1. Chapter 6
Chemical Reaction Engineering
Mutiple Reactions

2. Objectives  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.

3. Definition of Multiple Reaction
 Parallel rxns (competing rxns) B A C
 Series rxns (consecutive rxns)
A B C
 Complex rxns (Parallel + Series rxns)
A + B C + D
A + C E
 Independent rxns
A B + C
D E + F k1 k2

4. Examples of Multiple Reaction
 Parallel rxns (competing rxns)
2CO2 + 2H2O
CH2=CH2 +O2
CH2-CH2
 Series rxns (consecutive rxns) O O
CH2-CH2 + NH HOCH2CH2NH2
EO EO
(HOCH2CH2)2NH (HOCH2CH2)3NH2
 Complex Rxns (Combination of parallel & series rxns)
C2H5OH C2H4 + H2O
C2H5OH CH3CHO + H2
C2H4 + CH3CHO C4H6 + H2O
 Independent rxns (The cracking of crude oil to form gasoline)
C15H C12H26 + C3H6
C8H C6H14 + C2H4 k1 k2
Monoethanolamine
Dinoethanolamine
Trinoethanolamine

5. Desired and Undesired Reaction Parallel Reactions
D (Desired Product) A
U (Undesired Product)
The economic incentive
Maximize the formation of D
Minimize the formation of U kD kU
Competing
or side rxn
Total cost
Reaction cost
Separation cost
cost
Low High A D
A, U
reaction
separation
Rxn-separation system producing both D & U
Efficiency of a reactor system

6. Rate selectivity parameter, SDU Rate selectivity parameter, S
D (Desired Product) A
U (Undesired Product)
The rate laws are
Rate selectivity parameter = Instantaneous selectivity kD kU
We want to maximize SDU
Rate of formation of D =
Rate of formation of U

7. Overall Selectivity, SDU~
Overall Selectivity FD ~
Exit molar flow rate of desired product
SDU = = FU
Exit molar flow rate of undesired product
For Batch Reactor ND ~
SDU = NU

8. Example 6-1: Comparison between SDU and SDU in a CSTR~ ~
Mission: Develop a relationship between SDU and SDU in a CSTR
Solution rD FD ~
SDU =
and
SDU = rU FU
By the way, FD=rDV and FU=rUV
rDV rD FD ~
SDU
SDU = = = = FU
rUV rU

9. Instantaneous Yield and Overall Yield
Instantaneous Yield (Basis: Reaction Rate) rD
YD =
-rA
Overall Yield (Basis: Molar Flow Rate) ND
Mole of desired product
formed at the end of reaction ~
YD = =
For a batch system:
NAo-NA
Number of moles of
key reactant consumed FD ~
YD =
For a flow system:
FAo-FA

10. Overall Selectivity
Overall Yield

11. Note Different definitions for selectivity and yield
Check carefully to ascertain the definition intended by the author
From an economic standpoint,
overall selectivities and yields are important
in determining profits
The instantaneous selectivities give insights
in choosing reactors and reaction schemes
that will help maximize the profit

12. Maximizing SDU in Parallel Reactions
 Case 1: a1 > a2 , a = a1- a2
maximize SDU
- 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

13. Maximizing SDU for one reactant
 Case 2: a2 > a1 , a = a2- a1
maximize SDU
- 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)

14. Maximizing SDU for one reactant
Whether the reaction should be run at high or low T
 Case 3: ED > EU
- kD (rD) increases more rapidly with increasing temperature than does the kU.
- keeping the temperature as high as possible to maximize SDU.
 Case 4: EU > ED
- keeping the temperature as low as possible to maximize SDU
- not so low that the desired rxn does not proceed to any significant extent.

15. Example: Minimizing unwanted products (U & Q) for a single reactant
A U (2)
Q (3)
- 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)
(5)
- From (5), SDU is minimized at low concentration of A
1. High Temperatures (to minimize the formation of Q)
2. Low concentration of A (to minimize the formation of U)
- Adding inerts, using low pressures, using CSTR or recycled reactor
Textbook Example 6-2: 수업 후 해동에서 꼭 풀어보고 가세요

16. Reactor Selection and Operating Conditions
Rate selectivity parameter, S
D (Desired Product)
A + B
U (Undesired Product)
The rate laws are
Rate selectivity parameter=Instantaneous selectivity k1 k2

17. Different reactors and schemes for maximizing the desired product
Figure 6-3
Different reactors and schemes for maximizing the desired product A B B A A B A B A B
(a) CSTR (b) tubular reactor (c ) batch (d) semi-batch 1 (e) semi-batch 2 A B A B
(f) Tubular reactor with side streams (g) Tubular reactor with side streams B A A B
(i) Tubular reactor with recycle (h) Series of small CSTRs

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

19. Example 6-3: Minimizing unwanted products for two reactants
for the parallel reaction
D (Desired Product)
A + B
U (Undesired Product)
Case I : a1 > a2, b1 > b2, a = a1-a2 > 0, b = b1-b2 > 0
the rate selectivity parameter k1 k2
To maximize the SDU, maintain the concentration of both A and B as high as possible
 a tubular reactor (Figure 6.3 (b))
 a batch reactor (Figure 6.3 (c))
 high pressures (if gas phase), reduce inert

20. Example 6-3: Minimizing unwanted products for two reactants
for the parallel reaction
D (Desired Product)
A + B
U (Undesired Product)
Case II : a1 > a2, b1 < b2, a = a1-a2 > 0, b = b2-b1 > 0
the rate selectivity parameter k1 k2
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 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))

21. Example 6-3: Minimizing unwanted products for two reactants
for the parallel reaction
D (Desired Product)
A + B
U (Undesired Product)
Case III : a1 < a2, b1 < b2, a = a2-a1 > 0, b = b2-b1 > 0
the rate selectivity parameter k1 k2
To maximize the SDU, maintain the concentration of both A and B as low as possible
 a CSTR (Figure 6.3(a))
 a tubular reactor in which there is a large recycle ratio (Figure 6.3(i))
 a feed diluted with inert material
 low pressures (if gas phase)

22. Example 6-3: Minimizing unwanted products for two reactants
for the parallel reaction
D (Desired Product)
A + B
U (Undesired Product)
Case IV : a1 < a2, b1 > b2, a = a2-a1 > 0, b = b1-b2 > 0
the rate selectivity parameter k1 k2
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 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))

23. Maximizing the desired product in series reactionk1 k2
A B C
 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

24. Maximizing the desired product in series reactionk1 k2
A B C
Desired Product
 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.

25. Example 6-4: Maximizing the yield of the intermediate product
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. k1 k2
CH3CH2OH(g)
CH3CHO
2CO2

26. Solution

31. Reaction paths for different k’s in series reaction
A B C k1 k2
For k1/k2>1, a
Large quantity of B
Can be obtained B
For k1/k2<1, a
Little quantity of B
Can be obtained
1st rxn is slow
2nd rxn is fast A C
Long rxn time in batch or long tubular reactor
-> B will be converted to C

32. Algorithm for solution to complex reactions
Steps 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
Algorithm for solution to complex reactions

33. 1. Mole balances for multiple reactions
It is better to solve problems using moles (Nj) or molar flow rates (Fj) rather than conversion
Reactor Gas Liquid
Batch
CSTR
PFR/PBR
Semibatch

34. . 2. Net Rates of Reaction Reaction 1: A + B 3C + D
Key point: to write the net rate of each species (e.g. A, B, …)
Reaction 1: A + B C + D
Reaction 2: A + 2C 3E
Reaction 3: 2B + 3E 4F
Reaction q: A + ½C G .
k1A
k2A
K3B
kqA
Net rates of reaction of A and B:

35. Total rate of formation of each species from all reaction
Number Reaction Stoichiometry
q rxns taking place
The net rate of rxn for A:
species
reaction number

36. . 3. Rate laws rA=r1A+r2A= -k1ACACB-k2ACACC2 Reaction 1: A + B 3C + D
Reaction 2: A + 2C 3E
Reaction 3: 2B + 3E 4F .
k1A
k2A
k3B 1 1 B
rA=r1A+r2A= -k1ACACB-k2ACACC2
We need to determine the rate law for at least one species in each rxn.

37. 4. Stoichiometry: Relative Rates of Reaction
Generic reaction:
aA + bB cC + dD
Relative rate of reactions
Multiple rxns:
For example,
Relative rates of
Rxn in reaction i

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

39. 6. Stoichiometry: Concentrations as a function of molar flow rate
Liquid phase:
Ideal Gas:
Isothermal gas phase without DP:
Net rate of formation

40. Multiple Reactions in a PFR/PBR
(combining mole balance, rate law, and stoichiometry)
Mole balance
Rate law
j Coupled ODEs
Isothermal
gas phase
without DP: .

41. write the net rates of formation of NO, O2, and N2.
Example 6-5: Write the rate law for each species in each reaction and then
write the net rates of formation of NO, O2, and N2.
Reaction 1: 4NH3 + 6NO  5N2 + 6H2O
Reaction 2: 2NO  N2 + O2
Reaction 3: N2 + 2O2  2NO2
Solution

46. Example 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
Solution

53. Example 6-7: Hydrodealkylation of Mesitylene in a PFR
CH3
+ H CH4 k1 k2
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/rb) is 238 ft3.
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’ rb)
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., topt) .

54. Reaction 1: M + H  X + Me
Reaction 2: X + H  T + Me
Mole balances:
Hydrogen
Mesitylene
Xylene
Toluene
Methane
2. Rate laws:

55. 3. 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:
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:

56. Ci t (hr) topt Parameter evaluation:
We now solve these three equations simultaneously using POLYMATH.
2.0
1.5
1.0
0.5
0.0
t (hr) Ci
topt CH CM CX

57. Multiple reactions in a CSTR
For 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, Cj, as variables.
If the reactions are gas-phase and there is volume change, we use molar flow rates, Fj as variables.
q reactions in gas-phase with N different species to be solved

58. Example 6-8: Hydrodealkylation of Mesitylene in a CSTR
Calculate the conversion of hydrogen and mesitylene along with the exiting concentrations of mesitylene, hydrogen, and xylene in a CSTR.
1. Mole Balances:
3. Stoichiometry:
2. Rate laws:
4. Combining and letting yields:

59. Example 6-8: Hydrodealkylation of Mesitylene in a CSTR
We put these equations in a form such that they can be readily solved using POLYMATH.
For t=0.5, the exiting concentrations are CH=0.0089, CM= and CX=
The overall conversion is
2.0
1.5
1.0
0.5
0.0 Ci
tCSTR CH CX CM

60. Example 6-8: Hydrodealkylation of Mesitylene in a CSTR
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
Example 6-8: Hydrodealkylation of Mesitylene in a CSTR
The conversion of hydrogen in reaction 2 is
The yield of xylene from mesitylene based on molar flow rates exiting the CSTR for t=0.5 is
The overall selectivity of xylene relative to toluene is

61. Selectivity 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
The instantaneous selectivity
The overall selectivity
Mole balances of xylene and toluene yields
Substituting in the instantaneous selectivity equation for rX and rT
For all ideal CSTR

62. Homework
P6-9C (a)-(g)
P6-12A (a)-(h)
Due Date: Next Week

63. Take Home Examination (Term-Paper)
P6-14B
Due Date: June 9, 2009