APPLIED REACTION KINETICS [PDF]

Internet web.vscht.cz/bernauem/. • Textbooks. Fogler Scott H.: Elements of Chemical Reaction Engineering, 4th Edition,

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Idea Transcript


κίνησις "kinesis", movement or to move

APPLIED REACTION KINETICS Friday A01 09:00-11:00 Friday BS9 13:00-15:00

(Assoc. prof. Bohumil Bernauer, A27, [email protected], +420220444017 ) (Dr. Milan Bernauer, H08, [email protected], +420220444287)

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….)

Fischer-Tropsch (SASOL, RSA )

N2O decomposition (IPC AS, CZ)

WGS (BASF, FRG)

Chemical reactor(s)  heart of chemical process

Methane aromatization (ICTP, CZ)

Raw material separation  reaction  separation  product

r ( com position , T , P , ..., param s )

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

"      " the m aterial elem ent (P lato)

Stoichiometry

"   " the count, the quantity

2NO  N2 + O2

closed (batch) system t =0

Atoms of oxygen

n N O  2 nO2 

Atoms of nitrogen

n NO  2 n N 2 

o

t>0

o

o

n N O  2 nO2

o

o

n NO  2 n N 2

o

o

n N O  2 nO2  n N O  2 nO2 o

o

n N O  n N O   2( n O 2  n O 2 )

2

Symbols for species Stoichiometric coefficients

-2NO + NO = A1 1 = -2

o

o

n N O  n N O   2( n N 2  n N 2 ) o

o

n NO  n NO

o

n NO  2 n N 2  n NO  2 n N 2



( nO2  nO2 )

O2 + O2 = A2 2 = 1

1

o



N2 = 0 N2 = A 3 3 = 1

i  0

p ro d u cts

i  0

in erts

3



i

Ai  0

(nN2  nN2 )

i 1

i  0

reac tan ts

1

Molar extent of the reaction o

o

 

n NO  n NO 2



( nO2  nO2 ) 1



[ksi:] o



(nN2  nN2 ) 1

o

 

ni  ni

i

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

Reaction extent for a single reaction in closed (batch) system t=0

t>0

o

ni

ni

Closed system

i  0

p ro d u cts

i  0

in erts

N



i

Ai  0

i 1

i  0 o

 

ni  ni

i

ni  ni   i o

reac tan ts

Example

2NO  N2 + O2

A1  N O , A2  N 2 , A3  O 2

 1   2, 2  1, 1  1 ν   2 T

ν A   2 T

 A1    1 1  , A  A2   A   3  A1    1 1  A2  0   A   3

Matrix notation

Fractional conversion of key component, j

X

j



j

  m ax *

 max 

X

o

nj  nj



Number of moles of key component in limits (chemical equilibrium or 0)

j

o

*

o

nj  0

X

j



  max

ni  ni 

i o

 nj







Xj

i j

nj  nj

o i

n  ni

j

i

n

o j

X j  (0,1)

o

nj

o

X

j

j

 100

nj  nj n

X j  (0,100)

o j

o

  X j m ax   X

nj j



j

o i

ni  n 

i 

j

o

njX

j

Stoichiometric matrix in the case of several reactions N



ki

Ai  0

k  1, N R

i 1

reaction

component

Stoechimetric matrix has NR rows and N columns

  11   21  ν  .    N R ,1

 12

...

 1N 

 22

...

 2N

 N R ,2

...  N R , N

νA  0

   ,A        

A1   A2  .   AN 

Number of moles of i-th component consumed or created in k-th reaction : o

Molar extent of k-th reaction :

k 

n ki  n ki

 ki

NR

Number of moles of i-th components:

o i

ni  n 



ki

k

k 1

Matrix notation

o

T

nn ν ξ  n1  n2 n  ..   nN

 n1o    1  o    2 n o 2   ,n  ,ξ    ..    ..      no     NR  N 

     

o

n ki  n ki

Problem 1.1 Oxidation of ammonia on Pt-Rh catalyst NH3 + 1.25 O2  NO + 1.5H2O NH3 + O2  0.5 N2O + 1.5H2O NH3 + 0.75O2  0.5 N2 + 1.5H2O

(1) (2) (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 NH3 O2 NO N2O N2 H2O 

t=0

t >0

o

n1  n1  ( 1   2   3 )

o

n 2  n 2  1.25 1   2  0.75 3

o

n3  n3  1

n4

o

n 4  n 4  0.5 2

o

n 5  n 5  0.5 3

o

n 6  n 6  1.5( 1   2   3 )

o

n1

o

n2

o

n3

o

o

n5

o

n6

6

6

n i 1

o i

n

o i

 0.25   1   3 

i 1

e.g .

m o lar fractio n o f N H 3 n1  (  1   2   3 ) o

y1 

6



i 1

n i  0 .2 5   1   3  o

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 NH3 + O2  0.5 N2O + 1.5H2O NH3 + 0.75O2  0.5 N2 + 1.5H2O 2NO  N2 + O2

(1) (2) (3) (4)

Task: To calculate the number of independent reactions. We determine the rank of stoichiometric matrix by Gaussian elimination:

 1  1   1   0

 1.25

1

0

0

1

0

0.5

0

 0.75

0

0

0.5

1

2

0

1

 1  0    0   0

1.5   1   1.5 0    0 1.5    0   0

 1.25

1

0

0

0.25

1

0.5

0

0

1

1

0.5

0

2

2

1

Rank()=3  only 3 reactions are independent

 1.25

1

0

0

0.25

1

0.5

0

0.5

1

0

0.5

1

2

0

1

1.5   1   0 0    0 0    0   0

1.5   0  0   0 

 1.25

1

0

0

0.25

1

0.5

0

0

1

1

0.5

0

0

0

0

1.5   0  0   0 

Extent of the reaction in a flow system

Fi

o

Fi

Inlet molar flow rates [mole of i-th species/s]

REACTOR

Outlet molar flow rates [mole of i-th species/s]

N

 i 1

ki

Ai  0 NR

Fi  Fi    ki &k o

k  1, N R Open (flow) system in a steady state

k &   k

k 1



o

Fi , Fi -o u tlet, in let m o lar flo w rates o f i-th sp ecies [m o le/s]

&k  ex ten t o f k -th reactio n p er u n it o f tim e [m o le/s]   m ean resid en ce tim e [s] o T F = F + ν ξ&

 F1  F2  F   ..   FN

 F1o    o   , F o   F2  , ξ&  ..    o     FN 

 & 1  &  2  ..   &  NR

      

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



o j

 REACTOR

j

N



ki

Ai  0

k  1, N R

Open (flow) system at steady state

i 1

j 

o j

o

 j ,  j -outlet, inlet m olar flow s of j-th chem ical elem ent [m ole/s] Φ =Φ

o

 1  2  Φ   ..    NEL

  1o    o ,Φo    2  ..   o     NEL

     

NEL - Number of chemical elements

Formula matrix E

NEL elements  NEL=3 N components  N=6 NH3

N

H

O

1

3

0

O2

0

0

2

NO

1

0

1

N2O

2

0

1

N2

2

0

0

H2O

0

2

1

Formula vector of NH3

There are no creation or annihilation of chemical elements in chemical reactions:

νE = 0

  11    21  .    N R ,1

 12

...

 1 N   E 11

E 12

...

 22

...

 2N

...

...

E N ,2

...

 N R ,2

...  N R , N

 E   21  .     E N ,1

E 1, N E L   0   E 2, NEL   0  .   E N , N E L   0

0

...

0

...

0

...

NEL

0  0    0

NR

Molar* weight (relative molecular mass) of i-th species: NEL

Mi 

E

ij

mj

j 1

Atomic weights (relative atomic mass) of j-th element

M = Em  M1  M2  M   ..  MN

  m1   m2 ,m     ..     m NEL

     

and we have for molar weights of species

νM = νE m  0 because νE  0 *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 lo sed (b atch ) system : o

T

n =n +ν ξ T

E n = E n + E ν ξ  E n +  νE  ξ  E n T

T

o

T

T

T

o

T

o

O p en (flo w ) system at stead y state: o T F = F + ν ξ& T T o T o & E F = E F + E ν ξ  E F +  νE  ξ& E F T

T

o

T

T

b ecau se νE  0 Finally

E

T

E

T

n  n   E n F  F  = E F o

o

T

T



o

n  0 o



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

Fi

o

Fi

NO 4.563 2.2905

Fi [  m ole / m in] Fi [  m ole / m in] o

Steady state, unknown stoichiometry of reactions

NO2 0.1845 2.3355

C3H8

CO 0 0

CO2 0 x

2.943 2.898

H2O 0 x

O2 90 x

N2 0 x

Calculate the missing outlet molar flow rates

E

N O C H

T

F

o

- F  = E F  0 T

1-NO

2-NO2

3-CO

4-C3H8

5-CO2

6-H2O

7-O2

8-N2

1 1 0 0

1 2 0 0

0 1 1 0

0 0 3 8

0 2 1 0

0 1 0 2

0 2 0 0

2 0 0 0

T

E2

T

E1 T

T

E 1  F K N O W N  E 2  FU N K N O W N  0 T

T

E 2  FU N K N O W N   E 1  F K N O W N

 

 FU N K N O W N   E

T 2

1

T

E 1  FK N O W N  Q  FK N O W N

  T

Q   E2

  F1   F2    F3    F4  F 5    F6  F 7  1 T E 1   F8

            

known

 Fi

unknown

 Fi

We obtain using Excel (homework 1)

 0  0  Q    0.5    0.5 

and taking

 F2   E 2

Fi [  m ole / m in] Fi [  m ole / m in]

NO 4.563 2.2905

o

T



1

T

0

1

0

0

1

0.5

 0.5

0

E 1  F1  Q  F1

NO2 0.1845 2.3355

3   4  5   0 

we have CO 0 0

C3H8 2.943 2.898

CO2 0 0.135

H2O 0 0.18

O2 90 88.76

N2 0 0.061

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

Ammonia oxidation

Component NH3 O2 NO N2O N2 H2O

t=0

t >0

o

n1  n1  ( 1   2   3 )

o

n 2  n 2  1.25 1   2  0.75 3

o

n3  n3  1

n4

o

n 4  n 4  0.5 2

o

n 5  n 5  0.5 3

o

n 6  n 6  1.5( 1   2   3 )

o

n1

o

n2

o

n3

o

o

n5

o

n6

n3  1 o

S NO  NH 3  YNO  NH 3 

( 1   2   3 ) n n  1  o 3

o 1

n

o 3

0





 1 o

n1

1 ( 1   2   3 )

Reaction rate (IUPAC Gold Book = rate of conversion )



 m ax

r 

d dt

d dt

 i dt

m ole.s

1

dk

rk  0; 0



1 d ni

Closed system of uniform pressure, temperature and composition.

dt

t

Rate of generation (consumption) of component Ai

R Ai 

d ni dt NR

R Ai 

  i.

  ki . k 1

d dt

dk dt

  i .r NR



  ki .rk k 1

o n e reactio n



R = νr

several reactio n s



T

R = ν r

We measure usually the rates of generation (consumption) of components R and we want to calculate r

R

exp

T

 ν r

 O bjective function



 r   R

exp

T

ν r

 R T

exp

T

ν r



m inim ization of   r  = least squares solutio n of overdefined system (N > N R )



r  νν

T



1

νR

exp

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)  1 ν  0

1

1

3

1

1

1

0  1

 r1   0.94619      r2   1.99546 

 12 T νν    3

3  4

 νν  T

1

ν

-0.10256 -0.02564 0.179487 0.230769 -0.07692 0.076923 -0.23077 -0.38462 0.076923 0.307692

Specific reaction rate d

The reaction rate dt is, like  , an extensive property of the system, a specific rate (intensive property) is obtained by dividing d  by the total volume, mass, surface of the system: dt

R eaction rate per vo l um e rV 

1

r 

V

rV , k 

1 V

1 d V dt

rk 



1 1 dn i V  i dt

3

m ole.m . s

1

1 dk V

dt

R eaction rate per m ass rM 

1 m

rM , k 

r  1

m

1 d m dt

rk 



1 dk m dt

1 1 dn i m  i dt

1

m ole.kg . s

1

R eaction rate per surface rS 

1

r 

S

1

rS , k 

1 d

S

S dt

rk 



1 1 dn i S  i dt

m ole.m

2

.s

1

1 dk S dt

R eaction rate per active center (turn over num be r) rR S 

1 n RS

rR S , k 

r

1 n RS

1 d n R S dt

rk 



1 dk n R S dt

1 1 dn i n R S  i dt

s

1

C entral proble m of A P P LIE D C H E M IC A L K IN E T IC S r  function  T , c1 , c 2 , . ..c N , P , catalytic activity, transport param e ters,... . 

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 r  k (T ) g  c1 , c 2 , ...c N



Rule 3: The rate constant k depends on temperature (Svante Arrhenius, 1889): k (T )  A e



E

RT

Rule 4: The function g is independent of temperature: g  c1 , c 2 , ...c N

  c1

1

2

N

c 2 ...c N 

N



i

ci

i 1

Rule 5: When a reaction is reversible: r  r f  rb  k f (T ) g f  c1 , c 2 , ...c N   k b (T ) g b  c1 , c 2 , ...c N



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)

C O  A1 , H 2  A2 , C H 3 O H  A3  1   1      ν  2 , R  2 r      1   1      NR



o

Fi  Fi 

Fractional conversion of key com ponent 1  key com ponent X1 

&k  Fi  Fi   i & o

ki

& &M A X

k 1

rM 

& m CAT



Fi  Fi

o

, rS 

 i m CAT

& S



Fi  Fi

, rR S 

 iS

inlet 1-CO

F1

2-H2

2 F1



F

i  1, & rM  rR S 

o

o

o

y2 

o

 1.157  10

o o

3 F1 o 2 F1 3 F1

o

 

1

3

1

3 2

55  10  1200  10

 i n RS

o

y1  y2 

o F3  F3    1  & y  3

F  F    2  & o

 1.38839 m ol.s

3

55  10  1200

23 3

s

1

o F1    1  & o 3 F1    2  & o 2 F    2  & 1

3 F1    2  & & o

F1  F1

o

1 

X1

o

o

F2  2 F1  2 F1 X 1 o

F3  F1 X 1

3 F1    2  & o

o

o

F  3 F1  2 F1 X 1

1

1

1.38839

1

 1.267  10

yi

F2  F2    2  &

3

m ol.kg . s , rS 

19

o

o F1  F1    1  &

1

1.38839  6.0221413  10 3

F1

0

 3 F1

Fi  Fi

outlet

o

y1 

o

1

1200



o

 860  1000  / 0.02801 / 3600

1.38839

n RS

yi

0

3-CH3OH

&

o

 F  Fi o    F1o  i o o  i F1 X 1 /   Fi  Fi  1  i   1 

8

 2.104  10 m ol.m

2

.s

1

X1 

1000  860  1000

 0.14

inlet 1-CO

y1 

2-H2

y2 

F1

o

o

outlet

o

yi o

3 F1

o

2 F1 3 F1



o

o



1 3 2 3

y1 

y2 

yi

F1

o

1 

X1

o

o

3 F1  2 F1 X 1 o

o

o

o

2 F1  2 F1 X 1 3 F1  2 F1 X 1



o

3-CH3OH

0



1

y3 

F1 X 1 o

o

3 F1  2 F1 X 1

1





1 X1 3  2X1

2 1  X 1  3  2X1 X1 3  2X1

 0.3162

 0.6324

 0.05147

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