Lecture 3
Chemical Reaction Engineering (CRE) is the
field that studies the rates and mechanisms of
chemical reactions and the design of the reactors in
which they take place.
Lecture 3 – Thursday 1/13/2011
 Review of Lectures 1 and 2
 Block 1
(Review)
 Size CSTRs and PFRs given –rA= f(X)
 Conversion for Reactors in Series
 Block 2




2
Reaction Orders
Arrhenius Equation
Activation Energy
Effect of Temperature
The GMBE applied to the four major reactor types
(and the general reaction AB)
Reactor
Differential
Algebraic
Integral
NA
Batch
CSTR
PFR
PBR
3

dN A
t 
rV
N A0 A
dN A
 rAV
dt
V
dFA
 rA
dV

dFA
 rA
dW
FA 0  FA
rA
FA
dFA
V 
drA
FA 0
W
FA

FA 0
dFA
rA
NA
t
FA
V
FA
W
a A  bB
 c C  d D
Choose limiting reactant A as basis of calculatio n
b
c
d
A  B
 C  D
a
a
a
moles A reacted
X
moles A fed
4
Reactor
Differential
Algebraic
Integral
X
X
Batch
N A0
V
CSTR
PFR
dX
t  N A0 
 rAV
0
dX
  r AV
dt
dX
FA0
 rA
dV
t
FA0 X
 rA
X
dX
 rA
0
V  FA0 
X
X
PBR
5
dX
FA0
  rA
dW
dX
 rA
0
W  FA0 
W
6
moles of A reacted up to point i
Xi 
moles of A fed to first reactor
Only valid if there are no side streams
7
8
Lecture 3
Power Law Model


 rA  kCA CB
9
α order in A
β order in B
Overall Rection Order  α  β
2A  B  3C
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 AC A2CB
2nd order in A, 1st order in B, overall third order
10
2A+B3C
If  rA  kCA2
› Second Order in A
› Zero Order in B
› Overall Second Order
-rA = kA CA2 CB
-rB = kB CA2 CB
rC = kC CA2 CB
11
How to find
 rA  f  X 
Step 1: Rate Law
 rA  g Ci 
Step 2: Stoichiometry
Ci   h X 
Step 3: Combine to get  rA  f  X 
12
aA  bB  cC  dD
b
c
d
A B  C  D
a
a
a
rC rD
rA
rB

 
a b c
d
13
2A  B  3C
mol
 rA  10
dm 3  s
rA
rB rC


 2 1 3
 rA
mol
 rB 
5
2
dm 3  s
14
3
mol
rC 
rA  15
2
dm 3  s
A+2B
kA
k-A
3C
3


C
2
3
2
C
 rA  k AC AC B  k  ACC  k A C A C B 

k
k
A
A 

 2
CC3 
 k A C A C B 

K
e 

15
A+2B
kA
k-A
Reaction is:
3C
moles
 rA   3
dm s
First Order in A
Second Order in B
Overall third Order
  rA 
mole dm3 s
k    2  
3
3
C
C
mole
dm
mole
dm
 A B

16
moles
dm 3
CA 


2
dm 6

mole 2 s
17
k is the specific reaction rate (constant) and is given by the
Arrhenius Equation.
where:
k  Ae E RT
T  k  A
T 0 k 0
k
A  10
13
T
18
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)
19
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

20
21
22
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).
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.
23
We will use the Maxwell-Boltzmann Distribution of Molecular
Velocities. For a species af mass m, the Maxwell distribution of
velocities (relative velocities) is
32
 m   mU 2 2 k BT 2
 e
f U , T   dU  4 
U dU
 2k BT 
1
E  mU 2
2
24
A plot of the distribution function, f(U,T), is shown as a
function of U:
T1
T2
25
T2>T1
U
Maxwell-Boltzmann Distribution of velocities.
f(E,T)dE=fraction of molecules with energies between E+dE
One such distribution of energies is in the following figure:
26
End of Lecture 3
27
AB
BC
VBC
VAB
kJ/
Molecule
kJ/
Molecule
r0
rBC
rAB
Potentials (Morse or Lennard-Jones)
28
r0
One can also view the reaction coordinate as variation of the BC
distance for a fixed AC distance:
l
l
l











A B C
AB
C
A
BC
rAB
29
rBC
For a fixed AC distance as B moves away from C the distance of
separation of B from C, rBC increases as N moves closer to A. As rBC
increases rAB decreases and the AB energy first decreases then
increases as the AB molecules become close. Likewise as B moves
away from A and towards C similar energy relationships are found.
E.g., as B moves towards C from A, the energy first decreases due
to attraction then increases due to repulsion of the AB molecules as
they come closer together. We now superimpose the potentials for
AB and BC to form the following figure:
Reference
AB
S2
S1
BC
Energy
E*
Ea
E1P
E2P
r
30
∆HRx=E2P-E1P