Solutions for Chapter 6 – Multiple Reactions
P6-1 Individualized solution
P6-2 (a) Example 6-2
For PFR,
2
321 AA
ACkCkk
d
dC !!!=
“
1
k
d
dC X=
!
A
BCk
d
dC
2
=
!
2
3A
YCk
d
dC =
!
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(tau) = -k1-k2*Ca-k3*Ca^2
[2] d(Cx)/d(tau) = k1
Explicit equations as entered by the user
[1] Cao = 0.4
[2] X = 1-Ca/Cao
[3] k1 = 0.0001
[4] k2 = 0.0015
(2) Pressure increased by a factor of 100.
Now CA0 = P/RT = 0.4 × 100 = 40 mol/dm3
* does not change but
2
AAO
AAO
CC
r
CC
v
V
!
=
!
!
==
“
sec
0001.000168.00001.0
112.040
++
!
=
6-3
P6-2 (c) Example 6-4
(1) For k1 = k2, we get
( )
‘
1
exp
!
kCC AOA “=
and
( )
‘
111
‘exp
!
!
kCkCk
d
dC
AOB
B“=+
!
( )( )
AO
BCk
kCd
1
‘
‘
1
exp =
!
Now at
0
‘=
!
, CB = 0
!
( )
‘
1
‘
1exp
!!
kCkC AOB “=
Optimum yield:
( ) ( )
[ ]
‘
1
‘
1
‘
11
‘expexp0
!!!
kkkCk
dC
AO
B“““==
(2) For a CSTR:
A
AA
A
AA
Ck
CC
r
CC
1
0
‘
0!
=
!
!
=
“
#
01 AAA CCCk =+
!
“
( )
1
1
0
+
!
=k
C
CA
A
“
[ ]
[ ]
BA
B
BA
B
B
BB
CkCk
C
CkCk
C
r
CC
21
21
‘
00
!
=
!!
!
=
!
!
=
“
#
BBA CCkCk =
!
“
!21
##
( ) ( )( )
111 1
0
2
1
2
1
+
!
+
!
!
=
+
!
!
=k
C
k
k
k
Ck
CA
A
B
““
“
“
“
( )( )
[ ] [ ]
( )( )
[ ]2
12
2121011201
11
211
0
+
!
+
!
!
++
!
“+
!
+
!
==
!kk
kkkkCkkkCk
d
dC AA
B
##
####
#
( )( )
[ ] [ ]
!!!!
“
++
“
=+
“
+
“212112 211 kkkkkk
( ) ( )2
2121
2
2112 21
!!!!!!
“
+
“
+
“
=
“
+
“
+
“
+kkkkkkkk
( ) 1
2
12 =
!
“
kk
21
1
kk
OPT =
!
“
(3) T = 310.52oC
6-4
POLYMATH Results
No Title 02-14-2006, Rev5.1.233
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 100 100
Cao 5 5 5 5
tau 0.5 0.5 0.5 0.5
k2o 0.01 0.01 0.01 0.01
E1 10 10 10 10
R 0.001987 0.001987 0.001987 0.001987
k2 0.4 0.4 1757.3524 1757.3524
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(t) = 1
Explicit equations as entered by the user
[1] Cao = 5
[2] tau = .5
[3] k1o = .4
[4] k2o = .01
[8] k1 = k1o*exp((-E1/R)*(1/T-1/300))
[9] k2 = k1o*exp((-E2/R)*(1/T-1/300))
6-5
P6-2 (d) Example 6-5
KC = Ke = 0.25
!
“
#
$
%
&‘=
C
ON
NONN K
CC
Ckr 22
22
2
22
!
#
$
&‘‘‘=
ON
NONNONHNONO K
CC
CkCCkr 22
23
2
2
5.1
12
2
03
2
2
5.1
1222
22
232 2
1
6
5CCk
K
CC
CkCCkr NO
C
ON
NONNONHNON !
“
#
$
%
&
‘!+=
2
03
2
2222
22
22
CCk
K
CC
Ckr NO
ON
NONO !
“
$
%
‘!=
P6-2 (e) Example 6-6
For liquid phase,
joj CvF =
Equation (E 6-6.3):
2
2
5.1
23
2NONNONH
NO
O
NO CkCkC
dV
dC
v
dV
dF !!==
2
2
5.1
23
2NONNONH
NO CkCkC
d
dC !!=“
#
Equation (E6-6.8):
2
3222
22
ONO
NO
O
NO CCk
dV
dC
v
dV
dF
==
2
3222
2
ONO
NO CCk
d
dC
=!
“
P6-2 (f) Example 6-7
For equal molar feed in hydrogen and mesitylene.
we get
38.0=
opt
!
hr. At
5.0=
!
hr all of the H2 is reacted and only the decomposition of X takes place..
Ex 6-7
This question
XH
0.50
0.99
CH
0.0105
0.00016
CM
0.0027
0.0042
!
0.2hr
0.38hr
~
/X T
S
0.596
1.865
6-6
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
tau 0 0 0.43 0.43
CH 0.016 1.64E-06 0.016 1.64E-06
CM 0.016 0.0041405 0.016 0.0041405
k2 30.2 30.2 30.2 30.2
r1M -0.1117169 -0.1117169 –2.927E-04 -2.927E-04
r2T 0 0 0.0159818 2.986E-04
r2X 0 -0.0159818 0 -2.986E-04
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(CH)/d(tau) = r1H+r2H
[2] d(CM)/d(tau) = r1M
Explicit equations as entered by the user
[1] k1 = 55.2
[2] k2 = 30.2
Increasing θH decreases τopt and
~
S
X/T.
P6-2 (g) Example 6-8
Using equation from example 6-8:
Polymath code:
NLES Solution
Variable Value f(x) Ini Guess
CH 4.783E-05 –4.889E-11 1.0E-04
CX 0.0023222 –9.771E-12 0.002
K1 55.2
CHo 0.016
6-7
NLES Report (safenewt)
Nonlinear equations
[1] f(CH) = CH-CHo+K1*(CM*CH^.5+K2*CX*CH^.5)*tau = 0
[2] f(CM) = CM-CMo+K1*CM*CH^.5*tau = 0
[3] f(CX) = (K1*CM*CH^.5-K2*CX*CH^0.5)*tau–
CX = 0
Explicit equations
[5] CMo = 0.016
A plot using different values of
!
is given.
For
!
=0.5, the exit concentration are
CH = 4.8 ×10-5 lbmol/ft3 CM =0.0134 lbmol/ft3
The yield of xylene from mesitylene based on molar
flow rates exiting the CSTR for
!
=0.5:
reactedmesitylenemole
producedxylenemole
CC
C
FF
F
Y
MMO
X
MMO
X
MX !!
!!
=
“
=
“
=
“
=89.0
0134.0016.0
00232.0
The overall selectivity of xylene relative to toluene is:
producedtoluenemole
producedxylenemole
F
F
S
T
X
TX !!
!!
== 3.8
~
/
Ex 6-8
This Question
CH
0.0089
4.8 x 10-5
CM
0.0029
0.0134
!
0.5
0.5
YMX
0.41
0.89
SX/T
0.7
8.3
P6-2 (h) Example 6-9
(1)
SD/U Original Problem
SD/U P6-2 h
Doubling the incoming flow rate of species B lowers the selectivity.
6-8
P6-2 (i) Example 6-10
Original Case – Example 6-10 P6-2 i
The reaction does not go as far to completion when the changes are made. The exiting concentration of D,
P6-2 (j)
At the beginning, the reactants that are used to create TF-VIIa and TF-VIIaX are in high concentration. As
the two components are created, the reactant concentration drops and equilibrium forces the production to
slow. At the same time the reactions that consume the two components begin to accelerate and the
concentration of TF-VIIa and TF-VIIaX decrease. As those reactions reach equilibrium, the reactions that
are still producing the two components are still going and the concentration rises again. Finally the
6-9
P6-2 (k)
Base case Equal-molar feed
un-reacted mesitylene.
P6-2 (l) Individualized solution
P6-3 Solution is in the decoding algorithm given with the modules ( ICM problem )
P6-4 (a)
Assume that all the bites will deliver the standard volume of venom. This means that the initial
concentration increases by 5e-9 M for every bite.
After 11 bites, no amount of antivenom can keep the number of free sites above 66.7% of total sites. This
means that the initial concentration of venom would be 5.5e-8 M. The best result occurs when a dose of
antivenom such that the initial concentration of antivenom in the body is 5.7e-8 M, will result in a
P6-4 (b)
The victim was bitten by a harmless snake and antivenom was injected. This means that the initial
concentration of venom is 0. From the program below, we see that if an amount of antivenom such that the
6-10
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 0.5 0.5
fsv 0 0 0 0
fs 1 0.6655661 1 0.6655661
fsa 0 0 0.3344339 0.3344339
Cp 0 0 0 0
kia 1 1 1 1
Cso 5.0E-09 5.0E-09 5.0E-09 5.0E-09
ksa 6.0E+08 6.0E+08 6.0E+08 6.0E+08
kov 0 0 0 0
koa 0.3 0.3 0.3 0.3
g 0 0 0 0
h 0 0 0 0
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(fsv)/d(t) = kv * fs * Cv – ksv * fsv * Ca
[2] d(fs)/d(t) = -kv*fs*Cv – ka * fs * Ca + kia * fsa + g
[3] d(Cv)/d(t) = Cso * (-kv * fs * Cv – ksa * fsa * Cv) + h
Explicit equations as entered by the user
[1] kv = 2e8
[2] ksv = 6e8
[7] kp = 1.2e9
[8] kov = 0
[14] j = -Cso * ksv * fsv * Ca – kp * Cv * Ca – koa * Ca
6-11
P6-4 (c)
The latest time after being bitten that antivenom can successfully be administerd is 27.49 minutes. See the
P6-4 (d) Individualized Solution
P6-5 (a)
Plot of CA, CD and CU as a function of time (t):
See Polymath program P6-5-a.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 15 15
Ca 1 0.0801802 1 0.0801802
Cd 0 0 0.7995475 0.7995475
k2 100 100 100 100
K1a 10 10 10 10
K2a 1.5 1.5 1.5 1.5
Differential equations as entered by the user
[1] d(Ca)/d(t) = -(k1*(Ca-Cd/K1a)+k2*(Ca-Cu/K2a))
[2] d(Cd)/d(t) = k1*(Ca-Cd/K1a)
6-12
Explicit equations as entered by the user
[1] k1 = 1.0
[2] k2 = 100
[3] K1a = 10
To maximize CD stop the reaction after a long time. The concentration of D only increases with time
P6-5 (b)
Conc. Of U is maximum at t = 0.31 min.(CA = 0.53)
P6-5 (c)
Equilibrium concentrations:
CAe = 0.08 mol/dm3
CDe = 0.8 mol/dm3
CUe = 0.12 mol/dm3
P6-5 (d)
See Polymath program P6-5-d.pol.
NLES Solution
Variable Value f(x) Ini Guess
Ca 0.0862762 –3.844E-14 1
Cu 0.1293949 6.478E-14 0
K1a 10
K2a 1.5
6-13
NLES Report (safenewt)
Nonlinear equations
[1] f(Ca) = Ca0-t*(k1*(Ca-Cd/K1a)+k2*(Ca-Cu/K2a))-Ca = 0
[2] f(Cd) = t*k1*(Ca-Cd/K1a)-Cd = 0
Explicit equations
[1] Ca0 = 1
[2] k1 = 1
[5] K2a = 1.5
P6-6 (a)
XA !
2/1
1AX Ckr =
min004.0
2/1
3
1!
“
#
$
%
&
=dm
mol
k
BA !
AB Ckr 2
=
1
2min3.0 !
=k
YA !
2
3AY Ckr =
min.
25.0
3
3mol
dm
k=
Sketch SBX, SBY and SB/XY as a function of CA
1) SB/X =
2/1
1
2
2/1
1
2
A
A
A
X
BC
k
k
Ck
Ck
r
r==
2) SB/Y =
A
A
A
Y
B
Ck
k
Ck
Ck
r
r
3
2
2
3
2==
CAexit
0.295
0.133
0.0862
CDexit
0.2684
0.666
0.784
CUexit
0.436
0.199
0.129
3) SB/XY =
2
3
2/1
1
2
AA
A
YX
B
CkCk
Ck
rr
r
+
=
+
P6-6 (b)
Volume of first reactor can be found as follows
We have to maximize SB/XY
* = 0.040 mol/dm3
Also, CA0 = PA/RT = 0.162 mol/dm3
And
)( 2
32
2/1
1AAAYBXA CkCkCkrrrr ++=++=!
=>
3
2
*
3
*
2
2/1
*
1
*
00
*
00 4.92
))()((
)()( dm
CkCkCk
CCv
r
CCv
V
AAA
AA
A
AA =
++
!
=
!
!
=
P6-6 (c)
Effluent concentrations:
We know, τ = 9.24 min =>
3
*
11.0
mol
C
Ck
C
r
C
B
B
B=!==
“
6-15
Conversion of A in the first reactor:
74.0
00 =!=“XXCCC AAA
P6-6 (e)
A CSTR followed by a PFR should be used.
3
99.0
74.0 2
32
2/1
1
8.92
)(
162.010 dm
CkCkCk
dX
V
AAA
=
++
!!=“#
P6-6 (f)
If we notice that E2 is the smallest of the activation energies, we get a higher selectivity at lower
temperatures. However, the tradeoff is that the reaction rate of species B, and therefore production of B,
decrease as temperature drops. So we have to compromise between high selectivity and production. To do
this we need expressions for k1, k2, and k3 in terms of temperature. From the given data we know:
exp 1.98
i
i i
E
k A
T
!
” #
=$ %
& ‘
Since we have the constants given at T = 300 K, we can solve for Ai.
( )
1
.004 1.49 12
20000
exp 1.98 300
A e= =
! “
#
$ %
$ %
& ‘
( )
2
.3 5.79 6
10000
exp 1.98 300
A e= =
! “
#
$ %
$ %
& ‘
( )
3
.25 1.798 21
30000
exp 1.98 300
A e= =
! “
#
$ %
$ %
& ‘
Now we use a mole balance on species A
0A A
A
F F
V
r
!
=!
( )
0A A
A
v C C
V
r
!
=!
0 0
0.5 2
1 2 3
A A A A
A A A A
C C C C
r k C k C k C
!
” “
= =
“+ +
A mole balance on the other species gives us:
i i i
F vC rV= =
i i
C r
!
=
6-16
Using these equations we can make a Polymath program and by varying the temperature, we can find a
maximum value for CB at T = 306 K. At this temperature the selectivity is only 5.9. This may result in too
much of X and Y, but we know that the optimal temperature is not above 306 K. The optimal temperature
POLYMATH Results
NLE Solution
Variable Value f(x) Ini Guess
Ca 0.0170239 3.663E-10 0.05
T 306
R 1.987
k1 0.0077215
Cao 0.1
Cb 0.070957
k3 0.6707505
Cy 0.0019439
NLE Report (safenewt)
Nonlinear equations
[1] f(Ca) = (Cao-Ca)/(k1*Ca^.5+k2*Ca+k3*Ca^2)-10 = 0
Explicit equations
[1] T = 306
[2] R = 1.987
[3] k1 = 1.49e12*exp(-20000/R/T)
[6] Cb = 10*k2*Ca
[7] k3 = 1.798e21*exp(-30000/R/T)
[10] Cy = tau*k3*Ca^2
P6-6 (g)
Concentration is proportional to pressure in a gas-phase system. Therefore:
/2
B XY
P
S
P P+
which would suggest that a low pressure would be ideal. But as before the
P6-7
US legal limit: 0.8 g/l
Sweden legal limit: 0.5 g/l
6-17
1
A
A
dC k C
dt =!
1 2
B
A
dC k C k
dt =!
1
110k hr !
=
20.192 g
kL hr
=
Two tall martinis = 80 g of ethanol
Body fluid = 40 L
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 10 10
Ca 2 7.131E-44 2 7.131E-44
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = -k1*Ca
Explicit equations as entered by the user
[1] k1 = 10
P6-7 (a)
In the US the legal limit it 0.8 g/L.
P6-7 (b)
In Sweden CB = 0.5 g/l , t = 7.8 hrs.
P6-7 (c) In Russia CB = 0.0 g/l, t = 10.5 hrs
P6-7 (d)
For this situation we will use the original Polymath code and change the initial concentration of A to 1 g/L.
Then run the Program for 0.5 hours. This will give us the concentration of A and B at the time the second
martini is ingested. This means that 1 g/l will be added to the final concentration of A after a half an hour.
6-18
See Polymath program P6-7-d.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0.5 0.5 10 10
Ca 1.0067379 5.394E-42 1.0067379 5.394E-42
k1 10 10 10 10
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = -k1*Ca
[2] d(Cb)/d(t) = -k2+k1*Ca
Explicit equations as entered by the user
[1] k1 = 10
[2] k2 = 0.192
for the US t = 6.2 hours
P6-7 (e)
The mole balance on A changes if the drinks are consumed at a continuous rate for the first hour. 80 g of
ethanol are consumed in an hour so the mass flow rate in is 80 g/hr. Since volume is not changing the rate
of change in concentration due to the incoming ethanol is 2 g/L/hr.
For the first hour the differential equation for CA becomes:
12
A
A
dC k C t
dt =!+
after that it reverts back to the original equations.
See Polymath program P6-7-e.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 11 11
Ca 0 0 0.1785514 6.217E-45
Cb 0 -1.1120027 0.7458176 -1.1120027
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = if(t<1)then(-k1*Ca+2*t)else(-k1*Ca)
[2] d(Cb)/d(t) = -k2+k1*Ca
Explicit equations as entered by the user
[1] k1 = 10
[2] k2 = 0.192
US: CB never rises above 0.8 g/L so the is no time that it would be illegal.
Sweden: t = 2.6 hours
6-19
Russia: t = 5.2 hours
P6-7 (f)
60 g of ethanol immediately CA = 1.5 g/L
CB = 0.8 g/L at 0.0785 hours or 4.71 minutes.
P6-7 (g)
A heavy person will have more body fluid and so the initial concentration of CA would be lower. This
means a heavier person will reach the legal limit quicker. The opposite is true for a slimmer person. They
P6-8 (a)
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 4 4
Ca 6.25 0.3111692 6.25 0.3111692
k1 0.15 0.15 0.15 0.15
k4 0.2 0.2 0.2 0.2
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = -k1*Ca-k2*Ca
[2] d(Cb)/d(t) = k1*Ca-k3-k4*Cb
Explicit equations as entered by the user
[1] k1 = 0.15
[2] k2 = 0.6
6-20
[3] k4 = 0.2
[4] k3 = if(Cb<0)then (k1*Ca-k4*Cb) else (0.1)
P6-8 (b)
P6-8 (c)
If one takes initially two doses of Tarzlon, it is not recommended to take another dose within the first six