P10-18 (c)
10-42
10-43
P10-18 (d)
10-44
P10-18 (e)
P10-19(a)
D
da k
dt =!
S
W U t=
S
dW U dt=
S
10-46
If
0 1 D
S
k W
U
=!
then
.5 2.5
.2
S
D
U
W kg
k
= = =
P10-19(b)
P10-19(c)
For infinite catalyst loading a = 1.
P10-19(d)
10-47
1.5
S
kg
Us
=
P10-19(e)
P10-19(f)
a = 0 means there is no reaction is taking place. Activity can never be less than 0.
P10-19(g)
S
U U=!
Now find We.
0 1
D e D t
S S
k W k W
U U
= + !
10-48
( ) 0.2 * 5 25 6.25
0.2 5 2.5 1
1 0.5 2
X
X
! “
#
$ %
=# # +
& ‘
( )
#* +
, –
1.875
1
X
X=
!
0.65X=
P10-19(h)
0
$ 160 10
A S
F X U=!
To maximize profit, a maximum in profit is reached and so we set the differential of
profit equal to 0.
0
$0 160 10
A
S S
d dX
F
dU dU
= = !
P10-19(i) No solution will be given
P10-20 (a)
P10-20 (b)
10-50
P10-20 (c)
P10-20 (d)
10-51
P10-20 (e)
10-52
P10-21 (a)
A B Elementary reaction with 1st order decay.
P10-21 (b)
P10-21 (c)
Mole balance:
0 0 0 0 0
‘ ‘ ‘
AAA
A A A
r W r W r
dX W
dt N C v C v
! ! !
= = =
Stoichiometry:
( )
01
A A
C C X=!
Combine:
P10-21 (d)
( )( )
0.2 1
k W
P10-21 (e)
P10-21 (f)
( )
1
0
1 1
2exp 1500 0.57
310 400
D
k W T
k v
! “
# $
=% % =
& ‘
( )
* +
, –
P10-22 (a)
P10-22 (b)
P10-22 (c)
P10-22 (d)
P10-22 (e)
10-57
P10-23 (a)
P10-23 (b)
10-58
P10-24 (a)
P10-24 (b)