κίνησις "kinesis", movement or to move

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)

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 ) WGS (BASF, FRG) N2O decomposition (IPC AS, CZ)

Chemical reactor(s)  heart of chemical process

Methane aromatization (ICTP, CZ)

Raw material separation  reaction  separation  product 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 " " the material element (Plato) Stoichiometry " " the count, the quantity

2NO  N2 + O2 closed (batch) system

t =0 t > 0

Atoms of oxygen nnoo2 nn 2 NOO 2 NOO2 Atoms of nitrogen oo nnNON2 nn 2 2 NON 2

noo22 n  n  n noo22 n  n  n NOONOO22NONNON22

n noo  2( n  n ) n noo  2( n  n ) NONOOO22 NONONN22

o oo nn ()()nOONN n n n NONO 2 2 2 2 2 1 1

-2NO + O2 + N2 = 0 Symbols for species NO = A1 O2 = A2 N2 = A3 Stoichiometric coefficients 1 = -2 2 = 1 3 = 1

 i  0 products 3 A 0 0 inerts  i i i i 1

 i  0 reactants Molar extent of the reaction  [ksi:] o oo nn ()()nOONN n n n  NON O 2 2  2 2 2 1 1 n  n o   i i  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 Closed system n i ni

 i  0 products N A 0 0 inerts  i i i i 1

 i  0 reactants

nn o   ii  i

o nni i i Example

2NO  N2 + O2

AAA1N O , 2  N 2 , 3  O 2

1 2,  2  1,  1  1

A1 T  νA 2 1 1 ,  A   2  A3 Matrix notation A1 T  νA 2 1 1A  0   2  A3 Fractional conversion of key component, j X j  X j   max

* o n j  n j Number of moles of key  max   component in limits j (chemical equilibrium or 0)

o nnjj n *  0 o XX (0,1) j n i  n i ij jjo o n j    j n  n X   i  i i j o o  max  n j  i n j o nnjj  XX100 (0,100) j jjo n j

n o j o  i o XXjjmax   n  n  n X  i i j j j  j Stoichiometric matrix in the case of several reactions

N  A0 k 1, NR  ki i i 1 reaction component

Stoechimetric matrix has NR rows and N columns

11  12...  1 N A1  21  22...  2 N A νA, 2 . .   ...  A NRNRNRN,1 ,2 , N

νA 0 Number of moles of i-th component consumed or created in k-th reaction : o nnki ki nn o Molar extent of k-th reaction : ki ki  k   ki

NR Number of moles of i-th nno  components: i i ki k k 1

Matrix notation nnoTνξ

n o  11 n1       n n o  nn22 ,,o 2 ξ    .... ..       n o  NNR n N   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 o o NH3 n1 nn1 1 () 1   2   3

o o O2 n 2 nn2 2 1.25 1   2  0.75  3

o o NO n 3 nn3 3 1

o o N2O n 4 nn4 40.5 2

o o N2 n 5 nn5 50.5 3

o o H2O n 6 nn6 6 1.5( 1   2   3 )

6 6  o o n n 0.25   i  i  13 i 1 i 1

e.g. molar fraction of NH3 n o ()     y  1 1 2 3 1 6 n o 0.25   i  13 i 1 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  10.5 0 0     1  0.75 0 0 0.51.5   0 0.5  1 0 0.5 0      01 2010   01 2010  1  1.25 1 0 0 1.5  1  1.25 1 0 0 1.5  0 0.25 10.5 0 0 0 0.25 10.5 0 0  0 0 110.50 0 0 1 1 0.5 0  0 0 2 2 1 0 0 0 0 0 0 0 Rank()=3  only 3 reactions are independent

Extent of the reaction in a flow system

o Fi Fi REACTOR Inlet molar flow rates Outlet molar flow rates [mole of i-th species/s] [mole of i-th species/s] N  A0 k 1, NR Open (flow) system in a steady state  ki i i 1 NR  FFo  &&   k i i ki k k k 1  o FFii, -outlet, inlet molar flow rates of i-th species [mole/s] &  k  extent of k-th reaction per unit of time [m ole/s]   mean residence time [s] F o & 1 F1 1    o T & F F o & F = F + νξ FF2 ,,o 2 ξ&  2 .. .. ..    F o & N FN  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  j REACTOR

N  A0 k 1, NR Open (flow) system at steady state  ki i i 1

o jj  

o j ,j -outlet, inlet molar flows of j-th chemical element [mole/s]

o Φ = Φ  o 1  1     o ΦΦ2 , o 2 NEL - Number of .. .. chemical elements    o NEL  NEL Formula matrix E

NEL elements  NEL=3 N components  N H O N=6 Formula vector of NH NH3 1 3 0 3

O2 0 0 2 NO 1 0 1

N2O 2 0 1

N2 2 0 0

H2O 0 2 1

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

11  12......  1NNEL  EEE 11 12 1,  0 0 ... 0      21  22......  2NNELEE 21 2 , 0 0 ... 0      NR νE = 0 ..    .       ......   EEE  NRNRNRNNNNNEL,1 ,2 ,   ,1 ,2 ,  0 0 ... 0

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

NEL M E m i ij j j 1 M = E m Atomic weights (relative atomic mass) of j-th element

Mm11        Mm Mm22 ,   ....        MmNNEL   

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

Closed (batch) system: n = no + νξT

T ETTTTTT n = E no + E ν ξ E n o + νE ξ E n o Open (flow) system at steady state: F = Fo + νξT &

T ETTTTTT F = E Fo + E ν ξ&& E F o + νE ξ E F o because νE 0 Finally

TToo These equations are used E n n  E n  n  0 in data reconciliation tasks around chemical reactors. ETT F Foo = E F  F  0 The last equation (flow     system) is valid only at steady state. Problem 1.3 Steady state, Selective reduction of NOx by C3H8 unknown o F Fi i stoichiometry of reactions

NO NO2 CO C3H8 CO2 H2O O2 N2 o Fi [ mole / min] 4.563 0.1845 0 2.943 0 0 90 0

Fi [ m ole / m in] 2.2905 2.3355 0 2.898 x x x x

Calculate the missing outlet molar flow rates ET F o - F = E T  F 0

F1 1-NO 2-NO2 3-CO 4-C3H8 5-CO2 6-H2O 7-O2 8-N2 known N 1 1 0 0 0 0 0 2 F2 O 1 2 1 0 2 1 2 0  C 0 0 1 3 1 0 0 0 F 3 H 0 0 0 8 0 2 0 0  F T 4 E T E 1 2 F TT 5 E12 FKNOWNUNKNOWN  E  F  0 F6 TT EFEF21UNKNOWNKNOWN     unknown F7 11 TTTT F FEEFQFQEE         i UNKNOWNKNOWNKNOWN 2 1 2 1 F8

We obtain using Excel (homework 1)

0 0 1 3  0 0 0 4 Q   0.5 1 0.5 5  0.5 0.5 0 0

1 TT and taking  FEEFQF 2    2  1  1   1 we have

NO NO2 CO C3H8 CO2 H2O O2 N2 o Fi [ mole / min] 4.563 0.1845 0 2.943 0 0 90 0

Fi [ m ole / m in] 2.2905 2.3355 0 2.898 0.135 0.18 88.76 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 t = 0 t > 0 o o NH3 n1 nn1 1 () 1   2   3

o o O2 n 2 nn2 2 1.25 1   2  0.75  3

o o NO n 3 nn3 3 1

o o N2O n 4 nn4 40.5 2

o o N2 n 5 nn5 50.5 3

o o H2O n 6 nn6 6 1.5( 1   2   3 )

n o    S 31 1 NONH 3 ()()1  2   3  1   2   3

n o  0 n o    3   Y 31 1 NONH 3 oo nn11 Reaction rate

(IUPAC Gold Book = rate of conversion ) 

 d  1 dn m ax r  i m ole.s 1 d Closed system of uniform dt dt dt i pressure, temperature and d  k composition. rk  0; 0 t dt

Rate of generation (consumption) of component Ai

one reaction dni d  RA  i.  i . r  R=ν r i dt dt NRNRd  several reactions Rr.k  .  R=νrT Ai  ki ki k kk11dt We measure usually the rates of generation (consumption) of components R and we want to calculate r

T R expνTTT r  Objective function  r  R exp  ν r  R exp  ν r  minimization of  r  = least squares solution of overdefined system (N>NR)

1 r   ννT  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 1 1 3 0 12 3 1  TT   -0.10256 -0.02564 0.179487 0.230769 -0.07692 ν  νν     νν ν  0 1 1 1 1   3 4  0.076923 -0.23077 -0.38462 0.076923 0.307692

r1 0.94619   r2 1.99546 Specific reaction rate

d  The reaction rate d t 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: Reaction rate per vole um

1 1d  1 1 dni 31 rV  r   mole.m . s V V dt V i dt

1 1 d  k rrV, k k V V dt

Reaction rate per mass

1 1d  1 1 dni 11 rM  r   mole.kg . s m m dt m i dt

11d  k rrM, k k m m dt Reaction rate per surface

1 1d  1 1 dni 21 rS  r   mole.m . s S S dt S i dt

1 1 d  k rrS, k k S S dt

Reaction rate per active center (turn over num ber)

1 1d  1 1 dni 1 rrRS    s nRS n RS dt n RS i dt

1 1 d  k rrRS, k k nRSRS n dt C entral problem of APPLIED CHEMICAL KINETIC S r functionTc ,,12 c,... cN , P ,catalytic activity, transport parameters,....

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 c12 c c N 

Rule 3: The rate constant k depends on temperature (Svante Arrhenius, 1889): E  k() T Ae R T

Rule 4: The function g is independent of temperature: N   g c,,...... c c c12 c cNi c  1 2N 1 2 N i i 1 Rule 5: When a reaction is reversible:

rrf  r b  kTgcc f(),,...(),,... f 1 2 c N  kTgcc b b 1 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: -1 MCO= 28.010 kg.kmol 23 -1 NA=6.0221413x10 mol (Avogadro number)

COAHACHOHA1,, 2  2 3  3 11        νR 2 ,   2 r     Fractional conversion of key component     11    1 key component o o NR & FF  F  XFFFX i i/ 1  oo  i oo&& 1ii 1 1 FFFF         &    i i ki k i i i M AX i 11 k 1 &&&o o o FFFFFFi i  i  i  i  i rMSRS ,, r   r   mCAT iCAT m S  i S n RS  iRS n

y o y inlet i outlet i o F o 1 & o o F1 1 o & 1   o F y  FF11  1  y  FFX1 1-CO 1 1 o 1 o & 1 1 1  33F1 32F1    o o & 2 F 2 o 22F   oo o o 1 FF  2 & 1   2 F y  22  y  FFFX222 1 1 1 2-H2 1 2 o 2 o & 33F1 32F1    o & 0 FF  1 &  FFX o 33  y  3 1 1 3-CH3OH 3 o & 32F1    o oo FF  2 & oo FF 3 1   FFFX321 1 1  1

860 1000 /0.02801/3600 i 1,&   1.38839 mol.s 1 1

1.388393  1  1 1.38839  8  2  1 rMS 1.157  10 mol.kg . s , r   2.104  10 mol.m . s 1200 551012003 23 1.38839 6.0221413 10 31 rRS  1.267  10 s 55 103  1200  10 19 1000 860  X 1 0.14 1000

y o y inlet i outlet i

o F o 1 FX1   1  X y o 1 y 11 1  0.3162 1-CO 1 o 1 oo 33F1 3FFXX1 2 1 1 3 2 1

2 F o 2 22FFXoo 21  X  y o 1 y 1 1 1 1  0.6324 2-H2 2 o 2 oo 33F1 3FFXX1 2 1 1 3 2 1

FXXo y 1 1  1  0.05147 3-CH3OH 0 3 oo 3FFXX1 2 1 1 3 2 1

 1