APPLIED REACTION KINETICS Friday A01 09:00-11:00 (Assoc ...bernauem/ark/lectures/Chapter 1.pdf ·...
Transcript of APPLIED REACTION KINETICS Friday A01 09:00-11:00 (Assoc ...bernauem/ark/lectures/Chapter 1.pdf ·...
APPLIED REACTION KINETICS
Friday A01 09:00-11:00 (Assoc. prof. Bohumil Bernauer, A27, [email protected],
+420220444017 )
Friday BS9 13:00-15:00 (Dr. Milan Bernauer, H08, [email protected],
+420220444287)
κίνησις "kinesis", movement or to move
Resources and references
• Notes from lectures
• Internet
web.vscht.cz/bernauem/
• Textbooks
Fogler Scott H.: Elements of Chemical Reaction Engineering, 4th Edition, Prentice
Hall, 2006. (http://www.engin.umich.edu/~cre/)
Missen R.W., Mims C.A., Saville B.A., Introduction to Chemical Reaction
Engineering and Kinetics, J. Wiley&Sons, N.Y. 1999.
• Journals (on-line)
• Software
MS Excel, (Matlab, Octave, Athena Visual Studio, FORTRAN,
Maple….)
Chemical reactor(s) heart of
chemical process
Raw material separation reaction separation product
Fischer-Tropsch (SASOL, RSA ) WGS (BASF, FRG)
Methane aromatization (ICTP, CZ)
N2O decomposition (IPC AS, CZ)
( , , , ..., )r composition T P params
Summary of the 1st lecture • Stoichiometry • Extent of reaction • Fractional conversion of key component • Stoichiometric matrix • Balance of chemical elements • Selectivity, Yield • Reaction rate definition
2NO N2 + O2 closed (batch) system
Stoichiometry
t =0 t > 0
Atoms of oxygen
Atoms of nitrogen
2
2o o
NO On n
2
2NO On n
2
2o o
NO Nn n
2
2NO Nn n
2 2
2 2o o
NO O NO On n n n
2 2
2 2o o
NO N NO Nn n n n
2 2
2( )o o
NO NO O On n n n
2 2
2( )o o
NO NO N Nn n n n
2 2 2 2
( ) ( )
2 1 1
o oo
O O N NNO NOn n n nn n
-2NO + O2 + N2 = 0 Symbols for species NO = A1 O2 = A2 N2 = A3
Stoichiometric coefficients 1 = -2 2 = 1 3 = 1
3
1
0 produ cts
0 0 inerts
0 reactants
i
i i i
i
i
A
" " the m aterial elem ent (P lato)
" " the count, the quantity
Molar extent of the reaction [ksi:]
2 2 2 2
( ) ( )
2 1 1
o oo
O O N NNO N
i
O
o
i
i
n
n n n nn n
n
From the definition of the reaction extent follows:
1. The reaction extent has the dimension of moles (number of molecules)
2. The reaction extent value depends on stoichiometry of reaction
3. The reaction extent is an extensive variable
t = 0 t > 0
Reaction extent for a single reaction in closed (batch) system
ino
in
o
i i
i
n n
o
i i in n
Closed system
1
0 produ cts
0 0 inerts
0 reactan
ts
i
N
i i i
i
i
A
Example
2NO N2 + O2
1 2 2 3 2
1 2 1
1
2
3
1
2
3
N O , N , O
2, 1, 1
2 1 1 ,
2 1 1 0
T
T
A A A
A
A
A
A
A
A
ν A
ν A
Matrix notation
Fractional conversion of key component, j
m ax
jX
j
o
jjnn
*
max
o
j
i
o
i
i
j
j
o
j
i
o
ii
jn
nn
n
nn
X
max
jX
max
o
j
j j
j
nX X
j
o
j
j
io
iiXnnn
(0,1)
o
j j
j jo
j
n nX X
n
100 (0,100)
o
j j
j jo
j
n nX X
n
Number of moles of key
component in limits
(chemical equilibrium or 0)
*0
jn
i j
Stoichiometric matrix in the case of several reactions
1
0 1,
N
ki i
i
A k NR
11 12 1 1
21 22 2 2
,1 ,2 ,
...
...,
. .
...
N
N
NR NR NR N N
A
A
A
ν A
Stoechimetric matrix has NR rows and N columns
reaction component
νA 0
1
NR
o
i i ki k
k
n n
o
ki ki
k
ki
n n
Number of moles of i-th component consumed or created in k-th reaction :
o T n n ν ξ
1 11
2 22, ,
.. ....
o
o
o
o
N NRN
n n
n n
n n
n n ξ
o
ki kin n
Molar extent of k-th reaction :
Number of moles of i-th
components:
Matrix notation
Problem 1.1
Oxidation of ammonia on Pt-Rh catalyst
NH3 + 1.25 O2 NO + 1.5H2O (1)
NH3 + O2 0.5 N2O + 1.5H2O (2)
NH3 + 0.75O2 0.5 N2 + 1.5H2O (3)
Task: To write down the stoichiometric matrix.
A1 A2 A3 A4 A5 A6
Reaction NH3 O2 NO N2O N2 H2O
(1) -1 -1.25 1 0 0 1.5
(2) -1 -1 0 0.5 0 1.5
(3) -1 -0.75 0 0 0.5 1.5
Molar balance table of component in closed (batch) system
Component t = 0 t > 0
NH3 1
on
1 1 1 2 3( )
on n
O2 2
on
2 2 1 2 31.25 0.75
on n
NO 3
on
3 3 1
on n
N2O 4
on
4 4 20.5
on n
N2 5
on
5 5 30.5
on n
H2O 6
on
6 6 1 2 31.5( )
on n
6
1
o
i
i
n
6
1 3
1
0.25o
i
i
n
3
1 1 2 3
1 6
1 3
1
e.g . m olar fraction of N H
( )
0 .25
o
o
i
i
ny
n
Independent reactions
Set of NR reactions in reaction network is independent if
Rank()=NR
or
number of independent reactions = Rank()
Problem 1.2
NH3 + 1.25 O2 NO + 1.5H2O (1)
NH3 + O2 0.5 N2O + 1.5H2O (2)
NH3 + 0.75O2 0.5 N2 + 1.5H2O (3)
2NO N2 + O2 (4)
Task: To calculate the number of independent reactions.
We determine the rank of stoichiometric matrix by Gaussian elimination:
1 1.25 1 0 0 1.5 1 1.25 1 0 0 1.5
1 1 0 0.5 0 1.5 0 0.25 1 0.5 0 0
1 0.75 0 0 0.5 1.5 0 0.5 1 0 0.5 0
0 1 2 0 1 0 0 1 2 0 1 0
1 1.25 1 0 0 1.5 1 1.25 1 0 0 1.5
0 0.25 1 0.5 0 0 0 0.25 1 0.5 0 0
0 0 1 1 0.5 0 0 0 1
0 0 2 2 1 0
1 0.5 0
0 0 0 0 0 0
Rank()=3 only 3 reactions are independent
Extent of the reaction in a flow system
1
0 1,
N
ki i
i
A k NR
o
iF
Open (flow) system in a steady state
iF
1
, -outlet, inlet m olar flow rates of i-th species [m ole/s]
extent of k-th reaction per unit of tim e [m ole/s]
m ean residence tim e [s]
NR
o k
i i ki k k
k
o
i i
k
T
F F
F F
oF = F + ν ξ
& &
&
&1 1 1
2 2 2, ,
.. .. ..
o
o
o
o
N N NR
F F
F F
F F
F F ξ
&
&&
&
Inlet molar flow rates
[mole of i-th species/s]
Outlet molar flow rates
[mole of i-th species/s]
REACTOR
A reaction is at steady-state if the concentration of all species in each
element of the reaction space (i.e. volume in the case of homogeneous
reaction or surface of catalyst in the case of catalytic heterogeneous
reaction) does not change in time.
Balance of chemical elements
1
0 1,
N
ki i
i
A k NR
Open (flow) system at steady state
, -outlet, inlet m olar flow s
of j-th chem ical elem ent [m ole/s]
o
o
j
j j
j
oΦ = Φ
1 1
2 2,
.. ..
o
o
o
NEL NEL
oΦ Φ
o
j
j
REACTOR
NEL - Number of
chemical elements
Formula matrix E
N components
N=6
N H O
NH3 1 3 0
O2 0 0 2
NO 1 0 1
N2O 2 0 1
N2 2 0 0
H2O 0 2 1
NEL elements NEL=3
There are no creation or annihilation of chemical elements in chemical reactions:
νE = 0
11 12 1 11 12 1,
21 22 2 21 2 ,
,1 ,2 , ,1 ,2 ,
... ... 0 0 ... 0
... ... ... 0 0 ... 0
. . .
... ... 0 0 ... 0
N NEL
N NEL
NR NR NR N N N N NEL
E E E
E E
E E E
NR
NEL
Formula vector of NH3
Molar* weight (relative molecular mass) of i-th species:
1
1 1
2 2,
.. ..
NEL
i ij j
j
N NEL
M E m
M m
M m
M m
M = Em
M m
and we have for molar weights of species
0
because 0
νM = νEm
νE
Atomic weights (relative atomic mass) of j-th element
*The mole is the amount of substance of a system which contains as many elementary
entities as there are atoms in 0.012 kilogram of carbon 12.
Avogadro constant = 6.022 141 29(27) × 1023 mol−1 (http://www.nist.gov)
Balances of atoms in batch and flow systems
C losed (batch) system :
O pen (flow ) system at steady state:
because
T
TT T T T T T
T
TT T T T T T
o
o o o
o
o o o
n = n + ν ξ
E n = E n + E ν ξ E n + νE ξ E n
F = F + ν ξ
E F = E F + E ν ξ E F + νE ξ E F
νE 0
&
& &
Finally
T T
T T
o o
o o
E n n E n n 0
E F F = E F F 0
These equations are used
in data reconciliation tasks
around chemical reactors.
The last equation (flow
system) is valid only at
steady state.
Problem 1.3
Selective reduction of NOx by C3H8
NO NO2 CO C3H8 CO2 H2O O2 N2
4.563 0.1845 0 2.943 0 0 90 0
2.2905 2.3355 0 2.898 x x x x
[ / min]o
iF mole
o
iF i
F
[ / m in]iF mole
Calculate the missing outlet molar flow rates
Steady state,
unknown
stoichiometry
of reactions
T o T E F - F = E F 0
1-NO 2-NO2 3-CO 4-C3H8 5-CO2 6-H2O 7-O2 8-N2
N 1 1 0 0 0 0 0 2
O 1 2 1 0 2 1 2 0
C 0 0 1 3 1 0 0 0
H 0 0 0 8 0 2 0 0
1
2
5
6
3
4
7
8
F
F
F
F
F
F
F
F
unknown
iF
1
TE 2
TE
1 2
2 1
1 1
2 1 2 1
T T
KNOWN UNKNOWN
T T
UNKNOWN KNOWN
T T T T
UNKNOWN KNOWN KNOWN
E F E F 0
E F E F
F E E F Q F Q E E
known
iF
0 0 1 3
0 0 0 4
0.5 1 0.5 5
0.5 0.5 0 0
Q
We obtain using Excel (homework 1)
NO NO2 CO C3H8 CO2 H2O O2 N2
4.563 0.1845 0 2.943 0 0 90 0
2.2905 2.3355 0 2.898 0.135 0.18 88.76 0.061
and taking we have 1
2 2 1 1 1
T T
F E E F Q F
[ / min]o
iF mole
[ / m in]iF mole
Selectivity Moles of a particular product generated per mole of key reactant consumed
Yield Moles of a particular product generated per one initial mole of key reactant
Component t = 0 t > 0
NH3 1
on
1 1 1 2 3( )
on n
O2 2
on
2 2 1 2 31.25 0.75
on n
NO 3
on
3 3 1
on n
N2O 4
on
4 4 20.5
on n
N2 5
on
5 5 30.5
on n
H2O 6
on
6 6 1 2 31.5( )
on n
Ammonia oxidation
3
3
3
3 1 1
1 2 3 1 2 3
0
3 1 1
1 1
( ) ( )
o
o
NO NH
no
NO NH o o
nS
nY
n n
Reaction rate
(IUPAC Gold Book = rate of conversion )
11 m ole.s
i
i
k
k
dndr
dt dt
dr
dt
one reaction
several reactions
1 1
. .
. .
i
i
i
A i i
NR NR
Tk
A ki ki k
k k
dn dR r r
dt dt
dR r
dt
R = ν
R = ν r
Rate of generation (consumption) of component Ai
Closed system
of uniform
pressure,
temperature
and
composition.
t
m ax
0; 0
d
d t
exp exp exp O bjective function
m inim ization of = least squares solution of overdefined system (N >N R )
TT T T
R ν r r R ν r R ν r
r
We measure usually the rates of generation (consumption) of
components R and we want to calculate r
Problem 1.4
Steam reforming of methane (5 species, N=5, 2 reactions, NR=2)
1 2 3 4
CH4 + H2O CO + 3H2 (1)
5
CO + H2O CO2 + H2 (2)
Measured R : (-0.97572, -2.88778, -1.02127, 4.832804, 2.078537)T (mol/s)
11 1 1 3 0 12 3
0 1 1 1 1 3 4
T T
ν νν νν ν-0.10256 -0.02564 0.179487 0.230769 -0.07692
0.076923 -0.23077 -0.38462 0.076923 0.307692
1
2
0.94619
1.99546
r
r
1
expT
r νν νR
Specific reaction rate
The reaction rate is, like , an extensive property of the system, a specific rate (intensive property) is obtained by dividing by the total volume, mass, surface of the system:
3 1
,
1 1 1 1 m ole.m .
1
R eaction rate per vo um
1
l e
i
V
i
k
V k k
dndr r s
V V dt V dt
dr r
V V dt
1 1
,
1 1 1 1 m ole.kg .
1 1
R eaction rate per m ass
i
M
i
k
M k k
dndr r s
m m dt m dt
dr r
m m dt
d
d t
d
d t
2 1
,
1 1 1 1 m ole.m .
1
R eaction rate per surface
1
i
S
i
k
S k k
dndr r s
S S dt S dt
dr r
S S dt
1
,
R eaction rate per active center (turn
1 1 1 1 s
1
over num be
1
r)
i
RS
RS RS RS i
k
RS k k
RS RS
dndr r
n n dt n dt
dr r
n n dt
1 2, ,
C entral
, .
proble
.. , , catalytic activity, transport param e ters,...
m of APPLIED C H EM IC AL K IN ET IC
.
S
Nr function T c c c P
Rule 1: The rate function r at constant temperature generally
decreases in monotonic fashion with time (or extent or conversion).
Rule 2: The rate of irreversible reaction can be written as
1 2( ) , , ...
Nr k T g c c c
Rule 3: The rate constant k depends on temperature (Svante
Arrhenius, 1889):
R( )
E
Tk T Ae
Rule 4: The function g is independent of temperature:
1 2
1 2 1 2
1
, , ... ... N i
N
N N i
i
g c c c c c c c
Rule 5: When a reaction is reversible:
1 2 1 2( ) , , ... ( ) , , ...
f b f f N b b Nr r r k T g c c c k T g c c c
Problem 1.5 (homework 2)
In flow catalytic reactor the synthesis of methanol is carried out
CO(g) + 2H2(g) CH3OH(g)
The inlet mass flow rate of CO is 1000 kg.h-1 of CO and the inlet flow rate of
hydrogen is supplied so that the inlet molar ratio H2:CO is equal to 2:1. 1200 kg of
the catalyst is placed in the reactor.
The outlet mass flow of CO is 860 kg.h-1.
To determine:
1. Mean reaction rate per mass of catalyst in mol.kg-1.s-1 .
2. If specific internal surface of catalyst is 55 m2.g-1, calculate mean reaction rate
per surface of catalyst in mol.m-2.s-1.
3. If per 1 m2 of catalyst contains 1019 active sites, calculate mean reaction rate
per active site in s-1.
4. Calculate inlet and outlet gas mixture composition in molar fractions.
Data:
MCO= 28.010 kg.kmol-1
NA=6.0221413x1023 mol-1 (Avogadro number)
1 2 2 3 3
1
, ,
1 1
2 , 2
1 1
, ,
NR
o o
i i ki k i i i
k
o o o
i i i i i i
M S RS
CAT i CAT i RS i RS
CO A H A CH OH A
r
F F F F
F F F F F Fr r r
m m S S n n
ν R
& &
& & &
inlet outlet
1-CO
2-H2
3-CH3OH
1
oF
12
oF
0
13
o oF F
o
iy
1
1
1
1
3 3
o
o
o
Fy
F
1
2
1
2 2
3 3
o
o
o
Fy
F
0
1 11
oF F &
2 22
oF F &
3 31
oF F &
1 2o
F F &
iy
1
1
1
1
3 2
o
o
Fy
F
&
&
1
2
1
2 2
3 2
o
o
Fy
F
&
&
3
13 2
oy
F
&
&
1
1 1 11
oF F X
1
1 1 1
1 1
Fractional conversion of key com ponent
1 key com ponent
/
o o
o oi i i
i i
MAX i
F F FX F F F X
&
&
2 1 1 12 2
o oF F F X
3 1 1
oF F X
1 1 13 2
o oF F F X
1
3 1 1 8 2 1
3
23
3 1
3 19
860 1000 / 0.02801 / 36001, 1.38839 m ol.s
1
1.38839 1.388391.157 10 m ol.kg . , 2 .104 10 m ol.m .
1200 55 10 1200
1.38839 6.0221413 101.267 10 s
55 10 1200 10
M S
RS
i
r s r s
r
&
inlet outlet
1-CO
2-H2
3-CH3OH
o
iy
1
1
1
1
3 3
o
o
o
Fy
F
1
2
1
2 2
3 3
o
o
o
Fy
F
0
1
1 1 1
1
1 1 1 1
1 10.3162
3 2 3 2
o
o o
F X Xy
F F X X
1
1000 8600.14
1000X
11 1 1
2
1 1 1 1
2 12 20.6324
3 2 3 2
o o
o o
XF F Xy
F F X X
iy
1 1 1
3
1 1 1 1
0.051473 2 3 2
o
o o
F X Xy
F F X X
1