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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
P5-1 (d) Individualized solution
P5-1 (e) Individualized solution
P5-1 (f) Individualized solution
P5-1 (g) Individualized solution
P5-1 (h) Example 5-1
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
Ist order reaction
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 50 100 150 200 250 300
t(min.)
ln(1/CA)
zero order reaction
0
0.01
0.02
0.03
0.04
0.05
0.06
0 50 100 150 200 250 300
t(min.)
CA(mol/dm3)
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
CA(mol/dm3)
0.05
0.38
0.0306
0.0256
0.0222
0.0195
0.0174
Since none of these plots are straight lines, its is not Ist order reaction or second order reaction.
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
CHCl,0 (mol/dm3)
1.0
4.0
2.0
0.1
0.5
'' 2 7
,0 ( / . ) 10
HCl
r mol cm s! "
1.2
2.0
1.36
0.36
0.74
See Polymath program P5-1-k.pol.
POLYMATH Results
Nonlinear regression (L-M)
Model: r = k*(Ca^alfa)
Variable Ini guess Value 95% confidence
k 0.1 1.0672503 0.0898063
R^2 = 0.9812838
R^2adj = 0.9750451
Rmsd = 0.0341709
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)
5.2e-3
1
1
30e-3
4.08
1
5.25e-3
1
4
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
alfa 1 1.1381455 0.0850634
R^2 = 0.9869709
R^2adj = 0.9782849
Rmsd = 4.176E-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
R^2 = -0.8194508
R^2adj = -0.8194508
Rmsd = 0.0049354
P5-2 Solution is in the decoding algorithm given with the modules.
P5-3 Individualized solution
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
Mole balance:
( )
1n
n
AO AO
dX
F kC X
dV =!
c
dV A dz=
, where AC is the tube cross sectional area
( )
1
n
n
Ao c
Ao
kC A
dX X
dz F
=!
therefore,
( )
ln ln ln 1
,
n
Ao
dX a n X
dz
where a
F
! " = + #
$ %
& '
=
Electrode Position
1
2
3
4
5
6
7
Position (cm)
0
5
10
15
20
25
30
(1-XA)
1.0000
0.9807
0.9618
0.9432
0.9252
0.9075
0.8900
X
-
0.0193
0.0189
0.0186
0.018
0.0177
0.0175
X/ z (cm-1)
-
0.00386
0.00378
0.00372
0.00360
0.00354
0.00350
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 5.5477 1.0577 ln 1
dX X
dz
! " =#+#
$ %
& '
n = 1
ln(a) = -5.5
( )
ln ln ln 1
,
n
Ao c
Ao
dX a n X
dz
kC A
where a
F
! " = + #
$ %
& '
=
5-5
( )
6 3
6
6
3 1
6 3 2
2.3 10 /
45.7 10 /
45.7 10 / 4.1 10 4.1
2.3 10 / 0.0196
Ao
Ao
AO
n
AO C
C mol cm
F moles s
F a moles s
k cm s
C A moles cm cm
!
!
!
! !
!
="
="
"
= = "=
" "
Hence rate law is,
3
4.1
A A
mol
r C
dm s
!=
P5-4 (b)
First we fit a polynomial to the data. Using Polymath we use regression to find an expression for X(z)
See Polymath program P5-4-b.pol.
POLYMATH Results
Polynomial Regression Report
Variable Value 95% confidence
a0 2.918E-14 0
a2 -6.14E-05 0
a3 7.767E-06 0
a6 -1.6E-10 0
General
Order of polynomial = 6
Statistics
R^2 = 1
R^2adj = 0
Next we differentiate our expression of X(z) to find dX/dz and knowing that
( )
ln ln ln 1
,
n
Ao c
Ao
dX a n X
dz
kC A
where a
F
! " = + #
$ %
& '
=
Linear regression of
ln dX
dz
! "
# $
% &
as a function of
( )
ln 1 X!
gives us similar vaules of slope and intercept
as in the finite differences.
POLYMATH Results
Linear Regression Report
5-6
Variable Value 95% confidence
a0 -5.531947 0.0241574
General
Regression including free parameter
Number of observations = 7
Statistics
R^2 = 0.9482059
R^2adj = 0.9378471
Rmsd = 0.0044015
( )
3 1
6 3 2
45.7 10 / 3.96 10 4.0
2.3 10 / 0.0196
AO
n
AO C
F a moles s
k cm s
C A moles cm cm
! !
!
"
= = "=
" "
Hence rate law is,
1.28
3
4.0
A A
mol
r C
dm s
!=
P5-5 (a)
Liquid phase irreversible reaction:
A B + C ; CAO = 2 mole/dm3
AO A
A
C C kC
!
"
#=
ln ln ln
AO A
A
C C k C
!
"
#
$ % = +
& '
( )
Space time (
!
)min.
CA(mol/dm3)
ln(CA)
ln((CAO-CA)/
!
)
15
1.5
0.40546511
-3.4011974
38
1.25
0.22314355
-3.9252682
300
0.75
-0.28768207
-5.4806389
By using linear regression in polymath:
See Polymath program P5-5-a.pol.
POLYMATH Results
Linear Regression Report
Model: y = a0 + a1*lnCa
ln ln ln
AO A
A
C C k C
!
"
#
$ % = +
& '
Variable Value 95% confidence
a0 -4.6080579 0.0162119
R^2 = 0.9999443
R^2adj = 0.9999258
5-7
Hence,
3slope
!
="
ln(k) = intercept = -4.6
P5-5 (b) Individualized solution
P5-5 (c) Individualized solution
P5-6 (a)
Constant voume batch reactor: A B +C
Mole balance:
A
A
dC kC
dt
!
"=
Integrating with initial condition when t = 0 and CA = CAO for
1.0
!
"
mol/dm3.
t (min.)
CA (mol/dm3)
0
2
5
1.6
9
1.35
22
0.87
40
0.53
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)
Variable Ini guess Value 95% confidence
k 0.1 0.0329798 3.628E-04
alfa 2 1.5151242 0.0433727
R^2 = 0.9997773
R^2adj = 0.9997327
Rmsd = 0.1007934
Hence , rate law is
1.5 3
0.03 / .
A
A
dC C mol dm s
dt
!=
P5-6 (b) Individualized solution
5-8
P5-6 (c) Individualized solution
P5-6 (d) Individualized solution
P5-7 (a)
Liquid phase reaction of methanol and triphenyl in a batch rector.
CH3OH + (C6H5)3CCl (C6H5)3COCH3 + HCl
t (h)
CA(mol/dm3)
0
0.1
2
0.0735
10
0.0357
n
B
m
AA CkCr =!
For table 2 data: CAO
CBO =>
'm
A A
r k C!=
where
'n
BO
k kC=
Using eqn 5-21,
(1 ) (1 ) (1 ) (1 )
(0.01)
1 1
' (1 ) ' (1 )
m m m m
AO A A
C C C
t
k m k m
! ! ! !
!!
= =
! !
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
k 1 1.815656 0.0109025
Nonlinear regression settings
Precision
R^2 = 1
R^2adj = 0.9999999
Rmsd = 3.268E-04
Therefore, m = 2
For first set of data, equal molar feed => CA = CB
Hence, rate law becomes
2 (2 )n n
A A B A
r kC C kC +
!= =
t (h)
CA(mol/dm3)
0
1.0
5-9
1.389
0.816
2.78
0.707
16.66
0.37
(1 (2 )) (1 (2 )) ( 1 ) ( 1 )
(0.1)
1 1
(1 (2 )) ( 1 ))
n n n n
AO A A
C C C
t
k n k n
!+!+! ! ! !
!!
= =
!+! !
POLYMATH Results
Nonlinear regression (L-M)
Model: t = (1^(-1-n)-Ca^(-1-n))/(k*(-1-n))
Variable Ini guess Value 95% confidence
n 3 0.8319298 0.0913065
Nonlinear regression settings
Precision
R^2 = 0.9999078
R^2adj = 0.9998848
Rmsd = 0.0233151
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
there are 3 moles of gas produced, the final pressure should be 3 times that of the initial pressure.
P5-8 (b)
Constant volume reactor at T = 504°C = 777 K
Data for the decomposition of dimethylether in a gas phase:
Time
0
390
777
1195
3155
!
5-10
01
A
y=
3 1 2
!
="=
02
A
y
! "
= =
( )
0
0 0
1
P
V V X V
P
!
" #
=$=
% &
' (
because the volume is constant.
( )
01P P X
!
= +
at t = ∞, X = XAF = 1
0
0
1A
A
A
N
dN dX r
V dt V dt
=!=
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 =!
and
0
0
P P
X
P
!
"
=
Therefore:
0
1dX dP
dt P dt
!
=
[ ]
( )
0
0
0 0 0
11 1
P P
dP k
k P P
P dt P P
!
! ! !
" #
$
=$= + $
% &
' (
or
[ ]
( )
0
1
dP k P P
dt
!
= + "
[ ]
000
1
P t
P
dP kdt
P P
!
=
+"
# #
Integrating gives:
( )
0 0
0 0
2624
ln ln ln
1 3 936
P P kt
P P P P P
!
!
" # " # " #
= = =
$ % $ % $ %
+& & &
' (
$ % ' (
' (
Therefore, if a plot of ln
624
936 P!
versus time is linear, the reaction is first order. From the figure below,
Therefore the rate law is:
5-11
y = 0.00048x - 0.02907
-0.4
0
0.4
0.8
1.2
1.6
0 1000 2000 3000 4000
P5-8 (c) Individualized solution
P5-8 (d) The rate constant would increase with an increase in temperature. This would result in the
pressure increasing faster and less time would be need to reach the end of the reaction. The opposite is true
fro colder temperatures.
P5-9
Photochemical decay of bromine in bright sunlight:
P5-9 (a)
Mole balance: constant V
A
A A
dC r kC
dt
!
= = "
Differentiation
T (min)
10
20
30
40
50
60
Δt (min)
10
10
10
10
10
10
5-12
After plotting and differentiating by equal area
-dCA/dt
0.082
0.061
0.042
0.030
0.0215
0.014
ln(-dCA/dt)
-2.501
-2.797
-3.170
-3.507
-3.840
-4.269
ln CA
0.896
0.554
0.207
-0.128
-0.478
-0.821
P5-9 (b)
A
A B
dN Vr F
dt = =
0.0344 0.0344
min min
A
ppm mg
rl
=!=!
at CA = 1 ppm
5-13
( ) min 1 3.7851 1
25000 0.0344 60 0.426
min 1000 1 453.6
B
mg g l lbs lbs
F gal l hr mg gal g hr
! " ! "! "! "
! "
= =
# $ # $# $# $
# $
% &
% & % &% &% &
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
Assume the rate law as:
n
A
AkC
dt
dC !=
Integrating:
!
!
#
$
$
&'
='' 1
1
11
1
n
n
t
1
0.025
4.1
1.410987
-3.6888795
3
0.01
9.8
2.2823824
-4.6051702
5
0.075
1.3
0.26236426
-2.5902672
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
a1 -1.0128699 0.0492329
Regression including free parameter
Number of observations = 5
Statistics
R^2 = 0.9993004
R^2adj = 0.9990673
5-14
POLYMATH Results
Nonlinear regression (L-M)
Precision
R^2 = 0.9986943
R^2adj = 0.9982591
Rmsd = 0.0531391
5-15
P5-10 (b)
We know,
!
!
"
#
$
$
%
&
'
'
='
'
1
0
2/1
11
)1(
)12(
(
(
(
A
C
t
k
Solving for k at 110° C
( )
(2 1)
(2 1)
2 1 1
' 20
2(2 1) 0.025 .min
lt
kgmol
!
!
!" #
= =
$ %
!& '
From these values,
2
1
1 2
20
ln 8.314 ln
. 10.5
1 1
1 1
373 383
kJ
R
kmol K
E
K K
T T
! "
# $
% &
= = =
! " ! "
'
'# $
# $ % &
% &
76.5 kJ/mol
P5-11
The values of k1 and k2 may depend on your initial guess. Look for the lowest s2. You could try
"rO3=kCBu
COZ
O3 + wall loss of O3 k1
O3 + alkene products k2
Rate law:
3
O
r!
=
3
O
dC
dt
= k1 + k2
Bu
Oz
C
C
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
1 1.5e-7 0.01 1e-12
3 3.5e-7 0.015 1e-10
5 8.8e-7 0.001 1e-08
POLYMATH Results
Nonlinear regression (L-M)
Model: Ozra = k1+k2*Cbu/CO3
Variable Ini guess Value 95% confidence
5-16
Nonlinear regression settings
Max # iterations = 300
Precision
R^2 = 0.7572693
R^2adj = 0.6965866
Rmsd = 4.531E-08
Rate law:
( ) ( )
3
3
7 3
3.5 10 0.05 / .
Bu
O
O
C
r mol dm s
C
!
!="+
P5-12
Given: Plot of percent
decomposition of NO2 vs V/FA0
2
% Decomposition of NO
X= 100
Assume that
n
A A
r kC!=
X has a linear relationship with
0A
V
F
as
shown in the figure.
Therefore the reaction is zero order.
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
AC = cross-sectional area
ρS = silicon dioxide density
F
F
V
N = moles of HF=
100 MW
w
!
w = weight percentage of HF in solution
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
dt =
100
C S
S F
Ad wV
k V
MW dt V MW
!
"#
$ %
&=' (
) *
100
S
C S F
kMW
dVw
dt A MW
!
!
!
" #
#
$ %
&=' (
) *
dw
dt
!
"#
$=
where
100
S
C S F
kMW V
A MW
!
!
"
#"
$ %
=& '
( )
ln ln ln
dw
dt
!" #
$ %
&= +
' (
) *
ln d
dt
!
" #
$
% &
' (
-16.629
-15.425
-14.32
-13.816
-13.479
ln w
2.079
2.996
3.497
3.689
3.871
where
d
dt
!
" #
$
% &
' (
is in
m
min
From linear regression between
ln d
dt
!
" #
$
% &
and ln w we have:
1.775
1.775
100
S
C S F
kMW V
A MW
!
"!
# $
=% &
' (
( )( )( )( )
6 2
10 *10 10 2 sides 1000 wafers 0.2
C
A m m m
!
= =
