tutorial chemical reaction engineering - otto von guericke ...uebung+05... · 1institute of process...

1Institute of Process Engineering, G25-217, [email protected]

Tutorial Chemical Reaction Engineering:

Dipl.-Ing. Andreas Jörke1

22-May-14

5. Kinetics I

Tutorial CRE: Kinetics I

Equilibrium: Information about „what is possible“ but no time information

Kinetics answer the question „how fast“ a reaction reaches itsequilibrium state

Kinetic information is needed for reactor design and optimization

22-May-14CRE: Kinetics I 2

X eq

T

equilibrium kinetics

feasible region

eqc c t

t

c

kinetics equilibrium


Kinetic description requires knowledge about the reaction rate r

Definition: reaction rate

For systems with constant volume(liquid phase reactions)

Besides this definition there is also a kinetic approach

Depends on the system, catalyst, chemical mechanisms, …

Examples: power law, Langmuir-Hinshelwood, Eley-Rideal, Mars-vanKrevelen, Michaelis-Menten / Monod, …


vol mass

R cat

d d1 1 1 1

d d

i i

i i

n nr r

V t m t

t

c

vol

R R

d d1 1 1with

d d

i i ii

i i

n c nr c

V t t V

vol

d1

d

i

i

c tr

t


In this tutorial: power law for a liquid phase with constant volume

Example: overall order of reaction = 2.7

Task 5.1: Series reaction (e.q. selective oxidation sequence)

Power law approach

Questions:

Integrated kinetic expression for all three components

Concentration-time curves for different parameter sets


in

j j i

i

r k T c

Order of component i (= νi for elementary reactions)

Σni = overall order of the reaction

1 1.7

A Br k T c c

1 2A B Ck k

0

A A B C0 , 0 0, 0 0c t c c t c t


Kinetic approach: power law, first order for all reactants

Using the approach for the mass balances

Different solution strategies for the coupled ODE system possible

Linear system!

Laplace Transformation!


1 1 A

2 2 B

r k c

r k c

A1 1 A

B1 2 1 A 2 B

C2 2 B

d

d

d

d

d

d

cr k c

t

cr r k c k c

t

cr k c

t

R

,

1

d

d

ii j j

j

cr

t

Holds for a batch systemwith constant volume


Laplace transformation

Reduces ODE to AE

AE are easy to solve

Transformations andinverse transformationscan be found in tables!


0

1

0

d

for 01d with

2π 0 for 0

st

i

st

i

L f t F s f t e t

f t tL F s e F s s s

i t

ODE

System

Solution

in t

Algebraic

equations

Solution

in s

time domain

frequency domain

„hard“

„easy“


Laplace tables



Some simple rules for LPT

Application to our system yields

From (1) follows


d0

d

f t F s

f tsF s f t

t

0

A A 1 A

0

B B

1sC s c k C s

sC s c

0

1 A 2 B

0

C C

2k C s k C s

sC s c

0

2 B 3k C s

0

AA

1

cC s

s k

0

*AB 1 2 B

1

2c

sC s k k C ss k

0

1 AB

1 2

k cC s

s k s k

0

*1 AC 2

1 2

3k c

sC s ks k s k

0

1 2 AC

1 2

k k cC s

s s k s k


Inverse transformation generates the time functions!



Important transformations from the tables

Solution in time domain


1

1

1 11

at

at bt

at bt

es a

e e

a b s a s b

be ae

ab a b s s a s b

1

1 2

1 2

0

A A

0

B 1 A

2 1

0 2 1C A

2 1

1

k t

k t k t

k t k t

c t c e

e ec t k c

k k

k e k ec t c

k k

1

1

2 1

min 0.03

0.1

k

k k





1

1

1 11

at

at bt

at bt

es a

e e

a b s a s b

be ae

ab a b s s a s b

1

1 2

1 2

0

A A

0

B 1 A

2 1

0 2 1C A

2 1

1

k t

k t k t

k t k t

c t c e

e ec t k c

k k

k e k ec t c

k k

1

1

2 1

min 0.03

1

k

k k





1

1

1 11

at

at bt

at bt

es a

e e

a b s a s b

be ae

ab a b s s a s b

1

1 2

1 2

0

A A

0

B 1 A

2 1

0 2 1C A

2 1

1

k t

k t k t

k t k t

c t c e

e ec t k c

k k

k e k ec t c

k k

1

1

2 1

min 0.03

2

k

k k





1

1

1 11

at

at bt

at bt

es a

e e

a b s a s b

be ae

ab a b s s a s b

1

1 2

1 2

0

A A

0

B 1 A

2 1

0 2 1C A

2 1

1

k t

k t k t

k t k t

c t c e

e ec t k c

k k

k e k ec t c

k k

1

1

2 1

min 0.03

10

k

k k





1

1

1 11

at

at bt

at bt

es a

e e

a b s a s b

be ae

ab a b s s a s b

1

1 2

1 2

0

A A

0

B 1 A

2 1

0 2 1C A

2 1

1

k t

k t k t

k t k t

c t c e

e ec t k c

k k

k e k ec t c

k k

1

1

2 1

min 0.03

100

k

k k


If an intermediate is consumed „very fast“

Simplified solution


A1 1 A

B1 2 1 A 2 B

C2 2 B

d

d

d0

d

d

d

cr k c

t

cr r k c k c

t

cr k c

t

1

1

0

A A

1 A

B

2

0

C A 1

k t

k t

c t c e

k c tc t

k

c t c e

Bodenstein steady stateapproximation

1

1

2 1

min 0.03

2

k

k k



Simplified solution


A1 1 A

B1 2 1 A 2 B

C2 2 B

d

d

d0

d

d

d

cr k c

t

cr r k c k c

t

cr k c

t

1

1

0

A A

1 A

B

2

0

C A 1

k t

k t

c t c e

k c tc t

k

c t c e


1

1

2 1

min 0.03

5

k

k k



Simplified solution


A1 1 A

B1 2 1 A 2 B

C2 2 B

d

d

d0

d

d

d

cr k c

t

cr r k c k c

t

cr k c

t

1

1

0

A A

1 A

B

2

0

C A 1

k t

k t

c t c e

k c tc t

k

c t c e


1

1

2 1

min 0.03

10

k

k k



Simplified solution


A1 1 A

B1 2 1 A 2 B

C2 2 B

d

d

d0

d

d

d

cr k c

t

cr r k c k c

t

cr k c

t

1

1

0

A A

1 A

B

2

0

C A 1

k t

k t

c t c e

k c tc t

k

c t c e


1

1

2 1

min 0.03

100

k

k k


Task 5.2: Equilibrium reaction with constant volume

Question:

Derivation of an integrated kinetic expression

Stoichiometric mixture at t = 0

Using the extend of reaction yields

Only description of A is necessary!


A B C D

CA B D

A B C D

cc c c

A B

C D

0 1 mol/l, 0 1 mol/l

0 0 mol/l, 0 0 mol/l

c t c t

c t c t

A B

0

C D A A A

c t c t

c t c t c c c t


Power law approach and balance for A

Introducing conversion


A B

C D

22 0A

A B C D A A A

d

d

r k c c

r k c c

cr r k c c k c c k c k c c

t

0

A A

A 0

A

c c tX t

c

2 20 0 0A AA A A A A

20 0 2AA A A A

0 2 2

A A A A

d d1

d d

d1

d

11 2 1

c Xk c X k c X c

t t

Xk c X k c X

t

kk c X X X K

K k


Continued…

Solve ODE with separation of variables Integration


0 2 2AA A A A

0 2

A A A

0 2AA A A

d 11 2

d

11 2 1

d 11 2

d

Xk c X X X

t K

k c X XK

X Kk c X X

t K

A

A

0AA

20 0A A

ˆdˆd

1ˆ ˆ1 2

X t

X t

Xk c t

KX X

K


From integration tables (e.g. Bronstein) we find the standardintegral

Check, if Δ < 0 and evaluate conversion integral

Include this intermediate result in the ODE and integrate over t


2

20 0

ˆ ˆd 1 2ln with 4

ˆˆ ˆ 2

XX

X

X aX bac b

aX baX bX c

2 44 0ac b

K

A

A

A

A

20AA A

1 11ˆd 1

ln1 121 1ˆ ˆ 11 2

X

X

KX

X K KK K

KK KXX X

KK K


Continued…

Solve for XA

With the conversion


0A

A

A

1 11

21exp

1 1 11

KX

k cKK Kt

K K KX

K K

0

A

A 0

A

21 exp

1 21exp

1

k ct

KKX t

K k cKt

K K

0

A A

A 0

A

c c tX t

c


Continued…

The final result


0

A

0

A A 0

A

21 exp

11 21

exp1

k ct

KKc t c

K k cKt

K K

1

1

/ min 1

/ min 0.1

k

k

1

1

/ min 0.1

/ min 1

k

k

tutorial chemical reaction engineering - otto von guericke ...uebung+05... · 1institute of process...

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