ChE 452 Lecture 08
Analysis Of Data From A Batch Reactor
1
Objective

Data analysis from indirect measurements


Essen’s method (learned in p-chem)
 Does not usually work
Van’t Hoff’s method
 Accurate but amplifies errors in data
2
Background: Kinetic Data After
Measuring
Pressure, x2, torr
100
Figure 3.8 Typical batch data
for reaction (3.7). Data of
Tamaru[1955].
10
0
5
10
15
20
25
30
35
Time, Hours

Indirect method – a method where you measure
some other property (i.e. concentration vs time)
and infer a rate equation.
3
Objective For Today:
Analysis Of Rate Data



Derive basic equations
Essen’s method
Van’t Hoff’s method
4
Derivation Of Performance
Equation For A Batch Reactor
For A B, the moles of A
reacted/volume/time will equal the reaction
rate, i.e.
dC A
=rA
dτ
(1)
CA is the concentration of A,  is time,
and rA is the rate of reaction per unit
volume.
Figure 3.11 A batch reactor
5
Integration Yields The
Following
C0A

CfA
dCA
=τ
(-rA )
(3.31)
Memorize this equation
6
For A First Order Reaction
rA = -k1CA
(3.38)
Substituting equation (3.38) into equation
(3.31) and integrating yields:
 C0A 
1
Ln  f   
k1  C A 
Memorize this equation
(3.39)
Derivation
7
For An nth Order Reaction:
rA  k n C A 
n
(3.41)
Substituting equation (3.41) into equation (3.31),
integrating, and rearranging yields:
1
 
o n 1
n  1k n C A
 C o  n -1 
 A   1  
 C fA 



(3.42)
Memorize this equation
Derivation
9
Plots Of Equations
10
8
Time
6
4
Second
Order
2
0
0%
First Order
20%
40%
60%
80%
100%
conversion
11
Table 3.4 Rate Laws For A
Number Of Reactions
Rate Laws for a number of reactions
Reaction
Rate Law
Differential Equation
Integral Equation
A Products
A+B  Products
rA=kA
dXA
 kA
d
kA 
A Products
A+B  Products
rA=kA[A]
dXA
 k AXA
d
1  1 
k A  Ln 

  1  XA 
A Products
A+B  Products
rA=kA[A]n
dXA
 k A (CnA )n 1X nA
d
 1 n 1 

  1
kA 
0 n 1  1  X 
(n  1)(CA ) 

A
A+B  Products
rA=kA[A][B]
dXA
 k A (1  X A )(C0BA  C0A X A )
d
 C0B (1  X A ) 

kA 
Ln  0
0
0
0 

(CA  CB  CB  X A CA 
A+2B  Products
rA=kA[A][B]
dXA
 k A (1  X A )(C0B  2CA X A )
d
AB
rA=k1[A]k2[B]
dXA
 k A (1  X A )  k 2X A
d
XA

1
1
 C0B (1  X A ) 

kA 
Ln  0
0
0
0 

(2CA  CB  CB  2X A CA 
1

1 
1

(k1  k 2 )  Ln 
  Xe  X A 
12
Fitting Batch Data To A Rate
Law
Steps
• Start with a batch reactor and measure
concentrations vs time.
• Fit those data to a first order and a
second order rate law and see which
equation fits better.
• Whichever rate equation fits best is
assumed to be the correct rate equation
for the reaction.
13
Key Challenge: First And Second Order
Data Does Not Look That Much Different
1
.
8
0
.
6
0
.
0
.
Thir
dO
rder
Sec
ond
Ord
er
4
Fir
st
0
0
Same
k(CA0)n-1
CA /C A
0
CA /C A
1
.
8
0
.
6
0
.
Third
Orde
r
4
Or
de
r
Fir
st
Ha
lf O
rd
er
2
0
Vary k to fit
data
0
.
2
Ha
lf
0
Or
de
0
0
0
.
5
Time
1
1
.
5
0
0
.
5
1
Second Order
Or
de
r
r
1
.
5
Time
14
Essen’s Method
First order
 C0A 
1
Ln  f   
k1  C A 
(3.39)
nth order
1
 
o n 1
n  1k n C A
 C o  n -1 
 A   1  
 C fA 



(3.42)
15
Essen’s Method
1
.
6
0
.
4
0
.
2
r
de
r
O
ird
h
T
0
0
.
8
0
.
6
0
.
4
0
.
2
First
Orde
Sec
r
ond
Th
Ord
er
ird
O
rd
er
0
nd
o
c
Se
er
d
r
O
Half O
rder
8
Firs
tO
rde
r
.
(C A0/C A) - 1
ln(CA0/C A )
0
Half
Ord
er
1
0
0
0
.
5 1
1
.
5
0
Time
0
.
5 1
1
.
5
Time
Figure 3.15
16
Example: The Concentration Of
Dye As A Function Of Time
CA,m , Min
moles
/Lit
1
0
0.91
1
0.83
2
0.77
3
0.71
4
0.67
5
CA

CA
0.63 6 0.45
0.59 7 0.43
0.56 8 0.42
0.53 9 0.40
0.50 10 0.38
0.48 11 0.37

12
13
14
15
16
17
Table 3.5
17
Essen Plot For Example:
1
6
1.5
5
0.8
(C A /C A ) 2 - 1
0.6
0.4
r2=.984
3
0
(C A0/C A) - 1
ln(CA0 /CA )
4
1
r2=.999
0.5
2
r2=.981
0.2
1
0
0
0
5
10
Time, Mins
15
0
0
5
10
Time, Mins
15
0
5
10
15
Time, Mins
Figure 3.16
No statistically significant difference between results.
18
Example Shows Essen’s Method
Does Not Distinguish Between Models



In the literature, Essen’s method is often used.
Useful for impressing your boss since it always
fits with good r2 (given good data)
It often gives the incorrect answers.
19
Van’t Hoff’s Method
•
•
•
Take batch data as before.
Calculate kone (first order rate constant)
ktwo (second order rate).
kone should be constant for a first order
reaction, ktwo should be constant for a
second order reaction. (Use f test to
check).
20
Equations For kone And
ktwo Follow From Before
0 

1
C
Ln  A   
k1  C A 
(3.39)
Derived previously
Solve for k1
 C o  n -1 
1
 A   1  

o n 1  C f 
n  1k n C A  A

 
(3.42)
Derived previously
0 

1  CA 
k1  Ln
  C A 
(3.51)
 0 n 1 
1
CA 



kn 

1
0 n 1  CA 

(n  1) CA  

Solve for kn
 
(3.52)
21
Easy Solution: Define A VB
Module In Microsoft Excel
Public Function kone(ca0, ca, tau) As Variant
kone = Log(ca0 / ca) / tau
End Function
Public Function ktwo(ca0, ca, tau) As Variant
ktwo = ((1# / ca) - (1# / ca0)) / tau
End Function
Public Function kthree(ca0, ca, tau) As Variant
kthree = ((1# / ca) ^ 2 - (1# / ca0) ^ 2) / tau
End Function
22
Microsoft Excel/Visual Basic
Return Types
As Variant
As Single
General return type (can be an
integer, real, vector, matrix,
logical or text)
Single precision real
As Double
Double precision real
As Integer
Integer
23
The Formulas In The
Spreadsheet For Van’t Hoff’s
Method
B
C
1
2
D
E
Ca0=
3
4
time conc
5
6
7
8
9
0
1
2
3
4
Essen's
Method
first
ln(Ca0/Ca)
1 =kone(ca0,C5,B5)
0.91 =kone(ca0,C6,B6)
0.83 =kone(ca0,C7,B7)
0.77 =kone(ca0,C8,B8)
0.71 =kone(ca0,C9,B9)
F
1
second
(Ca0/Ca)-1
third
(CA0/CA)^
2-1
=ktwo(ca0,C5,B5) =kthree(ca0,C5,B5)
=ktwo(ca0,C6,B6) =kthree(ca0,C6,B6)
=ktwo(ca0,C7,B7) =kthree(ca0,C7,B7)
=ktwo(ca0,C8,B8) =kthree(ca0,C8,B8)
=ktwo(ca0,C9,B9) =kthree(ca0,C9,B9)
24
The Numerical Values For
Van’t Hoff’s Method
B
C
3 Time conc
4
0
5
6
7
8
9
10
1
2
3
4
5
6
D
E
k1
F
k2
k3
1 ln(1/Ca ((Ca0/C ((CA0/CA)
)/t
a)-1)/t
^2-1)/t/2
0.91 0.094
0.099
0.104
0.83 0.093
0.102
0.113
0.77 0.087
0.1
0.114
0.71 0.086
0.102
0.123
0.67
0.08
0.099
0.123
0.63 0.077
0.098
0.127
25
Van’t Hoff Plot
0.2
Rate Constant
Oxidation of
Red Dye
K
3
0.15
K
0.1
K
2
1
0.05
0
5
10
Time, Mins
15
Figure 3.18 Van’t Hoff plot of the data from tables 3.5 and 3.6
Van’t Hoff’s method is much more accurate than Essens’ method.
Essen’s is more common!
26
Discussion Problem: Use Van’t Hoff’s
Method To Determine The Order For The
Following Data
Table 4.1 Buchanan’s [1871] data for the reaction:
CICH2COOH + H2COOH + HCI at 100º C
Time Hours
[CICH2COOH] gms/liter
0
2
3
4
6
10
13
19
28
34.5
43
48
4
3.80
3.69
3.60
3.47
3.10
2.91
2.54
2.26
1.95
1.59
1.39
27
Solution:
Ca0=
4
Van't Hoff's
time
0
Conc
first
ln(ca0/Ca)
second
(Ca0/Ca)-1
third
(CA0/CA)^21
4 =kone(cao,B5 =ktwo(cao,B5 =kthree(cao
,A5)
,A5)
,B5,A5)
2
3.8 =kone(cao,B6 =ktwo(cao,B6 =kthree(cao
,A6)
,A6)
,B6,A6)
3
3.69 =kone(cao,B7 =ktwo(cao,B7 =kthree(cao
,A7)
,A7)
,B7,A7)
4
3.6 =kone(cao,B8 =ktwo(cao,B8 =kthree(cao
,A8)
,A8)
,B8,A8)
6
3.47 =kone(cao,B9 =ktwo(cao,B9 =kthree(cao
,A9)
,A9)
,B9,A9)
28
Solution Continued:
ca0= 4
Van't
time
0
2
3
4
6
10
13
19
25
34.5
43
48
conc
Hoff's
first
second
third
ln(ca0/
Ca)
(Ca0/Ca
)-1
(CA0/CA
)^2-1
4 #VALUE!
3.8
0.026
3.69
0.027
3.6
0.026
3.47
0.024
3.1
0.025
2.91
0.024
2.54
0.024
2.26
0.023
1.95
0.021
1.59
0.021
1.39
0.022
#VALUE!
0.007
0.007
0.007
0.006
0.007
0.007
0.008
0.008
0.008
0.009
0.01
#VALUE!
0.003
0.004
0.004
0.003
0.004
0.004
0.005
0.005
0.006
0.008
0.009
29
Van’t Hoff Plot
0.05
Rate Constant
Hydration of
Chloracetic Acid
K
3
0.04
K
2
0.03
K
1
0.02
0
10
20
30
Time, Mins
40
50
Figure 3.18 Van’t Hoff plot of the data from tables 3.5 and 3.6
30
Discussion Problem 2
Ammonium-dinitramide, (ADN) NH4N(NO2)2, is a oxidant used in solid
fuel rockets and plastic explosives. ADN is difficult to process because it
can blow up. Oxley et. Al., J. Phys chem A, 101 (1997) 5646, examined
the decomposition of ADN to try to understand the kinetics of the
explosion process. At 160º C they obtained the data in Table P3.20.
Table P3.20 Oxley's measurements of the decomposition of
dinitramide at 160 C
time,
fraction of
time,
fraction of
time,
fraction of
seconds the AND
seconds
the AND
seconds
the AND
remaining
remaining
remaining
0
1.0
900
0.58
300
0.84
1200
0.49
600
0.70
1500
0.41
2400
0.24
31
Discussion Problem 2
Continued:
a) Is this a direct or indirect measurement of the rate?
b) Use Van’t Hoff’s Method to fit this data to a rate
equation.
c) If you had to process ADN at 160° C, how long could
you run the process without blowing anything up?
Assume that there is an explosion hazard once 5% of
the ADN has reacted to form unstable intermediates.
32
This Is An Indirect Measurement! Use
Same Spreadsheet As Before To Fit
Data
Ca0=
=b5
Van't Hoff's
time
0
300
600
900
1200
1500
2400
Conc
first
1 ln(ca0/Ca)
0.84 =kone(cao,B5
,A5)
0.7 =kone(cao,B6
,A6)
0.58 =kone(cao,B7
,A7)
0.49 =kone(cao,B8
,A8)
0.41 =kone(cao,B9
,A9)
0.24 =kone(cao,B1
0,A10)
second
third
(Ca0/Ca)-1
(CA0/CA)^2-1
=ktwo(cao,B5
,A5)
=ktwo(cao,B6
,A6)
=ktwo(cao,B7
,A7)
=ktwo(cao,B8
,A8)
=ktwo(cao,B9
,A9)
=ktwo(cao,B1
0,A10)
=kthree(cao,
B5,A5)
=kthree(cao,
B6,A6)
=kthree(cao,
B7,A7)
=kthree(cao,
B8,A8)
=kthree(cao,
B9,A9)
=kthree(cao,
B10,A10)
33
Solution Cont.
Ca0=
1
Van't Hoff's
time
Conc
first
second
third
ln(ca0/Ca)
(Ca0/Ca)-1
(CA0/CA)^2-1
0
1
#VALUE!
#VALUE!
#VALUE!
300
0.84
0.000581
0.000635
0.001391
600
0.7
0.000594
0.000714
0.001735
900
0.58
0.000605
0.000805
0.002192
1200
0.49
0.000594
0.000867
0.002637
1500
0.41
0.000594
0.000959
0.003299
2400
0.24
0.000595
0.001319
0.006817
34
Solution Cont.
c) from equ 3.39
1  CoA 
Ln f  
k1  CA 
k
k1=0.0006/sec (from spreadsheet)
CoA =1 (given)
CfA =0.95 (what's left if 5% converted)
1  CoA 
1
 1 
  Ln f 
Ln 
 85

k1  CA  0.0006  0.95 
sec
35
Summary: Two Methods To Fit
Rate Data
• Essen’s Method
• Most common method
• Plots look the best
• Gives great looking results even with incorrect rate
equation
• Van’t Hoff’s Method
•
•
•
•
More accurate than Essen
Rare in literature
Plots noisier
Highlights weaknesses in rate equations
36
Class Question

What did you learn new today?
37