Chemical Reaction Engineering Chapter 6 Chapter 6 Mutiple Reactions.

Chemical Reaction Engineering

Chapter 6Chapter 6

Mutiple Reactions

Define different types of selectivity and yield

Choose a reaction system

that would maximize the selectivity of the desired product

given the rate laws for all the reactions occurring in the system.

Describe the algorithm used to design reactors

with multiple reactions.

Size reactors to maximize the selectivity

and to determine the species concentrations in a batch reactor,

semi-batch reactor, CSTR, PFR, and PBR, systems.

Objectives

Parallel rxns (competing rxns)Parallel rxns (competing rxns) B

AC

Series rxns (consecutive rxns)Series rxns (consecutive rxns)

A B C

Complex rxns (Parallel + Series rxns)Complex rxns (Parallel + Series rxns)

A + B C + DA + C E

Independent rxnsIndependent rxns

A B + CD E + F

k1

k2

k1 k2

k1

k2

k1

k2

Definition of Multiple Reaction

Parallel rxns Parallel rxns (competing rxns)

2CO2 + 2H2OCH2=CH2 +O2

CH2-CH2

Series rxns Series rxns (consecutive rxns) O O CH2-CH2 + NH3 HOCH2CH2NH2 EO EO

(HOCH2CH2)2NH2 (HOCH2CH2)3NH2

Complex Rxns (Combination of parallel & series rxns)Complex Rxns (Combination of parallel & series rxns)

C2H5OH C2H4 + H2O C2H5OH CH3CHO + H2

C2H4 + CH3CHO C4H6 + H2O

Independent rxns Independent rxns ((The cracking of crude oil to form gasoline)

C15H32 C12H26 + C3H6

C8H18 C6H14 + C2H4

k1

k2

Examples of Multiple Reaction

Monoethanolamine

Dinoethanolamine Trinoethanolamine

Parallel ReactionsParallel Reactions

D (Desired Product)A

U (Undesired Product)

kD

kU

Competing or side rxn

A

D

A, U

reaction

separation

Total cost

Reaction cost

Separation costcost

Low High

The economic incentiveThe economic incentive

Maximize the formation of DMaximize the formation of D

Minimize the formation of U Minimize the formation of U

Rxn-separation system producing both D & U Efficiency of a reactor system

Desired and Undesired Reaction

Rate selectivity parameter, SRate selectivity parameter, S

D (Desired Product)A


The rate laws are

Rate selectivity parameter = Instantaneous selectivity

kD

kU

21

2

1

21

AU

D

U

DDU

AUU

ADD

AUADA

Ck

k

r

rS

Ckr

Ckr

CkCkr

We want to maximize

SDU

Rate selectivity parameter, SDU

=Rate of formation of U

Rate of formation of D

Overall Selectivity, SDU

~

Overall Selectivity

SDU =~

FU

FD =Exit molar flow rate of undesired product

Exit molar flow rate of desired product

For Batch Reactor

SDU =~

NU

ND

Example 6-1: Comparison between SDU and SDU in a CSTR~

Mission: Develop a relationship between SDU and SDU in a CSTR~

Solution

SDU = rU

rDand SDU =

~

FU

FD

By the way, FD=rDV and FU=rUV

SDU =~

FU

FD =rDVrUV

=rD

rU

= SDU

Instantaneous Yield and Overall Yield

Instantaneous Yield (Basis: Reaction Rate)

YD = -rA

rD

Overall Yield (Basis: Molar Flow Rate)

For a batch system: YD = NAo-NA

ND

For a flow system: YD = FAo-FA

FD

= Number of moles of key reactant consumed

Mole of desired productformed at the end of reaction

~

~

Overall SelectivityOverall Selectivity

Uof rate flowmolar exit

D of rate flowmolar exitsystem flow DU,

~

U

D

F

FS

rxn time theof end at the Uof moles ofnumber the

rxn time theof end at the D of moles ofnumber thesystembatch DU,

~

U

D

N

NS

Overall YieldOverall Yield

consumedbeen have that A, reactant,key theof moles ofnumber the

rxn time theof end at the D of moles ofnumber the

0

systembatch D,

~

AA

D

NN

NY

consumedbeen haveA that of moles ofnumber the

exit at the D of moles ofnumber the

0

system flow D,

~

AA

D

FF

FY

From an economic standpoint, overall selectivities and yieldsoverall selectivities and yields are important

in determining profitsprofits

Different definitions for selectivity and yield

Check carefully to ascertain the definition intended by the author

NoteNote

The instantaneous selectivitiesinstantaneous selectivities give insightsin choosing reactors and reaction schemeschoosing reactors and reaction schemes

that will help maximize the profit

Case 1: Case 1: 1 1 > > 2 2 , a = , a = 11- - 2 2

aA

U

D

U

DDU C

k

k

r

rS

maximize Smaximize SDUDU

- keeping the concentration of reactant A as high as possible during the rxn

- in gas phase rxn, we should run it without inerts and at high pressures to keep CA high

- in liquid phase rxn, the use of diluents should be keep to a minimum

- use a batch or plug-flow reactor

Maximizing SDU in Parallel Reactions

Case 2: Case 2: 2 2 > > 1 1 , a = , a = 22- - 1 1

aAU

D

U

DDU Ck

k

r

rS

1

maximize Smaximize SDUDU

- keeping the concentration of reactant A as low as possible during the rxn

- in gas phase rxn, we should run it with inerts and at low pressures to keep CA low

- in liquid phase rxn, the use of diluents should be keep to a maximum

- use a CSTR or recycle reactor (product stream act as a diluent)

Maximizing SDU for one reactant

Case 3: ECase 3: ED D > E> EU U

- kD (rD) increases more rapidly with increasing temperature than does the kU.

- keeping the temperature as high as possible to maximize SDU.

RT

EE

U

D

U

DUD

eA

A

k

k

Case 4: ECase 4: EU U > E> ED D

- keeping the temperature as low as possible to maximize SDU

- not so low that the desired rxn does not proceed to any significant extent.

Maximizing SDU for one reactantWhether the reaction should be run at high or low T

- Pre-exponential factors are comparable- the activation energies of (1) & (2) are much greater than that of (3).- at high temperature rQ << rD, rU

(4)(4)

(5)(5)

- From (5), SDU is minimized at low concentration of A

Example: Minimizing unwanted products (U & Q) for a single reactantExample: Minimizing unwanted products (U & Q) for a single reactant

DD (1)(1)

AA U U (2)(2)

QQ (3)(3)

AT

D Cer

1

300

1000,36

0002.0

5.11

300

1000,25

0018.0 AT

U Cer

5.01

300

1000,5

00452.0 AT

Q Cer

5.0

1

300

1000,11

11.0

largevery

A

T

U

DDU

Q

DDQ

C

e

r

rS

r

rS

1. High Temperatures (to minimize the formation of Q)1. High Temperatures (to minimize the formation of Q)2. Low concentration of A (to minimize the formation of U)2. Low concentration of A (to minimize the formation of U)

- Adding inerts, using low pressures, using CSTR or recycled reactor- Adding inerts, using low pressures, using CSTR or recycled reactor

YX

DYXD rr

rS

)/(

Textbook Example 6-2: 수업 후 집에서 꼭 풀어 보세요

Rate selectivity parameter, SRate selectivity parameter, S

D (Desired Product) A + B


The rate laws are

Rate selectivity parameter=Instantaneous selectivity

k1

k2

2121

22

11

2211

2

1

2

1

21

BAU

DDU

BAU

BAD

BABAA

CCk

k

r

rS

CCkr

CCkr

CCkCCkr

Reactor Selection and Operating Conditions

Different reactors and schemes for maximizing the desired productDifferent reactors and schemes for maximizing the desired product

AB

AB A

B

B

A

A

B

(a) CSTR (b) tubular reactor (c ) batch (d) semi-batch 1 (e) semi-batch 2

AB

AB

(f) Tubular reactor with side streams (g) Tubular reactor with side streams

AB

(i) Tubular reactor with recycle (h) Series of small CSTRs

BA

Figure 6-3Figure 6-3

Figure 6-3Figure 6-3Different reactors and schemes for maximizing the desired productDifferent reactors and schemes for maximizing the desired product

Example 6-3: Minimizing unwanted products for two reactantsExample 6-3: Minimizing unwanted products for two reactants

for the parallel reaction



Case I : 1 > 2, 1 > 2, a = 1-2 > 0, b = 1-2 > 0

the rate selectivity parameter

k1

k2

bB

aA

U

DDU CC

k

k

r

rS

2

1

To maximize the SDU, maintain the concentration of both A and B as high as possible

a tubular reactor (Figure 6.3 (b))a tubular reactor (Figure 6.3 (b))

a batch reactor (Figure 6.3 (c))a batch reactor (Figure 6.3 (c))

high pressures (if gas phase), reduce inerthigh pressures (if gas phase), reduce inert





Case II : 1 > 2, 1 < 2, a = 1-2 > 0, b = 2-1 > 0


k1

k2

bB

aA

U

DDU Ck

Ck

r

rS

2

1

To maximize the SDU, maintain CA high and CB low.

a semibatch reactor in which B is fed slowly into A. (Figure 6.3(d))a semibatch reactor in which B is fed slowly into A. (Figure 6.3(d))

a tubular reactor with side stream of B continually (Figure 6.3(f))a tubular reactor with side stream of B continually (Figure 6.3(f))

a series of small CSTRs with A fed only to the first reactor (Figure 6.3(h))a series of small CSTRs with A fed only to the first reactor (Figure 6.3(h))





Case III : 1 < 2, 1 < 2, a = 2-1 > 0, b = 2-1 > 0


k1

k2

bB

aAU

DDU CCk

k

r

rS

2

1

To maximize the SDU, maintain the concentration of both A and B as low as possible

a CSTR (Figure 6.3(a))a CSTR (Figure 6.3(a))

a tubular reactor in which there is a large recycle ratio (Figure 6.3(i))a tubular reactor in which there is a large recycle ratio (Figure 6.3(i))

a feed diluted with inert materiala feed diluted with inert material

low pressures (if gas phase)low pressures (if gas phase)





Case IV : 1 < 2, 1 > 2, a = 2-1 > 0, b = 1-2 > 0


k1

k2

aA

bB

U

DDU Ck

Ck

r

rS

2

1

To maximize the SDU, maintain the concentration of both A and B as high as possible

a semibatch reactor in which A is slowly fed to B (Figure 6.3(e))a semibatch reactor in which A is slowly fed to B (Figure 6.3(e))

a tubular reactor with side stream of A (Figure 6.3(g))a tubular reactor with side stream of A (Figure 6.3(g))

a series of small CSTRs with fresh B fed to each reactor (Figure 6.3(h))a series of small CSTRs with fresh B fed to each reactor (Figure 6.3(h))

In parallel rxns, maximize the desired product

by adjusting the reaction conditions

by choosing the proper reactor

In series rxns, maximize the desired product

by adjusting the space-time for a flow reactor

by choosing real-time for a batch reactor

Maximizing the desired product in series reaction

k1 k2A B C

If the first reaction is slow and second reaction is fast, it will be extremely difficult to produce species B.

If the first reaction (formation of B) is fast and the reaction to form C is slow, a large yield of B can be achieved.

However, if the reaction is allowed to proceed for a long time in a batch reactor or if the tubular flow reactor is too long, the desired product B will be converted to C.

In no other type reaction is exactness in the calculation of the time needed to carry out the reaction more important than in series reactions.

Maximizing the desired product in series reaction

k1 k2A B CDesired Product

Example 6-4: Maximizing the yield of the intermediate productExample 6-4: Maximizing the yield of the intermediate product

k1 k2CH3CH2OH(g) CH3CHO 2CO2

OHO 222

1 OHO 22 2

2

5

The oxidation of ethanol to form acetaldehyde is carried out on a catalyst of 4 wt% Cu-2wt% Cr on Al2O3. Unfortunately, acetaldehyde is also oxidized on this catalyst to form carbon dioxide. The reaction is carried out in a dilute concentrations (ca. 0.1% ethanol, 1% O2, and 98.9% N2). Consequently, the volume change with the reaction can be neglected. Determine the concentration of acetaldehyde as a function of space time.

The rxns are irreversible and first-order in ethanol and acetaldehyde, respectively.

Solution

Reaction paths for different ks in series reaction Reaction paths for different ks in series reaction

A B Ck1 k2

1

1~

1

2

1

2

1

2

1

k

k

k

k

k

k

A C

B

'1

'2

For k1/k2>1, aLarge quantity of BCan be obtained

For k1/k2<1, aLittle quantity of BCan be obtained 1st rxn is slow

2nd rxn is fast

'3

Long rxn time in batch or long tubular reactor-> B will be converted to C

Algorithm for solution to complex reactionsAlgorithm for solution to complex reactions

Steps in in Analyzing Multiple ReactionSteps in in Analyzing Multiple Reaction

1. Number each reaction. 2. Mole balance for each species. 3. Determine rij

4. Relate to rate of reaction of each species to the species for which the rate law is given. 5. Determine the net rate of formation of each species 6. Express rate laws as a function of Cj when X<<1. 7. Express rate laws as a function of mole when X>>1. 8. Combine all the above and solve the resulting set of equations

1. Mole balances for multiple reactions1. Mole balances for multiple reactionsIt is better to solve problems using moles (Nj) or molar flow rates (Fj) rather than conversion

V

CCr

dt

dCFVr

dt

dN

V

Cr

dt

dCVr

dt

dN

rdV

dCr

dV

dF

r

CCV

r

FFV

rdt

dCVr

dt

dN

BBB

BBB

B

AA

AA

A

AA

AA

A

AA

A

AA

AA

AA

][ 000

0

0

00

0

Reactor Gas Liquid

Batch

CSTR

PFR/PBR

Semibatch

2. Net Rates of Reaction2. Net Rates of Reaction

Reaction 1: A + B 3C + D

Reaction 2: A + 2C 3E

Reaction 3: 2B + 3E 4F

Reaction q: A + ½C G

...

k1A

k2A

K3B

kqA

Key point: to write the net rate of each species (e.g. A, B, …)

q

iiBqBBBBB

q

iiAqAAAAA

rrrrrr

rrrrrr

1321

1321

Net rates of reaction of A and B:

0

00

Total rate of formation of each species from all reactionTotal rate of formation of each species from all reaction

5533

4243232121

1212111

2

1

AAAq

AAAi

AAAA

AAA

qqjqj

pipkikjij

jj

jj

Reaction Number Reaction Stoichiometry

q

iijj rr

1

The net rate of rxn for A:

q rxns taking place

species

reaction number

3. Rate laws3. Rate laws

Reaction 1: A + B 3C + D

Reaction 2: A + 2C 3E

Reaction 3: 2B + 3E 4F...

k1A

k2A

k3B

We need to determine the rate law for at least one species in each rxn.

222 CAAA CCkr

222 CAAA CCkr 1 1 B

rA=r1A+r2A= -k1ACACB-k2ACACC2

4. Stoichiometry: Relative Rates of Reaction4. Stoichiometry: Relative Rates of ReactionGeneric reaction:

aA + bB cC + dD

Relative rate of reactions

Multiple rxns:

For example,

Relative rates of Rxn in reaction i

d

r

c

r

b

r

a

r DCBA

i

iD

i

iC

i

iB

i

iA

d

r

c

r

b

r

a

r

2222 33 CAAAE CCkrr

ik

ik

ij

ij rr

5. Combining individual rate laws to find net law5. Combining individual rate laws to find net law

6. Stoichiometry:6. Stoichiometry: Concentrations as a function of molar flow rate Concentrations as a function of molar flow rate

T

T

P

P

F

FC

T

T

P

P

F

FFFC

T

jT

T

jTjj

0

00

0

00

0

0

0

00 RT

PCT

n

jjT FF

1

Ideal Gas:

Isothermal gas phase without P:

0j

j

FC Liquid phase:

T

jTj F

FCC 0

T

jT

TT

TT

T

jT

TT

TT

F

FC

F

FC

F

FCfnr

F

FC

F

FC

F

FCfnr

02

01

022

02

01

011

,,,

,,,

Net rate of formation

Multiple Reactions in a PFR/PBRMultiple Reactions in a PFR/PBR(combining mole balance, rate law, and stoichiometry)(combining mole balance, rate law, and stoichiometry)

T

jT

TT

TTj

q

iijj

j

T

jT

TT

TT

n

ii

T

jT

TT

TT

m

ii

F

FC

F

FC

F

FCfnrr

dV

dF

F

FC

F

FC

F

FCfnrr

dV

dF

F

FC

F

FC

F

FCfnrr

dV

dF

02

01

01

02

01

021

222

02

01

011

111

,,,

,,,

,,,

.

.

.

Rate lawMole balance

j Coupled ODEsIsothermal gas phase without P:

Example 6-5: Write the rate law for each species in each reaction and then Example 6-5: Write the rate law for each species in each reaction and then

write the net rates of formation of NO, Owrite the net rates of formation of NO, O2, and N, and N22..

233

222

5.111

2222

22

3

ONOO

NONN

NONHNONO

CCkr

Ckr

CCkr

Solution

Reaction 1: 4NH3 + 6NO 5N2 + 6H2O

Reaction 2: 2NO N2 + O2

Reaction 3: N2 + 2O2 2NO2

Example 6-6: Write mole balance on a PFR in terms of molar flow rates for each speciesExample 6-6: Write mole balance on a PFR in terms of molar flow rates for each species Reaction 1: 4NH3 + 6NO 5N2 + 6H2O

Reaction 2: 2NO N2 + O2

Reaction 3: N2 + 2O2 2NO2

233

222

5.111

2222

22

3

ONOO

NONN

NONHNONO

CCkr

Ckr

CCkr

Solution

Example 6-7: Hydrodealkylation of Mesitylene in a PFRExample 6-7: Hydrodealkylation of Mesitylene in a PFR

The hydrodealkylation of mesitylene is to be carried out isothermally at 1500 R and 35 atm in a packed-bed reactor in which the feed is 66.7 mol% hydrogen and 33.3 mol% mesitylene. The volumetric feed rate is 476 ft3/h and the reactor volume (i.e. V=W/b) is 238 ft3.

CH3

CH3CH3

CH3

CH3

CH3

+ H2 + CH4

CH3

CH3

+ H2 + CH4

k1

k2

@1500R/hmol) /lb(ft20.30

@1500R/hmol) /lb(ft20.550.53

25.0

22

0.531

5.011

kCCkr

kCCkr

HXT

HMM

M=mesitylene, X=xylene, T=toluene, Me=methane, H=hydrogen

The bulk density of the catalyst has been included in the specific reaction rate (i.e., k1=k1’ b) Plot the concentrations of hydrogen, mesitylene, and xylene as a function of space-time.Calculate the space-time where the production of xylene is a maximum (i.e., opt) .

Reaction 1:Reaction 1: M + H M + H X + Me X + Me

Reaction 2:Reaction 2: X + H X + H T + Me T + Me

1.1. Mole balances:Mole balances:

Hydrogen

Mesitylene

Xylene

Toluene

Methane

2. 2. Rate laws:Rate laws:

MeMeMe

TT

XXX

MM

HHH

rrdV

dF

rdV

dF

rrdV

dF

rdV

dF

rrdV

dF

21

2

21

1

21

XHT

MHH

CCkr

CCkr5.0

22

5.011

3. 3. Stoichiometry:Stoichiometry: (no volume change with reaction, v=v0)

a. Reaction rates:

Reaction 1: -r1H = -r1M = r1X = r1Me

Reaction 2: -r2H = -r2X = r2T = r2Me

b. Flow rates:

4.4. Combining Combining and substituting in terms of the space-time ( )

If we know CM, CH, and CX, then CMe and CT can be calculated from the reaction stoichiometry. Consequently, we need only to solve the following three equations:

)(

)(

000

0000

0

0

0

XMMXMMT

HHHHMeMe

XX

MM

HH

CCCFFFF

CCFFCF

CF

CF

CF

0 V

XHMHX

MHM

XHMHH

CCkCCkd

dC

CCkd

dC

CCkCCkd

dC

5.02

5.01

5.01

5.02

5.01

5.5. Parameter evaluation:Parameter evaluation:

We now solve these three equations simultaneously using POLYMATH.

h 0.5h/ ft 476

ft 238

0C

mol/ft lb 0105.02

1

mol/ft lb 021.0R) R)(1500 mol atm/lbft (0.73

atm) (0.667)(35

3

3

0

X0

300

33

000

V

CC

RT

PyC

HM

HH

0.0 0.1 0.2 0.3 0.4 0.5

2.0

1.5

1.0

0.5

0.0 (hr)

Ci opt

CH

CMCX

Multiple reactions in a CSTRMultiple reactions in a CSTRFor a CSTR, a coupled set of algebraic eqns analogous to PFR differential eqns must be solved.

Rearranging yields where

After writing a mole balance on each species in the reaction set, we substitute for concentrations in the respective rate laws.

If there is no volume change with reaction, we use concentrations, C j, as variables.

If the reactions are gas-phase and there is volume change, we use molar flow rates, F j as variables.

q reactions in gas-phase with N different species to be solved

j

jj

r

FFV

0

VrFF jjj 0 ),...,,( 21

1 N

q

ijijj CCCfrr

001

0

001

0

100

1111110

,,

,,

,,

TT

NT

TNNNN

TT

NT

Tjjjj

q

iT

T

NT

Ti

CF

FC

F

FfVVrFF

CF

FC

F

FfVVrFF

CF

FC

F

FfVrVVrFF

Example 6-8: Hydrodealkylation of Mesitylene in a CSTRExample 6-8: Hydrodealkylation of Mesitylene in a CSTRCalculate the conversion of hydrogen and mesitylene along with the exiting concentrations of mesitylene, hydrogen, and xylene in a CSTR.

VrrF

VrF

VrrF

VrFF

VrrFF

MeMeMe

TT

XXX

MMM

HHHH

)(

)(

)(

21

2

21

10

210

1. Mole Balances:

2. Rate laws:

XHMeTXH

MHMeXMH

CCkrrrr

CCkrrrr2/1

22222

2/111111

3. Stoichiometry:

)(

)(

000

00

0

0

0

0

HHMeMe

XMMTT

XX

MM

HH

CCCF

FFFCF

CF

CF

CF

4. Combining and letting yields:0/ V

)(

)(

)(

2/12

2/11

2/110

2/12

2/110

XHMHX

MHMM

XHMHHH

CCkCCkC

CCkCC

CCkCCkCC

Example 6-8: Hydrodealkylation of Mesitylene in a CSTRExample 6-8: Hydrodealkylation of Mesitylene in a CSTRWe put these equations in a form such that they can be readily solved using POLYMATH.

XXHMHX

MHMMM

XHMHHHH

CCCkCCkCf

CCkCCCf

CCkCCkCCCf

)(0)(

)(0)(

)(0)(

2/12

2/11

2/110

2/12

2/110

For =0.5, the exiting concentrations are CH=0.0089, CM=0.0029 and CX=0.0033.The overall conversion is

72.00105.0

0029.00105.0 :Mesitylene

58.0021.0

0089.0021.0 :Hydrogen

0

0

0

0

0

0

0

0

M

MM

M

MMM

H

HH

H

HHH

C

CC

F

FFX

C

CC

F

FFX

2.0

1.5

1.0

0.5

0.0

Ci

CSTR

0.0 1.0 2.0

CH

CX

CM

The moles of hydrogen consumed in reaction 1 are equal to the moles of mesitylene consumed. Therefore, the conversion of hydrogen in reaction 1 is

The conversion of hydrogen in reaction 2 is

22.036.058.0

36.0021.0

0029.00105.0

12

0

01

HHH

H

MMH

XXX

C

CCX

The yield of xylene from mesitylene based on molar flow rates exiting the CSTR for =0.5 is

produced toluenemole

produced xylenemole7.0

00313.00029.00105.0

00313.0~

reacted mesitylene mole

produced xylenemole41.0

0029.00105.0

00313.0~

0

00

XMM

X

T

X

T

XXT

MM

X

MM

XMX

CCC

C

C

C

F

FS

CC

C

FF

FY

The overall selectivity of xylene relative to toluene is

Example 6-8: Hydrodealkylation of Mesitylene in a CSTRExample 6-8: Hydrodealkylation of Mesitylene in a CSTR

Selectivity and overall selectivity in a CSTRSelectivity and overall selectivity in a CSTR

For a CSTR the instantaneous selectivity and the overall selectivity are the same. For example, the instantaneous selectivity of xylene w.r.t. toluene is

Mole balances of xylene and toluene yields

Substituting in the instantaneous selectivity equation for rX and rT

T

XXT

T

XXT

F

FS

r

rS

~

The instantaneous selectivity

The overall selectivity

VrF

VrF

TT

XX

XTT

X

T

X

T

XXT S

F

F

VF

VF

r

rS

~

/

/

XTXT SS~

For all ideal CSTR

Homework

P6-9C (a)-(g)

P6-12A (a)-(h)P6-14B

Due Date: Next Week

Chemical Reaction Engineering Chapter 6 Chapter 6 Mutiple Reactions.

Documents

Transcript of Chemical Reaction Engineering Chapter 6 Chapter 6 Mutiple Reactions.