Chapter 1

Introduction to Chemical Kinetics

& Chemical Reaction Engineering
and
Ideal Reactor Models

Chemical kinetics is concerned mainly

with the rates of chemical reactions:
• How fast chemical reactions occur
• The factors affecting the rates of
chemical reactions

1
 Chemical kinetics enables us (as a
chemical engineer) to, for instances,
• predict the rate at which chemical
substances react (consumed or
generated)
• control the rate of any given reaction
• choose suitable chemical reactions
to produce desirable chemical
products
• select an appropriate catalyst to
enhance the rate of a specific
reaction

2
 Chemical reaction engineering (CRE)
is concerned with the design and/or
analysis of the performance of chemical
reactors
A chemical reactor is a device in which
a chemical reaction takes place
The design of chemical reactors
includes determining
• type
• size
• configuration
• operating conditions
• cost
of chemical reactors

3
 Note that chemical reactors do not have
to be in the form of, e.g., tank, vessel, tube,
they can also be, for instances,
• human organs, such as stomach,
lung, heart, muscle, at which
chemical reactions take place

Reactions in Stomach
(from www.wikipedia.org)

4
 Reactions in Lungs
(from www.wikipedia.org)

• sky, at which the reaction between NO2

(or SO2) and H2O & O2 occurs
SO2 + H2O + 1/2 O2
Æ H2SO4

2NO + H2O + 3/2 O2

Æ 2HNO3

Acid Rain Formation on the Sky

(from www.epa.gov)
5
 • a surface of metal, at which the
reaction between Fe and O2 takes
place

(from www.rollanet.com) (from www.isa.org)

Rust Formation Reactions

6
• a garbage basin, in which an
anaerobic digestion occurs (turn
organic materials to CH4 and/or H2S)

(from www.ecofriend.org)

7
 Chemical engineers employ the
kinetic information (from chemists),
such as rate of reaction, dependence of
reaction rate on various factors (e.g.,
concentrations of reactants, temperature),
to determine, e.g., reactor size, flow
pattern, thermal configuration

Hence, in designing chemical reactors,

not only is chemical kinetic information
required, additional knowledge in other
subjects, such as
• fluid mechanics
• heat transfer
• mass transfer
• thermodynamics
is also needed
8
 This implies that

Chemical Kinetics + Material & Energy

Balances + Thermodynamics + Fluid
Mechanics + Heat Transfer + Mass
Transfer + Mathematics + Economics =
Chemical Reaction Engineering (CRE)

Before designing any chemical reactors,

(at least) two questions have to be
answered first:
1) How far the changes are expected
to take place, for a specific
chemical reaction?
2) How fast will the chemical reaction
occur?

9
 To answer the 1st question, you need to
learn “thermodynamics”
Let’s consider the following chemical
reaction:
C2H4 + H2O U C2H5OH
in which equilibrium constant ( K ) at

145 oC is 6.8 × 10 −2

Before we can design or choose a

suitable reactor for this reaction, we need
to answer the question that, at
equilibrium, how many moles of C2H5OH
are to be generated, if the initial
concentrations of C2H4, H2O, and C2H5OH
are 1, 5, and 0 moles/L, respectively?
Assume that the reaction takes place at
atmospheric pressure
10
 Stoichiometry
C2H4 + H2O U C2H5OH
Initial: 1 5 0
Consumed/
Generated: -x -x +x
@ equilibrium: 1-x 5-x +x

Thermodynamics
From General Chemistry or Physical
Chemistry or Chemical Thermodynamics,
we have learned that the “equilibrium
constant” equation can be written as
follows

K =
[ C2H5OH]
(1.1)
[CH4 ] [H2O]

It is given that, at 145 oC, K = 6.8 × 10-2

11
 Thus, the concentration of C2H5OH at
equilibrium can be calculated as follows

6.8 × 10 −2
=
[ x]
(1.2)
[1 − x ][5 − x ]
Solving Eq. 1.2 for x yields
x  0.243

The knowledge of thermodynamics

allow us to have a picture of how far a
specific reaction can go, at given T & P

However, it does not enable to know

how fast this reaction will take place

In other words, by using the knowledge

of thermodynamics alone, it is not possible
for us know how long we have to wait until
we obtain C2H5OH at the concentration of
0.243 M
12
 To answer the 2nd question (i.e. How
fast does the chemical reaction occur?),
we need to learn “Chemical Kinetics”

How fast a reaction takes place can be

expressed in term of “rate of reaction” (or
“reaction rate”)

Rate of reaction can be measured as the

rate of consumption (or depletion) of a
given reactant or rate of generation of a
specified product; e.g., for the following
reaction:
A+BÆC (1.3)
The rate of reaction (Rxn.) 1.3 may be
expressed as

13
 dN A
r1.3 = −rA = − (1.4)
dt
dN C
or r1.3 = +rC = + (1.5)
dt
(Note: N i = number (#) of moles of species i)

Rate of reaction 1.3 expressed as Eqs.

1.4 and 1.5 can be considered as
“extensive reaction rate”

Normally, the rate of reaction is

expressed in term of “intensive reaction
rate”

Examples of “intensive rate of reaction”

(for Rxn. 1.3) are as follows

14
 • Rate of the consumption of A per
volume (V ) of a system
1 ⎛ dN A ⎞
r1.3 =− ⎜ ⎟ (1.6)
V ⎝ dt ⎠
• Rate of generation of C per mass of
catalyst (Wcat )
1 ⎛ dN C ⎞
r1.3 =+ ⎜ ⎟ (1.7)
Wcat ⎝ dt ⎠

Re-arranging Eq. 1.6 results in

⎛ NA ⎞
d⎜ ⎟
V ⎠
r1.3 =− ⎝
dt
dC A
r1.3 = − (1.8)
dt
Note that C A = concentration of A

Is Eq. 1.8 ‘extensive’ or ‘intensive’

reaction rate equation?
15
 Consider another example
A + 3B Æ 2C (1.9)

When we want to express the rate of

reaction 1.9 as the consumption rate of A
per unit volume of a system, we can write
the following equation:
1 ⎛ dN A ⎞
−rA = − ⎜ ⎟ (1.10)
V ⎝ dt ⎠

The rate of Rxn. 1.9 can also be

expressed as the rate of consumption of B
or the generation rate of C, per unit volume
of the system, as follows
1 ⎛ dN B ⎞
−rB = − ⎜ ⎟ (1.11)
V ⎝ dt ⎠
1 ⎛ dN C ⎞
rC = + ⎜ ⎟ (1.12)
V ⎝ dt ⎠
16
 However, since when 1 mol of A is
consumed (–),
• 3 mol of B is consumed (–)
• 2 mol of C is generated (+)
we obtain the fact that
−rA ≠ −rB ≠ rC
rB rC
but −rA = − = + = r1.9
3 2

Thus, generally, we can write the

equation for the reaction rate as follows
ri
r =
νi
where
ri = rate of change (consumption (–) or
generation (+)) of species i
ν i = stoichiometric coef. of species i

17
Design Equations for Ideal Reactors

Ideal BATCH reactors

• The reactor is well-mixed
• Although a composition changes
with time, at any instant of time, the
composition is uniform
throughout the reactor

CA1
CB1 CA2
CC1 CB2
uniformly CC2
mixed

CA1 = CA2; CB1 = CB2; CC1 = CC2

18
Ideal continuous reactors
Ideal continuously stirred tank
reactor (CSTR)
• The reactor is well-mixed or
perfectly mixed – thus, the exit
stream from the reactor has the
same composition as the fluid
within the reactor
Hence, CSTR may be called “mixed
reactor” and “back-mixed reactor”

19
 CA1
CB1
CC1
Feed

Product
perfectly
CA2
mixed
CB2
CC2

CA1 = CA2; CB1 = CB2; CC1 = CC2

20
Ideal plug flow reactor (PFR)
• The flow in the reactor is in order –
no overtaking and/or mixing with
other elements ahead or behind
• No back-mixing

Feed Product

21
Design Equations for Ideal Batch
Reactors

Employing the principle of “mole

balance” (already learned from Material &
Energy Balances) for species i, which is a
reactant, results in the following equation:
[Molar flow rate of species i flowing into a batch reactor]
– [Molar flow rate of species i flowing out of a batch
reactor]
– [Consumption rate of species i in a batch reactor]1
= –[Rate of change in moles of species i in a batch reactor]2
Note:
1 if species i is a product, this term is for generation
rate, and of “positive (+)” sign
2 if species i is a product, this term is of “positive
(+)” sign

22
 The above expression can be written in
an equation form as follows
dN i
Fio − Fi − Gi = − (1.13)
dt

where
Fio = Molar flow rate of species i

flowing into a batch reactor

Fi = Molar flow rate of species i
flowing out of a batch reactor
Gi = Rate of consumption of species i
in a batch reactor
N i = # of moles of species i in a batch
reactor

23
 For batch reactors, since there are no
streams flowing in and out (i.e. Fio = 0 &

Fi = 0 ), Eq. 1.13 is reduced to

dN i
−Gi = − (1.14)
dt

We have just learned that the rate of

consumption (or depletion) of species i per
unit system volume (V ) is as follows:
1 dN i
−ri = − (1.15)
V dt

Thus, by re-arranging Eq 1.15, the rate

of consumption of species i for the whole
system volume ( −Gi ) is
dN i
−rV
i =− = −Gi (1.16)
dt

24
 Re-arranging Eq. 1.16 gives
−dN i
dt = (1.17)
−rV
i

Integrating Eq. 1.17 yields

Ni
−dN i
t= ∫
N io
−rV
i
(1.18)

where
N io = initial # of moles of species i

N i = final # of moles of species i

If we define “conversion” ( x i ) of species i

as
N io − N i
xi = (1.19)
N io

Re-arranging equation 1.19 results in

N i = N io (1 − x i ) (1.20)

25
 Substituting Eq. 1.20 into Eq. 1.18:
Ni
−dN i
t= ∫
N io
−rV
i
(1.18)

gives
Ni
−d ⎡⎣N io (1 − x i ) ⎤⎦
t= ∫
N io
−rV
i
(1.21)

Re-arranging Eq 1.21 yields

⎡ N io ⎤
Ni −d ⎢ (1 − xi )⎥
⎣V ⎦
t= ∫
N io
−ri
Ci
−d ⎡⎣Cio (1 − x i ) ⎤⎦
= ∫
Cio
−ri
xi
dx i
t = Cio ∫ (1.22)
0
−ri

26
 Re-arranging Eq. 1.18:
Ni
−dN i
t= ∫
N io
−rV
i
(1.18)

gives
⎛ Ni ⎞ ⎛ Ni ⎞
Ni −d ⎜ ⎟ Ni d ⎜ ⎟
V
⎝ ⎠ =t =− V
⎝ ⎠
t= ∫
N io
−ri ∫N −ri
i o

Ci
dCi
t =−∫ (1.23)
Cio
−ri

Eqs. 1.22 & 1.23 are design equations

for ideal batch reactors

Note that Eqs. 1.22 & 1.23 are valid (or

applicable) only for a CONSTANT volume
system (i.e. no change in system volume)

27
Example The data for a constant-V Rxn:
P Æ Q in a batch reactor are presented in
the following Table
⎡ ( mol / L ) ⎤
CP [mol/L] -rP ⎢ ⎥
⎣ s ⎦
0.1 0.10
0.3 0.50
0.5 0.55
0.7 0.10
0.8 0.06
1.0 0.05
1.3 0.045
2.0 0.042

If the initial concentration of P (C Po ) is

1.2 mol/L, determine the amount of time

needed to achieve the conversion ( x ) is
0.75 (or 75%)

28
 Since the volume of this system (V ) is
constant, and the data are presented in the
form of the concentration of P (C P ) , Eq.
1.23:
CP
dC P
t=−∫ (1.23)
CPo
−rP

should be used to calculate the amount of

time needed to obtain the conversion of
0.75 (75%)

Once again, since the system volume is

constant, Eq. 1.20 can be divided by the
system volume, and the following equation
can be obtained as follows
N i N io
= (1 − xi )
V V
Ci = Cio (1 − x i ) (1.24)

29
 Re-arranging Eq. 1.24 for species P
results in
C Po − C P
xP = (1.25)
C Po

The desired conversion ( x ) is 0.75

Hence, the final concentration of P can

be computed, using Eq. 1.25, as follows (it
is given that C Po = 1.2 mol/L)
1.2 − C P
0.75 =
1.2
C P = 1.2 (1 − 0.75 )
= 0.30

30
 The numerical value of the integration:
CP
dC P or, in fact, the amount of time required
− ∫
C Po
−rP

to obtain 75% conversion, can be obtained

from the area under the curve of a plot
1
between C P (X-axis) and (Y-axis) from C Po =
−rP
1.2 mol/L to C P = 0.30 mol/L , as shown below

25.0

20.0
1/-rP [s/(mol/L)]

15.0

10.0

5.0

0.0
0.0 0.5 1.0 1.5 2.0 2.5
CP [mol/L]

CP = 0.30 mol/L CPo = 1.2 mol/L

31
 The area under the curve (between
C Po = 1.2 mol/L and C P = 0.30 mol/L ) is found

to be 10.7, and the unit of the area is, in fact,

⎡ s ⎤
⎢ mol/L ⎥ × ( mol/L ) = s
⎣( )⎦

Thus, the amount of time required to

achieve the conversion of 0.75 for this Rxn. is
10.7 s

Note, however, that, this is only the

reaction time; in the real situation, there are
also
• time required to load reactants into a
reactor (preparation time)
• time needed to unload products from
the reactor (harvest time)
• clean-up time

32
Design Equations for Ideal CSTRs

Since a CSTR is a continuous reactor,

there are moles (mass) flowing into and out
of the reactor

The time interval between the initial

inlet of the substance flowing into the
reactor and the final product flowing out of
the reactor is called “space time” or
“residence time”, which is denoted as τ

Note that the term “residence time” (or

“space time”) can be used for any other
flow or continuous reactors

33
 In addition to space/residence time (τ ) ,
there is another technical term for a
continuous reactor, which is “space
velocity ( s ) ”

Space velocity is defined as

1
s= (1.26)
τ

The physical meaning of “space velocity”

is the number (fold) of the reactor volume
per unit time required for a particular Rxn.

For example, s = 4 h−1 means, for the

time of 1 h, the volume of substance(s)
flowing into the reactor is 4 times that of
the reactor

34
 In other words, the residence time
required for this reaction is
1
τ =
s
1
=
4 h−1
τ = 0.25 h = 15 min

Let
V = volume of a reactor
v o = volumetric flow rate of
a substance flowing into
the reactor

The residence time (τ ) can then be

calculated from the following equation
V
τ = (1.27)
vo
35
 In order to establish the design
equations for a CSTR, the principle of mole
balance for species i (a reactant) is
employed once again, as follows
dN i
Fio − Fi − Gi = − (1.13)
dt

At steady state, there is no change in #

of moles of species i in the reactor; thus,
dN i
=0
dt

Accordingly, Eq. 1.13 is reduced to

Fio − Fi − Gi = 0 (1.28)

We have learned that, for a reactant

−Gi = −rV
i (1.16)

36
 Combining Eq. 1.16 with Eq. 1.28 and
re-arranging results in
Fio − Fi − Gi = 0
Fio − Fi − rV
i =0

−rV
i = Fi − Fio (1.29)

Fio and Fi are molar flow rates of species

i flowing into and out of the reactor,

respectively; thus,
N i − N io
Fi − Fio = (1.30)
τ
but N i = N io (1 − x i ) (1.20)

Combining Eqs. 1.30 & 1.20 together

yields
N io (1 − x i ) − N io
Fi − Fio =
τ
37
 N io − N io x i − N io
Fi − Fio =
τ
−N io x i
Fi − Fio = (1.31)
τ

Since species i is specified to be a

reactant, the term Fi − Fio is of negative (–)

sign

Accordingly, Eq. 1.31 becomes

N io x i
Fi − Fio = (1.32)
τ

Combining Eq. 1.32 with Eq. 1.29 gives

N io x i
−rV = (1.33)
i
τ

38
Re-arranging Eq. 1.33 results in
N io x i
τ =
−rV
i

⎛ N io ⎞
⎜ V ⎟ xi
τ =⎝ ⎠
−ri
Cio x i
τ = (1.34)
−ri

Re-arranging Eq. 1.34 gives

τ xi
= (1.35)
Cio −ri

We have just learned that

V
τ = (1.27)
vo

39
 Substituting Eq. 1.27 into Eq. 1.35 and
re-arranging yields
⎛V ⎞
⎜v ⎟ x
⎝ o⎠= i
Cio −ri
V xi
= (1.36)
v oCio −ri

The dimensions of
L3
• v o , a volumetric flow rate, is or
t
volume
time
• Cio , an initial concentration of
N # of moles
species i, is 3 or
L volume

40
 Hence, the dimension of the term v oCio

is
L3 N N # of moles
× 3 = ≡
t L t time

Thus, the term v oCio is, in fact, Fio

Accordingly, Eq. 1.36 can be re-written

as follows
V x
= i (1.37)
Fio −ri

Combining Eqs. 1.35 & 1.37 together

results in
τ V x
= = i (1.38)
Cio Fio −ri

41
 Re-arranging Eq. 1.24
Ci = Cio (1 − x i ) (1.24)

yields
Ci = Cio − Cio x i

Cio x i = Cio − Ci (1.39)

Combining Eq. 1.34

Cio x i
τ = (1.34)
−ri
With Eq. 1.39 yields
Cio − Ci
τ = (1.40)
−ri

Eqs. 1.38 & 1.40 are design equations

for an ideal CSTR

42
 It is important to note that −ri is rate of
Rxn. at the exit point of a CSTR
This is because, for an ideal CSTR, the
composition inside the reactor is identical
to that at the exit of the reactor; since the
rate of reaction at the exit can be
measured easier than that in the reactor,
the rate of Rxn. is, thus, evaluated at the
exit point (instead of being evaluated at
any point inside the reactor)

43
Example The reaction rate data at various
concentration of A of the liquid-phase
Rxn.: A Æ C is shown in the following
Table

CA ⎡ kmol ⎤ 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
⎢⎣ m3 ⎥⎦
−rA
0.50 0.70 0.85 0.92 0.95 0.93 0.88 0.80 0.70 0.60
⎡ kmol ⎤
⎢⎣ m3 ⋅ h ⎥⎦

Calculate the volume of a CSTR

required to obtain a conversion of 0.80
Given C Ao = 1.0 kmol/m3 and FAo = 2

kmol/h

44
 Since this is a liquid-phase Rxn., it is
safe to assumed that this Rxn. is of
constant volume

The given data are in the form of

concentration and reaction rate

Thus, Eq. 1.40 must be employed (note

that Eq. 1.40 is valid only for a constant-V
system)
C Ao − C A
τ = (1.40)
( −rA )
which can be re-written as follows
1
τ = ⎡⎣C A − C A ⎤⎦ × (1.41)
o
( −rA )

45
 It is given that
• C Ao = 1.0 kmol/m3

• x A = 0.80

Hence, the final concentration of A can

be calculated, using Eq. 1.24, as follows
C A = C Ao (1 − x A )
⎛ kmol ⎞
3 ⎟(
= ⎜1.0 1 − 0.80 )
⎝ m ⎠
kmol
= 0.2
m3

We have learned that, for a CSTR, the

rate of Rxn. (i.e. the consumption rate of A
in this case) must be evaluated at the exit
point or at the final concentration

46
 From the given data, the final
concentration of A is 0.2 kmol/m3, which
is corresponded to the reaction rate of 0.50
kmol/(m3-h)

Substituting corresponding numerical

values into Eq. 1.41 yields
1
τ = ⎣C Ao − C A ⎦ ×
⎡ ⎤
( −r A )
⎡ kmol ⎤ 1
= ⎢(1.0 − 0.2) 3 ⎥
×
⎣ m ⎦ ⎛ kmol ⎞
⎜ 0.5 ⎟
⎝ m ⋅h⎠
3

τ = 1.6 h

The calculation of the residence time

can also be presented graphically as
follows

47
 CA -rA 1/-rA
0.2 0.50 2.000
0.4 0.70 1.429
0.6 0.85 1.176
0.8 0.92 1.087
1.0 0.95 1.053
1.2 0.93 1.075
1.4 0.88 1.136
1.6 0.80 1.250
1.8 0.70 1.429
2.0 0.60 1.667

2.50

2.00
1/-rA [h/(kmol/m )]
3

1.50

1.00

0.50

0.00
0.0 0.4 0.8 1.2 1.6 2.0 2.4
3
CA [kmol/m ]

48
 The residence time is the area of the
1
rectangular with the height of and the
−rA
width of C Ao − C A

In this Example,
1 m3 ⋅ h
• =2
−rA kmol
kmol
• C Ao − C A = 1 − 0.2 = 0.8
m3
which results in the area (or residence
time) of
m3 ⋅ h kmol
2 × 0.8 3
= 1.6 h
kmol m

49
 It is given that FAo = 2 kmol/h

Thus, the volume (size) of a CSTR can

be computed, using Eq. 1.38,
τ V xA
= = (1.38)
C Ao FAo −rA

as follows
FAo × τ
V =
C Ao
⎛ 2 kmol ⎞ × 1.6 h
⎜ ⎟ ( )
h ⎠
=⎝
⎛1.0 kmol ⎞
⎜ 3 ⎟
⎝ m ⎠
V = 3.2 m3

50
Design Equations for PFRs

Employing the principle of mole

balances of species i (a reactant),
dN i
Fio − Fi − Gi = − (1.13)
dt
for the change in a small volume, dV , of
an ideal PFR shown below
Cio Ci f
Fio Fi Fi − dFi Fi f
vo xi x i + dx i
vf
x io = 0 xi f
dV
gives
dN i
Fi − ( Fi − dFi ) − Gi = − (1.42)
dt

51
 dN i
At steady state, = 0 , and Eq. 1.16,
dt
−Gi = −rV
i (1.16)
can be written for a change in a small
volume (dV ) as
−Gi = −ri dV (1.43)

Therefore, Eq. 1.42 can be re-arranged

and then combined with Eq. 1.43 as
follows
Fi − ( Fi − dFi ) − Gi = 0
Fi − Fi + dFi − ri dV = 0
−ri dV = −dFi (1.44)

Once again, since species i is a

reactant, dFi or d ( Fiout − Fiin ) or d ( Fi − Fio ) is

of negative (–) sign; thus, Eq. 1.44 becomes

52
 −ri dV = dFi (1.45)

Re-arranging Eq. 1.45 gives

dFi
dV = (1.46)
−ri

We have learned that

N i − N io
Fi − Fio = (1.30)
τ
and N i = N io (1 − x i ) (1.20)

Thus, we can obtain the following

equation
Fi = Fio (1 − x i ) (1.47)

Differentiating Eq. 1.47 yields

dFi = −Fio dx i (1.48)

53
 For the case that species i is a reactant,
dFi is of negative (–) sign

Hence, Eq. 1.48 becomes

dFi = Fio dx i (1.49)

Substituting Eq. 1.49 into Eq. 1.46

results in
Fio dx i
dV = (1.49)
−ri

Integrating Eq. 1.49 yields

V xi f
dx i
∫ dV
0
= Fio ∫
0
−ri
xi f
dx i
V = Fio ∫
0
−ri

54
 xi f
V dx i
Fio
= ∫
0
−ri
(1.50)

We have learned that

V τ
= (1.38)
Fio Cio

Combining Eq. 1.38 with Eq. 1.50 gives

xi f
V τ dx i
=
Fio Cio
= ∫
0
−ri
(1.51)

Re-arranging Eq. 1.51 results in

xi f
dx i
τ = Ci o ∫
0
−ri
(1.52)

Ci
d (Cio x i )
or τ = ∫
Cio
−ri
(1.53)

55
 Combining Eq. 1.53 with Eq. 1.39
Cio x i = Cio − Ci (1.39)

yields
Ci
d (Cio − Ci )
τ = ∫
Cio
−ri
(1.54)

Since dCi = d (Ci − Cio ) , the term

d (Cio − Ci ) is, in fact, −dCi

Accordingly, Eq. 1.53 or 1.54 can be re-

written as
Ci
−dCi
τ = ∫
Cio
−ri
Ci
dCi
or τ =−∫ (1.55)
Cio
−ri

56
 Eqs. 1.51 & 1.55
xi f
V τ dx i
=
Fio Cio
= ∫
0
−ri
(1.51)

Ci
dCi
τ =−∫ (1.55)
Cio
−ri

are design equations for an ideal PFR

Re-arranging Eq. 1.51 gives

xi f
dx i
τ = Cio ∫
0
−ri
(1.56)

Note that Eqs. 1.56 & 1.55 are identical

to Eqs. 1.22 & 1.23, respectively

57
Example Redo the Example on Page 44
for a PFR

The residence time of a PFR can be

calculated using Eq. 1.55
CA
dC A
τ =− ∫
C Ao
−rA

The numerical answer of the integration

(i.e. Eq. 1.55) can be obtained from the
area under the curve between C A (X-axis)
1
and (Y-axis) from C Ao = 1.0 kmol/m3 to
−rA
C A = 0.2 kmol/m3

58
 2.50

2.00
1/-rA [h/(kmol/m )]
3

1.50

1.00

0.50

0.00
0.0 0.4 0.8 1.2 1.6 2.0 2.4
3
CA [kmol/m ]

The area under the curve or the

residence time (τ ) is found to be 1.04 h

59
 The volume of a PFR can, thus, be
calculated, using Eq. 1.38:
V τ
= (1.38)
Fio Cio

as follows
τ
V = Fio
Cio

= ( 2 kmol/h )
(1.04 h )
(1.0 kmol/m ) 3

V = 2.08 m3

It should be noted that, in this case,

• VCSTR ( 3.2 m3 ) > VPFR ( 2.08 m3 )

• τ CSTR (1.6 h ) > τ PFR (1.04 h )

for the same conversion ( x = 0.80 or 80%)

60