4-1$
Solutions)for)Chapter)4)–)Stoichiometry)
)
P4-1)(a))Example)4-4))
(i)$The$critical$value$is$around$0.5$atm$
(iv)$Individualized$Solution$
$
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
epsilon
–0.14
–0.14
–0.14
–0.14
2
Fao
3.
3.
3.
3.
3
k
9.7
9.7
9.7
9.7
4
KO2
38.5
38.5
38.5
38.5
5
KP
930.
930.
930.
930.
6
KSO3
42.5
42.5
42.5
42.5
7
PO2
2.214
1.476695
2.214
1.476695
8
PSO2
4.1
2.605932
4.1
2.605932
9
PSO20
4.1
4.1
4.1
4.1
10
PSO3
0
0
1.737288
1.737288
11
ra
–0.5651727
–0.5651727
–0.0045241
–0.0045241
12
t
0
0
0.4
0.4
13
thetaB
0.54
0.54
0.54
0.54
14
X
0
0
0.4
0.4
15
yaxis
5.308112
5.308112
663.1111
663.1111
Differential equations
1
d(X)/d(t) = 1
Explicit equations
1
Fao = 3
2
PSO20 = 4.1
3
epsilon = –0.14
4
PSO2 = PSO20* (1–X)/(1+epsilon*X)
5
PSO3 = PSO20* X/(1+epsilon* X)
6
thetaB = 0.54
7
PO2 = PSO20* (thetaB–X/2)/(1+epsilon* X)
4-2$
8
k = 9.7
9
KO2 = 38.5
10
KSO3 = 42.5
11
KP = 930
12
ra = –k *(PSO2* Sqrt(PO2)–PSO3/KP)/(1+Sqrt(PO2* KO2)+PSO3 *KSO3)^2
13
yaxis = Fao/(–ra)
General
Total number of equations
14
Number of differential equations
1
Number of explicit equations
13
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps
0.000001
$
(vi)$See$Polymath$program$P4-1-avi-1.pol.$
POLYMATH Report
Nonlinear Equation
05–Jun-2017
Calculated values of NLE variables
Variable
Value
f(x)
Initial Guess
1
X
0.997577
–1.251E–09
0.5 ( 0 < X < 1. )
Variable
Value
1
epsilon
–0.14
2
k
9.7
3
KO2
38.5
4
KP
930.
5
KSO3
42.5
6
PO2
0.1963961
7
PSO2
0.0115472
8
PSO20
4.1
9
PSO3
4.754015
10
thetaB
0.54
Nonlinear equations
1
f(X) = –k *(PSO2* Sqrt(PO2)–PSO3/KP)/(1+Sqrt(PO2* KO2)+PSO3 *KSO3)^2 = 0
Explicit equations
4-3$
1
PSO20 = 4.1
2
epsilon = –0.14
3
PSO2 = PSO20* (1–X)/(1+epsilon*X)
4
thetaB = 0.54
5
PSO3 = PSO20* X/(1+epsilon* X)
6
k = 9.7
7
KO2 = 38.5
8
KSO3 = 42.5
9
KP = 930
10
PO2 = PSO20* (thetaB–X/2)/(1+epsilon* X)
General Settings
Total number of equations
11
Number of implicit equations
1
Number of explicit equations
10
Elapsed time
0.0000 sec
Solution method
SAFENEWT
Max iterations
150
Tolerance F
0.0000001
Tolerance X
0.0000001
Tolerance min
0.0000001
$
Hence,$Xe$=$0.9976$
$
See$Polymath$program$P4-1-avi-2.pol.$
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
epsilon
–0.14
–0.14
–0.14
–0.14
2
Fao
1000.
1000.
1000.
1000.
3
k
9.7
9.7
9.7
9.7
4
KO2
38.5
38.5
38.5
38.5
5
KP
930.
930.
930.
930.
6
KSO3
42.5
42.5
42.5
42.5
7
PO21
2.214
1.669102
2.214
1.669102
8
PO22
2.214
0.2197559
2.214
0.2197559
9
PSO20
4.1
4.1
4.1
4.1
10
PSO21
4.1
2.995825
4.1
2.995825
11
PSO22
4.1
0.0588832
4.1
0.0588832
12
PSO31
0
0
1.283925
1.283925
13
PSO32
0
0
4.698973
4.698973
14
ra1
–0.5651727
–0.5651727
–0.0092831
–0.0092831
15
ra2
–0.5651727
–0.5651727
–5.276E–06
–5.276E–06
16
t
0
0
0.3
0.3
17
thetaB
0.54
0.54
0.54
0.54
18
X1
0
0
0.3
0.3
4-4$
19
X2
0
0
0.987624
0.987624
20
Xe
0.9976
0.9976
0.9976
0.9976
21
yaxis1
1769.371
1769.371
1.077E+05
1.077E+05
22
yaxis2
1769.371
1769.371
1.895E+08
1.895E+08
Differential equations
1
d(X1)/d(t) = 1
2
d(X2)/d(t) = 0.99*Xe/0.3
Explicit equations
1
Fao = 1000
2
PSO20 = 4.1
3
epsilon = –0.14
4
PSO21 = PSO20* (1–X1)/(1+epsilon*X1)
5
PSO31 = PSO20* X1/(1+epsilon* X1)
6
thetaB = 0.54
7
PO21 = PSO20* (thetaB–X1/2)/(1+epsilon* X1)
8
k = 9.7
9
KO2 = 38.5
10
KSO3 = 42.5
11
KP = 930
12
ra1 = –k *(PSO21* Sqrt(PO21)–PSO31/KP)/(1+Sqrt(PO21* KO2)+PSO31 *KSO3)^2
13
yaxis1 = Fao/(–ra1)
14
Xe = 0.9976
15
PSO22 = PSO20* (1–X2)/(1+epsilon*X2)
16
PSO32 = PSO20* X2/(1+epsilon* X2)
17
PO22 = PSO20* (thetaB–X2/2)/(1+epsilon* X2)
18
ra2 = –k *(PSO22* Sqrt(PO22)–PSO32/KP)/(1+Sqrt(PO22* KO2)+PSO32 *KSO3)^2
19
yaxis2 = Fao/(–ra2)
$
At$X$=$0.3$
FA0/-rA’$=$107,700$$
W$=$(FA0/-rA’)X$at$X$=$0.3$
(i)$
0.0
0.2
0.4
0.6
0.8
1.0
yA0
0.0
0.2
0.4
0.6
0.8
1.0
X
$$
4-6$
$
P4-3)(b))
For$a$liquid$phase$reaction,$CA$=$1.6$mol/dm3$$
P4-3)(c))
CA$=$CA0(1–X)/(1+$εX)$$
CB$=$CA0(0.5*X)/(1+$εX)$$$
P4-3)(d))
The$type$of$reactor$does$not$change$the$expression$for$-rA.$
)
P4-4)
$ $
1
1
1
1
2
1
2
1
→+
CBA
P4–4)(a))
CB0 =CA0 =0.1 mol
dm 3
$
)
P4–4)(b))
CA=CA0
1−X
( )
1+εX
( )=0.1
( ) 1−0.25
( )
1−1
2
X
⎛
⎝
⎜ ⎞
⎠
⎟
=
0.1
( ) 0.75
( )
0.875 =0.086 mol
dm 3
$
( ) ( ) ( )( )
03
0.25
0.1 0.125
22
0.1 0.0143
1
10.875
1
2
CA
X
mol
CC
Xdm
X
ε
⎛⎞ ⎛ ⎞
⎜⎟ ⎜ ⎟
⎝⎠ ⎝ ⎠
== ==
+⎛⎞
−
⎜⎟
⎝⎠
$
)
P4–4)(c))
( )
3
0
0
01.0
2
1
1
2
1
1
1
2
1
1
dm
mol
C
X
XC
X
X
CC A
A
AB ==
⎟
⎠
⎞
⎜
⎝
⎛−
⎟
⎠
⎞
⎜
⎝
⎛−
=
+
⎟
⎠
⎞
⎜
⎝
⎛−
==
ε
$
)
P4–4)(d))
4-7$
CB=0.1 mol
dm 3
$
P4–4)(e))
rA
−4=rC
2, rA=−2rC=−4mol dm 3min
$
)
P4–4)(f))
2A$+$B$!$C$
$
)
P4–5)(a)$$
Liquid$phase$reaction,)
))))))))))))))))))))))))))O$$$$$$$$$$$$$$$$$$$$$$CH2—OH$
))))))))))))))))))))))))))))))))))
)))))))))))))))CH2$-)CH2$$+$$H2O$→))CH2—OH))
$$$$$$$$$$$$$$$$$$$$$$$A$$$$$$$$+$$$$B$$$→$$$$$C$
CAO$=$16.13mol/dm3$ $ $ CBO$=$55.5$mol/dm3$
$
P4–5)(a)$Continued$
Stoichiometric$Table:$
Species$
Symbol$
Initial$
Change$
Remaining$
$$$$Ethylene$
oxide$
$$$$A$
CAO=16.13$mol/dm3$
-$CAOX$
CA=$CAO(1–X)$
$$$$$=$(1–X)$mol/dm3$
$$$$Water$
$$$$B$
CBO=$55.5$mol/dm3,$
B
θ
=3.441$
-CAOX$
CB$=$CAO(
B
θ
–X)$
$$$$$=(3.441-X)$mol/dm3$
$$$$Glycol$
$$$$C$
0$
$CAOX$
CC$=$CAOX$mol/dm3$
$
Rate$law:$ -rA$=$kCACB$
Therefore,$ -rA$=$k$
2
AO
C
(1–X)$(
B
θ
–X)$=$k$(16.13)2(1–X)$(3.441-X)$$$
At$300K$$$E$=$12500$cal/mol,$X$=$0.9,$
k$=$0.1dm3/mol.s$)
4-8$
CSTR
τ
$=$
AO
A
C X
r−
$=$
( )( )
( )( ) ( )( )
2
16.13 0.9
2.196sec
0.1 16.13 1 0.9 3.441 0.9
=
−−
)
and,$V$=$τ$*vo$=$2.196$sec$X$200$liters/sec$=$439.2$liters$
)
At$350K,$
k2$=$k$exp((E/R)(1/T–1/T2))=$0.1exp((12500/1.987)(1/300–1/350))$
$$$$$=$1.99$dm3/mol.s$
Therefore,$
$
CSTR
τ
$=$
AO
C X
r−
$=$
( )( )
2
16.13 0.9
0.110 sec
=
,$$
P4–5)(b)$$
Isothermal,$isobaric$gas–phase$pyrolysis,$
$$$$$$$$$$$C2H6$$$$$$C2H4$$+$$$H2$
Stoichiometric$table:$
Species$
symbol$
Entering$
Change$
Leaving$
C2H6$
$$$A$
FAO$
-FAOX$
FA=FAO(1–X)$
C2H4$
$$$B$
0$
+FAOX$
FB=FAOX$
H2$
$$$C$
0$
+FAOX$
FC=FAOX$
$
$
FTO=FAO$
$
FT=FAO(1+X)$
$
ε
$=$yao
δ
=$1(1+1–1)$=$1$
v$=$vo(1+
ε
X)$ $ =>$v$=$vo(1+X)$
P4–5)(b)$Continued$
CAO$=$yAO$CTO$=$yAO$
P
$
$
Rate$law:$
4-9$
-rA$=$kCA=$kCAO
( )
1
X
−
$$=0.067$k
( )
1
X
−
$
)
P4–5)(c))
Isothermal,$isobaric,$catalytic$gas$phase$oxidation,$
C2H4$$$+$$$
1
2
O2$$$“$$C2H4O$
pA$=$yA0P$=$yA0CRT$=$CART$
$
(1)$$For$a$fluidized$batch$reactor,$V$is$constant$
$ CA$=$CA0(1–X)$
$
P4–5)(c))Continued$
$(2)$Stoichiometric$table:$
Species$
Symbol$
Entering$
Change$
Leaving$
C2H4$
A$
FAO$
-FAOX$
FA=FAO(1–X)$
O2$
B$
FBO$
–
1
2
FAOX)
FB=FAO(
B
θ
-X/2)$
C2H4O$
C$
0$
+FAOX)
FC=FAOX$
)
B
θ
=$
1
1
2
2
AO
BO
AO AO
F
F
F F
= =
$$
2
3
AO AO
AO
TO AO BO
F F
yF F F
= = =
+
$
$
( )
( )
3
3
6
20.092
3.
0.082 533
.
AO AO TO AO
atm
P mol
C y C y RT dm
atm dm K
mol K
= = = =
⎛ ⎞
⎜ ⎟
⎝ ⎠
$
( )
( )
( )
( )
( )
( )
1 1 0.092 1
1 1 0.33 1 0.33
AO AO
A
A
O
F X C X X
F
C
v v X X X
ε
− − −
= = = =
+− −
$
4-10$
( )
( )
( )
0.046 1
2
1 1 0.33
AO B
B
B
O
X
FX
F
C
v v X X
θ
ε
⎛ ⎞
−
⎜ ⎟ −
⎝ ⎠
= = =
+−
$
( )
( )
( )
0.092
1 1 0.33
C AO
C
O
X
F F X
C
v v X X
ε
= = =
+−
$
If$the$reaction$follow$elementary$rate$law$
Rate$law:$-rA’$=$kACACB
0.5(RT)1.5,$T$=$533$K$
⇒
( )
( )
0.5
0.092 1 0.046 1
A
X X
r k
⎧ ⎫⎧ ⎫
− −
⎪ ⎪⎪ ⎪
−=⎨ ⎬⎨ ⎬
$
(RT)1.5$$
P4–5)(d)$$
Isothermal,$isobaric,$catalytic$gas–phase$reaction$in$a$PBR$
C6H6$+$2H2$→$C6H10$
)$$A$$$+$$2B$$→$$$$C)
Given:$
min/50 3
0dm=
ν
$
$
Stoichiometric)Table:)
Species$
Symbol$
Entering$
Change$
Leaving$
C6H6$
A$
FA0$
-FA0X$
FA=FA0(1–X)$
H2$
B$
FB0=2FA0$
–2FA0X$
FB=FA0(θB–2X)$
C6H10$
C$
0$
FA0X$
Fc=FA0X$
)
P4–5)(d)$Continued$
2
2
0
0
0
0===
A
A
A
B
BF
F
F
F
θ
)))))))
3
1
00
0
0
0
0=
+
==
BA
A
T
A
AFF
F
F
F
y
)))))))
3
2
)121(
3
1
0−=−−==
δε
A
y
)
3
000 055.0
3
1
)443(*0821.0
6
3
1
3dm
mol
K
atm
RT
P
yCC
Kmol
atmdm
ATA =
⎟
⎠
⎞
⎜
⎝
⎛
=
⎟
⎠
⎞
⎜
⎝
⎛
==
⋅
⋅
)
)1(
)1(
)1(
)1(
3
2
0
0
0
X
X
C
X
XF
F
CA
A
A
A−
−
=
+
−
==
ενν
$
)1(
)1(
*2
)1(
)22(
)1(
)2(
3
2
0
3
2
0
0
0
X
X
C
X
X
C
X
XF
F
CAA
BA
B
B−
−
=
−
−
=
+
−
==
εν
θ
ν
$
)1()1( 3
2
0
0
0
X
X
C
X
XFF
CA
AC
C−
=
+
==
ενν
$
)
Rate)Law:)
NOTE:$$For$gas–phase$reactions,$rate$laws$are$sometimes$written$in$terms$of$partial$pressures$instead$of$
concentrations.$$The$units$of$the$rate$constant,$k,$will$differ$depending$on$whether$partial$pressure$or$
concentration$units$are$used.$$See$below$for$an$example.$
2
‘BAA PkPr=−
$
[ ] 2
3**
min
min atmatm
atmkgcat
mol
kgcat
mol
⋅⋅
=
⋅
$
4-11$
Notice$that$if$you$use$concentrations$in$this$rate$law,$the$units$will$not$work$out.$
RTCRTCyPyP AAAA === *)( 00
$
3
3
3
2
3
3
0
3
22 )(
)1(
)1(
4)( RT
X
X
kCRTCkCPkPrABABAA −
−
===−
$
)
Design)Equation)for)a)fluidized)CSTR:)
‘
0
A
A
r
XF
W
−
=
)
33
3
0
3
3
2
0
)()1(4
)1(
RTXkC
XXF
W
A
A
−
−
=
)
33
2
0
3
3
2
0
)()1(4
)1(
RTXkC
XX
W
A−
−
=
ν
)
)
Evaluating)the)constants:)
Kat
atmkgcat
mol
k300
min
53 3
⋅⋅
=
$
P4–5)(d)$Continued$
At$170°C$(443K),$
Kmol
⎦
⎣
⋅
314.8
Plugging$in$all$the$constants$into$the$design$equation:$
X$=$0.8$
dm
atmkgat
min
⋅⋅
At$270°C$(543K),$
atmkgcat
mol
KK
Kmol
J
mol
J
TTR
E
kk A
⋅⋅
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛−
⋅
=
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛−=min
90790000
543
1
300
1
314.8
80000
exp53
11
exp
543300
300543
$
Plugging$in$all$the$constants$into$the$design$equation:$
X$=$0.8$
dm
atmkgat
min
⋅⋅
$
)
P4–6)(a)))
Let$$$$A$=$ONCB$C$=$Nibroanaline$
4-12$
$B$=$NH3$D$=$Ammonium$Chloride$$$$$$$
)
P4–6)(b))
Species)
Entering)
Change)
Leaving)
A$
FA0$
-$FA0X$
FA0(1–X)$
B$
FB0$=$ΘBFA0$
$$$$$$$$$=6.6/1.8$FA0$
–2$FA0X$
FB=$
FA0(ΘB$–$2X)$
C$
0$
FA0X)
FC=FA0X$
D$
0$
FA0X$
FD=FA0X$
)
P4–6)(c))
For$batch$system,$
$$$$$CA=NA/V$ $ $ -rA$=$kNANB/V2$
)
P4–6)(d))
–A A B
r kC C=
$
( ) ( ) ( )
0
0 0
0 0 0
1 1 , 1
A
A A A A
A A A A
N
N N F F
F X C X C C X
V V V v v
= = = −=−= = = −
$
( ) ( ) ( )
0
0 0
0 0 0
2 2 , 2
A
B B B
B B A B B A B
N
N N F
F X C X C C X
V V V v
θ θ θ
= = = −=−= = −
$
( )( )
2
01 2
A A B
r kC X X
θ
−=− −
)
0
0
6.6 3.67
1.8
B
B
A
C
C
θ
= = =
$
03
1.8
A
kmol
C
m
=
$
( ) ( )( )
2
1.8 1 3.67 2
A
r k X X−=− −
$
)
P4–6)(e))
1)$$At$X$=$0$and$T$=$188°C$=$461$K$
min
0202.067.38.1
min
0017.0 3
2
3
3
2
00 m
kmol
m
kmol
kmol
m
kCrBAA =
⎟
⎠
⎞
⎜
⎝
⎛
=Θ=−
$
03
kmol
0.0202
m min
A
r−=
)
)
2))At$X$=$0$and$T$=$250C$=$298K$
$$$
⎟
⎟
⎞
⎜
⎜
⎛
⎟
⎟
⎞
⎜
⎜
⎛−=TTR
E
kk
O
11
exp
$
4-13$
$$
min.
3
1003.2
298
1
461
1
.
987.1
11273
exp
min.
0017.0
6
3
kmol
m
Kmol
cal
mol
cal
kmol
m
k
−
×=
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛−=
$
2
3
6
33
m
2.03 10 1.8 3.67
kmol min
A
kmol kmol
r
mm
−⎛⎞⎛ ⎞
−=×⎜⎟⎜ ⎟
⎝⎠⎝ ⎠
$
-rAO$=$kCAO
aCBO$=$2.41$X$10-5$kmol/m3min$
)
3))
0
0
1 1
exp E
k k
R T T
⎡ ⎤
⎛ ⎞
=−
⎢ ⎥
⎜ ⎟
⎢ ⎥
⎝ ⎠
⎣ ⎦
$
P4-6)(e)$Continued$
311273
m 1 1
0.0017 exp
kmol min 461 561
1.987
cal
mol
kcal K K
mol K
⎡ ⎤
⎢ ⎥
⎛ ⎞
=⎢ − ⎥
⎜ ⎟
⎢ ⎥
⎝ ⎠
⎢ ⎥
⎣ ⎦
$
3
m
0.0152
kmol min
k=
)
2
000AAB
rkCC−=
$
2
3
33
m
0.0152 1.8 3.67
kmol min
A
kmol kmol
r
mm
⎛⎞⎛ ⎞
−=⎜⎟⎜ ⎟
⎝⎠⎝ ⎠
$
3
kmol
0.1807
m min
A
r−=
$
)
P4–6)(f))
rA$=$kCAO
2(1–X)(θB–2X)$
At$X$=$0.90$and$T$=$188C$=$461K$
1))at$T$=$188$C$=$461$K$
( ) ( )( )
min
00103.0
9.0267.39.018.1
min.
0017.0
3
2
3
3
m
kmol
m
kmol
kmol
m
rA
=
−−
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=−
$
2))at$X$=$0.90$and$T$=$25C$=$298K$
$$$$$$$$ $
( ) ( )( )
min
1023.1
9.0267.39.018.1
min.
1003.2
3
6
2
3
3
6
m
m
kmol
kmol
m
rA
−
−
×=
−−
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛×=−
$
4-14$
3))at$X$=$0.90$and$T$=$288C$=$561K$
( ) ( )( )
min
0092.0
9.0267.39.018.1
min.
0152.0
3
2
3
3
m
kmol
m
kmol
kmol
m
rA
=
−−
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=−
$
$
P4–6)(g)$
FAO$=$2$mol/min$
P4-6)(g)$Continued$
$$$$$$$$$$$
( )
3
9.0
1.0min/2
1
mol
r
XF
V
XA
AO
×
−
−
=
=M
$
$
P4–7)No$solution$will$be$given)$
$
)
P4-8)(a))
Isothermal$gas$phase$reaction.$
2 2 3
1 3
2 2
N H NH+→
$
Making$H2$as$the$basis$of$calculation:$
2 2 3
1 2
3 3
H N NH+→
$
1 2
3 3
A B C+→
$$$
Stoichiometric$table:$
Species$
Symbol$
Initial$
change$
Leaving$
H2$
A$
FAO$
-FAOX$
FA=FAO(1–X)$
N2$
B$
FBO=
B
θ
FAO$
-FAOX/3$
FB=FAO(
B
θ
–X/3)$
NH3$
C$
0$
+2FAOX/3$
FC=(2/3)FAOX$
$
P4-8)(b))
2 1 2
1
3 3 3
2 1
0.5
3 3
AO
y
δ
ε δ
⎛ ⎞
=− − =−
⎜ ⎟
⎝ ⎠
⎛ ⎞
= = × − =−
⎜ ⎟
⎝ ⎠
$
( )
( )
3
16.4
0.5
.
0.082 500
.
AO
atm
C
atm dm K
mol K
=⎛ ⎞
⎜ ⎟
⎝ ⎠
=$0.2$mol/dm3$
P4-8)(b)$Continued$
( )
( )
( )
( )
( )
( )
2
3
3
3
1 0.2 1 0.6
0.1 /
0.6
11
3
0.2 0.6
22 0.1 /
0.6
31 3 1
3
AO
HA
AO
NH C
CX
C C mol dm
X
CX
C C mol dm
X
ε
ε
−−
== = =
+⎛⎞
−
⎜⎟
⎝⎠
==×=×=
+⎛⎞
−
⎜⎟
⎝⎠
$
P4-8)(c))
kN2$=$40$dm3/mol.s$
))))(1)$For$Flow$system:$
))))))))))
2 2 2 2
3
1
2 2
3
1
2 2
11
N N N H
r k C C
X
⎡ ⎤ ⎡ ⎤
−=⎣ ⎦ ⎣ ⎦
⎡ ⎤ ⎡ ⎤
⎛ ⎞
−
⎜ ⎟
⎢ ⎥ ⎢ ⎥
$
1−X
3
⎝
⎜ ⎞
⎠
⎟
⎣
⎢
⎦
⎥
4-16$
)))))(2)$For$batch$system,$constant$volume.$
)))))))))))))))))))))))))
( )
( )
( ) ( )
[ ]
3
1
22
222 2
2
2
1
2
3
2
2
1
2
3
2
0
00
0
0
0
2
0
1
1
3
1
3
40 1 1
3
1.6 1 1
3
NNN H
A
AA
A
AA
AB
B
B
HO
NA
rkC C
NX
NN
C
VV V
CC X
X
N
N
C
VV
X
C
X
rC X
XX
⎡⎤⎡⎤
−=⎣⎦⎣⎦
−
===
=−
⎛⎞
Θ−
⎜⎟
⎝⎠
==
⎛⎞
=−
⎜⎟
⎝⎠
⎡⎤
⎛⎞
−=− −
⎡⎤
⎜⎟
⎢⎥
⎣⎦
⎝⎠
⎣⎦
⎡⎤
=− −
⎢⎥
⎣⎦
$
$
P4–9)$
$
(a)$P0$=$PA0$=$CA0RT$
PA$=$PA0(1-Xe)/(1$+$ɛXe)$
PB$=$PA0Xe/2(1$+$ɛXe)$
CA$=$PA$/$RT$
)
P4–10)$
No$solution$will$be$given.$
$
$
4-17$
P4–11$
No$solution$will$be$given.$
$
$
P4–12)
)
Given:$Gas$phase$reaction$A$+$B$“$8C$in$a$batch$reactor$fitted$with$a$piston$such$that$V$=$0.1P0$
( )2
3
2
1.0
sec
ft
klb mol
=
$
2
A A B
r kC C−=
$
NA0$=$NB0$at$t$=$0$
V0$=$0.15$ft3$
T$=$140°C$=$600°R$=$Constant$
$
P4–12)(a))
0
0
0 0
0.5
A
A
A B
N
yN N
= =
+
$
8 1 1 6
δ
=− − =
$
03
A
y
ε δ
= =
$
Now$$
( )
0 0
0
1
V P
V X
T
P
T
ε
= +
⎛ ⎞
⎜ ⎟
⎝ ⎠
$
$and$
0
1
T
T=
,$
0 0
10P V=
,$and$
10P V=
$
Therefore$$$$$$$$
( )
2
0
10 1
10
V
V X
V
ε
= +
$$$$$or$$$$$$
( )
2 2
01V V X
ε
= +
$
[ ]
01
A A
N N X=−
$ $
[ ]
0B A B
N N X
θ
=−
$ $
0
0
1
B
B
A
N
N
θ
= =
$
[ ]
( )
3
3
2
0
2
3
3
32
0
1
1
A
A B
A A B
kN X
kN N
r kC C
VV X
ε
−
−= = =
+
$$$$$$$$$$$$$$$$$$$$$$$
0 0
0 0
A
A
y P
N V
RT
⎛ ⎞
=⎜ ⎟
⎝ ⎠
$
Therefore$
[ ]
( )
3
0 0
3
2
1
1
A
A
X
y P
r k RT X
ε
−
⎛ ⎞
−=⎜ ⎟
⎝ ⎠ +
$
[ ]
( )
3
9
33
2
1
5.03*10
sec
1 3
A
Xlb mol
rft
X
−−
−=
+
$
$
P4–12)(b))
( )
2 2
01V V X
ε
= +
$
( )
2 2
0.2 0.15 1 X
ε
= +
$
4-18$
0.259X=
$
10
3
8.63*10
sec
A
lb mol
rft
−
−=
)
)
$