5-1
Solutions for Chapter 5 – Collection and Analysis of
Rate Data
P5-1 (a) Individualized solution
P5-1 (b) Individualized solution
P5-1 (c) Individualized solution
The graphical method requires estimations of the area under and above curves on a plot as well as in
reading the intersection of lines on the plot. This can lead to small inaccuracies in each data point.
P5-1 (i) Example 5-2
Assuming zero order reaction:
Rate law:
‘
A
dC k
dt
!=
CA = CAO – k’t
t(min.)
0
50
100
150
200
250
300
Ist order reaction
3.5
4
4.5
zero order reaction
0.05
0.06
5-2
Assuming first order reaction:
Rate law:
‘
A
A
dC k C
dt
!=
Or, ln(1/CA) = k’t + 3
t(min.)
0
50
100
150
200
250
300
P5-1 (j) Example 5-3 Because when α is set equal to 2, the best value of k must be found.
P5-1 (k) Example 5-4
POLYMATH Results
Nonlinear regression (L-M)
Model: r = k*(Ca^alfa)
Variable Ini guess Value 95% confidence
k 0.1 1.0672503 0.0898063
alfa 0.5 0.4461986 0.076408
Precision
P5-1 (l) Example 5-5
rate law:
4 2
CH CO H
r kP P
! “
=
Regressing the data
r’(gmolCH4/gcat.min)
PCO (atm)
PH2 (atm)
See Polymath program P5-1-l.pol.
5-3
POLYMATH Results
Nonlinear regression (L-M)
Model: r = k*(PCO^alfa)*(PH2^beta)
Variable Ini guess Value 95% confidence
k 0.1 0.0060979 6.449E-04
Therefore order of reaction = 1.14
Again regressing the above data putting
1
!
=
POLYMATH Results
Nonlinear regression (L-M)
Model: r = k*(PCO^0.14)*(PH2)
Variable Ini guess Value 95% confidence
k 0.1 0.0040792 0.0076284
Precision
P5-2 Solution is in the decoding algorithm given with the modules.
P5-4 (a)
The kinetics of this deoxygenation of hemoglobin in blood was studied with the aid of a tubular reactor.
HbO2 Hb + O2
Rate law: -rA=k
n
A
C
5-4
Electrode Position
1
2
3
4
5
6
7
Position (cm)
0
5
10
15
20
25
30
z (cm)
5
5
5
5
5
5
5
A histogram plot of X/ z vs. z is then produced. The values of dX/dz are evaluated using equal-area
graphical differentiation:
Using the values obtained above, a plot of ln(dXA/dz) vs. ln(1-XA) is produced and a line
is fit to the data
( )
ln ln ln 1
dX a n X
dz
! ” = + #
$ %
5-5
6 3
6
2.3 10 /
45.7 10 /
Ao
C mol cm
F moles s
!
!
=“
=“
P5-4 (b)
First we fit a polynomial to the data. Using Polymath we use regression to find an expression for X(z)
POLYMATH Results
Polynomial Regression Report
Model: X = a0 + a1*z + a2*z^2 + a3*z^3 + a4*z^4 + a5*z^5 + a6*z^6
Variable Value 95% confidence
a0 2.918E-14 0
General
Order of polynomial = 6
Regression including free parameter
Number of observations = 7
Statistics
R^2 = 1
Next we differentiate our expression of X(z) to find dX/dz and knowing that
as in the finite differences.
POLYMATH Results
Linear Regression Report
Model: ln(dxdz) = a0 + a1*ln(1-X)
5-6
Variable Value 95% confidence
General
Regression including free parameter
Number of observations = 7
Statistics
P5-5 (a)
Liquid phase irreversible reaction:
A B + C ; CAO = 2 mole/dm3
Space time (
!
)min.
CA(mol/dm3)
ln(CA)
ln((CAO–CA)/
!
)
15
1.5
0.40546511
-3.4011974
By using linear regression in polymath:
See Polymath program P5-5-a.pol.
POLYMATH Results
Linear Regression Report
Model: y = a0 + a1*lnCa
5-7
Hence,
3slope
!
=“
P5-5 (b) Individualized solution
P5-5 (c) Individualized solution
P5-6 (a)
Constant voume batch reactor: A B +C
Mole balance:
mol/dm3.
t (min.)
CA (mol/dm3)
0
2
See Polymath program P5-6-a.pol.
POLYMATH Results
Nonlinear regression (L-M)
Model: t = (1/k)*((2^(1-alfa))-(Ca^(1-alfa)))/(1-alfa)
K= 0.03 (mol/dm3)-0.5.s-1 and
1.5
!
=
P5-6 (b) Individualized solution
5-8
P5-6 (c) Individualized solution
P5-6 (d) Individualized solution
P5-7 (a)
t (h)
CA(mol/dm3)
Rate law:
n
B
m
AA CkCr =!
For table 2 data: CAO
CBO =>
‘m
A A
r k C!=
where
‘n
BO
k kC=
POLYMATH Results
Nonlinear regression (L-M)
Model: t = (1/k)*((0.1^(1-m))-(Ca^(1-m)))/(1-m)
Variable Ini guess Value 95% confidence
Precision
t (h)
CA(mol/dm3)
0
1.0
0.278
0.95
5-9
See Polymath program P5-7-a-2.pol.
POLYMATH Results
Nonlinear regression (L-M)
Model: t = (1^(-1-n)-Ca^(-1-n))/(k*(-1-n))
Variable Ini guess Value 95% confidence
Therefore, n = 0.8
Hence rate law is:
2 0.8
3
0.17
A A B
mol
r C C
dm h
!=
P5-7 (b) Individualized solution
P5-8 (a)
At t = 0, there is only (CH3)2O. At t = ∞, there is no(CH3)2O. Since for every mole of (CH3)2O consumed
P5-8 (b)
Constant volume reactor at T = 504°C = 777 K
Data for the decomposition of dimethylether in a gas phase:
5-10
01
A
y=
3 1 2
!
=“=
02
A
y
! “
= =
0
Assume
A A
r kC!=
(i.e. 1st order)
( )
01
A A
C C X=!
(V is constant)
Then:
( )
0 0 1
A A
dX
C kC X
dt =!
[ ]
000
1
P t
P
dP kdt
P P
!
=
+“
# #
A A
5-11
y = 0.00048x – 0.02907
1.2
1.6
P5-8 (c) Individualized solution
P5-8 (d) The rate constant would increase with an increase in temperature. This would result in the
P5-9
Photochemical decay of bromine in bright sunlight:
t (min)
10
20
30
40
50
60
P5-9 (a)
Mole balance: constant V
A
A A
dC r kC
dt
!
= = “
5-12
After plotting and differentiating by equal area
P5-9 (b)
A
A B
dN Vr F
5-13
! “ ! “! “! “
P5-9 (c) Individualized solution
P5-10 (a)
Gas phase decomposition
A B +2C
Determine the reaction order and specific reaction rate for the reaction
Run #
CA0 (gmol/lt)
t (min.)
ln t
ln CA0
1
0.025
4.1
1.410987
-3.6888795
2
0.0133
7.7
2.0412203
-4.3199912
See Polymath program P5-10-a.pol.
POLYMATH Results
Linear Regression Report
Model: lnt = a0 + a1*lnCa0
Variable Value 95% confidence
a0 -2.3528748 0.1831062
5-14
From linearization, n = 1- slope = 2.103 ≈ 2
See Polymath program P5-10-a.pol.
POLYMATH Results
Nonlinear regression (L-M)
Model: t = ((2^(a-1))-1)/(k*(a-1))*(1/Ca0^(a-1))
Variable Ini guess Value 95% confidence
5-15
P5-10 (b)
We know,
!
!
“
#
$
$
%
&
‘
‘
=‘
‘
1
0
2/1
11
)1(
)12(
(
(
(
A
C
t
k
P5-11
The values of k1 and k2 may depend on your initial guess. Look for the lowest s2. You could try
“rO3=kCBu
Using polymath nonlinear regression we can find the values of k1 and k2
Run ozone rate Ozone concentration Butene concentration
# (mol/s.dm3) (mol/s.dm3) (mol/s.dm3)
Ozra CO3 Cbu
POLYMATH Results
Nonlinear regression (L-M)
Model: Ozra = k1+k2*Cbu/CO3
5-16
Nonlinear regression settings
Max # iterations = 300
Precision
R^2 = 0.7572693
P5-12
Given: Plot of percent
decomposition of NO2 vs V/FA0
2
% Decomposition of NO
X= 100
P5-13
2 2 6 2
6 2SiO HF H SiF H O+!+
C S
S 2
S
A ñ ä
N = moles of SiO =
MW
5-17
MWF = molecular weight of HF = 20.0
Assume the rate law is
S F
r kC
!
“=
Mole balance:
S
S
dN r V
ln d
dt
!
” #
$
% &
‘ (
-16.629
-15.425
-14.32
-13.816
-13.479
where
d
dt
!
” #
$
% &
‘ (
is in
m
min