ERT 208/4 REACTION
ENGINEERING:
Bioreaction in Bioreactors
By; Mrs Hafiza Binti Shukor
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Students should be able to;
APPLY pseudo-steady-state hypothesis (PSSH) in
gas-phase reactions and in order to DEVELOP
rate laws.
DESCRIBE reaction mechanism, chain reaction &
reaction pathways utilizing biomolecular reaction
(yeast fermentation)
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Common form of the RATE LAW
•homogeneous reaction,
 rA  kC
n
A
If n=1 (interger), reaction was 1 order
If n=not interger number?
CH 3CHO
Eg,
The rate law for the decomposition of acetaldehyde
•Reaction involving active intermediate
k1k3CI2 CH2
H 2  I 2  2 HI
rHI 
k 2  k 3C H 2
 CH 4  CO
3
2
CH3CHO
 rCH3CHO  kC
Reaction Order
cannot de defined
(polynomial fuction)
NON ELEMTARY REACTION
No direct correspondence
between reaction order &
stoichiometry
Reaction Order are described only for
limiting values of reactant and/or
product conc.
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Fundamentals of Nonelementary Reaction
•For Gas-Phase Decomposition of azomethane, AZO
CH 
3 2
N 2  C2 H 6  N 2
EXPERIMENTAL OBSERVATIONS SHOWS ;
rN2  C AZO
1st order at;
•high conc
•pressure >1atm
rN2  C 2 AZO
2nd order at;
•Low conc
•Pressure < 50mmHg
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Fundamentals of Nonelementary Reaction…cont…
•Active Intermediates
change in reaction order can be explained by the theory
developed by Lindemann
‘ an active molecule,  CH  N  @[ A*]
results from collision or interaction
between molecules’
3 2
2
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Fundamentals of Nonelementary Reaction…cont…
k1


CH 3 2 N 2  CH 3 2 N 2 *
N

CH
N


2
3 2
2
2
CH 
3
k1
A* M
A M 

•Lindemann Theory
the decomposition of intermediate does not occur
instantaneously after internal activation of the molecule
…rather, there is a time lag although infinitesimally small
during which the species remains activated.
•Other types of active intermediates that can be formed are;
a) Free radicals (one @ > unpaired electrons like H)
b) Ionic internidiates (eg. Carbonium ion)
c) Enzymes substrate complexes
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Fundamentals of Nonelementary Reaction…cont…
k1


CH 3 2 N 2  CH 3 2 N 2 *
N

CH
N


2
3 2
2
2
CH 
3
rAZO*  k1C
2
AZO
where.,
AZO  CH 3 2 N 2
2 reaction path that active intermediate may follow;
CH 
3
k2



CH 3 2 N 2  CH 3 2 N 2
N
*

CH
N


2
3 2
2
2
rAZO*  k 2 C AZOC AZO*
CH 
3
Activated molecule become
deactivated through collision
with another molecule
k3

N
*


C2 H 6  N 2
2
2
rAZO*  k3C AZO*
where.,
Activated molecule decomposes
spontaneously to form ethane &
nitrogen
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Fundamentals of Nonelementary Reaction…cont…
CH 
3 2
1.
N 2  C2 H 6  N 2
The overall reaction is NON ELEMENTARY
consist of sequence of ELEMENTARY reactions
k1


CH 3 2 N 2  CH 3 2 N 2 *
N

CH
N


2
3 2
2
2
CH 
3
2 AZO molecules collide & the kinetic energy of one AZO molecule is transferred to internal
rotational & vibrational energies of the other AZO molecule & it becomes activated & highly
reactive.
2.
CH 
3
k2



CH 3 2 N 2  CH 3 2 N 2
N
*

CH
N


2
3 2
2
2
Activated AZO* is deactivated through collision with another AZO
3.
CH 
3
k3

N
*


C2 H 6  N 2
2
2
Activated AZO* is widely vibrating, spontaneously decomposes into ethane & nitrogen
Nitrogen & Ethane only form from 3rd equation.
The net rate of formation of nitrogen is;
rN2  k 3C AZO*
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Fundamentals of Nonelementary Reaction…cont…
Rate of formation of active intermediate = sum of the rates of formation of all reaction
rAZO*  rAZO*1  rAZO*2  rAZO*3
Where,
rAZO*1  k1C 2 AZO
rAZO*2  k 2 C AZO C AZO*
rAZO*3  k 3C AZO*
rAZO*  k1C 2 AZO  k 2 C AZO C AZO*  k 3C AZO*
The concentration of the active intermediate, AZO* is very difficult to measure because it is
highly reactive and very short lived about 10 -9 s
To express CAZO* in term of MEASURABLE CONC, we have to use STEADY STATE
HYPOTHESIS (PSSH)
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Pseudo-Steady-State Hypothesis (PSSH)..
 Its not possible to eliminate the concentration of active
intermediate
 Active intermediate molecule has a very short lifetime
because of its high reactivity (large specific reaction rates).
 Have to consider it present at very low concentrations
Pseudo-Steady-State
approximation
The rate of formation = is assumed to be equal to its rate of disappearance.
As a results, the net rate of formation of the active intermediate r* is ZERO
Rate of formation of product, nitrogen;
Rate of formation of AZO*;
Using PSSH;
r*  0
rN2  k 3C AZO*
rAZO*  k1C 2 AZO  k 2 C AZO C AZO*  k 3C AZO*
rAZO*  0
rAZO*  k1C 2 AZO  k 2 C AZO C AZO*  k 3C AZO*  0
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Pseudo-Steady-State Hypothesis (PSSH)cont..
rAZO*  k1C 2 AZO  k 2 C AZO C AZO*  k 3C AZO*  0
k1C 2 AZO  k 2 C AZO C AZO*  k 3C AZO*  0
k1C 2 AZO  C AZO* (k 2 C AZO  k 3 )  0
C AZO*
k1C 2 AZO

k 3  k 2 C AZO
k 2 C AZO k3
At low conc
azomethane;
rN2  k1C 2 AZO
rN2  k 3C AZO*
2
The final
form of rate law
rN2
2nd order
k1 k 3C AZO

k 3  k 2 C AZO
At high conc
azomethane;
rN2
k 2 C AZO  k3
1st order
k1 k 3

C AZO  k CAZO
k2
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Pseudo-Steady-State Hypothesis (PSSH)cont..
Rules of Thumb For Development of Mechanism
1. Species having the conc(s) appearing in the
denominator of the rate law probably collide
with the active intermediate.
A  A*  collision __ products
2. If a constant in the denominator, one of the
reaction steps is probably the spontaneous
decomposition of the active intermediate.
A*  decomposition __ products
3. Species having the conc(s) appearing in the
numerator of the rate law probably produce the reac tan t   A * other __ products
active intermediate in one of the reaction step
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Pseudo-Steady-State Hypothesis (PSSH)cont..
Exercise 1;
Mechanism For Azomethane????
A  A*  collision __ products
A*  decomposition __ products
reac tan t   A * other __ products
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Pseudo-Steady-State Hypothesis (PSSH)cont..
Ans;
Mechanism For Azomethane
A  A*  collision __ products
AZO  AZO*  AZO  AZO
A*  decomposition __ products
reac tan t   A * other __ products
AZO*  C2 H 6  N 2 
AZO AZO  AZO* AZO
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Pseudo-Steady-State Hypothesis (PSSH)cont..
Exercise 2;
By assuming the main product for the reaction below is
ethane, write down the final form equation for rate of
formation for ethane.
CH 
3 2
N 2  C2 H 6  N 2
Ans;
rC2H6  k 3C AZO*
rC2H6
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
k1 k 3C 2 AZO

k 3  k 2 C AZO
Pseudo-Steady-State Hypothesis (PSSH)cont..
CHAIN REACTIONS
Initiation……….
Formation of an active
intermediate
Propagation / Chain
Transfer……….
Interaction of an active
intermediate with the
reactant/product to produce
another active intermediate
Termination……….
Deactivation of the active
intermediate
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
EXAMPLE 1; PSSH Applied to Thermal Cracking
of Ethane (Gas-Phase Reaction)
The thermal decomposition of ethane to ethylene, methane, butane and
hydrogen is believed to proceed in the following sequence;
Initiation;
1_ 2 6
C2 H 6 

 2CH 3 
k C H
Propagation ;
k2
CH 3  C2 H 6 
CH 4  C2 H 5 
k3
C2 H 5  
C2 H 4  H 
k4
H  C2 H 6 
C2 H 5   H 2
Termination ;
r1 _ C2H6  k1 _ C2H6 C2 H 6 
Lets _ k1  k1 _ C2H6
r2 _ C2H6  k 2 CH 3 C2 H 6 
r3 _ C2H 4  k 3 C 2 H 5 
r4 _ C2H6  k 4 H C2 H 6 
r5 _ C2H5  k5 _ C2H5  C2 H 5 
2
k5
2C 2 H 5  
C4 H 1 0
Let _ k5  k5 _ C2H5 
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
EXAMPLE 1; PSSH Applied to Thermal Cracking
of Ethane ….cont
a)Use PSSH to derive a rate law for the RATE OF FORMATION OF
ETHYLENE & RATE OF DISAPPEARANCE OF ETHANE….
Solutions……
Rate of formation of ethylene (Reaction 3) is,
r3 _ C2H 4  k 3 C 2 H 5 
k3
C2 H 5  
C2 H 4  H 
Active intermediates :
The net of reactions are:
C2 H 5 
C H  : r
CH 3 
CH  : r
H
2
5
3
C2 H 5
CH3
H  : r
H
  r2C2H5  r3C2H5  r4C2H5  r5C2H5  0
  r1CH3   r2CH3  2r1C2H6  r2C2H6  0
  r3 H   r4 H   r3C2H 4  r4C2H6  0
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
EXAMPLE 1; PSSH Applied to Thermal Cracking
of Ethane ….cont
From reaction stoichiometry, we have;
r3C2H5   r3C2H4
r2C2H5   r2C2H6
C H  : r
  r2C2H5  r3C2H5  r4C2H5  r5C2H5  0
C H  : r
  r2C2H6  r3C2H 4  r4C2H6  r5C2H5  0
2
Then,
r4C2H5   r4C2H6
2
5
C2 H 5
5
C2 H 5
Finally got
C H  : r
CH  : r
2
5
C2 H 5
3
CH3
H  : r
H
  r2C2H6  r3C2H 4  r4C2H6  r5C2H5  0
  r1CH3   r2CH3  2r1C2H6  r2C2H6  0
  r3 H   r4 H   r3C2H 4  r4C2H6  0
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
EXAMPLE 1; PSSH Applied to Thermal Cracking
of Ethane ….cont
CH  : r
3
CH3
  r1CH3   r2CH3  2r1C2H6  r2C2H6  0
2r1C2H6  r2 C2H6  0
From substituting the concentrations into the elementary equation gives;
 2k1 C2 H 6   k 2 CH 3 C2 H 6   0
Solving for the conc of the free radical
CH  ,
Where,
r1 _ C2H6  k1 _ C2H6 C 2 H 6 
r2 _ C2H6  k 2 CH 3 C2 H 6 
3
2k1
CH 3  
k2
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
EXAMPLE 1; PSSH Applied to Thermal Cracking
of Ethane ….cont
Adding this 2 equations…..
C H  : r
H  : r
2
5
C2 H 5
H
  r2C2H6  r3C2H 4  r4C2H6  r5C2H5  0
  r3 H   r4 H   r3C2H 4  r4C2H6  0
get…
r2C2H6  r5C2H5   0
Substituting for conc in the rate laws…..
k2 CH 3 C2 H 6   k5 C2 H 5   0 where… r2 _ C2H6  k 2 CH 3 C2 H 6 
2
r5 _ C2H5  k5 _ C2H5  C2 H 5 
2
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
EXAMPLE 1; PSSH Applied to Thermal Cracking
of Ethane ….cont
PSSH solution…..
from
Solving for
k2 CH 3 C2 H 6   k5 C2 H 5   0
2
C H  gives us,
2
5
1/ 2
 k2

C2 H 5    CH 3 C2 H 6  

 k5
1/ 2
 2 k1 k 2

C2 H 5   
C2 H 6  

 k 2 k5
where…
1/ 2
 2k1

C2 H 5    C2 H 6  

 k5
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
CH   2k
1
3
k2
EXAMPLE 1; PSSH Applied to Thermal Cracking
of Ethane ….cont
Substituting for
C H  in equation
2
5
r3 _ C2H 4  k 3 C 2 H 5 
yields the rate of formation of ethylene;
rC2H 4
 2 k1 

 k 3 C 2 H 5   k 3 

 k5 
1/ 2
C H 
1/ 2
2
6
where…
1/ 2
 2k1

C2 H 5    C2 H 6  

 k5
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
EXAMPLE 1; PSSH Applied to Thermal Cracking
of Ethane ….cont
Next, we write the net rate of
In terms of concentration,
H  : r
H
H  formation in
H  : r
  r3 H   r4 H   r3C2H 4  r4C2H6  0
  r3 H   r4 H   r3C2H 4  r4C2H6  0
r3C2H 4  r4 C2H6  0
k3 C2 H 5   k 4 H C2 H 6   0
1/ 2
Using eq
H
 2k1

C2 H 5    C2 H 6  

 k5
to substitute for
where…
r3 _ C2H 4  k 3 C 2 H 5 
r4 _ C2H6  k 4 H C2 H 6 
C H 
2
5
conc of the hydrogen radical
1/ 2
 2k

k 3  1 C 2 H 6    k 4 H C 2 H 6   0
 k 5

ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
gives the
EXAMPLE 1; PSSH Applied to Thermal Cracking
of Ethane ….cont
from
1/ 2
 2k

k 3  1 C 2 H 6    k 4 H C 2 H 6   0
 k 5

1/ 2
 2k k

1 3
C2 H 6    k 4 H C2 H 6   0

 k 5

H   k3
k4
 2k1 


 k 
 5 
1/ 2
C H 
1 / 2
2
6
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
EXAMPLE 1; PSSH Applied to Thermal Cracking
of Ethane ….cont
Rate of disappearance of ethane is
rC2H6  k1 _ C2H6 C2 H 6   k 2 CH 3 C2 H 6   k 4 H C2 H 6 
Where,
r1 _ C2H6  k1 _ C2H6 C 2 H 6 
r2 _ C2H6  k 2 CH 3 C2 H 6 
r4 _ C2H6  k 4 H C2 H 6 
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
EXAMPLE 1; PSSH Applied to Thermal Cracking
of Ethane ….cont
Substituting for the concentration of free radicals, the rate law of disappearance of ethane is….
rC2H6  k1 _ C2H6 C2 H 6   k 2 CH 3 C2 H 6   k 4 H C2 H 6 
rC2H6
 2k1 
 k3
 k1 _ C2H6 C 2 H 6   k 2 
C 2 H 6   k 4 
 k2 
 k4
 rC2H6
 2 k1 

 k1  2k1 C 2 H 6   k 3 

k
5


1/ 2
Where,
 2k1 


 k 
 5 
1/ 2
C H  C H 
1 / 2
2
6
2
6
C H 
1/ 2
2
6
CH   2k
1
3
k2
k 3  2k1 
H    
k 4  k5 
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
1/ 2
C H 
1 / 2
2
6
Conclusions
Reaction that not follow elementary rate law (NON
ELEMENTARY involve a number of reaction steps, each
of which is ELEMENTARY
After finding net rates of reaction for each species, we use
PSSH to derive a rate law of the reaction.
PSSH not only can be used in gas-phase reaction, but also
can be used in biological reactions (enzymatic reactions).
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
BIOLOGICAL REACTIONS
•BIOREACTORS
Lab Scale Bioreactor
Industrial Scale Bioreactor
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Fermentation Process
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)
Major Functions of a Bioreactor
1) Provide operation free from contamination;
2) Maintain a specific temperature;
3) Provide adequate mixing and aeration;
4) Control the pH of the culture;
5) Allow monitoring and/or control of dissolved oxygen;
6) Allow feeding of nutrient solutions and reagents;
7) Provide access points for inoculation and sampling;
8) Minimize liquid loss from the vessel;
9) Facilitate the growth of a wide range of organisms.
Ref;(Allman A.R., 1999: Fermentation Microbiology and Biotechnology)
ERT 208/4 REACTION ENGINEERING
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Biotechnological Processes Of Growing
Microorganisms In A Bioreactor
1) Batch culture: microorganisms are inoculated into a fixed volume of
medium and as growth takes place nutrients are consumed and
products of growth (biomass, metabolites) accumulate.
2) Semi-continuous: fed batch-gradual addition of concentrated nutrients
so that the culture volume and product amount are increased (e.g.
industrial production of baker’s yeast);
Perfusion-addition of medium to the culture and withdrawal of an equal
volume of used cell-free medium (e.g. animal cell cultivations).
3) Continuous: fresh medium is added to the bioreactor at the
exponential phase of growth with a corresponding withdrawal of
medium and cells. Cells will grow at a constant rate under a constant
condition.
ERT 208/4 REACTION ENGINEERING
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Biotechnological processes of growing
microorganisms in a bioreactor
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Batch Culture VS Continuous Culture
Continuous systems: limited to single cell protein,
ethanol productions, and some forms of waste-water
treatment processes.
Batch cultivation: the dominant form of industrial usage
due to its many advantages.
Ref;(Smith J.E, 1998: Biotechnology)
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Advantages of Batch Culture VS
Continuous Culture
1)
2)
3)
4)
Products may be required only in a small quantities at any given time.
Market needs may be intermittent.
Shelf-life of certain products is short.
High product concentration is required in broth for optimizing
downstream processes.
5) Some metabolic products are produced only during the stationary
phase of the growth cycle.
6) Instability of some production strains require their regular renewal.
7) Compared to continuous processes, the technical requirements for
batch culture is much easier.
ERT 208/4 REACTION ENGINEERING
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Fermentation Technology
• What is it important to know the kinetics of the reaction in the
fermenter?
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Cell Growth
Typical pattern of growth cycle during batch
fermentation
I.
II.
III.
IV.
V.
VI.
VII.
VIII.
Lag phase
Acceleration phase
Exponential (logarithmic) phase
Deceleration phase
Stationary phase
Accelerated death phase
Exponential death phase
Survival phase
From: EL-Mansi and Bryce (1999)
Fermentation Microbiology
and Biotechnology.
ERT 208/4 REACTION ENGINEERING
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Cell Growth...cont...
Lag Phase
•Little increase in
cell conc.
•Cell adjusting their
new environment,
synthesizing
enzymes & ready to
reproducing
Exponential Growth
Phase
•Cell are dividing at
max rate
•Cell able to use the
nutrients most
efficiently
Stationary Phase
•Cell reach a
minimum biological
space (lack of 1@>
nutrients limits cell
growth)
•Net growth = 0
•Fermentation
product produce.
ERT 208/4 REACTION ENGINEERING
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Death Phase
•Decrease in live cell
conc occur.
•Results of toxic byproduct
Rate Laws
Rate law for the cell growth rate of new cells,
Cells + Substrate
More Cells + Product
The most commonly used expression is the Monod equation for exponential growth;
rg  Cc
Where,
rg  cell _ growth _ rate( g / dm3 .s)
  specific _ growth _ rate( s 1 )
Cc  cell _ concentration( g / dm3 )
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Rate Laws...cont...
Specific cell growth rate can be expressed as,
   max
Cs
1
,s
K s  Cs
Where,
 max  a _ max_ specific _ growth _ reaction _ rate( s 1 )
K s  the _ monod _ cons tan t ( g / dm3 )
C s  substrate(nutrient)concentration( g / dm3 )
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Rate Laws...cont...
Combine ,
rg  Cc
Will get,
rg 
and
   max
 max C s Cc
K s  Cs
 max  1.3h 1
K s  2.2 X 10 5 mol / dm3
Cs
, s 1
K s  Cs
Monod equation for bacterial cell
growth rate
Parameter value for the E.coli
growth on glucose.
Ks is small for a numb of
different bacteria in which case
the rate law reduce to,
Plz refer ERT 104 Bioprocess Eng Principle
rg   max C c
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Thank You
ERT 208/4 REACTION ENGINEERING
SEM 2 (2009/2010)