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Lecture 11
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.
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Lecture 11 – Thursday 2/14/2013
Block 1:
Mole Balances
Block 2:
Rate Laws
Block 3:
Stoichiometry
Block 4:
Combine
Determining the
Rate Law
from Experimental Data Integral Method Differential (Graphical) Method Nonlinear Least Regression
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Integral Method
Consider the following reaction that occurs in a constant volume Batch Reactor : (We will withdraw samples and record the concentration of A as a function of time.)
A
Products Mole Balances:
dN A dt
r A V
Rate Laws:
r A
kC
A
Combine:
V
V
0
dC A dt
kC A
4 Finally we should also use the formula to plot reaction rate data in terms of conversion vs. time for 0, 1st and 2nd order reactions.
Derivation equations used to plot 0th, 1st and 2nd order reactions.
These types of plots are usually used to determine the values k for runs at various temperatures and then used to determine the activation energy.
Zeroth order
dC A dt
r A
k at t
0,
C A
C A
0
C A
C A
0
kt
First Order
dC A dt
r A
kC A
Second Order
dC A dt
r A
kC A
2
at t
0,
C A
C A
0 ln
C A
0
C A
kt at t
0,
C A
C
1
C A
1
C A
0
kt A
0
5 C A
Integral Method
Guess and check for α = 0, 1, 2 and check against experimental plot.
0 1 2
r A
C A
0
kt
ln
C A
0
C A
kt C
1
A
1
C A
0
kt
ln(C A0 /C A ) 1/C A t t t
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Differential Method
Taking the natural log of
dC A dt
kC
A
ln
dC A dt
ln
k
ln
C A
The reaction order can be found from a ln-ln plot of: dC A dt
ln
dC A dt
vs C A
dC A dt P Slope = α
k
dC A dt C
Ap p
C AP
ln
C A
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Methods for finding the slope of log-log and semi-log graph papers may be found at http://www.physics.uoguelph.ca/tutorials/GLP/ However, we are usually given concentration as a function of time from batch reactor experiments: time (s) concentration (moles/dm 3 ) 0 C A0 t 1 C A1 t 2 C A2 t 3 C A3
Three ways to determine (-dC A /dt) from concentration-time data Graphical differentiation Numerical differentiation formulas Differentiation of a polynomial fit to the data
1. Graphical
C A
t
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t
C A
t
9
dC A dt
0
dC A dt
dC A dt t
2
t
1 0 t 1 t 2 t The method accentuates measurement error!
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Example – Finding the Rate Law
t(min) 0 C A (mol/L)
C A
t
1 0.3
C A
t
.3
.2
.1
1 0.7
0.2
2 0.5
0.15
3 0.35
Areas equal for both sides of the histogram t 1 2 3
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Example – Finding the Rate Law
C
t
C A -dC A /dt 1 0.35
0.7
0.25
0.5
0.175
Plot ( –dC A /dt) as a function of C A ln 0.35
0.12
dC A /dt Slope = α ln C A
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Example – Finding the Rate Law
Choose a point, p, and find the concentration and derivative at that point to determine k.
k
dC A dt C
Ap p
ln dC A /dt
dC A dt p C A p
Slope = α ln C A
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Non-Linear Least-Square Analysis
We want to find the parameter values ( α, k, E) for which the sum of the squares of the differences, the measured rate (r m ), and the calculated rate (r c ) is a minimum.
2
i n
1
C im N
C ic K
2
N S
2
K
That is, we want 2 to be a minimum.
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Non-Linear Least-Square Analysis
For concentration-time data, we can combine the mole balance equation for
r A
kC
A
to obtain:
dC A
kC dt
0
C A
A C A
0
C
1
A
0
C A
1 )
kt
Rearranging to obtain the calculated concentration as a function of time, we obtain:
C Ac
C A
[
C
1
A
0 ) ] 1/(1 )
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Non-Linear Least-Square Analysis
Now we could use Polymath or MATLAB to find the values of α and k that would minimize the sum of squares of differences between the measured (C Am ) and calculated (C Ac ) concentrations. That is, for N data points, Similarly one can calculate the time at a specified concentration, t c and compare it with the measured time, t m , at that same concentration. That is, we find the values of k and α that minimize:
Non-Linear Least Squares Analysis
Guess values for α and k and solve for measured data points then sum squared differences:
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C Am C Ac (C Ac -C Am ) (C Ac -C Am ) 2 1 1 0 0 0.7
0.5
-0.2
0.04
for α= 2, k = 1 → s 2 = 0.07
for α = 2, k = 2 → s 2 = 0.27
etc. until s 2 is a minimum 0.5
0.33
-0.17
0.029
0.35
0.25
-0.10
0.01
0.07
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Non-Linear Least Squares Analysis
Non-Linear Least Squares Analysis
s
2
N
C Ami i
1
C Aci
2
N
i
1
C Ami
C
1
A
0 1
kt i
1 1 2 We find the values of alpha and k which minimize s 2 18
Minimum Sum of Squares
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Residuals
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