pset03_soln - 10.37 HW 3 Spring 2007 Problem 1 A CSTR of...

10.37 HW 3
Spring 2007
Problem 1. A CSTR of volume 0.602 liters (constant density liquid phase) is operated in
which the following reaction occurs:
A + B → C + D
The feed rate of A is 1.16 liters/hr of a solution at concentration 5.87 mmol/L. The feed
rate of B is 1.20 liters/hr at a concentration of 38.9 mmol/L. The outlet concentration of
species A is 1.094 mmol/L.
Calculate the rate constant for this reaction assuming a mass-action rate law of the form:
r = k [ A ][ B ].
First draw a diagram of the problem:
q
A
[A]
0
q
B
[B]
0
q
out
[A] [B]
The only unknown quantities in the above figure are [B] and
q
tot
.
A volume balance on the carrier solvent gives the following
relationship (constant fluid density):
q
A
+ q
B
= q
out
= 1.16 L/h + 1.20 L/h = 2.36 L/h
Define the extent of reaction:
ξ
&
= rV [ = ]
mol
time
A material balance on component A yields the following:
dn
A
= F
A 0
− F
A
− ξ
&
= [ A ]
0
q
A
− [ A ] q
out
− ξ
&
.
dt
Setting the accumulation term to zero (steady state
operation) and solving for the reaction rate we find:
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

ξ
&
= [ A ]
0
q
A
− [ A ] q
out
ξ
&
=
(
5.87mmol/L
)(
1.16L/h
)
−
(
1.094mmol/L
)(
2.36L/h
)
= 4.23mmol/h
A material balance on component B yields the following:
dn
B
= F
B 0
− F
B
− ξ
&
= [ B ]
0
q
B
− [ B ] q
out
− ξ
&
.
dt
Setting accumulation to zero, and solving for [B] we find:
[ B ] =
[ B ]
0
q
B
− ξ
&
.
q
out
(
38.9 mmol/L
)(
1.20 L/h
)
−
(
4.23 mmol/h
)
[ B ] = = 18.0 mmol/L
(
2.36 L/h
)
Using the given rate law and knowing the extent of
reaction, we can now calculate the rate constant:
ξ
&
= rV = k [ A ][ B ] V
ξ
&
k =
[ A ][ B ] V
(
4.23 mmol/h
)
L
− 1 − 1 k = = 0.357 = 357 M h
(
1.094 mmol/L
)(
18.0 mmol/L
)(
0.602 L
)
h mmol
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

Problem 2. Consider the catalyzed reaction:
A + B → B + C
with the second-order rate constant 1.15 x 10
-3
m
3
/mol/ksec. The rate law is
r = k [ A ][ B ].
What volume of CSTR would be necessary to give 40% conversion of species A if the
feed concentration of A is 96.5 mol/m
3
, the feed concentration of B is 6.63 mol/m
3
, and
the flow rate is 0.5 m
3
/ksec?
First draw a diagram of the problem:
q [A]
0
[B]
0
[A] [B] [C]
Define the extent of reaction:
ξ
&
= rV [ = ]
mol
time
Define the conversion in terms of the variables of the
problem:
mol A reacted − r
A
V ξ
&
X
A
= = =
mol A fed q [ A ]
0
q [ A ]
0
A material balance on species A gives the following
equation:
dn
A = F
A 0
− F
A
− ξ
&
= [ A ]
0
q − [ A ] q − ξ
&
dt
At steady state:
0 = [ A ]
0
q − [ A ] q − ξ
&
ξ
&
= [ A ]
0
q − [ A ] q
Thus our conversion definition is equivalent to:
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

X
A
=
[ A ]
0
q − [ A ] q
= 1 −
[ A ]
q [ A ]
0
[ A ]
0
[ A ] =
(
1 − X
A
)
[ A ]
0
A material balance on species [B] gives the following:
dn
B
dt
= F
B 0
− F
B
+ 0 ξ
&
= [ B ]
0
q − [ B ] q