3-21
( )( ) 3
0.305 1 0.39 0.19
A
mol
C
dm
=!=
( )( ) 3
0.305 0.39 0.36
C
mol
C
dm
= =
P3-16 (d)
Gas phase reaction in a constant pressure, batch reactor
Rate law (reversible reaction):
3
C
A A
C
C
r k C
K
! “
#=#
$ %
& ‘
Stoichiometry:
( )( )
01 3 1 2
A
y
! “
= = #=
and
0
C
!
=
( )
( )
( )
( )
0
0
0
1 1
1 1 2
A
A
A A
N X X
N
C C
V V X X
!
” “
= = =
+ +
( ) ( )
0
0
0
33
1 1 2
C A
C A
N N X X
C C
V V X X
!
= = =
+ +
Combine and solve for Xe:
0.58
e
X=
Equilibrium concentrations:
( )
( ) 3
0.305 1 0.58
0.059
1 2 0.58
A
mol
C
dm
!
= =
+
( )( )
( ) 3
3 0.305 0.58
0.246
1 2 0.58
C
mol
C
dm
= =
+
P3-17
Given: Gas phase reaction A + B 8C in a batch reactor fitted with a piston such that
V = 0.1P0
( )2
3
2
1.0
sec
ft
klb mol
=
2
A A B
r kC C!=
NA0 = NB0 at t = 0
V0 = 0.15 ft3
T = 140°C = 600°R = Constant
3-22
P3-17 (a)
0
0
0 0
0.5
A
A
A B
N
yN N
= =
+
8 1 1 6
!
=” “ =
03
A
y
! “
= =
Now
( )
0 0
0
1
V P
V X
T
P
T
!
= +
” #
$ %
& ‘
and
0
1
T
T=
,
0 0
10P V=
, and
10P V=
Therefore
( )
2
0
10 1
10
V
V X
V
!
= +
or
( )
2 2
01V V X
!
= +
[ ]
01
A A
N N X=!
[ ]
0B A B
N N X
!
=“
0
0
1
B
B
A
N
N
!
= =
( )
0
1
!
+
Therefore
[ ]
( )
3
0 0
3
2
1
1
A
A
X
y P
r k RT X
!
“
# $
“=% &
‘ ( +
[ ]
( )
3
9
33
2
1
5.03*10
sec
1 3
A
Xlb mol
rft
X
!!
!=
+
P3-17 (b)
( )
2 2
01V V X
!
= +
( )
2 2
0.2 0.15 1 X
!
= +
0.259X=
10
3
8.63*10
sec
A
lb mol
rft
!
!=
P3-18 No solution will be given.
P3-19 No solution will be given.
3-23
P3-20 No solution will be given.
CDP3-A
CDP3-B
Polanyi equation: E = C – α(-ΔHR)
We have to calculate E for the reaction
From the given data table, we get
6.8 = C – α(17.5)
CDP3-C (a)
A B
Rate law at low temperature:
“r
A=kCA
The rate law t higher temperature must:
1) Satisfy thermodynamics relationships at equilibrium, and
Also, We know,
KC=CBe
CAe
Rearranging, we get
CAe “CBe
KC
=0
3-24
So, lets assume rate law as
“r
A=kACA“CB
KC
#
%
&
(
Also when CB = 0, it satisfies the given rate law. Hence the proposed rate law is correct.
CDP3-C (b)
A + 2B 2D
Rate law at low temperature:
“r
A=kCA
1/ 2CB
Here,
KC=CDe
2
2
But it does not satisfy the irreversible rate law at low temperatures.
Hence it is not correct
So, taking square root of KC
KC=CDe
CAe
1/ 2CBe
,
CAe
1/ 2CBe “CDe
KC
=0
Which satisfies the irreversible rate law.
Hence it is the required rate law.
CDP3-C (c)
A + B C + D
Irreversible rate law:
“r
A‘=kPAP
B
1+KAP
A+KBP
B
We know,
KP=P
CP
D
P
or
P
BP
A“P
CP
D
KP
=0
Hence assume rate law as:
“r
A=
k P
AP
B“P
CP
D
KP
#
$
%
&
‘
(
1+KAP
A+KBP
B+KCP
C+KDP
D
Which satisfies both the above mentioned conditions.
CDP3-D
3-25
3-26
3-27
CDP3-E
3-28
3-29
CDP3-F
3-30
CDP3-G
3-31
CDP3-H
3-32
3-33
CDP3-I
3-34
3-35
3-36
CDP3-J
Species
Symbol
Entering
Change
Leaving
SiH4
A
FA0
–3FA0
FA0(1–X)
NH3
B
FB0 = ΘBFA0
–4FA0X
FA0(Θ–4X/3)
“B=FB0
FA0
=1
yA0=0.5
#
=yA0
$
=0.5 12 %3%4
( )
=2.5
CA0=yA0CT0=0.5 P0
RT0
&
‘
(
)
*
+ =0.5 1Pa
8314 Pa ,dm3
mol ,k,973K
&
‘
(
(
(
(
)
*
+
+
+
+
=6.18 ,10%8
CA=CA01%X
( )
1+
#
X=6.18 ,10%81%X
( )
1+2.5X
CB=
1.68 ,10%81%4X
3
&
‘
( )
*
+
1+2.5X
CC=2.06 ,10%8X
1+2.5X
CD=2.47 ,10%7X
1+2.5X
CDP3-K
3-38
3-39