3-1
Solutions for Chapter 3 – Rate Law and Stoichiometry
P3-1 Individualized solution.
P3-2 (a) Example 3-1
0
0.001
0.002
0.003
0.004
0.005
0.006
0.007
0.008
310 315 320 325 330 335
T (K)
k (1/s)
E = 240 kj/mol
0
5E-23
1E-22
1.5E-22
2E-22
2.5E-22
3E-22
3.5E-22
0.00295 0.003 0.00305 0.0031 0.00315 0.0032 0.00325
1/T (1/K)
k (1/s)
E = 60 kj/mol
0
1000000
2000000
3000000
4000000
5000000
6000000
0.00295 0.003 0.00305 0.0031 0.00315 0.0032 0.00325
1/T (1/K)
k (1/s)
P3-2 (b) Example 3-2 Yes, water is already considered inert.
P3-2 (c) Example 3-3
The solution to the example at a conversion of 20% would remain unchanged. For 90 % conversions of the
3-2
P3-2 (d) Example 3-4
A+b
a
B“c
a
C+d
a
D
So, the minimum value of ΘB = b/a =
1/ 3
1=0.33
P3-2 (e) Example 3-5
For the concentration of N2 to be constant, the volume of reactor must be constant. V = VO.
Plot:
2
1 0.5(1 0.14 )
(1 )(0.54 0.5 )
X
r X X
!
=
! ! !
1/(-ra) vs X
0
20
40
60
80
100
120
140
160
180
0 0.2 0.4 0.6 0.8 1 1.2
X
1/(-ra)
The rate of reaction decreases drastically with increase in conversion at higher conversions.
P3-2 (f) Example 3-6
For a given conversion, concentration of B is lower in flow reactor than a constant volume batch reactor.
Therefore the reverse reaction decreases.
CT0 = constant and inerts are varied.
N2O4“2NO2
A ↔ 2B
Equilibrium rate constant is given by:
KC=CB,e
2
CA,e
Stoichiometry:
“
=yA0
#
=yA0(2 $1) =yA0
CAO =yAO P
O
RT
O
=yAO 0.07176
( )
mol /dm3
Combining: For constant volume batch:
See Polymath program P3-2-f.pol.
POLYMATH Results
NLES Report (safenewt)
Nonlinear equations
Explicit equations
[1] yao = 1
[2] kc = 0.1
[3] Cao = 0.07174*yao
3-4
Yinert
Yao
Xeb
Xef
0
1
0.44
0.508
0.1
0.9
0.458
0.5217
0.3
0.7
0.5
0.5547
0.5
0.5
0.556
0.601
0.7
0.3
0.6435
0.6743
0.9
0.1
0.8112
0.8212
0.956
0.044
0.893
0.896
P3-2 (g) No solution will be given
P3-2 (h)
A +
2
1
B
2
1
C
Rate law:
“r
A=kACA
2CB
and
kA=25 1
s
dm3
mol
“
$
%
‘
2
P3-2 (i)
A+3B“2C
Rate law:
“r
A=kACACB
at low temperatures.
At equilibrium,
But at t = 0, CC = 0
So the rate law is not valid at t = 0.
Next guess:
KC
2=CC,e
2
CA,eCB,e
3
, or
CACB“CC
2
KC
2CB
2=0
!
“r
A=kACACB“CC
2
KC
2CB
2
#
$
%
%
&
‘
(
(
which satisfies both the initial conditions and equilibrium rate law.
3-5
Hence
“r
A=kACACB“CC
2
2CB
2
#
%
%
&
(
(
is the required rate law.
P3-3 Solution is in the decoding algorithm available separately from the author.
P3-4 (a)
Note: This problem can have many solutions as data fitting can be done in many ways.
Using Arrhenius Equation
For Fire flies:
T(in
K)
1/T
Flashes/
min
ln(flashe
s/min)
294
0.003401
9
2.197
298
0.003356
12.16
2.498
T(in K)
1/T
x103
chrips/
min
ln(chirps/
min)
287.2
3.482
80
4.382
293.3
3.409
126
4.836
Plotting ln(chirps/min) Vs 1/T,
we get a straight line.
Both, Fireflies and Crickets data
follow the Arrhenius Model.
ln y = A + B/T , and have the same activation
P3-4 (b)
For Honeybee:
T(in K)
1/T
x103
V(cm/s)
ln(V)
298
3.356
0.7
-0.357
303
3.300
1.8
0.588
Plotting ln(V) Vs 1/T, almost straight line.
ln(V) = 44.6 – 1.33E4/T
3-6
P3-4 (c)
For ants:
T(in K)
1/T x103
V(cm/s)
ln(V)
283
3.53
0.5
-0.69
293
3.41
2
0.69
303
3.30
3.4
1.22
311
3.21
6.5
1.87
Plotting ln(V) Vs 1/T,
almost straight line.
So activity of bees, ants, crickets and fireflies follow
Insect
Activation Energy
Cricket
52150
Firefly
54800
P3-4 (d)
There is a limit to temperature for which data for any one of he insect can be extrapolate. Data which would
be helpful is the maximum and the minimum temperature that these insects can endure before death.
P3-5
There are two competing effects that bring about the maximum in the corrosion rate: Temperature and
HCN-H2SO4 concentration. The corrosion rate increases with increasing temperature and increasing
concentration of HCN-H2SO4 complex. The temperature increases as we go from top to bottom of the
column and consequently the rate of corrosion should increase. However, the HCN concentrations (and the
P3-6 Antidote did not dissolve from glass at low temperatures.
P3-7 (a)
If a reaction rate doubles for an increase in 10°C, at T = T1 let k = k1 and at T = T2 = T1+10, let k = k2 =
3-7
2 1
1 1
2
1
E
R T T
ke
k
! “
# #
$ %
& ‘
=
or
( )
2 2
1 1
1 2
1 2
2 1
ln ln
1 1
k k
k k
E
T T
R
T T
T T
! “ ! “
# $ # $
% & % &
=‘=‘‘
! “
‘
# $
% &
Therefore:
P3-7 (b)
Equation 3-18 is
E
RT
k Ae!
=
From the data, at T1 = 0°C,
1
/
1
E RT
k Ae!
=
, and at T2 = 100°C,
2
/
2
E RT
k Ae!
=
Dividing gives
2 1
1 1
2
E
R T T
ke
k
! “
# #
$ %
& ‘
=
, or
( )
1.99 273
mol K
$ %
( )
* +
, –
P3-7 (c) Individualized solution
P3-8
When the components inside air bag are ignited, following reactions take place,
2NaN3 → 2Na + 3N2 ……………………………………………..(1)
10Na + 2KNO3 → K2O + 5Na2O + N2 …………………………(2)
Species
Symbol
Initial
Change
Final
NaN3
A
NA
-NAX
NA(1-X)
KNO3
B
NA
B
!
-0.2X NA
NA(
B
!
– 0.2X)
SiO2
C
NA
C
!
-0.1XNA
NA(
C
!
– 0.1X)
1 moles of NaN3 requires 0.2 mole of KNO3
=> Moles of B, KNO3 = 0.2(2.3) = 0.46 moles Mwt of KNO3 = 101.1
1 moles of NaN3 requires 0.1 mole of SiO2.
Following proposals are given to handle all the un-detonated air bags in cars piling up in the junkyards:
• Store cars in cool, dry, ventilated areas.
• Avoid Physical damage of the bag in car.
P3-9 (a)
From the web module we know that
(1 )
dX k x
dt =!
and that k is a function of temperature, but not a
3-9
P3-10 (a)
1) C2H6 → C2H4 + H2 Rate law: -rA =
62 HC
kC
2) C2H4 + 1/2O2 → C2H4O Rate law: -rA =
2/1
0242
CkC HC
4) n-C4H10 ↔ I- C4H10 Rate law: -rA = k[
104 HnC
C
–
104 HiC
C
/Kc]
5) CH3COOC2H5 + C4H9OH ↔ CH3COOC4H9 + C2H5OH
P3-10 (b)
2A + B → C
2
P3-10 (c)
(1) C2H6 → C2H4 + H2 Rate law: -rA =
62 HC
kC
P3-11 (a)
Liquid phase reaction,
O CH2–OH
Stoichiometric Table:
Species
Symbol
Initial
Change
Remaining
Ethylene
oxide
A
CAO=1 lbmol/ft3
– CAOX
CA= CAO(1-X)
= (1-X) lbmol/ft3
3-10
Therefore, -rA = k
2
AO
C
(1-X)(
B
!
-X) = k(1-X)(3.47-X)
At 350K,
k2 = k exp((E/R)(1/T-1/T2))= 1.6018exp((12500/1.987)(1/300-1/350))
= 32.034 dm3/mol.s
P3-11 (b)
Isothermal, isobaric gas-phase pyrolysis,
Stoichiometric table:
Species
symbol
Entering
Change
Leaving
C2H6
A
FAO
-FAOX
FA=FAO(1-X)
C2H4
B
0
+FAOX
FB=FAOX
!
v = vo(1+
!
X) => v = vo(1+X)
CAO = yAO CTO = yAO
RT
=
( )( )
( )
3
1 6
0.082 1100
.
atm
K kmol
# $
% &
= 0.067 kmol/m3 = 0.067 mol/dm3
CA =
( )
( )
1
(1 )
(1 ) 1
AO
A
AO
O
X
F X
FC
v v X X
!
!
= =
+ +
mol/dm3
CB =
( )
(1 ) 1
AO
B
AO
F X
FX
C
v v X X
= =
+ +
mol/dm3
Rate law:
-rA = kCA = kCAO
( )
( )
1
1
X
X
!
+
=0.067 k
( )
( )
1
1
X
X
!
+
If the reaction is carried out in a constant volume batch reactor, =>(
!
= 0)
P3-11 (c)
3-11
Isothermal, isobaric, catalytic gas phase oxidation,
C2H4 +
1
2
O2 C2H4O
A +
1
2
B C
Stoichiometric table:
Species
Symbol
Entering
Change
Leaving
C2H4
A
FAO
-FAOX
FA=FAO(1-X)
B
!
=
1
1
2
2
AO
BO
AO AO
F
F
F F
= =
2
3
AO AO
AO
TO AO BO
F F
yF F F
= = =
+
2 1
1 1 0.33
3 2
AO
y
! “
# $
= = % % =
& ‘
( )
( )
( )
3
3
6
20.092
3.
0.082 533
.
AO AO TO AO
atm
P mol
C y C y RT dm
atm dm K
mol K
= = = =
! “
# $
% &
( )
( )
( )
( )
( )
( )
1 1 0.092 1
1 1 0.33 1 0.33
AO AO
A
A
O
F X C X X
F
C
v v X X X
!
” “ “
= = = =
+” “
( )
( )
( )
0.046 1
2
1 1 0.33
AO B
B
B
O
X
FX
F
C
v v X X
!
“
# $
%
& ‘ %
( )
= = =
+%
( )
( )
( )
0.092
1 1 0.33
C AO
C
O
X
F F X
C
v v X X
!
= = =
+“
If the reaction follow elementary rate law
Rate law:
0.5
A A B
r kC C!=
!
( )
( )
0.5
0.092 1 0.046 1
1 0.33 1 0.33
A
X X
r k
X X
! “! “
# #
$ $$ $
#=% &% &
# #
P3-11 (d)
Isothermal, isobaric, catalytic gas phase reaction in a PBR
C6H6 + 2H2 C6H10
Species
Symbol
Entering
Change
Leaving
Benzene
A
FAO
-FAOX
FA=FAO(1-X)
H2
B
FBO=2FAO
-2FAOX
FB=FAO(
B
!
-2X)
( )
22
1
3
1 2
1 2 1
3 3
BO AO
B
AO AO
AO AO
AO
TO AO BO
AO
F F
F F
F F
yF F F
y
!
” #
= = =
= = =
+
= = $ $ =$
=
AO
C
( )
3
3
1 6 1 0.055 /
3 3
.
0.082 443.2
.
TO AO
P atm
C y mol dm
RT atm dm K
mol K
! “ ! “
= = =
# $ # $
! “
% & % &
# $
% &
( )
( )
( ) ( )
1 1 0.055 1
2 2
11 1
3 3
AO AO
A
A
O
F X C X X
F
C
v v X X X
!
” “ “
= = = =
+# $ # $
” “
% & % &
‘ ( ‘ (
( )
( )
( )
2 0.11 1
2
11
3
AO B
B
B
O
F X X
F
C
v v X X
!
“
# #
= = =
+$ %
#
& ‘
( )
( )
0.055
2 2
11 1
3 3
C AO AO
C
O
F F X C X X
C
v v X X X
!
= = = =
+“ # “ #
$ $
% & % &
‘ ( ‘ (
If the reaction follow elementary rate law.
Rate law:
( )
2
3
3
‘
1
‘ 0.0007
2
1
3
A A B
A
r kC C
X
r k
X
!=
!
!=
” #
!
$ %
& ‘
For a fluidized CSTR:
0
‘
A
A
F X
W
r
=!
( )
0
3
3
1
0.0007
2
1
3
A
F X
W
X
k
X
=!
” #
!
$ %
& ‘
k = 53
3
min
mol
kgcat atm
at 300K
@ T = 170oC
1
1
1 1 80000 1 1
exp 53exp 8.314 300 443
E
k k
R T T
! “
! “ ! “
! “
=#=#
$ %
$ % $ %
$ %
$ % & ‘
& ‘
& ‘
& ‘
= 1663000
3
min
mol
kgcat atm
3-13
FA0 = CA0* v0
( )
0 0
3
3
1
0.0007
2
1
3
A
C v X
W
X
k
X
=!
” #
!
$ %
& ‘
v0 = 5 dm3/min
( )
0 0
3
3
1
0.0007
2
1
3
A
C v X
W
X
k
X
=!
” #
!
$ %
& ‘
at X = 0.8
5
104.4 !
“=W
kg of catalyst
P3-12 C2H4 +
1
2
O2 C2H4O
A +
1
2
B C
Stoichiometric table for the given problem will be as follows
Assuming gas phase
Species
Symbol
Entering
Change
Leaving
C2H4
A
FA0
– FA0X
FA0(1-X)
O2
B
FB0 = ΘBFA0
-1/2 FA0X
FA0(ΘB – X/2)
3-14
yA0=FA0
FT0
=0.30
,
“
=yA0
#
=$0.15
CA0=yA0P
RT =0.041 mol
dm3
CA=FA0
v=CA0
(1“X)
(1+
#
X)=0.041(1“X)
1“0.15X
CB=FB
v=CA0
(1
2“1
2
X)
1“0.15X=0.020(1“X)
1“0.15X
CC=F
C
v=CA0X
1“0.15X=0.041X
1“0.15X
P3-13 (a)
Let A = ONCB C = Nibroanaline
B = NH3 D = Ammonium Chloride
P3-13 (b)
Species
Entering
Change
Leaving
A
FA0
– FA0X
FA0(1-X)
P3-13 (c)
For batch system,
P3-13 (d)
–A A B
r kC C=
( ) ( ) ( )
0
0 0
0 0 0
1 1 , 1
A
A A A A
A A A A
N
N N F F
F X C X C C X
V V V v v
= = = !=!= = = !
( ) ( ) ( )
0
0 0
0 0 0
2 2 , 2
A
B B B
B B A B B A B
N
N N F
F X C X C C X
V V V v
! ! !
= = = “=“= = “
( )( )
2
01 2
A A B
r kC X X
!
“=” “
0
0
6.6 3.67
1.8
B
B
A
C
C
!
= = =
3-15
03
1.8
A
kmol
C
m
=
( ) ( )( )
2
1.8 1 3.67 2
A
r k X X!=! !
P3-13 (e)
1) At X = 0 and T = 188°C = 461 K
min
0202.067.38.1
min
0017.0 3
2
3
3
2
00 m
kmol
m
kmol
kmol
m
kCr BAA =
!
“
#
$
%
&
=‘=(
03
kmol
0.0202
m min
A
r!=
2) At X = 0 and T = 25C = 298K
!
!
“
#
$
$
%
&
!
!
“
#
$
$
%
&‘=TTR
E
kk
O
O
11
exp
298
1
461
1
.
987.1
11273
exp
min.
0017.0
3
Kmol
cal
mol
cal
kmol
m
k
#
#
#
#
$
%
&
&
&
&
‘
(
#
$
%
&
‘
(!=
3)
0
0
1 1
exp E
k k
R T T
! “
# $
=%
& ‘
( )
& ‘
* +
, –
311273
m 1 1
0.0017 exp
kmol min 461 561
1.987
cal
mol
kcal K K
mol K
! “
# $
% &
=# ‘ $
( )
# $
* +
# $
, –
3
m
0.0152
kmol min
k=
0 0 0A A B
r kC C!=
3
3 3
m
0.0152 1.8 6.6
kmol min
A
kmol kmol
r
m m
! “! “
#=$ %$ %
& ‘& ‘
3
kmol
0.1806
m min
A
r!=
P3-13 (f)
rA = kCAO
2(1-X)(θB–2X)
3-16
At X = 0.90 and T = 188C = 461K
1) at T = 188 C = 461 K
( ) ( )( )
min
00103.0
9.0267.39.018.1
min.
0017.0
3
2
3
3
m
m
kmol
kmol
m
rA
=
!!
“
#
$
%
&
‘
“
“
#
$
%
%
&
‘
=!
2)
At X = 0.90 and T = 25C = 298K
( ) ( )( )
min
1023.1
9.0267.39.018.1
min.
1003.2
3
6
2
3
3
6
m
m
kmol
kmol
m
rA
!
!
“=
!!
#
$
%
&
‘
(
#
#
$
%
&
&
‘
(“=!
3)
At X = 0.90 and T = 288C = 561K
( ) ( )( )
min
0092.0
9.0267.39.018.1
min.
0152.0
3
2
3
3
m
kmol
m
kmol
kmol
m
rA
=
!!
“
#
$
%
&
‘
“
“
#
$
%
%
&
‘
=!
P3-13 (g)
FAO = 2 mol/min
1) For CSTR at 25oC -rA
min
1023.1 3
6
m
kmol
!
“=
( )
3
3
3
9.0
min
1023.1
1.0min/2
1
m
mol
mol
r
XF
V
XA
AO
!
!
“
“
=
“
=M
2)At 288oC, -rA
min
0092.0 3
m
kmol
=
( )
3
3
9.0
739.21
min
0092.0
1.0min/2
1
m
m
mol
mol
r
XF
V
XA
AO
=
!
=
“
“
=
=M
P3-14
C6H12O6 + aO2 + bNH3 → c(C4.4H7.3N0.86O1.2) + dH2O + eCO2
To calculate the yields of biomass, you must first balance the reaction equation by finding the coefficients
3-17
proceeds to completion and calculating the ending mass of the cells.
P3-14 (a)
Apply mass balance
For C 6 = 4.4c + e For O 6 + 2a = 1.2c + d + 2e
Next we solve for e using the other carbon balance
6 = 4.4 (0.909) + e
We can solve for b using the nitrogen balance
Next we use the hydrogen balance to solve for d
12 + 3b = 7.3c + 2d
Finally we solve for a using the oxygen balance
6 + 2a = 1.2c + d + 2e
6 + 2a = 1.2(0.909) + 3.85 + 2(2)
P3-14 (b)
Assume 1 mole of glucose (180 g) reacts:
mass of cells = 83.12 g
Yc/o2 = mass of cells / mass of O2
If we assume 1 mole of glucose reacted, then 1.47 moles of O2 are needed and 83.12 g of cells are
produced.
P3-15 (a)
Isothermal gas phase reaction.
2 2 3
1 3
2 2
N H NH+!
Making H2 as the basis of calculation:
2 2 3
1 2
3 3
H N NH+!
1 2
3 3
A B C+!
Stoichiometric table:
Species
Symbol
Initial
change
Leaving
H2
A
FAO
-FAOX
FA=FAO(1-X)
N2
B
FBO=
B
!
FAO
-FAOX/3
FB=FAO(
B
!
-X/3)
P3-15 (b)
2 1 2
1
3 3 3
2 1
0.5
3 3
AO
y
!
” !
# $
=% % =%
& ‘
( )
# $
= = * % =%
& ‘
( )
( )
( )
3
16.4
0.5
.
0.082 500
.
AO
atm
C
atm dm K
mol K
=! “
# $
% &
= 0.2 mol/dm3
( )
( )
( )
( )
( )
( )
2
3
3
3
1 0.2 1
0.1 /
11
3
1 0.2
2 2 0.1 /
3 1 3 1
3
AO
H A
AO
NH C
C X X
C C mol dm
X
X
C X X
C C mol dm
X
X
!
!
” “
= = = =
+# $
“
% &
‘ (
“
= = )=)=
+# $
“
% &
‘ (
P3-15 (c)
kN2 = 40 dm3/mol.s
(1) For Flow system:
2 2 2 2
3
1
2 2
3
1
2 2
2
11
3
N N N H
r k C C
X
X
! ” ! “
#=$ % $ %
! “ ! “
& ‘
#
( )
* + * +
#
“r
N2=1.6 1“X
( )
1“X
3
#
$
% &
‘
(
)
*
+
+
+
+
,
–
.
.
.
.
3
2
(2) For batch system, constant volume.
( ) ( )
2 2 2 2
3
1
2 2
1
23
22
40 1 1
3
N N N H
AO
r k C C
X
C X
! ” ! “
#=$ % $ %
! “
& ‘
=# #
! “
( )
* + $ %
3-20
( )( )
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.
( )
( ) ( )
3
0 0
13
1 2 1 2
ee
C A A
e e
XX
K C C
X X
! “
#=$ %
+ +
$ %
& ‘
( )( )22 3
0
1 1 2 27
C e e A e
K X X C X!+ =
2
3
0
27
4 3 1 0
A
e e
C
CX X
K
! “
#+ + + =
$ %
& ‘
0.58
e
X=
Equilibrium concentrations:
( )
0
03
3
0
10 0.305
400 0.082
A
Patm mol
C
RT dm
dm atm
K
mol K
= = =
! “
# $
% &
( )
( )
( ) 3
1 0.58
0.305 0.059
1 2 0.58
A
mol
C
dm
!
= =
+
( )( )
( )
( ) 3
3 0.58 0.305
0.246
1 2 0.58
C
mol
C
dm
= =
+
P3-16 (c)
Same reaction, rate law, and initial concentration as part (b) gas phase, batch reaction.
Stoichiometry:
Combine and solve for Xe
Equilibrium concentrations