Corresponding Solutions for Chemical Reaction Engineering
CHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING .................................................. 1
CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS ...................................................................... 3
CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA ................................................................ 7
CHAPTER 4 INTRODUCTION TO REACTOR DESIGN ............................................................................ 19
CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR ..................................................................... 22
CHAPTER 6 DESIGN FOR SINGLE REACTIONS ...................................................................................... 26
CHAPTER 10 CHOOSING THE RIGHT KIND OF REACTOR .................................................................. 32
CHAPTER 11 BASICS OF NON-IDEAL FLOW ............................................................................................ 34
CHAPTER 18 SOLID CATALYZED REACTIONS........................................................................................ 43
Chapter 1 Overview of Chemical Reaction Engineering
1.1 Municipal waste water treatment plant. Consider a municipal water treatment plant for a
small community (Fig.P1.1). Waste water, 32000 m3/day, flows through the treatment
plant with a mean residence time of 8 hr, air is bubbled through the tanks, and microbes in
the tank attack and break down the organic material
(organic waste) +O2
microbes


CO2 + H2O
A typical entering feed has a BOD (biological oxygen demand) of 200 mg O 2/liter, while
the effluent has a megligible BOD. Find the rate of reaction, or decrease in BOD in the
treatment tanks.
Waste water
32,000 m3/day
200 mg O2
needed/liter
Waste water
Treatment plant
Mean residence
time t =8 hr
Clean water
32,000 m3/day
Zero O2
needed
Figure P1.1
Solution:
 rA  
dN A
Vdt
1 m3
mg 1g 1000 L 32 g
 32000  (
 day )  (200  0)
/
1
3 day
L 1000mg m 3
mol

3
1
m
1
day
32000
 day
3
day 3
 18.75mol /( m 3  day )  2.17  10  4 mol /( m 3  s )
1.2 Coal burning electrical power station. Large central power stations (about 1000 MW
electrical) using fluiding bed combustors may be built some day (see Fig.P1.2). These
giants would be fed 240 tons of coal/hr (90% C, 10%H2), 50% of which would burn
within the battery of primary fluidized beds, the other 50% elsewhere in the system. One
suggested design would use a battery of 10 fluidized beds, each 20 m long, 4 m wide, and
containing solids to a depth of 1 m. Find the rate of reaction within the beds, based on the
oxygen used.
1
Solution:
V  (20  4  1)  10  800m3
N c
coal
kgc
kgc
 240  10 3
 0.5  0.9
 108  10 3
 9000molc /(bed  hr )
t
hr
kgcoal
hr
rO2  rc  
1 N c
1  9000


 11.25kmolO2 /( m 3  hr )
V t
800
1
dO2
12000
 9000  1 
 12000mol /(bed  hr )
dt
4
 rO2 
1 dO2 1.5  10 4 mol /(bed  hr )

 4.17mol /( m 3  s)
V dt
800
2
Chapter 2 Kinetics of Homogeneous Reactions
2.1 A reaction has the stoichiometric equation A + B =2R . What is the order of reaction?
Solution:
Because we don’t know whether it is an elementary reaction or not, we can’t tell
the index of the reaction.
2.2 Given the reaction 2NO2 + 1/2
O2 = N2O5 , what is the relation between the rates of
formation and disappearance of the three reaction components?
Solution:
 rNO2  4rO2  2rN2O5
2.3 A reaction with stoichiometric equation 0.5 A + B = R +0.5 S has the following rate
expression
-rA = 2 C0.5
A CB
What is the rate expression for this reaction if the stoichiometric equation is written as
A + 2B = 2R + S
Solution:
No change. The stoichiometric equation can’t effect the rate equation, so it
doesn’t change.
2.4 For the enzyme-substrate reaction of Example 2, the rate of disappearance of substrate is
given by
-rA =
1760[A ][E 0 ]
,
6 C A
mol/m3·s
What are the units of the two constants?
Solution:
[rA ] 
mol [k ][ A][ E0 ]

m 3  s [6]  [C A ]
[6]  [C A ]  mol / m3
[k ] 
2.5
mol
mol / m 3
1


3
3
3
m  s (mol / m )(mol / m ) s
For the complex reaction with stoichiometry A + 3B → 2R + S and with second-order
rate expression
-rA = k1[A][B]
3
are the reaction rates related as follows: rA= rB= rR?
then how are they related?
1
1
Solution:  rA   rB  rR
3
2
If the rates are not so related,
Please account for the sings , + or - .
2.6 A certain reaction has a rate given by
mol/cm3·min
-rA = 0.005 C2A ,
If the concentration is to be expressed in mol/liter and time in hours, what would be
the value and units of the rate constant?
Solution:
(rA' ) 
 rA' 
mol
mol
 (rA )  3
L  hr
cm  min
L  hr
mol
 3
(rA )  10 4  6  rA  6  10 4  0.005C A2  300C A2
mol cm  min
mol
mol
 (C A )  3
L
cm
L
mol
 C A' 
 3  C A  10 3 C A
mol cm
(C A' ) 
 (rA' )  300C A2  300  (10 3 C A' ) 2  3  10 4 C A'
2
k '  3  10 4
2.7
For a gas reaction at 400 K the rate is reported as
-
dp A
= 3.66 p2A, atm/hr
dt
(a) What are the units of the rate constant?
(b) What is the value of the rate constant for this reaction if the rate equation is
expressed as
-rA = -
1 dN A
V dt
= k C2A ,
mol/m3·s
Solution:
(a) The unit of the rate constant is [1 / atm  hr ]
(b)  rA  
1 dN A
V dt
Because it’s a gas reaction occuring at the fined terperatuse, so V=constant, and
T=constant, so the equation can be reduced to
4
 rA  
dP
V dPA
1
3.66 2 3.66

( A ) 
PA 
(C A RT ) 2
VRT dt
RT
dt
RT
RT
 (3.66RT )C A2  kCA2
So we can get that the value of
k  3.66RT  3.66  0.08205  400  120.1
2.9 The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol.
How much faster the decomposition at 650℃ than at 500℃?
Solution:
Ln
r2
k
E 1 1
300kJ / mol
1
1
 Ln 2  (  ) 
(

)  7.586
3
r1
k1 R T1 T2
8.314kJ /(10 mol  K ) 173K 923K

2.11
r2
 1970.7
r1
In the mid-nineteenth century the entomologist Henri Fabre noted that French ants
(garden variety) busily bustled about their business on hot days but were rather sluggish on
cool days. Checking his results with Oregon ants, I find
Running speed, m/hr
150
160
230
295
370
Temperature, ℃
13
16
22
24
28
What activation energy represents this change in bustliness?
Solution:
r  k0e

E
RT
f (concentrat ion )let  f (concentrat ion )  cons tan t  ak 0 e
 LnrA  Lnk ' 
E
RT
 k 'e

E
RT
1E
T R
Suppose y  LnrA , x 
so slope  

1
,
T
E
, intercept  Lnk '
R
5
150
160
230
295
370
-3.1780
-3.1135
-2.7506
-2.5017
-2.2752
T /o C
13
16
22
24
28
1
 10 3
T
3.4947
3.4584
3.3881
3.3653
3.3206
rA /( m  h 1 )
LnrA
4
-y = 5417.9x - 15.686
R2 = 0.97
-Ln r
3
2
1
0
0.0033
0.00335
0.0034
0.00345
0.0035
1/T
-y = -5147.9 x + 15.686
Also slope  
E
 5147.9 K , intercept  Lnk ' = 15.686 ,
R
E  5147.9K  8.3145J /( mol  K )  42.80kJ / mol
6
Chapter 3 Interpretation of Batch Reactor Data
3.1
If -rA = - (dCA/dt) =0.2 mol/liter·sec
when CA = 1 mol/liter, what is the rate of
reaction when CA = 10 mol/liter?
Note: the order of reaction is not known.
Solution: Information is not enough, so we can’t answer this kind of question.
3.2
Liquid a sedomposes by first-order kinetics, and in a batch reactor 50% of A is
converted in a 5-minute run. How much longer would it take to reach 75% conversion?
Solution:
Because the decomposition of A is a 1st-order reaction, so we can express the rate
equation as:
 rA  kCA
We know that for 1st-order reaction, Ln
Ln
C Ao
 kt ,
CA
C Ao
 kt1 ,
C A1
Ln
C Ao
 kt2
C A2
C A1  0.5C Ao , C A2  0.25C Ao
So t 2  t1 
t1 
C
C
1
1
1
( Ln Ao  Ln Ao )  ( Ln4  Ln2)  Ln2
k
C A2
C A1
k
k
C
1
1
Ln( Ao )  Ln2  5 min
k
C A1
k
equ(1)
equ(2)
So t 2  t1  t1  5 min
3.3
Repeat the previous problem for second-order kinetics.
Solution:
We know that for 2nd-order reaction,
1
1

 kt ,
C A C A0
So we have two equations as follow:
1
1
2
1
1




 kt1  k 5 min , equ(1)
C A1 C A0 C Ao C Ao C Ao
7
1
1
4
1
1



 3(
)  3kt1  kt2 , equ(2)
C A2 C Ao C Ao C Ao
C Ao
So t 2  3t1  15 min , t 2  t1  10 min
3.4 A 10-minute experimental run shows that 75% of liquid reactant is converted to product
1
by a
-order rate. What would be the fraction converted in a half-hour run?
2
Solution:
In a
dC
1
 order reaction:  rA   A  kCA0.5 ,
2
dt
kt
0.5
 C Ao
 C A0.15 ,
2
So we have two equations as follow:
1 0.5
0.5
0.5
0.5
C Ao
 C A0.15  C Ao
 ( C Ao
)  0.5C Ao
 kt1  k (10 min) , equ(1)
4
After integration, we can get:
0.5
C Ao
 C A0.25  kt2  k (30 min) ,
equ(2)
0.5
Combining these two equations, we can get: 1.5C Ao
 kt2 , but this means C A0.25  0 , which is
impossible, so we can conclude that less than half hours, all the reactant is consumed up. So
the fraction converted X A  1 .
3.5
In a hmogeneous isothermal liquid polymerization, 20% of the monomer disappears in
34 minutes for initial monomer concentration of 0.04 and also for 0.8 mol/liter. What rate
equation represents the disappearance of the monomer?
Solution:
The rate of reactant is independent of the initial concentration of monomers, so
we know the order of reaction is first-order,
 rmonomer  kCmonomer
And Ln
Co
 (34 min) k
0.8Co
k  0.00657 min 1
 rmonomer  (0.00657 min 1 )Cmonomer
8
3.6 After 8 minutes in a batch reactor, reactant (CA0 = 1 mol/liter) is 80% converted; after 18
minutes, conversion is 90%. Find a rate equation to represent this reaction.
Solution:
1
1
Ln
t
k 1  X A2
Ln1
In 1st order reaction, 2 

 1.43 , dissatisfied.
1
1
t1
Ln5
Ln
k 1  X A1
1 1
1
1
1
(

)

k C A2 C Ao
0.1C Ao C Ao
9 / C Ao 9
t
In 2nd order reaction, 2 


 , satisfied.
1 1
1
1
1
t1
4 / C Ao 4
(

)

k C A1 C Ao
0.2C Ao C Ao
According to the information, the reaction is a 2nd-order reaction.
3.7 nake-Eyes Magoo is a man of habit. For instance, his Friday evenings are all alike—into
the joint with his week’s salary of ＄180, steady gambling at “2-up” for two hours, then
home to his family leaving ＄45 behind. Snake Eyes’s betting pattern is predictable. He
always bets in amounts proportional to his cash at hand, and his losses are also
predictable—at a rate proportional to his cash at hand. This week Snake-Eyes received a raise
in salary, so he played for three hours, but as usual went
home with ＄135. How much
was his raise?
Solution:
n Ao  180 , n A  13 , t  2h ,
n A'  135 , t ;  3h ,  rAknA
Ln
So we obtain
n Ao
 kt ,
nA
Ln
'
n Ao
n Ao
Ln(
) Ln( ' )
nA
nA

t
t'
n'
180
Ln Ao
13 
135 , n '  28
A
2
3
3.9 The first-order reversible liquid reaction
A
 R , CA0 = 0.5 mol/liter, CR0=0
takes place in a batch reactor. After 8 minutes, conversion of A is 33.3% while equilibrium
conversion is 66.7%. Find the equation for the this reaction.
Solution:
Liquid reaction, which belongs to constant volume system,
9
1st order reversible reaction, according to page56 eq. 53b, we obtain
t
t   dt  
0
XA
0
dX A
k1
1

Ln
k1  (k1  k 2 ) X A k1  k 2
k1  (k1  k 2 ) X A
1
 t  480 sec  8 min , X A  0.33 , so we obtain
480 sec  8 min
Kc 
k1
1
Ln
k1  k 2
k1  (k1  k 2 )0.33
eq(1)
eq(1)
C
k1 C Re M  X Ae
, M  Ro  0 , so we obtain eq(2)


k 2 C Ae
1  X Ae
C Ao
Kc 
X Ae
k1


k 2 1  X Ae
2
3
1
2
3
 2,
 k1  2k 2 eq(2)
Combining eq(1) and eq(2), we obtain
k 2  0.02888 min 1  4.8  10 4 sec 1
k1  2k 2  0.05776 min 1  9.63  10 4 sec 1
So the rate equation is  rA  
dC A
 k1C A  k 2 (C Ao  C A )
dt
 4.8  10 4 sec 1 C A  9.63  10 4 s e c1 (C A0  C A )
3.10 Aqueous A reacts to form R (A→R) and in the first minute in a batch reactor its
concentration drops from CA0 = 2.03 mol/liter to CAf = 1.97 mol/liter. Find the rate
equation from the reaction if the kinetics are second-order with respect to A.
Solution: It’s a irreversible second-order reaction system, according to page44 eq 12, we
obtain
1
1

 k1  1 min ,
1.97 2.03
so k1  0.015
L
mol  min
so the rate equation is  rA  (0.015 min 1 )C A2
3.15 At room temperature sucrose is hydrolyzed by the catalytic action of the enzyme
sucrase as follows:
10
Aucrose
sucrase
 products
Starting with a sucrose concentration CA0 = 1.0 millimol/liter and an enzyme concentration
CE0= 0.01 millimol/liter, the following kinetic data are obtained in a batch reactor
(concentrations calculated from optical rotation measurements):
CA, millimol/liter
t,hr
0.84
0.68
0.53
0.38
0.27
0.16
0.09
0.04
0.018
0.006
0.0025
1
2
3
4
5
6
7
8
9
10
11
Determine whether these data can be reasonably fitted by a knietic equation of the
Michaelis-Menten type, or
-rA =
k 3C AC E 0
C A  CM
where CM = Michaelis constant
If the fit is reasonable, evaluate the constants k3 and CM. Solve by the integral method.
Solution:
Solve the question by the integral method:
 rA  
dC A k 3C Eo C A
kC

 4 A ,
dt
C A  C M 1 k 5 C A
k4 
k 3C Eo
1
, k5 
CM
CM
Ln
k
t
1
 5  
C Ao  C A k 4 k 4 C Ao
t, hr
C A ,mmol/L
Ln
C Ao
C Ao
CA
 CA
C Ao
CA
 CA
t
C Ao  C A
1
0.84
1.0897
6.25
2
0.68
1.2052
6.25
3
0.53
1.3508
6.3830
4
0.38
1.5606
6.4516
5
0.27
1.7936
6.8493
6
0.16
2.1816
7.1428
7
0.09
2.6461
7.6923
8
0.04
3.3530
8.3333
9
0.018
4.0910
9.1650
10
0.006
5.1469
10.0604
11
11
0.0025
Ln
Suppose y=
t
, x=
C Ao
C Ao  C A
6.0065
11.0276
C Ao
CA
, thus we obtain such straight line graph
 CA
12
t/(Cao-Ca)
10
8
6
y = 0.9879x + 5.0497
4
R2 = 0.998
2
0
0
1
Slope 
2
3
4
5
Ln(Cao/Ca)/(Cao-Ca)
6
7
k
C
1
 M  0.9879 , intercept= 5  5.0497
k 4 k 3C Eo
k4
So C M 
k3 
1 0.9879

 0.1956(mmol / L) ,
k 5 5.0497
k 4CM
0.1956

 19.80hr 1
C Eo
0.9879  0.01
3.18 Enzyme E catalyzes the transformation of reactant A to product R as follows:
A enzyme
 R,
-rA =
200C A C E 0
mol
2  C A liter  min
If we introduce enzyme (CE0 = 0.001 mol/liter) and reactant (CA0 = 10 mol/liter)
into a batch rector and let the reaction proceed, find the time needed for the
concentration of reactant to drop to 0.025 mol/liter. Note that the concentration of
enzyme remains unchanged during the reaction..
Solution:

2  CA
1
dt
10



5
rA
dC A 200  0.001C A C A
Rearranging and integrating, we obtain:
12
10
t
t   dt   
0
0.025
10


C
10
(
 5)dC A  10 Ln Ao  5(C Ao  C A )
CA
CA

 0.025
 10 Ln
10
 5(C Ao  C A )  109.79 min
0.025
3.20 M.Hellin and J.C. Jungers, Bull. soc. chim. France, 386(1957), present the data in
Table P3.20 on thereaction of sulfuric acid with diethylsulfate in a aqueous solution at 22.9℃:
H2SO4 + (C2H5)2SO4 → 2C2H5SO4H
Initial concentrations of H2SO4 and (C2H5)2SO4 are each 5.5 mol/liter. Find a rate equation for
this reaction.
Table P3.20
t, min
C2H5SO4H,
t, min
C2H5SO4H,
mol/liter
Solution:
mol/liter
0
0
180
4.11
41
1.18
194
4.31
48
1.38
212
4.45
55
1.63
267
4.86
75
2.24
318
5.15
96
2.75
368
5.32
127
3.31
379
5.35
146
3.76
410
5.42
162
3.81
∞
(5.80)
It’s a constant-volume system, so we can use XA solving the problem:
i) We postulate it is a 2nd order reversible reaction system A  B  2R
The rate equation is:  rA  
dC A
 k1C AC B  k 2 C R2
dt
C Ao  C Bo  5.5mol / L , C A  C Ao (1  X A ) ,
C B  C Bo  C Ao X A  C A , C R  2C Ao X A
When t   , CRe  2C Ao X Ae  5.8mol / L
So X Ae 
5.8
 0.5273 ,
2  5.5
C Ae  C Be  C Ao (1  X Ae )  5.5  (1  0.5273)  2.6m o /l L
After integrating, we obtain
13
Ln
X Ae  (2 X Ae  1) X A
1
 2k1 (
 1)C Ao t eq (1)
X Ae  X A
X Ae
The calculating result is presented in following Table.
t,
Ln
X Ae  (2 X Ae  1) X A
X Ae  X A
Ln(1 
XA
)
X Ae
C R , mol / L
C A , mol / L
XA
0
0
5.5
0
0
0
41
1.18
4.91
0.1073
0.2163
-0.2275
48
1.38
4.81
0.1254
0.2587
-0.2717
55
1.63
4.685
0.1482
0.3145
-0.3299
75
2.24
4.38
0.2036
0.4668
-0.4881
96
2.75
4.125
0.25
0.6165
-0.6427
127
3.31
3.845
0.3009
0.8140
-0.8456
146
3.76
3.62
0.3418
1.0089
-1.0449
162
3.81
3.595
0.3464
1.0332
-1.0697
180
4.11
3.445
0.3736
1.1937
-1.2331
194
4.31
3.345
0.3918
1.3177
-1.3591
212
4.45
3.275
0.4045
1.4150
-1.4578
267
4.86
3.07
0.4418
1.7730
-1.8197
318
5.15
2.925
0.4682
2.1390
-2.1886
368
5.32
2.84
0.4836
2.4405
-2.4918
379
5.35
2.825
0.4864
2.5047
-2.5564
410
5.42
2.79
0.4927
2.6731
-2.7254

5.8
2.6
0.5273
—
—
min
Draw Ln
X Ae  (2 X Ae  1) X A
~ t plot, we obtain a straight line:
X Ae  X A
14
3
y = 0.0067x - 0.0276
2.5
2
R = 0.9988
Ln
2
1.5
1
0.5
0
0
100
200
Slope  2k1 (
 k1 
t
300
400
500
1
 1)C Ao  0.0067 ,
X Ae
0.0067
 6.794  10 4 L /( mol  min)
1
2(
 1)  5.5
0.5273
When approach to equilibrium, K c 
2
C Re
k1

,
k 2 C Ae C Be
k1C Ae C Be 6.794  10 4  2.6 2

 1.364  10 4 L /( mol  min)
so k 2 
2
2
C Re
5.8
So the rate equation is
 rA  (6.794  10 4 C AC B  1.364  10 4 C R2 )mol /( L  min)
ii) We postulate it is a 1st order reversible reaction system, so the rate equation is
 rA  
dC A
 k1C A  k 2 C R
dt
After rearranging and integrating, we obtain
Ln(1 
Draw Ln(1 
XA
1 '
)
k1t
X Ae
X Ae
eq (2)
XA
) ~ t plot, we obtain another straight line:
X Ae
15
3
-y = 0.0068x - 0.0156
2.5
R2 = 0.9986
-Ln
2
1.5
1
0.5
0
0
100
200
x
300
400
500
k1'
Slope 
 0.0068 ,
X Ae
So k1'  0.0068  0.5273  3.586  10 3 min 1
k 2' 
k1' C Ae  3.586  10 3  2.6

 1.607  10 3 min 1
C Re
5.8
So the rate equation is
 rA  (3.586  10 3 C A  1.607  10 3 C R )mol /( L  min)
We find that this reaction corresponds to both a 1st and 2nd order reversible reaction
system, by comparing eq.(1) and eq.(2), especially when XAe =0.5 , the two equations are
identical. This means these two equations would have almost the same fitness of data when
the experiment data of the reaction show that XAe =0.5.(The data that we use just have XAe
=0.5273 approached to 0.5, so it causes to this.)
3.24
In the presence of a homogeneous catalyst of given concentration, aqueous reactant A
is converted to product at the following rates, and CA alone determines this rate:
CA,mol/liter
1
2
4
6
7
9
12
-rA, mol/liter·hr
0.06
0.1
0.25
1.0
2.0
1.0
0.5
We plan to run this reaction in a batch reactor at the same catelyst concentration as used in
getting the above data. Find the time needed to lower the concentration of A from C A0 = 10
mol/liter to CAf = 2 mol/liter.
Solution:
By using graphical integration method, we obtain that the shaped area is 50 hr.
16
20
16
12
-1/Ra
8
4
0
0
2
4
6
Ca
8
10
12
14
3.31 The thermal decomposition of hydrogen iodide
2HI → H2 + I2
is reported by M.Bodenstein [Z.phys.chem.,29,295(1899)] as follows:
T,℃
508
427
393
356
283
k,cm3/mol·
0.1059
0.00310
0.000588
80.9×10-6
0.942×10-6
s
Find the complete rate equation for this reaction. Use units of joules, moles, cm 3, and
seconds.
According to Arrhenius’ Law,
k = k0e-E/R T
transform it,
- In(k) = E/R·(1/T) －In(k0)
Drawing the figure of the relationship between k and T as follows:
16
y = 7319.1x - 11.567
2
-Ln(k)
12
R = 0.9879
8
4
0
0.001
0.002
0.003
0.004
1/T
From the figure, we get
slope = E/R = 7319.1
intercept = - In(k0) = -11.567
17
E = 60851 J/mol
k0 = 105556 cm3/mol·s
From the unit [k] we obtain the thermal decomposition is second-order reaction, so the rate
expression is
- rA = 105556e-60851/R T·CA2
18
Chapter 4 Introduction to Reactor Design
4.1 Given a gaseous feed, CA0 = 100, CB0 = 200, A +B→ R + S, XA = 0.8. Find XB,CA,CB.
Solution: Given a gaseous feed, C Ao  100 , C Bo  200 , A  B  R  S
X A  0 , find X B , C A , C B
 A   B  0 , C A  C Ao (1  X A )  100  0.2  20
XB 
bC Ao X A 1  100  0.8

 0.4
C Bo
200
C B  C Bo (1  X B )  200  0.6  120
4.2 Given a dilute aqueous feed, CA0 = CB0 =100, A +2B→ R + S, CA = 20. Find XA, XB,
CB.
Solution: Given a dilute aqueous feed, C Ao  C Bo  100 ,
A  2B  R  S , C A  20 , find X A , X B , C B
Aqueous reaction system, so  A   B  0
When X A  0 , V  200
When X A  1 , V  100
 C
1
1
So  A   ,  B  A Bo  
2
bC Ao
4
X A  1
XB 
CA
20
 1
 0.8 ,
C Ao
100
b C Ao X A 2 100  0.8

 
 1.6  1 , which is impossible.
a C Bo
1
100
So X B  1 , C B  C Bo  100
4.3
Given a gaseous feed, CA0 =200, CB0 =100, A +B→ R, CA = 50. Find XA, XB, CB.
Solution: Given a gaseous feed, C Ao  200 , C Bo  100 ,
19
A  B  R , C A  50 .find X A , X B , C B
X A  1
XB 
CA
50
 1
 0.75 ,
C Ao
200
bC Ao X A
 1.5  1 , which is impossible.
C Bo
So C B  C Bo  100
4.4
Given a gaseous feed, CA0 = CB0 =100, A +2B→ R, CB = 20. Find XA, XB, CA.
Solution: Given a gaseous feed, C Ao  C Bo  100 ， A  2B  R , C Bo  20 ,
Find X A , X B , C A
X B  0 , V  100 A  100B  200
X B  1, V  50 A  100R  150
150  200
 0.25  100
 0.25 ,   A 
 0.5
1
200
 100
2
1
 100  0.842
100  20
 0.421
XB 
 0.842 , X A  2
100
100  0.25  20
B 
C A  C Ao
4.6
1 X A
1  0.421
 100 
 73.34
1  A X A
1  0.5  0.421
Given a gaseous feed, T0 =1000 K, π0＝5atm, CA0=100, CB0=200, A +B→5R,T =400 K,
π=4atm, CA =20. Find XA, XB, CB.
Solution:
Given a gaseous feed, To  1000K ,  0  5atm , C Ao  100 , C Bo  200
A  B  5R , T  400K ,   4atm , C A  20 ,
A 
find X A , X B , C B .
a C
T 0
400  5
600  300

 0.5
 1 ,  B  A Bo  2 ,
bC Ao
T0 1000  4
300
According to eq page 87,
20
C A T 0
20
1
 0.5
C Ao T0
100
XA 

 0.818
20
C A T 0
1  1
 0.5
1  A
100
C Ao T0
1
XB 
bC Ao X A 100  0.818

 0.409
aC Bo
200
C Bo b
T
 X A ) 0 C Ao ( 200  0.818)  200
C
a
T 0
C B  Ao
 100
 130
1  A X A
1  1  0.818
(
4.7 A Commercial Popcorn Popping Popcorn Popper. We are constructing a 1-liter popcorn
to be operated
in steady flow. First tests in this unit show that 1 liter/min of raw corn feed stream
produces 28 liter/min
of mixed exit stream. Independent tests show that when raw corn pops its volume goes
from 1 to 31.
With this information determine what fraction of raw corn is popped in the unit.
31  1
1
1
Solution:  A 
 30 , C Ao  1a.u. , C A  C Ao 
a.u.
1
28
28
1
1
C Ao  C A
28  46.5%
XA 

1
C Ao   A C A
1  30 
28
21
Chapter 5 Ideal Reactor for a single Reactor
5.1 Consider a gas-phase reaction 2A → R + 2S with unknown kinetics. If a space velocity
of 1/min is needed for 90% conversion of A in a plug flow reactor, find the
corresponding space-time and mean residence time or holding time of fluid in the plug
flow reactor.
1
Solution:    1 min ,
s
Varying volume system, so t can’t be found.
5.2
In an isothermal batch reactor 70% of a liquid reactant is converted in 13 min. What
space-time and space-velocity are needed to effect this conversion in a plug flow
reactor and in a mixed flow reactor?
Solution:
Liquid reaction system, so  A  0
According to eq.4 on page 92, t  C Ao 
XA
0
Eq.13,  M .F .R 
C Ao  C A C Ao X A
,  M .F .R can’t be certain.

 rA
 rA
Eq.17,  P.F .R  C Ao 
XA
0
5.4
dC A
 13 min
 rA
dX A
, so  P.F .R  t B.R  13 min
 rA
We plan to replace our present mixed flow reactor with one having double the bolume.
For the same aqueous feed (10 mol A/liter) and the same feed rate find the new
conversion. The reaction are represented by
A → R,
Solution:
-rA = kC1.5
A
Liquid reaction system, so  A  0
C  CA
X
XA
V


 A , Ao

FAo C Ao  rA C Ao (rA ) k[C Ao (1  X A )]1.5


Now we know: V   2V , FAo  FAo , C Ao  C Ao , X A  0.7
So we obtain
22

XA
2X A
2V



1
.
5
1.5
 FAo

FAo
kC1Ao.5 (1  X A )1.5 kCAo ()1  X A )
V
XA



(1  X A )

1.5
2  0.7
 8.52
(1  0.7)1.5

X A  0.794
5.5
An aqueous feed of A and B (400liter/min, 100 mmol A/liter, 200 mmol B/liter) is to be
converted to product in a plug flow reactor. The kinetics of the reaction is represented
by
mol
liter  min
Find the volume of reactor needed for 99.9% conversion of A to product.
A +B→ R,
-rA = 200CACB
Solution: Aqueous reaction system, so  A  0
V
t
1


FAo C Ao C Ao
According to page 102 eq.19,

V
o
 C Ao 
0
V  C Ao o 
X Af
0
5.9
X Af

X Af
0
X Af dX
dC A
A

0
 rA
 rA
dX A
,  o  400liter / min ,
 rA
0.999 dX
dX A
A
 0.1  400
 124.3L
0
 rA
 rA
A specific enzyme acts as catalyst in the fermentation of reactant A. At a given enzyme
concentration in the aqueous feed stream (25 liter/min) find the volume of plug flow
reactor needed for 95% conversion of reactant A (CA0 =2 mol/liter ). The kinetics of the
fermentation at this enzyme concentration is given by
A
Solution:
enzyme
 R ,
-rA =
0.1C A
mol
1  0.5C A min  liter
P.F.R, according to page 102 eq.18, aqueous reaction,   0
X A dX
V
A

0
FAo
 rA
V  FAo 
XA
0
1  0.5C A
1
dX A  125  2( Ln
 X A)\
0.1C A
1 X A
23
 125( Ln
1
 0.95)  986.4 L
0.05
5.11 Enzyme E catalyses the fermentation of substrate A (the reactant) to product R. Find
the size of mixed flow reactor needed for 95% conversion of reactant in a feed stream
(25 liter/min ) of reactant (2 mol/liter) and enzyme. The kinetics of the fermentation at
this enzyme concentration are given by
A
enzyme
 R ,
-rA =
0.1C A
mol
1  0.5C A min  liter
Solution:  o  25L / min , C Ao  2mol / L , FAo  50mol / min , X A  0.95
Constant volume system, M.F.R., so we obtain

V
o

C Ao X A

 rA
2  0.95
 199.5 min ,
0.1  0.05  2
1  0.5  0.05  2
V   o  199.5 min 25L / min  4.9875m3
5.14 A stream of pure gaseous reactant A (CA0 = 660 mmol/liter) enters a plug flow reactor at
a flow rate of
FA0 = 540 mmol/min and polymerizes the as follows
mmol
3A → R,
-rA = 54
liter  min
How large a reactor is needed to lower the concentration of A in the exit stream to C Af
= 330 mmol/liter?
C
330
1
1 A
1
1
C Ao
2
660  0.75
Solution:  A  3
  , XA 

CA
2 330
1
3
1 
1  A
3 660
C Ao
0-order homogeneous reaction, according to page 103 eq.20
k  k
V
o
k
VC Ao
 C Ao X A
FAo
So we obtain
V
FAo
1
0.75
C Ao X A  540 
 7.5L
C Ao
k
54
5.16 Gaseous reactant A decomposes as follows:
24
A → 3 R,
-rA = (0.6min-1)CA
Find the conversion of A in a 50% A – 50% inert feed (  0 = 180 liter/min, CA0 =300
mmol/liter) to a 1 m3 mixed flow reactor.
42
Solution: V  1m 3 , M.F.R.  A 
1
2
According to page 91 eq.11,  
V
o


C Ao X A

 rA
C Ao X A
1 X A
0.6C Ao
1 X A
X A (1  X A )
1000 L

0.6(1  X A ) 180 L / m i n
So we obtain X A  0.667
25
Chapter 6 Design for Single Reactions
6.1
A liquid reactant stream (1 mol/liter) passes through two mixed flow reactors in a
series. The concentration of A in the exit of the first reactor is 0.5 mol/liter. Find the
concentration in the exit stream of the second reactor. The reaction is second-order
with respect to A and V2/V1 =2.
Solution:
 1=
V2/V1 = 2,
V1
 01
C A0  C A1
,
 rA
=
CA0=1mol/l , CA1=0.5mol/l ,
-rA2=kC2A2 (2nd-order) , 2×
2 =
V2
 02
C A1  C A2
 rA2
=
 01   02 , -rA1=kC2A1 ,
C A0  C A1
kCA1
2
=
C A1  C A2
kCA2
2
2×(1-0.5)/(k0.52)=(0.5-CA2)/(kCA22)
So we obtain
CA2= 0.25 mol/l
6.2
Water containing a short-lived radioactive species flows continuously through a
well-mixed holdup tank. This gives time for the radioactive material to decay into
harmless waste. As it now operates, the activity of the exit stream is 1/7 of the feed
stream. This is not bad, but we’d like to lower it still more.
One of our office secretaries suggests that we insert a baffle down the middle of the
tank so that the holdup tank acts as two well-mixed tanks in series. Do you think this
would help? If not, tell why; if so calculate the expected activity of the exit stream
compared to the entering stream.
Solution:
1st-order reaction, constant volume system. From the information offered about
the first reaction,
we obtain
1 =
V1
 01
C  C A1
 A0

kCA1
1
C A0  C A0
7
1
k  C A0
7
If a baffle is added,
 2   21   22 
V21

V22
=
V1
 021  022  01
=
C A0  C A21 C A2  C A22

kCA21
kCA22
6
C A0
7
=
=6/k ……
1
kC A0
7
①
26
1
V
C A0  C A21 2 1
C  C A22
=3/k= A21

kCA21
 021
kCA22
……
②
Combining equation ① and ② we obtain:
CA21= 0.25CA0 ;
1
C A0
16
So it will help, and the expected activity of the exit stream is 1/16 of the feed.
CA22=0.25CA21=
6.3
An aqueous reactant stream (4 mol A/liter) passes through a mixed flow reactor
followed by a plug flow reactor. Find the concentration at the exit of the plug flow
reactor if in the mixed flow reactor CA = 1 mol/liter. The reaction is second-order with
respect to A, and the volume of the plug flow unit is three times that of the mixed flow
unit.
Solution:
m 
Vm
p 
Vp
=
0
0
Constant volume system and 2nd-order reaction:

C A0  C A1 C A0  C A1 4  1
=
>
=3/k
2
k
 rA1
kCA1
3Vm
0
=9/k= - 
CA f
C A1
……
①
……
②
2
C Af C
dC A
1 1
A
 1)
= -
dC A = (
1
k C Af
 rA
k
Combining equation. ① and ② we obtain:
CAf= 0.1 mol/liter
6.4
Reactant A (A → R,CA0=26 mol/m3) passes in steady flow through four equal-size
mixed flow reactors in series ( 
total
=2 min). When steady state is achieved the
concentration of A is found to be 11, 5, 2, 1 mol/m3 in the four units. For this reaction,
what must be 
plug
so as to reduce CA from CA0 = 26 to CAf = 1 mol/m3?
Solution:
 m   m1   m2   m3   m4 =
C A0  C A1 C A1  C A2 C A2  C A3 C A3  C A4
=
=
=
 rA1
 rA2
 rA3
 rA4
CA0=26mol/liter, CA1=11 mol/liter, CA2=5 mol/liter, CA3= 2mol/liter, CA4=1mol/liter
So we abtain:
15/(-rA1) = 6/(-rA2) = 3/(-rA3) = 1/(-rA4)
We postalate the reaction rate is 1 unit when CA4=1 mol/liter
So we obtain
27
CA, mol
11
5
2
1
-rA
30
12
6
2
1/(-rA)
1/30
1/12
1/6
1/2
C A 0 dC
dC A
A
=
C A0  r
C Af  r
A
A
 p  
C Af
So we obtain  p  2.63 min.
6.6 At 100℃ pure gaseous A reacts away with stoichiometry 2A → R + S in a constant
volume batch reactor as follows:
t, sec
0
20
40
60
80
100
120
140
160
pA, atm
1.00
0.96
0.80
0.56
0.32
0.18
0.08
0.04
0.02
What size of plug flow reactor operating at 100℃ and 1 atm can treat 100 moles A/hr in a
feed consisting of 20% inserts to obtain 95% conversion of A?
Solution:
FA0= 100 mol/hr = 0.0278 mol A/s, CA0 = (1 atm×101.3)/(8.314×37.3)=0.0327mol/l
t, sec
0
20
40
60
80
100
120
140
160
pA, atm
1.00
0.96
0.80
0.56
0.32
0.18
0.08
0.04
0.02
CA,mol/l
0.0327
0.03136
0.0261
0.0285
0.01045
0.00588
0.00261
0.001307
0.0006533

Now , p A0 =0.8 atm,

C A0 =0.0261 mol/l


When the conversion of A is 90%, C A =0.001305mol/l , p A =0.04 atm
6.7
So
 =  0.04   0.8 =140-40=100 s
So
V =  0 = 
FA 0
=100×0.027/0.0261=106.4 L
C A0
We wish to treat 10 liters/min of liquid feed containing 1mol A/liter to 99% conversion.
The stoichiometry and kinetics of the reaction are given by
A → R,
-rA =
CA
mol
0.2  C A liter  min
Suggest a good arrangement for doing this using two mixed flow reactors, and find
the size of the two units needed. Sketch the final design chosen.
28
Solution:
 0 =10l/min , CA0=1 mol/l,
when
Vi
=
(1) two for equal-she M.F.R.
0
-rA1=CA1/(0.2+CA1) ,
So we obtain
=
ZA=0.99
CAf =0.91mol/l
C A0  C A1 C A1  C Af
=
 rAf
 rA1
-rAf = CAf/(0.2+CAf) = 0.0476,
 i=2.333min
CA1=0.121mol/l,
Vt=2Vi=2  i 0 =2×2.333×10=46.65 L
(2)
-/rA
L
K
N
M
0
Z2
Z1
ZA
When the area of rectangle KLMN is maximum, the volume of reactors needed its minimum.
=
So
V1
0

V2
0
=
C A0  C A1 C A1  C A 2

=2.7+1.89=4.59
 rA1
 rA2
V =4.59×10=45.9
1
)
rA
=0.2/(1-ZA)2
dx
d (
Slope of curse =
(
Slope of
So when slope of curse = slope of
LN =
1
)
rA
Z A1
0.2
21  (
 1)
1  Z Ai
=
Z Ai
LN, ZA1=0.9,
CA1=0.1 mol/L, -rA1=1/3 , 1/(-rA) = 3
6.8
From steady-state kinetics runs in a mixed flow reactor, we obtain the following data
on the reaction A → R.
29
 , sec CA0, mmol/liter CA, mmol/liter
60
50
20
35
100
40
11
100
60
20
200
80
11
200
100
Find the space time needed to treat a feed of
C A0= 100 mmol/liter to 80%
conversion
(a) in a plug flow reactor.
(b) in a mixed flow reactor.
Solution: From the data offered, we obtain
(a) in a P.F.R:
So
(b)
CA0=100 mmol/L, XA=0.8
 p =area of the shaded
CA=20 mmol/L
in a M.F.R.
m 
C A0  C Af
 rAf
=(100-20)/(-rA20)=2×80=160s.
6.20 Reactant A decomposes with stoichiometry A → R and with rate dependent only on
CA. The following data on this aqueous decomposition are obtained in a mixed flow reactor:
 , sec
CA0
CA
14
200
100
25
190
90
29
180
80
30
170
70
29
160
60
27
150
50
24
140
40
19
130
30
15
120
20
12
110
10
20
101
1
Determine which setup, plug flow, mixed flow, or any two-reactor combination gives
minimum  for 90% conversion of a feed consisting of CA0 = 100. Also find this
 minimum. If a two-reactor scheme is found to be optimum, give CA between
stages and  for each stage.
30
Solution :
Draw a
1
～ CA curse
 rA
CA0=100, ZA=0.9,
From the
so CA=10
1
1
～ CA curse, we know that when CA∈(10,70),
increases as CA
 rA
 rA
increases and that when CA∈(70,100),
1
decrease as CA increase.
 rA
So when CA∈(10,70),
we plan to use a P.F.R  p = area of the shaded region followed with a M.F.R. to treat the feed
CA∈(10,70),  m = area of the shaded region.
31
Chapter 10 Choosing the Right Kind of Reactor
10.1 Given the two reactions
1
A+ B 
R

-r1 = k1CACB
2
R+B 
S
-r2 =k2CACB

Where R is the desired product and is to be maximized. Rate the four sohemes shown in
Fig. P10.1 ---either “good” or “not so good,” please, no complicated calculations, just
reason it out.
Solution:
Number (d) is good for the formation of intermediate –R which causes no
maximum occurring there. rR  (r1 )  (r2 )  (k1  k 2 )C AC B , the same order, so R can’t
affect the D.
10.2 Repeat Problem 1 with just one change
-r2 = k2CRC2B
Solution:
-r1 = k1CACB
n1=2
-r2 = k2CRCB2
n2=3
n1< n2, therefore, low CB favors the reaction of lowest order, mixed flow reactor should be
used.
So (a) Not so good.
(c) Not so good.
(b) Good.
(d) Not so good
10.3 Repeat Problem 1 with just one change
-r2 = k2C2R CB
Solution:
-r1 = k1CACB
n1=2
-r2 = k2CR2CB
n2=3
32
n1< n2, and the order is 2 for R, low CB and low CR is helpful to get more desired product R.
So (a) Not so good.
(c) Good.
(b) Not so good.
(d) Not so good
33
Chapter 11 Basics of Non-ideal Flow
11.1 A pulses input to a vessel gives the results shown in Fig.P11.1.
(a) Check the material balance with the tracer curve to see whether the results are consistent.
(b) If the result is consistent, determine t , V and sketch the E curve.
Solution:
(a) Act  0.05  5  0.25 ,
 Ac t 
M



1
 0.25 ,
4
, so the results are consistent.

5
0

0
 tCdt   t  0.05dt  2.5 min
t
 Cdt  0.05dt
(b)
0
t
M
V

E
5
0
, so V  t   2.5 min 4L / min  10L
C pulse
M /
 0.05  4  0.2 min 1
11.2 Repeat problem P11.1 with one change: the tracer curve is now as shown in Fig.P11.2.
Solution: Ac t  0.05  6  0.30 
M

So the results are not consistent.
11.3
A pulse input to a vessel gives the results shown in Fig. P11.3.
(a) Are the results consistent? (check the material balance with the experimental tracer
curve.)
34
(b) If the results are consistent, determine the amount of tracer introduced M, and the E
curve.
Solution:
1
9
h  (25  16)  h ,
2
2
19C
25 C
19
25
max
max
(
t

16
)
tdt

(t  25)tdt
tCdt

tCdt


16
19


3
6
19
t  1619

 20 sec
25
19C
25 C
max
max
16 Cdt  19 Cdt 16 3 (t  16)dt  19 6 (t  25)dt
C
C
16  19
19  25
 3  max 
 6  max
2
2
2
2  20.5 sec
or 
C max
 (25  16)
2
V 60
t'  
 15 sec  t ,

4
So the results are not consistent.
(a) Ac t 
11.4 A step experiment is made on a reactor. The results are shown in Fig. P11.4.
(a) Is the material balance consistent with the tracer curve?
(b) If so, determine the vessel volume V, t ,the F curve and the E curve.
Solution:
(a) If C max 
m


0.5
1 3
 0.125mol / L , A  C max t 
C max  2C max
4
2
so t  2 min
(b) F 

m
C
C
,
C max
35
0, t  [0,1)
1

then F curse is F   (t  1), t  [1,3]
2
1, t  (3, )
A 2
16
V
 2  0.125 
 8liters
m
0.5
0, t  [0,1)  (3, )
dF

, so E   1
E
dt
 2 , t  [1,3]
E curse is
0.8
E
0.6
0.4
0.2
0
0
1
2 t,min 3
4
5
11.5 A batch of radioactive material is dumped into the Columbia River at Hanford,
Washington. At Bonneville Dam, about 400 km downstream the flowing waters (6000 m3/s)
are monitored for a particular radioisotope (t1/2>10 yr) and the data of Fig.P11.5 are obtained.
(a) How many units of this tracer were introduced into the river?
(b) What is the volume of Columbia River waters between Bonneville Dam and the point
of introduction of tracer?
Solution:
(a) Ac t 
1
r
M
,
 10 6  105  5.25  10 5 3  day 
2

m
so we obtain M  5.25  10
5
r
m3
 day  6000
 27216units of radioisotope
s
m3
36
(b) t  

0

tCdt

Cdt

35
20
t
125
10 6
10 6
(t  20)d   t
(t  125)dt
35
35  20
35  125
1
 (125  20)  10 6
2
3.15  10 3
 60day
52.5  10 6
m3
V  t   60day  6000
 3.11  1010 m 3
s
11.6 A pipeline (10 cm I.D., 19.1 m long) simultaneously transports gas and liquid from
here to there. The volumetric flow rate of gas and liquid are 60 000 cm 3/s and 300 cm3/s,
respectively. Pulse tracer tests on the fluids flowing through the pipe give results as shown in
Fig.P11.6. What fraction of the pipe is occupied by gas and what fraction by liquid?
Solution:
t g  2s , t l  100s , so Vg   g t g  6000  2  1.2  10 5 cm 3
Vl   l t l  300  100  3  10 4 cm 3
Vtotal 

4
D2 L 

4
 10 2  19.1  10 2  1.5  10 5 cm 3
1.2  10 5
So we obtain %G 
 100%  80% , %L  20%
1.5  10 5
A liquid macrofluid reacts according to A → R as it flows through a vessel. Find the
conversion of A for the flow patterns of Fig.P11.7 to P11.11 and kinetics as shown.
0.5
11.7 CA0 = 1mol/liter, -rA = kC0.5
/liter0.5·min
A , k = 2 mol
37
Figure 11.7
Solution:
CA
   
C Ao
CA
1
dC A  C Ao  C A  C Ao (1 
)
rA
C Ao
CA
 2
 (1 
)  (1   ) 2
C Ao
C Ao
So we obtain
 X A  1
11.8
 C
L
CA
1
1
  ( A ) batch  Edt   (1  t ) 2 dt 
0
0
C Ao
C Ao
2
6
1
 0.833
6
CA0 = 2mol/liter, -rA = kC2A, k = 2 liter / mol·min
Figure 11.8
Solution:
According to page274 eq.15,
CA
1

C Ao 1  kCAo t
4t  1, t  [0,0.5]
AE t  1, so Emax  3 min 1 , E  
0, t  (0.5,)
So
 C
0.5
CA
1
  ( A ) element E  dt  
(4t  1)dt  0.5 ,
0 C
0 1  4t
C Ao
Ao
X A  1  0.5  0.5
11.9 CA0 = 6mol/liter, -rA = k, k = 3 mol/liter·min
38
Figure 11.9
Solution:
kt

, t  2 min
C A 1 
  C Ao
C Ao 
0, t  2 min
 AE t
So
1
 , t  [0,3]
 1 , so E   3
0, t  (3,)
 C
3
CA
kt 1
1
  ( A ) e Edt   (1 
)  dt  ,
0 C
0
C Ao
C Ao 3
3
Ao
X A  1
1
 0.667
3
11.10 CA0 = 4mol/liter, -rA = k, k = 1 mol/liter·min
Figure 11.10
Solution:
kt
t

 1  , t  4 min
, t  3
C A 1 
4
  C Ao
, E
C Ao 
0, t  3
0, t  4 min
So
 C
CA
3 
1
  ( A ) e E d t (1  )  E d t ,
0 C
C Ao
4 0
4
Ao
X A  1
1
 0.75
4
11.11 CA0 = 0.1mol/liter, -rA = k, k =0.03 mol/liter·min
39
Figure 11.11
Solution:
For the zero-order reaction,
(
1  0.3t , t  3.3 min
CA
kt
) element  1 

C Ao
C Ao 0, t  3.3 min
( X A ) element  1  (
So
0.3t , t  3.3 min
CA
) element  
C Ao
1, t  3.3 min

X A   (X A )
0
element
E (t )dt (liquid macrofluid reacts)
From Fig11.11, we know
0,0  t  4 min
0.1(t  4),4  t  6 min

E (t )  
0.025(14  t ),6  t  14
0, t  14 min
14
X A   1  E (t )dt  1
4
Thus, the conversion of A is 100%.
11.12—11.14 Hydrogen sulfide is removed form coal gas by contact with a moving bed of
iron oxide
particles which convert to the sulfide as follows:
Fe2O3 → FeS
In our reactor the fraction of oxide converted in any particle is any particle is
determined by its residence time t and the time needed for complete conversion of the
particle  and this is given by
t 3
1 – X = (1)
when t < 1 hr, and with  = 1 hr

and
X=1
when t  1 hr
Find
the conversion of iron oxide to sulfide if the RTD of solides in the
contactor is approximated by the curve of Fig.P11.12,P11.13,P11.14.
40
11.12
Figure 11.12
Solution:
From Fig.11.12, we know
1,0  t  1hr
E (t )  
0, t  1hr
When t<1hr, 1  ( X A ) element  (1  t ) 3
So

1
0
0
X A   ( X A ) element E (t )dt  1 [1  (1  t ) 3 ]dt  0.75
11.13
Figure 11.13
Solution:
From Fig.11.13, we know
, t  0.5hr
E (t )  
0, t  0.5hr
When t<1hr, 1  ( X A ) element  (1  t ) 3
also ( X A ) element  1  (1  t ) 3
So


0
0
X A   ( X A ) element E (t )dt   [1  (1  t ) 3 ]   (t  0.5)dt  0.875
Thus, the conversion of A is 87.5%.
11.14
41
Figure 11.14
From Fig.11.13, we know
0, t  0.5hr

E (t )  1,0.5  t  1.5hr
0, t  1.5hr

When t<1hr, 1  ( X A ) element  (1  t ) 3
also ( X A ) element  1  (1  t ) 3
When t>1hr, ( X A ) element  1
So

1
1.5
0
0.5
1
X A   ( X A ) element E (t )dt   [1  (1  t ) 3 ]  1dt   1 1dt  0.9844
Thus, the conversion of A is 98.44%.
42
Chapter 18 Solid Catalyzed Reactions
18.22
In the absence of pore diffusion resistance a particular first-order gasphase reaction
proceeds as reported below.
- rA = 10-6 mol/cm3cat·s
at
CA = 10-5 mol/cm3, at 1 atm and 400℃
what size of spherical catalyst pellets (e = 10-3 cm-3/cm cat·s) would ensure that
pore resistance effects do not intrude to slow the rate of reaction?
Solution:

The rate expression of first-order reaction is  rA  k C A
And with  rA , CA replaced by numerical values, we obtain
k  
10 6
 0.1s 1
5
10
To judge what type of the pore resistance, we need to calculate the Thiele modulus
k  dp 10 1 5
MT  L

 dp
De
6 10 3 3
When MT, the pore resistance effects could be neglected.
3
Thus,
dp  0.4 , also dp  2.4mm
5
So the diameter of spherical catalyst pellets should be less than 2.4mm.
18.24
The first-order decomposition of A is run in an experimental mixed flow reactor.
Find the role played by pore diffusion in these runs; in effect determine whether the runs were
made under diffusion-free, strong resistance, or intermediate conditions.
dp
W
CA0

XA
4
1
300
60
0.8
8
3
100
160
0.6
Solution:
For the first run,
 
 rAout
 1
 rAout
A→R
C AoX A o u t
W
300  60  0.8
 14400
1
43
For the second run,
 2
 rAout
100  160  0.6
 3200
3
 1 14400
 rAout

 4.5
 2
 rAout
3200
While
R2 dP2

 2,
R1 dP1
Thus, the runs were running under intermediate pore resistance.
44