Lecture 2
Kjemisk reaksjonsteknikk
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Review of Lecture 1 and 2 (chapter 1 and 2)
Rate Laws
Reaction Orders
Arrhenius Equation
Activation Energy
Effect of Temperature
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General Mole Balance
System
Volume, V
FA0
FA
GA
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General Mole Balance on System Volume V
In
 Out  Generation  Accumulation
FA 0  FA

 r dV
A
dN A

dt
2
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Reactor Mole Balance Summary
The GMBE applied to the four major reactor types
(and the general reaction AB)
Reactor
Batch
Differential
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PBR

dFA
 rA
dW
NA

N A0
V 
dFA
 rA
dV

Integral
t
dN A
 rAV
dt
CSTR
PFR
Algebraic
FA 0  FA
rA
dN A
rAV
FA

V
FA 0
W
FA

FA 0
dFA
drA
dFA
rA
NA
t
FA
V
FA
W
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Reactor Mole Balances in terms of
conversion
FA=FA0-FA0X=FA0(1-X)
Reactor
Differential
Algebraic
Integral
X
X
Batch
N A0
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0
V
CSTR
PFR
t  N A0 
dX
  r AV
dt
FA 0
dX
  rA
dV
dX
 rA V
FA 0 X
 rA
t
X
X
dX
V  FA 0 
 rA
0
X
PBR
FA 0
dX
  rA
dW
X
W  FA 0 
0
dX
 rA
W
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Reactor Mole Balances in terms of
residence time and conversion
FA=CAΦV , CA=CA0(1-X), τ=V/ ΦV
X
Batch
reactor
 rA  
dC A
dx
 C A0
dt
dt
t
X
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C A0 x
CSTR
 rA 
PFR
dx
dx
 rA  
 C A0
d (V / FA0 )
d

X
τ
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Two types of problems
 Design problem: Design a reactor to achieve certain conversion
Levenspiel Plots
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 Operating problem: in a existing reactor to find conversion or the residence time
to reach the certain conversion
PFR
x
dx
dx
 
 C A0 
d (V / FA0 )
 rA
0
CSTR
C A0 x

 rA
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V1 
FA0 x1
 rA, x1
Reactors in Series
FA0 x1
V1 
 rA, x1
FA0 ( x2  x1 )
V2 
 rA, x2
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x1
dx
V1  FA0 
 rA
0
x2
dx
V2  FA0 
 rA
x1
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PFR vs CSTR in series
…..
FA0
 rA
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FA0 xi
 rA 
Vi
x
Is the PFR always better than the CSTR in terms of reactor size
to achieve a identical conversion?
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Reactors in Series
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moles of A reacted up to point i
Xi 
moles of A fed to first reactor
Only valid if there are no side streams
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Reactors in Series
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Exothermic reaction in an adiabatic reactor
How can we minimize the reaction size?
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Basic Definitions
 Homogeneous reactions involve only one phase
 Heterogeneous reactions involve more than one phase,
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and reactions occur at interfaces of two phases
 Irreversible reactions occur at only one direction
 Reversible reactions occur at both directions, depending
one the approach to equilibrium
 Molecularity of the reaction is the number of the atoms,
ions or molecules involved in a reaction step. Unimolecular,
bimolecular and termolecular refer to reactions involving
one, two and three atoms or molecules in one reaction step,
respectively
 Elementary reaction involves only one bond breaking or
formation
 Non-elementary reaction could involve multi elementary
reaction steps
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Kinetics
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Kinetics - Power Law Model

A

B
 rA  kC C
α order in A
β order in B
Overall Rection Order  α  β
2A  B  3C
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A reactor follows an elementary rate law if the reaction orders just
happens to agree with the stoichiometric coefficients for the reaction
as written.
e.g. If the above reaction follows an elementary rate law
 rA  k A C2A CB
2nd order in A, 1st order in B, overall third order
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"Everyone has Problems Department of Chemical Engineering
but Chemists have Solutions"
Chemical Engineers have Simple Solutions !!!
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Example : Ammonia decomposition
 2NH3=3H2+N2
11 kJ/mol
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The kinetic study was performed in a fixed bed reactor (6
mm diameter) on Fe/Al2O3 catalysts at atmospheric
pressure and total flow of 100 ml/min with Ar as the
balance. 100 mg catalysts mixed with 1 g SiC were
loaded in the reactor.
FNH3.s (ml/min)
20
40
80
FAr,s (ml/min)
80
60
20
XNH3
0.050 0.051 0.050
1) Can we determine the reactor order? (n=0,1,2 ? )
2) How can we reduce the conversion from 5.0 % to 2.5 %
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Relative Rates of Reaction
aA  bB  cC  dD
b
c
d
A B C D
a
a
a
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rC rD
rA
rB



a b c
d
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Relative Rates of Reaction
2A  B  3C
mol
 rA  10
dm 3  s
rC
rA
rB


 2 1 3
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 rA
mol
 rB 
5
2
dm 3  s
3
mol
rC 
rA  15
2
dm 3  s
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2A+B3C
If
 rA  kC2A
› Second Order in A
› Zero Order in B
› Overall Second Order
 rA  k AC A2CB
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 rB  k B C A2 C B
2
3
rC  kC C A2 C B
2
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Reversible Reaction
2A  B  3C
Elementary
 rA  k A C A C 2B  k  A C3C
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3


C
2
C
 k A C A C B 

k A k A 

3

CC 
2
 k A C A C B 

Ke 

This equation is thermodynamically consistent.
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Reversible Reaction
A+2B
kA
3C
k-A
Reaction is: First Order in A
Second Order in B
Overall third Order
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moles
 rA   3
m s
moles
CA 
m3
  rA 
mole m3 s
m6
k   


2
2 
2
3
3
C
C
mole
s
 A B  mole m mole m 
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Arrhenius Equation
k is the specific reaction rate (constant) and is given by the Arrhenius
Equation.
where:
k  Ae  E RT
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T  k  A
T 0 k 0
k
A  1013
Svante August Arrhenius
was a Swedish scientist,
received the Nobel Prize
for Chemistry in 1903
T
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Arrhenius Equation
where:
E = Activation energy (cal/mol)
R = Gas constant (cal/mol*K)
T = Temperature (K)
A = Frequency factor (same units as rate constant k)
(units of A, and k, depend on overall reaction order)
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Reaction Coordinate
The activation energy can be thought of as a barrier to the reaction.
One way to view the barrier to a reaction is through the reaction
coordinates. These coordinates denote the energy of the system as a
function of progress along the reaction path. For the reaction:
A  BC  A ::: B ::: C  AB  C
The reaction coordinate is
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
Transition state theory
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Why is there an Activation Energy?
We see that for the reaction to occur, the reactants must overcome
an energy barrier or activation energy EA. The energy to overcome
their barrier comes from the transfer to the kinetic energy from
molecular collisions and internal energy (e.g.Vibrational Energy).
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1. The molecules need energy to disort or stretch their bonds in
order to break them and thus form new bonds
2. As the reacting molecules come close together they must
overcome both stearic and electron repulsion forces in order to
react.
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Collision probability
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f(E,T)dE=fraction of molecules with energies between E+dE
One such distribution of energies is in the following figure:
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Rate expression – Gas phase reaction
A+B = 2C
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Rate expression – Catalysed reaction
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