7-1$
Solutions)for)Chapter)7)–)Collection)and)Analysis)of)Rate)Data)
)
P7-1)(a))Example)7-4)
rate$law:$
4 2
CH CO H
r kP P
α β
=
$
Regressing$the$data$
r’(gmolCH4/gcat.min)
PCO (atm)
PH2 (atm)
5.2e-3
1
1
13.2e-3
1.8
1
30e–3
4.08
1
4.95e-3
1
0.1
7.42e-3
1
0.5
5.25e-3
1
4
)
See$Polymath$program$P7-1-a.pol.$
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
R^2 = 0.9869709
$
Therefore$order$of$reaction$=$1.14$
)
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
R^2 = –0.8194508
$
)
P7-2)Individualized$solution$
$
)
P7-3)The$key$for$decoding$the$algorithm$to$arrive$at$a$numerical$score$for$the$Interaction$Computer$
Games$(ICGs)$is$given$at$the$front$of$this$Solutions$Manual.$
P7-4)(a))
For$the$reaction,$HbO2$→$Hb$+$O2$
Limiting$Reactant$ $ A,$or$HbO2$
7-3$
P7–4)(b))Continued$
POLYMATH)Results
Polynomial)Regression)Report
a0 2.918E–14 0
a1 0.0040267 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
Model: ln(dxdz) = a0 + a1*ln(1–X)
Variable Value 95% confidence
n$=$1.28$
$
7-4$
P7-5)
Constant$volume$batch$reactor:))))))))))))))))))))))))A!)B)
Mole$balance:$
$
A
A
dC kC
α
−=
$
Integrating$with$initial$condition$when$t$=$0$and$CA$=$CAO$for$
1.0
α
≠
$
(1 ) (1 )
1
(1 )
AO A
C C
t
k
α α
α
− −
−
=−
……………substituting$for$initial$concentration$CAO$=$4$mol/dm3.$
t (min.)
CA (mol/dm3)
0
4
5
2.25
8
1.45
10
1
12
0.65
15
0.25
17.5
0.06
20
0.008
$
See$Polymath$program$P7-5.pol.$
POLYMATH Report
Nonlinear Regression (L–M)
Model: t = ((4^(1–alpha))–Ca^(1–alpha))/(k*(1–alpha))
Variable
Initial guess
Value
95% confidence
alpha
2.
0.532479
0.034828
k
1.
0.1963504
0.0031774
Nonlinear regression settings
Max # iterations = 64
Precision
R^2
0.9989802
R^2adj
0.9988102
Rmsd
0.0699673
Variance
0.0522179
General
Sample size
8
Model vars
2
Indep vars
1
Iterations
11
7-5$
P7–5)Continued$
Source data points and calculated data points
Ca
t
t calc
Delta t
1
4
0
0
0
2
2.25
5
4.912288
0.0877122
3
1.45
8
7.867621
0.1323793
4
1
10
9.934279
0.0657208
5
0.65
12
11.92141
0.0785877
6
0.25
15
15.13018
–0.1301777
7
0.06
17.5
17.90411
–0.4041085
8
0.008
20
19.688
0.3120008
$
Hence,$
k=$0.2$(mol/dm3)0.5min-1$and$α$=$0.5$
$
$
P7-6)(a))
Liquid$phase$irreversible$reaction:$
A$$!$$B$$+$$C$;$CAO$=$2$mole/dm3$
Space time (
τ
)min.
CA(mol/dm3)
ln(CA)
ln((CAO–CA)/
τ
)
15
1.5
0.40546511
–3.4011974
38
1.25
0.22314355
–3.9252682
100
1.0
0
–4.6051702
300
0.75
–0.28768207
–5.4806389
1200
0.5
–0.69314718
–6.6846117
$
By$using$linear$regression$in$polymath:$
See$Polymath$program$P7-5-a.pol.$
$
)
7-6$
P7-6)(a))Continued$
POLYMATH)Results
$Linear)Regression)Report
Model: y = a0 + a1*lnCa
ln ln ln
AO A
A
C C k C
α
−
⎛ ⎞ = +
⎜ ⎟
R^2 = 0.9999443
3slope
α
=≡
$
ln(k)$=$intercept$=$–4.6$
$
P7-6)(b))Individualized$solution$
$
P7-6)(c))Individualized$solution$
$
)
P7-7)(a))
Constant$voume$batch$reactor:))))))))))))))))))))))))A!)B)+C)
Mole$balance:$
$
A
A
dC kC
α
−=
$
Integrating$with$initial$condition$when$t$=$0$and$CA$=$CAO$for$
1.0
α
≠
$
(1 ) (1 ) (1 ) (1 )
(2)
1 1
(1 ) (1 )
AO A A
C C C
t
k k
α α α α
α α
− − − −
−−
= =
− −
……………substituting$for$initial$concentration$CAO$=$2$mol/dm3.$
$
t (min.)
CA (mol/dm3)
0
2
5
1.6
9
1.35
15
1.1
22
0.87
30
0.70
40
0.53
60
0.35
)
See$Polymath$program$P7-7-a.pol.$
7-7$
P7-7)(a))Continued$
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=$0.03$(mol/dm3)–0.5min-1$and$
1.5
α
=
$$
$
P7-7)(b))Individualized$solution)
P7-7)(c))Individualized$solution$
P7-7)(d))Individualized$solution$
$
$
P7-8)(a))
At$t$=$0,$there$is$only$(CH3)2O.$At$t$=$∞,$there$is$no(CH3)2O.$Since$for$every$mole$of$(CH3)2O$consumed$
there$are$3$moles$of$gas$produced,$the$final$pressure$should$be$3$times$that$of$the$initial$pressure.$
$
P7-8)(b))
Constant$volume$reactor$at$T$=$504°C$=$777$K$
Data$for$the$decomposition$of$dimethylether$in$a$gas$phase:$
Time
0
390
777
1195
3155
∞
PT(mm Hg)
312
408
488
562
799
931
$
COHCHOCH ++→2423 )(
$$
01
A
y=
$
3 1 2
δ
=−=
$
02
A
y
ε δ
= =
$
( )
0
00
1
P
VV X V
P
ε
⎛⎞
=+=
⎜⎟
⎝⎠
$$because$the$volume$is$constant.$
( )
01P P X
ε
= +
$
at$t$=$∞,$X$=$XAF$=$1$
0
0
1A
A
A
N
dN dX r
V dt V dt
=−=
$
7-8$
P7-8)(b))Continued$
Assume$
A A
r kC−=
$(i.e.$1st$order)$
000
Integrating$gives:$
( )
0 0
0 0
2624
ln ln ln
1 3 936
P P kt
P P P P P
ε
ε
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= = =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
+− − −
⎣ ⎦
⎢ ⎥ ⎣ ⎦
⎣ ⎦
$
Therefore,$if$a$plot$of$ln
624
$versus$time$is$linear,$the$reaction$is$first$order.$From$the$figure$below,$
$
y = 0.00048x – 0.02907
-0.4
0
0.4
0.8
1.2
1.6
0 1000 2000 3000 4000
7-9$
P7-8)(c))Individualized$solution$
$
P7-8)(d))The$rate$constant$would$increase$with$an$increase$in$temperature.$This$would$result$in$the$
)
P7-9)(a))
Photochemical$decay$of$bromine$in$bright$sunlight:$
t (min)
10
20
30
40
50
60
CA (ppm)
2.45
1.74
1.23
0.88
0.62
0.44
Mole$balance:$constant$V$
dCA
dt =rA=−kCA
α
$
ln −dCA
dt
“
#
$
$
%
&
‘
‘=ln k
( )
+αln CA
( )
$
Differentiation$
T (min)
10
20
30
40
50
60
Δt (min)
10
10
10
10
10
10
CA (ppm)
2.45
1.74
1.23
0.88
0.62
0.44
ΔCA (ppm)
–0.71
–0.51
–0.35
–0.26
–0.18
ppm
min
A
C
t
Δ⎛ ⎞
⎜ ⎟
Δ⎝ ⎠
–0.071
–0.051
–0.035
–0.026
–0.018
$
$
$
$
$
$
7-10$
P7-9)(a))continued$
$
$
After$plotting$and$differentiating$by$equal$area$
–dCA/dt
0.082
0.061
0.042
0.030
0.0215
0.014
ln(–dCA/dt)
–2.501
–2.797
–3.170
–3.507
–3.840
–4.269
ln CA
0.896
0.554
0.207
–0.128
–0.478
–0.821
$
Using$linear$regression:$α$=$1.0$
ln$k$=$–3.3864$
$
P7-9)(b))
dNA
dt =−VrA−F
B=0**,**F
B=–rAV
$
0.0344 0.0344
min min
A
ppm mg
rl
=−=−
$at$CA$=$1$ppm$
( ) min 1 3.7851 1
25000 0.0344 60 0.426
min 1000 1 453.6
B
mg g l lbs lbs
F gal l hr mg gal g hr
⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞
⎛ ⎞
= =
⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠
$
P7-9)(c)$Individualized$solution$
$
)
P7-10)
For$the$reaction,$$ Oz$+$wall$!$loss$of$O3$$$$$$$k1$
$Oz$+$alkene$!$products$$$$$$k2$
7-11$
P7–10)Continued$
See$Polymath$program$P7-10.pol.$
$
)
)
7-12$
P7-11)
Given:$Plot$of$percent$decomposition$of$NO2$vs$V/FA0$
X= %DecompositionofNO2
100
$
Assume$that$
n
A A
r kC−=
$
Therefore$the$reaction$is$zero$order.$
$
)
P7-12))
)
$$$See$Polymath$program$P7-12.pol)
)
)
7-13$
P7-12)Continued$
$