CHE411 Fall 2010-Chemical Reaction Engineeirng-Ahmed a Abdala

CHEG411: Chemical Reaction Engineering

Ahmed Abdala

Chemical Engineering program The Petroleum Institute

Abu Dhabi, UAE

Chemical Reaction Engineering CHEG 411 Fall 2010


Introduction


Chemical Plants: The Refinery EXAample

A refinery converts crude oil into valuable petroleum products such as gasoline, kerosene, diesel, aviation fuel, .etc

$$$

Distillation Cracking, reforming, Blending

Petrochemical Feedstock,

Coke, Sulfur

LPG

Crude Oil

Catalyst/H2

Gasoline Kerosine Diesel


Processes in Chemical Plants

The aim of any chemical plant is to produce valuable products form available raw materials

A Chemical plant involves chemical and physical processes

A chemical process is a process that involves chemical reactions

A physical process includes separation, miXAing, or blending of different materials

$$$

Physical Processes

Chemical Processes

Physical Processes

products Products

Raw Materials

Reactants Catalyst

products


Course Overview

This course focuses in the chemical processes part where chemical reactions take place

It focuses mainly on the design of the vessels where these reactions occur, i.e. the chemical reactors

We do not, however, focus on the mechanical design of the reactor


Course Objectives

The aim of the course is to answer the following questions:

1. Will a certain reaction take place under certain conditions (temperature, pressure and catalyst)?

2. If the answer for question 1 is yes, how fast will the reaction proceed?

3. What type of reactor(s) should we use?

4. What is the size of the reactor(s) and, if more than one reactor is to be used, how those reactors should be arranged?


Approach

1. To answer the first question we need information about thermodynamic functions for the products and reactants at the reaction conditions Chemical reaction equilibria [thermodynamics]

2. To answer the second question we need to have the necessary information to determine the reaction rate equation Kinetics

3. To answer the third and fourth questions we need information about the reaction rate and reactants/products flow rates Kinetics, transport (fluid, heat and mass)


Chemical Reaction Equilibria Consider the gas phase reaction:

From the Law of Mass Action:

The true (dimensionless) equilibrium constant K is:

ai is activity of species i

fi is the fugacity of species i fi0 is the fugacity of species i at the standard state

For gases, standard state is 1 atm i is the activity coefficient

++ ++ b c dAA bB Dd BcC D Ca a a

a

=c d

bC D

A B

K ( 1 )a a

a

a aa a

ii i0

i

fP ( 2 )

f= =ia


Chemical Reaction Equilibria

The true equilibrium constant, K

K=1 for ideal gas

The pressure equilibrium constant, KP

Pi is the partial pressure of species i, Pi=Ci RT

The concentration equilibrium constant, KC

For ideal gases:

+ +

= = =c d c d c d

d c bd c b 11 s a as

b b b

a a atmatmactivity pressureequilibrium constant equilibrium const

C D C D C DP

A B A B A B

ant

P PK K K ( 3 )P P

a a a a a a

a a a

a aa a

( ) = = + P Cc d bK K RT ; ( 6 )1a a a

=c d

bC D

PA B

P PK ( 4 )P P

a a

a

=c d

bC D

CA B

C CK ( 5 )C C

a a

a


Equilibrium Constant versus Temperature

KP is a function of temperature only

Integrating

If CP is assumed to be constant (CP=0), then

Rx ,TP2

Hd ln K van' t Hoff ' s equation (7 )dT RT

=

( )R

0PRx ,T RP

2

H C T Td ln K ( 8 )dT RT

+ =

( ) R

1

0P PRx ,T RP ,T

P ,T R R

H C T TK 1 1 C Tln ln ( 9 )K R T T R T

= +

R

R

Rx ,TP ,T P ,T

R

H 1 1K K exp ( 10 )R T T

=


Equilibrium Constant versus Temperature

For eXAothermic Reactions: As T increases, the equilibrium

shifts to the left. i.e. both K and XAe decreases

For endothermic Reactions: As T increases, the equilibrium

shifts to the right. i.e. both K and XAe increases

T

K

endothermic reactions

T

K

exothermic reactions


Equilibrium Constant and Free Energy

The equilibrium constant at temperature T can be calculated from the change in the Gibbs free energy as follow:

where

G is related to the reaction enthalpy and entropy as follow:

13 = ( )G H T S

( ) ( )0 11 = ln ( )RxRT K T G T0 0 0 0 0 12Rx C D A BG cG dG aG bG = + - - ( )


Example: Calculation of Equilibrium Constant

Acetic acid is esterified in the liquid phase with ethanol at 100 C and atmospheric pressure to produce ethyl acetate and water according to the reaction:

For this reaction G0298 = -4.65 kJ/mol and H0298 = 3.64 kJ/mol

Estimate: 1. The equilibrium constant for the reaction at 100 C. For this

reaction G0298 =-4.65 kJ and H0298 = 3.64 kJ 2. The mole fraction of the reaction components at equilibrium,

if initially there is one mole of each of acetic acid and ethanol.

3 2 5 3 2 5 2CH COOH C H OH CH COOC H H O+ +


Solution: Equilibrium Constant

Calculation of the equilibrium constant: K298

K373

Assume constant Cp between 298 and 373 K

( ) ( ) = = = 0RxG 298 4650ln K 298 1.876Rx 298 8.314 x 298.5

( ) ( )= =K 298 exp 1.876 6.527

( )( )

= = =

0298K 373 H 1 1 3640 1 1ln 0.295

K 298 R 298 373 8.314 298 373

( ) ( ) ( )= =K 373 K 298 exp 0.295 8.77

( ) ( ) ( )

=

Rx RP P 1

1

H T 1 1K T K T expR T T


Solution: Equilibrium Composition

The activity coefficient () is assumed to be 1

Calculation of equilibrium composition

3 2 5 3 2 5 2CH COOH C H OH CH COOC H H O+ +

( ) ( )3 2 5 2

3 2 5

22CH COOC H H O

C 2CH COOH C H OH

C C X XKC C 1 X1 X

= = =

( )

2X8.77 X 0.75

1 X

= =

3 2 5 2

3 2 5

CH COOC H H O

CH COOH C H OH

K =a aa a

3 2 5 2

3 2 5

CH COOC H H OC

CH COOH C H OH

C CK K

C C= =

CH3COOH C2H5OH CH3COOC2H5 H2O

Initial, Ni0 1 1 0 0

Change, dNi -XA -XA XA XA

Equilibrium, Nie

1-XA 1-XA XA XA

Mole fraction

(1-XA)/2 (1-XA)/2 (XA)/2 (XA)/2

(1-0.75)/2 = 12.5%

(1-0.75)/2 = 12.5%

(0.75)/2 = 37.5%

(0.75)/2 = 37.5%


EXample 2: Gas phase

The water-gas shift reaction: is carried out under different conditions. Calculate the equilibrium conversion of steam following cases:

a) The reactants consist of 1 mol of H2O vapors and 1 mol of CO. the temperature is 1100 K and the pressure is 1 bar

b) Same as in a but the total pressure is 10 bar c) The reactants consist of 2 mol of H2O vapors and 1 mol of CO. the

temperature is 1100 K and the pressure is 1 bar d) The reactants consist of 1 mol of H2O vapors and 2 mol of CO. the

temperature is 1100 K and the pressure is 1 bar e) The reactants consist of 1 mol of H2O vapors, 1 mol of CO, and 1

mol of CO2. the temperature is 1100 K and the pressure is 1 bar

The equilibrium constant (K) for this reaction at 1100 K is 1. Assume ideal gas.

2 2 2CO H O CO H+ +

Chemical Reaction Engineering

Chapter 2: Conversion and Reactor Sizing


Batch Reactor Design Equation

The reaction

Can be rewritten as:

Conversion, Xi, is used to quantify how far the reaction proceeds in the right direction

Conversion, XA, is the number of moles of A that have reacted per mole of A fed to the system

The maximum conversion For irreversible reaction is 1 For reversible reaction is the equilibrium conversion, Xe

aA bB cC dD+ +

+ +c dbA B C D

a a a

AMoles of A reactedX

Moles of A fed=


Assume number of moles of A initially fed to the reactor is NA0 Number of moles of A reacted after a time t

Number of moles of A that remain in the reactor after a time t (NA):


0A

Mole of AMole of A Mole of Areactedreacted

fed Moles of A Fed( consumed )

=N X

=

= =

Mole of AMole of A Mole of Athat

initially fedin reactor have been consumed

to reactor atat time t by chemical reaction

t 0

( )= = A A0 A0 A A0N N N X N N 1 X


For ideal batch reactor No spatial variations in concentration nor reaction rate

The mole Balance for species A:

Since


=A AdN

r Vdt

( )= = AA A0 A0dN dXN N 1 X 0 Ndt dt

= =A A A0 AdN dXr V N r Vdt dt

= A0 AdXN r Vdt Design equation ( differential form )

=

X

A0 0A

dXt Nr V Design equation ( Integral )


For constant volume batch reactor V=V0

The Design Equation becomes: The differential form

The integral form

Constant Volume Batch Reactor

X X

A0 A00 0A A

dX dXt N t Cr V r

= =

( )= = =A 0A A A 0 A

d N / VdNr V r V r

dt dt

AA

dC rdt

=

A0 A A0 AdX dXN r V C rdt dt

= =


Conversion for Flow Reactors

For flow reactors

=

= =A0Moles of A fedF

tim

Mole of A reactedXMoles of A fed

Mole of A reacted Mole of A reactedXMoles of A fede time

=

Molar Flow rate Molar rate at Molar flow rateat which A is which A is consumed at which A

fed to the system within the system leaves the system

A0 A0 A F F X = F

( )= A A0 F F 1 X


For ideal gas system

Molar Flow Rate for Ideal Gas System

=A A0 0F C

A0 A0 0A0

0 0

P y PC ideal gas systemRT RT

= =

A0 A0 0A0 A0 0 0 0

0 0

P y PF C ideal gas systemRT RT

= = =


Design Equation for CSTR

For the reaction:

The volume of batch reactor, V

With

The reactor volume becomes

+ +c dbA B C D

a a a

=

A0 A

A

F FV

r

( )= A A0 F F 1 X

( ) =

A0 A0 A0

A

F F F XV

r

( )=

A0

A exit

F X V

r


Tubular Flow Reactor PFR

The differential Equation for PFR:

But Hence, the differential Equation becomes

The integral Equation for PFR:

=Ar dVA-dF

( )= = A A0 A A0 F F 1 X dF F dX

=A A0r F dVdX

j

j 0

F

Fj

Vr

= jdF

X XA0A00 0

A A

V Fr r

= =

-F dx dx


PFR volume = area under curve (dashed area)

Volume of Plug Flow Reactor

The volume of PFR can be obtained by plotting The parameter FA0/(-rA) versus conversion X

The volume correspond to the area under curve Dashed Area

The area of equivalent CSTR reactor is the area under curve plus the dotted area CSTR volume= the area of

the rectangle


Packed-bed Reactor (PBR)

The differential form of the design Equation for PBR:

But Hence, the differential Equation becomes

The integral Equation for PBR:

='Ar dWA-dF

( )= = A A0 A A0 F F 1 X dF F dX

='A A0r F dWdX

= j

j 0

F

'FA

Wr

jdF

X XA0A0' '0 0

A A

W Fr r

= =

-F dx dx


CSTRs in Series

Consider two CSTRs connected in series: For the first reactor

Mole Balance around the second

reactor

The volume of 2 CSTRs connected in series is smaller than the volume of one CSTR to achieve the same conversion

( )=

A0 1

1A 1

F X V

r

( ) ( )

+ = + =

+ =A1 A 2 2 A 2

A0 A0 2 2 A 2

In Out Generation 0F F V r 0F 1 X F 1 X V r 0

( )( )

=

A0 2 1

2A 2

F X X V

r


CSTRs in Series

For n CSTRs connected in series we have:

As the number of CSTRs increases: The volume of the CSTRs

becomes equivalent to that of PFR to achieve the same conversion

PFR can be modeled with a large number of CSTRs connected in series

( ) ( )=

A0

n n n 1A 1

F V X X

r


PFRs in Series

For two PFRs in series we have: The volume of the first reactor

The volume of the second reactor

The total volume of the two reactors

The volume of n PFRs in series is the same as the volume of one PFR to achieve the same conversion

1X

1 A0 0A

V Fr

=dx

2

1

X

2 A0 XA

V Fr

=dx

1 2 2

1

X X X

1 2 A0 A0 A00 X 0A A A

V V F F Fr r r

+ = + = dx dx dx


Combination of CSTRs and PFRs in Series

CSTRs and PFRs can be connected in series

The arrangement of the reactors in this case is of high importance

For the case of two CSTRs and one PFR: Option 1: CSTR-CSTR-PFR

Option 2: CSTR-PFR-CSTR

Option 3: PFR-CSTR-C,STR FA1, X1 FA2, X2

FA0, X=0 FA3, X3

FA1, X1 FA2, X2

FA0, X=0

FA1, X1

FA2, X2 FA0, X=0

FA3, X3


Combination of CSTRs and PFRs in Series

FA1, X1 FA2, X2

FA0, X=0

FA3, X3

X

F A0/

(-r A

)

X1 X1 X1

X

F A0/

(-r A

)

X1 X1 X1

V1 V2

V3 V1

V2 V3

FA1, X1 FA2, X2

FA0, X=0 FA3, X3


Space Time and Space Velocity

Space time is the time necessary to process one reactor volume of fluid based on the entrance conditions

The space time is equal to the mean residence time, tm

Space Velocity (SV) is the reciprocal of the space time

0

Vv Reactor

Mean Residence Time

Batch 1 5 min - 20 h

CSTR 1 0 min- 4 h

Tubular 0. 5 s - 1 h

=0

v 1SVV


Space Velocity

Space velocity commonly reported as either: Liquid Hourly Space Velocity, LHSV

Gas Hourly Space Velocity, GHSV

0 liquid @60F

vLHSV

V

0 Gas @STP

vLHSV

V

Chemical Reaction Engineering CHEG 411 Fall 2010

Chapter 3: Rate Laws and Stoichiometry

Ahmed Abdala


Abu Dhabi, UAE


Rate Laws and Stoichiometry

Chapter 3


Types of Reactions

Reaction can be classified according to:

Phases

Homogeneous Involves one

phase Heterogeneous Involves more

than one phase Interfacial Occurs on the

interface of two or more phases

Direction

Irreversible Reversible

Molecularity

Unimolecular Bimolecular Termolecular


The Rate of Reaction and the Rate Law

The rate of reaction is usually reported as the rate of disappearance of the limiting species A, -rA

-rA is function of temperature and concentration

k(T) is the rate constant and is a function of temperature

The rate equation is an algebraic equation that relates rA to the species concentration

For the reaction: The rate constant for different species are related as follow:

.

( ) ( )A A Br k T fn C ,C , ..... =

aA bB cC dD+ +

b C iA D

i

k k kk ka b c d v

= = = =b C iA Di

r r rr ra b c d

= = = =


Reaction Rate Equation: Power Law Models

Zero Order

First Order

Second Order

n-th Order

[ ]; = =A

moler k kvolume time

[ ] 1; = =A Ar k C k Time

[ ]2 ; = =A A

Volumer k C kMole Time

[ ]1 1;

nn

A AMoler k C k

Volume Time

= =


Reaction Rate Constant: k

k is the specific reaction rate (constant) and is given by the Arrhenius Equation: A is the frequency factor (same units as k)

E is the activation energy, kcal/mol

R is the gas constant

T is the Absolute temperature

ERTk A e

=

T

k

Reaction coordinates

Reac

tant

s

Prod

ucts

E

H

r

Pote

ntia

l Ene

rgy

Ln(k

)

1/T

High E

Low E

Slope=-E/R


Reaction Rate Constant: In Class Exercise

The rule of thumb that the rate of reaction doubles for a 10 C increase in temperature occurs only at a specific temperature for a given activation energy. Develop a relationship between the temperature and activation

energy for which the rule of thumb holds.

Determine the activation energy and the frequency factor from the following data:

k (min-1) 0.001 0.050 0.500 2.00

T ( C) 0 100 200 300


Power Law Model and Elementary Laws

The dependence of reaction rate on concentration is usually determined by experimental observation

It can also be postulated from theory The rate law is the product of concentration of the

individual reacting species each of which is raised to a power : Where

is the reaction order with respect to species A is the reaction order with respect to species A n= + is the overall reaction order

The dimension of the rate constant k will vary depending on the reaction order

A A A Br k C C =

{ } ( )1 nConcentration

kTime


Elementary Reactions

Elementary reaction are reactions that involves a single step

The stoichiometric coefficients in elementary reaction are identical to the powers of the Rate Law

The equation can be written based on the stoichiometric equation For the reaction:

The rate equation is:

aA bB cC dD+ +

a bA A A Br k C C =


Pseudo Elementary Reactions

The powers of the rate law for some reactions are identical to the stoichiometric coefficients but it involves more than one step These reactions are nonelementary but follow elementary rate

law

Example is the oxidation of nitric oxide:

which has the rate equation

but the mechanism involves more than one step

22 NO O 2 NO+

2

2NO NO NO Or k C C =


Nonelementary Reactions

Many reactions does not follow simple rate laws: Reactions with non-integer reaction order

Reaction with rate equation that cannot be separated into temperature

dependent term and concentration dependent term:

The overall reaction order cannot be stated At the beginning of the reaction:

low concentration of O2, the reaction is apparent first order

1st order w.r.t. N2O and zero order w.r.t. O2

At the end of reaction: high concentration of O2 and low concentration of N2O The reaction has an apparent order of 0

1st order w.r.t. N2O and order of -1 w.r.t. O2

32

22 2 CO CO CO ClCO Cl COCl ; r k C C+ =

2 2

2

2

N O N O2 2 2 N O '

O

k C2N O 2N O ; r

1 k C + =

+


Heterogeneous Reaction Rate

For catalyzed reactions the reaction rate is expressed as the rate of disappearance of species A per unit mass of the catalyst:

For gas catalyzed reaction the rate law is usually written in terms of partial pressure

Where k is the rate constant, [mole/(kg cat.s.kPa2)]

KB, KT are the adsorption constant, [kPa-1]

-rA and -rA are related through the bulk density b

'A

molesrcatalyst mass time

+ + =+ +

2

6 5 3

H TCatalyst '6 5 3 2 6 6 4 C H CH

B B T T

kP PC H CH H C H CH ; r

1 K P K P

( ) = 'A b Ar r


Reversible Reactions

For the reversible reaction:

The equilibrium constant, KC

Assuming elementary reaction The rate of disappearance of A:

The rate of formation of A:

The overall rate of formation of A,

The overall rate of disappearance of A:

f

r

k

kaA bB cC dD+ +

c dCe De A

C a bAe Be A

C C kKC C k

= =

a bA ,forward A A Br k C C =

c dA ,reverse A C Dr k C C=

( ) c d a bA A ,reverse A ,forward A C D A A Br r r k C C k C C= =

a b c dA A A B A C Dr k C C k C C =



Similarly

Remember:

a b c d a b c dAA A A B A C D A A B C D

A

c da b C D

A A BC

kr k C C k C C k C C C Ck

C Ck C CK

= =

=

c da b C D

B B A BC

c da

c da b C D

C C A BC

b C DD D A B

C

C

C Cr k C CK

C Cr k

CrK

CK

k

C

C C

=

=

=

CA B D CA B Dkk k k ; kk k kaa c cb d b d

= = == = =

CA B Drr r ra b c d

= = =


In Class Exercise:

Write the rate law for the following reactions assuming each reaction follows an elementary rate law:

4 9 3 2 5 3 4 9 2 5C H OH CH COOC H CH COOC H C H OH+ +

2 6 2 4 2C H C H H +

2 212 4 22

CH CHC H O

O

+

( ) ( )3 3 2 6 3 33 3CH COOC CH C H 2CH COCH +

4 10 4 10n C H i C H


Stoichiometry

For the reaction

The relative rates of reaction are related through the stoichiometric coefficients as follow

We can rewrite the reaction as follow:

Everything is put on a basis of per mole of A

We can now setup a stoichiometric table for different reaction systems

CA B Drr r ra b c d

= = =

+ +aA bB cC dD

+ +b c dA B C Da a a


Stoichiometric Table: Batch System

For the reaction

Which can be written As

The Stoichiometric is as follow:


Species Initial Moles Change Remaining moles Ni0 Ni Ni

A NA0 - (NA0X)

B NB0 - b/a (NA0X)

C NC0 c/a (NA0X)

D ND0 d/a (NA0X)

I, inert NI0 0

Totals NT0 NA0X(d/a+c/a- b/a- 1 )

t=0 NA0 NB0 NC0 ND0 NIO

NA NB NC ND NI

t=t

+ +aA bB cC dD

= A A0 A0N N N X

= B B 0 A0bN N N Xa

= +C C 0 A0cN N N Xa

= +D D0 A0dN N N Xa

=i i 0N N = + +

T T 0 A0d c bN N 1 N Xa a a


Stoichiometric Table

For the reaction Moles reacted of A

NA = NA0X Moles reacted of species i

Ni= (i/A)NA0X Total number of moles, NT

Where ,

= +T T 0 A0N N N X

+ +aA bB cC dD

= + d c b 1a a a


Concentration Equations for Batch System

Let

= = A0 A0AA

N N XNC

V V

= AiN

CV

= =

B 0 A0B

B

bN N XN aCV V

+= =

C 0 A0C

C

cN N XN aCV V

+= =

D0 A0D

D

dN N XN aCV V

= = =i 0 i 0 i 0iA0 A0 A0

N C yN C y

=

iA0 B

Ai

N XC

V

= = A0 A0AA

N N XNC

V V = =

A0 BB

B

bN XN aCV V

+ = =A0 C

CC

cN XN aCV V

+ = =A0 D

DD

dN XN aCV V


Concentration Equations for Constant Volume Batch System For constant volume system, V=V0

Liquid phase reaction Gas phase reaction with =0

The concentration of species i becomes, Ci=Ni/V0

Now the rate equation can be reported in terms of CA0 and X

( ) ( )

= = = A0AA A00

N 1 XNC C 1 X

V V

= = =

A0 BB

B A0 B0 0

bN XN baC C XV V a

+ = = = +

A0 DD

D A0 D0 0

dN XN daC C XV V a

+ = = = +

A0 CC

C A0 C0 0

cN XN caC C XV V a


Stoichiometric Table: Flow Systems

For the reaction:

Entering

FA0 FB0 FC0 FD0 FI0

FA FB FC FD FI

Leaving

a c dA B C D

b a a+ +

Species Feed rate to reactor Change within reactor Effluent rate from reactor

A

B

C

D

I

Total

= + + = +

T T A T A

d c bF F F X F F X

a a a0 0 0 01

+ =

A A

d c bF X F X

a a a0 01

Ab

F Xa 0

AF X0

+ Ac

F Xa 0

+ Ad

F Xa 0

0

( )= A AF F X0 1

=

B A B

bF F X

a0

= + C A C

cF F X

a0

= + D DA

dF F X

a0

=I A IF F 0

( )= + + + +T A B C D IF F 0 0 1

=B A BF F 0 0

=C A CF F 0 0

=D A DF F 0 0

=I A IF F 0 0

AF 0



Stoichiometric Table: Flow Systems

For the reaction:

Species Feed rate to reactor Change Effluent rate from reactor

Effluent Concentration

Effluent Conc, v=constant

A

B

C

D

I

Total

= +T T AF F F X0 0AF X 0

Ab

F Xa 0

AF X0

A

cF X

a 0

A

dF X

a 0

0

( )= A AF F X0 1

=

B A B

bF F X

a0

C A C

cF F X

a

= +

0

D A D

dF F X

a

= +

0

=I A IF F 0

= +

T A i

i

F F 0 0 1

=B A BF F 0 0

=C A CF F 0 0

=D A DF F 0 0

=I A IF F 0 0

AF 0( )

= AAAF xF

C

0 1

( )( )= =

baA BB

B

F XFC

0

( )( )caA CCC

F XFC

+= =

0

( )( )d aA DDD

F XFC

+= =

0

= = A IIIFF

C

0

( )= A AC C x0 1

( )( )= b aB A BC C X0( )( )c aC A CC C X= +0( )( )d aD A DC C X= +0

=I A IC C 0

( )( )= =

A ii

i

iF XF aC

0

( )( ) ( )( )= = = =Ai i A iF i iC X C X ; constanta a 0 0 00



In Class Exercise

For the following liquid-phase elementary reaction:

Write the rate law in terms of conversion, XA

+ 2 A B 2C

Species Feed rate to reactor Change Effluent rate from reactor

Effluent Concentration

Effluent Conc, v=constant

A

B

C

I

Total


Variable Volume Batch Reactor

For gas- phase reaction in a batch reactor where the volume of reactor is not constant

The concentration of individual component can be determined by expressing the volume at any time t as follow:

At time t=0 this equation becomes

The reaction volume at time t, V, as function of the initial volume, V0, becomes

TPV ZN RT=

0 0 0 T 0 0P V Z N RT=

0 T0

0 0 T 0

P NT ZV V

P T Z N

=


Variable Volume Batch Reactor

But Then

Where

At X=1, NT=NTf and becomes

And the volume at time t becomes (assuming Z Z0)

T A0A0

T 0 T 0

N N1 X 1 y X 1 X

N N = + = + = +

( )000

P TV V 1 X

P T

= +

T T 0 A0N N N X= +

T T 0A0

T 0

N Ny

N X

= =

Tf T 0

T 0

N N

NChnagein total numbers of Moles at completeconversion

total moles fed

=

=


Variable Volumetric Flow Rate Flow Reactors

The total concentration at any point of the reactor, CT is given by

The concentration at the entrance of the reactor, CT0

The volumetric flow rate at some point along the reactor, v as function of the volumetric flow rate at the entrance, v0

TT

F PC

ZRT= =

T 0 0T 0

0 0 0

F PC

Z RT= =

0T0

T 0 0

PF TF P T

=


Concentrations in Terms Other Than Conversion

The concentration of species j at any point of the reactor, Cj is given by

The concentration at the entrance of the reactor, CT0

j jj

0T0

T 0 0

F FC

PF TF P T

= =

jT 0 0j

0 T 0

FF TPC

F P T

=

j 0j T 0

T 0

F TPC C

F P T

=


Variable Volumetric Flow Rate Reactors

To express the concentration as function of conversion for variable v reactors We have and

Then

The concentration of species j, Cj becomes

0T0

T 0 0

PF TF P T

=

T T 0 A0F F F X= +

( )T 0 A0 0 00 A0T 0 0 0

F F X P PT T1 y X

F P T P T

+

= = +

( ) 000

P T1 X

P T

== +

( )( )

( )( )

A0 j j A0 j jj 0j

000

0

F X C XF TPC

P T1 XP T1 X

P T

+ + = = =

+ +


Summary


The gas-phase reaction

is to be carried out isothermally. The molar feed is 50% H2, and 5O% N2, at a pressure of 16.4 atm and 227 K.

a) Construct a complete stoichiometric table. b) What are CA0, , and ?. Calculate the concentrations of

ammonia and hydrogen when the conversion of H2 is 60%. c) Suppose by chance the reaction is elementary with kN2 =

40 dm3/mol/s. Write the rate of reaction solely as a function of conversion for:

i. A flow system ii. A constant volume batch system

In-Class Exercise (Fogler P3-13)

312 2 32 2 H NHN +


Chapter 3 Review Rate Laws and Stoichiometry


Reaction Rate Equation

-rA is function of temperature and concentration

The rate constant for different species are related as follow:

Rate constant

Units

( ) ( )A A Br k T fn C ,C , ..... =

aA bB cC dD+ +

b C iA D

i

k k kk ka b c d v

= = = =b C iA Di

r r rr ra b c d

= = = =

[ ]1 1 =

nmolekvolume time

E 1ln k ln AR

EkT

A exp ;RT =

=

Ln(k

)

1/T

High E

Low E

Slope=-E/R


Rate Law

For the reaction:

The power-law rate equation:

If the reaction is elementary or pseudo-elementary

Non-power-law forms

A A A Br k C C =

aA bB cC dD+ +

a bA A A Br k C C =

2 2

2

2

N O N O2 2 2 N O '

O

k C2N O 2N O ; r

1 k C + =

+


Heterogeneous Reaction Rate

Gas catalyzed reaction the rate law is usually written in terms of partial pressure

Where k is the rate constant

KB, KT are the adsorption constant, [kPa-1]

-rA and -rA are related through the bulk density b

'A

molesrcatalyst mass time

+ + =+ +

2

6 5 3

H TCatalyst '6 5 3 2 6 6 4 C H CH

B B T T

kP PC H CH H C H CH ; r

1 K P K P

( ) = 'A b Ar r

'A b Ak k=



For the reversible reaction:

The equilibrium constant, KC

Assuming elementary reaction The rate of disappearance of A:

The rate of formation of A:

The overall rate of formation of A,

The overall rate of disappearance of A:

f

r

k

kaA bB cC dD+ +

c dCe De A

C a bAe Be A

C C kKC C k

= =

a bA ,forward A A Br k C C =

c dA ,reverse A C Dr k C C=

( ) c d a bA A ,reverse A ,forward A C D A A Br r r k C C k C C= =

a b c dA A A B A C Dr k C C k C C =


Stoichiometry

For the reaction

We can rewrite the reaction as follow:

Everything is put on a basis of per mole of A

Stoichiometric

+ +aA bB cC dD


i A0 iiN N Xa

= +

i A0 iiF F Xa

= +

i A0iC C Xa

= + liquid phase

Gas Phase0T

0T 0 0 0

PF T ZV VF P T Z

=

T 0 0 0i A0T 0

F T Zi PC C Xa F P T Z

= +

0T0T 0 0 0

PF T ZF P T Z

=


Stoichiometric Table

+ +aA bB cC dD

= + d c b 1a a a

T T 0 A0F F F X= +

T T 0 A0N N N X= +

A0TA0

T 0 T 0

FF1 X 1 y X 1 X

F F = + = + = +

A0TA0

T 0 T 0

NN1 X 1 y X 1 X

N N = + = + = +

Chemical Reaction Engineering CHEG 411 Fall 2010 Chapter 4

Ahmed Abdala


Abu Dhabi, UAE


Isothermal Reactor Design

Chapter 4

Start

VA

A A A

dNF F r dV

dt + =0

Mole Balance to specific reactor design equations in terms of conversion

Evaluate the design equations numerically or analytically to

Estimate the reactor volume the reaction time, or conversion

Determine the rate law in terms of concentration

Obtain rA=f(X)

Combine mole balance, rate law and stoichiometry and transport

law, and P terms in an ODE solver

Using stoichiometry, express concentration as function of X

End

General mole balance equation

Chapter 1

Chapter 2

Chapter 2

Yes

No

Chapter 3 Elementary reaction

Chapter 3

Chapter 4

Chapter 4

is rA=f(X)

given?

Liquid phase or gas phase with P=P0

Yes

No


Algorithm for Reaction Design

The solution of reactor design involve the following steps: 1. Mole balance

Based on the limiting species

2. Rate law

Based on reaction type, mechanism, or experimental data

3. Stoichiometry

4. Combine the mole balance, the rate equation, and the Stoichiometry relations

5. Evaluate to get reactor volume, reaction time, or conversion

Analytical, numerically, graphically, or using software

Mole Balance

Rate Law Stoichiometry

Combine Evaluate


French Menu


Scale UP of Liquid Phase Batch Reactor Data to the Design of a CSTR To design of a full-scale plant:

1. Microplant, laboratory-bench-scale unit is used to collect data about the process

2. A pilot-plant maybe designed based on the laboratory data

3. The full-plant is designed by scaling up the pilot plant data

Step 2 can be surpassed and a full scale plant may be designed based on laboratory data. However, this requires thorough understanding and careful analysis of the laboratory data


Liquid Phase-Batch Operation

For a batch system, the mole balance equation is:

For a liquid phase reaction, the volume is constant and the mole balance equation can be written in terms of concentration as follow:

This form will be used to analyze the reaction rate data

AA

dNr

V dt

=

1

( )AA A Ad N VdN dN dCV dt V dt dt dt

= = =

0

0

1 1

AA

dCr

dt=


For batch system, the time calculated to achieve certain conversion is the reaction time

The total time required to process one batch, total cycle time, tt

The reaction time is usually optimized with the processing time to produce the maximum number of batches per day

Batch Operation

c f e R c

Total Cycle time Filling time Heat time time Cleaning timet t t t t

= + + +

= + + +

Reaction


The volume of CSTR reactor to achieve conversion X

And the space time,

For a first order liquid phase reaction:

-rA = k CA=k CA0(1-X)

Design of CSTR

( )= =

A A

AA exit

F X C XV

rr0 0 0

= =A

A

C XVr

0

0

=+

AA

CC

k0

1

( )=

X

k X

1 = +k

Xk

1


Design of CSTR

Damkohler number, Da, is the ratio of the rate of reaction at entrance to the entering flow rate of A

For a first order reaction: The conversion is related to Da as follow

At high Da, Da 10, X is greater

than 90% At low Da, Da 0.1, X is less

than 10%

For a second order reaction:

A

A

r V ReatcionrateDa

F Convectionrate

= =00

A A

A A

r V kC VDa k

F C

= = =0 00 0 0

A AA

A A

r V kC VDa kC

F C

= = =2

0 00

0 0 0

k DaX

k Da

= =

+ +1 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 10 20 30Da

X

f irst order reaction


CSTR in Series: First Order Reactions

For two CSTRs in series:

If both reactors are of equal size ( 1=1= ) and operates at same temperature (k1=k2=k)

For n reactors connected in series

If the final conversion based on reactor n is X

CA0

CA1 X1

CA2 X2

-rA2 V2

-rA1 V1

A AA A

C CC and C

k k = =

+ +0 1

1 21 1 2 21 1

( )( ) ( )A AA

A

C CCC

k k k k = = =

+ + + +0 01

2 22 2 1 1 2 21 1 1 1

( ) ( )= =

+ +A A

An n n

C CC

k Da0 0

1 1

( )( )

AAn A n

CC C X

Da= =

+0

0 11 ( ) ( )n n

XDa k

= = + +

1 11 11 1


CSTR in Series: First Order Reactions

For n CSTRs connected in series

The rate of disappearance of species A the nth reactor is

( ) ( )n nX

Da k= =

+ +

1 11 11 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14n

X

Da = 0.1

Da = 0.5Da = 1

( )A

A An n

Cr kC k

k = =

+0

1


CSTR in Parallel

For n equal size CSTRs connected in parallel, operates at the same temperature and the feed is distributed equally among them

The rate in each reactor is

The volume of each reactor

And the total volume V,

A ii

Ai

F XVV

n n r

= =

0

FA0

Xn

-rA1 V1

FA01

FA02

FA0n

X1

nX X ................ X X= = = =1 2

A A An Ar r ................ r r = = = = 1 2

A i A

Ai A

F X F XV

r r= =

0 0


Design of PFR

Recall the assumption for PFR: No radial variation in velocity, temperature, concentration, nor

reaction rate Plug flow assumption is valid when Re > 2300

The design equations of PFR Differential:

Integral:

Reactants Products A0 A

dXF rdV

=

X

A0 0A

dXV Fr

=


Second Order Liquid Phase Reaction in PFR

For the second order reaction The rate law is The differential form of the design equation becomes

1 . liquid phase isothermal reaction

From stoichiometry Combine the design equation with the stoichiometry

Evaluate

A Products

2A Ar kC =

2

2

DaX

1 Da=

+

( )A A0C C 1 hyX w=

2A

A0

kCdXdV F

=

( ) ( )2 22 2A0 A0A0 A0 0

kC 1 X kC 1 XdXdV F C

= =

( ) ( )V X XA0 0

2 220 0 0A0 A0

F dx dxV dVkC kC1 X 1 X

= = =

( )X

A0 2 A0 200

V dxkC Da kC1 X

= = =


Second Order Gas Phase Reaction in PFR

For the second order reaction The rate law is The differential form of the design equation becomes

2. Gas phase isothermal reaction with no pressure drop

(P=P0) From stoichiometry

Combine the design equation with the stoichiometry Evaluate

A Products

2A Ar kC =

( )( )

( )( )

( )( )

A0 A0AA A0

0 0

F 1 X 1 X 1 XFFC C1 X 1 X 1 X

== = = =

+ + +

2A

A0

kCdXdV F

=

( )( )

( ) ( ) ( )

2V XA0

220 0A0

220

A0

1 XFV dV dxkC 1 X

1 XV 2 1 ln 1 X X

kC 1 X

+= =

+ = + + +


Gas Phase reactions in PFR

For gas phase reaction, the volume of the PFR reactor depends not only on X but also with

Therefore, the conversion depends on both the reactor volume and

( ) ( ) ( )2

20

A0

1 XV 2 1 ln 1 X X

kC 1 X

+ = + + +

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 2 4 6 8 10 12

X

V, m3

=-0.5

=0

=1

=2


Pressure Drop in Reactors

The effect of pressure on the concentration of reactants in liquid phase reaction is insignificant Effect of pressure drop can be ignored

In gas phase reactions, the concentration of reacting species is proportional to the total pressure Proper accounting for pressure drop is crucial This is very important in packed bed system, in particular

microreactors and reaction in membranes

Recall that the concentration of Species i for variable density, variable pressure system is expressed as:

( )( )

+ = +

i i 0A0

0

X TPC C

1 X P T


Pressure Drop in Packed Bed Reactors The gas phase second order reaction:

is to be carried in a packed bed reactor (PBR) The mole balance for Packed Bed Reactor is:

The rate equation is :

From Stoichiometry for gas phase reactions:

And the rate equation becomes

Combining the rate equation and the mole balance: To solve this equation, the relation between the pressure drop and the

catalyst weight must be known

=

'A

A0

rdXdW F

( )( )

= + 0

A00

T1 X PC C

1 X P T

2 A B C +

( )( )

2

' 0A A0

0

T1 X Pr kC

1 X P T

= +

( )( )A0

22

2A0

0

dX 1 X PF kC

dW 1 X P

= +

( )( )

( )22

A01

0 0

kCdX 1 X PF X ,P

dW 1 X P

= = +

' 2A Ar kC =


Flow Through a Packed Bed

Many gas phase reactions are carried in a tubular reactor packed with catalyst, Packed Bed reactor

The pressure drop for flow through a packed porous bed is given by Ergun Equation:

The only parameter that varies with pressure in the right hand side is density,

.

= = =

= =

=

... .

00 0

0

00 0 0

0 T0

0 T 0

mm; and m m

TP FP T F

( )

= +

3

c p P

150 1dP G 11.75G

dZ g D D

z L

=

= = =

=

=

=

==

2 2c m mf f

P

P pressurevolumeof void

porosity void fractiontotal bed volume

volumeof solid1

total bed volume

g conversionfactor (32.174lb .ft / s .lb or1kg .m/ s .kg )

D Particlediametervis cos ity of gas

z lengthdownthe pack

= 2ed bed

G superficial mass velocity ,kg / m .s

( ) ( )

= +

0

0 T3

c p P0 0 T 0

G 1 150 1 PdP T F1.75G

dZ g D D P T F

=

0 T0

0 T 0

PdP T FdZ P T F


Flow Through a Packed Bed

To relate the pressure drop to the catalyst weight rather than z

By solving the two differential equation we can get the relation between W and X

z L

=

0 T0

0 T 0

PdP T FdZ P T F

( )

=

=

c c

CrossSection Area density of catalyst

c b

bulk density

W 1 A z

W A z

( )

= 0 0 T

c c 0 T 0

PdP T FdW A 1 P T F

( ) = +02PdP

1 XdW 2 P

( )= 2dP

F X ,PdW

( )= 1dX

F X ,PdW

( )= 1dX

F X ,PdW


Analytical Solution for Reaction with Pressure Drop

for a second order reaction we had:

For =0 we get W, kg

P

=-0.5

=0=1

=2

( )= 12

0P P 1 W

( )( )

( )

= = +

22

A01

0 0

kCdX 1 X PF X ,P

dW 1 X P

( ) ( ) = + = 20

dP P1 X F X ,P

dW 2 P ( ) ( )

= + =2

02

PdP1 X F X ,P

dW 2 P

( ) ( )

= =

22A0

10 0

kCdX P1 X F X ,P

dW P

( )= =02

2

PdPF P

dW 2 P

( ) ( )

= 2A00

kCdX1 X 1 W

dW( )

( )

W XA0

200 0

kC 11 W dW dX

1 X

=

=

W X2A0

000

kCW 1W

2 1 X

= =

2A0 A0

0 0

kC kCW 1 XW 1

2 1 X 1 X


In-Class Exercise

Ethyl acetate is an extensively used solvent and can be formed by the vapor- phase esterification of acetic acid and ethanol. The reaction was studied using a microporous resin as a catalyst in a packed bed reactor [Ind. Eng. Chem. Res. , 26(2), 1 98 (1 987)]. The reaction is first- order in ethanol and pseudo first- order in acetic acid. For an equal molar feed rate of acetic acid and ethanol, the specific reaction rate is 1 . 2 dm3/g cat/min. The total molar feed rate is 1 0 mol/min, the initial pressure is 1 0 atm, the temperature is 1 1 8 C, and the pressure drop parameter, , equals 0. 01 g- 1 . Calculate the maximum weight of catalyst that one could use and

maintain an exit pressure above 1 atm. (Ans: W = 99 g) Determine the catalyst weight necessary to achieve 90%

conversion. What is the ratio of catalyst needed to achieve the last 5% (85 to 90%) conversion to the weight necessary to achieve the first 5% (0 to 5%) conversion in the reactor? Vary and write a few sentences describing and explaining your findings. What generalizations can you make?


Solution

Acetic Acid + Ethanol Ethyl Acetate +Water A+B C+D

1 . Mole Balance PBR FA0 dX/dW=-rA

2. Rate equation -rA=k CA=k CA (why?)

3. Stoichiometry CA=CA0(1-X)/(1+X) (P/P0)(T0/T)=CA0(1-X)(P/P0)

=0 and isothermal conditions CB=CA0 (FB0/FA0-X) (1+X) (P/P0)(T0/T)=CA

4. Combine 2 and 3 1. -rA=kCA0(1-X)(P/P0)

5. Combine 4 and 1 FA0 dX/dW=kCA0 (1-X) (P/P0) dX/dW=kCA0/FA0 (1-X) (P/P0) =F1(X,P)

P0=10 atm T=118 C F0=2FA0=10 mol/min


Solution, continue

1 . X versus W

6. Pressure drop function

Separating the variables and integrating:

( ) ( ) = + = =2 2

0 02

P PdP1 X F X ,P

dW 2 P 2 P

( ) ( )

= =

22A0

10 0

kCdX P1 X F X ,P

dW P

=

0

P W20

P 0

P2PdP dW

2

=

22 2 0

0

PP P W

2

=2 2

02

0

P PW

P

= =

100 1W 99 g

0.01x100


Solution, Part B

We have And from the pressure drop equation

Substitute (1 - W) for (P/P0)2

Separating the variables

Integrating

( )A0A0 0

kCdX P1 X

dW F P

=

( )

= =

12

2 20

20 0

P P PW 1 W

P P

( ) ( )A0A0

kCdX1 X 1 W

dW F=

( )( )

0.5 A0

A0

F 11 W dW dX

kC 1 X =

( )

W XA0

A00 0

F 11 WdW dX

kC 1 X =

( )

W3 XA020

0 A0

F2w w ln(1 X )

3 kC

=

( ) ( )3

A02

A0

F21 w ln(1 X )

3 kC

=

( )3

A0 2

A0

2kCX 1 exp 1 W 1

3 F

=


Analytical Solution for Reaction with Pressure Drop

W, kg

P

=0

W, kg

P

=0


Ethylene Glycol is produced by the first order reaction of ethylene oxide and water in presence of catalyst (H2SO4)

It is required to design a CSTR reactor to produce 200 million pounds of EG per year

The reaction will be carried at 55 C to avoid formation of by-products

A laboratory experiment is carried to estimate the reaction rate equation and use that equation to design the CSTR reactor (find the reactor volume)

The laboratory data was obtained using 500 ml batch reactor using 2 mol/liter of ethylene oxide in water which is mixed with mixed with 2 obtained

Production of Ethylene Glycol


Chapter 4: Sizing of Isothermal reactor

Mole Balance Rate Law

Stoichiometry Combine Evaluate

Reaction Liquid System

Gas System

Constant Density

Constant density, no pressure drop effect

Variable Density, Constant Pressure

Variable Density, Constant Pressure

Variable Density, Variable Pressure


Chapter 4: Sizing of Isothermal reactor

Reaction Types Liquid phase, V=V0 or = 0, and no pressure effect Gas phase [V=V0 (1+X) (P0/P) or v=v0 (1+X) (P0/P)]

1. n = 0, no pressure drop (same as liquid phase) 2. n = 0, pressure drop (V=V0(P0/P) or v= v0 (P0/P) 3. n 0, no pressure drop (V=V0 (1+X) or v=v0 (1+X)) 4. n 0, pressure drop (V=V0 (1+X) (P0/P) or

v=v0 (1+X) (P0/P))

Mole Balance Rate Law

Stoichiometry Combine Evaluate


Sections 4.6 4.10

Ahmed Abdala


Abu Dhabi, UAE


Design of Chemical Plant

Microreactors

Membrane Reactors

Unsteady State Operation of CSTR

Overview


Most chemical Plants involve number of reaction and separation steps

For example, production of polyethylene glycol requires the following reaction and separation steps

Some of the reactors shown will include multiple reactors connected in series or in parallel

For production of 200 tons/year of polyethylene glycol requires the following flow rates and reactor sizes

Production of Ethylene Glycol

CH3 CH3 CH2 CH2 H2+

CH2 CH2 + O2O

CH2 CH21/2Ag

+

CH2 CH2OH OH

+ H2OO

CH2 CH2H2SO4


Ethylene is produced in 100 parallel PFR reactors

Ethylene oxide is produced in 1000 parallel tubes packed with catalyst coated with silver

Ethylene glycol is produced in 197 ft3 CSTR reactor

Profit can be increased by optimizing separation and conversion and using recycles

Ethylene Glycol Plant

= Profit value of products cost of reactants operating cost Separation costs


Microreactors are characterized by their high surface area-to-volume ratios in their microstructured regions that contain tubes or channels.

A typical channel width might be100 m with a length of 20000 m

The resulting high surface area-to volume ratio (ca. 10000 m2/m3 reduces or even eliminates heat and mass transfer resistance

Microreactors are also used for the production of special chemicals, combinatorial chemical screening, lab-on-a-chip, and chemical sensors

Microreactor


Design of Microreactors

The gas-phase reaction

is carried out at 425C and 1641 kPa (16.7 atm). Pure NOCl is to be fed and the reaction follows an elementary rate. It is desired to produce 20 tons of NO per year in a microreactor system using a bank of ten microrractors in parallel. Each microreactor has 100 channels with each channel 0.2 mm square and 250 mm length.

Assume 85% conversion, plot the molar flow rates as a function of volume down the length of the reactor.

The reaction rate constant, k, is 0.29 dm3/mol/s at 500 K and the activation energy, E, is 24 kcal/mole

22 NOCl 2 NO Cl +


Membrane reactors are used to: Increase conversion of thermodynamically limited

reactions Increase selectivity for multiple reactions

The membrane can either provide a barrier to certain components, preventing certain components such as particulates from contacting the catalyst or contacting reactive sites and being catalyst in itself

Membrane Reactors


Like reactive distillation, the membrane reactor is another technique for driving reversible reactions the right toward completion in order to achieve high conversions.

These high conversion can he achieved having one of the reaction products diffuse out of a semipermeable membrane surrounding the reacting mixture. As a result, the reverse reaction will not be able to take place

and the reaction will continue to proceed to the right toward completion.

Types f catalytic membrane reactors Inert membrane reactor with catalyst pellets on the feed side

(IMRCF) Catalytic membrane reactor (CMR)

Membrane Reactors


Types of Catalytic Membrane Reactors


The following reaction

is to be carried out isothermally in a membrane reactor with no pressure drop. The membrane is permeable to Product C, but it is impermeable to all other species. The total initial concentration is 0.2 mol/dm3 and the molar flow of A is 10 mol/s. The reaction rate constant, kA is 10 dm3/kg-cat/s. the equilibrium constant, KC, is 200 mol3/dm9 and the mass transfer coefficient for H2, 0.5 dm3/kg.cat/s.

Plot the concentration of the reacting species as function of catalyst mass.

Design of Membrane Reactor

+C H C H H6 12 6 6 23


Example of Unsteady state Operations Batch Reactor Startup of CSTR Semibatch reactor

Unsteady State Operation


Estimate the time required to reach steady state after the startup of CSTR reactor for a first-order liquid-phase reaction

Solution: Mole balance

Rate Equation

Startup of CSTR: Steady State time

( )AA A AdN

F F r V unsteady statedt

+ =0

A Ar kC =

A A AA

F F dCr

V V dt + =0 A A A

A

C C dCr

dt + =0


The production of methyl bromide is an irreversible liquid-phase reaction that follows laws an elementary rate Paw. The reaction is carried out isothermally in a semibatch reactor. An aqueous solution of methyl amine (B)at a concentration of 0.025 mol/dm3 is to be fed at a rate of 0.05 dm3/s to an aqueous solution of bromine cyanide (A) contained in a glass-lined reactor

The initial Volume of fluid in the vessel is to be 5 dm3 with a bromine cyanide concentration of 0.05 mol/dm3. The specific reaction rate constant is k = 2.2 dm3/mol/s

Solve for the concentrations of bromine cyanide and methyl bromide and the rate reaction as a function of time.

Semibatch Reactor



Ahmed Abdala


Abu Dhabi, UAE


Collection and Analysis of Rate Data

Chapter 5 Lecture 1: Differential Method


The Rate Expression

The rate law is an algebraic equation that relates the reaction rate of reaction to the concentration of reacting species

The temperature dependent term can be evaluated by knowing the equilibrium constant k at different temperatures

A 0 A B A B

Concentation DependenttermTemperature Dependent

Term

Er A exp C C k C CRT

= =

k

1T

( ) ( )

0

0

Ek A expRT

E 1ln k ln AR T

=

=


Collecting and Analyzing of Experimental Data

Experimental data obtained from Batch reactor or flow reactor system

Data obtained are: Concentration-time measurement (Batch Reactor) Concentration measurements (flow reactor)

Data Analysis Differential Method Integral Method Method of half-lives Method of initial rates Linear and nonlinear regression


Batch Reactor Data

Batch reactor is preferred for collecting data for estimating of the rate equation parameters

For the irreversible reaction Which is carried at constant volume batch reactor Assume that the rate law takes a power-law form:

Rate equation of the form

can also be converted to the previous form 1. Using excess concentration of component B 2. Using equimolar concentration of A and B

For Batch reactor and constant density reaction we know that:

Combining with the rate expression

A B Product+

A

AA A

dCr k Cdt

= =

AA A Br k C C =

AA

dCrd

( h wt

o ?) =

AA

dC kCdt

=


Analysis of Batch Reactor Data: Integral Method

Using the integral method the reaction order and rate constant can be determined:

1. Guess the reaction order 2. Substitute the rate expression into the mole balance equation 3. Integrate the differential design equation 4. Linearize the time concentration relation 5. Plot the appropriate concentration function versus time 6. Linear relation?

Yes, find k

No, repeat steps 2-4

A Ar kC =2

AA

dCkC

dt = 2

A

A

C tA

C A

dCk dt

C = 2

0 A A

ktC C

= 0

1 1

Guess Mole balance Integrate Linearize Plot Check


Analyzing Batch Reactor Data: 1. Differential Method For constant volume batch reactor:

The concentration of a reactant species is measured as function of time Taking the natural logarithm for both sides of the equation

Plotting ln(-dCA/dt) versus ln(CA) gives straight line

= the slope k can be calculated = the intercept

-dCA/dt is obtained from the CA versus t data using one of the following methods: Graphical method Numerical method

For equally spaced data point Polynomial fit

CA= a0 + a1t + a2t2 + a3t3 ++ antn

-dCA/dt = -[ a1+ 2a2t + 3a3t2++ nantn-1]

AA

dC kCdt

=

( ) ( )A AdCln ln k ln Cdt = +

Ln(CA)

Ln(-

dCA/d

t)


Numerical Estimation of dCA/dt

For equally spaced data points ti = constant i.e. (t1-t0)=t2-t1=t3-t2=..ti+1-ti

First Point

Interior points:

Last Point:

t, min t0 t1 t2 t3 tn

CA, mol/liter CA0 CA1 CA2 CA3 CAn

-dCA/dt first Interior points last

0

A0 A1 A 2A

t

3C 4C CdCdt 2 t

+ =

i

A( i 1 ) A( i 1 )A

t

C CdCdt 2 t

+ =

n

A( n 2 ) A( n 1 ) AnA

t

C 4C 3CdCdt 2 t

+ =

t, min 0 5 10 15 20 25

CA, mol/liter 1.0 0.92 0.85 0.78 0.74 0.68

-dCA/dt


Determination of Reaction order and Rate Constant

The following reaction was carried out in a constant volume batch reactor and the following concentration measurement was estimated

Determine the reaction order and the rate constant using the differential Method

t, min 0 5 10 15 20 25

CA, mole/liter 2 1.6 1.35 1.1 0.87 0.7

A B C +


Solution: Differential Method

1. Let: 2. Hence: 3. Estimation of dCA/dt graphically:

1. Calculate

2. Plot versus time

t, min 0 5 10 15 20 30 40 60 CA,

mole/liter 2 1.6 1.35 1.1 0.87 0.7 0.53 0.35

i 1 i

A i i 1

i i 1t ,t

C C Ct t t

=

t, min 0 5 10 15 20 30 40 60

CA, mole/liter 2.0 1.6 1.35 1.1 0.87 0.7 0.53 0.35

CA/t= -(Ci-Ci-1)/(ti-ti-1)

0.08 0.063 0.042 0.033 0.021 0.017 0.009

A B C +

AA A

dCr kCdt

= =

( ) ( )A AdCln ln k ln Cdt = +

1 , 2

A

t t

C 1.6 2.0 0.08t 5 0

= =

ACt

0

0.02

0.04

0.06

0.08

0.1

0.12

0 10 20 30 40 50 60 70t, min

-DCA

/Dt


0.006 0.016 0.0024 0.035 0.053 0.07 0.092 - dC A /dt

0.009 0.017 0.021 0.033 0.042 0.063 0.08 C A / t = - (C i - C i - 1 )/(t i - t i - 1 )

0.35 0.53 0.7 0.87 1.1 1.35 1.6 2.0 C A, mole/liter

60 40 30 22 15 10 5 0 t, min

0.006 0.016 0.0024 0.035 0.053 0.07 0.092 - dC A /dt

0.009 0.017 0.021 0.033 0.042 0.063 0.08 C A / t = - (C i - C i - 1 )/(t i - t i - 1 )

0.35 0.53 0.7 0.87 1.1 1.35 1.6 2.0 C A, mole/liter

60 40 30 20 15 5 0 t, min

0

0.02

0.04

0.06

0.08

0.1

0.12

0 10 20 30 40 50 60 70

t, min

-DC

A/D

t .

Solution, Continued

3. Using equal-area differentiation, the value of dCA/dt can be estimated


Solution, continued

Plot ln(-dCA/dt) versus ln(CA)

Fit to a straight line

Slope = =1.58 ln(k)=-3.45

k = 0.032 (what units?)

y = 1.58x - 3.45R2 = 1.00

-6

-5

-4

-3

-2

-1

0

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8

ln(CA), mole/literln

(-d

CA

/dt)

y = 0.032x1.580

R2 = 0.996

0.01

0.1

0.1 1 10

CA, mole/liter

-dC

A/d

t


Collection and Analysis of Rate Data

Chapter 5 Lecture 2: Integral Method


Integral Method: Zero Order Reactions For zero order reaction:

Substitute the rate expression into the design equation for

batch reactor

Separate variables and integrate

The concentration time relation

Plot CA versus t Straight line?

k = -slope Assume different order and repeat

0A Ar kC k = =

=AdC kdt

= A

A0

C t

AC 0

dC k dt

= A A0C C kt

time C A

slope=-k


Integral Method: First Order Reactions For first order reaction:

Substitute the rate expression into the design equation for batch

reactor



Plot ln(CA0/CA) versus t Straight line?

k = Slope Assume different order and repeat

=A Ar kC

=A AdC kCdt

= A

A0

C tA

AC 0

dC k dtC

=

A0

A

Cln ktC

time Ln

(CA

0/C A

)

slope=k


Integral Method: Second Order Reactions For second order reaction:

Substitute the rate expression into the design equation for batch

reactor



Plot 1/CA versus t

Straight line? k = Slope Assume different order and repeat

=A

2Ar kC

=A

2AdC kCdt

= A

AA0

C tA

2C 0

dC k dtC

A A0

1 1 ktC C

= +

time 1/

C A

slope=k


Integral Method: Example

The irreversible isomerization was carried out in a batch reactor and the following concentration-time data were obtained:

Determine the reaction order and the reaction rate constant.

t, min 0 3 4 8 10 12 15 17.5

CA, mol/dm3 4.0 2.89 2.25 1.45 1.0 0.65 0.25 0.07

A B


Solution

1. Assume zero order reaction: CA=CA0-kt Plot CA versus time The data does not follow

a zero order relation

A B

t, min 0 3 4 8 10 12 15 17.5

CA, mol/dm3 4.0 2.89 2.25 1.45 1.0 0.65 0.25 0.07


2. Assume first order reaction: ln(CA0/CA)= k t Plot ln(CA0/CA) versus time The relation is not linear The reaction does not follow

a first order kinetics

y = 0.214x - 0.3673 R = 0.9226

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 5 10 15 20

Ln(C

A,/C

A)

time, min

Solution, continued

A B

t, min 0 3 4 8 10 12 15 17.5

CA, mol/dm3 4.0 2.89 2.25 1.45 1.0 0.65 0.25 0.07 ln(CA0/CA) 0.00 0.33 0.58 1.01 1.39 1.82 2.77 4.05


3. Assume second order reaction:

1/CA0 = 1/CA0+ k t Plot 1/CA0 versus time The relation is not linear The reaction does not follow

second order kinetics

The reaction does not follow zero, first, or second order reaction

Try differential method or nonlinear regression method

Solution, continued

A B

t, min 0 3 4 8 10 12 15 17.5

CA, mol/dm3 4.0 2.89 2.25 1.45 1.0 0.65 0.25 0.07

1/CA 0.25 0.35 0.44 0.69 1.00 1.54 4.00 14.29


Integral Method: n-order Reaction

For reaction of n order

Substitute in the mole balance equation

Integrate

Because n is not know, this expression cannot be plotted

Using regression method, least squares, the reaction order, n, and the rate constant (k) can be estimated

=A

nAr kC

=A

nAdC kCdt

= A

A0

C tA

nAC 0

dC k tC

=

A0

1 n 1 nAC C1t

k 1 n


Nonlinear Regression Analysis

Assuming the rate expression is:

The integral methods yield the following relation:

Using regression analysis we can search for values for n and k that minimize the sum of the square of the error (S2):

This can be done numerically using one of software packages such as Matlab, Polymath, etc.

=A

nAr kC

( )A0

1 n 1 nAC C 1 n kt

=

( ) ( )A ,m A ,m21N N

2 1 n 1 nA ,c A0 i

i 1 i 1S C C C C 1 n kt

= =

= =



Ahmed Abdala


Abu Dhabi, UAE


Multiple Reaction

Chapter 6 Lecture 1: Parallel Reactions


Multiple Reactions

Parallel Reactions A B

A C

Series Reaction

A B C

Complex Reactions A + B C + D

A + C E

Independent Reactions A B + C

D E + F


Multiple Reactions

Parallel Reactions CH2=CH2 + 3 O2 2 CO2 + 2 H2O CH2=CH2 + 0.5 O2 Ethylene Oxide

Series Reaction

Paraffins Naphthenes Aromatics

Complex Reactions Ethylene Oxide + NH3 HOCH2CH2NH2 Ethylene Oxide + HOCH2CH2NH2 (HOCH2CH2)2NH

Independent Reactions C15H32 C12H26 + C3H6 C8H18 C6H14 + C2H4

-H2 -H2


Selectivity

For the following Parallel Reaction: A D (Desired )

A U (Undesired)

Minimization of undesired compound U relative to desired product D is always desirable

Selectivity is used to quantify formation of desired product D with respect to undesired

Instantaneous Selectivity, SD/U:

Overall selectivity,

DD

UU

rateof formationofDrS

r Rateof formationof U= =

~

FlowReactorBatchReactor

DU

D D

U U

N FS

N F= =

~

:DU

S


Reaction Yield

Instantaneous Yield, YD:

Overall Yield,

For CSTR reactor: The instantaneous yield is equivalent to the overall yield

The instantaneous selectivity is equivalent to the overall

yield

DD

A

rateof formationof DrY

r Rateof reactionof keyreactant= =

D

~D D

A A A A

Batchreactor FlowReactor

N FY

N N F F= =

0 0

~

:DU

Y


Parallel Reactions:

For the following Parallel Reactions

Rate of disappearance of reactant A, -rA

The instantaneous selectivity

1

1

A D r

A U r

D

U

kD D A

kU U A

k C

k C

=

=

1 1rA D U D A U Ar r k C k C = + = +

11 2

2

D D A DD A

U UU U A

r k C kS C

r kk C

= = =


Parallel Reactions: Maximizing Selectivity

For the parallel reactions:

We can maximize the selectivity by choosing the right reactor system (reactor type, arrangement, feed)

Case 1: 1 > 2 (1 - 2 = + ve) To make SD/U large , CA must be maintained as high as

possible Inert or diluents should be kept minimum

A Batch or plug-flow reactor should be used

1 2=DU

DA

U

kS C

k

1

2

=

=

A D

A U

D D A

U U A

r k C

r k C


Parallel Reactions: Maximizing Selectivity

Case 2: 1 < 2 (1 - 2 = -ve) To make SD/U large , CA must be maintained as low as

possible Use Inert or diluents and/or use recycle to dilute reactant

with products

Low pressure for gas phase

Use CSTR

Case 3: 1 = 2 Case 3-1: ED >EU

Operate at highest possible T

Case 3-2: ED < EU Operate at low T (high enough to have significant rate)

( )

eD U

DU

E E

RTD D

U U

k AS

k A

= =


Parallel Reactions: Example Trambouze Reaction (P6-6) Consider the following system of gas-phase reactions:

The reactions are to be carried at 27 C

and 4 atm. Pure A enters the system at volumetric flow rate of 10 dm3/min.

a. Sketch the instantaneous selectivity SB/X, SB/R, SB/XR as function of CA b. Consider a series of reactors. What should be the volume of the first

reactor? c. What is the conversion of A in the first reactor? d. What are the effluent concentration of A, B, X, and R from the first

reactor. e. If 99% conversion of A is desired, what reaction scheme and reactor sizes

should we use to maximize SB/XR. f. Suppose that E1=20 kcal/mol, E2=10 kcal/mol, E3=30 kcal/mol. What

temperature would you recommend for a single CSTR with a space time of 10 min and entering concentration of A of 0.1 mol/dm3?

g. If you could vary the pressure between 1 and 100 atm, what pressure would you choose?

1

2

2

12

3

3

23

0 004

0 3

0 35

A X (Undesired) r . .min

A B (Desired) r ..min

A R (Undesired) r ..min

kX A

kB A

kR A

molC

dmmol

Cdm

molC

dm

=

=

=


Solution: Part a

note that SB/XR curve have

a maxima at CA*

We can find CA* by setting dSB/XS/dCA=0

CA*=0.032 mol/dm3

1

2

2

12

3

3

23

0 004

0 3

0 35

A X (Undesired) r . .min

A B (Desired) r ..min

A R (Undesired) r ..min

kX A

kB A

kR A

molC

dmmol

Cdm

molC

dm

=

=

=

2 1 1 0 5 0 52

1

0 3 750 004

. ...

BX A A A

kS C C C

k = = =

2 3 1 2 12

3

0 3 0 860 35

..

.B

R A A A

kS C C C

k = = =

0 5 2

0 30 004 0 35.

.

. .B

XR

B A

X R A A

r CS

r r C C= =

+ +

0

5

10

15

20

25

30

35

0

20

40

60

80

100

120

140

160

180

200

-5.55E-170.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16

S B/X

or

S B/X

R

S B/R

CA

SB/R

SB/X

SB/XR

( ) ( )0 5 2 0 50 3 0 004 0 35 0 3 0 5 0 004 2 35 0. .. . . . . * . * .B XR A A A A AA

dSC C C C C

dC= + + + =


0

2

4

6

8

10

00.020.040.060.080.10.120.140.16

CA, mol/liter

S B/X

R

CA0

CSTR

Solution: Part b, c

To maximize the selectivity SB/XR, we need to operate at the maximum selectivity Use the first reactor as CSTR operates at final CA=CA*

Reaction Direction

PFR

A

* *A0 A0

CSTR 0.5 2A* 1 2 A 3 A

F X F XVr k C k C k C

= = + +

= = =

** A0 A

A0

C C 0.16 0.032X 0.8C 0.16

= = =A0 0 A0F C 10* 0.16 1.6mol / min

=+ +

=

CSTR 0.5 2

3

1.6* 0.8V0.004* 0.032 0.3* 0.032 0.35* 0.032119 dm

( )= = =

+3A0

A0P 1* 4C 0.16 mol / dmRT 0.082* 27 273


Solution: Part d

CA=0.032 mol/dm3

CB, CR, CX?

= =

00

B BA ACSTR

A

F FF FV

r

= 0

R R

B

F F

r

= 0

X X

R

F F

r Xr

+ + +A B X R A0Check : C C C X must equal toC

= = =0 B 0 R 0 XCSTR 2 0.5

A A A

C C CV0.3C 0.35C 0.004c

= = =A

0.5 0.53

X0

0.004C V 0.004* 0.032 * 119C 0.0085 mol / dm10

= = = 3AB

0

0.3C V 0.3* 0.032* 119C 0.114 mol / dm10

= = =

2 23A

R0

0.35C V 0.35* 0.032 * 119C 0.004 mol / dm10


Part e

For 99% conversion: The first reactor is CSTR (X=0 to X=0.8, V=119 dm3) The second reactor is PFR:

( )

( ) A0A 2

AA A1 0

1 0.8 CF0 AA

PFR 0.5 2A 1 2 A 3 AF 1 0.99 C

v dCdFVr k C k C k C

= =+ +

( )

( )

= = =+ +

A0A 2

AA A1 0

1 0.99 CF3A A

PFR 0.5 2A A AF 1 0.8 C

dF dCV 10 9.1dmr 0.004C 0.3C 0.35C


Solution Part f

For single CSTR:

For max selectivity

20 5 2

1 3.B XR

AfB

X R Af Af

k CrS

r r k C k C= =

+ +

( ) ( )0 5 2 0 52 1 3 2 1 30 5 2 0= + + =. .. * *B XR A A A A AA

dSk k C k C k C k C k C

dC

0 5 2 0 5 21 3 1 30 5 2 0+ =

. .. * *AA A A

k C k C k C k C

0 5 21 30 5 0 =

.. A Ak C k C

( )( )

2233

01 11

3 03 3

0 5 exp /.exp /Af

A E RTkC

k A E RT

= =


Solution Part f

For max selectivity

( ) ( )0 533

1201 1 1

20 100 0041 987 300

= = =

./

exp / . exp 1.49x10. min

dm molxA k E RT

x

( )3

6 102 2 2

10 100 31 987 300

= = =

exp / . exp 5.79x10 min.

xA k E RT

x

( )3

21 3 1 103 3 3

30 100 351 987 300

= = =

exp / . exp 2.52x10 min.

xA k E RT dm mol

x

( )( )

2 223 33

01 1 01 3 11

3 0303 3

0 5 0 50 5 = = =

. exp / ..exp

exp /AfA E RT A E Ek

Ck A RTA E RT

2 3

1

30 00 5 2 1 3 2 3 4 3

1 2 3 1 1 11 2 3

3 3 3

0 12

10

2 2 2

= = = = + +

+ +

/

. / / /

.A A A A

A A A A

kkC C C C

r k C k C k C k k kk k k

k k k

( )2 3 2 33 11 2 34 3 1 3 2 3 4 3

1 3 2 1 1 3

0 210

=

+ +

/ /

/ / / /

.expressions , ,

k ksubstitute for k k and k as functionof T and solve for T

k k k k k k

315=usingSolving for T Polymath givesT K


Solution part f and g

The optimum pressure should be selected such that the initial concentration of A, CA0=0.1 mol/dm3

This will ensure that the exit condition of the CSTR reactor will correspond to maximum selectivity

0 0 0 1 0 08205 315 2 6= = =. . .AP C RT x x atm


For the following parallel reactions

The instantaneous Selectivity

Depending on the values of 1, 2, 1, and 2 we have the following cases Case 1: 1 > 2 and 1 > 2 Case 2: 1 > 2 and 1 < 2 Case 3: 1 < 2 and 1 < 2 Case 4: 1 < 2 and 1 > 2

Parallel Reactions: Reactor Selection

1 1 1

2 2 2

1

2

A + B D r

A + B U r

kD A B

kU A B

k C C

k C C

=

=

/D

D U A BU

r kS C C

r k = = 1 2 1 21

2


Case 1: 1 > 2, 1 > 2

High CA and High CB is preferred

Use PFR or Batch Reactor

Use high Pressure (gas reaction)

Do not use inert or diluents

/D

D U A BU

r kS C C

r k = = 1 2 1 21

2

A

B

1 11

2 22

1

2

A + B D r

A + B U r

kD A B

kU A B

k C C

k C C

=

=

B A


High CA and Low CB is preferred Semibatch reactor

A present initially and B is continuously fed to the reactor

PFR with side streams or membrane Series of Small CSTR with A fed to the first reactor and B is divided

between the reactors

.

Case 2: 1 > 2, 1 < 2

A

B /D

D U A BU

r kS C C

r k = = 1 2 1 21

2

1 11

2 22

1

2

A + B D r

A + B U r

kD A B

kU A B

k C C

k C C

=

=

B A

A B


Low CA and Low CB is preferred CSTR PFR with large recycle Diluted feed or inert Low pressure

Case 3: 1 < 2, 1 < 2

/D

D U A BU

r kS C C

r k = = 1 2 1 21

2

1 11

2 22

1

2

A + B D r

A + B U r

kD A B

kU A B

k C C

k C C

=

=

B

A

Recycle

B

A

B Recycle


Low CA and High CB is preferred Semibatch reactor

B present initially

A is continuously fed to the reactor

PFR with side streams or membrane

Series of Small CSTR with B fed to the first reactor and A is divided between the reactors

Case 4: 1 < 2, 1 > 2

/D

D U A BU

r kS C C

r k = = 1 2 1 21

2

B

A

B

A

A B

1 11

2 22

1

2

A + B D r

A + B U r

kD A B

kU A B

k C C

k C C

=

=


For the series reactions:

The exact reaction time to maximize B is very important

For Batch System:

Series Reactions: Batch System

1 2

1 21 2

A B( ) C ( )A A BC

B A B

k k

r k C r k Cr k C k C

Desired undesired = =

=

1 0 1AdC exp( )

dt A A A Ar k C C C k t = = =

1 2 1 0 1 2BdC exp( )

dt B A B A Br k C k C k C k t k C= = =

1 2

1 02 1

BCk t k t

A

e ek C

k k

=

CA

CC

CB

k1/k2=2

( ) ( )1 20 2 12 1

1 1CCk t k tAC k e k e

k k =

1 2

2 2 1 02 1

CdC dt

k t k t

B AC

e er k C k k C

k k

= = =

0A A BCremember C C C C=


The maximum CB is at dCB/dt=0

Series Reactions: Maximum Yield of B

( )1 21 0 1 21 2

0BdCdt

k t k tAk C k e k ek k

= = +

CA

CC

CB

k1/k2=2

topt

C 1

1 2 2

1lnopt

kt

k k k

=

1 2

1 02 1

BCk t k t

A

e ek C

k k

=


Series Reaction: Example

For the following liquid-phase series reaction is carried in a 500-dm3 batch reactor. The initial concentration of A is 1.6 mol/dm3. The desired product is B, and separation of the undesired product C is very difficult and costly. The given rate constants are in h-1 and are at 100C. Plot the CA, CB, CC as function of time for batch system. What is the optimum reaction time? Assuming the reaction is carried in CSTR of a space time 0.5

h. What temperature would you recommend to maximize B? (E1=10 kcal/mol, E2=20 kcal/mol)

0 010 4 ..A C ( )BCA A r Cr C B undesired= =


Solution: Batch System

Mole Balance for Batch Reactor

0 010 4 ..A C ( )BCA A r Cr C B undesired= = 0 40 010 01 0 4

===

.

.

. .

A A

C B

C B A

r Cr Cr C C

( )0 4 1 = = .A A AdC

r Cdt

( )0 01 0 4 2= = . .B B B AdC

r C Cdt

( )0 01 3= = .C C BdC

r Cdt

0 1 1 6 0 4= = exp( ) . exp( . )A AC C k t t

( )1 2

0 01 0 41 0

2 1

= =

. .

BC 4 1.6k t k t

t tA

e ek C e e

k k

( ) ( )

( ) ( )

1 202 1

2 1

0 4 0 01

1 1

1 6 0 01 1 0 4 10 01 0 4

=

=

C

. .

C

.. .

. .

k t k tA

t t

Ck e k e

k k

e e

0

0.5

1

1.5

2

0 20 40 60

Ci

time, min

CA

CB

CC 1

1 2 2

1 1 0 4 9 460 4 0 01 0 01

= = =

.ln ln . min

. . .optk

tk k k


Solution Batch System Using PolyMath


Solution: CSTR

Mole Balance for CSTR

( )0 00 0 = = =

A AA A A

A A A

C CF X F FV

r r r

0 1 1

1 6 1 61

. .AA

A

V CC

k C k

= = =+

0

2 11

51 61

.B B B

BB

C C Cr k C k

k

= = =

+

( )( )1 1

21 1 2

8 851 5 1 5 1 5B B B

k kk C C C

k k k = =

+ + +

0.000

0.200

0.400

0.600

0.800

1.000

1.200

200 300 400 500T, K

CB

388 115= = optimumT K C

T k1 k2 CB

200 3.4E-06 7.3E-13 2.7E-05

225 5.6E-05 2.0E-10 4.5E-04

250 5.2E-04 1.7E-08 4.2E-03

275 3.3E-03 6.7E-07 2.6E-02

300 1.5E-02 1.4E-05 1.1E-01

325 5.5E-02 1.9E-04 3.4E-01

350 1.6E-01 1.7E-03 7.2E-01

375 4.3E-01 1.2E-02 1.

CHE411 Fall 2010-Chemical Reaction Engineeirng-Ahmed a Abdala

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