Solutions for Chapter 3 – Rate Law and Stoichiometry
P3-1 Individualized solution.
P3-2 (a) Example 3-1
0.006
0.007
0.008
For E = 60kJ/mol For E1 = 240kJ/mol
E = 240 kj/mol
1/T (1/K)
3.5E-22
E = 60 kj/mol
5000000
6000000
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
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.
1/(-ra) vs X
X
140
160
180
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
3-3
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
[1] f(Xeb) = Xeb – (kc*(1-Xeb)/(4*Cao))^0.5 = 0
0.7
0.8
0.9
1
3-4
Yinert
Yao
Xeb
Xef
0
1
0.44
0.508
0.1
0.9
0.458
0.5217
0.2
0.8
0.4777
0.537
0.3
0.7
0.5
0.5547
P3-2 (g) No solution will be given
P3-2 (h)
A +
2
1
B
2
1
C
P3-2 (i)
A+3B“2C
Rate law:
“r
A=kACACB
at low temperatures.
At equilibrium,
KC=CC,e
CA,e
1/ 2CB,e
3 / 2
!
CA
1/ 2CB
3 / 2 “CC
KC
=0
3-5
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)
For Crickets:
T(in K)
1/T
x103
chrips/
min
ln(chirps/
min)
287.2
3.482
80
4.382
P3-4 (b)
For Honeybee:
T(in K)
1/T
x103
V(cm/s)
ln(V)
298
3.356
0.7
-0.357
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
So activity of bees, ants, crickets and fireflies follow
Arrhenius model. So activity increases with an
increase in temperature. Activation energies for
fireflies and crickets are almost the same.
Insect
Activation Energy
P3-4 (d)
There is a limit to temperature for which data for any one of he insect can be extrapolate. Data which would
P3-5
There are two competing effects that bring about the maximum in the corrosion rate: Temperature and
P3-6 Antidote did not dissolve from glass at low temperatures.
P3-7 (a)
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!
=
( )
13 1 1
1
7960
10 min exp 2100 min
1.99 273
E
RT
cal
mol
A k e
cal K
mol K
! ! !
” #
$ %
$ %
= = =
$ %
& ‘
$ %
( )
* +
, –
P3-7 (c) Individualized solution
P3-8
NaN3 + 0.2KNO3 + 0.1SiO2 0.4Na2O + 1.6N2 + complex/10………(4)
Stoichiometric table:
Species
Symbol
Initial
Change
Final
Given weight of NaN3 = 150g Mwt of NaN3 = 65
Therefore, no. of moles of NaN3 = 2.3
Following proposals are given to handle all the un-detonated air bags in cars piling up in the junkyards:
P3-9 (a)
From the web module we know that
(1 )
dX k x
and that k is a function of temperature, but not a
P3-9 (b)
When you boil the potato in water, the heat transfer coefficient is much larger, but the temperature can only
P3-9 (c) No solution will be given
3-9
P3-10 (a)
1) C2H6 → C2H4 + H2 Rate law: -rA =
62 HC
kC
P3-10 (b)
2A + B → C
(1) –rA = kCACB
2
P3-10 (c)
(1) C2H6 → C2H4 + H2 Rate law: -rA =
62 HC
kC
P3-11 (a)
Liquid phase reaction,
O CH2–OH
CH2 – CH2 + H2O → CH2–OH
3-10
Therefore, -rA = k
2
AO
C
(1-X)(
B
!
-X) = k(1-X)(3.47-X)
At 300K E = 12500 cal/mol, X = 0.9,
k = 0.1dm3/mol.s =1.6018
slbmolft ./
3
P3-11 (b)
Isothermal, isobaric gas-phase pyrolysis,
C2H6 C2H4 + H2
A B + C
Stoichiometric table:
CAO = yAO CTO = yAO
P
RT
=
( )( )
( )
3
1 6
0.082 1100
.
atm
m atm K
K kmol
! “
# $
% &
= 0.067 kmol/m3 = 0.067 mol/dm3
Rate law:
P3-11 (c)
3-11
Isothermal, isobaric, catalytic gas phase oxidation,
2
Stoichiometric table:
Species
Symbol
Entering
Change
Leaving
C2H4
A
FAO
-FAOX
FA=FAO(1-X)
O2
B
FBO
–
B
!
FAOX
FB=FAO(
B
!
-X)
C2H4O
C
0
+FAOX
FC=FAOX
( )
( )
( )
( )
( )
( )
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
!
“
# $
%
& ‘ %
( )
= = =
+%
‘ (‘ (
P3-11 (d)
Isothermal, isobaric, catalytic gas phase reaction in a PBR
C6H6 + 2H2 C6H10
A + 2B C
Stoichiometric table:
Species
Symbol
Entering
Change
Leaving
3-12
22
BO AO
B
AO AO
F F
F F
!
= = =
( )
( )
( ) ( )
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
!
” “ “
= = = =
+# $ # $
” “
% & % &
‘ ( ‘ (
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:
3-13
FA0 = CA0* v0
v0 = 5 dm3/min
3
104.2 !
“=W
kg of catalyst
@ T = 270oC
at X = 0.8
5
104.4 !
“=W
kg of catalyst
P3-12 C2H4 +
1
2
O2 C2H4O
Assuming gas phase
Species
Symbol
Entering
Change
Leaving
C2H4
A
FA0
– FA0X
FA0(1-X)
yA0=FA0
FT0
=0.30
,
“
=yA0
#
=$0.15
P3-13 (a)
Let A = ONCB C = Nibroanaline
B = NH3 D = Ammonium Chloride
A + 2B C + D
–A A B
r kC C
!!“
=
P3-13 (b)
Species
Entering
Change
Leaving
A
FA0
– FA0X
FA0(1-X)
P3-13 (c)
For batch system,
CA=NA/V -rA = kNANB/V2
P3-13 (d)
–A A B
r kC 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
2) At X = 0 and T = 25C = 298K
!
!
“
#
$
$
%
&
!
!
“
#
$
$
%
&‘=TTR
E
kk
O
O
11
exp
3)
0
0
1 1
exp E
k k
R T T
! “
# $
=%
& ‘
( )
& ‘
* +
, –
P3-13 (f)
rA = kCAO
2(1-X)(θB–2X)
At X = 0.90 and T = 188C = 461K
1) at T = 188 C = 461 K
2)
At X = 0.90 and T = 25C = 298K
3)
At X = 0.90 and T = 288C = 561K
min
m
P3-13 (g)
FAO = 2 mol/min
1) For CSTR at 25oC -rA
1023.1 3
6
kmol
!
“=
P3-14
C6H12O6 + aO2 + bNH3 → c(C4.4H7.3N0.86O1.2) + dH2O + eCO2
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
b = 0.86c = 0.86* (0.909)
b = 0.78
d = 3.85
Finally we solve for a using the oxygen balance
P3-14 (b)
Assume 1 mole of glucose (180 g) reacts:
Yc/s= mass of cells / mass of glucose = mass of cells / 180 g
mass of cells = c*(molecular weight) = 0.909 mol* (91.34g/mol)
P3-15 (a)
Isothermal gas phase reaction.
3-18
2 2 3
1 3
2 2
N H NH+!
Making H2 as the basis of calculation:
Stoichiometric table:
Species
Symbol
Initial
change
Leaving
H2
A
FAO
-FAOX
FA=FAO(1-X)
P3-15 (b)
2 1 2
1
3 3 3
2 1
0.5
3 3
AO
y
!
” !
# $
=% % =%
& ‘
( )
# $
= = * % =%
& ‘
( )
P3-15 (c)
kN2 = 40 dm3/mol.s
(1) For Flow system:
3-19
(2) For batch system, constant volume.
P3-16 (a)
Liquid phase reaction assume constant volume
Rate Law (reversible reaction):
To find the equilibrium concentrations, substitute the equilibrium conversion into the stiochiometric
relations.
P3-16 (b)
Stoichiometry:
Combine and solve for Xe.
Equilibrium concentrations:
( )
0
03
3
0
10 0.305
400 0.082
A
Patm mol
C
RT dm
dm atm
K
mol K
= = =
! “
# $
% &
P3-16 (c)
Same reaction, rate law, and initial concentration as part (b) gas phase, batch reaction.
Combine and solve for Xe
Equilibrium concentrations