Second Order

Issues With “Effectiveness Factor”

Best catalysts have low “effectiveness factors” Effectiveness goes down as rate goes

up High rate implies low selectivity

Often want mass transfer limitations for selectivity

I prefer “mass transfer factor” not “effectiveness factor”

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Issues, Continued: Effective Diffusivity, De,

Unknown

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Figure 14.3 A cross sectional diagram of a typical catalyst support.

Knudsen Diffusion

Diffusion rate in gas phase controlled by gas – gas collisions

Diffusion rate in small poles controlled by gas – surface collisions

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Knudsen Diffusion

From kinetic theory

applies when

36

)pathfreemeanλ(λv3

1Dgas

diameterporeaa2V3

1Dk

gasa2

Knudsen Diffusion

compared to 0.86 in the gas phase

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21

212

3K M

AMU

K300

T

R1

a

sec

CM10x68.1D

22120100 Hforsec/cm.D poreÅ K

Simple Models Never Work

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Notice one molecule interferes with diffusion of second molecule

Summary

Catalytic reactions follow a catalytic cycle reactants + S adsorbed reactantsAdsorbed reactants products + S

Different types of reactionsLangmuir HinshelwoodRideal-Eley

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Summary

Calculate kinetics via Hougan and Watson; Identify rate determining step (RDS) Assume all steps before RDS in

equilibrium with reactants All steps after RDS in equilibrium with

products Plug into site balance

Predicts non-linear behavior also seen experimentally

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Query

What did you learn new in this lecture?

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