Solutions for Chapter 11 – External Diffusion Effects on
Heterogeneous Reactions
P11-1 Individualized solution
P11-2 (a)
WA=cDAB
dyA
dz +yAWA+WB
( )
Taking the ratio of Equation (1) to Equation (2) to eliminate WA and solving for yA
ln 1+yA
( )
” # z=ln 1+yA0
( )
“
11-2
P11-2 (b)
(a)
T
then
kc2
kc1
=“1
“2
#
$
%
&
‘
(
5 6
dP1
dP 2
#
$
%
&
‘
(
1 2
U2
U1
#
$
%
&
‘
(
1 2
At T1 = 300K µ1
“0.883cP
At T2 = 350K µ2
“0.380cP
Assume density doesn’t change that much,
“=µ
P11-2 (c)
A 50-50 mixture of hydrazine and helium would only affect the kinematic viscosity to a small
kc2 =kc1
dP1
dP2
“
#
$
%
&
‘
1 2
=2.9m s 1
5
“
#
$ %
&
‘
1 2
P11-2 (d)
P11-2 (e)
“2
“1
=e#4,000 1
T
1
#1
T2
$
%
&
‘
(
)
e#4,000 1
773 #1
873
$
%
& ‘
(
)
=e#0.59 =0.55
Assume
11-4
P11-2 (f)
Assume concentration in blood is negligible
CA2 =0 at “2
( )
. Assume quasi steady state
( )
P
P A
A P
d V C
W A
2 1
2 1
A
A
AB AB
C
W
D D
! !
=
+
A AP
C HC=
11-5
If
CAD
varies
R
P11-3
11-6
11-7
P11-4
11-9
P11-5 (a)
11-12
11-14
P11-5 (b)
P11-5 (c)
11-15
P11-5 (d)
11-16
P11-5 (e)
P11-6
Given,
• Minimum respiration rate of chipmunk,
Assuming,
Minimum flow rate of oxygen to the bottom, FAL
= Minimum respiration rate of chipmunk
= 1.5 μmol of O2/min
Area sectional–Cross A ofFlux
hole down theA of rate Flow
“=
11-17
P11-6 (a) At Pasadena, California
25
0
) level seaat situated ( California Pasadena,at /10013.1
,
law) gas (Ideal
mNP
where
y
RT
P
C
T
A
T
A
“=
“=
Now,
Flow rate of A = (Concentration of A at the bottom) x (Volumetric intake of gas)
0
AL AL
F C v
=!
Solving for the length from (1),
P11-6 (b) At Boulder, Colorado
Boulder, Colorado is 5430 feet above sea-level. The corresponding atmospheric pressure is 0.829
x 10-5 N/m2.
11-18
CA0=0.829 “105N/m2
8.314 J
mol.K“298K
#
$
% &
‘
(
“0.21
) CA0=7.03 mol /m3
Solving for the length from (1),
P11-6 (c)
During winter at Ann Arbor, Michigan
0 0
0 17.78 255.37
T F C K
= = !=
Solving for the length from (1),
L=DAB
CA0“CAL
FAL
#AC
1.75 3
4 2 4 2
8
255 (10.02 0.03) /
0.18 10 / 7.069 10
298 2.5 10 /
3.87
mol m
L m s m
mol s
L m
! !
!
” #
!
” #
=$ $ $ $ $
% &
% & $
‘ ( ‘ (
)=
During winter at Boulder, Colorado
Solving for the length from (1),
L=DAB
CA0“CAL
FAL
#AC
P11-6 (d) Individualized solution
P11-7 (a)
Given : Pv = 510 mm Hg @ 35 oC (from plot of ln Pv vs 1/T )
where VA and VB are the Fuller molecular diffusion volumes which are calculated by summing the atomic
contributions. This also lists some special diffusion volumes for simple molecules.
Fuller diffusion volumes
Atomic and structural diffusion volume increments
C 15.9 F 14.7
H 2.31 Cl 21.0
P11-7 (b)
By CS2 molar flow rate balance
dNA
dz =0
And
Z=0 cm
Equating the results of Fick’s First Law and molar flow balance, then rearranging and writing in the
integral form
K1dz
0
z2
“=CDAB dln(1#XA)
X0
X2
“
P11-7 (c)
For any value of z between z = 0 and z = 20
z
ln 1“XA
1“X0
#
$
%
&
‘
(
Z=20 cm