Summary • Simple systems of elementary reactions • Open x Closed sequence of elementary steps • Quasi Steady State Hypothesis • Practically important examples

Elementary (one step)

reactions between stable

molecules are very rare.

Rather, a sequence of

elementary steps is

necessary.

3. Systems of reactions. Reversible, parallel, consecutive reactions. Complex reaction systems.

Examples Polymerization Catalytic and enzymatic reactions Combustion Catalytic reactions (homogeneous, heterogeneous) Basic characteristics large number of species (N > 106 ) complex mechanism effect of environment (e.g. effect of solid surface on reaction rate) highly exothermic or endothermic processes Basic types of complex reactions 1) Reversible reactions 2) Parallel reactions 3) Consecutive reactions

Homework 3 To find c1(t), c2(t),c3(t) in the closed isotherm

constant volume reaction system, in which the first

order irreversible consecutive reactions take place

1 2 1 1 1 1 2 3

2 3 2 2 2

1 1

1 2

1

1 2

2

0, 1 m ol/l, 0

(0; 5)

) 2 m in , 1 m in

) 1 m in

) for

V

V

opt

A A r k c t c c c

A A r k c t

a k k

b k k

c t c

0

0.1

0.2

0.3

0.4

0.5

0.6

0 2 4 6

c2

mole/l

t/min

1 2

1 2 1 1 1 1 2 3

2 3 2 2 2

1 1

1 2

1

1 2

2

1

2 1

2 1

2 1 1

0, 1 m ol/l, 0

(0; 5)

) 2 m in , 1 m in

) 1 m in

) for

)

)

V

V

opt

k t k to

o

A A r k c t c c c

A A r k c t

a k k

b k k

c t c

ka c c e e

k k

b c c k t

Solution

1

2

2 1

2

1 1

1 2 2 1

2 1 2

1

1 2 2

1

ln1

) , 0 .693, ( ) m ol/l2

1, 1, ( ) 0.368 m ol/l

k t

k

k ko

opt opt

o

opt opt

e

k

k kc k k t c t c

k k k

ck k t c t

k e

k1=2, k2=1

k1=1, k2=1

HW 3

LINEAR 1ST ORDER DIFFERENTIAL

EQUATION INTEGRATION FACTOR

• In the sequence of elementary steps, the reactants and products of these are not stable reactants or products but are highly reactive intermediates. • The reactive intermediates can be of several different chemical types : free radicals, free ions, solvated ions, complexes at solid surfaces, complexes in homogeneous phase, complexes with enzymes. • Many intermediates may be involved in a given reaction, however the advancement of the reaction can still be described by means of a single parameter – extent of reaction or fractional conversion of key component. • There are two types of sequences leading from reactants to products through intermediates: OPEN or CLOSED. • An open sequence is one in which an intermediate is not reproduced in any other step of the sequence.

• A closed sequence is one in which an intermediate is reproduced so that a cyclic reaction pattern repeats itself and a large number of molecules of products can be made through only one intermediate. (Catalysis)

Open x Closed sequence of elementary steps

3 ( ) 2 ( )

3 ( ) 2 ( )

3 ( ) 2

(

( )

( )

)2

2 3

g g

g g

g

gg g

O O O

O O

O O

O

Open

( )

(

( )

(

3 ( ) 2 ( )

3 ( ) 2 ( )

3 ( )

)

) 2 (

)2

2 3

g

g

g g

g g

g

g

g g

ClO

C lO

O O

O O

Cl

C l

O O

Closed

(catalytic)

Quasi Steady State Hypothesis – QSSH The concentrations of the intermediates remain low and constant these intermediate concentrations can be expressed using reactant and product concentrations

Example Fosgen (COCl2) is manufactured by gas phase reaction between CO and Cl2

CO(g) + Cl2(g) COCl2(g)

It follows from experimental data :

2

3/ 2 3 1 [mol.m . ]

V CO Clr kc c s

The proposed mechanism involves 2 intermediates Cl and COCl.

Reaction Kinetic equation

1. Cl2(g) 2Cl(g)

2. CO(g) + Cl(g) COCl(g)

3. COCl(g) + Cl2(g) COCl2(g) + Cl(g)

2

2

1 ,1 ,1f Cl b Clr k c k c

2 ,2 ,2f CO Cl b COClr k c c k c

23 ,3f COCl C lr k c c

1 2 3

2 3

2 0

0

Cl

COCl

R r r r

R r r

2

2

2

1 1 ,1 ,1

1/ 2

,1

,1

2 0f C l b C l

f

C l C l

b

r r k c k c

kc c

k

2

2 2

2 2 2

2 3 ,2 ,2 ,3

1/ 2 1/ 2

1/ 2,1 ,1

,2 ,2

,1 ,1,2

,2 ,3 ,2 ,3 ,2 ,3

0COCl f CO Cl b COCl f COCl Cl

f f

f CO Cl f CO Cl

b bf CO Cl

COCl

b f C l b f C l b f C l

R r r k c c k c k c c

k kk c c k c c

k kk c cc

k k c k k c k k c

The balances of intermediates at steady state: Cl2 Cl CO COCl COCl2

By adding above equations

Concentration of COCl:

1 2 0 0 0

0 1 1 1 0

1 1 0 1 1

ν

2

2 2

2

1/ 2

3 / 2,1

,3 ,2

,1

3 ,3

,2 ,3

f

f f CO Cl

b

COCl f COCl C l

b f C l

kk k c c

kR r k c c

k k c

Rate of COCl2 production is given by equation (3) in which we substitute for COCl concentration

If Cl2 concentration is low or is valid, we get ,2 ,3b f

k k

2 2 2

1/ 2

3 / 2 3 / 2,3 ,2 ,1

,2 ,1

1/ 2

,3 ,2 ,1

,2 ,1

f f f

COCl CO Cl CO Cl

b b

f f f

b b

k k kR c c kc c

k k

k k kk

k k

Isothermal batch constant volume reactor

Validity of QSSH

Kinetic parameters used in numerical simulation

1/ 2

,3 ,2 ,1 3 -1 3/2 -1

,2 ,1

0.07 (m m ol ) sf f f

b b

k k kk

k k

Concentration profiles of COCl2

Homework 4 (due after Chapter 4) Calculate volumes of CSTR a PFR working at 150 oC and 300 kPa to produce 1 t COCl2/day with CO conversion equal to 95 %. A mixture of CO and Cl2 (molar ratio 1:1) is fed at 300 kPa and 150 oC. Data k(423 K) = 0.07 (m3mol-1)3/2.s-1 MCOCl2 = 98.92 kg/kmol.

Answer: VCSTR = 0.053 m3 VPFR = 0.0021 m3

Example

2NO(g) + O2(g) 2NO2(g)

Mechanism:

2NO(g) (NO)2(g) equlibrium(1)

(NO)2(g) + O2(g) 2NO2(g) (2)

2

2 2 2 2 2 2 2

,2 1,2 1 1

1

( )

1 2

2 2 2 2 2

2 ,2 ( ) ,2 ,2 1 ,2

,2 1 ,2 ,2

,2 ,2 1,

oo o

f rf fr r

o

r

NO

NO

f NO O b NO f NO O b NO f NO O b NO

E HE EG S

RT RT R RT RT

f f of of of

S

oR

of of f f r

cK

c

r k c c k c k K c c k c k c c k c

k k K A e e A e e A e

A A e E E H

Max Bodenstein, 1941

Radical polymerization

Inicialization

1

2

PMR

RI

i

d

k

k

MRiinit

Idd

cckr

ckr

Propagation

.

.

.

1

32

21

i

k

i

k

k

PMP

PMP

PMP

p

p

p

.

.

.

1

2

1

1

2

1

MPpi

MPp

MPp

cckr

cckr

cckr

i

Termination

tk

k l k lP P P

, k lt k l t P P

r k c c

Intermediates

0iP R

dc dc

dt dt

2 0

2

2

d init

d I i R M

d I

R

i M

r r

k c k c c

k cc

k c

1PBalance of

1 1 11 ,1

1

0jP init t i R M p P M t P P

j

R r r r k c c k c c k c c

Balance of initiator

Balance of

1

1

0k k k k jP p P M p P M t P P

j

R k c c k c c k c c

, 2, 3, ...kP k

0

2

1

j

PtMRi jckcck

By summing the last equation from k=2,3,… and adding balance

equation of

1

2

j

i R M d I

P

j t t

k c c k cc

k k

1P

Monomer consumption

1 1

22

2

j jM

d

init p M P i R M p M P

j j

d I

d I p M

t

I

p M

t

r k c c k c c k c cR

k ck c

k

k ck c k c

k

Polymer Pn production

1

1

1

2n n k k

n

P t P P

k

R k c c

1

2

k

k

p P M

P

p M t d I

k c cc

k c k k c

From balance of kP

1

1

2

2 2 2

k

k

k

p P M p Md I

P

p M t d I p M t d I p M t d I

k c c k ck cc

k c k k c k c k k c k c k k c

1

2

1

2

( 1) 2

22 2 2

1

n kn k

n

t P P

n

p Mt d I

P

p M t d I p M t dk I

k cn k k cR

k c k k c k c k k ck c c

Homework 5: to justify these equations

0.0E+00

2.0E-06

4.0E-06

6.0E-06

8.0E-06

1.0E-05

1.2E-05

1.4E-05

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

0.E+00 5.E+03 1.E+04 2.E+04 2.E+04 3.E+04 3.E+04 4.E+04 4.E+04

CPn [mol/dm3]

cM [mol/dm3]

t [s]

cM

n=25

n=45

n=70

n=100

3.0 1.0x10-2

kd [s-1] 1.45x10-5 kt [dm3.mol-1.s-1] 1.2x106

kp [dm3.mol-1.s-1] 4.4x102

3[m ol/dm ]

o

Mc

3[m ol/dm ]o

Ic

Summary • Simple systems of elementary reactions • Open x Closed sequence of elementary steps • Quasi Steady State Hypothesis • Practically important examples