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10-1
Solutions for Chapter 10 – Catalysis and Catalytic
Reactors
P10-1 Individualized solution
P10-2 (a) Example 10-1
(1) Pentane isomerization
Pt
nP iP!!"
Assume that Pt is the catalyst used.
Minimum:
( )
3 7
1 1
3*10 0.5 7.69 *10
195 100
P
mol
rs gcat
! !
" #
!= =
$ %
& '
Maximum:
f = 100 molecules/site/sec
( )
2
1 1
100 0.5 0.00849
58.9 100
H
mol
rs gcat
! "
#= =
$ %
& '
Minumum:
f = 0.01
10-2
P10-2 (b) Example 10-2
For 60% conversion
6.3% of the sites are vacant
(2) X = 0.8
9% of the sites are covered by toluene
(3) Linearize the rate law to:
P10-2 (c) Example 10-3
Increasing the pressure will increase the rate law.
P10-2 (d) Example 10-4
With the new data, model (a) best fits the data
10-3
(a)
POLYMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/(1+Kea*Pea+Ke*Pe)
Variable Ini guess Value 95% confidence
k 3 3.5798145 0.0026691
(b)
POLYMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/(1+Ke*Pe)
Variable Ini guess Value 95% confidence
(c)
POLYMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/((1+Ke*Pe)^2)
Variable Ini guess Value 95% confidence
k 3 1.9496445 0.319098
(d)
POLYMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe^a*Ph^b
Variable Ini guess Value 95% confidence
k 3 0.7574196 0.2495415
a 1 0.2874239 0.0955031
10-4
Model (e) at first appears to work well but not as well as model (a). However, the 95%
(e)
POLYMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/((1+Ka*Pea+Ke*Pe)^2)
Variable Ini guess Value 95% confidence
k 3 2.113121 0.2375775
(f)
POLYMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/(1+Ka*Pea)
Variable Ini guess Value 95% confidence
k 3 44.117481 7.1763989
Ka 1 101.99791 16.763192
Precision
P10-2 (e) Example 10-5
0
'
A
A
dX W
r
dt N
=!
2
' '
A A
r ak C!=
[ ]
exp d
a k t=!
1
d
d
k
k
Xk
k
=
+
(3) First order reaction with first order decay
( ) [ ]
0
'1 exp
A d
dX Wk C X k t
dt N
=! !
P10-2 (f) Example 10-6
Increasing the space time makes the minimum disappear. Decreasing the space time
moves the minimum to the left and the concentration is higher.
P10-2 (g) Example 10-7
(1) If the solids and reactants are fed from opposite ends,
d
S
k a
da
dW U
=
at W = We, a = 1
( )2
2
0 0 1
A A
dX
F kC X a
dW =!
This gives the same expression for conversion as in the example.
(2) Second order decay
1
1d
S
ak W
U
=
+
(3) If ε = 2
P10-2 (h) Example 10-8
P10-2 (i)
For EA = 10 and Ed = 35, for first order decay we rearrange Eq 10-120 to:
0
0
1 1
ln 1 d d d
A
k tE E
E R T T
! "
! "
#=#
$ %
$ %
& ' & '
10-8
P10-2 (j) Individualized solution
P10-3 Solution is in the decoding algorithm given with the modules
P10-4
P10-4 (a)
10-9
P10-4 (b)
Adsorption of isobutene limited
P10-4 (c)
P10-4 (d)
10-10
P10-4 (f) Individualized solution
P10-5 (a)
P10-5 (b) Individualized solution
P10-5 (c)
22 2O S O S
!!"
+ •
#!!
22 2A S A S
!!"
+ •
#!!
3 6 3 5
C H O S C H OH S+ • !•
B A S C S+ • !•
10-11
[ ]
C V
C S D C S D C S C C V
D
P C
r k C k C K P C
K
• • •
! "
=#=#
$ %
& '
P10-6 (a)
P10-6 (b)
P10-6 (c)
P10-6 (d) Individualized solution
P10-6 (e) Individualized solution
P10-7
10-14
P10-8
P10-8 (a)
10-15
P10-8 (b)
P10-8 (c)
P10-9
10-16
P10-9 (a)
10-17
P10-9 (b)
10-18
Substituting the expressions for CV and CA·S into the equation for –r’A
P10-9 (c) Individualized solution
P10-9 (d)
First we need to calculate the rate constants involved in the equation
for –r’A in part (a). We can rearrange the equation to give the
following
10-19
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
W 0 0 23 23
X 0 0 0.9991499 0.9991499
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(W) = -ra/Fao
Explicit equations as entered by the user
[1] e = 1
P10-9 (f)
See Polymath program P10-9-f.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
W 0 0 23 23
X 0 0 0.9997919 0.9997919
ODE Report (RKF45)
Differential equations as entered by the user
