View
223
Download
0
Category
Preview:
Citation preview
7/24/2019 Lecture 4- Stoichiometry
1/50
Stoichiometry
andRate Laws
Suchithra Thangalazhy Gopakumar
H83RED Reactor Design
Autumn Semester 2015
7/24/2019 Lecture 4- Stoichiometry
2/50
Outline
Rate Law and Reaction Rate Constant
Arrhenius Law
Elementary reaction
Order of Reaction
Rate Law for reversible reactions Stoichiometric Table
Batch systems:
Constant volume and
Constant pressure
Flow systems:
Liquid phase
Gas phase reactions that have a change in the total number of moles
Reactors with variable volumetric flowrates
7/24/2019 Lecture 4- Stoichiometry
3/50
Chemical kinetics deals with reaction rates
Types of Reactions
Homogeneous reaction
involves substances of only one phase
Heterogeneous reaction
involves more than one phase, and thereaction usually occurs at or very near the
interface between the phases
Phase
Irreversible reaction
proceeds in only one direction and continues
until the reactants are exhausted
Reversible reaction
proceeds in both forward and reverse
directions; continues until the equilibrium state
is reached
Equilibrium
A B
A B
Types of Reactions
7/24/2019 Lecture 4- Stoichiometry
4/50
The Rate Law and the Reaction Rate Constant
General form of the rate law, which is an algebraic equation representing the
reaction rate as a function of the concentrations (activities) of various species:
(1)
k = reaction rate constant
~ independent of the species concentrations
,...)C,C(f)T(kr BAA
The specific reaction rate
constant, kA, is always referred to
a particular species
Exception
OHNaClHClNaOH kkkkk 2
NaOH + HCl NaCl + H2O
Liquid Phase:
Reaction rate constantis an intensive parameter concentration-independent
Gas Phase:
k=f(temperature, total pressure, ionic strength, choice of solvent )
k=f(temperature, catalysts, total pressure )
Rate Law
7/24/2019 Lecture 4- Stoichiometry
5/50
A = pre-exponential factor or frequency factorE= activation energy (J/mol or cal/mol),
R = gas constant (8.314 J/mol.K=1.987 cal/mol.K)
T = absolute temperature (K)
(2)
Activation energy, E characterise energy state of molecules which could
result in their chemical interaction
Minimum energy that molecules must have to react
(3)
Temperature Dependency from Arrhenius Law
RT/EAe)T(k
1/T, K-1
ln
k,sec-1
slope= -E/R
1/T, K-1
ln
k,sec-1
Low E
High E
TR
EAkA
1lnlnActivation energy from experimental results
E , k~ sensitive to the temperature
E , k~ less sensitive to thetemperature
Rate Law
7/24/2019 Lecture 4- Stoichiometry
6/50
Temperature rise needed to double the rate of reaction
Average
TemperatureActivation Energy, E
40 kJ/mol 160 kJ/mol 280 kJ/mol 400 kJ/mol
0o
C 11o
C 2.7o
C 1.5o
C 1.1o
C400 oC 65 oC 16 oC 9.3 oC 6.5 oC
1000 oC 233 oC 58 oC 33 oC 23 oC
Reactions with high activation energies are very temperature-sensitive
Any given reaction is more temperature-sensitive at lower temperatures thanat higher temperatures
Rules of thumb:
Rate of reaction doubles for every 10o
Cincrease in temperature
~ Only true for a specific combination
of activation energy and temperature1/T, K-1
ln k, sec-1
Low E
High E
Arrhenius Law
7/24/2019 Lecture 4- Stoichiometry
7/50
Task
Consider the following elementary reactions :
Solut ion
1) Which reaction has the higher activation energy?
2) Which reactions have the same activation energy?
3) Which reaction is virtually temperature insensitive?
Arrhenius Law
7/24/2019 Lecture 4- Stoichiometry
8/50
Solut ion
4) Which reaction will dominate (i.e. take place the fastest) at high temperatures?
5) Which reaction will take place the fastest at low temperatures?
Arrhenius Law
7/24/2019 Lecture 4- Stoichiometry
9/50
Finding k(T) given k(T0)
0/0)( RTEAeTk
RTEAeTk /)(
k(T)k(T0)exp E
R
1
T0
1
T
If we know:
The specific reaction rate k0(T
0) at a temperature T
0 The activation energy, E,
We can find the specific reaction rate constant k(T) at any other
temperature,
T, for that reaction
0/
0RTEeTkA substitute
RT
E
RT
E
eeTkTk
0)()( 0
Arrhenius Law
7/24/2019 Lecture 4- Stoichiometry
10/50
Task
Determine the activation energy, E and frequency factor, A from the following data:
Solut ion
Arrhenius equation in logarithm form:
k (min-1) 0.001 0.050
T (oC) 0.0 100.0
TRE
AkA1
lnln
RT/EAe)T(k
Arrhenius Law
7/24/2019 Lecture 4- Stoichiometry
11/50
Elementary and Non- elementary Reactions
Assumption from Collision theory:
Reactant particles must collide with each other.
Reaction is a result of the collision of a single molecule A with a singlemolecule B
E: Energy required to activate the molecules into a state from whichreactant bonds can change into product bonds.
Only those collisions with enough energy to exceed E can lead toreaction.
Rate law can be derived from collision theory
(1),...)C,C(f)T(kr BAA
Rate of reaction is proportional to the number of such collisions.
The number of collisions is proportional to the concentration ofreactant.
Elementary Reaction
7/24/2019 Lecture 4- Stoichiometry
12/50
The dependence of possible collisions on the reactant concentrations.
A
A
B
B
A
A
B
B
A
4 collisions
Add anothermolecule of A
6 collisions
Add anothermolecule of B
A
A
B
B
A B
9 collisions
Elementary Reaction
7/24/2019 Lecture 4- Stoichiometry
13/50
Elementary Reactions
corresponds to the stoichiometric
equationBAA CkCr
If kinetic experiments results in a rate law, which corresponds tothe stoichiometric equation, we call such reactions elementary.
Molecularity Elementary step Rate law
1. Unimolecular A -> products rate = k CA
2. Bimolecular2A -> productsA + B -> products
rate = k C2Arate = k C
AC
B
3. Termolecular3 A -> productsA + 2 B -> productsA + B + C -> products
rate = k C3Arate = k CAC
2B
rate = k CACBCC
A + B R
The molecularity of an elementary reaction is the number of molecules
(atoms) involved (colliding) in the reaction.
Elementary Reaction
7/24/2019 Lecture 4- Stoichiometry
14/50
Rate law reflects a limiting step or combination of limiting steps
The rate limiting step is the slowest step in the sequence of stepsleading to product formation
The overall rate of reaction depends on this slowest step
Non-elementary reactions involve more complex, multi-step reactions
H2 + I2 2HIElementary Reaction
222 IHH CkCr
H2 + Br2 2HBr
2
22
BrHBr2
21BrH1
HBrC/Ck
CCkr
/
Non-ElementaryReaction
Elementary and Non-Elementary Reactions
Elementary Reaction
7/24/2019 Lecture 4- Stoichiometry
15/50
The Reaction Order and Rate Law
General form of the rate law: (4)
2 NO + O2 2 NO2
...CCkr BAAA
Temperature-dependentterm
Concentration-dependentterms
The order of a reaction refers to the power to which the concentrations are rais
The overall order of the reaction: n = + +
2O
2
NONONO CCkr
CO + Cl2 COCl2
23CO
/ClCO CkCr
H2 + Br2 2HBr
2
22
BrHBr2
21
BrH1
HBr
C/Ck
CCkr
/
2N2O 2N2 + O2
2
22
2
O
ONON
ON1 Ck
Ckr
2OCk1>> Pseudo-first order with
respect to nitrous oxide
2OCk1
7/24/2019 Lecture 4- Stoichiometry
16/50
Examples of Forms of Rate Law and Units of Reaction Rate Constant
Zero-order:AA kr
secdm
mol
secdm
mol33 AA
kr secdm
mol3
k
First-order:AAA Ckr
[k]=sec-1
secdm
mol
dm
mol
sec
1
secdm
mol333 AAA Ckr
Second-order: 2
AAA Ckr
secmol
dm3k
secdm
mol
dm
mol
secmol
dm
secdm
mol36
22
3
3 AAA Ckr
Third-order: 3
AAA Ckr
secmol
dm2
6
k
secdm
mol
dm
mol
secmol
dm
secdm
mol39
33
2
6
3 AAA Ckr
Rate Law
7/24/2019 Lecture 4- Stoichiometry
17/50
TaskWhat is the reaction rate law for the reaction:if the reaction is elementary?Calculate the rates of A, B, and C in a CSTR where the concentrationsareCA = 1.5 mol/dm
3, CB = 9 mol/dm3 and kA = 2 (dm
3/mol)(1/2)(1/s).
C B2
1A
Rate Law
7/24/2019 Lecture 4- Stoichiometry
18/50
(6)
One more limit condition for reversible reactions: Equilibrium State
The rate of reaction is identical for all species and equal zero
(5)
Rate Law for Reversible Reactions
aA + bB k
cC + dDk-1
At thermodynamic equilibrium:
b
Be
a
Ae
d
De
c
Ce
CCC
CCK
KC is an equilibrium constant Units
abcd
3
dm
mol
The equilibrium constant characterise: equilibrium concentration of all species
ratio between rates of forward andreverse reactions
1-reverse
forward
k
k
k
k
KC
-rA 0
k= Forward reaction rate constantk-1 = Reverse reaction rate constant
The state in which the concentrations of the reactants and productshave no net change over time
Rate Law (Reversible Reaction)
7/24/2019 Lecture 4- Stoichiometry
19/50
Kinetic rate law for reversiblereaction must be consistentwith
thermodynamicequilibrium relationship
additional limit conditionwith specified rate law offorward and reversereactions
AkA
Bk-A
An elementary reversible reaction
(7)
formation of species A (8)
Reaction rate with respect to species A
AAA Ckr disappearance of species A
AAA Ckr
Reaction rate with respect to species A
BAA Ckr (9)formation of species A
b
Be
a
Ae
d
De
c
CeC
CCCCK
Forward reaction
Reverse reaction
A BkA
ABk-A
Rate Law (Reversible Reaction)
7/24/2019 Lecture 4- Stoichiometry
20/50
The overall rate of formation ofspecies A
(10)BAAA
reverseAforwardAA
CkCk
rrr
,,
(11)
The overall rate of disappearance of species A:
)( BA
AAABAAA
reverseAforwardAA
C
k
kCkCkCk
rrr
,,
(12)
Rate law for elementary monomolecular reversible reaction:
)(C
BAAA
K
CCkr
Check of the consistency with thermodynamic equilibrium:
)(0C
BeAeAA
KCCkr
After rearranging:C
BeAe
K
CC
Ae
Be
C C
CK
Ae
Be
C C
CK
Thermodynamics tells that:
At equilibrium:
AkA
Bk-A
1-reverse
forward
k
k
k
kKC
Rate Law (Reversible Reaction)
7/24/2019 Lecture 4- Stoichiometry
21/50
TaskWrite the rate law for the elementary reaction
Here kfA and krA are the forward and reverse specific reaction rates bothdefined with respect to A.
Rate Law (Reversible Reaction)
7/24/2019 Lecture 4- Stoichiometry
22/50
Stoichiometry provides information on relative quantities of reactants
and products in chemical reactions. We have already addressed related topics such as limiting reactant
and excess in our previous lecture.
Stoichiometry
Stoichiometry
Rate of reaction ri is a variable in all the design equations for
reactors. Rate of reaction is a function of thermodynamic quantities
such as temperature and also concentration of the species.
Conversion of the limiting reactant is a function of the concentrationand therefore rate of reaction is a function of the conversion: ri=f(Xi).
To evaluate the integrals associated with the design equations ofthe form
the functional form of ri(X) has to be found.
Stoichiometry provides a solution for this challenge.
)( jA Cfr )(XhC jj )(XgrA
rate law From a stoichiometric table
7/24/2019 Lecture 4- Stoichiometry
23/50
Mole Balance for Different Reactors
Reactor Type Differential Algebraic Integral
Batch
CSTR
PFR
PBR
Vrdt
dXN AA 0
)(
0
0
tX
A
A
A
Vr
dXNt
exit
0
)( A
A
r
XFV
A
A r
dV
dXF0
ou t
in
X
X A
Ar
dXFV 0
A
A rdW
dXF0
ou t
in
X
X A
Ar
dXFW 0
exit)r(
XC
A
A
0
ou t
in
X
X A
Ar
dXC 0
Conversion: Revision of concepts
7/24/2019 Lecture 4- Stoichiometry
24/50
Relative quantities (extent)
Stoichiometry: Extent
+ + 0 0 0 0
0 0 0 0
Reaction
Initial quantities
Quantities remaining at
t=t
Quantities reacted
Amount of B react with A = 0
0 = 0 0
= 0
0 =
0 =
0 =
0 =
Therefore the relative quantities are given by the following equation
[13]
The extent E of the reaction is the relative number of moles reacted/ generat
7/24/2019 Lecture 4- Stoichiometry
25/50
Relative reaction rates
Stoichiometry: Relative rates
+ + Reaction Rates
Rate of disappearance of A =
Rate of disappearance of B = Rate of appearance of C =
=
=
=
Therefore the relative reaction rates are given by
[14]
7/24/2019 Lecture 4- Stoichiometry
26/50
Constructing stoichiometric table
Stoichiometry: Stoichiometric table
Consider the reaction
t=0
NA0NB0NC0ND0NI0
t=t
NANBNCNDNI
+ +
I stands for inert substances, e.g.solventsReaction was terminated at time t when
the conversion of the limiting reactantA
reachedX. Composition of the product:
NA, NB, NC, ND and NI moles of substances
A, B, C, D and I, respectivelyFrom the definition of conversion
= 0 0
Xreachedconversion
whenreacted
Aofmoles
volumesystem
theintointroduced
Aofmoles
Xconversionat
volumesystema
inAofmoles
For every mole of A reacted, b/a moles of B must react
)(reactedmolesreactedBmoles 0XNa
bA
a
bA
7/24/2019 Lecture 4- Stoichiometry
27/50
Stoichiometry: Stoichiometric table
Species Initially
(mol)
Change
(mol)
Remaining
(mol)
A NA0
B NB0
C NC0
D ND0
I NI 0
Total NT0
)( 0XNA
)( 0XNab A
)( 0XNa
cA
)( 0XN
a
dA
XNNN AAA 00
XNabNN ABB 00
XNa
cNN ACC 00
XN
a
dNN ADD 00
XNa
b
a
c
a
dNN ATT 00 1
0II NN
A + b/a B c/a C + d/a D
XNa
d
a
c
a
bA01
)(
0
0
tX
A
A
A
Vr
dXNt
i i i i i
7/24/2019 Lecture 4- Stoichiometry
28/50
Total number of moles changed
Stoichiometry: Stoichiometric table
The total number of moles of all active species (reactants and products) at
conversionX of speciesA:
= 0+ +
1 0
The change in the total number of moles per mole of A
reacted
=+
1 =0 0
= 0+ 0 where
= + = /
0 = 0+ 0
We will introduce concentrations in thestoichiometric table using:
=
[15]
[16]
St i hi t St i hi t i t bl
7/24/2019 Lecture 4- Stoichiometry
29/50
Stoichiometry: Stoichiometric table
In order to define all concentrations on a basis per mole of A we
introduce the parameter :
The final equation for concentration of species B:
0
0
0
0
0
0
A
i
A
i
A
ii
y
y
C
C
N
N
V
XN)a/b(N
V
XN)a/b(NC AABABB
0000
V
Xa/bNC BAB
)(0 [18]
Expressing all in terms of NA0
We now need Vas a function of conversionX in order to obtain )(XhC jj
[17]
St i hi t St i hi t i t bl
7/24/2019 Lecture 4- Stoichiometry
30/50
Stoichiometry: Stoichiometric table
= 00
()
Initially
(mol)
Change
(mol)
Remaining
(mol) (mol/dm3)
A NA0
B
C
D
I
Total NT0
)( 0XNA
)( 0XNa
bA
)( 0XNa
cA
)( 0XNad
A
XNNN AAA 00
Xa
bNN BAB 0
Xa
cNN CAC 0
X
adNN DAD 0
0II NN
XNNN ATT 00
V
X1NC 0AA
)(
V
NC II
0
V
Xa/bNC BAB
)(0
V
Xa/cNC CAC
)(0
V
Xa/dNC DAD)(0
A + b/a B c/a C + d/a D
00 ABB NN
00 ACC NN
00 ADD NN
00 AII NN
St i hi t St i hi t i t bl
7/24/2019 Lecture 4- Stoichiometry
31/50
Stoichiometry: Stoichiometric table
Constant-Volume (or Density) Reaction Systems
Sealed vessel (gas phase reaction)
~Example:
Laboratory bomb calorimeter
(to measure the heat of combustion
of a particular reaction)
Isothermal gas-phase reaction with no change in the total number of moles
(ideal gas law applies)
~ Example: coal gasification
Liquid phase reaction in an incompressible solvent
(except polymerizations)
~ Solvent dominates
~ Changes in the density of the solute do not affect
the overall density of the solution significantly
CO + H2O CO2 + H2P
T
nRV
nRTPV
St i hi t St i hi t i t bl
http://upload.wikimedia.org/wikipedia/commons/3/34/Bombenkalorimeter_mit_bombe.jpghttp://upload.wikimedia.org/wikipedia/commons/3/34/Bombenkalorimeter_mit_bombe.jpg7/24/2019 Lecture 4- Stoichiometry
32/50
Stoichiometry: Stoichiometric table
= 0
0
()
Initially
(mol)
Change
(mol)
Remaining
(mol) (mol/dm3)
A NA0
B NB0
C NC0
D ND0
I NI 0
Total NT0
)( 0XNA
)( 0XNa
bA
)( 0XNa
cA
)( 0XNad
A
XNNN AAA 00
Xa
bNN BAB 0
Xa
cNN CAC 0
X
adNN DAD 0
0II NN
XNNN ATT 00
)1(0 XCC AA
XabCC BAB )( /0
Xa/cCC CAC )(0
XadCC DAD )/(0
0II CC
Stoichiometric Table for a Batch, Constant-Volume System
A + b/a B c/a C + d/a D
St i hi t St i hi t i t bl
7/24/2019 Lecture 4- Stoichiometry
33/50
Stoichiometry: Stoichiometric table
)( XCC AA 10
A + b/a B c/a C + d/a D
BAA CkCr
XabCXkCr BAAA )( /).1( 00
XabCC BAB )( /0
XabXkCr BAA )( /)1(02 )(XfrA
exit)(
0
A
A
r
XFV
Rate law
Design Equation ou t
in
X
X A
Ar
dXFV 0
Sizing of the batch, CSTR, PFR system
St i hi t i T bl E l
7/24/2019 Lecture 4- Stoichiometry
34/50
Task
A component of soap (sodium stearate) is produced by saponification of glyceryl
stearate with aqueous caustic soda according the following equation:
3 NaOH + (C17H35COO)3C3H5 3 C17H35COONa + C3H5(OH)3
Production is carried out in a batch reactor. The initial mixture consists solely of sodium
hydroxide at a concentration of 10 mol/dm3 and glyceryl stearate at a concentration
of 2 mol/dm3. What is the concentration of glycerine when the conversion of sodium
hydroxide is (a) 20% and (b) 90%?
Stoichiometric Table: Example
St i hi t i T bl Fl t
7/24/2019 Lecture 4- Stoichiometry
35/50
Stoichiometric Table for Flow Systems
EnteringFA0FB0FC0FD0FI0
LeavingFAFBFCFDFI
A + b/a B c/a C + d/a D
Batch System
)1(000 XNXNNN AAAA
[19]
Concentration of each species in a stream:
volume
moles
timevolume
timemoles
q
FC
j
j
/
/
Flow systems
)X(FXFFF AAAA 1000
Nj0
Nj
Fj0
Fj
In flow systems we deal with molar
fluxes
Replaced by
Stoichiometric Table: Flow system
St i hi t i T bl Fl t
7/24/2019 Lecture 4- Stoichiometry
36/50
Concentrations of species A, B, C, and D in terms of entering molarflow rate (FA, FB, FC, FD), the conversion X, and the volumetric flowrate
[20]
)1(0 Xq
F
q
FC AAA
q
XFabF
q
FC ABBB
00 )/(
q
XFacF
q
FC ACCC
00 )/(q
XFadF
q
FC ADDD
00 )/(
0
0
0
0
0
0
0
0
A
i
A
i
A
i
A
ii
y
y
C
C
vC
vC
F
F
The parameter :
Expressing all in terms of FA0 + +
Stoichiometric Table: Flow system
Stoichiometric Table: Flow system
7/24/2019 Lecture 4- Stoichiometry
37/50
Species Initially(mol / time) Change(mol / time) Remaining(mol / time)
A FA0
B
C
D
I
Total FT0
00 ABB FF
00 ACC FF
00 ADD FF
00 AII FF
)( 0XFA
)( 0XF
a
bA
)( 0XFa
cA
)( 0XFa
dA
)1(0 XFF AA
Xa
bFF BAB 0
Xa
cFF CAC 0
Xa
dFF DAD 0
IAI FF 0
XFa
b
a
c
a
dFF ATT 00 1
XFFF ATT 00
Stoichiometric Table for Flow Systems
Stoichiometric Table: Flow system
Stoichiometric Table: Flow system
7/24/2019 Lecture 4- Stoichiometry
38/50
For liquid-phase reactions, the volumetric flow rate does not practically
change in the course of reaction: v =v0 = Const
)1(0 XFF AA
X
a
bFF BAB 0
Xa
cFF CAC 0
Xa
dFF DAD 0
Stoichiometric Table for Flow Systems (Liquid-Phase Reactions)
)1()1( 00
0 XCXq
FC A
AA
X
a
bCC BAB 0
Xa
cCC CAC 0
Xa
dCC DAD 0
Therefore for a given rate law we have:)(Xgr
A
Stoichiometric Table: Flow system
Stoichiometric Table
7/24/2019 Lecture 4- Stoichiometry
39/50
For most batch and liquid-phase reactions and some gas-phase reactions
the volumetric flow rate or reaction volume does not change, however:
Volume Change with Reaction
volume (or volumetric flow rate) will change in the following situations:
Flow system with gas-phase reaction with a change in the total
number of moles
Flow system with non-isothermal gas-phase reaction
Batch system with variable volume
01a
b
a
c
a
d
Stoichiometric Table
+ +
N2 + 3H2 2NH3
Stoichiometric Table
7/24/2019 Lecture 4- Stoichiometry
40/50
Equation of State is a fundamental equation for gas systems:
RTZNPV T [21]Z = a compressibility factorNT= an overall number of moles in a gas mixture
[22]
By dividing equation (21) by equation (22):
00000 T
T
N
N
Z
Z
VP
PV
T
T
Volume Change with Reaction
The first four columns of the stoichiometric tables are always valid
Assumptions concerning a volume change were made when expressing
concentration in terms of conversion
Expressing volume as a function of conversion V = f (X) (Batch Systems)
000
00
T
T
N
N
Z
Z
P
PVV
T
T [23]
00000
RTNZVPT
time = 0
Stoichiometric Table
Stoichiometric Table
7/24/2019 Lecture 4- Stoichiometry
41/50
Total number of moles in system as a function of
conversion:
yA0 is the molar fraction of
species A in an initial
mixture
[24]
[25]
000
0
0T
T
N
N
Z
Z
P
PVV
T
T
XNNN ATT 00
XyXN
N
N
NA
T
A
T
T
0
0
0
0
11
Let the parameter :
[26]
reactorthetofedmolesofnumbertotal
conversioncompleteformolesofnumberin totalchange
00
0
0
0
1 AT
A
T
A
yN
N
N
N
a
b
a
c
a
d
Equation [25] now becomes:
XXy
N
NA
T
T 11 00
Equation [23]:
)1(
00
00 X
T
T
Z
Z
P
PVV [28]
Volume Change with Reaction in a Batch System
Stoichiometric Table
[27]
Stoichiometric Table
7/24/2019 Lecture 4- Stoichiometry
42/50
In gas-phase systems, at the certain ranges of temperatures and pressuresthe compressibility factor does not change significantly during the course of
the reaction: Z0Z
[29]
The volume of gas mixture for a variable volume batch reactorat anytime/conversion:
)1(00
0
0 XTT
ZZ
PPVV
0
00 )1(
T
TX
P
PVV
Volume Change with Reaction in a Batch System
If the reactor is a rigid steel containerof constant volume, then:
0VV
and equation [29] can be used to calculate the pressure change as afunction of TandX
Stoichiometric Table
Stoichiometric Table: Flow system
7/24/2019 Lecture 4- Stoichiometry
43/50
Consider the total concentration at
any point within the reactor:
[30]ZRT
P
qZRT
Pq
q
FC TT
[31]
At the entrance to the
reactor:
00
0
0RTZ
PCT
Taking the ratio of the equations(31) and (30) and assuming that
Z0Z:
[32]0
0
0
0T
T
P
P
F
Fqq
T
T
XXyX
F
F
F
FA
T
A
T
T 111 00
0
0
Total molar flux rate as a
function of conversion in acontinuous reactor:
[33]XFFF ATT 00
[34]
Volumetric flow rate as a functionof conversion can be obtained bycombining the equations (32) and
(34) :
0
00 )1(
T
T
P
PXqq [35]
Volume Change with Reaction in a Flow System
Stoichiometric Table: Flow system
Stoichiometric Table: Flow system
7/24/2019 Lecture 4- Stoichiometry
44/50
If the volumetric flow rate is known as a function of conversion, then theconcentration of any species in a flow system can be defined as a functionof conversion:
Thus:
[36]
0
00 )1(
T
T
P
PXqq
q
FC
j
j
Molar flow rate of each species as a function of conversion is:
XSFF jjAj 0
T
T
P
P
Xq
XSFC
jjA
j 0
00
0
)1(
)(
T
T
P
P
X
XSCC
jjA
j0
0
0
)1(
)(
Concentrations as a Function of Conversion
Stoichiometric Table: Flow system
Stoichiometric Table: Flow system
7/24/2019 Lecture 4- Stoichiometry
45/50
T
T
P
P
X
XsCC
jjA
j0
0
0
)1(
)(
Initially
(mol/time)
Change
(mol/time)
Remaining
(mol/time) (mol/volume. time)
A FA0
B
C
D
I
Stoichiometric Table for a Variable-Volume Gas Flow System
T
T
P
P
X
XCC AA
0
0
0)1(
1
T
T
P
P
X
Xa/bCC BAB
0
0
0)1(
)(
T
T
P
P
X
Xa/c
CC
C
AC
0
00 )1(
)(
T
T
P
P
X
Xa/dCC DAD
0
0
0)1(
)(
T
T
P
P
X
CC IAI
0
0
0
)1(
IAI FF 0
Xa
dFF DAD 0
Xa
cFF
CAC 0
Xa
bFF BAB 0
)1(0 XFF AA
00 ABB FF
00 ACC FF
00 ADD FF
00 AII FF
)( 0XFA
)( 0XFa
bA
)( 0XFa
cA
)( 0XFa
dA
Stoichiometric Table: Flow system
+ +
Stoichiometric Table: Flow system-Example
7/24/2019 Lecture 4- Stoichiometry
46/50
Task
A mixture of 28% SO2 and 72% air is charged to a flow reactor in which SO2 is
oxidized.
2 SO2 + O2 2 SO3
Calculate concentrations of all species at the following conversion of SO2: 0; 0.25;
0.5; 0.75; 1, if the total pressure is 1485 kPa and the temperature is constant at
227oC. Note that air composition is 21% O2 and 79% N2.
Solution
Taking SO2 as the basis of calculations:
+12
Stoichiometric Table: Flow system Example
2 SO2 + O2 2 SO3
Stoichiometric Table: Flow system-Example
7/24/2019 Lecture 4- Stoichiometry
47/50
Task
The elementary gas phase reaction
3A + 2B
3 C +5 D
is carried out in a flow reactor operated isothermally at 427C and 28.7 atmospheres.
Pressure drop can be neglected. Express the rate law and the concentration of each
species as a function of conversion and as a function of the total molar flow rates. The
entering volumetric flow rate is 10 dm3/s and the specific reaction rate,k is 200
dm12/mol4s. The feed is equal molar in A and B.
Solut ion
k1
A is the limiting reactant
Stoichiometric Table: Flow system Example
Volume change (Flow system) -Example
7/24/2019 Lecture 4- Stoichiometry
48/50
Task
The elementary gas phase reaction
3A + 2B
3 C +5 D
is carried out in a flow reactor operated isothermally at 427C and 28.7 atmospheres.
Pressure drop can be neglected. Express the rate law and the concentration of each
species as a function of conversion and as a function of the total molar flow rates. The
entering volumetric flow rate is 10 dm3/s and the specific reaction rate is 200 dm12/mol4s.
The feed is equal molar in A and B.
k1
Volume change (Flow system) Example
Volume change (Flow system) -Example
7/24/2019 Lecture 4- Stoichiometry
49/50
Task
The elementary gas phase reaction
3A + 2B
3 C +5 D
is carried out in a flow reactor operated isothermally at 427C and 28.7 atmospheres.
Pressure drop can be neglected. Express the rate law and the concentration of each
species as a function of conversion and as a function of the total molar flow rates. The
entering volumetric flow rate is 10 dm3/s and the specific reaction rate is 200 dm12/mol4s.
The feed is equal molar in A and B.
k1
Volume change (Flow system) Example
Summary
7/24/2019 Lecture 4- Stoichiometry
50/50
Summary
Recommended