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Chemical Reaction

A + B C + D

Reaction rate law

- rA = k(T) fn (CA,CB)

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Individual steps in heterogeneous catalysis ( A P )

CAb

CAs

Cat-A

Cat-P

A

A

AA

PP

P

4

Boundary layer

P

Catalyst pellet

Catalyst pore

1

2 3

56

7

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Solid catalysed chemical reaction

- How many step involves in a heterogeneousreaction?

* 7 steps.

- What are they?

* 1. External mass transfer (Reactant)* 2. Internal mass transfer (Reactant)

* 3. Reactant adsorption

* 4. Surface reaction

* 5. Product desorption* 6. Internal mass transfer (product)

* 7. External mass transfer (product)

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e

Can we develop the reaction rate law including 7 steps?

* Non-practical very complex

Can we simplify our work?* Yes

How can we do it?

* Avoid external and internal mass transfer by working at high

fluid velocity and small particle

How to find the reaction rate law?1. Select a mechanism

2. Assume a rate-limiting step3. Find the expression for concentration of the adsorbed species Ci,S4. Write a site balance

5. Derive the rate law

6. Compare with data

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2. Find the reaction rate law of the reaction

SO2 + O2 SO32.1 Select a mechanism

O2 + 2 S 2 SO (1)

SO + SO2 SO3 + S (2)

2.2 Assume a rate limiting step

Reaction 2 is slow, so that it is the rate limiting step. If so then (r1/k1=0)

2.3 Find the expression for concentration of the adsorbed species

From (1)

k1

k-1

2

1

2

112

OSVOCkCPkr

k2

k-2

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From (2)

01

1

1

11

k

rand

k

kK

2

1

2

2 vOOSCPKC

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2.4 Write a site balance

2.5 Derive the rate law

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2.6 Compare with the experimental data- Differential reactor

- Measureable data

- Keep PSO2 constant, change PO2

- Keep PO2 constant, change PSO2

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Synthesizing a rate law

Mechanism and rate limiting step

Foglers Example (page 671): Reaction

C6H5CH(CH3)2 C6H6 + C3H6Cumene Benzene Propylene

C B P

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Reaction mechanism

C + S CS (Adsorption)

CS BS + P(gas) (Surface reaction)

BS B + S (Desorption)

kA

k-A

kS

k-S

kD

k-D

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We assume the surface reaction is rate limiting (slowest step).

What is the rate law if it is true?

Surface reaction

* kS : small

* kA : large or rAD/kA 0

* kD : large or rD/kD 0

S

SBP

SCSS K

CP

Ckr

We get these concentrations from

adsorption and desorption reaction.

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At t = 0 min

At low partial pressure of C we have 1 >> KC

PC,0

Initial rate increase linearly with the partial pressure of C

0,

0,,

0,

1 CC

C

SC

PK

Pkrr

tCS CKkkwith

0,

,

0, CSC Pkrr

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At high initial partial pressure of C we have KC PC,0 >>1

tconsK

k

rrC

SC tan,

0,

'

,OCr

0,CP

k

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Problem with solid catalyst.

* Catalyst decay

- Effective reaction rate- Catalyst life time

- Catalyst regeneration

Reactor selection and design

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We want to know how the catalyst decay affect the effective reaction rate.

Introduction of catalyst activity a(t)

Definition

Catalyst activity a changes with operation time.

)0(

)()(

'

'

tr

trta

A

A

),....,,()()('

, PBAreffA CCCFnTktar

Catalytic activity, time dependent

Specific reaction rate, temperature dependent

Gas phase concentration of reactants or poisons

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Deactivation by sintering

Example:

Second order decay reaction can be used

With a =1 at t = 0 min

dt

daakr dd

2

tk

ta

d

1

1)(

kd : sintering decay constant

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Stoichiometry

Combining

)1()1( 00 XV

NXCC AAA

)1)((' Xtak

V

W

dt

dX

'k

V

Wk

dttakX

dX )(1

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Substituting for a(t)

Solving for the conversion X

t

d

X

tk

dt

kX

dX

0011

tkk

k

X

d

d

1ln

1

1ln

dkk

dtkX

/)1(

11

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Temperature Time Trajectory

Slow catalyst decay rate

Increase feed temperature to maintain a constant conversion

First -order reactionA P

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Temperature Time Trajectory

Slow catalyst decay rate

A

d

d

dAA

E

Enk

TTR

EnEE

t

1

11exp1

0

0

First -order reactionA P

Order of decay rate law

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Notes: reaction rate units

smmolCkdtdNVr AA

A3/1

sgmolCk

dt

dN

wr

catA

A

A

/1 ''

smmolCkdt

dN

Sr A

AA

2'''' /1

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Chapter 11: External diffusion effects on heterogeneous reaction

Effective reaction rate constant

Two steps involve in the reaction (External mass transfer and

surface reaction )

Catalyst pellet

1

4Gas

Boundary layer

CAb

CAs

A

P

4

1

P

Reaction first-order

A P

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External mass transfer

Ficks law

with

- kc : Mass transfer coefficient (m/s)

- DAB : Diffusion coefficient (m2/s) (is a function of T and P)

- : Boundary layer thickness (m) (is unknown, depends on fluid velocity,

particle diameter, viscosity, density, temperature.)

- JAZ: average molar flux from the bulk fluid to the surface (mol/m2

.s)

AsAbc

A

ABAzCCk

dz

dCDJ

AB

c

Dk

Concentration

Positional coordinate

Gas

Cat

CAb

CAs

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Determination of kc by using correlation equation

Sherwood number

Flow over a sphere

Reynolds and Schmidt number

u : fluid flow velocity (m/s)

l : characteristic length (particle diameter) (m)

: kinematic viscosity (m2/s)

AB

c

D

lkSh

3/12/1Re6.02 ScSh

ABDSc

luRe

Frossling correlation

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Effect of fluid velocity on the effective reaction rate

reff

u

With mass transfer effect

Without mass transfer effect

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Example: Rapid reaction on a catalyst surface

Determine the effective rate of reaction per unit surface area ofcatalyst.

l = dp = 1 cmCAs = 0 mol/L

CAb = 1 mol/L

u = 0.1 m/s

= 0.5x10-6 m2/s

Catalyst pellet

1

4

Fluid

Boundary layer

CAb

CAs

A

P

P

Reaction first-order

A P

DAB = 10-10

m2

/s

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Solution

AsAbcMT CCkr

7.460Re6.023/12/1 Sc

D

dkSh

AB

pc

2000105.0

1.0()01.0(Re 126

1

sm

msmudp

500010

1051210

127

sm

sm

DSc

AB

161210

1061.401.0

7.46010

msm

sm

d

ShDk

p

ABc

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CAb = 103 mol/m3

rMT

= 4.61x10-6 m/s x (103 0) mol/m3 = 4.61x10-3 mol/m2s

This is the effective reaction rate.

smmolr effA23''

, /106.4

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Mass transfer-limited reaction in packed bed

Reaction A + B P

At steady state

z z+z

Mass balance for this sliceof the catalyst bed

0''\\ zAarFF ccAzzAzzAz

Molar rate in

Molar rate out

Molar rate of accumulation

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Determine the reactor length L necessary to achieve a conversion X

At z = L

z

U

akCkr ccAcA exp0

''

Reaction rate along the length of the reactor

0

0

A

ALA

C

CCX

LU

ak

X

cc11

ln

Ch t 12 Diff i d ti

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Chapter 12: Diffusion and reaction

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In order to calculate the effective reaction rate we need to know the

concentration CA in the catalyst pore.Lit. Octave Levenspiel; Chemical Reaction Engineering; 3rd ed. P.381

Single cylindrical pore First order reaction

0

0

L

CAS

C

x

xin xout

2 r

Characteristic length

Pore radius

Material balance for this slice of

catalyst pore

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Effective reaction rate and effectiveness factor

Effectiveness factor

We want to know how the concentration profile CAaffects the rA,eff

As

As

AeffA Catratereaction

Catratereaction

Catratereactionr

' ,

AseffACkr 1

'

, First-order reaction

Reaction rate in catalyst pore

Reaction rate on catalyst surface

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Effective reaction rate and effectiveness factor

mL

mLdx

mL

xLm

LLCk

dxxCkl

Asr

L

x

Ar)tanh(

)cosh(

))(cosh(1)(

0

0

1

1tanh

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Effect of pellet size on effective rate constant

keff

Particle radius

With internal mass transfer

effect

Without internal

Mass transfer effect

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Fogler page 849

One more problem

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One more problem.

Pressure drop in reactor

Liquid reaction : ignore the effect of pressure drop

Gas reaction: effect of pressure drop is important

Pressure drop and the rate law

T

T

P

P

X

XCC

ii

Ai

0

0

0

1

Concentration is a function of pressureWe need to relate the pressure drop to

the reactor design equation

Fogler page 114

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Example:

Packed Bed Reactor (PBR)

For second order gas phase reaction

2A B + C

Design equation for PBR

2'

AA Ckr

'

0 AAr

dW

dXF

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Isothermal reactor T=T0

T

T

P

P

X

XCC AA

0

0

0

1

1

2

0

2

0

0

1

1

P

P

X

XC

kdW

dX

F

A

A

2

0

2

0

0

1

1

P

P

X

X

v

kC

dW

dXA

We need to relate the pressure drop to

the reactor design equation

Ergun equation

Fogler page 114

Design equation for PBR

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Catalyst weight in tubular packed bed reactor

With

cczAW 1 Density of solid catalyst

1cb

00

00

1 T

T

cc F

F

T

T

P

P

AdW

dP

00

1

2

PAcc

0P

Py

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002 T

T

F

F

T

T

ydW

dy

XF

F

T

T 10

Fogler P. 113

012 TT

XydW

dy

For isothermal reaction and = 0

dwydy

y W

1 02

WP

Py 1

0

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2

0

2

0

0

1

1

P

P

X

X

v

kC

dW

dX A

0P

Py

012 TT

XydW

dy

00

1

2

PAcc